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6.9. RECTANGULAR WAVEGUIDE: TE MODES 113 expression for eEz must account for all non-trivial modes. Summarizing: eEz = ∞ X m=1 ∞ X n=1 eE(m,n) z (6.186) where eE(m,n) z ≜E(m,n) 0 sin mπ a x sin nπ b y e−jk(m,n) z z (6.187) where E(m,n) 0 is an arbitrary constant (consolidating the constants B and D), and, since k2 x + k2 y = k2 ρ ≜β2 −k2 z: k(m,n) z = r ω2µǫ − mπ a 2 − nπ b 2 (6.188) Summarizing: The TM ( eHz = 0) component of the unidi- rectional (+ˆz-traveling) wave in a rectangular waveguide is completely determined by Equa- tion 6.186, and consists of modes as defined by Equations 6.187, 6.188, 6.184, and 6.185. The remaining non-zero field components can be de- termined using Equations 6.151–6.154. It is customary and convenient to refer to the TM modes in a rectangular waveguide using the notation “TMmn.” For example, the mode TM12 is given by Equation 6.187 with m = 1 and n = 2. Finally, note that values of k(m,n) z obtained from Equation 6.188 are not necessarily real-valued. It is apparent that for any given value of m, k(m,n) z will be imaginary-valued for all values of n greater than some value. Similarly, it is apparent that for any given value of n, k(m,n) z will be imaginary-valued for all values of m greater than some value. This phenomenon is common to both TM and TE components, and so is addressed in a separate section (Section 6.10). Additional Reading: • “Waveguide (radio frequency)” on Wikipedia. • “Separation of variables” on Wikipedia. 6.9 Rectangular Waveguide: TE Modes [m0225] A rectangular waveguide is a conducting cylinder of rectangular cross section used to guide the propagation of waves. Rectangular waveguide is commonly used for the transport of radio frequency signals at frequencies in the SHF band (3–30 GHz) and higher. The fields in a rectangular waveguide consist of a number of propagating modes which depends on the electrical dimensions of the waveguide. These modes are broadly classified as either transverse magnetic (TM) or transverse electric (TE). In this section, we consider the TE modes. Figure 6.7 shows the geometry of interest. Here the walls are located at x = 0, x = a, y = 0, and y = b; thus, the cross-sectional dimensions of the waveguide are a and b. The interior of the waveguide is presumed to consist of a lossless material exhibiting real-valued permeability µ and real-valued permittivity ǫ, and the walls are assumed to be perfectly-conducting. Let us limit our attention to a region within the waveguide which is free of sources. Expressed in phasor form, the magnetic field intensity within the waveguide is governed by the wave equation: ∇2 eH + β2 eH = 0 (6.189) where β = ω√µǫ (6.190) Figure 6.7: Geometry for analysis of fields in a rect- angular waveguide. | Electromagnetics_Vol2_Page_126_Chunk2301 |
114 CHAPTER 6. WAVEGUIDES Equation 6.189 is a partial differential equation. This equation, combined with boundary conditions imposed by the perfectly-conducting plates, is sufficient to determine a unique solution. This solution is most easily determined in Cartesian coordinates, as we shall now demonstrate. First we express eH in Cartesian coordinates: eH = ˆx eHx + ˆy eHy + ˆz eHz (6.191) This facilitates the decomposition of Equation 6.189 into separate equations governing the ˆx, ˆy, and ˆz components of eH: ∇2 eHx + β2 eHx = 0 (6.192) ∇2 eHy + β2 eHy = 0 (6.193) ∇2 eHz + β2 eHz = 0 (6.194) Next we observe that the operator ∇2 may be expressed in Cartesian coordinates as follows: ∇2 = ∂2 ∂x2 + ∂2 ∂y2 + ∂2 ∂z2 (6.195) so the equations governing the Cartesian components of eH may be written as follows: ∂2 ∂x2 eHx + ∂2 ∂y2 eHx + ∂2 ∂z2 eHx + β2 eHx = 0 (6.196) ∂2 ∂x2 eHy + ∂2 ∂y2 eHy + ∂2 ∂z2 eHy + β2 eHy = 0 (6.197) ∂2 ∂x2 eHz + ∂2 ∂y2 eHz + ∂2 ∂z2 eHz + β2 eHz = 0 (6.198) In general, we expect the total field in the waveguide to consist of unidirectional waves propagating in the +ˆz and −ˆz directions. We may analyze either of these waves; then the other wave is easily derived via symmetry, and the total field is simply a linear combination (superposition) of these waves. With this in mind, we limit our focus to the wave propagating in the +ˆz direction. In Section 6.7, it is shown that all components of the electric and magnetic fields can be easily calculated once eEz and eHz are known. The problem is further simplified by decomposing the unidirectional wave into TM and TE components. In this decomposition, the TE component is defined by the property that eEz = 0; i.e., is transverse (perpendicular) to the direction of propagation. Thus, the TE component is completely determined by eHz. Equations 6.136–6.139 simplify to become: eEx = −j ωµ k2ρ ∂eHz ∂y (6.199) eEy = +j ωµ k2ρ ∂eHz ∂x (6.200) eHx = −j kz k2ρ ∂eHz ∂x (6.201) eHy = −j kz k2ρ ∂eHz ∂y (6.202) where k2 ρ ≜β2 −k2 z (6.203) and kz is the phase propagation constant; i.e., the wave is assumed to propagate according to e−jkzz. Now let us address the problem of finding eHz, which will then completely determine the TE field. As in Section 6.7, we recognize that eHz can be represented as the propagation factor e−jkzz times a factor that describes variation with respect to the remaining spatial dimensions x and y: eHz = ehz(x, y)e−jkzz (6.204) Substitution of this expression into Equation 6.198 and dividing out the common factor of e−jkzz yields: ∂2 ∂x2 ehz + ∂2 ∂y2 ehz −k2 zehz + β2ehz = 0 (6.205) The last two terms may be combined using Equation 6.203, yielding: ∂2 ∂x2 ehz + ∂2 ∂y2 ehz + k2 ρehz = 0 (6.206) This is a partial differential equation for ehz in the variables x and y. This equation may be solved using the technique of separation of variables. In this technique, we recognize that ehz(x, y) can be written as the product of a function X(x) which depends only on x, and a function Y (y) that depends only on y. That is, ehz(x, y) = X(x)Y (y) (6.207) | Electromagnetics_Vol2_Page_127_Chunk2302 |
6.9. RECTANGULAR WAVEGUIDE: TE MODES 115 Substituting this expression into Equation 6.206, we obtain: Y ∂2 ∂x2 X + X ∂2 ∂y2 Y + k2 ρXY = 0 (6.208) Next dividing through by XY , we obtain: 1 X ∂2 ∂x2 X + 1 Y ∂2 ∂y2 Y + k2 ρ = 0 (6.209) Note that the first term depends only on x, the second term depends only on y, and the remaining term is a constant. Therefore, the sum of the first and second terms is a constant; namely −k2 ρ. Since these terms depend on either x or y, and not both, the first term must equal some constant, the second term must equal some constant, and these constants must sum to −k2 ρ. Therefore, we are justified in separating the equation into two equations as follows: 1 X ∂2 ∂x2 X + k2 x = 0 (6.210) 1 Y ∂2 ∂y2 Y + k2 y = 0 (6.211) where the new constants k2 x and k2 y must satisfy k2 x + k2 y = k2 ρ (6.212) Now multiplying Equations 6.210 and 6.211 by X and Y , respectively, we find: ∂2 ∂x2 X + k2 xX = 0 (6.213) ∂2 ∂y2 Y + k2 yY = 0 (6.214) These are familiar one-dimensional differential equations. The solutions are:5 X = A cos (kxx) + B sin (kxx) (6.215) Y = C cos (kyy) + D sin (kyy) (6.216) where A, B, C, and D – like kx and ky – are constants to be determined. At this point, we observe that the wave we seek can be expressed as follows: eHz = ehz(x, y)e−jkzz = X(x) Y (y) e−jkzz (6.217) 5Students are encouraged to confirm that these are correct by confirming that they are solutions to Equations 6.213 and 6.213, re- spectively. The solution is essentially complete except for the values of the constants A, B, C, D, kx, and ky. The values of these constants are determined by applying the relevant electromagnetic boundary condition. In this case, it is required that any component of eE that is tangent to a perfectly-conducting wall must be zero. Therefore: eEy (x = 0) = 0 (6.218) eEy (x = a) = 0 (6.219) eEx (y = 0) = 0 (6.220) eEx (y = b) = 0 (6.221) Referring to Equation 6.217 and employing Equations 6.199–6.202, we obtain: ∂ ∂xX (x = 0) = 0 (6.222) ∂ ∂xX (x = a) = 0 (6.223) ∂ ∂y Y (y = 0) = 0 (6.224) ∂ ∂y Y (y = b) = 0 (6.225) Evaluating the partial derivatives and dividing out common factors of kx and ky, we find: −A sin(kx · 0) + B cos(kx · 0) = 0 (6.226) −A sin(kx · a) + B cos(kx · a) = 0 (6.227) −C sin(ky · 0) + D cos(ky · 0) = 0 (6.228) −C sin(ky · b) + D cos(ky · b) = 0 (6.229) Evaluating: −A · 0 + B · 1 = 0 (6.230) −A sin (kxa) + B cos (kxa) = 0 (6.231) −C · 0 + D · 1 = 0 (6.232) −C sin (kyb) + D cos (kyb) = 0 (6.233) Equations 6.230 and 6.232 can be satisfied only if B = 0 and D = 0, respectively. Subsequently, Equations 6.231 and 6.233 reduce to: sin (kxa) = 0 (6.234) sin (kyb) = 0 (6.235) This in turn requires: kx = mπ a , m = 0, 1, 2... (6.236) ky = nπ b , n = 0, 1, 2... (6.237) | Electromagnetics_Vol2_Page_128_Chunk2303 |
116 CHAPTER 6. WAVEGUIDES Each positive integer value of m and n leads to a valid expression for eHz known as a mode. Summarizing: eHz = ∞ X m=0 ∞ X n=0 eH(m,n) z (6.238) where eH(m,n) z ≜H(m,n) 0 cos mπ a x cos nπ b y e−jk(m,n) z z (6.239) where H(m,n) 0 is an arbitrary constant (consolidating the constants A and C), and, since k2 x + k2 y = k2 ρ ≜β2 −k2 z: k(m,n) z = r ω2µǫ − mπ a 2 − nπ b 2 (6.240) Summarizing: The TE ( eEz = 0) component of the unidi- rectional (+ˆz-traveling) wave in a rectangular waveguide is completely determined by Equa- tion 6.238, and consists of modes as defined by Equations 6.239, 6.240, 6.236, and 6.237. The remaining non-zero field components can be de- termined using Equations 6.199–6.202. It is customary and convenient to refer to the TE modes in a rectangular waveguide using the notation “TEmn.” For example, the mode TE12 is given by Equation 6.239 with m = 1 and n = 2. Although Equation 6.238 implies the existence of a TE00 mode, it should be noted that this wave has no non-zero electric field components. This can be determined mathematically by following the procedure outlined above. However, this is also readily confirmed as follows: eEx is constant for the TE00 mode because kρ = 0 for this mode, however, eEx must be zero to meet the boundary conditions on the walls at y = 0 and y = b. Similarly, eEy is constant for the TE00 mode, however, eEy must be zero to meet the boundary conditions on the walls at x = 0 and x = b. Finally, note that values of k(m,n) z obtained from Equation 6.240 are not necessarily real-valued. It is apparent that for any given value of m, k(m,n) z will be imaginary-valued for all values of n greater than some value. Similarly, it is apparent that for any given value of n, k(m,n) z will be imaginary-valued for all values of m greater than some value. This phenomenon is common to both TE and TM components, and so is addressed in a separate section (Section 6.10). Additional Reading: • “Waveguide (radio frequency)” on Wikipedia. • “Separation of variables” on Wikipedia. | Electromagnetics_Vol2_Page_129_Chunk2304 |
6.10. RECTANGULAR WAVEGUIDE: PROPAGATION CHARACTERISTICS 117 6.10 Rectangular Waveguide: Propagation Characteristics [m0224] In this section, we consider the propagation characteristics of TE and TM modes in rectangular waveguides. Because these modes exhibit the same phase dependence on z, findings of this section apply equally to both sets of modes. Recall that the TM modes in a rectangular waveguide are given by: eE(m,n) z = E(m,n) 0 sin mπ a x sin nπ b y e−jk(m,n) z z (6.241) where E(m,n) 0 is an arbitrary constant (determined in part by sources), and: k(m,n) z = r ω2µǫ − mπ a 2 − nπ b 2 (6.242) The TE modes in a rectangular waveguide are: eH(m,n) z = H(m,n) 0 cos mπ a x cos nπ b y e−jk(m,n) z z (6.243) where H(m,n) 0 is an arbitrary constant (determined in part by sources). Cutoff frequency. First, observe that values of k(m,n) z obtained from Equation 6.242 are not necessarily real-valued. For any given value of m, (k(m,n) z )2 will be negative for all values of n greater than some value. Similarly, for any given value of n, (k(m,n) z )2 will be negative for all values of m greater than some value. Should either of these conditions occur, we find: k(m,n) z 2 = ω2µǫ − mπ a 2 − nπ b 2 < 0 = − ω2µǫ − mπ a 2 − nπ b 2 = −α2 (6.244) where α is a positive real-valued constant. So: k(m,n) z = ±jα (6.245) Subsequently: e−jk(m,n) z z = e−j(±jα)z = e±αz (6.246) The “+” sign option corresponds to a wave that grows exponentially in magnitude with increasing z, which is non-physical behavior. Therefore: e−jk(m,n) z z = e−αz (6.247) Summarizing: When values of m or n are such that (k(m,n) z )2 < 0, the magnitude of the associated wave is no longer constant with z. Instead, the magnitude of the wave decreases exponentially with increasing z. Such a wave does not effectively convey power through the waveguide, and is said to be cut off. Since waveguides are normally intended for the efficient transfer of power, it is important to know the criteria for a mode to be cut off. Since cutoff occurs when (k(m,n) z )2 < 0, cutoff occurs when: ω2µǫ > mπ a 2 + nπ b 2 (6.248) Since ω = 2πf: f > 1 2π√µǫ rmπ a 2 + nπ b 2 (6.249) = 1 2√µǫ rm a 2 + n b 2 (6.250) Note that 1/√µǫ is the phase velocity vp for the medium used in the waveguide. With this in mind, let us define: vpu ≜ 1 √µǫ (6.251) This is the phase velocity in an unbounded medium having the same permeability and permittivity as the interior of the waveguide. Thus: f > vpu 2 rm a 2 + n b 2 (6.252) In other words, the mode (m, n) avoids being cut off if the frequency is high enough to meet this criterion. Thus, it is useful to make the following definition: fmn ≜vpu 2 rm a 2 + n b 2 (6.253) The cutoff frequency fmn (Equation 6.253) is the lowest frequency for which the mode (m, n) is able to propagate (i.e., not cut off). | Electromagnetics_Vol2_Page_130_Chunk2305 |
118 CHAPTER 6. WAVEGUIDES Example 6.4. Cutoff frequencies for WR-90. WR-90 is a popular implementation of rectangular waveguide. WR-90 is air-filled with dimensions a = 22.86 mm and b = 10.16 mm. Determine cutoff frequencies and, in particular, the lowest frequency at which WR-90 can be used. Solution. Since WR-90 is air-filled, µ ≈µ0, ǫ ≈ǫ0, and vpu ≈1√µ0ǫ0 ∼= 3.00 × 108 m/s. Cutoff frequencies are given by Equation 6.253. Recall that there are no non-zero TE or TM modes with m = 0 and n = 0. Since a > b, the lowest non-zero cutoff frequency is achieved when m = 1 and n = 0. In this case, Equation 6.253 yields f10 = 6.557 GHz; this is the lowest frequency that is able to propagate efficiently in the waveguide. The next lowest cutoff frequency is f20 = 13.114 GHz. The third lowest cutoff frequency is f01 = 14.754 GHz. The lowest-order TM mode that is non-zero and not cut off is TM11 (f11 = 16.145 GHz). Phase velocity. The phase velocity for a wave propagating within a rectangular waveguide is greater than that of electromagnetic radiation in unbounded space. For example, the phase velocity of any propagating mode in a vacuum-filled waveguide is greater than c, the speed of light in free space. This is a surprising result. Let us first derive this result and then attempt to make sense of it. Phase velocity vp in the rectangular waveguide is given by vp ≜ ω k(m,n) z (6.254) = ω q ω2µǫ −(mπ/a)2 −(nπ/b)2 (6.255) Immediately we observe that phase velocity seems to be different for different modes. Dividing the numerator and denominator by β = ω√µǫ, we obtain: vp = 1 √µǫ 1 r 1 −(ω2µǫ)−1 h (mπ/a)2 + (nπ/b)2i (6.256) Note that 1/√µǫ is vpu, as defined earlier. Employing Equation 6.253 and also noting that ω = 2πf, Equation 6.256 may be rewritten in the following form: vp = vpu q 1 −(fmn/f)2 (6.257) For any propagating mode, f > fmn; subsequently, vp > vpu. In particular, vp > c for a vacuum-filled waveguide. How can this not be a violation of fundamental physics? As noted in Section 6.1, phase velocity is not necessarily the speed at which information travels, but is merely the speed at which a point of constant phase travels. To send information, we must create a disturbance in the otherwise sinusoidal excitation presumed in the analysis so far. The complex field structure creates points of constant phase that travel faster than the disturbance is able to convey information, so there is no violation of physical principles. Group velocity. As noted in Section 6.1, the speed at which information travels is given by the group velocity vg. In unbounded space, vg = vp, so the speed of information is equal to the phase velocity in that case. In a rectangular waveguide, the situation is different. We find: vg = ∂k(m,n) z ∂ω !−1 (6.258) = vpu q 1 −(fmn/f)2 (6.259) which is always less than vpu for a propagating mode. Note that group velocity in the waveguide depends on frequency in two ways. First, because fmn takes on different values for different modes, group velocity is different for different modes. Specifically, higher-order modes propagate more slowly than lower-order modes having the same frequency. This is known as modal dispersion. Secondly, note that the group velocity of any given mode depends on frequency. This is known as chromatic dispersion. | Electromagnetics_Vol2_Page_131_Chunk2306 |
6.10. RECTANGULAR WAVEGUIDE: PROPAGATION CHARACTERISTICS 119 The speed of a signal within a rectangular waveg- uide is given by the group velocity of the as- sociated mode (Equation 6.259). This speed is less than the speed of propagation in unbounded media having the same permittivity and perme- ability. Speed depends on the ratio fmn/f, and generally decreases with increasing frequency for any given mode. Example 6.5. Speed of propagating in WR-90. Revisiting WR-90 waveguide from Example 6.4: What is the speed of propagation for a narrowband signal at 10 GHz? Solution. Let us assume that “narrowband” here means that the bandwidth is negligible relative to the center frequency, so that we need only consider the center frequency. As previously determined, the lowest-order propagating mode is TE10, for which f10 = 6.557 GHz. The next-lowest cutoff frequency is f20 = 13.114 GHz. Therefore, only the TE10 mode is available for this signal. The group velocity for this mode at the frequency of interest is given by Equation 6.259. Using this equation, the speed of propagation is found to be ∼= 2.26 × 108 m/s, which is about 75.5% of c. [m0212] | Electromagnetics_Vol2_Page_132_Chunk2307 |
120 CHAPTER 6. WAVEGUIDES Image Credits Fig. 6.1: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Geometry for analysis of fields in parallel plate waveguide.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/), modified. Fig. 6.2: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:TE component of electric field in parallel plate waveguide.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 6.3: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Magnitude of the two lowest-order TE modes in parallel plate waveguide.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 6.4: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:TM component of electric field in parallel plate waveguide.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 6.5: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:TM component of electric field in parallel plate waveguide.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/), modified. | Electromagnetics_Vol2_Page_133_Chunk2308 |
Chapter 7 Transmission Lines Redux 7.1 Parallel Wire Transmission Line [m0188] A parallel wire transmission line consists of wires separated by a dielectric spacer. Figure 7.1 shows a common implementation, commonly known as “twin lead.” The wires in twin lead line are held in place by a mechanical spacer comprised of the same low-loss dielectric material that forms the jacket of each wire. Very little of the total energy associated with the electric and magnetic fields lies inside this material, so the jacket and spacer can usually be neglected for the purposes of analysis and electrical design. Parallel wire transmission line is often employed in radio applications up to about 100 MHz as an alternative to coaxial line. Parallel wire line has the advantages of lower cost and lower loss than coaxial line in this frequency range. However, parallel wire line lacks the self-shielding property of coaxial cable; i.e., the electromagnetic fields of coaxial line are isolated by the outer conductor, whereas those of c⃝SpinningSpark, Inductiveload CC BY SA 3.0 (modified) Figure 7.1: Twin lead, a commonly-encountered form of parallel wire transmission line. c⃝C. Wang CC BY SA 4.0 Figure 7.2: Parallel wire transmission line structure and design parameters. parallel wire line are exposed and prone to interaction with nearby structures and devices. This prevents the use of parallel wire line in many applications. Another discriminator between parallel wire line and coaxial line is that parallel wire line is differential.1 The conductor geometry is symmetric and neither conductor is favored as a signal datum (“ground”). Thus, parallel wire line is commonly used in applications where the signal sources and/or loads are also differential; common examples are the dipole antenna and differential amplifiers.2 Figure 7.2 shows a cross-section of parallel wire line. Relevant parameters include the wire diameter, d; and the center-to-center spacing, D. 1The references in “Additional Reading” at the end of this section may be helpful if you are not familiar with this concept. 2This is in contrast to “single-ended” line such as coaxial line, which has conductors of different cross-sections and the outer con- ductor is favored as the datum. Electromagnetics Vol. 2. c⃝2020 S.W. Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics-vol-2 | Electromagnetics_Vol2_Page_134_Chunk2309 |
122 CHAPTER 7. TRANSMISSION LINES REDUX c⃝S. Lally CC BY SA 4.0 Figure 7.3: Structure of the electric and magnetic fields for a cross-section of parallel wire line. In this case, the wave is propagating away from the viewer. The associated field structure is transverse electromagnetic (TEM) and is therefore completely described by a single cross-section along the line, as shown in Figure 7.3. Expressions for these fields exist, but are complex and not particularly useful except as a means to calculate other parameters of interest. One of these parameters is, of course, the characteristic impedance since this parameter plays an important role in the analysis and design of systems employing transmission lines. The characteristic impedance may be determined using the “lumped element” transmission line model using the following expression: Z0 = s R′ + jωL′ G′ + jωC′ (7.1) where R′, G′, C′, and L′ are the resistance, conductance, capacitance, and inductance per unit length, respectively. This analysis is considerably simplified by neglecting loss; therefore, let us assume the “low-loss” conditions R′ ≪ωL′ and G′ ≪ωC′. Then we find: Z0 ≈ r L′ C′ (low loss) (7.2) and the problem is reduced to determining inductance and capacitance of the transmission line. These are L′ = µ0 π ln (D/d) + q (D/d)2 −1 (7.3) C′ = πǫ ln (D/d) + q (D/d)2 −1 (7.4) Because the wire separation D is typically much greater than the wire diameter d, D/d ≫1 and so q (D/d)2 −1 ≈D/d. This leads to the simplified expressions L′ ≈µ0 π ln (2D/d) (D ≫d) (7.5) C′ ≈ πǫ ln (2D/d) (D ≫d) (7.6) Now returning to Equation 7.2: Z0 ≈1 π rµ0 ǫ ln (2D/d) (7.7) Noting that ǫ = ǫrǫ0 and p µ0/ǫ0 ≜η0, we obtain Z0 ≈1 π η0 √ǫr ln (2D/d) (7.8) The characteristic impedance of parallel wire line, assuming low-loss conditions and wire spac- ing much greater than wire diameter, is given by Equation 7.8. Observe that the characteristic impedance of parallel wire line increases with increasing D/d. Since this ratio is large, the characteristic impedance of parallel wire line tends to be large relative to common values of other kinds of TEM transmission line, such as coaxial line and microstrip line. An example follows. Example 7.1. 300 Ωtwin-lead. A commonly-encountered form of parallel wire transmission line is 300 Ωtwin-lead. Although implementations vary, the wire diameter is usually about 1 mm and and the wire spacing is usually about 6 mm. The relative permittivity of the medium ǫr ≈1 for the purposes of calculating transmission line parameters, since the jacket and spacer have only a small effect on the fields. For these values, Equation 7.8 gives Z0 ≈298 Ω, as expected. Under the assumption that the wire jacket/spacer material has a negligible effect on the electromagnetic | Electromagnetics_Vol2_Page_135_Chunk2310 |
7.2. MICROSTRIP LINE REDUX 123 fields, and that the line is suspended in air so that ǫr ≈1, the phase velocity vp for a parallel wire line is approximately that of any electromagnetic wave in free space; i.e., c. In practical twin-lead, the effect of a plastic jacket/spacer material is to reduce the phase velocity by a few percent up to about 20%, depending on the materials and details of construction. So in practice vp ≈0.8c to 0.9c for twin-lead line. Additional Reading: • “Twin-lead” on Wikipedia. • “Differential signaling” on Wikipedia. • Sec. 8.7 (“Differential Circuits”) in S.W. Ellingson, Radio Systems Engineering, Cambridge Univ. Press, 2016. 7.2 Microstrip Line Redux [m0186] A microstrip transmission line consists of a narrow metallic trace separated from a metallic ground plane by a slab of dielectric material, as shown in Figure 7.4. This is a natural way to implement a transmission line on a printed circuit board, and so accounts for an important and expansive range of applications. The reader should be aware that microstrip is distinct from stripline, which is a distinct type of transmission line; see “Additional Reading” at the end of this section for disambiguation of these terms. A microstrip line is single-ended3 in the sense that the conductor geometry is asymmetric and the one conductor – namely, the ground plane – also normally serves as ground for the source and load. The spacer material is typically a low-loss dielectric material having permeability approximately equal to that of free space (µ ≈µ0) and relative permittivity ǫr in the range of 2 to about 10 or so. The structure of a microstrip line is similar to that of a parallel plate waveguide (Section 6.6), with the obvious difference that one of the “plates” has finite length in the direction perpendicular to the direction of propagation. Despite this difference, the parallel plate waveguide provides some useful insight into the operation of the microstrip line. Microstrip line is nearly always operated below the cutoff frequency of 3The reference in “Additional Reading” at the end of this section may be helpful if you are not familiar with this concept. h W l r t 0 c⃝7head7metal7 CC BY SA 3.0 (modified) Figure 7.4: Microstrip transmission line structure and design parameters. | Electromagnetics_Vol2_Page_136_Chunk2311 |
124 CHAPTER 7. TRANSMISSION LINES REDUX Figure 7.5: Approximate structure of the electric and magnetic fields within microstrip line, assuming TM0 operation. The fields outside the line are possibly sig- nificant, complicated, and not shown. In this case, the wave is propagating away from the viewer. all but the TM0 mode, whose cutoff frequency is zero. This guarantees that only the TM0 mode can propagate. The TM0 mode has the form of a uniform plane wave, as shown in Figure 7.5. The electric field is oriented in the direction perpendicular to the plates, and magnetic field is oriented in the direction parallel to the plates. The direction of propagation E × H for the TM0 mode always points in the same direction; namely, along the axis of the transmission line. Therefore, microstrip lines nominally exhibit transverse electromagnetic (TEM) field structure. The limited width W of the trace results in a “fringing field” – i.e., significant deviations from TM0 field structure in the dielectric beyond the edges of the trace and above the trace. The fringing fields may play a significant role in determining the characteristic impedance Z0. Since Z0 is an important parameter in the analysis and design of systems using transmission lines, we are motivated to not only determine Z0 for the microstrip line, but also to understand how variation in W affects Z0. Let us address this issue by considering the following three special cases, in order: • W ≫h, which we shall refer to as the “wide” case. • W ≪h, which we shall refer to as the “narrow” case. • W ∼h; i.e., the intermediate case in which h and W equal to within an order of magnitude or so. Wide case. If W ≫h, most of the energy associated with propagating waves lies directly underneath the trace, and Figure 7.5 provides a relatively accurate impression of the fields in this region. The characteristic impedance Z0 may be determined using the “lumped element” transmission line model using the following expression: Z0 = s R′ + jωL′ G′ + jωC′ (7.9) where R′, G′, C′, and L′ are the resistance, conductance, capacitance, and inductance per unit length, respectively. For the present analysis, nothing is lost by assuming loss is negligible; therefore, let us assume R′ ≪ωL′ and G′ ≪ωG′, yielding Z0 ≈ r L′ C′ (7.10) Thus the problem is reduced to determining capacitance and inductance of the transmission line. Wide microstrip line resembles a parallel-plate capacitor whose plate spacing d is very small compared to the plate area A. In this case, fringing fields may be considered negligible and one finds that the capacitance C is given by C ≈ǫA d (parallel plate capacitor) (7.11) In terms of wide microstrip line, A = Wl where l is length, and d = h. Therefore: C′ ≜C l ≈ǫW h (W ≫h) (7.12) To determine L′, consider the view of the microstrip line shown in Figure 7.6. Here a current source applies a steady current I on the left, and a resistive load closes the current loop on the right. Ampere’s law for magnetostatics and the associated “right hand rule” require that H is directed into the page and is approximately uniform. The magnetic flux between the trace and the ground plane is Φ = Z S B · ds (7.13) where S is the surface bounded by the current loop and B = µ0H. In the present problem, the area of S is hl and H is approximately constant over S, so: Φ ≈µ0Hhl (7.14) | Electromagnetics_Vol2_Page_137_Chunk2312 |
7.2. MICROSTRIP LINE REDUX 125 c⃝C. Wang CC BY SA 4.0 Figure 7.6: View from the side of a microstrip line, used to determine L′. where H is the magnitude of H. Next, recall that: L ≜Φ I (7.15) So, we may determine L if we are able to obtain an expression for I in terms of H. This can be done as follows. First, note that we can express I in terms of the current density on the trace; i.e., I = JsW (7.16) where Js is the current density (SI base units of A/m) on the trace. Next, we note that the boundary condition for H on the trace requires that the discontinuity in the tangent component of H going from the trace (where the magnetic field intensity is H) to beyond the trace (i.e., outside the transmission line, where the magnetic field intensity is approximately zero) be equal to the surface current. Thus, Js ≈H and I = JsW ≈HW (7.17) Returning to Equation 7.15, we find: L ≜Φ I ≈µ0Hhl HW = µ0hl W (7.18) Subsequently, L′ ≜L l ≈µ0h W (W ≫h) (7.19) Now the characteristic impedance is found to be Z0 ≈ r L′ C′ ≈ s µ0h/W ǫW/h = rµ0 ǫ h W (7.20) The factor p µ0/ǫ is recognized as the wave impedance η. It is convenient to express this in terms c⃝S. Lally CC BY SA 4.0 Figure 7.7: Electric and magnetic fields lines for nar- row (W ≪h) microstrip line. of the free space wave impedance η0 ≜ p µ0/ǫ0, leading to the expression we seek: Z0 ≈η0 √ǫr h W (W ≫h) (7.21) The characteristic impedance of “wide” (W ≫ h) microstrip line is given approximately by Equation 7.21. It is worth noting that the characteristic impedance of wide transmission line is proportional to h/W. The factor h/W figures prominently in the narrow- and intermediate-width cases as well, as we shall soon see. Narrow case. Figure 7.5 does not accurately depict the fields in the case that W ≪h. Instead, much of the energy associated with the electric and magnetic fields lies beyond and above the trace. Although these fields are relatively complex, we can develop a rudimentary model of the field structure by considering the relevant boundary conditions. In particular, the tangential component of E must be zero on the surfaces of the trace and ground plane. Also, we expect magnetic field lines to be perpendicular to the electric field lines so that the direction of propagation is along the axis of the line. These insights lead to Figure 7.7. Note that the field lines shown in Figure 7.7 are quite similar to those for parallel wire line (Section 7.1). If we choose diameter d = W and wire spacing D = 2h, then the fields of the microstrip line in the dielectric spacer must be similar to those of the parallel wire line between the lines. This is shown in | Electromagnetics_Vol2_Page_138_Chunk2313 |
126 CHAPTER 7. TRANSMISSION LINES REDUX Figure 7.8: Noting the similarity of the fields in nar- row microstrip line to those in a parallel wire line. Figure 7.9: Modeling the fields in narrow microstrip line as those of a parallel wire line, now introducing the ground plane. Figure 7.8. Note that the fields above the dielectric spacer in the microstrip line will typically be somewhat different (since the material above the trace is typically different; e.g., air). However, it remains true that we expect most of the energy to be contained in the dielectric, so we shall assume that this difference has a relatively small effect. Next, we continue to refine the model by introducing the ground plane, as shown in Figure 7.9: The fields in the upper half-space are not perturbed when we do this, since the fields in Figure 7.8 already satisfy the boundary conditions imposed by the new ground plane. Thus, we see that the parallel wire transmission line provides a pretty good guide to the structure of the fields for the narrow transmission line, at least in the dielectric region of the upper half-space. We are now ready to estimate the characteristic impedance Z0 ≈ p L′/C′ of low-loss narrow microstrip line. Neglecting the differences noted above, we estimate that this is roughly equal to the characteristic impedance of the analogous (d = W and D = 2h) parallel wire line. Under this assumption, we obtain from Equation 7.8: Z0 ∼1 π η0 √ǫr ln (4h/W) (W ≪h) (7.22) This estimate will exhibit only “order of magnitude” accuracy (hence the “∼” symbol), since the contribution from the fields in half of the relevant space is ignored. However we expect that Equation 7.22 will accurately capture the dependence of Z0 on the parameters ǫr and h/W. These properties can be useful for making fine adjustments to a microstrip design. The characteristic impedance of “narrow” (W ≪ h) microstrip line can be roughly approximated by Equation 7.22. Intermediate case. An expression for Z0 in the intermediate case – even a rough estimate – is difficult to derive, and beyond the scope of this book. However, we are not completely in the dark, since we can reasonably anticipate that Z0 in the intermediate case should be intermediate to estimates provided by the wide and narrow cases. This is best demonstrated by example. Figure 7.10 shows estimates of Z0 using | Electromagnetics_Vol2_Page_139_Chunk2314 |
7.2. MICROSTRIP LINE REDUX 127 Figure 7.10: Z0 for FR4 as a function of h/W, as determined by the “wide” and “narrow” approxima- tions, along with the Wheeler 1977 formula. Note that the vertical and horizontal axes of this plot are in log scale. the wide and narrow expressions (blue and green curves, respectively) for a particular implementation of microstrip line (see Example 7.2 for details). Since Z0 varies smoothly with h/W, it is reasonable to expect that simply averaging the values generated using the wide and narrow assumptions would give a pretty good estimate when h/W ∼1. However, it is also possible to obtain an accurate estimate directly. A widely-accepted and broadly-applicable formula is provided in Wheeler 1977 (cited in “Additional Reading” at the end of this section). This expression is valid over the full range of h/W, not merely the intermediate case of h/W ∼1. Here it is: Z0 ≈42.4 Ω √ǫr + 1 × ln " 1 + 4h W ′ K + r K2 + 1 + 1/ǫr 2 π2 !# (7.23) where K ≜14 + 8/ǫr 11 4h W ′ (7.24) and W ′ is W adjusted to account for the thickness t of the microstrip line. Typically t ≪W and t ≪h, for which W ′ ≈W. Although complicated, this formula should not be completely surprising. For example, note the parameters h and W appear in this formula as the factor 4h/W, which we encountered in the “narrow” approximation. Also, we see that the formula indicates Z0 increases with increasing h/W and decreasing ǫr, as we determined in both the “narrow” and “wide” cases. The following example provides a demonstration for the very common case of microstrip implementation in FR4 circuit board material. Example 7.2. Characteristic impedance of microstrip lines in FR4. FR4 printed circuit boards consist of a substrate having h ∼= 1.575 mm and ǫr ≈4.5. Figure 7.10 shows Z0 as a function of h/W, as determined by the wide and narrow approximations, along with the value obtained using the Wheeler 1977 formula. The left side of the plot represents the wide condition, whereas the right side of this plot represents the narrow condition. The Wheeler 1977 formula is an accurate estimate over the entire horizontal span. Note that the wide approximation is accurate only for h/W < 0.1 or so, improving with decreasing h/W as expected. The narrow approximation overestimates Z0 by a factor of about 1.4 (i.e., about 40%). This is consistent with our previous observation that this approximation should yield only about “order of magnitude” accuracy. Nevertheless, the narrow approximation exhibits approximately the same rate of increase with h/W as does the Wheeler 1977 formula. Also worth noting from Figure 7.10 is the important and commonly-used result that Z0 ≈50 Ωis obtained for h/W ≈0.5. Thus, a 50 Ωmicrostrip line in FR4 has a trace width of about 3 mm. A useful “take away” from this example is that the wide and narrow approximations serve as useful guides for understanding how Z0 changes as a function of h/W and ǫr. This is useful especially for making adjustments to a microstrip line design once an accurate value of Z0 is obtained from some other | Electromagnetics_Vol2_Page_140_Chunk2315 |
128 CHAPTER 7. TRANSMISSION LINES REDUX method; e.g., using the Wheeler 1977 formula or from measurements. FR4 circuit board construction is so common that the result from the previous example deserves to be highlighted: In FR4 printed circuit board construction (sub- strate thickness 1.575 mm, relative permittivity ≈4.5, negligible trace thickness), Z0 ≈50 Ω requires a trace width of about 3 mm. Z0 scales roughly in proportion to h/W around this value. Simpler approximations for Z0 are also commonly employed in the design and analysis of microstrip lines. These expressions are limited in the range of h/W for which they are valid, and can usually be shown to be special cases or approximations of Equation 7.23. Nevertheless, they are sometimes useful for quick “back of the envelope” calculations. Wavelength in microstrip line. An accurate general formula for wavelength λ in microstrip line is similarly difficult to derive. A useful approximate technique employs a result from the theory of uniform plane waves in unbounded media. For such waves, the phase propagation constant β is given by β = ω√µǫ (7.25) It turns out that the electromagnetic field structure for the guided wave in a microstrip line is similar in the sense that it exhibits TEM field structure, as does the uniform plane wave. However, the guided wave is different in that it exists in both the dielectric spacer and the air above the spacer. Thus, we presume that β for the guided wave can be approximated as that of a uniform plane wave in unbounded media having the same permeability µ0 but a different relative permittivity, which we shall assign the symbol ǫr,eff (for “effective relative permittivity”). Then: β ≈ω√µ0 ǫr,eff ǫ0 (low-loss microstrip) = β0√ǫr,eff (7.26) In other words, the phase propagation constant in a microstrip line can be approximated as the free-space phase propagation β0 ≜ω√µ0ǫ0 times a correction factor √ǫr,eff. Next, ǫr,eff is crudely approximated as the average of the relative permittivity of the dielectric slab and the relative permittivity of free space; i.e.,: ǫr,eff ≈ǫr + 1 2 (7.27) Various refinements exist to improve on this approximation; however, in practice, variations in the value of ǫr for the dielectric due to manufacturing processes typically make a more precise estimate irrelevant. Using this concept, we obtain λ = 2π β = 2π β0√ǫr,eff = λ0 √ǫr,eff (7.28) where λ0 is the free-space wavelength c/f. Similarly the phase velocity vp can be estimated using the relationship vp = ω β = c √ǫr,eff (7.29) i.e., the phase velocity in microstrip is slower than c by a factor of √ǫr,eff. Example 7.3. Wavelength and phase velocity in microstrip in FR4 printed circuit boards. FR4 is a low-loss fiberglass epoxy dielectric that is commonly used to make printed circuit boards (see “Additional Reading” at the end of this section). For FR4, ǫr ≈4.5. The effective relative permittivity is therefore: ǫr,eff ≈(4.5 + 1)/2 = 2.75 Thus, we estimate the phase velocity for the wave guided by this line to be about c/ √ 2.75; i.e., 60% of c. Similarly, the wavelength of this wave is about 60% of the free space wavelength. In practice, these values are found to be slightly less; typically 50%–55%. The difference is attributable to the crude approximation of Equation 7.27. Additional Reading: • “Microstrip” on Wikipedia. | Electromagnetics_Vol2_Page_141_Chunk2316 |
7.3. ATTENUATION IN COAXIAL CABLE 129 • “Printed circuit board” on Wikipedia. • “Stripline” on Wikipedia. • “Single-ended signaling” on Wikipedia. • Sec. 8.7 (“Differential Circuits”) in S.W. Ellingson, Radio Systems Engineering, Cambridge Univ. Press, 2016. • H.A. Wheeler, “Transmission Line Properties of a Strip on a Dielectric Sheet on a Plane,” IEEE Trans. Microwave Theory & Techniques, Vol. 25, No. 8, Aug 1977, pp. 631–47. • “FR-4” on Wikipedia. 7.3 Attenuation in Coaxial Cable [m0189] In this section, we consider the issue of attenuation in coaxial transmission line. Recall that attenuation can be interpreted in the context of the “lumped element” equivalent circuit transmission line model as the contributions of the resistance per unit length R′ and conductance per unit length G′. In this model, R′ represents the physical resistance in the inner and outer conductors, whereas G′ represents loss due to current flowing directly between the conductors through the spacer material. The parameters used to describe the relevant features of coaxial cable are shown in Figure 7.11. In this figure, a and b are the radii of the inner and outer conductors, respectively. σic and σoc are the conductivities (SI base units of S/m) of the inner and outer conductors, respectively. Conductors are assumed to be non-magnetic; i.e., having permeability µ equal to the free space value µ0. The spacer material is assumed to be a lossy dielectric having relative permittivity ǫr and conductivity σs. Resistance per unit length. The resistance per unit length is the sum of the resistances of the inner and outer conductor per unit length. The resistance per unit length of the inner conductor is determined by σic and the effective cross-sectional area through which the current flows. The latter is equal to the circumference 2πa times the skin depth δic of the ϵr σs σic σoc b a Figure 7.11: Parameters defining the design of a coax- ial cable. | Electromagnetics_Vol2_Page_142_Chunk2317 |
130 CHAPTER 7. TRANSMISSION LINES REDUX inner conductor, so: R′ ic ≈ 1 (2πa · δic) σic for δic ≪a (7.30) This expression is only valid for δic ≪a because otherwise the cross-sectional area through which the current flows is not well-modeled as a thin ring near the surface of the conductor. Similarly, we find the resistance per unit length of the outer conductor is R′ oc ≈ 1 (2πb · δoc) σoc for δoc ≪t (7.31) where δoc is the skin depth of the outer conductor and t is the thickness of the outer conductor. Therefore, the total resistance per unit length is R′ = R′ ic + R′ oc ≈ 1 (2πa · δic) σic + 1 (2πb · δoc) σoc (7.32) Recall that skin depth depends on conductivity. Specifically: δic = p 2/ωµσic (7.33) δoc = p 2/ωµσoc (7.34) Expanding Equation 7.32 to show explicitly the dependence on conductivity, we find: R′ ≈ 1 2π p 2/ωµ0 1 a√σic + 1 b√σoc (7.35) At this point it is convenient to identify two particular cases for the design of the cable. In the first case, “Case I,” we assume σoc ≫σic. Since b > a, we have in this case R′ ≈ 1 2π p 2/ωµ0 1 a√σic = 1 2πδicσic 1 a (Case I) (7.36) In the second case, “Case II,” we assume σoc = σic. In this case, we have R′ ≈ 1 2π p 2/ωµ0 1 a√σic + 1 b√σic = 1 2πδicσic 1 a + 1 b (Case II) (7.37) A simpler way to deal with these two cases is to represent them both using the single expression R′ ≈ 1 2πδicσic 1 a + C b (7.38) where C = 0 in Case I and C = 1 in Case II. Conductance per unit length. The conductance per unit length of coaxial cable is simply that of the associated coaxial structure at DC; i.e., G′ = 2πσs ln (b/a) (7.39) Unlike resistance, the conductance is independent of frequency, at least to the extent that σs is independent of frequency. Attenuation. The attenuation of voltage and current waves as they propagate along the cable is represented by the factor e−αz, where z is distance traversed along the cable. It is possible to find an expression for α in terms of the material and geometry parameters using: γ ≜ p (R′ + jωL′) (G′ + jωC′) = α + jβ (7.40) where L′ and C′ are the inductance per unit length and capacitance per unit length, respectively. These are given by L′ = µ 2π ln (b/a) (7.41) and C′ = 2πǫ0ǫr ln (b/a) (7.42) In principle we could solve Equation 7.40 for α. However, this course of action is quite tedious, and a simpler approximate approach facilitates some additional insights. In this approach, we define parameters αR associated with R′ and αG associated with G′ such that e−αRze−αGz = e−(αR+αG)z = e−αz (7.43) which indicates α = αR + αG (7.44) Next we postulate αR ≈KR R′ Z0 (7.45) | Electromagnetics_Vol2_Page_143_Chunk2318 |
7.3. ATTENUATION IN COAXIAL CABLE 131 where Z0 is the characteristic impedance Z0 ≈η0 2π 1 √ǫr ln b a (low loss) (7.46) and where KR is a unitless constant to be determined. The justification for Equation 7.45 is as follows: First, αR must increase monotonically with increasing R′. Second, R′ must be divided by an impedance in order to obtain the correct units of 1/m. Using similar reasoning, we postulate αG ≈KGG′Z0 (7.47) where KG is a unitless constant to be determined. The following example demonstrates the validity of Equations 7.45 and 7.47, and will reveal the values of KR and KG. Example 7.4. Attenuation constant for RG-59. RG-59 is a popular form of coaxial cable having the parameters a ∼= 0.292 mm, b ∼= 1.855 mm, σic ∼= 2.28 × 107 S/m, σs ∼= 5.9 × 10−5 S/m, and ǫr ∼= 2.25. The conductivity σoc of the outer conductor is difficult to quantify because it consists of a braid of thin metal strands. However, σoc ≫σic, so we may assume Case I; i.e., σoc ≫σic, and subsequently C = 0. Figure 7.12 shows the components αG and αR computed for the particular choice KR = KG = 1/2. The figure also shows αG + αR, along with α computed using Equation 7.40. We find that the agreement between these values is very good, which is compelling evidence that the ansatz is valid and KR = KG = 1/2. Note that there is nothing to indicate that the results demonstrated in the example are not generally true. Thus, we come to the following conclusion: The attenuation constant α ≈αG + αR where αG ≜R′/2Z0 and αR ≜G′Z0/2. Minimizing attenuation. Let us now consider if there are design choices which minimize the attenuation of coaxial cable. Since α = αR + αG, we Figure 7.12: Comparison of α = Re {γ} to αR, αG, and αR + αG for KR = KG = 1/2. The result for α has been multiplied by 1.01; otherwise the curves would be too close to tell apart. may consider αR and αG independently. Let us first consider αG: αG ≜1 2G′Z0 ≈1 2 · 2πσs ln (b/a) · 1 2π η0 √ǫr ln (b/a) = η0 2 σs √ǫr (7.48) It is clear from this result that αG is minimized by minimizing σs/√ǫr. Interestingly the physical dimensions a and b have no discernible effect on αG. Now we consider αR: αR ≜R′ 2Z0 = 1 2 (1/2πδicσic) [1/a + C/b] (1/2π) | Electromagnetics_Vol2_Page_144_Chunk2319 |
132 CHAPTER 7. TRANSMISSION LINES REDUX Here we see that αR is minimized by minimizing ǫr/σic. It’s not surprising to see that we should maximize σic. However, it’s a little surprising that we should minimize ǫr. Furthermore, this is in contrast to αG, which is minimized by maximizing ǫr. Clearly there is a tradeoff to be made here. To determine the parameters of this tradeoff, first note that the result depends on frequency: Since αR dominates over αG at sufficiently high frequency (as demonstrated in Figure 7.12), it seems we should minimize ǫr if the intended frequency of operation is sufficiently high; otherwise the optimum value is frequency-dependent. However, σs may vary as a function of ǫr, so a general conclusion about optimum values of σs and ǫr is not appropriate. However, we also see that αR – unlike αG – depends on a and b. This implies the existence of a generally-optimum geometry. To find this geometry, we minimize αR by taking the derivative with respect to a, setting the result equal to zero, and solving for a and/or b. Here we go: ∂ ∂aαR = 1 2 √ 2 · η0 rωµ0ǫr σic · ∂ ∂a [1/a + C/b] ln (b/a) (7.51) This derivative is worked out in an addendum at the end of this section. Using the result from the addendum, the right side of Equation 7.51 can be written as follows: 1 2 √ 2 · η0 rωµ0ǫr σic · −1 a2 ln (b/a) + 1/a + C/b a ln2 (b/a) (7.52) In order for ∂αR/∂a = 0, the factor in the square brackets above must be equal to zero. After a few steps of algebra, we find: ln (b/a) = 1 + C b/a (7.53) In Case I (σoc ≫σic), C = 0 so: b/a = e ∼= 2.72 (Case I) (7.54) In Case II (σoc = σic), C = 1. The resulting equation can be solved by plotting the function, or by a few iterations of trial and error; either way one quickly finds b/a ∼= 3.59 (Case II) (7.55) Summarizing, we have found that α is minimized by choosing the ratio of the outer and inner radii to be somewhere between 2.72 and 3.59, with the precise value depending on the relative conductivity of the inner and outer conductors. Substituting these values of b/a into Equation 7.46, we obtain: Z0 ≈59.9 Ω √ǫr to 76.6 Ω √ǫr (7.56) as the range of impedances of coaxial cable corresponding to physical designs that minimize attenuation. Equation 7.56 gives the range of characteristic impedances that minimize attenuation for coaxial transmission lines. The precise value within this range depends on the ratio of the conductivity of the outer conductor to that of the inner conductor. Since ǫr ≥1, the impedance that minimizes attenuation is less for dielectric-filled cables than it is for air-filled cables. For example, let us once again consider the RG-59 from Example 7.4. In that case, ǫr ∼= 2.25 and C = 0, indicating Z0 ≈39.9 Ωis optimum for attenuation. The actual characteristic impedance of Z0 is about 75 Ω, so clearly RG-59 is not optimized for attenuation. This is simply because other considerations apply, including power handling capability (addressed in Section 7.4) and the convenience of standard values (addressed in Section 7.5). Addendum: Derivative of a2 ln(b/a). Evaluation of Equation 7.51 requires finding the derivative of a2 ln(b/a) with respect to a. Using the chain rule, we find: ∂ ∂a a2 ln b a = ∂ ∂aa2 ln b a +a2 ∂ ∂a ln b a (7.57) Note ∂ ∂aa2 = 2a (7.58) | Electromagnetics_Vol2_Page_145_Chunk2320 |
7.4. POWER HANDLING CAPABILITY OF COAXIAL CABLE 133 and ∂ ∂a ln b a = ∂ ∂a [ln (b) −ln (a)] = −∂ ∂a ln (a) = −1 a (7.59) So: ∂ ∂a a2 ln b a = [2a] ln b a + a2 −1 a = 2a ln b a −a (7.60) This result is substituted for a2 ln(b/a) in Equation 7.51 to obtain Equation 7.52. 7.4 Power Handling Capability of Coaxial Cable [m0190] The term “power handling” refers to maximum power that can be safely transferred by a transmission line. This power is limited because when the electric field becomes too large, dielectric breakdown and arcing may occur. This may result in damage to the line and connected devices, and so must be avoided. Let Epk be the maximum safe value of the electric field intensity within the line, and let Pmax be the power that is being transferred under this condition. This section addresses the following question: How does one design a coaxial cable to maximize Pmax for a given Epk? We begin by finding the electric potential V within the cable. This can be done using Laplace’s equation: ∇2V = 0 (7.61) Using the cylindrical (ρ, φ, z) coordinate system with the z axis along the inner conductor, we have ∂V/∂φ = 0 due to symmetry. Also we set ∂V/∂z = 0 since the result should not depend on z. Thus, we have: 1 ρ ∂ ∂ρ ρ∂V ∂ρ = 0 (7.62) Solving for V , we have V (ρ) = A ln ρ + B (7.63) where A and B are arbitrary constants, presumably determined by boundary conditions. Let us assume a voltage V0 measured from the inner conductor (serving as the “+” terminal) to the outer conductor (serving as the “−” terminal). For this choice, we have: V (a) = V0 → A ln a + B = V0 (7.64) V (b) = 0 → A ln b + B = 0 (7.65) Subtracting the second equation from the first and solving for A, we find A = −V0/ ln (b/a). Subsequently, B is found to be V0 ln (b) / ln (b/a), and so V (ρ) = −V0 ln (b/a) ln ρ + V0 ln (b) ln (b/a) (7.66) | Electromagnetics_Vol2_Page_146_Chunk2321 |
134 CHAPTER 7. TRANSMISSION LINES REDUX The electric field intensity is given by: E = −∇V (7.67) Again we have ∂V/∂φ = ∂V/∂z = 0, so E = −ˆρ ∂ ∂ρV (7.68) = −ˆρ ∂ ∂ρ −V0 ln (b/a) ln ρ + V0 ln (b) ln (b/a) (7.69) = +ˆρ V0 ρ ln (b/a) (7.70) Note that the maximum electric field intensity in the spacer occurs at ρ = a; i.e., at the surface of the inner conductor. Therefore: Epk = V0 a ln (b/a) (7.71) The power transferred by the line is maximized when the impedances of the source and load are matched to Z0. In this case, the power transferred is V 2 0 /2Z0. Recall that the characteristic impedance Z0 is given in the “low-loss” case as Z0 ≈1 2π η0 √ǫr ln b a (7.72) Therefore, the maximum safe power is Pmax = V 2 0 2Z0 (7.73) ≈ E2 pk a2 ln2 (b/a) 2 · (1/2π) | Electromagnetics_Vol2_Page_147_Chunk2322 |
7.5. WHY 50 OHMS? 135 Using the chain rule, we find: ∂ ∂a 1/a + C/b ln (b/a) = ∂ ∂a 1 a + C b ln−1 b a + 1 a + C b ∂ ∂a ln−1 b a (7.79) Note ∂ ∂a 1 a + C b = −1 a2 (7.80) To handle the quantity in the second set of square brackets, first define v = ln u, where u = b/a. Then: ∂ ∂av−1 = ∂ ∂v v−1 ∂v ∂u ∂u ∂a = −v−2 1 u −ba−2 = −ln−2 b a ha b i −ba−2 = 1 a ln−2 b a (7.81) So: ∂ ∂a 1/a + C/b ln (b/a) = −1 a2 ln−1 b a + 1 a + C b 1 a ln−2 b a (7.82) This result is substituted in Equation 7.75 to obtain Equation 7.76. 7.5 Why 50 Ohms? [m0191] The quantity 50 Ωappears in a broad range of applications across the field of electrical engineering. In particular, it is a very popular value for the characteristic impedance of transmission line, and is commonly specified as the port impedance for signal sources, amplifiers, filters, antennas, and other RF components. So, what’s special about 50 Ω? The short answer is “nothing.” In fact, other standard impedances are in common use – prominent among these is 75 Ω. It is shown in this section that a broad range of impedances – on the order of 10s of ohms – emerge as useful values based on technical considerations such as minimizing attenuation, maximizing power handling, and compatibility with common types of antennas. Characteristic impedances up to 300 Ωand beyond are useful in particular applications. However, it is not practical or efficient to manufacture and sell products for every possible impedance in this range. Instead, engineers have settled on 50 Ωas a round number that lies near the middle of this range, and have chosen a few other values to accommodate the smaller number of applications where there may be specific compelling considerations. So, the question becomes “what makes characteristic impedances in the range of 10s of ohms particularly useful?” One consideration is attenuation in coaxial cable. Coaxial cable is by far the most popular type of transmission line for connecting devices on separate printed circuit boards or in separate enclosures. The attenuation of coaxial cable is addressed in Section 7.3. In that section, it is shown that attenuation is minimized for characteristic impedances in the range (60 Ω) /√ǫr to (77 Ω) /√ǫr, where ǫr is the relative permittivity of the spacer material. So, we find that Z0 in the range 60 Ωto 77 Ωis optimum for air-filled cable, but more like 40 Ωto 50 Ωfor cables using a plastic spacer material having typical ǫr ≈2.25. Thus, 50 Ωis clearly a reasonable choice if a single standard value is to be established for all such cable. Coaxial cables are often required to carry high power signals. In such applications, power handling | Electromagnetics_Vol2_Page_148_Chunk2323 |
136 CHAPTER 7. TRANSMISSION LINES REDUX capability is also important, and is addressed in Section 7.4. In that section, we find the power handling capability of coaxial cable is optimized when the ratio of radii of the outer to inner conductors b/a is about 1.65. For the air-filled cables typically used in high-power applications, this corresponds to a characteristic impedance of about 30 Ω. This is significantly less than the 60 Ωto 77 Ωthat minimizes attenuation in air-filled cables. So, 50 Ω can be viewed as a compromise between minimizing attenuation and maximizing power handling in air-filled coaxial cables. Although the preceding arguments justify 50 Ωas a standard value, one can also see how one might make a case for 75 Ωas a secondary standard value, especially for applications where attenuation is the primary consideration. Values of 50 Ωand 75 Ωalso offer some convenience when connecting RF devices to antennas. For example, 75 Ωis very close to the impedance of the commonly-encountered half-wave dipole antenna (about 73 + j42 Ω), which may make impedance matching to that antenna easier. Another commonly-encountered antenna is the quarter-wave monopole, which exhibits an impedance of about 36 + j21 Ω, which is close to 50 Ω. In fact, we see that if we desire a single characteristic impedance that is equally convenient for applications involving either type of antenna, then 50 Ωis a reasonable choice. A third commonly-encountered antenna is the folded half-wave dipole. This type of antenna is similar to a half-wave dipole but has better bandwidth, and is commonly used in FM and TV systems and land mobile radio (LMR) base stations. A folded half-wave dipole has an impedance of about 300 Ω and is balanced (not single-ended); thus, there is a market for balanced transmission line having Z0 = 300 Ω. However, it is very easy and inexpensive to implement a balun (a device which converts the dipole output from balanced to unbalanced) while simultaneously stepping down impedance by a factor of 4; i.e., to 75 Ω. Thus, we have an additional application for 75 Ωcoaxial line. Finally, note that it is quite simple to implement microstrip transmission line having characteristic impedance in the range 30 Ωto 75 Ω. For example, 50 Ωon commonly-used 1.575 mm FR4 requires a width-to-height ratio of about 2, so the trace is about 3 mm wide. This is a very manageable size and easily implemented in printed circuit board designs. Additional Reading: • “Dipole antenna” on Wikipedia. • “Monopole antenna” on Wikipedia. • “Balun” on Wikipedia. [m0187] | Electromagnetics_Vol2_Page_149_Chunk2324 |
7.5. WHY 50 OHMS? 137 Image Credits Fig. 7.1: c⃝SpinningSpark, Inductiveload, https://commons.wikimedia.org/wiki/File:Twin-lead cable dimension.svg, CC BY-SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/). Minor modifications. Fig. 7.2: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Parallel wire transmission line structure and design parameters.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 7.3: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Electric and Magnetic Field of Wire Cross-Section.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 7.4: c⃝7head7metal7, https://commons.wikimedia.org/wiki/File:Microstrip scheme.svg, CC BY-SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/). Minor modifications from the original. Fig. 7.6: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:View from the side of a microstrip line.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 7.7: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Electric and Magnetic Fields for Microstrip.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). | Electromagnetics_Vol2_Page_150_Chunk2325 |
Chapter 8 Optical Fiber 8.1 Optical Fiber: Method of Operation [m0178] In its simplest form, optical fiber consists of concentric regions of dielectric material as shown in Figure 8.1. A cross-section through the fiber reveals a circular region of transparent dielectric material through which light propagates. This is surrounded by a jacket of dielectric material commonly referred to as cladding. A characteristic of the design of any optical fiber is that the permittivity of the fiber is greater than the permittivity of the cladding. As explained in Section 5.11, this creates conditions necessary for total internal reflection. The mechanism of total internal reflection contains the light within the fiber. In the discipline of optics, the permittivity of a c⃝S. Lally CC BY-SA 4.0 Figure 8.1: Construction of the simplest form of opti- cal fiber. material is commonly quantified in terms of its index of refraction. Index of refraction is the square root of relative permittivity, and is usually assigned the symbol n. Thus, if we define the relative permittivities ǫr,f ≜ǫf/ǫ0 for the fiber and ǫr,c ≜ǫc/ǫ0 for the cladding, then nf ≜√ǫr,f (8.1) nc ≜√ǫr,c (8.2) and nf > nc (8.3) Figure 8.2 illustrates total internal reflection in an optical fiber. In this case, a ray of light in the fiber is incident on the boundary with the cladding. We may treat the light ray as a uniform plane wave. To see why, consider that optical wavelengths range from 120 nm to 700 nm in free space. Wavelength is slightly shorter than this in fiber; specifically, by a factor equal to the square root of the relative permittivity. The fiber is on the order of millimeters in diameter, which is about 4 orders of magnitude greater than the wavelength. Thus, from the perspective of the light ray, the fiber appears to be an unbounded half-space sharing a planar boundary with the cladding, which also appears to be an unbounded half-space. Continuing under this presumption, the criterion for total internal reflection is (from Section 5.11): θi ≥arcsin r ǫr,c ǫr,f = arcsin nc nf (8.4) As long as rays of light approach from angles that satisfy this criterion, the associated power remains in Electromagnetics Vol. 2. c⃝2020 S.W. Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics-vol-2 | Electromagnetics_Vol2_Page_151_Chunk2326 |
8.1. OPTICAL FIBER: METHOD OF OPERATION 139 c⃝S. Lally CC BY-SA 4.0 Figure 8.2: Total internal reflection in optical fiber. the fiber, and is reflected onward. Otherwise, power is lost into the cladding. Example 8.1. Critical angle for optical fiber. Typical values of nf and nc for an optical fiber are 1.52 and 1.49, respectively. What internal angle of incidence is required to maintain total internal reflection? Solution. Using Equation 8.4, θi must be greater than about 78.8◦. In practice this is quite reasonable since we desire light to be traveling approximately parallel the axis of the fiber (θi ≈90◦) anyway. The total internal reflection criterion imposes a limit on the radius of curvature of fiber optic cable. If fiber optic cable is bent such that the radius of curvature is too small, the critical angle will be exceeded at the bend. This will occur even for light rays which are traveling perfectly parallel to the axis of the fiber before they arrive at the bend. Note that the cladding serves at least two roles. First, it determines the critical angle for total internal reflection, and subsequently determines the minimum radius of curvature for lossless operation of the fiber. Second, it provides a place for the evanescent surface waves associated with total internal reflection (Section 5.12) to exist without interference from objects in contact with the fiber. Additional Reading: • “Optical fiber” on Wikipedia. | Electromagnetics_Vol2_Page_152_Chunk2327 |
140 CHAPTER 8. OPTICAL FIBER 8.2 Acceptance Angle [m0192] In this section, we consider the problem of injecting light into a fiber optic cable. The problem is illustrated in Figure 8.3. In this figure, we see light incident from a medium having index of refraction n0, with angle of incidence θi. The light is transmitted with angle of transmission θ2 into the fiber, and is subsequently incident on the surface of the cladding with angle of incidence θ3. For light to propagate without loss within the cable, it is required that sin θ3 ≥nc nf (8.5) since this criterion must be met in order for total internal reflection to occur. Now consider the constraint that Equation 8.5 imposes on θi. First, we note that θ3 is related to θ2 as follows: θ3 = π 2 −θ2 (8.6) therefore sin θ3 = sin π 2 −θ2 (8.7) = cos θ2 (8.8) so cos θ2 ≥nc nf (8.9) Squaring both sides, we find: cos2 θ2 ≥n2 c n2 f (8.10) c⃝S. Lally CC BY-SA 4.0 Figure 8.3: Injecting light into a fiber optic cable. Now invoking a trigonometric identity: 1 −sin2 θ2 ≥n2 c n2 f (8.11) so: sin2 θ2 ≤1 −n2 c n2 f (8.12) Now we relate the θ2 to θi using Snell’s law: sin θ2 = n0 nf sin θi (8.13) so Equation 8.12 may be written: n2 0 n2 f sin2 θi ≤1 −n2 c n2 f (8.14) Now solving for sin θi, we obtain: sin θi ≤1 n0 q n2 f −n2c (8.15) This result indicates the range of angles of incidence which result in total internal reflection within the fiber. The maximum value of θi which satisfies this condition is known as the acceptance angle θa, so: θa ≜arcsin 1 n0 q n2 f −n2c (8.16) This leads to the following insight: In order to effectively launch light in the fiber, it is necessary for the light to arrive from within a cone having half-angle θa with respect to the axis of the fiber. The associated cone of acceptance is illustrated in Figure 8.4. It is also common to define the quantity numerical aperture NA as follows: NA ≜1 n0 q n2 f −n2c (8.17) Note that n0 is typically very close to 1 (corresponding to incidence from air), so it is common to see NA defined as simply q n2 f −n2c. This parameter is commonly used in lieu of the acceptance angle in datasheets for fiber optic cable. | Electromagnetics_Vol2_Page_153_Chunk2328 |
8.3. DISPERSION IN OPTICAL FIBER 141 Example 8.2. Acceptance angle. Typical values of nf and nc for an optical fiber are 1.52 and 1.49, respectively. What are the numerical aperture and the acceptance angle? Solution. Using Equation 8.17 and presuming n0 = 1, we find NA ∼= 0.30. Since sin θa = NA, we find θa = 17.5◦. Light must arrive from within 17.5◦from the axis of the fiber in order to ensure total internal reflection within the fiber. Additional Reading: • “Optical fiber” on Wikipedia. • “Numerical aperture” on Wikipedia. c⃝S. Lally CC BY-SA 4.0 Figure 8.4: Cone of acceptance. 8.3 Dispersion in Optical Fiber [m0193] Light may follow a variety of paths through a fiber optic cable. Each of the paths has a different length, leading to a phenomenon known as dispersion. Dispersion distorts signals and limits the data rate of digital signals sent over fiber optic cable. In this section, we analyze this dispersion and its effect on digital signals. Figure 8.5 shows the variety of paths that light may take through a straight fiber optic cable. The nominal path is shown in Figure 8.5(a), which is parallel to the axis of the cable. This path has the shortest associated propagation time. The path with the longest associated propagation time is shown in Figure 8.5(c). In this case, light bounces within the fiber, each time approaching the core-clad interface at the critical angle for total internal reflection. Any ray approaching at a greater angle is not completely reflected, and so would likely not survive to the end of c⃝S. Lally CC BY-SA 4.0 Figure 8.5: Paths that light may take through a straight fiber optic cable: (a) Along axis, corresponding to minimum path length; (b) Intermediate between (a) and (c); (c) Maximum length, corresponding to thresh- old angle for total internal reflection. | Electromagnetics_Vol2_Page_154_Chunk2329 |
142 CHAPTER 8. OPTICAL FIBER c⃝S. Lally CC BY-SA 4.0 Figure 8.6: A digital signal that might be applied to the input of a fiber optic cable. the cable. Figure 8.5(b) represents the continuum of possibilities between the extreme cases of (a) and (c), with associated propagation times greater than that of case (a) but less than that of case (c). Regardless of how light is inserted into the fiber, all possible paths depicted in Figure 8.5 are likely to exist. This is because fiber is rarely installed in a straight line, but rather follows a multiply-curved path. Each curve results in new angles of incidence upon the core-cladding boundary. The existence of these paths leads to dispersion. To see this, consider the input signal shown in Figure 8.6. This signal has period T, during which time a pulse of length ton < T may be present. Let the minimum propagation time through the fiber (as in Figure 8.5(a)) be τmin. Let the maximum propagation time through the fiber (as in Figure 8.5(c)) be τmax. If τmax −τmin ≪ton, we see little degradation in the signal output from the fiber. Otherwise, we observe a “smearing” of pulses at the output of the fiber, as shown in Figure 8.7. Any smearing may pose problems for whatever device is detecting pulses at the receiving end. However, as τmax −τmin becomes a larger fraction of ton, we see that it becomes possible for adjacent pulses to overlap. This presents a much more serious challenge in detecting individual pulses, so let us examine the c⃝S. Lally CC BY-SA 4.0 Figure 8.7: The effect of dispersion on the signal out- put from the fiber optic cable: Top: Arriving wave- form corresponding to minimum path length, Middle: Arriving waveform corresponding to maximum path length, and Bottom: Sum of all arriving waveforms. | Electromagnetics_Vol2_Page_155_Chunk2330 |
8.3. DISPERSION IN OPTICAL FIBER 143 overlap problem in greater detail. To avoid overlap: τmax −τmin + ton < T (8.18) Let us define the quantity τ ≜τmax −τmin. This is sometimes referred to as the delay spread.1 Thus, we obtain the following requirement for overlap-free transmission: τ < T −ton (8.19) Note that the delay spread imposes a minimum value on T, which in turn imposes a maximum value on the rate at which information can be transmitted on the cable. So, we are motivated to calculate delay spread. The minimum propagation time tmin is simply the length l of the cable divided by the phase velocity vp of the light within the core; i.e., tmin = l/vp. Recall phase velocity is simply the speed of light in free space c divided by √ǫr where ǫr is the relative permittivity of the core material. Thus: tmin = l vp = l c/√ǫr = l c/nf = lnf c (8.20) The maximum propagation time tmax is different because the maximum path length is different. Specifically, the maximum path length is not l, but rather l/ cos θ2 where θ2 is the angle between the axis and the direction of travel as light crosses the axis. So, for example, θ2 = 0 for tmin since light in that case travels along the axis. The value of θ2 for tmax is determined by the threshold angle for total internal reflection. This was determined in Section 8.2 to be given by: cos θ2 = nc nf (threshold value) (8.21) so we obtain the following relationship: tmax = l/ cos θ2 vp = lnf/nc c/nf = ln2 f cnc (8.22) and subsequently we find: τ = lnf c nf nc −1 (8.23) 1Full disclosure: There are many ways to define “delay spread,” but this definition is not uncommon and is useful in the present anal- ysis. Note that τ increases linearly with l. Therefore, the rate at which pulses can be sent without overlap decreases linearly with increasing length. In practical applications, this means that the maximum supportable data rate decreases as the length of the cable increases. This is true independently of media loss within the cable (which we have not yet even considered!). Rather, this is a fundamental limitation resulting from dispersion. Example 8.3. Maximum supportable data rate in multimode fiber optic cable. A multimode fiber optic cable of length 1 m is used to transmit data using the scheme shown in Figure 8.6, with ton = T/2. Values of nf and nc for this fiber are 1.52 and 1.49, respectively. What is the maximum supportable data rate? Solution. Using Equation 8.23, we find the delay spread τ ∼= 102 ps. To avoid overlap, T > 2τ; therefore, T greater than about 204 ps is required. The modulation scheme allows one bit per period, so the maximum data rate is 1/T ∼= 4.9 × 109 bits per second; i.e., ∼= 4.9 Gb/s. The finding of 4.9 Gb/s may seem like a pretty high data rate; however, consider what happens if the length increases to 1 km. It is apparent from Equation 8.23 that the delay spread will increase by a factor of 1000, so the maximum supportable data rate decreases by a factor of 1000 to a scant 4.9 Mb/s. Again, this is independent of any media loss within the fiber. To restore the higher data rate over this longer path, the dispersion must be reduced. One way to do this is to divide the link into smaller links separated by repeaters which can receive the dispersed signal, demodulate it, regenerate the original signal, and transmit the restored signal to the next repeater. Alternatively, one may employ “single mode” fiber, which has intrinsically less dispersion than the multimode fiber presumed in our analysis. Additional Reading: • “Optical fiber” on Wikipedia. [m0209] | Electromagnetics_Vol2_Page_156_Chunk2331 |
144 CHAPTER 8. OPTICAL FIBER Image Credits Fig. 8.1: c⃝S. Lally, https://commons.wikimedia.org/wiki/File:Figure 8.1-01.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Modified by author. Fig. 8.2: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Internal Reflection in Optical Fiber.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 8.3: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Injecting Light into Optical Fiber.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 8.4: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Cone of Acceptance.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 8.5: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Paths of Light through Optic Cable.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 8.6: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Digital Signal on Fiber Optic Cable.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 8.7: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Dispersion on Signal Output from Fiber Optic Cable.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). | Electromagnetics_Vol2_Page_157_Chunk2332 |
Chapter 9 Radiation 9.1 Radiation from a Current Moment [m0194] In this section, we begin to address the following problem: Given a distribution of impressed current density J(r), what is the resulting electric field intensity E(r)? One route to an answer is via Maxwell’s equations. Viewing Maxwell’s equations as a system of differential equations, a rigorous mathematical solution is possible given the appropriate boundary conditions. The rigorous solution following that approach is relatively complicated, and is presented beginning in Section 9.2 of this book. If we instead limit scope to a sufficiently simple current distribution, a simple informal derivation is possible. This section presents such a derivation. The advantage of tackling a simple special case first is that it will allow us to quickly assess the nature of the solution, which will turn out to be useful once we do eventually address the more general problem. Furthermore, the results presented in this section will turn out to be sufficient to tackle many commonly-encountered applications. The simple current distribution considered in this section is known as a current moment. An example of a current moment is shown in Figure 9.1 and in this case is defined as follows: ∆J(r) = ˆz I ∆l δ(r) (9.1) where I ∆l is the scalar component of the current moment, having units of current times length (SI base units of A·m); and δ(r) is the volumetric sampling function1 defined as follows: δ(r) ≜0 for r ̸= 0; and (9.2) Z V δ(r) dv ≜1 (9.3) where V is any volume which includes the origin (r = 0). It is evident from Equation 9.3 that δ(r) has SI base units of m−3. Subsequently, ∆J(r) has SI base units of A/m2, confirming that it is a volume current density. However, it is the simplest possible form of volume current density, since – as indicated by Equation 9.2 – it exists only at the origin and nowhere else. Although some current distributions approximate the current moment, current distributions encountered in common engineering practice generally do not exist 1Also a form of the Dirac delta function; see “Additional Read- ing” at the end of this section. c⃝C. Wang CC BY-SA 4.0 Figure 9.1: A +ˆz-directed current moment located at the origin. Electromagnetics Vol. 2. c⃝2020 S.W. Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics-vol-2 | Electromagnetics_Vol2_Page_158_Chunk2333 |
146 CHAPTER 9. RADIATION in precisely this form. Nevertheless, the current moment turns out to be generally useful as a “building block” from which practical distributions of current can be constructed, via the principle of superposition. Radiation from current distributions constructed in this manner is calculated simply by summing the radiation from each of the constituent current moments. Now let us consider the electric field intensity ∆E(r) that is created by this current distribution. First, if the current is steady (i.e., “DC”), this problem falls within the domain of magnetostatics; i.e., the outcome is completely described by the magnetic field, and there can be no radiation. Therefore, let us limit our attention to the “AC” case, for which radiation is possible. It will be convenient to employ phasor representation. In phasor representation, the current density is ∆eJ(r) = ˆz eI ∆l δ(r) (9.4) where eI ∆l is simply the scalar current moment expressed as a phasor. Now we are ready to address the question “What is ∆eE(r) due to ∆eJ(r)?” Without doing any math, we know quite a bit about ∆eE(r). For example: • Since electric fields are proportional to the currents that give rise to them, we expect ∆eE(r) to be proportional to eI ∆l . • If we are sufficiently far from the origin, we expect ∆eE(r) to be approximately proportional to 1/r where r ≜|r| is the distance from the source current. This is because point sources give rise to spherical waves, and the power density in a spherical wave would be proportional to 1/r2. Since time-average power density is proportional to ∆eE(r) 2 , ∆eE(r) must be proportional to 1/r. • If we are sufficiently far from the origin, and the loss due to the medium is negligible, then we expect the phase of ∆eE(r) to change approximately at rate β where β is the phase propagation constant 2π/λ. Since we expect spherical phasefronts, ∆eE(r) should therefore contain the factor e−jβr. • Ampere’s law indicates that a ˆz-directed current at the origin should give rise to a ˆφ-directed magnetic field in the z = 0 plane.2 At the same time, Poynting’s theorem requires the cross product of the electric and magnetic fields to point in the direction of power flow. In the present problem, this direction is away from the source; i.e., +ˆr. Therefore, ∆eE(z = 0) points in the −ˆz direction. The same principle applies outside of the z = 0 plane, so in general we expect ∆eE(r) to point in the ˆθ direction. • We expect ∆eE(r) = 0 along the z axis. Subsequently ∆eE(ˆr) must increase from zero at θ = 0 and return to zero at θ = π. The symmetry of the problem suggests ∆eE(ˆr) is maximum at θ = π/2. This magnitude must vary in the simplest possible way, leading us to conclude that ∆eE(ˆr) is proportional to sin θ. Furthermore, the radial symmetry of the problem means that ∆eE(ˆr) should not depend at all on φ. Putting these ideas together, we conclude that the radiated electric field has the following form: ∆eE(r) ≈ˆθC eI ∆l (sin θ) e−jβr r (9.5) where C is a constant which accounts for all of the constants of proportionality identified in the preceding analysis. Since the units of ∆eE(r) are V/m, the units of C must be Ω/m. We have not yet accounted for the wave impedance of the medium η, which has units of Ω, so it would be a good bet based on the units that C is proportional to η. However, here the informal analysis reaches a dead end, so we shall simply state the result from the rigorous solution: C = jηβ/4π. The units are correct, and we finally obtain: ∆eE(r) ≈ˆθjηβ 4π eI ∆l (sin θ) e−jβr r (9.6) Additional evidence that this solution is correct comes from the fact that it satisfies the wave equation ∇2∆eE(r) + β2∆eE(r) = 0.3 2This is sometimes described as the “right hand rule” of Am- pere’s law. 3Confirming this is straightforward (simply substitute and evalu- ate) and is left as an | Electromagnetics_Vol2_Page_159_Chunk2334 |
exercise for the student. | Electromagnetics_Vol2_Page_159_Chunk2335 |
9.2. MAGNETIC VECTOR POTENTIAL 147 Note that the expression we have obtained for the radiated electric field is approximate (hence the “≈”). This is due in part to our presumption of a simple spherical wave, which may only be valid at distances far from the source. But how far? An educated guess would be distances much greater than a wavelength (i.e., r ≫λ). This will do for now; in another section, we shall show rigorously that this guess is essentially correct. We conclude this section by noting that the current distribution analyzed in this section is sometimes referred to as a Hertzian dipole. A Hertzian dipole is typically defined as a straight infinitesimally-thin filament of current with length which is very small relative to a wavelength, but not precisely zero. This interpretation does not change the solution obtained in this section, thus we may view the current moment and the Hertzian dipole as effectively the same in practical engineering applications. Additional Reading: • “Dirac delta function” on Wikipedia. • “Dipole antenna” (section entitled “Hertzian Dipole”) on Wikipedia. 9.2 Magnetic Vector Potential [m0195] A common problem in electromagnetics is to determine the fields radiated by a specified current distribution. This problem can be solved using Maxwell’s equations along with the appropriate electromagnetic boundary conditions. For time-harmonic (sinusoidally-varying) currents, we use phasor representation.4 Given the specified current distribution eJ and the desired electromagnetic fields eE and eH, the appropriate equations are: ∇· eE = eρv/ǫ (9.7) ∇× eE = −jωµ eH (9.8) ∇· eH = 0 (9.9) ∇× eH = eJ + jωǫeE (9.10) where eρv is the volume charge density. In most engineering problems, one is concerned with propagation through media which are well-modeled as homogeneous media with neutral charge, such as free space.5 Therefore, in this section, we shall limit our scope to problems in which eρv = 0. Thus, Equation 9.7 simplifies to: ∇· eE = 0 (9.11) To solve the linear system of partial differential Equations 9.8–9.11, it is useful to invoke the concept of magnetic vector potential. The magnetic vector potential is a vector field that has the useful property that it is able to represent both the electric and magnetic fields as a single field. This allows the formidable system of equations identified above to be reduced to a single equation which is simpler to solve. Furthermore, this single equation turns out to be the wave equation, with the slight difference that the equation will be mathematically inhomogeneous, with the inhomogeneous part representing the source current. 4Recall that there is no loss of generality in doing so, since any other time-domain variation in the current distribution can be repre- sented using sums of time-harmonic solutions via the Fourier trans- form. 5A counter-example would be propagation through a plasma, which by definition consists of non-zero net charge. | Electromagnetics_Vol2_Page_160_Chunk2336 |
148 CHAPTER 9. RADIATION The magnetic vector potential eA is defined by the following relationship: eB ≜∇× eA (9.12) where eB = µ eH is the magnetic flux density. The magnetic field appears in three of Maxwell’s equations. For Equation 9.12 to be a reasonable definition, ∇× eA must yield reasonable results when substituted for µ eH in each of these equations. Let us first check for consistency with Gauss’ law for magnetic fields, Equation 9.9. Making the substitution, we obtain: ∇· ∇× eA = 0 (9.13) This turns out to be a mathematical identity that applies to any vector field (see Equation B.24 in Appendix B.3). Therefore, Equation 9.12 is consistent with Gauss’ law for magnetic fields. Next we check for consistency with Equation 9.8. Making the substitution: ∇× eE = −jω ∇× eA (9.14) Gathering terms on the left, we obtain ∇× eE + jω eA = 0 (9.15) Now, for reasons that will become apparent in just a moment, we define a new scalar field eV and require it to satisfy the following relationship: −∇eV ≜eE + jω eA (9.16) Using this definition, Equation 9.15 becomes: ∇× −∇eV = 0 (9.17) which is simply ∇× ∇eV = 0 (9.18) Once again we have obtained a mathematical identity that applies to any vector field (see Equation B.25 in Appendix B.3). Therefore, eV can be any mathematically-valid scalar field. Subsequently, Equation 9.12 is consistent with Equation 9.8 (Maxwell’s curl equation for the electric field) for any choice of eV that we are inclined to make. Astute readers might already realize what we’re up to here. Equation 9.16 is very similar to the relationship E = −∇V from electrostatics,6 in which V is the scalar electric potential field. Evidently Equation 9.16 is an enhanced version of that relationship that accounts for the coupling with H (here, represented by A) in the time-varying (decidedly non-static) case. That assessment is correct, but let’s not get too far ahead of ourselves: As demonstrated in the previous paragraph, we are not yet compelled to make any particular choice for eV , and this freedom will be exploited later in this section. Next we check for consistency with Equation 9.10. Making the substitution: ∇× 1 µ∇× eA = eJ + jωǫeE (9.19) Multiplying both sides of the equation by µ: ∇× ∇× eA = µeJ + jωµǫeE (9.20) Next we use Equation 9.16 to eliminate eE, yielding: ∇× ∇× eA = µeJ + jωµǫ −∇eV −jω eA (9.21) After a bit of algebra, we obtain ∇× ∇× eA = ω2µǫ eA −jωµǫ∇eV + µeJ (9.22) Now we replace the left side of this equation using vector identity B.29 in Appendix B.3: ∇× ∇× eA ≡∇ ∇· eA −∇2 eA (9.23) Equation 9.22 becomes: ∇ ∇· eA −∇2 eA = ω2µǫ eA −jωµǫ∇eV + µeJ (9.24) Now multiplying both sides by −1 and rearranging terms: ∇2 eA + ω2µǫ eA = ∇ ∇· eA + jωµǫ∇eV −µeJ (9.25) Combining terms on the right side: ∇2 eA+ω2µǫ eA = ∇ ∇· eA + jωµǫeV −µeJ (9.26) Now consider the expression ∇· eA + jωµǫeV appearing in the parentheses on the right side of the 6Note: No tilde in this expression. | Electromagnetics_Vol2_Page_161_Chunk2337 |
9.2. MAGNETIC VECTOR POTENTIAL 149 equation. We established earlier that eV can be essentially any scalar field – from a mathematical perspective, we are free to choose. Invoking this freedom, we now require eV to satisfy the following expression: ∇· eA + jωµǫeV = 0 (9.27) Clearly this is advantageous in the sense that Equation 9.26 is now dramatically simplified. This equation becomes: ∇2 eA + ω2µǫ eA = −µeJ (9.28) Note that this expression is a wave equation. In fact it is the same wave equation that determines eE and eH in source-free regions, except the right-hand side is not zero. Using mathematical terminology, we have obtained an equation for eA in the form of an inhomogeneous partial differential equation, where the inhomogeneous part includes – no surprise here – the source current eJ. Now we have what we need to find the electromagnetic fields radiated by a current distribution. The procedure is simply as follows: 1. Solve the partial differential Equation 9.28 for eA along with the appropriate electromagnetic boundary conditions. 2. eH = (1/µ)∇× eA 3. eE may now be determined from eH using Equation 9.10. Summarizing: The magnetic vector potential eA is a vector field, defined by Equation 9.12, that is able to represent both the electric and magnetic fields simultane- ously. Also: To determine the electromagnetic fields radiated by a current distribution eJ, one may solve Equa- tion 9.28 for eA and then use Equation 9.12 to determine eH and subsequently eE. Specific techniques for performing this procedure – in particular, for solving the differential equation – vary depending on the problem, and are discussed in other sections of this book. We conclude this section with a few comments about Equation 9.27. This equation is known as the Lorenz gauge condition. This constraint is not quite as arbitrary as the preceding derivation implies; rather, there is some deep physics at work here. Specifically, the Lorenz gauge leads to the classical interpretation of eV as the familiar scalar electric potential, as noted previously in this section. (For additional information on that idea, recommended starting points are included in “Additional Reading” at the end of this section.) At this point, it should be clear that the electric and magnetic fields are not merely coupled quantities, but in fact two aspects of the same field; namely, the magnetic vector potential. In fact modern physics (quantum mechanics) yields the magnetic vector potential as a description of the “electromagnetic force,” a single entity which constitutes one of the four fundamental forces recognized in modern physics; the others being gravity, the strong nuclear force, and the weak nuclear force. For more information on that concept, an excellent starting point is the video “Quantum Invariance & The Origin of The Standard Model” referenced at the end of this section. Additional Reading: • “Lorenz gauge condition” on Wikipedia. • “Magnetic potential” on Wikipedia. • PBS Space Time video “Quantum Invariance & The Origin of The Standard Model,” available on YouTube. | Electromagnetics_Vol2_Page_162_Chunk2338 |
150 CHAPTER 9. RADIATION 9.3 Solution of the Wave Equation for Magnetic Vector Potential [m0196] The magnetic vector potential eA due to a current density eJ is given by the following wave equation: ∇2 eA −γ2 eA = −µeJ (9.29) where γ is the propagation constant, defined in the usual manner7 γ2 ≜−ω2µǫ (9.30) Equation 9.29 is a partial differential equation which is inhomogeneous (in the mathematical sense) and can be solved given appropriate boundary conditions. In this section, we present the solution for arbitrary distributions of current in free space. The strategy is to first identify a solution for a distribution of current that exists at a single point. This distribution is the current moment. An example of a current moment is shown in Figure 9.2. A general expression for a current moment located at the origin is: eJ(r) = ˆl eI ∆l δ(r) (9.31) where eI has units of current (SI base units of A), ∆l has units of length (SI base units of m), ˆl is the 7Alternatively, γ2 ≜−ω2µǫc accounting for the possibility of lossy media. c⃝C. Wang CC BY-SA 4.0 Figure 9.2: An example of a current moment located at the origin. In this case, ˆl = ˆz. direction of current flow, and δ(r) is the volumetric sampling function8 defined as follows: δ(r) ≜0 for r ̸= 0; and (9.32) Z V δ(r) dv ≜1 (9.33) where V is any volume which includes the origin (r = 0). It is evident from Equation 9.33 that δ(r) has SI base units of m−3; i.e., inverse volume. Subsequently J(r) has SI base units of A/m2, indicating that it is a volume current density.9 – as indicated by Equation 9.32 – it exists only at the origin and nowhere else. Substituting Equation 9.31 into Equation 9.29, we obtain: ∇2 eA −γ2 eA = −µˆl eI ∆l δ(r) (9.34) The general solution to this equation presuming homogeneous and time-invariant media (i.e., µ and ǫ constant with respect to space and time) is: eA(r) = ˆl µ eI ∆l e±γr 4πr (9.35) To confirm that this is a solution to Equation 9.34, substitute this expression into Equation 9.29 and observe that the equality holds. Note that Equation 9.35 indicates two solutions, corresponding to the signs of the exponent in the factor e±γr. We can actually be a little more specific by applying some common-sense physics. Recall that γ in general may be expressed in terms of real and imaginary components as follows: γ = α + jβ (9.36) where α is a real-valued positive constant known as the attenuation constant and β is a real-valued positive constant known as the phase propagation constant. Thus, we may rewrite Equation 9.35 as follows: eA(r) = ˆl µ eI ∆l e±αre±jβr 4πr (9.37) 8Also a form of the Dirac delta function; see “Additional Read- ing” at the end of this section. 9Remember, the density of a volume current is with respect to the area through which it flows, therefore the units are A/m2. | Electromagnetics_Vol2_Page_163_Chunk2339 |
9.3. SOLUTION OF THE WAVE EQUATION FOR MAGNETIC VECTOR POTENTIAL 151 Now consider the factor e±αr/r, which determines the dependence of magnitude on distance r. If we choose the negative sign in the exponent, this factor decays exponentially with increasing distance from the origin, ultimately reaching zero at r →∞. This is precisely the expected behavior, since we expect the magnitude of a radiated field to diminish with increasing distance from the source. If on the other hand we choose the positive sign in the exponent, this factor increases to infinity as r →∞. That outcome can be ruled out on physical grounds, since it implies the introduction of energy independently from the source. The requirement that field magnitude diminishes to zero as distance from the source increases to infinity is known as the radiation condition, and is essentially a boundary condition that applies at r →∞. Invoking the radiation condition, Equation 9.35 becomes: eA(r) = ˆl µ eI ∆l e−γr 4πr (9.38) In the loss-free (α = 0) case, we cannot rely on the radiation condition to constrain the sign of γ. However, in practical engineering work there is always some media loss; i.e., α might be negligible but is not quite zero. Thus, we normally assume that the solution for lossless (including free space) conditions is given by Equation 9.38 with α = 0: eA(r) = ˆl µ eI ∆l e−jβr 4πr (9.39) Now consider the factor e−jβr in Equation 9.39. This factor exclusively determines the dependence of the phase of eA(r) with increasing distance r from the source. In this case, we observe that surfaces of constant phase correspond to spherical shells which are concentric with the source. Thus, eA(r) is a spherical wave. Let us now consider a slightly more complicated version of the problem in which the current moment is no longer located at the origin, but rather is located at r′. This is illustrated in Figure 9.3. This current distribution can be expressed as follows: eJ(r) = ˆl eI ∆l δ(r −r′) (9.40) The solution in this case amounts to a straightforward modification of the existing solution. To see this, note c⃝C. Wang CC BY-SA 4.0 Figure 9.3: Current moment displaced from the ori- gin. that eA depends only on r, and not at all on θ or φ. In other words, eA depends only on the distance between the “field point” r at which we observe eA, and the “source point” r′ at which the current moment lies. Thus we replace r with this distance, which is |r −r′|. The solution becomes: eA(r) = ˆl µ eI ∆l e−γ|r−r′| 4π |r −r′| (9.41) Continuing to generalize the solution, let us now consider a scenario consisting of a filament of current following a path C through space. This is illustrated in Figure 9.4. Such a filament may be viewed as a collection of a large number N of discrete current moments distributed along the path. The contribution of the nth current moment, located at rn, to the total magnetic vector potential is: ∆eA(r; rn) = ˆl(rn) µ eI(rn) ∆l e−γ|r−rn| 4π |r −rn| (9.42) Note that we are allowing both the current eI and current direction ˆl to vary with position along C. Assuming the medium is linear, superposition applies. So we add these contributions to obtain the total | Electromagnetics_Vol2_Page_164_Chunk2340 |
152 CHAPTER 9. RADIATION c⃝C. Wang CC BY-SA 4.0 Figure 9.4: A filament of current lying along the path C. This current distribution may be interpreted as a collection of current moments lying along C. magnetic vector potential: eA(r) ≈ N X n=1 ∆eA(r; rn) (9.43) ≈µ 4π N X n=1 ˆl(rn) eI(rn) e−γ|r−rn| |r −rn| ∆l (9.44) Now letting ∆l →0 so that we may replace ∆l with the differential length dl: eA(r) = µ 4π Z C ˆl(r′) eI(r′) e−γ|r−r′| |r −r′| dl (9.45) where we have also replaced rn with the original notation r′ since we once again have a continuum (as opposed to a discrete set) of source locations. The magnetic vector potential corresponding to radiation from a line distribution of current is given by Equation 9.45. Given eA(r), the magnetic and electric fields may be determined using the procedure developed in Section 9.2. Additional Reading: • “Magnetic potential” on Wikipedia. • “Dirac delta function” on Wikipedia. 9.4 Radiation from a Hertzian Dipole [m0197] Section 9.1 presented an informal derivation of the electromagnetic field radiated by a Hertzian dipole represented by a zero-length current moment. In this section, we provide a rigorous derivation using the concept of magnetic vector potential discussed in Sections 9.2 and 9.3. A review of those sections is recommended before tackling this section. A Hertzian dipole is commonly defined as an electrically-short and infinitesimally-thin straight filament of current, in which the density of the current is uniform over its length. The Hertzian dipole is commonly used as a “building block” for constructing physically-realizable distributions of current as exhibited by devices such as wire antennas. The method is to model these relatively complex distributions of current as the sum of Hertzian dipoles, which reduces the problem to that of summing the contributions of the individual Hertzian dipoles, with each Hertzian dipole having the appropriate (i.e., different) position, magnitude, and phase. To facilitate use of the Hertzian dipole as a building block suitable for constructing physically-realizable distributions of current, we choose to represent the Hertzian dipole using an essentially equivalent current distribution which is mathematically more versatile. This description of the Hertzian dipole replaces the notion of constant current over finite length with the notion of a current moment located at a single point. This is shown in Figure 9.5, and is given by: ∆eJ(r) = ˆl eI ∆l δ(r) (9.46) where the product eI∆l (SI base units of A·m) is the current moment, ˆl is the direction of current flow, and δ(r) is the volumetric sampling function defined as follows: δ(r) ≜0 for r ̸= 0; and (9.47) Z V δ(r) dv ≜1 (9.48) where V is any volume which includes the origin (r = 0). In this description, the Hertzian dipole is located at the origin. | Electromagnetics_Vol2_Page_165_Chunk2341 |
9.4. RADIATION FROM A HERTZIAN DIPOLE 153 The solution for the magnetic vector potential due to a ˆz-directed Hertzian dipole located at the origin was presented in Section 9.3. In the present scenario, it is: eA(r) = ˆz µ eI ∆l e−γr 4πr (9.49) where the propagation constant γ = α + jβ as usual. Assuming lossless media (α = 0), we have eA(r) = ˆz µ eI ∆l e−jβr 4πr (9.50) We obtain the magnetic field intensity using the definition of magnetic vector potential: eH ≜(1/µ)∇× eA (9.51) = eI ∆l 4π ∇× ˆz e−jβr r (9.52) To proceed, it is useful to convert ˆz into the spherical coordinate system. To do this, we find the component of ˆz that is parallel to ˆr, ˆθ, and ˆφ; and then sum the results: ˆz = ˆr (ˆr · ˆz) + ˆθ ˆθ · ˆz + ˆφ ˆφ · ˆz (9.53) = ˆr cos θ −ˆθ sin θ + 0 (9.54) Equation 9.52 requires computation of the following quantity: ∇× ˆze−jβr r = ∇× ˆr (cos θ) e−jβr r −ˆθ (sin θ) e−jβr r (9.55) c⃝C. Wang CC BY-SA 4.0 Figure 9.5: A Hertzian dipole located at the origin, represented as a current moment. In this case, ˆl = ˆz. At this point, it is convenient to make the following definitions: Cr ≜(cos θ) e−jβr r (9.56) Cθ ≜−(sin θ) e−jβr r (9.57) These definitions allow Equation 9.55 to be written compactly as follows: ∇× ˆze−jβr r = ∇× h ˆrCr + ˆθCθ i (9.58) The right side of Equation 9.58 is evaluated using Equation B.18 (Appendix B.2). Although the complete expression consists of 6 terms, only 2 terms are non-zero.10 This leaves: ∇× ˆze−jβr r = ˆφ1 r ∂ ∂r (rCθ) −∂ ∂θCr (9.59) = ˆφ (sin θ) e−jβr r jβ + 1 r (9.60) Substituting this result into Equation 9.52, we obtain: eH = ˆφ eI ∆l 4π (sin θ) e−jβr r jβ + 1 r (9.61) Let us further limit our scope to the field far from the antenna. Specifically, let us assume r ≫λ. Now we use the relationship β = 2π/λ and determine the following: jβ + 1 r = j 2π λ + 1 r (9.62) ≈j 2π λ = jβ (9.63) Equation 9.61 becomes: eH ≈ˆφj eI · β∆l 4π (sin θ) e−jβr r (9.64) where the approximation holds for low-loss media and r ≫λ. This expression is known as a far field approximation, since it is valid only for distances “far” (relative to a wavelength) from the source. Now let us take a moment to interpret this result: 10Specifically, two terms are zero because there is no ˆφ compo- nent in the argument of the curl function; and another two terms are zero because the argument of the curl function is independent of φ, so partial derivatives with respect to φ are zero. | Electromagnetics_Vol2_Page_166_Chunk2342 |
154 CHAPTER 9. RADIATION • Notice the factor β∆l has units of radians; that is, it is electrical length. This tells us that the magnitude of the radiated field depends on the electrical length of the current moment. • The factor e−jβr/r indicates that this is a spherical wave; that is, surfaces of constant phase correspond to concentric spheres centered on the source, and magnitude is inversely proportional to distance. • The direction of the magnetic field vector is always ˆφ, which is precisely what we expect; for example, using the Biot-Savart law. • Finally, note the factor sin θ. This indicates that the field magnitude is zero along the direction in which the source current flows, and is a maximum in the plane perpendicular to this direction. Now let us determine the electric field radiated by the Hertzian dipole. The direct method is to employ Ampere’s law. That is, eE = 1 jωǫ∇× eH (9.65) where eH is given by Equation 9.64. At field points far from the dipole, the radius of curvature of the spherical phasefronts is very large and so appear to be locally planar. That is, from the perspective of an observer far from the dipole, the arriving wave appears to be a plane wave. In this case, we may employ the plane wave relationships. The appropriate relationship in this case is: eE = −ηˆr × eH (9.66) where η is the wave impedance. So we find: eE ≈ˆθjη eI · β∆l 4π (sin θ) e−jβr r (9.67) Summarizing: The electric and magnetic fields far (i.e., ≫λ) from a ˆz-directed Hertzian dipole having con- stant current eI over length ∆l, located at the ori- gin, are given by Equations 9.67 and 9.64, respec- tively. Additional Reading: • “Dipole antenna” (section entitled “Hertzian Dipole”) on Wikipedia. | Electromagnetics_Vol2_Page_167_Chunk2343 |
9.5. RADIATION FROM AN ELECTRICALLY-SHORT DIPOLE 155 9.5 Radiation from an Electrically-Short Dipole [m0198] The simplest distribution of radiating current that is encountered in common practice is the electrically-short dipole (ESD). This current distribution is shown in Figure 9.6. The two characteristics that define the ESD are (1) the current is aligned along a straight line, and (2) the length L of the line is much less than one-half of a wavelength; i.e., L ≪λ/2. The latter characteristic is what we mean by “electrically-short.”11 The current distribution of an ESD is approximately triangular in magnitude, and approximately constant in phase. How do we know this? First, note that distributions of current cannot change in a complex or rapid way over such distances which are much less than a wavelength. If this is not immediately apparent, recall the behavior of transmission lines: 11A potential source of confusion is that the Hertzian dipole is also a “dipole” which is “electrically-short.” The distinction is that the current comprising a Hertzian dipole is constant over its length. This condition is rarely and only approximately seen in practice, whereas the triangular magnitude distribution is a relatively good ap- proximation to a broad class of commonly-encountered electrically- short wire antennas. Thus, the term “electrically-short dipole,” as used in this book, refers to the triangular distribution unless noted otherwise. c⃝C. Wang CC BY-SA 4.0 Figure 9.6: Current distribution of the electrically- short dipole (ESD). c⃝C. Wang CC BY-SA 4.0 Figure 9.7: Current distribution of the electrically- short dipole (ESD) approximated as a large number of Hertzian dipoles. The current standing wave on a transmission line exhibits a period of λ/2, regardless the source or termination. For the ESD, L ≪λ/2 and so we expect an even simpler variation. Also, we know that the current at the ends of the dipole must be zero, simply because the dipole ends there. These considerations imply that the current distribution of the ESD is well-approximated as triangular in magnitude.12 Expressed mathematically: eI(z) ≈I0 1 −2 L |z| (9.68) where I0 (SI base units of A) is a complex-valued constant indicating the maximum current magnitude and phase. There are two approaches that we might consider in order to find the electric field radiated by an ESD. The first approach is to calculate the magnetic vector potential eA by integration over the current distribution, calculate eH = (1/µ)∇× eA, and finally calculate eE from eH using Ampere’s law. We shall employ a simpler approach, shown in Figure 9.7. Imagine the ESD as a collection of many shorter segments of current that radiate independently. The total field is then the sum of these short segments. Because these segments are very short relative to the 12A more rigorous analysis leading to the same conclusion is pos- sible, but is beyond the scope of this book. | Electromagnetics_Vol2_Page_168_Chunk2344 |
156 CHAPTER 9. RADIATION length of the dipole as well as being short relative to a wavelength, we may approximate the current over each segment as approximately constant. In other words, we may interpret each of these segments as being, to a good approximation, a Hertzian dipole. The advantage of this approach is that we already have a solution for each of the segments. In Section 9.4, it is shown that a ˆz-directed Hertzian dipole at the origin radiates the electric field eE(r) ≈ˆθjη eI · β∆l 4π (sin θ) e−jβr r (9.69) where eI and ∆l may be interpreted as the current and length of the dipole, respectively. In this expression, η is the wave impedance of medium in which the dipole radiates (e.g., ≈377 Ωfor free space), and we have presumed lossless media such that the attenuation constant α ≈0 and the phase propagation constant β = 2π/λ. This expression also assumes field points far from the dipole; specifically, distances r that are much greater than λ. Repurposing this expression for the present problem, the segment at the origin radiates the electric field: eE(r; z′ = 0) ≈ˆθjη I0 · β∆l 4π (sin θ) e−jβr r (9.70) where the notation z′ = 0 indicates the Hertzian dipole is located at the origin. Letting the length ∆l of this segment shrink to differential length dz′, we may describe the contribution of this segment to the field radiated by the ESD as follows: deE(r; z′ = 0) ≈ˆθjη I0 · βdz′ 4π (sin θ) e−jβr r (9.71) Using this approach, the electric field radiated by any segment can be written: deE(r; z′) ≈ˆθ′jηβ eI(z′) 4π (sin θ′) e−jβ|r−ˆzz′| |r −ˆzz′| dz′ (9.72) Note that θ is replaced by θ′ since the ray r −ˆzz′ forms a different angle (i.e., θ′) with respect to ˆz. Similarly, ˆθ is replaced by ˆθ′, since it also varies with z′. The electric field radiated by the ESD is obtained by integration over these contributions: eE(r) ≈ Z +L/2 −L/2 deE(ˆr; z′) (9.73) c⃝C. Wang CC BY-SA 4.0 Figure 9.8: Parallel ray approximation for an ESD. yielding: eE(r) ≈j ηβ 4π Z +L/2 −L/2 ˆθ′eI(z′) (sin θ′) e−jβ|r−ˆzz′| |r −ˆzz′| dz′ (9.74) Given some of the assumptions we have already made, this expression can be further simplified. For example, note that θ′ ≈θ since L ≪r. For the same reason, ˆθ′ ≈ˆθ. Since these variables are approximately constant over the length of the dipole, we may move them outside the integral, yielding: eE(r) ≈ˆθj ηβ 4π (sin θ) Z +L/2 −L/2 eI(z′)e−jβ|r−ˆzz′| |r −ˆzz′| dz′ (9.75) It is also possible to simplify the expression |r −ˆzz′|. Consider Figure 9.8. Since we have already assumed that r ≫L (i.e., the distance to field points is much greater than the length of the dipole), the vector r is approximately parallel to the vector r −ˆzz′. Subsequently, it must be true that |r −ˆzz′| ≈r −ˆr · ˆzz′ (9.76) Note that the magnitude of r −ˆr · ˆzz′ must be approximately equal to r, since r ≫L. So, insofar as |r −ˆzz′| determines the magnitude of eE(r), we may use the approximation: |r −ˆzz′| ≈r (magnitude) (9.77) Insofar as |r −ˆzz′| determines phase, we have to be a bit more careful. The part of the integrand of | Electromagnetics_Vol2_Page_169_Chunk2345 |
9.5. RADIATION FROM AN ELECTRICALLY-SHORT DIPOLE 157 Equation 9.75 that exhibits varying phase is e−jβ|r−ˆzz′|. Using Equation 9.76, we find e−jβ|r−ˆzz′| ≈e−jβre+jβˆr·ˆzz′ (9.78) The worst case in terms of phase variation within the integral is for field points along the z axis. For these points, ˆr · ˆz = ±1 and subsequently |r −ˆzz′| varies from z −L/2 to z + L/2 where z is the location of the field point. However, since L ≪λ (i.e., because the dipole is electrically short), this difference in lengths is much less than λ/2. Therefore, the phase βˆr · ˆzz′ varies by much less than π radians, and subsequently e−jβˆr·ˆzz′ ≈1. We conclude that under these conditions, e−jβ|r−ˆzz′| ≈e−jβr (phase) (9.79) Applying these simplifications for magnitude and phase to Equation 9.75, we obtain: eE(r) ≈ˆθj ηβ 4π (sin θ) e−jβr r Z +L/2 −L/2 eI(z′)dz′ (9.80) The integral in this equation is very easy to evaluate; in fact, from inspection (Figure 9.6), we determine it is equal to I0L/2. Finally, we obtain: eE(r) ≈ˆθjη I0 · βL 8π (sin θ) e−jβr r (9.81) Summarizing: The electric field intensity radiated by an ESD lo- cated at the origin and aligned along the z axis is given by Equation 9.81. This expression is valid for r ≫λ. It is worth noting that the variation in magnitude, phase, and polarization of the ESD with field point location is identical to that of a single Hertzian dipole having current moment ˆzI0L/2 (Section 9.4). However, the magnitude of the field radiated by the ESD is exactly one-half that of the Hertzian dipole. Why one-half? Simply because the integral over the triangular current distribution assumed for the ESD is one-half the integral over the uniform current distribution that defines the Hertzian dipole. This similarly sometimes causes confusion between Hertzian dipoles and ESDs. Remember that ESDs are physically realizable, whereas Hertzian dipoles are not. It is common to eliminate the factor of β in the magnitude using the relationship β = 2π/λ, yielding: eE(r) ≈ˆθj ηI0 4 L λ (sin θ) e−jβr r (9.82) At field points r ≫λ, the wave appears to be locally planar. Therefore, we are justified using the plane wave relationship eH = 1 ηˆr × eE to calculate eH. The result is: eH(r) ≈ˆφj I0 4 L λ (sin θ) e−jβr r (9.83) Finally, let us consider the spatial characteristics of the radiated field. Figures 9.9 and 9.10 show the result in a plane of constant φ. Figures 9.11 and 9.12 show the result in the z = 0 plane. Note that the orientations of the electric and magnetic field vectors indicate a Poynting vector eE × eH that is always directed radially outward from the location of the dipole. This confirms that power flow is always directed radially outward from the dipole. Due to the symmetry of the problem, Figures 9.9–9.12 provide a complete characterization of the relative magnitudes and orientations of the radiated fields. | Electromagnetics_Vol2_Page_170_Chunk2346 |
158 CHAPTER 9. RADIATION c⃝S. Lally CC BY-SA 4.0 Figure 9.9: Magnitude of the radiated field in any plane of constant φ. c⃝S. Lally CC BY-SA 4.0 Figure 9.10: Orientation of the electric and magnetic fields in any plane of constant φ. c⃝S. Lally CC BY-SA 4.0 Figure 9.11: Magnitude of the radiated field in any plane of constant z. c⃝S. Lally CC BY-SA 4.0 Figure 9.12: Orientation of the electric and magnetic fields in the z = 0 plane. | Electromagnetics_Vol2_Page_171_Chunk2347 |
9.6. FAR-FIELD RADIATION FROM A THIN STRAIGHT FILAMENT OF CURRENT 159 9.6 Far-Field Radiation from a Thin Straight Filament of Current [m0199] A simple distribution of radiating current that is encountered in common practice is the thin straight current filament, shown in Figure 9.13. The defining characteristic of this distribution is that the current filament is aligned along a straight line, and that the maximum dimension of the cross-section of the filament is much less than a wavelength. The latter characteristic is what we mean by “thin” – i.e., the filament is “electrically thin.” Physical distributions of current that meet this description include the electrically-short dipole (Section 9.5) and the half-wave dipole (Section 9.7) among others. A gentler introduction to this distribution is via Section 9.5 (“Radiation from an Electrically-Short Dipole”), so students may want to review that section first. This section presents the more general case. Let the magnitude and phase of current along the filament be given by the phasor quantity eI(z) (SI base units of A). In principle, the only constraint on eI(z) c⃝C. Wang CC BY-SA 4.0 Figure 9.13: A thin straight distribution of radiating current. c⃝C. Wang CC BY-SA 4.0 Figure 9.14: Current distribution approximated as a set of Hertzian dipoles. that applies generally is that it must be zero at the ends of the distribution; i.e., eI(z) = 0 for |z| ≥L/2. Details beyond this constraint depend on L and perhaps other factors. Nevertheless, the characteristics of radiation from members of this class of current distributions have much in common, regardless of L. These become especially apparent if we limit our scope to field points far from the current, as we shall do here. There are two approaches that we might consider in order to find the electric field radiated by this distribution. The first approach is to calculate the magnetic vector potential eA by integration over the current distribution (Section 9.3), calculate eH = (1/µ)∇× eA, and finally calculate eE from eH using the differential form of Ampere’s law. In this section, we shall employ a simpler approach, shown in Figure 9.14. Imagine the filament as a collection of many shorter segments of current that radiate independently. The total field is then the sum of these short segments. Because these segments are very short relative to the length of the filament as well as being short relative to a wavelength, we may approximate the current over each segment as constant. In other words, we may interpret each segment as being, to a good approximation, a Hertzian dipole. The advantage of this approach is that we already have a solution for every segment. In Section 9.4, it is | Electromagnetics_Vol2_Page_172_Chunk2348 |
160 CHAPTER 9. RADIATION shown that a ˆz-directed Hertzian dipole at the origin radiates the electric field eE(r) ≈ˆθjη eI (β∆l) 4π (sin θ) e−jβr r (9.84) where eI and ∆l may be interpreted as the current and length of the filament, respectively. In this expression, η is the wave impedance of medium in which the filament radiates (e.g., ≈377 Ωfor free space), and we have presumed lossless media such that the attenuation constant α ≈0 and the phase propagation constant β = 2π/λ. This expression also assumes field points far from the filament; specifically, distances r that are much greater than λ. Repurposing this expression for the present problem, we note that the segment at the origin radiates the electric field: eE(r; z′ = 0) ≈ˆθjη eI(0) (β∆l) 4π (sin θ) e−jβr r (9.85) where the notation z′ = 0 indicates the Hertzian dipole is located at the origin. Letting the length ∆l of this segment shrink to differential length dz′, we may describe the contribution of this segment to the field radiated by the ESD as follows: deE(r; z′ = 0) ≈ˆθjη eI(0) (βdz′) 4π (sin θ) e−jβr r(9.86) Using this approach, the electric field radiated by any segment can be written: deE(r; z′) ≈ˆθ′jηβ eI(z′) 4π (sin θ′) e−jβ|r−ˆzz′| |r −ˆzz′| dz′ (9.87) Note that θ is replaced by θ′ since the ray r −ˆzz′ forms a different angle (i.e., θ′) with respect to ˆz. Subsequently, ˆθ is replaced by ˆθ′, which varies similarly with z′. The electric field radiated by the filament is obtained by integration over these contributions, yielding: eE(r) ≈ Z +L/2 −L/2 deE(ˆr; z′) (9.88) Expanding this expression: eE(r) ≈j ηβ 4π Z +L/2 −L/2 ˆθ′eI(z′) (sin θ′) e−jβ|r−ˆzz′| |r −ˆzz′| dz′ (9.89) c⃝C. Wang CC BY-SA 4.0 Figure 9.15: Parallel ray approximation. Given some of the assumptions we have already made, this expression can be further simplified. For example, note that θ′ ≈θ since L ≪r. For the same reason, ˆθ′ ≈ˆθ. Since these variables are approximately constant over the length of the filament, we may move them outside the integral, yielding: eE(r) ≈ˆθj ηβ 4π (sin θ) Z +L/2 −L/2 eI(z′)e−jβ|r−ˆzz′| |r −ˆzz′| dz′ (9.90) It is also possible to simplify the expression |r −ˆzz′|. Consider Figure 9.15. Since we have already assumed that r ≫L (i.e., the distance to field points is much greater than the length of the filament), the vector r is approximately parallel to the vector r −ˆzz′. Subsequently, it must be true that |r −ˆzz′| ≈r −ˆr · ˆzz′ (9.91) ≈r −z′ cos θ (9.92) Note that the magnitude of r −ˆr · ˆzz′ must be approximately equal to r, since r ≫L. So, insofar as |r −ˆzz′| determines the magnitude of eE(r), we may use the approximation: |r −ˆzz′| ≈r (magnitude) (9.93) Insofar as |r −ˆzz′| determines phase, we have to be more careful. The part of the integrand of Equation 9.90 that exhibits varying phase is e−jβ|r−ˆzz′|. Using Equation 9.91, we find e−jβ|r−ˆzz′| ≈e−jβre+jβz′ cos θ (9.94) | Electromagnetics_Vol2_Page_173_Chunk2349 |
9.7. FAR-FIELD RADIATION FROM A HALF-WAVE DIPOLE 161 These simplifications are known collectively as a far field approximation, since they are valid only for distances “far” from the source. Applying these simplifications for magnitude and phase to Equation 9.90, we obtain: eE(r) ≈ˆθj ηβ 4π e−jβr r (sin θ) · Z +L/2 −L/2 eI(z′)e+jβz′ cos θdz′ (9.95) Finally, it is common to eliminate the factor of β in the magnitude using the relationship β = 2π/λ, yielding: eE(r) ≈ˆθj η 2 e−jβr r (sin θ) · " 1 λ Z +L/2 −L/2 eI(z′)e+jβz′ cos θdz′ # (9.96) The electric field radiated by a thin, straight, ˆz- directed current filament of length L located at the origin and aligned along the z axis is given by Equation 9.96. This expression is valid for r ≫L and r ≫λ. At field points satisfying the conditions r ≫L and r ≫λ, the wave appears to be locally planar. Therefore, we are justified using the plane wave relationship eH = (1/η)ˆr × eE to calculate eH. As a check, one may readily verify that Equation 9.96 yields the expected result for the electrically-short dipole (Section 9.5). 9.7 Far-Field Radiation from a Half-Wave Dipole [m0200] A simple and important current distribution is that of the thin half-wave dipole (HWD), shown in Figure 9.16. This is the distribution expected on a thin straight wire having length L = λ/2, where λ is wavelength. This distribution is described mathematically as follows: eI(z) ≈I0 cos π z L for |z| ≤L 2 (9.97) where I0 (SI base units of A) is a complex-valued constant indicating the maximum magnitude of the current and its phase. Note that the current is zero at the ends of the dipole; i.e., eI(z) = 0 for |z| = L/2. Note also that this “cosine pulse” distribution is very similar to the triangular distribution of the ESD, and is reminiscent of the sinusoidal variation of current in a standing wave. Since L = λ/2 for the HWD, Equation 9.97 may equivalently be written: eI(z) ≈I0 cos 2π z λ (9.98) The electromagnetic field radiated by this distribution c⃝C. Wang CC BY-SA 4.0 Figure 9.16: Current distribution of the half-wave dipole (HWD). | Electromagnetics_Vol2_Page_174_Chunk2350 |
162 CHAPTER 9. RADIATION of current may be calculated using the method described in Section 9.6, in particular: eE(r) ≈ˆθj η 2 e−jβr r (sin θ) · " 1 λ Z +L/2 −L/2 eI(z′)e+jβz′ cos θdz′ # (9.99) which is valid for field points r far from the dipole; i.e., for r ≫L and r ≫λ. For the HWD, the quantity in square brackets is I0 λ Z +λ/4 −λ/4 cos 2π z′ λ e+jβz′ cos θdz′ (9.100) The evaluation of this integral is straightforward, but tedious. The integral reduces to I0 π cos [(π/2) cos θ] sin2 θ (9.101) Substitution into Equation 9.99 yields eE(r) ≈ˆθj ηI0 2π cos [(π/2) cos θ] sin θ e−jβr r (9.102) The magnetic field may be determined from this result using Ampere’s law. However, a simpler method is to use the fact that the electric field, magnetic field, and direction of propagation ˆr are mutually perpendicular and related by: eH = 1 ηˆr × eE (9.103) This relationship indicates that the magnetic field will be +ˆφ-directed. The magnitude and polarization of the radiated field is similar to that of the electrically-short dipole (ESD; Section 9.5). A comparison of the magnitudes in any radial plane containing the z-axis is shown in Figure 9.17. For either current distribution, the maximum magnitude of the fields occurs in the z = 0 plane. For a given terminal current I0, the maximum magnitude is greater for the HWD than for the ESD. Both current distributions yield zero magnitude along the axis of the dipole. The polarization characteristics of the fields of both current distributions are identical. Additional Reading: • “Dipole antenna” (section entitled “Half-wave dipole”) on Wikipedia. c⃝C. Wang CC BY-SA 4.0 Figure 9.17: Comparison of the magnitude of the ra- diated field of the HWD to that of an electrically-short dipole also oriented along the z-axis. This result is for any radial plane that includes the z-axis. 9.8 Radiation from Surface and Volume Distributions of Current [m0221] In Section 9.3, a solution was developed for radiation from current located at a single point r′. This current distribution was expressed mathematically as a current moment, as follows: eJ(r) = ˆl eI ∆l δ(r −r′) (9.104) where ˆl is the direction of current flow, eI has units of current (SI base units of A), ∆l has units of length (SI base units of m), and δ(r) is the volumetric sampling function (Dirac “delta” function; SI base units of m−3). In this formulation, eJ has SI base units of A/m2. The magnetic vector potential radiated by this current, observed at the field point r, was found to be eA(r) = ˆl µ eI ∆l e−γ|r−r′| 4π |r −r′| (9.105) This solution was subsequently generalized to obtain the radiation from any distribution of line current; i.e., | Electromagnetics_Vol2_Page_175_Chunk2351 |
9.8. RADIATION FROM SURFACE AND VOLUME DISTRIBUTIONS OF CURRENT 163 any distribution of current that is constrained to flow along a single path through space, as along an infinitesimally-thin wire. In this section, we derive an expression for the radiation from current that is constrained to flow along a surface and from current which flows through a volume. The solution in both cases can be obtained by “recycling” the solution for line current as follows. Letting ∆l shrink to the differential length dl, Equation 9.104 becomes: deJ(r) = ˆl eI dl δ(r −r′) (9.106) Subsequently, Equation 9.105 becomes: d eA(r; r′) = ˆl µ eI dl e−γ|r−r′| 4π |r −r′| (9.107) where the notation d eA(r; r′) is used to denote the magnetic vector potential at the field point r due to the differential-length current moment at the source point r′. Next, consider that any distribution of current can be described as a distribution of current moments. By the principle of superposition, the radiation from this distribution of current moments can be calculated as the sum of the radiation from the individual current moments. This is expressed mathematically as follows: eA(r) = Z d eA(r; r′) (9.108) where the integral is over r′; i.e., summing over the source current. If we substitute Equation 9.107 into Equation 9.108, we obtain the solution derived in Section 9.3, which is specific to line distributions of current. To obtain the solution for surface and volume distributions of current, we reinterpret the definition of the differential current moment in Equation 9.106. Note that this current is completely specified by its direction (ˆl) and the quantity eI dl, which has SI base units of A·m. We may describe the same current distribution alternatively as follows: deJ(r) = ˆl eJs ds δ(r −r′) (9.109) where eJs has units of surface current density (SI base units of A/m) and ds has units of area (SI base units of m2). To emphasize that this is precisely the same current moment, note that eJs ds, like eI dl, has units of A·m. Similarly, deJ(r) = ˆl eJ dv δ(r −r′) (9.110) where eJ has units of volume current density (SI base units of A/m2) and dv has units of volume (SI base units of m3). Again, eJ dv has units of A·m. Summarizing, we have found that eI dl = eJs ds = eJ dv (9.111) all describe the same differential current moment. Thus, we may obtain solutions for surface and volume distributions of current simply by replacing eI dl in Equation 9.107, and subsequently in Equation 9.108, with the appropriate quantity from Equation 9.111. For a surface current distribution, we obtain: eA(r) = µ 4π Z S eJs(r′) e−γ|r−r′| |r −r′| ds (9.112) where eJs(r′) ≜ˆl(r′) eJs(r′) and where S is the surface over which the current flows. Similarly for a volume current distribution, we obtain: eA(r) = µ 4π Z V eJ(r′) e−γ|r−r′| |r −r′| dv (9.113) where eJ(r′) ≜ˆl(r′) eJ(r′) and where V is the volume in which the current flows. The magnetic vector potential corresponding to radiation from a surface and volume distribution of current is given by Equations 9.112 and 9.113, respectively. Given eA(r), the magnetic and electric fields may be determined using the procedure developed in Section 9.2. [m0217] | Electromagnetics_Vol2_Page_176_Chunk2352 |
164 CHAPTER 9. RADIATION Image Credits Fig. 9.1: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:A z-directed current moment located at the origin.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 9.2: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:A z-directed current moment located at the origin.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 9.3: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Current moment displaced from the origin.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 9.4: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:A filament of current lying along the path c.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 9.5: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:A z-directed current moment located at the origin.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 9.6: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Current distribution of the esd.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 9.7: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Esd approximated as a large number of hertzian dipoles.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 9.8: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Parallel ray approximation for an esd.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 9.9: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Magnitude of the Radiated Field.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 9.10: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Electric and Magnetic Fields in Plane of Constant Theta.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 9.11: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:FHplaneMag.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 9.12: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:FHplanePol.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 9.13: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:A thin straight distribution of radiating current.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 9.14: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Esd approximated as a large number of hertzian dipoles.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). | Electromagnetics_Vol2_Page_177_Chunk2353 |
9.8. RADIATION FROM SURFACE AND VOLUME DISTRIBUTIONS OF CURRENT 165 Fig. 9.15: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Parallel ray approximation for an esd.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 9.16: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Current distribution of the hwd.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 9.17: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Comparison radiated field magnitude of hwd to esd.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). | Electromagnetics_Vol2_Page_178_Chunk2354 |
Chapter 10 Antennas 10.1 How Antennas Radiate [m0201] An antenna is a transducer; that is, a device which converts signals in one form into another form. In the case of an antenna, these two forms are (1) conductor-bound voltage and current signals and (2) electromagnetic waves. Traditional passive antennas are capable of this conversion in either direction. In this section, we consider the transmit case, in which a conductor-bound signal is converted into a radiating electromagnetic wave. Radiation from an antenna is due to the time-varying current that is excited by the bound electrical signal applied to the antenna terminals. Why ideal transmission lines don’t radiate. To describe the process that allows an antenna to transmit, it is useful to first consider the scenario depicted in Figure 10.1. Here a sinusoidal source is c⃝S. Lally CC BY-SA 4.0 Figure 10.1: Twin lead transmission line terminated into an open circuit, giving rise to a standing wave. applied to the input of an ideal twin lead transmission line. The spacing between conductors is much less than a wavelength, and the output of the transmission line is terminated into an open circuit. Without any additional information, we already know two things about the current on this transmission line. First, we know the current must be identically zero at the end of the transmission line. Second, we know the current on the two conductors comprising the transmission line must be related as indicated in Figure 10.1. That is, at any given position on the transmission line, the current on each conductor is equal in magnitude and flows in opposite directions. Further note that Figure 10.1 depicts only the situation over an interval of one-half of the period of the sinusoidal source. For the other half-period, the direction of current will be in the direction opposite that depicted in Figure 10.1. In other words: The source is varying periodically, so the sign of the current is changing every half-period. Generally, time-varying currents give rise to radiation. Therefore, we should expect the currents depicted in Figure 10.1 might radiate. We can estimate the radiation from the transmission line by interpreting the current as a collection of Hertzian dipoles. For a review of Hertzian dipoles, see Section 9.4; however, all we need to know to address the present problem is that superposition applies. That is, the total radiation is the sum of the radiation from the individual Hertzian dipoles. Once again looking back at Figure 10.1, note that each Hertzian dipole representing current on one conductor has an associated Hertzian dipole representing the current on the other conductor, and is only a tiny fraction of a wavelength distant. Furthermore, these pairs of Electromagnetics Vol. 2. c⃝2020 S.W. Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics-vol-2 | Electromagnetics_Vol2_Page_179_Chunk2355 |
10.1. HOW ANTENNAS RADIATE 167 Hertzian dipoles are identical in magnitude but opposite in sign. Therefore, the radiated field from any such pair of Hertzian dipoles is approximately zero at distances sufficiently far from the transmission line. Continuing to sum all such pairs of Hertzian dipoles, the radiated field remains approximately zero at distances sufficiently far from the transmission line. We come to the following remarkable conclusion: The radiation from an ideal twin lead transmis- sion line with open circuit termination is negligi- ble at distances much greater than the separation between the conductors. Before proceeding, let us be clear about one thing: The situation at distances which are not large relative to separation between the conductors is quite different. This is because the Hertzian dipole pairs do not appear to be quite so precisely collocated at field points close to the transmission line. Subsequently the cancellation of fields from Hertzian dipole pairs is less precise. The resulting sum fields are not negligible and depend on the separation between the conductors. For the present discussion, it suffices to restrict our attention to the simpler “far field” case. A simple antenna formed by modifying twin lead transmission line. Let us now consider a modification to the previous scenario, shown in Figure 10.2. In the new scenario, the ends of the twin lead are bent into right angles. This new section has an overall length which is much less than one-half wavelength. To determine the current distribution on the modified section, we first note that the current must still be precisely zero at the ends of the conductors, but not necessarily at the point at which c⃝S. Lally CC BY-SA 4.0 Figure 10.2: End of the twin lead transmission line fashioned into an electrically-short dipole (ESD). they are bent. Since the modified section is much shorter than one-half wavelength, the current distribution on this section must be very simple. In fact, the current distribution must be that of the electrically-short dipole (ESD), exhibiting magnitude which is maximum at the center and decreasing approximately linearly to zero at the ends. (See Section 9.5 for a review of the ESD.) Finally, we note that the current should be continuous at the junction between the unmodified and modified sections. Having established that the modified section exhibits the current distribution of an ESD, and having previously determined that the unmodified section of transmission line does not radiate, we conclude that this system radiates precisely as an ESD does. Note also that we need not interpret the ESD portion of the transmission line as a modification of the transmission line: Instead, we may view this system as an unmodified transmission line attached to an antenna, which in this case in an ESD. The general case. Although we developed this insight for the ESD specifically, the principle applies generally. That is, any antenna – not just dipoles – can be viewed as a structure that can support a current distribution that radiates. Elaborating: A transmitting antenna is a device that, when driven by an appropriate source, supports a time- varying distribution of current resulting in an electromagnetic wave that radiates away from the device. Although this may seem to be simply a restatement of the definition at the beginning of this section – and it is – we now see how this can happen, and also why transmission lines typically aren’t also antennas. | Electromagnetics_Vol2_Page_180_Chunk2356 |
168 CHAPTER 10. ANTENNAS 10.2 Power Radiated by an Electrically-Short Dipole [m0207] In this section, we determine the total power radiated by an electrically-short dipole (ESD) antenna in response to a sinusoidally-varying current applied to the antenna terminals. This result is both useful on its own and necessary as an intermediate result in determining the impedance of the ESD. The ESD is introduced in Section 9.5, and a review of that section is suggested before attempting this section. In Section 9.5, it is shown that the electric field intensity in the far field of a ˆz-oriented ESD located at the origin is eE(r) ≈ˆθjη I0 · βL 8π (sin θ) e−jβr r (10.1) where r is the field point, I0 is a complex number representing the peak magnitude and phase of the sinusoidally-varying terminal current, L is the length of the ESD, and β is the phase propagation constant 2π/λ where λ is wavelength. Note that L ≪λ since this is an ESD. Also note that Equation 10.1 is valid only for the far-field conditions r ≫L and r ≫λ, and presumes propagation in simple (linear, homogeneous, time-invariant, isotropic) media with negligible loss. Given that we have already limited scope to the far field, it is reasonable to approximate the electromagnetic field at each field point r as a plane wave propagating radially away from the antenna; i.e., in the ˆr direction. Under this assumption, the time-average power density is S(r) = ˆr eE(r) 2 2η (10.2) where η is the wave impedance of the medium. The total power Prad radiated by the antenna is simply S(r) integrated over any closed surface S that encloses the antenna. Thus: Prad = I S S(r) · ds (10.3) where ds is the outward-facing differential element of surface area. In other words, power density (W/m2) integrated over an area (m2) gives power (W). Anticipating that this problem will be addressed in spherical coordinates, we note that ds = ˆrr2 sin θ dθ dφ (10.4) and subsequently: Prad = Z π θ=0 Z 2π φ=0 ˆr eE(r) 2 2η · | Electromagnetics_Vol2_Page_181_Chunk2357 |
10.3. POWER DISSIPATED BY AN ELECTRICALLY-SHORT DIPOLE 169 that is, the length of the antenna expressed in radians, where 2π radians is one wavelength. Thus, we see that the power radiated by the antenna increases as the square of electrical length. Example 10.1. Power radiated by an ESD. A dipole is 10 cm in length and is surrounded by free space. A sinusoidal current having frequency 30 MHz and peak magnitude 100 mA is applied to the antenna terminals. What is the power radiated by the antenna? Solution. If no power is dissipated within the antenna, then all power is radiated. The wavelength λ = c/f ∼= 10 m, so L ∼= 0.01λ. This certainly qualifies as electrically-short, so we may use Equation 10.9. In the present problem, η ∼= 376.7 Ω(the free space wave impedance), I0 = 100 mA, and β = 2π/λ ∼= 0.628 rad/m. Thus, we find that the radiated power is ≈98.6 µW. 10.3 Power Dissipated by an Electrically-Short Dipole [m0208] The power delivered to an antenna by a source connected to the terminals is nominally radiated. However, it is possible that some fraction of the power delivered by the source will be dissipated within the antenna. In this section, we consider the dissipation due to the finite conductivity of materials comprising the antenna. Specifically, we determine the total power dissipated by an electrically-short dipole (ESD) antenna in response to a sinusoidally-varying current applied to the antenna terminals. (The ESD is introduced in Section 9.5, and a review of that section is suggested before attempting this section.) The result allows us to determine radiation efficiency and is a step toward determining the impedance of the ESD. Consider an ESD centered at the origin and aligned along the z axis. In Section 9.5, it is shown that the current distribution is: eI(z) ≈I0 1 −2 L |z| (10.10) where I0 (SI base units of A) is a complex-valued constant indicating the maximum current magnitude and phase, and L is the length of the ESD. This current distribution may be interpreted as a set of discrete very-short segments of constant-magnitude current (sometimes referred to as “Hertzian dipoles”). In this interpretation, the nth current segment is located at z = zn and has magnitude eI(zn). Now let us assume that the material comprising the ESD is a homogeneous good conductor. Then each segment exhibits the same resistance Rseg. Subsequently, the power dissipated in the nth segment is Pseg(zn) = 1 2 eI(zn) 2 Rseg (10.11) The segment resistance can be determined as follows. Let us assume that the wire comprising the ESD has circular cross-section of radius a. Since the wire is a good conductor, the segment resistance may be | Electromagnetics_Vol2_Page_182_Chunk2358 |
170 CHAPTER 10. ANTENNAS calculated using Equation 4.17 (Section 4.2, “Impedance of a Wire”): Rseg ≈1 2 r µf πσ · ∆l a (10.12) where µ is permeability, f is frequency, σ is conductivity, and ∆l is the length of the segment. Substitution into Equation 10.11 yields: Pseg(zn) ≈1 4a r µf πσ eI(zn) 2 ∆l (10.13) Now the total power dissipated in the antenna, Ploss, can be expressed as the sum of the power dissipated in each segment: Ploss ≈ N X n=1 " 1 4a r µf πσ eI(zn) 2 ∆l # (10.14) where N is the number of segments. This may be rewritten as follows: Ploss ≈1 4a r µf πσ N X n=1 eI(zn) 2 ∆l (10.15) Reducing ∆l to the differential length dz′, we may write this in the following integral form: Ploss ≈1 4a r µf πσ Z +L/2 z′=−L/2 eI(z′) 2 dz′ (10.16) It is worth noting that the above expression applies to any straight wire antenna of length L. For the ESD specifically, the current distribution is given by Equation 10.10. Making the substitution: Ploss ≈1 4a r µf πσ Z +L/2 z′=−L/2 I0 1 −2 L |z′| 2 dz′ ≈1 4a r µf πσ |I0|2 Z +L/2 z′=−L/2 1 −2 L |z′| 2 dz′ (10.17) The integral is straightforward to solve, albeit a bit tedious. The integral is found to be equal to L/3, so we obtain: Ploss ≈1 4a r µf πσ |I0|2 · L 3 ≈ L 12a r µf πσ |I0|2 (10.18) Now let us return to interpretation of the ESD as a circuit component. We have found that applying the current I0 to the terminals results in the dissipated power indicated in Equation 10.18. A current source driving the ESD does not perceive the current distribution of the ESD nor does it perceive the varying power over the length of the ESD. Instead, the current source perceives only a net resistance Rloss such that Ploss = 1 2 |I0|2 Rloss (10.19) Comparing Equations 10.18 and 10.19, we find Rloss ≈L 6a r µf πσ (10.20) The power dissipated within an ESD in response to a sinusoidal current I0 applied at the terminals is 1 2 |I0|2 Rloss where Rloss (Equation 10.20) is the resistance perceived by a source applied to the ESD’s terminals. Note that Rloss is not the impedance of the ESD. Rloss is merely the contribution of internal loss to the impedance of the ESD. The impedance of the ESD must also account for contributions from radiation resistance (an additional real-valued contribution) and energy storage (perceived as a reactance). These quantities are addressed in other sections of this book. Example 10.2. Power dissipated within an ESD. A dipole is 10 cm in length, 1 mm in radius, and is surrounded by free space. The antenna is comprised of aluminum having conductivity ≈3.7 × 107 S/m and µ ≈µ0. A sinusoidal current having frequency 30 MHz and peak magnitude 100 mA is applied to the antenna terminals. What is the power dissipated within this antenna? Solution. The wavelength λ = c/f ∼= 10 m, so L = 10 cm ∼= 0.01λ. This certainly qualifies as electrically-short, so we may use Equation 10.20. In the present problem, a = 1 mm and σ ≈3.7 × 107 S/m. Thus, we | Electromagnetics_Vol2_Page_183_Chunk2359 |
10.4. REACTANCE OF THE ELECTRICALLY-SHORT DIPOLE 171 find that the loss resistance Rloss ≈9.49 mΩ. Subsequently, the power dissipated within this antenna is Ploss = 1 2 |I0|2 Rloss ≈47.5 µW (10.21) We conclude this section with one additional caveat: Whereas this section focuses on the limited conductivity of wire, other physical mechanisms may contribute to the loss resistance of the ESD. In particular, materials used to coat the antenna or to provide mechanical support near the terminals may potentially absorb and dissipate power that might otherwise be radiated. 10.4 Reactance of the Electrically-Short Dipole [m0210] For any given time-varying voltage appearing across the terminals of an electrically-short antenna, there will be a time-varying current that flows in response. The ratio of voltage to current is impedance. Just as a resistor, capacitor, or inductor may be characterized in terms of an impedance, any antenna may be characterized in terms of this impedance. The real-valued component of this impedance accounts for power which is radiated away from the antenna (Section 10.2) and dissipated within the antenna (Section 10.3). The imaginary component of this impedance – i.e., the reactance – typically represents energy storage within the antenna, in the same way that the reactance of a capacitor or inductor represents storage of electrical or magnetic energy, respectively. In this section, we determine the reactance of the electrically-short dipole (ESD). Reactance of a zero-length dipole. We begin with what might seem initially to be an absurd notion: A dipole antenna having zero length. However, such a dipole is certainly electrically-short, and in fact serves as an extreme condition from which we can deduce some useful information. Consider Figure 10.3, which shows a zero-length dipole being driven by a source via a transmission line. It is immediately apparent that such a dipole is equivalent to an open circuit. So, the impedance of a zero-length dipole is equal to that of an open circuit. c⃝S. Lally CC BY-SA 4.0 Figure 10.3: Transmission line terminated into a zero- length dipole, which is equivalent to an open circuit. | Electromagnetics_Vol2_Page_184_Chunk2360 |
172 CHAPTER 10. ANTENNAS What is the impedance of an open circuit? One might be tempted to say “infinite,” since the current is zero independently of the voltage. However, we must now be careful to properly recognize the real and imaginary components of this infinite number, and signs of these components. In fact, the impedance of an open circuit is 0 −j∞. One way to confirm this is via basic transmission line theory; i.e., the input impedance of transmission line stubs. A more direct justification is as follows: The real part of the impedance is zero because there is no transfer of power into the termination. The magnitude of the imaginary part of the impedance must be infinite because the current is zero. The sign of the imaginary component is negative because the reflected current experiences a sign change (required to make the total current zero), whereas the reflected voltage does not experience a sign change (and so is not canceled at the terminals). Thus, the impedance of an open circuit, and the zero-length dipole, is 0 −j∞. Reactance of a nearly-zero-length dipole. Let us now consider what happens as the length of the dipole increases from zero. Since such a dipole is short compared to a wavelength, the variation with length must be very simple. The reactance remains large and negative, but can only increase (become less negative) monotonically with increasing electrical length. This trend continues until the dipole is no longer electrically short. Thus, we conclude that the reactance of an ESD is always large and negative. The reactance of an ESD is very large and nega- tive, approaching the reactance of an open circuit termination as length decreases to zero. Note that this behavior is similar to that of a capacitor, whose impedance is also negative and increases toward zero with increasing frequency. Thus, the reactance of an ESD is sometimes represented in circuit diagrams as a capacitor. However, the specific dependence of reactance on frequency is different from that of a capacitor (as we shall see in a moment), so this model is generally valid only for analysis at one frequency at time. An approximate expression for the reactance of an ESD. Sometimes it is useful to be able to estimate the reactance XA of an ESD. Although a derivation is beyond the scope of this text, a suitable expression is (see, e.g., Johnson (1993) in “Additional References” at the end of this section): XA ≈−120 Ω πL/λ ln L 2a −1 (10.22) where L is length and a ≪L is the radius of the wire comprising the ESD. Note that this expression yields the expected behavior; i.e., XA →−∞as L →0, and increases monotonically as L increases from zero. Example 10.3. Reactance of an ESD. A dipole is 10 cm in length, 1 mm in radius, and is surrounded by free space. What is the reactance of this antenna at 30 MHz? Solution. The wavelength λ = c/f ∼= 10 m, so L = 10 cm ∼= 0.01λ. This certainly qualifies as electrically-short, so we may use Equation 10.22. Given a = 1 mm, we find the reactance XA ≈−11.1 kΩ. For what it’s worth, this antenna exhibits approximately the same reactance as a 0.47 pF capacitor at the same frequency. The reactance of the ESD is typically orders of magnitude larger than the real part of the impedance of the ESD. The reactance of the ESD is also typically very large compared to the characteristic impedance of typical transmission lines (usually 10s to 100s of ohms). This makes ESDs quite difficult to use in practical transmit applications. Additional Reading: • R.C. Johnson (Ed.), Antenna Systems Handbook (Ch. 4), McGraw-Hill, 1993. | Electromagnetics_Vol2_Page_185_Chunk2361 |
10.5. EQUIVALENT CIRCUIT MODEL FOR TRANSMISSION; RADIATION EFFICIENCY 173 10.5 Equivalent Circuit Model for Transmission; Radiation Efficiency [m0202] A radio transmitter consists of a source which generates the electrical signal intended for transmission, and an antenna which converts this signal into a propagating electromagnetic wave. Since the transmitter is an electrical system, it is useful to be able to model the antenna as an equivalent circuit. From a circuit analysis point of view, it should be possible to describe the antenna as a passive one-port circuit that presents an impedance to the source. Thus, we have the following question: What is the equivalent circuit for an antenna which is transmitting? We begin by emphasizing that the antenna is passive. That is, the antenna does not add power. Invoking the principle of conservation of power, there are only three possible things that can happen to power that is delivered to the antenna by the transmitter:1 • Power can be converted to a propagating electromagnetic wave. (The desired outcome.) • Power can be dissipated within the antenna. • Energy can be stored by the antenna, analogous to the storage of energy in a capacitor or inductor. We also note that these outcomes can occur in any combination. Taking this into account, we model the antenna using the equivalent circuit shown in Figure 10.4. Since the antenna is passive, it is reasonable to describe it as an impedance ZA which is (by definition) the ratio of voltage eVA to current eIA at the terminals; i.e., ZA ≜ eVA eIA (10.23) 1Note that “delivered” power means power accepted by the an- tenna. We are not yet considering power reflected from the antenna due to impedance mismatch. In the phasor domain, ZA is a complex-valued quantity and therefore has, in general, a real-valued component and an imaginary component. We may identify those components using the power conservation argument made previously: Since the real-valued component must represent power transfer and the imaginary component must represent energy storage, we infer: ZA ≜RA + jXA (10.24) where RA represents power transferred to the antenna, and XA represents energy stored by the antenna. Note that the energy stored by the antenna is being addressed in precisely the same manner that we address energy storage in a capacitor or an inductor; in all cases, as reactance. Further, we note that RA consists of components Rrad and Rloss as follows: ZA = Rrad + Rloss + jXA (10.25) where Rrad represents power transferred to the antenna and subsequently radiated, and Rloss represents power transferred to the antenna and subsequently dissipated. To confirm that this model works as expected, consider what happens when a voltage is applied across the antenna terminals. The current eIA flows and the time-average power PA transferred to the c⃝S. Lally CC BY-SA 4.0 Figure 10.4: Equivalent circuit for an antenna which is transmitting. | Electromagnetics_Vol2_Page_186_Chunk2362 |
174 CHAPTER 10. ANTENNAS antenna is PA = 1 2Re n eVAeI∗ A o (10.26) where we have assumed peak (as opposed to root mean squared) units for voltage and current. Since eVA = ZAeIA, we have: PA = 1 2Re n (Rrad + Rloss + jXA) eIAeI∗ A o (10.27) which reduces to: PA = 1 2 eIA 2 Rrad + 1 2 eIA 2 Rloss (10.28) As expected, the power transferred to the antenna is the sum of Prad ≜1 2 eIA 2 Rrad (10.29) representing power transferred to the radiating electromagnetic field, and Ploss ≜1 2 eIA 2 Rloss (10.30) representing power dissipated within the antenna. The reactance XA will play a role in determining eIA given eVA (and vice versa), but does not by itself account for disposition of power. Again, this is exactly analogous to the role played by inductors and capacitors in a circuit. The utility of this equivalent circuit formalism is that it allows us to treat the antenna in the same manner as any other component, and thereby facilitates analysis using conventional electric circuit theory and transmission line theory. For example: Given ZA, we know how to specify the output impedance ZS of the transmitter so as to minimize reflection from the antenna: We would choose ZS = ZA, since in this case the voltage reflection coefficient would be Γ = ZA −ZS ZA + ZS = 0 (10.31) Alternatively, we might specify ZS so as to maximize power transfer to the antenna: We would choose ZS = Z∗ A; i.e., conjugate matching. In order to take full advantage of this formalism, we require values for Rrad, Rloss, and XA. These quantities are considered below. Radiation resistance. Rrad is referred to as radiation resistance. Equation 10.29 tells us that Rrad = 2Prad eIA −2 (10.32) This equation suggests the following procedure: We apply current eIA to the antenna terminals, and then determine the total power Prad radiated from the antenna in response. For an example of this procedure, see Section 10.2 (“Total Power Radiated by an Electrically-Short Dipole”). Given eIA and Prad, one may then use Equation 10.32 to determine Rrad. Loss resistance. Loss resistance represents the dissipation of power within the antenna, which is usually attributable to loss intrinsic to materials comprising or surrounding the antenna. In many cases, antennas are made from good conductors – metals, in particular – so that Rloss is very low compared to Rrad. For such antennas, loss is often so low compared to Rrad that Rloss may be neglected. In the case of the electrically-short dipole, Rloss is typically very small but Rrad is also very small, so both must be considered. In many other cases, antennas contain materials with substantially greater loss than metal. For example, a microstrip patch antenna implemented on a printed circuit board typically has non-negligible Rloss because the dielectric material comprising the antenna exhibits significant loss. Antenna reactance. The reactance term jXA accounts for energy stored by the antenna. This may be due to reflections internal to the antenna, or due to energy associated with non-propagating electric and magnetic fields surrounding the antenna. The presence of significant reactance (i.e., |XA| comparable to or greater than |RA|) complicates efforts to establish the desired impedance match to the source. For an example, see Section 10.4 (“Reactance of the Electrically-Short Dipole”). Radiation efficiency. When Rloss is non-negligible, it is useful to characterize antennas in terms of their radiation efficiency erad, defined as the fraction of power which is radiated compared to the total power delivered to the antenna; i.e., erad ≜Prad PA (10.33) | Electromagnetics_Vol2_Page_187_Chunk2363 |
10.6. IMPEDANCE OF THE ELECTRICALLY-SHORT DIPOLE 175 Using Equations 10.28–10.30, we see that this efficiency can be expressed as follows: erad = Rrad Rrad + Rloss (10.34) Once again, the equivalent circuit formalism proves useful. Example 10.4. Impedance of an antenna. The total power radiated by an antenna is 60 mW when 20 mA (rms) is applied to the antenna terminals. The radiation efficiency of the antenna is known to be 70%. It is observed that voltage and current are in-phase at the antenna terminals. Determine (a) the radiation resistance, (b) the loss resistance, and (c) the impedance of the antenna. Solution. From the problem statement, Prad = 60 mW, eIA = 20 mA (rms), and erad = 0.7. Also, the fact that voltage and current are in-phase at the antenna terminals indicates that XA = 0. From Equation 10.32, the radiation resistance is Rrad ≈ 2 · (60 mW) √ 2 · 20 mA 2 = 150 Ω (10.35) Solving Equation 10.34 for the loss resistance, we find: Rloss = 1 −erad erad Rrad ∼= 64.3 Ω (10.36) Since ZA = Rrad + Rloss + jXA, we find ZA ∼= 214.3 + j0 Ω. This will be the ratio of voltage to current at the antenna terminals regardless of the source current. 10.6 Impedance of the Electrically-Short Dipole [m0204] In this section, we determine the impedance of the electrically-short dipole (ESD) antenna. The physical theory of operation for the ESD is introduced in Section 9.5, and additional relevant aspects of the ESD antenna are addressed in Sections 10.1–10.4. The concept of antenna impedance is addressed in Section 10.5. Review of those sections is suggested before attempting this section. The impedance of any antenna may be expressed as ZA = Rrad + Rloss + jXA (10.37) where the real-valued quantities Rrad, Rloss, and XA represent the radiation resistance, loss resistance, and reactance of the antenna, respectively. Radiation resistance. The radiation resistance of any antenna can be expressed as: Rrad = 2Prad |I0|−2 (10.38) where |I0| is the magnitude of the current at the antenna terminals, and Prad is the resulting total power radiated. For an ESD (Section 10.2): Prad ≈η |I0|2 (βL)2 48π (10.39) so Rrad ≈η (βL)2 24π (10.40) It is useful to have an alternative form of this expression in terms of wavelength λ. This is derived as follows. First, note: βL = 2π λ L = 2π L λ (10.41) where L/λ is the antenna length in units of wavelength. Substituting this expression into Equation 10.40: Rrad ≈η 2π L λ 2 1 24π ≈η π 6 L λ 2 (10.42) | Electromagnetics_Vol2_Page_188_Chunk2364 |
176 CHAPTER 10. ANTENNAS Assuming free-space conditions, η ∼= 376.7 Ω, which is ≈120π Ω. Subsequently, Rrad ≈20π2 L λ 2 (10.43) This remarkably simple expression indicates that the radiation resistance of an ESD is very small (since L ≪λ for an ESD), but increases as the square of the length. At this point, a warning is in order. The radiation resistance of an “electrically-short dipole” is sometimes said to be 80π2 (L/λ)2; i.e., 4 times the right side of Equation 10.43. This higher value is not for a physically-realizable ESD, but rather for the Hertzian dipole (sometimes also referred to as an “ideal dipole”). The Hertzian dipole is an electrically-short dipole with a current distribution that has uniform magnitude over the length of the dipole.2 The Hertzian dipole is quite difficult to realize in practice, and nearly all practical electrically-short dipoles exhibit a current distribution that is closer to that of the ESD. The ESD has a current distribution which is maximum in the center and goes approximately linearly to zero at the ends. The factor-of-4 difference in the radiation resistance of the Hertzian dipole relative to the (practical) ESD is a straightforward consequence of the difference in the current distributions. Loss resistance. The loss resistance of the ESD is derived in Section 10.3 and is found to be Rloss ≈L 6a r µf πσ (10.44) where a, µ, and σ are the radius, permeability, and conductivity of the wire comprising the antenna, and f is frequency. Reactance. The reactance of the ESD is addressed in Section 10.4. A suitable expression for this reactance is (see, e.g., Johnson (1993) in “Additional References” at the end of this section): XA ≈−120 Ω πL/λ ln L 2a −1 (10.45) 2See Section 9.4 for additional information about the Hertzian dipole. where a ≪L is assumed. Example 10.5. Impedance of an ESD. A thin, straight dipole antenna operates at 30 MHz in free space. The length and radius of the dipole are 1 m and 1 mm respectively. The dipole is comprised of aluminum having conductivity ≈3.7 × 107 S/m and µ ≈µ0. What is the impedance and radiation efficiency of this antenna? Solution. The free-space wavelength at f = 30 MHz is λ = c/f ∼= 10 m. Therefore, L ∼= 0.1λ, and this is an ESD. Since this is an ESD, we may compute the radiation resistance using Equation 10.43, yielding Rrad ≈1.97 Ω. The radius a = 1 mm and σ ≈3.7 × 107 S/m; thus, we find that the loss resistance Rloss ≈94.9 mΩusing Equation 10.44. Using Equation 10.45, the reactance is found to be XA ≈−1991.8 Ω. The impedance of this antenna is therefore ZA = Rrad + Rloss + jXA ≈2.1 −j1991.8 Ω (10.46) and the radiation efficiency of this antenna is erad = Rrad Rrad + Rloss ≈95.4% (10.47) In the preceding example, the radiation efficiency is a respectable 95.4%. However, the real part of the impedance of the ESD is much less than the characteristic impedance of typical transmission lines (typically 10s to 100s of ohms). Another problem is that the reactance of the ESD is very large. Some form of impedance matching is required to efficiently transfer power from a transmitter or transmission line to this form of antenna. Failure to do so will result in reflection of a large fraction of the power incident on antenna terminals. A common solution is to insert series inductance to reduce – nominally, cancel – the negative reactance of the ESD. In the preceding example, about 10 µH would be needed. The remaining mismatch in the real-valued components of impedance is then relatively easy to mitigate using common “real-to-real” impedance matching | Electromagnetics_Vol2_Page_189_Chunk2365 |
10.7. DIRECTIVITY AND GAIN 177 techniques. Additional Reading: • R.C. Johnson (Ed.), Antenna Systems Handbook (Ch. 4), McGraw-Hill, 1993. 10.7 Directivity and Gain [m0203] A transmitting antenna does not radiate power uniformly in all directions. Inevitably more power is radiated in some directions than others. Directivity quantifies this behavior. In this section, we introduce the concept of directivity and the related concepts of maximum directivity and antenna gain. Consider an antenna located at the origin. The power radiated in a single direction (θ, φ) is formally zero. This is because a single direction corresponds to a solid angle of zero, which intercepts an area of zero at any given distance from the antenna. Since the power flowing through any surface having zero area is zero, the power flowing in a single direction is formally zero. Clearly we need a different metric of power in order to develop a sensible description of the spatial distribution of power flow. The appropriate metric is spatial power density; that is, power per unit area, having SI base units of W/m2. Therefore, directivity is defined in terms of spatial power density in a particular direction, as opposed to power in a particular direction. Specifically, directivity in the direction (θ, φ) is: D(θ, φ) ≜ S(r) Save(r) (10.48) In this expression, S(r) is the power density at (r, θ, φ); i.e., at a distance r in the direction (θ, φ). Save(r) is the average power density at that distance; that is, S(r) averaged over all possible directions at distance r. Since directivity is a ratio of power densities, it is unitless. Summarizing: Directivity is ratio of power density in a specified direction to the power density averaged over all directions at the same distance from the antenna. Despite Equation 10.48, directivity does not depend on the distance from the antenna. To be specific, directivity is the same at every distance r. Even though the numerator and denominator of Equation 10.48 both vary with r, one finds that the distance dependence always cancels because power density and average power density are both | Electromagnetics_Vol2_Page_190_Chunk2366 |
178 CHAPTER 10. ANTENNAS proportional to r−2. This is a key point: Directivity is a convenient way to characterize an antenna because it does not change with distance from the antenna. In general, directivity is a function of direction. However, one is often not concerned about all directions, but rather only the directivity in the direction in which it is maximum. In fact it is quite common to use the term “directivity” informally to refer to the maximum directivity of an antenna. This is usually what is meant when the directivity is indicated to be a single number; in any event, the intended meaning of the term is usually clear from context. Example 10.6. Directivity of the electrically-short dipole. An electrically-short dipole (ESD) consists of a straight wire having length L ≪λ/2. What is the directivity of the ESD? Solution. The field radiated by an ESD is derived in Section 9.5. In that section, we find that the electric field intensity in the far field of a ˆz-oriented ESD located at the origin is: eE(r) ≈ˆθjη I0 · βL 8π (sin θ) e−jβr r (10.49) where I0 represents the magnitude and phase of the current applied to the terminals, η is the wave impedance of the medium, and β = 2π/λ. In Section 10.2, we find that the power density of this field is: S(r) ≈η |I0|2 (βL)2 128π2 (sin θ)2 1 r2 (10.50) and we subsequently find that the total power radiated is: Prad ≈η |I0|2 (βL)2 48π (10.51) The average power density Save is simply the total power divided by the area of a sphere centered on the ESD. Let us place this sphere at distance r, with r ≫L and r ≫λ as required for the validity of Equations 10.49 and 10.50. Then: Save = Prad 4πr2 ≈η |I0|2 (βL)2 192π2r2 (10.52) Finally the directivity is determined by applying the definition: D(θ, φ) ≜ S(r) Save(r) (10.53) ≈1.5 (sin θ)2 (10.54) The maximum directivity occurs in the θ = π/2 plane. Therefore, the maximum directivity is 1.5, meaning the maximum power density is 1.5 times greater than the power density averaged over all directions. Since directivity is a unitless ratio, it is common to express it in decibels. For example, the maximum directivity of the ESD in the preceding example is 10 log10 1.5 ∼= 1.76 dB. (Note “10 log10” here since directivity is the ratio of power-like quantities.) Gain. The gain G(θ, φ) of an antenna is its directivity modified to account for loss within the antenna. Specifically: G(θ, φ) ≜ S(r) for actual antenna Save(r) for identical but lossless antenna (10.55) In this equation, the numerator is the actual power density radiated by the antenna, which is less than the nominal power density due to losses within the antenna. The denominator is the average power density for an antenna which is identical, but lossless. Since the actual antenna radiates less power than an identical but lossless version of the same antenna, gain in any particular direction is always less than directivity in that direction. Therefore, an equivalent definition of antenna gain is G(θ, φ) ≜eradD(θ, φ) (10.56) where erad is the radiation efficiency of the antenna (Section 10.5). Gain is directivity times radiation efficiency; that is, directivity modified to account for loss within the antenna. | Electromagnetics_Vol2_Page_191_Chunk2367 |
10.8. RADIATION PATTERN 179 The receive case. To conclude this section, we make one additional point about directivity, which applies equally to gain. The preceding discussion has presumed an antenna which is radiating; i.e., transmitting. Directivity can also be defined for the receive case, in which it quantifies the effectiveness of the antenna in converting power in an incident wave to power in a load attached to the antenna. Receive directivity is formally introduced in Section 10.13 (“Effective Aperture”). When receive directivity is defined as specified in Section 10.13, it is equal to transmit directivity as defined in this section. Thus, it is commonly said that the directivity of an antenna is the same for receive and transmit. Additional Reading: • “Directivity” on Wikipedia. 10.8 Radiation Pattern [m0205] The radiation pattern of a transmitting antenna describes the magnitude and polarization of the field radiated by the antenna as a function of angle relative to the antenna. A pattern may also be defined for a receiving antenna, however, we defer discussion of the receive case to a later section. The concept of radiation pattern is closely related to the concepts of directivity and gain (Section 10.7). The principal distinction is the explicit consideration of polarization. In many applications, the polarization of the field radiated by a transmit antenna is as important as the power density radiated by the antenna. For example, a radio communication link consists of an antenna which is transmitting separated by some distance from an antenna which is receiving. Directivity determines the power density delivered to the receive antenna, but the receive antenna must be co-polarized with the arriving wave in order to capture all of this power. The radiation pattern concept is perhaps best explained by example. The simplest antenna encountered in common practice is the electrically-short dipole (ESD), which consists of a straight wire of length L that is much less than one-half of a wavelength. In Section 9.5, it is shown that the field radiated by an ESD which is located at the origin and aligned along the z-axis is: eE(r) ≈ˆθjη I0 · βL 8π (sin θ) e−jβr r (10.57) where I0 represents the magnitude and phase of the current applied to the terminals, η is the wave impedance of the medium, and β = 2π/λ is the phase propagation constant of the medium. Note that the reference direction of the electric field is in the ˆθ direction; therefore, a receiver must be ˆθ-polarized relative to the transmitter’s coordinate system in order to capture all available power. A receiver which is not fully ˆθ-polarized will capture less power. In the extreme, a receiver which is ˆφ-polarized with respect to the transmitter’s coordinate system will capture zero power. Thus, for the ˆz-oriented ESD, we refer to the ˆθ-polarization of the transmitted field as | Electromagnetics_Vol2_Page_192_Chunk2368 |
180 CHAPTER 10. ANTENNAS “co-polarized” or simply “co-pol,” and the ˆφ-polarization of the transmitted field as “cross-pol.” At this point, the reader may wonder what purpose is served by defining cross polarization, since the definition given above seems to suggest that cross-pol should always be zero. In common engineering practice, cross-pol is non-zero when co-pol is different from the intended or nominal polarization of the field radiated by the antenna. In the case of the ESD, we observe that the electric field is always ˆθ-polarized, and therefore we consider that to be the nominal polarization. Since the actual polarization of the ESD in the example is precisely the same as the nominal polarization of the ESD, the cross-pol of the ideal ESD is zero. If, on the other hand, we arbitrarily define ˆθ to be the nominal polarization and apply this definition to a different antenna that does not produce a uniformly ˆθ-polarized electric field, then we observe non-zero cross-pol. Cross-pol is similarly used to quantify effects due to errors in position or orientation, or due to undesired modification of the field due to materials (e.g., feed or mounting structures) near the transmit antenna. Summarizing: Co-pol is commonly defined to be the intended or nominal polarization for a particular applica- tion, which is not necessarily the actual polariza- tion radiated by the antenna under consideration. Cross-pol measures polarization in the orthogo- nal plane; i.e., deviation from the presumed co- pol. Returning to the ESD: Since eE(r) depends only on θ and not φ, the co-pol pattern is the same in any plane that contains the z axis. We refer to any such plane as the E-plane. In general: The E-plane is any plane in which the nominal or intended vector eE lies. Since the ESD is ˆθ-polarized, the E-plane pattern of the ESD is simply: eE(r) ≈η I0 · βL 8π (sin θ) 1 r (10.58) This pattern is shown in Figure 10.5. Equation 10.58 is referred to as an unnormalized c⃝S. Lally CC BY-SA 4.0 Figure 10.5: E-plane co-pol pattern for the ˆz-oriented ESD. In the unnormalized pattern scaling, the dashed circle indicates the maximum value of Equation 10.58. pattern. The associated normalized pattern is addressed later in this section. Similarly, we define the H-plane as follows: The H-plane is any plane in which the nominal or intended vector eH lies, and so is perpendicular to both the E-plane and the direction of propagation. In the case of the ESD, the one and only H-plane is the z = 0 plane. The H-plane pattern is shown in Figure 10.6. It is often useful to normalize the pattern, meaning that we scale the pattern so that its maximum magnitude corresponds to a convenient value. There are two common scalings. One common scaling sets the maximum value of the pattern equal to 1, and is therefore independent of source magnitude and distance. This is referred to as the normalized pattern. The normalized co-pol pattern can be defined as follows: F(θ, φ) ≜ ˆe · eE(r) ˆe · eE(r) max (10.59) where ˆe is the co-pol reference direction, and the | Electromagnetics_Vol2_Page_193_Chunk2369 |
10.8. RADIATION PATTERN 181 c⃝S. Lally CC BY-SA 4.0 Figure 10.6: H-plane co-pol pattern for the ˆz-oriented ESD. In the unnormalized pattern scaling, the radius of the circle is the maximum value of Equation 10.58. denominator is the maximum value of the electric field at distance r. A normalized pattern is scaled to a maximum magnitude of 1, using the definition expressed in Equation 10.59. Note that the value of r is irrelevant, since numerator and denominator both scale with r in the same way. Thus, the normalized pattern, like directivity, does not change with distance from the antenna. For the ESD, ˆe = ˆθ, so the normalized co-pol pattern is: F(θ, φ) = ˆθ · eE(r) ˆθ · eE(r) max (ESD) (10.60) which yields simply F(θ, φ) ≈sin θ (ESD) (10.61) Thus, the E-plane normalized co-pol pattern of the ESD is Figure 10.5 where the radius of the maximum value circle is equal to 1, which is 0 dB. Similarly, the H-plane normalized co-pol pattern of the ESD is Figure 10.6 where the radius of the circle is equal to 1 (0 dB). c⃝T. Truckle CC BY-SA 4.0 (modified) Figure 10.7: Co-pol pattern typical of a highly direc- tional antenna, such as a yagi. The other common scaling for patterns sets the maximum value equal to maximum directivity. Directivity is proportional to power density, which is proportional to eE(r) 2 . Therefore, this “directivity normalized” pattern can be expressed as: Dmax |F(θ, φ)|2 (10.62) where Dmax is the directivity in whichever direction it is maximum. For the ESD considered previously, Dmax = 1.5 (Section 10.7). Therefore, the co-pol pattern of the ESD using this particular scaling is 1.5 sin2 θ (ESD) (10.63) Thus, the E-plane co-pol pattern of the ESD using this scaling is similar to Figure 10.5, except that it is squared in magnitude and the radius of the maximum value circle is equal to 1.5, which is 1.76 dB. The H-plane co-pol pattern of the ESD, using this scaling, is Figure 10.6 where the radius of the circle is equal to 1.5 (1.76 dB). Pattern lobes. Nearly all antennas in common use exhibit directivity that is greater than that of the ESD. The patterns of these antennas subsequently exhibit more complex structure. Figure 10.7 shows an example. The region around the direction of maximum directivity is referred to as the main lobe. | Electromagnetics_Vol2_Page_194_Chunk2370 |
182 CHAPTER 10. ANTENNAS c⃝S. Lally CC BY-SA 4.0 Figure 10.8: Half-power beamwidth (HPBW). The main lobe is bounded on each side by a null, where the magnitude reaches a local minimum, perhaps zero. Many antennas also exhibit a lobe in the opposite direction, known as a backlobe. (Many other antennas exhibit a null in this direction.) Lobes between the main lobe and the backlobe are referred to as sidelobes. Other commonly-used pattern metrics include first sidelobe level, which is the ratio of the maximum magnitude of the sidelobe closest to the main lobe to that of the main lobe; and front-to-back ratio, which is the ratio of the maximum magnitude of the backlobe to that of the main lobe. When the main lobe is narrow, it is common to characterize the pattern in terms of half-power beamwidth (HPBW). This is shown in Figure 10.8. HPBW is the width of the main lobe measured between two points at which the directivity is one-half its maximum value. As a pattern metric, HPBW depends on the plane in which it is measured. In particular, HPBW may be different in the E- and H-planes. For example, the E-plane HPBW of the ESD is found to be 90◦, whereas the H-plane HPBW is undefined since the pattern is constant in the H-plane. Omnidirectional and isotropic antennas. The ESD is an example of an omnidirectional antenna. An omnidirectional antenna is an antenna whose pattern magnitude is nominally constant in a plane containing the maximum directivity. For the ESD, this plane is the H-plane, so the ESD is said to be omnidirectional in the H-plane. The term “omnidirectional” does not indicate constant pattern in all directions. (In fact, note that the ESD exhibits pattern nulls in two directions.) An omnidirectional antenna, such as the ESD, ex- hibits constant and nominally maximum directiv- ity in one plane. An antenna whose pattern is uniform in all directions is said to be isotropic. No physically-realizable antenna is isotropic; in fact, the ESD is about as close as one can get. Nevertheless, the “isotropic antenna” concept is useful as a standard against which other antennas can be quantified. Since the pattern of an isotropic antenna is constant with direction, the power density radiated by such an antenna in any direction is equal to the power density averaged over all directions. Thus, the directivity of an isotropic antenna is exactly 1. The directivity of any other antenna may be expressed in units of “dBi,” which is simply dB relative to that of an isotropic antenna. For example, the maximum directivity of the ESD is 1.5, which is 10 log10 (1.5/1) = 1.76 dBi. Summarizing: An isotropic antenna exhibits constant directiv- ity in every direction. Such an antenna is not physically-realizable, but is nevertheless useful as a baseline for describing other antennas. Additional Reading: • “Radiation pattern” on Wikipedia. | Electromagnetics_Vol2_Page_195_Chunk2371 |
10.9. EQUIVALENT CIRCUIT MODEL FOR RECEPTION 183 10.9 Equivalent Circuit Model for Reception [m0206] In this section, we begin to address antennas as devices that convert incident electromagnetic waves into potentials and currents in a circuit. It is convenient to represent this process in the form of a Th´evenin equivalent circuit. The particular circuit addressed in this section is shown in Figure 10.9. The circuit consists of a voltage source eVOC and a series impedance ZA. The source potential eVOC is the potential at the terminals of the antenna when there is no termination; i.e., when the antenna is open-circuited. The series impedance ZA is the output impedance of the circuit, and so determines the magnitude and phase of the current at the terminals once a load is connected. Given eVOC and the current through the equivalent circuit, it is possible to determine the power delivered to the load. Thus, this model is quite useful, but only if we are able to determine eVOC and ZA. This section provides an informal derivation of these quantities that is sufficient to productively address the subsequent important topics of effective aperture and impedance matching of receive antennas.3 Vector effective length. With no derivation required, we can deduce the following about eVOC: • eVOC must depend on the incident electric field 3Formal derivations of these quantities are provided in subse- quent sections. The starting point is the section on reciprocity. c⃝S. Lally, CC BY-SA 4.0 (modified) Figure 10.9: Th´evenin equivalent circuit for an an- tenna in the presence of an incident electromagnetic wave. intensity eEi. Presumably the relationship is linear, so eVOC is proportional to the magnitude of eEi. • Since eEi is a vector whereas eVOC is a scalar, there must be some vector le for which eVOC = eEi · le (10.64) • Since eEi has SI base units of V/m and eVOC has SI base units of V, le must have SI base units of m; i.e., length. • We expect that eVOC increases as the size of the antenna increases, so the magnitude of le likely increases with the size of the antenna. • The direction of le must be related to the orientation of the incident electric field relative to that of the antenna, since this is clearly important yet we have not already accounted for this. It may seem at this point that le is unambiguously determined, and we need merely to derive its value. However, this is not the case. There are in fact multiple unique definitions of le that will reduce the vector eEi to the observed scalar eVOC via Equation 10.64. In this section, we shall employ the most commonly-used definition, in which le is referred to as vector effective length. Following this definition, the scalar part le of le = ˆlle is commonly referred to as any of the following: effective length (the term used in this book), effective height, or antenna factor. In this section, we shall merely define vector effective length, and defer a formal derivation to Section 10.11. In this definition, we arbitrarily set ˆl, the real-valued unit vector indicating the direction of le, equal to the direction in which the electric field transmitted from this antenna would be polarized in the far field. For example, consider a ˆz-oriented electrically-short dipole (ESD) located at the origin. The electric field transmitted from this antenna would have only a ˆθ component, and no ˆφ component (and certainly no ˆr component). Thus, ˆl = ˆθ in this case. Applying this definition, eEi ·ˆl yields the scalar component of eEi that is co-polarized with electric field radiated by the antenna when transmitting. Now | Electromagnetics_Vol2_Page_196_Chunk2372 |
184 CHAPTER 10. ANTENNAS le is uniquely defined to be the factor that converts this component into eVOC. Summarizing: The vector effective length le = ˆlle is defined as follows: ˆl is the real-valued unit vector cor- responding to the polarization of the electric field that would be transmitted from the antenna in the far field. Subsequently, the effective length le is le ≜ eVOC eEi ·ˆl (10.65) where eVOC is the open-circuit potential induced at the antenna terminals in response to the inci- dent electric field intensity eEi. While this definition yields an unambiguous value for le, it is not yet clear what that value is. For most antennas, effective length is quite difficult to determine directly, and one must instead determine effective length indirectly from the transmit characteristics via reciprocity. This approach is relatively easy (although still quite a bit of effort) for thin dipoles, and is presented in Section 10.11. To provide an example of how effective length works right away, consider the ˆz-oriented ESD described earlier in this section. Let the length of this ESD be L. Let eEi be a ˆθ-polarized plane wave arriving at the ESD. The ESD is open-circuited, so the potential induced in its terminals is eVOC. One observes the following: • When eEi arrives from anywhere in the θ = π/2 plane (i.e., broadside to the ESD), eEi points in the −ˆz direction, and we find that le ≈L/2. It should not be surprising that le is proportional to L; this expectation was noted earlier in this section. • When eEi arrives from the directions θ = 0 or θ = π – i.e., along the axis of the ESD – eEi is perpendicular to the axis of the ESD. In this case, we find that le equals zero. Taken together, these findings suggest that le should contain a factor of sin θ. We conclude that the vector effective length for a ˆz-directed ESD of length L is le ≈ˆθL 2 sin θ (ESD) (10.66) Example 10.7. Potential induced in an ESD. A thin straight dipole of length 10 cm is located at the origin and aligned with the z-axis. A plane wave is incident on the dipole from the direction (θ = π/4, φ = π/2). The frequency of the wave is 30 MHz. The magnitude of the incident electric field is 10 µV/m (rms). What is the magnitude of the induced open-circuit potential when the electric field is (a) ˆθ-polarized and (b) ˆφ-polarized? Solution. The wavelength in this example is c/f ∼= 10 m, so this dipole is electrically-short. Using Equation 10.66: le ≈ˆθ10 cm 2 sin π 4 ≈ˆθ (3.54 cm) Thus, the effective length le = 3.54 cm. When the electric field is ˆθ-polarized, the magnitude of the induced open-circuit voltage is eVOC = eEi · le ≈(10 µV/m) ˆθ · ˆθ (3.54 cm) ≈354 nV rms (a) When the electric field is ˆφ-polarized: eVOC ≈(10 µV/m) ˆφ · ˆθ (3.54 cm) ≈0 (b) This is because the polarization of the incident electric field is orthogonal to that of the ESD. In fact, the answer to part (b) is zero for any angle of incidence (θ, φ). Output impedance. The output impedance ZA is somewhat more difficult to determine without a formal derivation, which is presented in Section 10.12. For the purposes of this section, it suffices to jump directly to the result: | Electromagnetics_Vol2_Page_197_Chunk2373 |
10.9. EQUIVALENT CIRCUIT MODEL FOR RECEPTION 185 The output impedance ZA of the equivalent cir- cuit for an antenna in the receive case is equal to the input impedance of the same antenna in the transmit case. This remarkable fact is a consequence of the reciprocity property of antenna systems, and greatly simplifies the analysis of receive antennas. Now a demonstration of how the antenna equivalent circuit can be used to determine the power delivered by an antenna to an attached electrical circuit: Example 10.8. Power captured by an ESD. Continuing with part (a) of Example 10.7: If this antenna is terminated into a conjugate-matched load, then what is the power delivered to that load? Assume the antenna is lossless. Solution. First, we determine the impedance ZA of the equivalent circuit of the antenna. This is equal to the input impedance of the antenna in transmission. Let RA and XA be the real and imaginary parts of this impedance; i.e., ZA = RA + jXA. Further, RA is the sum of the radiation resistance Rrad and the loss resistance. The loss resistance is zero because the antenna is lossless. Since this is an ESD: Rrad ≈20π2 L λ 2 (10.67) Therefore, RA = Rrad ≈4.93 mΩ. We do not need to calculate XA, as will become apparent in the next step. A conjugate-matched load has impedance Z∗ A, so the potential eVL across the load is eVL = eVOC Z∗ A ZA + Z∗ A = eVOC Z∗ A 2RA (10.68) The current eIL through the load is eIL = eVOC ZA + Z∗ A = eVOC 2RA (10.69) Taking eVOC as an RMS quantity, the power PL delivered to the load is PL = Re {VLI∗ L} = eVOC 2 4RA (10.70) In part (a) of Example 10.7, eVOC is found to be ≈354 nV rms, so PL ≈6.33 pW. Additional Reading: • “Th´evenin’s Theorem” on Wikipedia. • W.L. Stutzman & G.A. Thiele, Antenna Theory and Design, 3rd Ed., Wiley, 2012. Sec. 4.2 (“Receiving Properties of Antennas”). | Electromagnetics_Vol2_Page_198_Chunk2374 |
186 CHAPTER 10. ANTENNAS 10.10 Reciprocity [m0214] The term “reciprocity” refers to a class of theorems that relate the inputs and outputs of a linear system to those of an identical system in which the inputs and outputs are swapped. The importance of reciprocity in electromagnetics is that it simplifies problems that would otherwise be relatively difficult to solve. An example of this is the derivation of the receiving properties of antennas, which is addressed in other sections using results derived in this section. As an initial and relatively gentle introduction, consider a well-known special case that emerges in basic circuit theory: The two-port device shown in Figure 10.10. The two-port is said to be reciprocal if the voltage v2 appearing at port 2 due to a current applied at port 1 is the same as v1 when the same current is applied instead at port 2. This is normally the case when the two-port consists exclusively of linear passive devices such as ideal resistors, capacitors, and inductors. The fundamental underlying requirement is linearity: That is, outputs must be proportional to inputs, and superposition must apply. This result from basic circuit theory is actually a special case of a more general theorem of electromagnetics, which we shall now derive. Figure 10.11 shows a scenario in which a current distribution eJ1 is completely contained within a volume V. This current is expressed as a volume current density, having SI base units of A/m2. Also, the current is expressed in phasor form, signaling that we are considering a single frequency. This current distribution gives rise to an electric field intensity eE1, Inductiveload, public domain (modified) Figure 10.10: Two-port device. c⃝C. Wang CC BY-SA 4.0 Figure 10.11: An electromagnetic system consisting of a current distribution eJ1 radiating an electric field eE1 in the presence of linear matter. c⃝C. Wang CC BY-SA 4.0 Figure 10.12: The same electromagnetic system as shown in Figure 10.11 (including the same distribution of matter), but now with a different input, eJ2, resulting in a different output, eE2. | Electromagnetics_Vol2_Page_199_Chunk2375 |
10.10. RECIPROCITY 187 having SI base units of V/m. The volume may consist of any combination of linear time-invariant matter; i.e., permittivity ǫ, permeability µ, and conductivity σ are constants that may vary arbitrarily with position but not with time. Here’s a key idea: We may interpret this scenario as a “two-port” system in which eJ1 is the input, eE1 is the output, and the system’s behavior is completely defined by Maxwell’s equations in differential phasor form: ∇× eE1 = −jωµ eH1 (10.71) ∇× eH1 = eJ1 + jωǫeE1 (10.72) along with the appropriate electromagnetic boundary conditions. (For the purposes of our present analysis, eH1 is neither an input nor an output; it is merely a coupled quantity that appears in this particular formulation of the relationship between eE1 and eJ1.) Figure 10.12 shows a scenario in which a different current distribution eJ2 is completely contained within the same volume V containing the same distribution of linear matter. This new current distribution gives rise to an electric field intensity eE2. The relationship between eE2 and eJ2 is governed by the same equations: ∇× eE2 = −jωµ eH2 (10.73) ∇× eH2 = eJ2 + jωǫeE2 (10.74) along with the same electromagnetic boundary conditions. Thus, we may interpret this scenario as the same electromagnetic system depicted in Figure 10.11, except now with eJ2 as the input and eE2 is the output. The input current and output field in the second scenario are, in general, completely different from the input current and output field in the first scenario. However, both scenarios involve precisely the same electromagnetic system; i.e., the same governing equations and the same distribution of matter. This leads to the following question: Let’s say you know nothing about the system, aside from the fact that it is linear and time-invariant. However, you are able to observe the first scenario; i.e., you know eE1 in response to eJ1. Given this very limited look into the behavior of the system, what can you infer about eE2 given eJ2, or vice-versa? At first glance, the answer might seem to be nothing, since you lack a description of the system. However, the answer turns out to be that a surprising bit more information is available. This information is provided by reciprocity. To show this, we must engage in some pure mathematics. Derivation of the Lorentz reciprocity theorem. First, let us take the dot product of eH2 with each side of Equation 10.71: eH2 · ∇× eE1 = −jωµ eH1 · eH2 (10.75) Similarly, let us take the dot product of eE1 with each side of Equation 10.74: eE1 · ∇× eH2 = eE1 · eJ2 + jωǫeE1 · eE2 (10.76) Next, let us subtract Equation 10.75 from Equation 10.76. The left side of the resulting equation is eE1 · ∇× eH2 −eH2 · ∇× eE1 (10.77) This can be simplified using the vector identity (Appendix B.3): ∇· (A × B) = B · (∇× A) −A · (∇× B) (10.78) Yielding eE1 · ∇× eH2 −eH2 · ∇× eE1 = ∇· eH2 × eE1 (10.79) So, by subtracting Equation 10.75 from Equation 10.76, we have found: ∇· eH2 × eE1 = eE1 ·eJ2 +jωǫeE1 · eE2 +jωµ eH1 · eH2 (10.80) Next, we repeat the process represented by Equations 10.75–10.80 above to generate a complementary equation to Equation 10.80. This time we take the dot product of eE2 with each side of Equation 10.72: eE2 · ∇× eH1 = eE2 · eJ1 + jωǫeE1 · eE2 (10.81) Similarly, let us take the dot product of eH1 with each side of Equation 10.73: eH1 · ∇× eE2 = −jωµ eH1 · eH2 (10.82) | Electromagnetics_Vol2_Page_200_Chunk2376 |
188 CHAPTER 10. ANTENNAS Next, let us subtract Equation 10.82 from Equation 10.81. Again using the vector identity, the left side of the resulting equation is eE2 · ∇× eH1 −eH1 · ∇× eE2 = ∇· eH1 × eE2 (10.83) So we find: ∇· eH1 × eE2 = eE2 ·eJ1 +jωǫeE1 · eE2 +jωµ eH1 · eH2 (10.84) Finally, subtracting Equation 10.84 from Equation 10.80, we obtain: ∇· eH2 × eE1 −eH1 × eE2 = eE1 · eJ2 −eE2 · eJ1 (10.85) This equation is commonly known by the name of the theorem it represents: The Lorentz reciprocity theorem. The theorem is a very general statement about the relationship between fields and currents at each point in space. An associated form of the theorem applies to contiguous regions of space. To obtain this form, we simply integrate both sides of Equation 10.85 over the volume V: Z V ∇· eH2 × eE1 −eH1 × eE2 dv = Z V eE1 · eJ2 −eE2 · eJ1 dv (10.86) We now take the additional step of using the divergence theorem (Appendix B.3) to transform the left side of the equation into a surface integral: I S eH2 × eE1 −eH1 × eE2 · ds = Z V eE1 · eJ2 −eE2 · eJ1 dv (10.87) where S is the closed mathematical surface which bounds V. This is also the Lorentz reciprocity theorem, but now in integral form. This version of the theorem relates fields on the bounding surface to sources within the volume. The integral form of the theorem has a particularly useful feature. Let us confine the sources to a finite region of space, while allowing V to grow infinitely large, expanding to include all space. In this situation, the closest distance between any point containing non-zero source current and S is infinite. Because field magnitude diminishes with distance from the source, the fields (eE1, eH1) and (eE2, eH2) are all effectively zero on S. In this case, the left side of Equation 10.87 is zero, and we find: Z V eE1 · eJ2 −eE2 · eJ1 dv = 0 (10.88) for any volume V which contains all the current. The Lorentz reciprocity theorem (Equa- tion 10.88) describes a relationship between one distribution of current and the resulting fields, and a second distribution of current and resulting fields, when both scenarios take place in identical regions of space filled with identical distributions of linear matter. Why do we refer to this relationship as reciprocity? Simply because the expression is identical when the subscripts “1” and “2” are swapped. In other words, the relationship does not recognize a distinction between “inputs” and “outputs;” there are only “ports.” A useful special case pertains to scenarios in which the current distributions eJ1 and eJ2 are spatially disjoint. By “spatially disjoint,” we mean that there is no point in space at which both eJ1 and eJ2 are non-zero; in other words, these distributions do not overlap. (Note that the currents shown in Figures 10.11 and 10.12 are depicted as spatially disjoint.) To see what happens in this case, let us first rewrite Equation 10.88 as follows: Z V eE1 · eJ2 dv = Z V eE2 · eJ1 dv (10.89) Let V1 be the contiguous volume over which eJ1 is non-zero, and let V2 be the contiguous volume over which eJ2 is non-zero. Then Equation 10.89 may be written as follows: Z V2 eE1 · eJ2 dv = Z V1 eE2 · eJ1 dv (10.90) The utility of this form is that we have reduced the region of integration to just those regions where the current exists. | Electromagnetics_Vol2_Page_201_Chunk2377 |
10.10. RECIPROCITY 189 c⃝C. Wang CC BY-SA 4.0 Figure 10.13: A two-port consisting of two dipole antennas. Reciprocity of two-ports consisting of antennas. Equation 10.90 allows us to establish the reciprocity of two-ports consisting of pairs of antennas. This is most easily demonstrated for pairs of thin dipole antennas, as shown in Figure 10.13. Here, port 1 is defined by the terminal quantities (eV1, eI1) of the antenna on the left, and port 2 is defined by the terminal quantities (eV2, eI2) of the antenna on the right. These quantities are defined with respect to a small gap of length ∆l between the perfectly-conducting arms of the dipole. Either antenna may transmit or receive, so (eV1, eI1) depends on (eV2, eI2), and vice-versa. The transmit and receive cases for port 1 are illustrated in Figure 10.14(a) and (b), respectively. Note that eJ1 is the current distribution on this antenna when transmitting (i.e., when driven by the impressed current eIt 1) and eE2 is the electric field incident on the antenna when the other antenna is transmitting. Let us select V1 to be the cylindrical volume defined by the exterior surface of the dipole, including the gap between the arms of the dipole. We are now ready to consider the right side of Equation 10.90: Z V1 eE2 · eJ1 dv (10.91) The electromagnetic boundary conditions applicable to the perfectly-conducting dipole arms allow this integral to be dramatically simplified. First, recall that all current associated with a perfect conductor must flow on the surface of the material, and therefore c⃝C. Wang CC BY-SA 4.0 Figure 10.14: The port 1 antenna (a) transmitting and (b) receiving. eJ1 = 0 everywhere except on the surface. Therefore, the direction of eJ1 is always tangent to the surface. The tangent component of eE2 is zero on the surface of a perfectly-conducting material, as required by the applicable boundary condition. Therefore, eE2 · eJ1 = 0 everywhere eJ1 is non-zero. There is only one location where the current is non-zero and yet there is no conductor: This location is the gap located precisely at the terminals. Thus, we find: Z V1 eE2 · eJ1 dv = Z gap eE2 · eIt 1ˆl1 dl (10.92) where the right side is a line integral crossing the gap that defines the terminals of the antenna. We assume the current eIt 1 is constant over the gap, and so may be factored out of the integral: Z V1 eE2 · eJ1 dv = eIt 1 Z gap eE2 ·ˆl1 dl (10.93) Recall that potential between two points in space is given by the integral of the electric field over any path between those two points. In particular, we may calculate the open-circuit potential eV r 1 as follows: eV r 1 = − Z gap eE2 ·ˆl1 dl (10.94) Remarkably, we have found: Z V1 eE2 · eJ1 dv = −eIt 1 eV r 1 (10.95) | Electromagnetics_Vol2_Page_202_Chunk2378 |
190 CHAPTER 10. ANTENNAS Applying the exact same procedure for port 2 (or by simply exchanging subscripts), we find: Z V2 eE1 · eJ2 dv = −eIt 2 eV r 2 (10.96) Now substituting these results into Equation 10.90, we find: eIt 1 eV r 1 = eIt 2 eV r 2 (10.97) At the beginning of this section, we stated that a two-port is reciprocal if eV2 appearing at port 2 due to a current applied at port 1 is the same as eV1 when the same current is applied instead at port 2. If the present two-dipole problem is reciprocal, then eV r 2 due to eIt 1 should be the same as eV r 1 when eIt 2 = eIt 1. Is it? Let us set eIt 1 equal to some particular value I0, then the resulting value of eV r 2 will be some particular value V0. If we subsequently set eIt 2 equal to I0, then the resulting value of eV r 1 will be, according to Equation 10.97: eV r 1 = eIt 2 eV r 2 eIt 1 = I0V0 I0 = V0 (10.98) Therefore, Equation 10.97 is simply a mathematical form of the familiar definition of reciprocity from basic circuit theory, and we have found that our system of antennas is reciprocal in precisely this same sense. The above analysis presumed pairs of straight, perfectly conducting dipoles of arbitrary length. However, the result is readily generalized – in fact, is the same – for any pair of passive antennas in linear time-invariant media. Summarizing: The potential induced at the terminals of one an- tenna due to a current applied to a second antenna is equal to the potential induced in the second an- tenna by the same current applied to the first an- tenna (Equation 10.97). Additional Reading: • “Reciprocity (electromagnetism)” on Wikipedia. • “Two-port network” on Wikipedia. 10.11 Potential Induced in a Dipole [m0215] An electromagnetic wave incident on an antenna will induce a potential at the terminals of the antenna. In this section, we shall derive this potential. To simplify the derivation, we shall consider the special case of a straight thin dipole of arbitrary length that is illuminated by a plane wave. However, certain aspects of the derivation will apply to antennas generally. In particular, the concepts of effective length (also known as effective height) and vector effective length emerge naturally from this derivation, so this section also serves as a stepping stone in the development of an equivalent circuit model for a receiving antenna. The derivation relies on the transmit properties of dipoles as well as the principle of reciprocity, so familiarity with those topics is recommended before reading this section. The scenario of interest is shown in Figure 10.15. Here a thin ˆz-aligned straight dipole is located at the origin. The total length of the dipole is L. The arms of the dipole are perfectly-conducting. The terminals consist of a small gap of length ∆l between the arms. The incident plane wave is described in terms of its electric field intensity eEi. The question is: What is eVOC, the potential at the terminals when the terminals are open-circuited? There are multiple approaches to solve this problem. A direct attack is to invoke the principle that potential is equal to the integral of the electric field intensity over a path. In this case, the path begins at the “−” terminal and ends at the “+” terminal, crossing the gap that defines the antenna terminals. Thus: eVOC = − Z gap eEgap · dl (10.99) where eEgap is the electric field in the gap. The problem with this approach is that the value of eEgap is not readily available. It is not simply eEi, because the antenna structure (in particular, the electromagnetic boundary conditions) modify the electric field in the vicinity of the antenna.4 4Also, if this were true, then the antenna itself would not matter; | Electromagnetics_Vol2_Page_203_Chunk2379 |
10.11. POTENTIAL INDUCED IN A DIPOLE 191 Fortunately, we can bypass this obstacle using the principle of reciprocity. In a reciprocity-based strategy, we establish a relationship between two scenarios that take place within the same electromagnetic system. The first scenario is shown in Figure 10.16. In this scenario, we have two dipoles. The first dipole is precisely the dipole of interest (Figure 10.15), except that a current eIt 1 is applied to the antenna terminals. This gives rise to a current distribution eI(z) (SI base units of A) along the dipole, and subsequently the dipole radiates the electric field (Section 9.6): eE1(r) ≈ˆθj η 2 e−jβr r (sin θ) · " 1 λ Z +L/2 −L/2 eI(z′)e+jβz′ cos θdz′ # (10.100) The second antenna is a ˆθ-aligned Hertzian dipole in the far field, which receives eE1. (For a refresher on the properties of Hertzian dipoles, see Section 9.4. A key point is that Hertzian dipoles are vanishingly small.) Specifically, we measure (conceptually, at least) the open-circuit potential eV r 2 at the terminals of the Hertzian dipole. We select a Hertzian dipole for this purpose because – in contrast to essentially all only the relative spacing and orientation of the antenna terminals would matter! c⃝C. Wang CC BY-SA 4.0 Figure 10.15: A potential is induced at the terminals of a thin straight dipole in response to an incident plane wave. c⃝C. Wang CC BY-SA 4.0 Figure 10.16: The dipole of interest driven by current eIt 1 radiates electric field eE1, resulting in open-circuit potential eV r 2 at the terminals of a Hertzian dipole in the far field. c⃝C. Wang CC BY-SA 4.0 Figure 10.17: The Hertzian dipole driven by current eIt 2 radiates electric field eE2, resulting in open-circuit potential eV r 1 at the terminals of the dipole of interest. | Electromagnetics_Vol2_Page_204_Chunk2380 |
192 CHAPTER 10. ANTENNAS other antennas – it is simple to determine the open circuit potential. As explained earlier: eV r 2 = − Z gap eEgap · dl (10.101) For the Hertzian dipole, eEgap is simply the incident electric field, since there is negligible structure (in particular, a negligible amount of material) present to modify the electric field. Thus, we have simply: eV r 2 = − Z gap eE1 · dl (10.102) Since the Hertzian dipole is very short and very far away from the transmitting dipole, eE1 is essentially constant over the gap. Also recall that we required the Hertzian dipole to be aligned with eE1. Choosing to integrate in a straight line across the gap, Equation 10.102 reduces to: eV r 2 = −eE1(r2) · ˆθ∆l (10.103) where ∆l is the length of the gap and r2 is the location of the Hertzian dipole. Substituting the expression for eE1 from Equation 10.100, we obtain: eV r 2 ≈−j η 2 e−jβr2 r2 (sin θ) · " 1 λ Z +L/2 −L/2 eI(z′)e+jβz′ cos θdz′ # ∆l (10.104) where r2 = |r2|. The second scenario is shown in Figure 10.17. This scenario is identical to the first scenario, with the exception that the Hertzian dipole transmits and the dipole of interest receives. The field radiated by the Hertzian dipole in response to applied current eIt 2, evaluated at the origin, is (Section 9.4): eE2(r = 0) ≈ˆθjη eIt 2 · β∆l 4π (1) e−jβr2 r2 (10.105) The “sin θ” factor in the general expression is equal to 1 in this case, since, as shown in Figure 10.17, the origin is located broadside (i.e., at π/2 rad) relative to the axis of the Hertzian dipole. Also note that because the Hertzian dipole is presumed to be in the far field, E2 may be interpreted as a plane wave in the region of the receiving dipole of interest. Now we ask: What is the induced potential eV r 1 in the dipole of interest? Once again, Equation 10.99 is not much help, because the electric field in the gap is not known. However, we do know that eV r 1 should be proportional to eE2(r = 0), since this is presumed to be a linear system. Based on this much information alone, there must be some vector le = ˆlle for which eV r 1 = eE2(r = 0) · le (10.106) This does not uniquely define either the unit vector ˆl nor the scalar part le, since a change in the definition of the former can be compensated by a change in the definition of the latter and vice-versa. So at this point we invoke the standard definition of le as the vector effective length, introduced in Section 10.9. Thus, ˆl is defined to be the direction in which an electric field transmitted from the antenna would be polarized. In the present example, ˆl = −ˆθ, where the minus sign reflects the fact that positive terminal potential results in terminal current which flows in the −ˆz direction. Thus, Equation 10.106 becomes: eV r 1 = −eE2(r = 0) · ˆθle (10.107) We may go a bit further and substitute the expression for eE2(r = 0) from Equation 10.105: eV r 1 ≈−jη eIt 2 · β∆l 4π e−jβr2 r2 le (10.108) Now we invoke reciprocity. As a two-port linear time-invariant system, it must be true that: eIt 1 eV r 1 = eIt 2 eV r 2 (10.109) Thus: eV r 1 = eIt 2 eIt 1 eV r 2 (10.110) Substituting the expression for eV r 2 from Equation 10.104: eV r 1 ≈− eIt 2 eIt 1 · j η 2 e−jβr2 r2 (sin θ) · " 1 λ Z +L/2 −L/2 eI(z′)e+jβz′ cos θdz′ # ∆l (10.111) Thus, reciprocity has provided a second expression for eV r 1 . We may solve for le by setting this expression | Electromagnetics_Vol2_Page_205_Chunk2381 |
10.11. POTENTIAL INDUCED IN A DIPOLE 193 equal to the expression from Equation 10.108, yielding le ≈2π βeIt 1 " 1 λ Z +L/2 −L/2 eI(z′)e+jβz′ cos θdz′ # sin θ (10.112) Noting that β = 2π/λ, this simplifies to: le ≈ " 1 eIt 1 Z +L/2 −L/2 eI(z′)e+jβz′ cos θdz′ # sin θ (10.113) Thus, you can calculate le using the following procedure: 1. Apply a current eIt 1 to the dipole of interest. 2. Determine the resulting current distribution eI(z) along the length of the dipole. (Note that precisely this is done for the electrically-short dipole in Section 9.5 and for the half-wave dipole in Section 9.7.) 3. Integrate eI(z) over the length of the dipole as indicated in Equation 10.113. Then divide (“normalize”) by eIt 1 (which is simply eI(0)). Note that the result is independent of the excitation eIt 1, as expected since this is a linear system. 4. Multiply by sin θ. We have now determined that the open-circuit terminal potential eVOC in response to an incident electric field eEi is eVOC = eEi · le (10.114) where le = ˆlle is the vector effective length defined previously. This result is remarkable. In plain English, we have found that: The potential induced in a dipole is the co- polarized component of the incident electric field times a normalized integral of the transmit cur- rent distribution over the length of the dipole, times sine of the angle between the dipole axis and the direction of incidence. In other words, the reciprocity property of linear systems allows this property of a receiving antenna to be determined relatively easily if the transmit characteristics of the antenna are known. Example 10.9. Effective length of a thin electrically-short dipole (ESD). A explained in Section 9.5, the current distribution on a thin ESD is eI(z) ≈I0 1 −2 L |z| (10.115) where L is the length of the dipole and I0 is the terminal current. Applying Equation 10.113, we find: le ≈ " 1 I0 Z +L/2 −L/2 I0 1 −2 L |z′| e+jβz′ cos θdz′ # · sin θ (10.116) Recall β = 2π/λ, so βz′ = 2π (z′/λ). Since this is an electrically-short dipole, z′ ≪λ over the entire integral, and subsequently we may assume e+jβz′ cos θ ≈1 over the entire integral. Thus: le ≈ "Z +L/2 −L/2 1 −2 L |z′| dz′ # sin θ (10.117) The integral is easily solved using standard methods, or simply recognize that the “area under the curve” in this case is simply one-half “base” (L) times “height” (1). Either way, we find le ≈L 2 sin θ (10.118) Example 10.7 (Section 10.9) demonstrates how Equation 10.114 with the vector effective length determined in the preceding example is used to obtain the induced potential. | Electromagnetics_Vol2_Page_206_Chunk2382 |
194 CHAPTER 10. ANTENNAS 10.12 Equivalent Circuit Model for Reception, Redux [m0216] Section 10.9 provides an informal derivation of an equivalent circuit model for a receiving antenna. This model is shown in Figure 10.18. The derivation of this model was informal and incomplete because the open-circuit potential Ei · le and source impedance ZA were not rigorously derived in that section. While the open-circuit potential was derived in Section 10.11 (“Potential Induced in a Dipole”), the source impedance has not yet been addressed. In this section, the source impedance is derived, which completes the formal derivation of the model. Before reading this section, a review of Section 10.10 (“Reciprocity”) is recommended. The starting point for a formal derivation is the two-port model shown in Figure 10.19. If the two-port is passive, linear, and time-invariant, then the potential v2 is a linear function of potentials and currents present at ports 1 and 2. Furthermore, v1 must be proportional to i1, and similarly v2 must be proportional to i2, so any pair of “inputs” consisting of either potentials or currents completely determines the two remaining potentials or currents. Thus, we may write: v2 = Z21i1 + Z22i2 (10.119) where Z11 and Z12 are, for the moment, simply constants of proportionality. However, note that Z11 and Z12 have SI base units of Ω, and so we refer to c⃝S. Lally CC BY-SA 4.0 Figure 10.18: Th´evenin equivalent circuit model for an antenna in the presence of an incident electric field Ei. Inductiveload, public domain (modified) Figure 10.19: Two-port system. these quantities as impedances. Similarly we may write: v1 = Z11i1 + Z12i2 (10.120) We can develop expressions for Z12 and Z21 as follows. First, note that v1 = Z12i2 when i1 = 0. Therefore, we may define Z12 as follows: Z12 ≜v1 i2 i1=0 (10.121) The simplest way to make i1 = 0 (leaving i2 as the sole “input”) is to leave port 1 open-circuited. In previous sections, we invoked special notation for these circumstances. In particular, we defined eIt 2 as i2 in phasor representation, with the superscript “t” (“transmitter”) indicating that this is the sole “input”; and eV r 1 as v1 in phasor representation, with the superscript “r” (“receiver”) signaling that port 1 is both open-circuited and the “output.” Applying this notation, we note: Z12 = eV r 1 eIt 2 (10.122) Similarly: Z21 = eV r 2 eIt 1 (10.123) In Section 10.10 (“Reciprocity”), we established that a pair of antennas could be represented as a passive linear time-invariant two-port, with v1 and i1 representing the potential and current at the terminals of one antenna (“antenna 1”), and v2 and i2 representing the potential and current at the terminals of another antenna (“antenna 2”). Therefore for any pair of antennas, quantities Z11, Z12, Z22, and Z21 | Electromagnetics_Vol2_Page_207_Chunk2383 |
10.12. EQUIVALENT CIRCUIT MODEL FOR RECEPTION, REDUX 195 can be identified that completely determine the relationship between the port potentials and currents. We also established in Section 10.10 that: eIt 1 eV r 1 = eIt 2 eV r 2 (10.124) Therefore, eV r 1 eIt 2 = eV r 2 eIt 1 (10.125) Referring to Equations 10.122 and 10.123, we see that Equation 10.125 requires that: Z12 = Z21 (10.126) This is a key point. Even though we derived this equality by open-circuiting ports one at a time, the equality must hold generally since Equations 10.119 and 10.120 must apply – with the same values of Z12 and Z21 – regardless of the particular values of the port potentials and currents. We are now ready to determine the Th´evenin equivalent circuit for a receiving antenna. Let port 1 correspond to the transmitting antenna; that is, i1 is eIt 1. Let port 2 correspond to an open-circuited receiving antenna; thus, i2 = 0 and v2 is eV r 2 . Now applying Equation 10.119: v2 = Z21i1 + Z22i2 = eV r 2 /eIt 1 eIt 1 + Z22 · 0 = eV r 2 (10.127) We previously determined eV r 2 from electromagnetic considerations to be (Section 10.11): eV r 2 = eEi · le (10.128) where eEi is the field incident on the receiving antenna, and le is the vector effective length as defined in Section 10.11. Thus, the voltage source in the Th´evenin equivalent circuit for the receive antenna is simply eEi · le, as shown in Figure 10.18. The other component in the Th´evenin equivalent circuit is the series impedance. From basic circuit theory, this impedance is the ratio of v2 when port 2 is open-circuited (i.e., eV r 2 ) to i2 when port 2 is short-circuited. This value of i2 can be obtained using Equation 10.119 with v2 = 0: 0 = Z21eIt 1 + Z22i2 (10.129) Therefore: i2 = −Z21 Z22 eIt 1 (10.130) Now using Equation 10.123 to eliminate eIt 1, we obtain: i2 = − eV r 2 Z22 (10.131) Note that the reference direction for i2 as defined in Figure 10.19 is opposite the reference direction for short-circuit current. That is, given the polarity of v2 shown in Figure 10.19, the reference direction of current flow through a passive load attached to this port is from “+” to “−” through the load. Therefore, the source impedance, calculated as the ratio of the open-circuit potential to the short circuit current, is: eV r 2 +eV r 2 /Z22 = Z22 (10.132) We have found that the series impedance ZA in the Th´evenin equivalent circuit is equal to Z22 in the two-port model. To determine Z22, let us apply a current i2 = eIt 2 to port 2 (i.e., antenna 2). Equation 10.119 indicates that we should see: v2 = Z21i1 + Z22eIt 2 (10.133) Solving for Z22: Z22 = v2 eIt 2 −Z21 i1 eIt 2 (10.134) Note that the first term on the right is precisely the impedance of antenna 2 in transmission. The second term in Equation 10.134 describes a contribution to Z22 from antenna 1. However, our immediate interest is in the equivalent circuit for reception of an electric field eEi in the absence of any other antenna. We can have it both ways by imagining that eEi is generated by antenna 1, but also that antenna 1 is far enough away to make Z21 – the factor that determines the effect of antenna 1 on antenna 2 – negligible. Then we see from Equation 10.134 that Z22 is the impedance of antenna 2 when transmitting. | Electromagnetics_Vol2_Page_208_Chunk2384 |
196 CHAPTER 10. ANTENNAS Summarizing: The Th´evenin equivalent circuit for an antenna in the presence of an incident electric field eEi is shown in Figure 10.18. The series impedance ZA in this model is equal to the impedance of the an- tenna in transmission. Mutual coupling. This concludes the derivation, but raises a follow-up question: What if antenna 2 is present and not sufficiently far away that Z21 can be assumed to be negligible? In this case, we refer to antenna 1 and antenna 2 as being “coupled,” and refer to the effect of the presence of antenna 1 on antenna 2 as coupling. Often, this issue is referred to as mutual coupling, since the coupling affects both antennas in a reciprocal fashion. It is rare for coupling to be significant between antennas on opposite ends of a radio link. This is apparent from common experience. For example, changes to a receive antenna are not normally seen to affect the electric field incident at other locations. However, coupling becomes important when the antenna system is a dense array; i.e., multiple antennas separated by distances less than a few wavelengths. It is common for coupling among the antennas in a dense array to be significant. Such arrays can be analyzed using a generalized version of the theory presented in this section. Additional Reading: • “Th´evenin’s Theorem” on Wikipedia. 10.13 Effective Aperture [m0218] When working with systems involving receiving antennas, it is convenient to have a single parameter that relates incident power (as opposed to incident electric field) to the power delivered to the receiver electronics. This parameter is commonly known as the effective aperture or antenna aperture. From elementary circuit theory, the power delivered to a load ZL is maximized when the load is conjugate matched to the antenna impedance; i.e., when ZL = Z∗ A where ZA is the impedance of the antenna. Thus, a convenient definition for effective aperture Ae employs the relationship: PR,max ≜Si coAe (10.135) where Si co is the incident power density (SI base units of W/m2) that is co-polarized with the antenna, and PR,max is the power delivered to a load that is conjugate matched to the antenna impedance. Summarizing: Effective aperture (SI base units of m2) is the ra- tio of power delivered by an antenna to a conju- gate matched load, to the incident co-polarized power density. As defined, a method for calculation of effective aperture is not obvious, except perhaps through direct measurement. In fact, there are at least three ways we can calculate effective aperture: (a) via effective length, (b) via thermodynamics, and (c) via reciprocity. Each of these methods yields some insight and are considered in turn. Effective aperture via effective length. The potential and current at the load of the receive antenna can be determined using the equivalent circuit model shown in Figure 10.20, using the Th´evenin equivalent circuit model for the antenna developed in Section 10.9. In this model, the voltage source is determined by the incident electric field intensity eEi and the vector effective length le = ˆlele of the antenna. We determine the associated effective aperture as follows. Consider a co-polarized | Electromagnetics_Vol2_Page_209_Chunk2385 |
10.13. EFFECTIVE APERTURE 197 sinusoidally-varying plane wave eEi co incident on the antenna. Further, let eEi co = Eiˆe (10.136) where ˆe is the reference direction of eEi co. The co-polarized power density incident on the antenna is: Si co = Ei 2 2η (10.137) where η is the wave impedance of the medium (e.g., ∼= 377 Ωfor an antenna in free space). In response, the time-average power delivered to the load is PR = 1 2Re n eVLeI∗ L o (10.138) where eVL and eIL are the potential and current phasors, respectively, at the load. We may determine eVL and eIL from the equivalent circuit model of Figure 10.20 using basic circuit theory. First, note that the magnitude and phase of the voltage source is: eEi co · le = Eile ˆe ·ˆle (10.139) and since we have defined ˆle to be equal to ˆe, eEi co · le = Eile (10.140) Next, note that the load impedance creates a voltage divider with the source (antenna) impedance such that eVL = | Electromagnetics_Vol2_Page_210_Chunk2386 |
198 CHAPTER 10. ANTENNAS wire antennas. For the electrically-short dipole (ESD) of length L, le ≈(L/2) sin θ and Rrad ≈20π2 (L/λ)2. Thus, we find the effective aperture assuming free space (i.e., η = η0) is: Ae ≈0.119λ2 |sin θ|2 (lossless ESD) (10.149) Remarkably, the effective aperture of the ESD does not depend on its length. In fact, it depends only on the frequency of operation. Also worth noting is that maximum effective aperture (i.e., the effective aperture in the θ = π/2 direction) is Ae ≈0.119λ2 (lossless ESD, max.) (10.150) Dipoles which are not electrically-short exhibit radiation resistance which is proportional to Lp, where p > 2. Therefore, the effective aperture of non-electrically-short dipoles does increase with L, as expected. However, this increase is not dramatic unless L is significantly greater than λ/2. Here’s an example: Example 10.10. Effective aperture of a half-wave dipole. The electrically-thin half-wave dipole exhibits radiation resistance ∼= 73 Ωand effective length λ/π. Assuming the dipole is lossless and in free space, Equation 10.148 yields: Ae ≈0.131λ2 (half-wave dipole, max.) (10.151) Again, this is the effective aperture for a wave incident from broadside to the dipole. Effective aperture via thermodynamics. A useful insight into the concept of effective aperture can be deduced by asking the following question: How much power is received by an antenna when waves of equal power density arrive from all directions simultaneously? This question is surprisingly easy to answer using basic principles of thermodynamics. Thermodynamics is a field of physics that addresses heat and its relation to radiation and electrical energy. No previous experience with thermodynamics is assumed in this derivation. Consider the scenario depicted in Figure 10.21. In this scenario, the antenna is completely enclosed in a chamber whose walls do not affect the behavior of the antenna and which have uniform temperature T. The load (still conjugate matched to the antenna) is completely enclosed by a separate but identical chamber, also at temperature T. Consider the load chamber first. Heat causes random acceleration of constituent charge carriers in the load, giving rise to a randomly-varying current at the load terminals. (This is known as Johnson-Nyquist noise.) This current, flowing through the load, gives rise to a potential; therefore, the load – despite being a passive device – is actually a source of power. The power associated with this noise is: Pload = kTB (10.152) where k ∼= 1.38 × 10−23 J/K is Boltzmann’s constant and B is the bandwidth within which Pload is measured. Similarly, the antenna is a source of noise power Pant. Pant can also be interpreted as captured thermal radiation – that is, electromagnetic waves stimulated by the random acceleration of charged particles comprising the chamber walls. These waves radiate from the walls and travel to the antenna. The Rayleigh-Jeans law of thermodynamics tells us that this radiation should have power density (i.e., SI base units of W/m2) equal to 2kT λ2 B (10.153) Chetvorno, public domain (modified) Figure 10.21: Thermodynamic analysis of the power exchanged between an antenna and its load. | Electromagnetics_Vol2_Page_211_Chunk2387 |
10.13. EFFECTIVE APERTURE 199 per steradian of solid angle.5 The total power accessible to the antenna is one-half this amount, since an antenna is sensitive to only one polarization at a time, whereas the thermal radiation is equally distributed among any two orthogonal polarizations. PA is the remaining power, obtained by integrating over all 4π steradians. Therefore, the antenna captures total power equal to6 Pant = I Ae(θ′, φ′) 1 2 2kT λ2 B sin θ′dθ′dφ′ = kT λ2 B I Ae(θ′, φ′) sin θ′dθ′dφ′ (10.154) Given the conditions of the experiment, and in particular since the antenna chamber and the load chamber are at the same temperature, the antenna and load are in thermodynamic equilibrium. A consequence of this equilibrium is that power Pant captured by the antenna and delivered to the load must be equal to power Pload generated by the load and delivered to the antenna; i.e., Pant = Pload (10.155) Combining Equations 10.152, 10.154, and 10.155, we obtain: kT λ2 B I Ae(θ′, φ′) sin θ′dθ′dφ′ = kTB (10.156) which reduces to: I Ae(θ′, φ′) sin θ′dθ′dφ′ = λ2 (10.157) Let ⟨Ae⟩be the mean effective aperture of the antenna; i.e., Ae averaged over all possible directions. This average is simply Equation 10.157 divided by the 4π sr of solid angle comprising all possible directions. Thus: ⟨Ae⟩= λ2 4π (10.158) Recall that a isotropic antenna is one which has the same effective aperture in every direction. Such 5Full disclosure: This is an approximation to the exact expres- sion, but is very accurate at radio frequencies. See “Additional Read- ing” at the end of this section for additional details. 6Recall sin θ′dθ′dφ′ is the differential element of solid angle. antennas do not exist in practice, but the concept of an isotropic antenna is quite useful as we shall soon see. Since the effective aperture of an isotropic antenna must be the same as its mean effective aperture, we find: Ae = λ2 4π ≈0.080λ2 (isotropic antenna) (10.159) Note further that this must be the minimum possible value of the maximum effective aperture for any antenna. Effective aperture via reciprocity. The fact that all antennas have maximum effective aperture greater than that of an isotropic antenna makes the isotropic antenna a logical benchmark against which to compare the effective aperture of antennas. We can characterize the effective aperture of any antenna as being some unitless real-valued constant D times the effective aperture of an isotropic antenna; i.e.: Ae ≜D λ2 4π (any antenna) (10.160) where D must be greater than or equal to 1. Astute readers might notice that D seems a lot like directivity. Directivity is defined in Section 10.7 as the factor by which a transmitting antenna increases the power density of its radiation over that of an isotropic antenna. Let us once again consider the ESD, for which we previously determined (via the effective length concept) that Ae ∼= 0.119λ2 in the direction in which it is maximum. Applying Equation 10.160 to the ESD, we find: D ≜Ae 4π λ2 ∼= 1.50 (ESD, max.) (10.161) Sure enough, D is equal to the directivity of the ESD in the transmit case. This result turns out to be generally true; that is: The effective aperture of any antenna is given by Equation 10.160 where D is the directivity of the antenna when transmitting. A derivation of this result for the general case is possible using analysis similar to the thermodynamics | Electromagnetics_Vol2_Page_212_Chunk2388 |
200 CHAPTER 10. ANTENNAS presented earlier, or using the reciprocity theorem developed in Section 10.10. The fact that effective aperture is easily calculated from transmit directivity is an enormously useful tool in antenna engineering. Without this tool, determination of effective aperture is limited to direct measurement or calculation via effective length; and effective length is generally difficult to calculate for antennas which are not well-described as distributions of current along a line. It is usually far easier to calculate the directivity of an antenna when transmitting, and then use Equation 10.160 to obtain effective aperture for receive operation. Note that this principle is itself an expression of reciprocity. That is, one may fairly say that “directivity” is not exclusively a transmit concept nor a receive concept, but rather is a single quantifiable characteristic of an antenna that applies in both the transmit and receive case. Recall that radiation patterns are used to quantify the way (transmit) directivity varies with direction. Suddenly, we have found that precisely the same patterns apply to the receive case! Summarizing: As long as the conditions required for formal reciprocity are satisfied, the directivity of a re- ceiving antenna (defined via Equation 10.160) is equal to directivity of the same antenna when transmitting, and patterns describing receive di- rectivity are equal to those for transmit directiv- ity. Finally, we note that the equivalence of transmit and receive directivity can, again via Equation 10.160, be used to define an effective aperture for the transmit case. In other words, we can define the effective aperture of a transmitting antenna to be λ2/4π times the directivity of the antenna. There is no new physics at work here; we are simply taking advantage of the fact that the concepts of effective aperture and directivity describe essentially the same characteristic of an antenna, and that this characteristic is the same for both transmit and receive operation. Additional Reading: • “Antenna aperture” on Wikipedia. • “Boltzmann constant” on Wikipedia. • “Rayleigh-Jeans law” on Wikipedia. • “Johnson-Nyquist noise” on Wikipedia. • “Thermodynamics” on Wikipedia. | Electromagnetics_Vol2_Page_213_Chunk2389 |
10.14. FRIIS TRANSMISSION EQUATION 201 10.14 Friis Transmission Equation [m0219] A common task in radio systems applications is to determine the power delivered to a receiver due to a distant transmitter. The scenario is shown in Figure 10.22: A transmitter delivers power PT to an antenna which has gain GT in the direction of the receiver. The receiver’s antenna has gain GR. As always, antenna gain is equal to directivity times radiation efficiency, so GT and GR account for losses internal to the antenna, but not losses due to impedance mismatch. A simple expression for PR can be derived as follows. First, let us assume “free space conditions”; that is, let us assume that the intervening terrain exhibits negligible absorption, reflection, or other scattering of the transmitted signal. In this case, the spatial power density at range R from the transmitter which radiates this power through a lossless and isotropic antenna would be: PT 4πR2 (10.162) that is, total transmitted power divided by the area of a sphere of radius R through which all the power must flow. The actual power density Si is this amount times the gain of the transmit antenna, i.e.: Si = PT 4πR2 GT (10.163) The maximum received power is the incident co-polarized power density times the effective Figure 10.22: Radio link accounting for antenna gains and spreading of the transmitted wave. aperture Ae of the receive antenna: PR,max = AeSi co = Ae PT 4πR2 GT (10.164) This assumes that the receive antenna is co-polarized with the incident electric field, and that the receiver is conjugate-matched to the antenna. The effective aperture can also be expressed in terms of the gain GR of the receive antenna: Ae = λ2 4π GR (10.165) Thus, Equation 10.164 may be written in the following form: PR,max = PT GT λ 4πR 2 GR (10.166) This is the Friis transmission equation. Summarizing: The Friis transmission equation (Equa- tion 10.166) gives the power delivered to a conjugate-matched receiver in response to a dis- tant transmitter, assuming co-polarized antennas and free space conditions. The factor (λ/4πR)2 appearing in the Friis transmission equation is referred to as free space path gain. More often this is expressed as the reciprocal quantity: Lp ≜ λ 4πR −2 (10.167) which is known as free space path loss. Thus, Equation 10.166 may be expressed as follows: PR,max = PT GT L−1 p GR (10.168) The utility of the concept of path loss is that it may also be determined for conditions which are different from free space. The Friis transmission equation still applies; one simply uses the appropriate (and probably significantly different) value of Lp. A common misconception is that path loss is equal to the reduction in power density due to spreading along | Electromagnetics_Vol2_Page_214_Chunk2390 |
202 CHAPTER 10. ANTENNAS the path between antennas, and therefore this “spreading loss” increases with frequency. In fact, the reduction in power density due to spreading between any two distances R1 < R2 is: PT /4πR2 1 PT /4πR2 2 = R1 R2 2 (10.169) which is clearly independent of frequency. The path loss Lp, in contrast, depends only on the total distance R and does depend on frequency. The dependence on frequency reflects the dependence of the effective aperture on wavelength. Thus, path loss is not loss in the traditional sense, but rather accounts for a combination of spreading and the λ2 dependence of effective aperture that is common to all receiving antennas. Finally, note that Equation 10.168 is merely the simplest form of the Friis transmission equation. Commonly encountered alternative forms include forms in which GT and/or GR are instead represented by the associated effective apertures, and forms in which the effects of antenna impedance mismatch and/or cross-polarization are taken into account. Example 10.11. 6 GHz point-to-point link. Terrestrial telecommunications systems commonly aggregate large numbers of individual communications links into a single high-bandwidth link. This is often implemented as a radio link between dish-type antennas having gain of about 27 dBi (that’s dB relative to a lossless isotropic antenna) mounted on very tall towers and operating at frequencies around 6 GHz. Assuming the minimum acceptable receive power is −120 dBm (that’s −120 dB relative to 1 mW; i.e., 10−15 W) and the required range is 30 km, what is the minimum acceptable transmit power? Solution. From the problem statement: GT = GR = 1027/10 ∼= 501 (10.170) λ = c f ∼= 3 × 108 m/s 6 × 109 Hz ∼= 5.00 cm (10.171) R = 30 km, and PR ≥10−15 W. We assume that the height and high directivity of the antennas yield conditions sufficiently close to free space. We further assume conjugate-matching at the receiver, and that the antennas are co-polarized. Under these conditions, PR = PR,max and Equation 10.166 applies. We find: PT ≥ PR,max GT (λ/4πR)2 GR (10.172) ∼= 2.26 × 10−7 W ∼= 2.26 × 10−4 mW ∼= −36.5 dBm Additional Reading: • “Free-space path loss” on Wikipedia. • “Friis transmission equation” on Wikipedia. [m0211] | Electromagnetics_Vol2_Page_215_Chunk2391 |
10.14. FRIIS TRANSMISSION EQUATION 203 Image Credits Fig. 10.1: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Standing Wave Creation.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 10.2: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Electrically-Short Dipole.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 10.3: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Zero-Length Dipole Open Circuit.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 10.4: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Transmitting Antenna Circuit.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 10.5: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Magnitude of the Radiated Field.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 10.6: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:FHplaneMag.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 10.7: c⃝T. Truckle, https://en.wikipedia.org/wiki/File:Sidelobes en.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Modified. Fig. 10.8: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Half-Power Beamwidth.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 10.9: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Antenna Equivalent Circuit.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Modified. Fig. 10.10: Inductiveload, https://commons.wikimedia.org/wiki/File:Two Port Circuit.svg, public domain. Modified. Fig. 10.11: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:An electromagnetic system consisting of a current distribution radiating an electric field.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 10.12: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:An electromagnetic system consisting of a different current distribution radiating an electric field.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 10.13: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:A two-port consisting of two dipole antennas.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 10.14: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:The port 1 antenna a transmitting and b receiving.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). | Electromagnetics_Vol2_Page_216_Chunk2392 |
204 CHAPTER 10. ANTENNAS Fig. 10.15: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Thin straight dipole respond to incident plane wave.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 10.16: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Dipole of interest driven by current.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 10.17: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Hertzian dipole drive by current.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 10.18: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Antenna Equivalent Circuit.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 10.19: Inductiveload, https://commons.wikimedia.org/wiki/File:Two Port Circuit.svg, public domain. Modified. Fig. 10.20: c⃝Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:Antenna Equivalent Circuit.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Modified. Fig. 10.21: Chetvorno, https://en.wikipedia.org/wiki/File:Antenna and resistor in cavity.svg, public domain. Modified. | Electromagnetics_Vol2_Page_217_Chunk2393 |
Appendix A Constitutive Parameters of Some Common Materials A.1 Permittivity of Some Common Materials [m0135] The values below are relative permittivity ǫr ≜ǫ/ǫ0 for a few materials that are commonly encountered in electrical engineering applications, and for which permittivity emerges as a consideration. Note that “relative permittivity” is sometimes referred to as dielectric constant. Here we consider only the physical (real-valued) permittivity, which is the real part of the complex permittivity (typically indicated as ǫ′ or ǫ′ r) for materials exhibiting significant loss. Permittivity varies significantly as a function of frequency. The values below are representative of frequencies from a few kHz to about 1 GHz. The values given are also representative of optical frequencies for materials such as silica that are used in optical applications. Permittivity also varies as a function of temperature. In applications where precision better than about 10% is required, primary references accounting for frequency and temperature should be consulted. The values presented here are gathered from a variety of references, including those indicated in “Additional References.” Free Space (vacuum): ǫr ≜1 Solid Dielectrics: Material ǫr Common uses Styrofoam1 1.1 Teflon2 2.1 Polyethylene 2.3 coaxial cable Polypropylene 2.3 Silica 2.4 optical fiber3 Polystyrene 2.6 Polycarbonate 2.8 Rogers RO3003 3.0 PCB substrate FR4 (glass epoxy laminate) 4.5 PCB substrate 1 Properly known as extruded polystyrene foam (XPS). 2 Properly known as polytetrafluoroethylene (PTFE). 3 Typically doped with small amounts of other materials to slightly raise or lower the index of refraction (= √ǫr). Non-conducting spacing materials used in discrete capacitors exhibit ǫr ranging from about 5 to 50. Semiconductors commonly appearing in electronics – including carbon, silicon, geranium, indium phosphide, and so on – typically exhibit ǫr in the range 5–15. Glass exhibits ǫr in the range 4–10, depending on composition. Gasses, including air, typically exhibit ǫr ∼= 1 to within a tiny fraction of a percent. Liquid water typically exhibits ǫr in the range 72–81. Distilled water exhibits ǫr ≈81 at room temperature, whereas sea water tends to be at the Electromagnetics Vol. 2. c⃝2020 S.W. Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics-vol-2 | Electromagnetics_Vol2_Page_218_Chunk2394 |
206 APPENDIX A. CONSTITUTIVE PARAMETERS OF SOME COMMON MATERIALS lower end of the range. Other liquids typically exhibit ǫr in the range 10–90, with considerable variation as a function of temperature and frequency. Animal flesh and blood consists primarily of liquid matter and so also exhibits permittivity in this range. Soil typically exhibits ǫr in the range 2.5–3.5 when dry and higher when wet. The permittivity of soil varies considerably depending on composition. Additional Reading: • CRC Handbook of Chemistry and Physics. • “Relative permittivity” on Wikipedia. • “Miscellaneous Dielectric Constants” on microwaves101.com. A.2 Permeability of Some Common Materials [m0136] The values below are relative permeability µr ≜µ/µ0 for a few materials that are commonly encountered in electrical engineering applications, and for which µr is significantly different from 1. These materials are predominantly ferromagnetic metals and (in the case of ferrites) materials containing significant ferromagnetic metal content. Nearly all other materials exhibit µr that is not significantly different from that of free space. The values presented here are gathered from a variety of references, including those indicated in “Additional References” at the end of this section. Be aware that permeability may vary significantly with frequency; values given here are applicable to the frequency ranges for applications in which these materials are typically used. Also be aware that materials exhibiting high permeability are also typically non-linear; that is, permeability depends on the magnitude of the magnetic field. Again, values reported here are those applicable to applications in which these materials are typically used. Free Space (vacuum): µr ≜1. Iron (also referred to by the chemical notation “Fe”) appears as a principal ingredient in many materials and alloys employed in electrical structures and devices. Iron exhibits µr that is very high, but which decreases with decreasing purity. 99.95% pure iron exhibits µr ∼200, 000. This decreases to ∼5000 at 99.8% purity and is typically below 100 for purity less than 99%. Steel is an iron alloy that comes in many forms, with a correspondingly broad range of permeabilities. Electrical steel, commonly used in electrical machinery and transformers when high permeability is desired, exhibits µr ∼4000. Stainless steel, encompassing a broad range of alloys used in mechanical applications, exhibits µr in the range 750–1800. Carbon steel, including a broad class of alloys commonly used in structural applications, exhibits µr on the order of 100. | Electromagnetics_Vol2_Page_219_Chunk2395 |
A.3. CONDUCTIVITY OF SOME COMMON MATERIALS 207 Ferrites include a broad range of ceramic materials that are combined with iron and various combinations of other metals and are used as magnets and magnetic devices in various electrical systems. Common ferrites exhibit µr in the range 16–640. Additional Reading: • CRC Handbook of Chemistry and Physics. • “Magnetic Materials” on microwaves101.com. • “Permeability (electromagnetism)” on Wikipedia. • “Iron” on Wikipedia. • “Electrical steel” on Wikipedia. • “Ferrite (magnet)” on Wikipedia. A.3 Conductivity of Some Common Materials [m0137] The values below are conductivity σ for a few materials that are commonly encountered in electrical engineering applications, and for which conductivity emerges as a consideration. Note that materials in some applications are described instead in terms of resistivity, which is simply the reciprocal of conductivity. Conductivity may vary significantly as a function of frequency. The values below are representative of frequencies from a few kHz to a few GHz. Conductivity also varies as a function of temperature. In applications where precise values are required, primary references accounting for frequency and temperature should be consulted. The values presented here are gathered from a variety of references, including those indicated in “Additional References” at the end of this section. Free Space (vacuum): σ ≜0. Commonly encountered elements: Material σ (S/m) Copper 5.8 × 107 Gold 4.4 × 107 Aluminum 3.7 × 107 Iron 1.0 × 107 Platinum 0.9 × 107 Carbon 1.3 × 105 Silicon 4.4 × 10−4 Water exhibits σ ranging from about 6 µS/m for highly distilled water (thus, a very poor conductor) to about 5 S/m for seawater (thus, a relatively good conductor), varying also with temperature and pressure. Tap water is typically in the range of 5–50 mS/m, depending on the level of impurities present. Soil typically exhibits σ in the range 10−4 S/m for dry soil to about 10−1 S/m for wet soil, varying also due to chemical composition. | Electromagnetics_Vol2_Page_220_Chunk2396 |
208 APPENDIX A. CONSTITUTIVE PARAMETERS OF SOME COMMON MATERIALS Non-conductors. Most other materials that are not well-described as conductors or semiconductors and are dry exhibit σ < 10−12 S/m. Most materials that are considered to be insulators, including air and common dielectrics, exhibit σ < 10−15 S/m, often by several orders of magnitude. Additional Reading: • CRC Handbook of Chemistry and Physics. • “Conductivity (electrolytic)” on Wikipedia. • “Electrical resistivity and conductivity” on Wikipedia. • “Soil resistivity” on Wikipedia. | Electromagnetics_Vol2_Page_221_Chunk2397 |
Appendix B Mathematical Formulas B.1 Trigonometry [m0138] ejθ = cos θ + j sin θ (B.1) cos θ = 1 2 | Electromagnetics_Vol2_Page_222_Chunk2398 |
210 APPENDIX B. MATHEMATICAL FORMULAS Gradient in spherical coordinates: ∇f = ˆr∂f ∂r + ˆθ1 r ∂f ∂θ + ˆφ 1 r sin θ ∂f ∂φ (B.12) Divergence Divergence in Cartesian coordinates: ∇· A = ∂Ax ∂x + ∂Ay ∂y + ∂Az ∂z (B.13) Divergence in cylindrical coordinates: ∇· A = 1 ρ ∂ ∂ρ (ρAρ) + 1 ρ ∂Aφ ∂φ + ∂Az ∂z (B.14) Divergence in spherical coordinates: ∇· A = 1 r2 ∂ ∂r | Electromagnetics_Vol2_Page_223_Chunk2399 |
B.3. VECTOR IDENTITIES 211 B.3 Vector Identities [m0140] Algebraic Identities A · (B × C) = B · (C × A) = C · (A × B) (B.22) A × (B × C) = B (A · C) −C (A · B) (B.23) Identities Involving Differential Operators ∇· (∇× A) = 0 (B.24) ∇× (∇f) = 0 (B.25) ∇× (fA) = f (∇× A) + (∇f) × A (B.26) ∇· (A × B) = B · (∇× A) −A · (∇× B) (B.27) ∇· (∇f) = ∇2f (B.28) ∇× ∇× A = ∇(∇· A) −∇2A (B.29) ∇2A = ∇(∇· A) −∇× (∇× A) (B.30) Divergence Theorem: Given a closed surface S enclosing a contiguous volume V, Z V (∇· A) dv = I S A · ds (B.31) where the surface normal ds is pointing out of the volume. Stokes’ Theorem: Given a closed curve C bounding a contiguous surface S, Z S (∇× A) · ds = I C A · dl (B.32) where the direction of the surface normal ds is related to the direction of integration along C by the “right hand rule.” | Electromagnetics_Vol2_Page_224_Chunk2400 |
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