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2.2. MAGNETIC FORCE ON A CURRENT-CARRYING WIRE 13 The net force on a current-carrying loop of wire in a uniform magnetic field is zero. Note that this does not preclude the possibility that the rigid loop rotates; for example, the force on opposite sides of the loop may be equal and opposite. What we have found is merel... | Electromagnetics_Vol2_Page_26_Chunk2201 |
14 CHAPTER 2. MAGNETOSTATICS REDUX When the currents I1 and I2 flow in the same direction (i.e., when the product I1I2 is positive), then the magnetic force exerted by the current on wire 2 pulls wire 1 toward wire 2. We are now able to summarize the results as follows: If currents in parallel wires flow in the same di- ... | Electromagnetics_Vol2_Page_27_Chunk2202 |
2.3. TORQUE INDUCED BY A MAGNETIC FIELD 15 2.3 Torque Induced by a Magnetic Field [m0024] A magnetic field exerts a force on current. This force is exerted in a direction perpendicular to the direction of current flow. For this reason, current-carrying structures in a magnetic field tend to rotate. A convenient descriptio... | Electromagnetics_Vol2_Page_28_Chunk2203 |
16 CHAPTER 2. MAGNETOSTATICS REDUX zero (Section 2.2). However, this does not preclude the possibility of different translational forces acting on each of the loop segments resulting in a rotation of the shaft. Let us first calculate these forces. The force FA on segment A is FA = IlA × B0 (2.15) where lA is a vector wh... | Electromagnetics_Vol2_Page_29_Chunk2204 |
2.3. TORQUE INDUCED BY A MAGNETIC FIELD 17 c⃝Abnormaal CC BY-SA 3.0 Figure 2.6: This DC electric motor uses brushes (here, the motionless leads labeled “+” and “−”) combined with the motion of the shaft to periodically alternate the direction of current between two coils, thereby cre- ating nearly constant torque. If s... | Electromagnetics_Vol2_Page_30_Chunk2205 |
18 CHAPTER 2. MAGNETOSTATICS REDUX 2.4 The Biot-Savart Law [m0066] The Biot-Savart law (BSL) provides a method to calculate the magnetic field due to any distribution of steady (DC) current. In magnetostatics, the general solution to this problem employs Ampere’s law; i.e., Z C H · dl = Iencl (2.26) in integral form or ... | Electromagnetics_Vol2_Page_31_Chunk2206 |
2.4. THE BIOT-SAVART LAW 19 the ˆφ direction. Find the magnetic field intensity along the z axis. Solution. The source current position is given in cylindrical coordinates as r′ = ˆρa (2.31) The position of a field point along the z axis is r = ˆzz (2.32) Thus, ˆRR ≜r −r′ = −ˆρa + ˆzz (2.33) and R ≜|r −r′| = p a2 + z2 (2... | Electromagnetics_Vol2_Page_32_Chunk2207 |
20 CHAPTER 2. MAGNETOSTATICS REDUX velocity v (SI base units of m/s), the relevant quantity is qv since C·m/s = (C/s)·m = A·m. In all of these cases, Equation 2.28 applies with the appropriate replacement for I dl. Note that the quantities qv, I dl, JS ds, and J dv, all having the same units of A·m, seem to be referrin... | Electromagnetics_Vol2_Page_33_Chunk2208 |
2.5. FORCE, ENERGY, AND POTENTIAL DIFFERENCE IN A MAGNETIC FIELD 21 The magnetic field does no work. Instead, the change of potential energy associated with the magnetic field must be completely due to a change in position resulting from other forces, such as a mechanical force or the Coulomb force. The presence of a mag... | Electromagnetics_Vol2_Page_34_Chunk2209 |
22 CHAPTER 2. MAGNETOSTATICS REDUX and y = y0 + l, respectively, we obtain V21 = Z y0+l y0 [ˆyBv] · ˆydy = Bvl (2.50) Thus, we see that endpoint 2 is at an electrical potential of Bvl greater than that of endpoint 1. This “voltage” exists even though the wire is perfectly-conducting, and therefore cannot be attributed ... | Electromagnetics_Vol2_Page_35_Chunk2210 |
2.5. FORCE, ENERGY, AND POTENTIAL DIFFERENCE IN A MAGNETIC FIELD 23 Astute readers will notice that this analysis seems to have a lot in common with Faraday’s law, V = −∂ ∂tΦ (2.54) which says the potential induced in a single closed loop is proportional to the time rate of change of magnetic flux Φ, where Φ = Z S B · d... | Electromagnetics_Vol2_Page_36_Chunk2211 |
24 CHAPTER 2. MAGNETOSTATICS REDUX Image Credits Fig. 2.1: c⃝Stannered, https://en.wikipedia.org/wiki/File:Charged-particle-drifts.svg, CC BY 2.5 (https://creativecommons.org/licenses/by/2.5/deed.en). Modified by Maschen, author. Fig. 2.2: c⃝M. Biaek, https://en.wikipedia.org/wiki/File:Cyclotron motion.jpg, CC BY-SA 4.0... | Electromagnetics_Vol2_Page_37_Chunk2212 |
Chapter 3 Wave Propagation in General Media 3.1 Poynting’s Theorem [m0073] Despite the apparent complexity of electromagnetic theory, there are in fact merely four ways that electromagnetic energy can be manipulated. Electromagnetic energy can be: • Transferred; i.e., conveyed by transmission lines or in waves; • Store... | Electromagnetics_Vol2_Page_38_Chunk2213 |
26 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA components of E and D, respectively. Subsequently, 1 ǫ ∂ ∂t (D · D) = 1 ǫ ∂ ∂tD2 = 1 ǫ 2D ∂ ∂tD = 2E · ∂ ∂tD (3.3) Summarizing: ∂ ∂t (E · D) = 2E · ∂ ∂tD (3.4) which is the expression we seek. It is worth noting that the expressions on both sides of the equation have the ... | Electromagnetics_Vol2_Page_39_Chunk2214 |
3.1. POYNTING’S THEOREM 27 Equation 3.16 is Poynting’s theorem. Each of the four terms has the particular physical interpretation identified in Equation 3.1, as we will now demonstrate. Power dissipated by ohmic loss. The first term of the right side of Equation 3.16 is PΩ≜ Z V E · J dv (3.17) Equation 3.17 is Joule’s la... | Electromagnetics_Vol2_Page_40_Chunk2215 |
28 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA Poynting’s theorem (Equation 3.24, with Equa- tions 3.23, 3.17, 3.19, and 3.21) states that the net electromagnetic power flowing into a region of space may be either dissipated, or used to change the energy stored in electric and magnetic fields within that region. Since w... | Electromagnetics_Vol2_Page_41_Chunk2216 |
3.2. POYNTING VECTOR 29 where Pin and Pout indicate power flow explicitly into and explicitly out of V as separate quantities. Proceeding, let’s ignore what we know about power flow in plane waves, and instead see where Poynting’s theorem takes us. Here, Pnet,in ≜− I S (E × H) · ds = 0 (3.28) The surface S consists of th... | Electromagnetics_Vol2_Page_42_Chunk2217 |
30 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA 3.3 Wave Equations for Lossy Regions [m0128] The wave equations for electromagnetic propagation in lossless and source-free media, in differential phasor form, are: ∇2 eE + ω2µǫeE = 0 (3.32) ∇2 eH + ω2µǫ eH = 0 (3.33) The constant ω2µǫ is labeled β2, and β turns out to be... | Electromagnetics_Vol2_Page_43_Chunk2218 |
3.3. WAVE EQUATIONS FOR LOSSY REGIONS 31 component of the permittivity. The “ǫ′′” notation allows us to accommodate both effects – nonlinearity and conductivity – using common notation. In this section, however, we remain focused exclusively on conductivity. Complex permittivity ǫc (SI base units of F/m) describes the ... | Electromagnetics_Vol2_Page_44_Chunk2219 |
32 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA First, note: γ2 = (α + jβ)2 = α2 −β2 + j2αβ (3.62) Expanding Equation 3.54 using Equations 3.45–3.47, we obtain: γ2 = −ω2µ ǫ −j σ ω = −ω2µǫ + jωµσ (3.63) The real and imaginary parts of Equations 3.62 and 3.63 must be equal. Enforcing this equality yields the followin... | Electromagnetics_Vol2_Page_45_Chunk2220 |
3.4. COMPLEX PERMITTIVITY 33 3.4 Complex Permittivity [m0134] The relationship between electric field intensity E (SI base units of V/m) and electric flux density D (SI base units of C/m2) is: D = ǫE (3.68) where ǫ is the permittivity (SI base units of F/m). In simple media, ǫ is a real positive value which does not depe... | Electromagnetics_Vol2_Page_46_Chunk2221 |
34 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA c⃝K.A. Mauritz (modified) Figure 3.3: The relative contributions of the real and imaginary components of permittivity for a typical di- electric material (in this case, a polymer). Additional Reading: • “Permittivity” on Wikipedia. 3.5 Loss Tangent [m0132] In Section 3.3, ... | Electromagnetics_Vol2_Page_47_Chunk2222 |
3.5. LOSS TANGENT 35 Comparing Equation 3.80 to Equation 3.76, we see loss tangent can equivalently be calculated as tan δ = ǫ′′ ǫ (3.81) and subsequently interpreted as shown in Figure 3.5. The discussion in this section has assumed that ǫc is complex-valued solely due to ohmic loss. However, it is explained in Sectio... | Electromagnetics_Vol2_Page_48_Chunk2223 |
36 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA 3.6 Plane Waves in Lossy Regions [m0130] The electromagnetic wave equations for source-free regions consisting of possibly-lossy material are (see Section 3.3): ∇2 eE −γ2 eE = 0 (3.82) ∇2 eH −γ2 eH = 0 (3.83) where γ2 ≜−ω2µǫc (3.84) We now turn our attention to the questi... | Electromagnetics_Vol2_Page_49_Chunk2224 |
3.7. WAVE POWER IN A LOSSY MEDIUM 37 eE = −ηˆk × eH (3.92) where ˆk is the direction of propagation and η is the wave impedance. In the lossless case, η = p µ/ǫ; however, in the possibly-lossy case we must replace ǫ = ǫ′ with ǫc = ǫ′ −jǫ′′. Thus: η →ηc = r µ ǫc = r µ ǫ′ −jǫ′′ = r µ ǫ′ s 1 1 −j (ǫ′′/ǫ′) (3.93) Thus: ηc ... | Electromagnetics_Vol2_Page_50_Chunk2225 |
38 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA where α and β are the attenuation constant and phase propagation constant, respectively, and ηc is the complex-valued wave impedance. As written, these expressions describe a wave which is +ˆx-polarized and propagates in the +ˆz direction. We make these choices for conven... | Electromagnetics_Vol2_Page_51_Chunk2226 |
3.8. DECIBEL SCALE FOR POWER RATIO 39 3.8 Decibel Scale for Power Ratio [m0154] In many disciplines within electrical engineering, it is common to evaluate the ratios of powers and power densities that differ by many orders of magnitude. These ratios could be expressed in scientific notation, but it is more common to us... | Electromagnetics_Vol2_Page_52_Chunk2227 |
40 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA However, note that this is not true if R1 ̸= R0. A power ratio in dB is equal to 20 log10 of the voltage ratio only if the associated impedances are equal. Adding to the potential for confusion on this point is the concept of voltage gain Gv: Gv ≜20 log10 V1 V0 (dB) (3.11... | Electromagnetics_Vol2_Page_53_Chunk2228 |
3.10. POOR CONDUCTORS 41 The attenuation rate is ∼= 8.69α ∼= 0.0738 dB/m The loss in 100 m of this cable is ∼= (0.0738 dB/m) (100 m) ∼= 7.4 dB Note that it would be entirely appropriate, and equivalent, to state that the attenuation rate for this cable is 7.4 dB/(100 m). The concept of attenuation rate is used in preci... | Electromagnetics_Vol2_Page_54_Chunk2229 |
42 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA material (for example, polyethylene) typically used in coaxial cables. The loss of these materials may or may not be significant, depending on the particulars of the application. The imprecise definition of Equation 3.120 is sufficient to derive some characteristics exhibite... | Electromagnetics_Vol2_Page_55_Chunk2230 |
3.11. GOOD CONDUCTORS 43 Applying the same approximation applied to γ earlier, this may be written ηc ≈ r µ ǫ′ · 1 −j ǫ′′ 2ǫ′ (poor conductor) (3.135) We see that for a poor conductor, Re{ηc} ≈η and that Im{ηc} ≪Re {ηc}. The usual approximation in this case is simply ηc ≈η (poor conductor) (3.136) Additional Readin... | Electromagnetics_Vol2_Page_56_Chunk2231 |
44 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA The imprecise definition of Equation 3.138 is sufficient to derive some characteristics that are common to materials over a wide range of conductivity. To derive these characteristics, first recall that the propagation constant γ is given in general as follows: γ2 = −ω2µǫc (... | Electromagnetics_Vol2_Page_57_Chunk2232 |
3.11. GOOD CONDUCTORS 45 Now multiplying numerator and denominator by 1 + j, we obtain ηc ≈ r µ 2ǫ′′ · (1 + j) (3.152) In the special case that ǫc is determined entirely by conductivity loss and is not accounting for delayed polarization response, then ǫ′′ = σ/ω, and we find: ηc ≈ rµω 2σ · (1 + j) (3.153) There are at l... | Electromagnetics_Vol2_Page_58_Chunk2233 |
46 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA 3.12 Skin Depth [m0158] The electric and magnetic fields of a wave are diminished as the wave propagates through lossy media. The magnitude of these fields is proportional to e−αl where α ≜Re {γ} is the attenuation constant (SI base units of m−1), γ is the propagation const... | Electromagnetics_Vol2_Page_59_Chunk2234 |
3.12. SKIN DEPTH 47 Image Credits Fig. 3.1: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Poynting%E2%80%99s theorem illustration.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 3.2: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Uniform plane wave incident cylin... | Electromagnetics_Vol2_Page_60_Chunk2235 |
Chapter 4 Current Flow in Imperfect Conductors 4.1 AC Current Flow in a Good Conductor [m0069] In this section, we consider the distribution of current in a conductor which is imperfect (i.e., a “good conductor”) and at frequencies greater than DC. To establish context, consider the simple DC circuit shown in Figure 4.... | Electromagnetics_Vol2_Page_61_Chunk2236 |
4.1. AC CURRENT FLOW IN A GOOD CONDUCTOR 49 axis of the wire and perpendicular to the axis of the wire. Waves propagating in any other direction may be expressed as a linear combination of waves traveling in the principal directions, so we need only consider the principal directions to obtain a complete picture. Consid... | Electromagnetics_Vol2_Page_62_Chunk2237 |
50 CHAPTER 4. CURRENT FLOW IN IMPERFECT CONDUCTORS 4.2 Impedance of a Wire [m0159] The goal of this section is to determine the impedance – the ratio of potential to current – of a wire. The answer to this question is relatively simple in the DC (“steady current”) case: The impedance is found to be equal to the resista... | Electromagnetics_Vol2_Page_63_Chunk2238 |
4.2. IMPEDANCE OF A WIRE 51 c⃝Y. Zhao CC BY-SA 4.0 Figure 4.4: Choice of S for calculating net current I. The dimensions of S are width W in the y dimension and extending to infinity in the z direction. Then we have eI ≈ Z W y=0 Z ∞ z=0 ˆxσE0e−(1+j)z/δs · (ˆx dy dz) = σE0W Z ∞ z=0 e−(1+j)z/δsdz (4.6) For convenience,... | Electromagnetics_Vol2_Page_64_Chunk2239 |
52 CHAPTER 4. CURRENT FLOW IN IMPERFECT CONDUCTORS justified in replacing W with the circumference 2πa. Thus, we obtain the following expressions: Z ≈1 + j σδs · l 2πa (δs ≪a) (4.13) and so R ≈ l σ(δs2πa) (δs ≪a) (4.14) The impedance of a wire of length l and radius a ≫δs is given by Equation 4.13. The resistance of suc... | Electromagnetics_Vol2_Page_65_Chunk2240 |
4.2. IMPEDANCE OF A WIRE 53 a ∼= 0.292 mm. Using Equation 4.17, we find: R′ ic ≜Ric l = 1 2 r µf πσ · 1 a ∼= 227 µΩ· m−1 · Hz−1/2 p f (4.19) Using Expression 4.16, we find this is valid only for f ≫130 kHz. So, for example, we may be confident that R′ ic ≈0.82 Ω/m at 13 MHz. At the other extreme (f ≪130 kHz), Equation ... | Electromagnetics_Vol2_Page_66_Chunk2241 |
54 CHAPTER 4. CURRENT FLOW IN IMPERFECT CONDUCTORS associated with skin effect is as important as the magnetostatic inductance in the kHz regime, and becomes gradually less important with increasing frequency. Recall that the phase velocity in a low-loss transmission line is approximately 1/ √ L′C′. This means that ski... | Electromagnetics_Vol2_Page_67_Chunk2242 |
4.3. SURFACE IMPEDANCE 55 Image Credits Fig. 4.1: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Current flow in cylinder new.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 4.2: Biezl, https://commons.wikimedia.org/wiki/File:Skin depth.svg, public domain. Modified from original. ... | Electromagnetics_Vol2_Page_68_Chunk2243 |
Chapter 5 Wave Reflection and Transmission 5.1 Plane Waves at Normal Incidence on a Planar Boundary [m0161] When a plane wave encounters a discontinuity in media, reflection from the discontinuity and transmission into the second medium is possible. In this section, we consider the scenario illustrated in Figure 5.1: a u... | Electromagnetics_Vol2_Page_69_Chunk2244 |
5.1. PLANE WAVES AT NORMAL INCIDENCE ON A PLANAR BOUNDARY 57 where B is a complex-valued constant that remains to be determined. Since the direction of propagation for the reflected wave is −ˆz, we have from the plane wave relationships that eHr(z) = −ˆy B η1 e+jβ1z , z ≤0 (5.4) Similarly, we infer the existence of a “t... | Electromagnetics_Vol2_Page_70_Chunk2245 |
58 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION incidence from Region 1 toward Region 2. We may now solve for C by substituting Equation 5.15 into Equation 5.10. We find: C = (1 + Γ12) Ei 0 (5.17) Now summarizing the solution: eEr(z) = ˆxΓ12Ei 0e+jβ1z , z ≤0 (5.18) eEt(z) = ˆx (1 + Γ12) Ei 0e−jβ2z , z ≥0 (5.19) Equations... | Electromagnetics_Vol2_Page_71_Chunk2246 |
5.1. PLANE WAVES AT NORMAL INCIDENCE ON A PLANAR BOUNDARY 59 equal to the incident power density minus the reflected power density. Thus: St ave = Si ave −Sr ave = | Electromagnetics_Vol2_Page_72_Chunk2247 |
60 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION 5.2 Plane Waves at Normal Incidence on a Material Slab [m0162] In Section 5.1, we considered what happens when a uniform plane wave is normally incident on the planar boundary between two semi-infinite media. In this section, we consider the problem shown in Figure 5.2: a u... | Electromagnetics_Vol2_Page_73_Chunk2248 |
5.2. PLANE WAVES AT NORMAL INCIDENCE ON A MATERIAL SLAB 61 complex-valued constant that remains to be determined. Now let us consider the boundary between Regions 2 and 3. Note that eEt2 is incident on this boundary in precisely the same manner as eEi is incident on the boundary between Regions 1 and 2. Therefore, we i... | Electromagnetics_Vol2_Page_74_Chunk2249 |
62 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION Electric Field Intensity Magnetic Field Intensity Region of Validity Region 1 eEi(z) = ˆxEi 0e−jβ1z eHi(z) = +ˆy | Electromagnetics_Vol2_Page_75_Chunk2250 |
5.2. PLANE WAVES AT NORMAL INCIDENCE ON A MATERIAL SLAB 63 yielding: ηeq = η2 1 + Γ23e−j2β2d 1 −Γ23e−j2β2d (5.55) Equation 5.55 is the wave impedance in the re- gion to the right of the boundary in the equiva- lent scenario shown in Figure 5.3. “Equivalent” in this case means that the incident and reflected fields in Reg... | Electromagnetics_Vol2_Page_76_Chunk2251 |
64 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION The ratio of reflected power density to incident power density is simply the squared magnitude of this reflection coefficient, i.e.: Sr ave Siave = |Γ1,eq|2 = 0.292 ∼= 29.2% (5.63) where Sr ave and Si ave are the reflected and incident power densities, respectively. Since B = ... | Electromagnetics_Vol2_Page_77_Chunk2252 |
5.3. TOTAL TRANSMISSION THROUGH A SLAB 65 the reflection coefficient Γ1,eq = ηeq −η1 ηeq + η1 (5.67) where ηeq is given by ηeq = η2 1 + Γ23e−j2β2d 1 −Γ23e−j2β2d (5.68) and where Γ23 is given by Γ23 = η3 −η2 η3 + η2 (5.69) Total transmission requires that Γ1,eq = 0. From Equation 5.67 we see that Γ1,eq is zero when η1 = η... | Electromagnetics_Vol2_Page_78_Chunk2253 |
66 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION used? Solution. This is a good application for half-wave matching because the material on either side of the slab is the same (presumably free space) whereas the material used for the slab is unspecified. The phase velocity in the slab is vp = c √ǫr ∼= 1.5 × 108 m/s (5.77) ... | Electromagnetics_Vol2_Page_79_Chunk2254 |
5.4. PROPAGATION OF A UNIFORM PLANE WAVE IN AN ARBITRARY DIRECTION 67 Thus, the minimum possible thickness of the radome panel is d = λ2/4 ∼= 1.05 mm, and the relative permittivity of the radome panel must be ǫr ∼= 1.41. Additional Reading: • “Radome” on Wikipedia. 5.4 Propagation of a Uniform Plane Wave in an Arbitrar... | Electromagnetics_Vol2_Page_80_Chunk2255 |
68 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION c⃝C. Wang CC BY-SA 4.0 Figure 5.5: The same plane wave described in a ro- tated coordinate system, yielding Equation 5.87. c⃝C. Wang CC BY-SA 4.0 Figure 5.6: The same plane wave described in yet an- other rotation of the coordinate system, yielding Equa- tion 5.88. propaga... | Electromagnetics_Vol2_Page_81_Chunk2256 |
5.4. PROPAGATION OF A UNIFORM PLANE WAVE IN AN ARBITRARY DIRECTION 69 Thus, k · r = βz, as expected. In ray-fixed coordinates, a wave can be represented by one – and only one – expression, which is the same expression regardless of the orientation of the “global” coordinate system. Moreover, only two basis directions (n... | Electromagnetics_Vol2_Page_82_Chunk2257 |
70 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION 5.5 Decomposition of a Wave into TE and TM Components [m0166] A broad range of problems in electromagnetics involve scattering of a plane wave by a planar boundary between dissimilar media. Section 5.1 (“Plane Waves at Normal Incidence on a Planar Boundary Between Lossless... | Electromagnetics_Vol2_Page_83_Chunk2258 |
5.5. DECOMPOSITION OF A WAVE INTO TE AND TM COMPONENTS 71 The electric field vector is always perpendicular to the direction of propagation, so ˆei · ˆki = 0. This leaves: ˆei = | Electromagnetics_Vol2_Page_84_Chunk2259 |
72 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION 5.6 Plane Waves at Oblique Incidence on a Planar Boundary: TE Case [m0167] In this section, we consider the problem of reflection and transmission from a planar boundary between semi-infinite media for a transverse electric (TE) uniform plane wave. Before attempting this sec... | Electromagnetics_Vol2_Page_85_Chunk2260 |
5.6. PLANE WAVES AT OBLIQUE INCIDENCE ON A PLANAR BOUNDARY: TE CASE 73 total field in Region 1 is the sum of incident and reflected fields, so eE1(r) = eEi T E(r) + eEr(r) (5.117) The total field in Region 2 is simply eE2(r) = eEt(r) (5.118) Next, note that all electric field components are already tangent to the boundary. ... | Electromagnetics_Vol2_Page_86_Chunk2261 |
74 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION In the global coordinate system: ˆkr = ˆx sin ψr −ˆz cos ψr (5.128) Thus: eHr(r) = (ˆz sin ψr + ˆx cos ψr) B η1 e−jkr·r (5.129) The transmitted magnetic field has the form: eHt(r) = 1 η2 ˆkt × eEt (5.130) In the global coordinate system: ˆkt = ˆx sin ψt + ˆz cos ψt (5.131) ... | Electromagnetics_Vol2_Page_87_Chunk2262 |
5.6. PLANE WAVES AT OBLIQUE INCIDENCE ON A PLANAR BOUNDARY: TE CASE 75 simply state the result, and in Section 5.8 we shall perform this part of the derivation in detail and with greater attention to the implications. One finds: ψr = ψi (5.145) and ψt = arcsin β1 β2 sin ψi (5.146) Equation 5.145 is the unsurprising r... | Electromagnetics_Vol2_Page_88_Chunk2263 |
76 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION 5.7 Plane Waves at Oblique Incidence on a Planar Boundary: TM Case [m0164] In this section, we consider the problem of reflection and transmission from a planar boundary between semi-infinite media for a transverse magnetic (TM) uniform plane wave. Before attempting this sec... | Electromagnetics_Vol2_Page_89_Chunk2264 |
5.7. PLANE WAVES AT OBLIQUE INCIDENCE ON A PLANAR BOUNDARY: TM CASE 77 where ˆkt is the unit vector indicating the direction of propagation and β2 = ω√µ2ǫ2 is the phase propagation constant in Region 2. At this point, the unknowns in this problem are the constants B and C, as well as the unknown directions ˆkr and ˆkt.... | Electromagnetics_Vol2_Page_90_Chunk2265 |
78 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION The transmitted magnetic field has the form: eEt(r) = −η2ˆkt × eHt (5.173) In the global coordinate system: ˆkt = ˆx sin ψt + ˆz cos ψt (5.174) Thus: eEt(r) = | Electromagnetics_Vol2_Page_91_Chunk2266 |
5.7. PLANE WAVES AT OBLIQUE INCIDENCE ON A PLANAR BOUNDARY: TM CASE 79 can be found using Equation 5.165. Here we shall simply state the result, and in Section 5.8 we shall perform this part of the derivation in detail and with greater attention to the implications. One finds: ψr = ψi (5.189) i.e., angle of reflection eq... | Electromagnetics_Vol2_Page_92_Chunk2267 |
80 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION c⃝C. Wang CC BY-SA 4.0 Figure 5.13: A TM uniform plane wave incident from air to glass. |ΓT M|2 ∼= 0.021; i.e., about 2.1%. 1 −|ΓT M|2 ∼= 97.9% of the power is transmitted into the glass. Note that the result obtained in the preceding example is different from the result f... | Electromagnetics_Vol2_Page_93_Chunk2268 |
5.8. ANGLES OF REFLECTION AND REFRACTION 81 respectively; and β1 and β2 are the phase propagation constants in Region 1 (from which the wave is incident) and Region 2, respectively. Equation 5.199 is essentially a boundary condition that enforces continuity of the phase of the electric and magnetic fields across the bou... | Electromagnetics_Vol2_Page_94_Chunk2269 |
82 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION c⃝Z. S´andor CC BY-SA 3.0 Figure 5.15: Angles of reflection and refraction for a light wave incident from air onto glass. As expected, the angle of reflection ψr is observed to be equal to ψi. The angle of refraction ψt is observed to be 35◦. What is the relative permittivit... | Electromagnetics_Vol2_Page_95_Chunk2270 |
5.9. TE REFLECTION IN NON-MAGNETIC MEDIA 83 c⃝D-Kuru CC BY-SA 3.0 Figure 5.17: A typical triangular prism. white light into its constituent colors since each color will be refracted by the same amount. Additional Reading: • “Prism” on Wikipedia. • “Refraction” on Wikipedia. • “Refractive index” on Wikipedia. • “Snell’s... | Electromagnetics_Vol2_Page_96_Chunk2271 |
84 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION Since permittivity ǫ can be expressed as ǫ0 times the relative permittivity ǫr, we may reduce further to: β1 β2 = rǫr1 ǫr2 (5.220) Now Equation 5.218 reduces to: sin ψt = rǫr1 ǫr2 sin ψi (5.221) Next, note that for any value ψ, one may write cosine in terms of sine as foll... | Electromagnetics_Vol2_Page_97_Chunk2272 |
5.10. TM REFLECTION IN NON-MAGNETIC MEDIA 85 5.10 TM Reflection in Non-magnetic Media [m0172] Figure 5.21 shows a TM uniform plane wave incident on the planar boundary between two semi-infinite material regions. In this case, the reflection coefficient is given by: ΓT M = −η1 cos ψi + η2 cos ψt +η1 cos ψi + η2 cos ψt (5.22... | Electromagnetics_Vol2_Page_98_Chunk2273 |
86 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION -1 -0.5 0 0.5 1 0 10 20 30 40 50 60 70 80 90 2 10 100 Reflection Coefficient angle of incidence [deg] Figure 5.22: The reflection coefficient ΓT M as a func- tion of angle of incidence ψi for various media combi- nations, parameterized as ǫr2/ǫr1. Using Equation 5.241, we ca... | Electromagnetics_Vol2_Page_99_Chunk2274 |
5.10. TM REFLECTION IN NON-MAGNETIC MEDIA 87 component of the transmitted wave will be TM, but the TM component of the reflected wave will be zero. Thus, the total (TE+TM) reflected wave will be purely TE, regardless of the TM component of the incident wave. This principle can be exploited to suppress the TM component of... | Electromagnetics_Vol2_Page_100_Chunk2275 |
88 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION 5.11 Total Internal Reflection [m0169] Total internal reflection refers to a particular condition resulting in the complete reflection of a wave at the boundary between two media, with no power transmitted into the second region. One way to achieve complete reflection with zer... | Electromagnetics_Vol2_Page_101_Chunk2276 |
5.11. TOTAL INTERNAL REFLECTION 89 So, when ψi > ψi c, we see that ǫr2/ǫr1 −sin2 ψi < 0 (ψi > ψi c) (5.263) and therefore q ǫr2/ǫr1 −sin2 ψi = jB (ψi > ψi c) (5.264) where B is a positive real-valued number. Now we may write Equation 5.261 as follows: ΓT E = A −jB A + jB (ψi > ψi c) (5.265) where A ≜cos ψi is also a po... | Electromagnetics_Vol2_Page_102_Chunk2277 |
90 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION The presence of an imaginary component in the reflection coefficient is odd for two reasons. First, we are not accustomed to seeing a complex-valued reflection coefficient emerge when the wave impedances of the associated media are real-valued. Second, the total reflection of t... | Electromagnetics_Vol2_Page_103_Chunk2278 |
5.12. EVANESCENT WAVES 91 We begin by postulating a complex-valued angle of transmission ψtc. Although the concept of a complex-valued angle may seem counterintuitive, there is mathematical support for this concept. For example, consider the well-known trigonometric identities: sin θ = 1 j2 | Electromagnetics_Vol2_Page_104_Chunk2279 |
92 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION components in this region is described by the factor e−jkt·r where kt = β2ˆkt = β2 | Electromagnetics_Vol2_Page_105_Chunk2280 |
5.12. EVANESCENT WAVES 93 Image Credits Fig. 5.1: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Upw incident on planar boundary.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.2: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Upw incident on a slab.svg, CC BY-S... | Electromagnetics_Vol2_Page_106_Chunk2281 |
94 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION Fig. 5.15: c⃝Z. S´andor, https://commons.wikimedia.org/wiki/File:F%C3%A9nyt%C3%B6r%C3%A9s.jpg, CC BY-SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/). Fig. 5.16: c⃝G. Saini, https://kids.kiddle.co/Image:Refractionn.jpg, CC BY-SA 4.0 (https://creativecommons.org/lic... | Electromagnetics_Vol2_Page_107_Chunk2282 |
Chapter 6 Waveguides 6.1 Phase and Group Velocity [m0176] Phase velocity is the speed at which a point of constant phase travels as the wave propagates.1 For a sinusoidally-varying wave, this speed is easy to quantify. To see this, consider the wave: A cos (ωt −βz + ψ) (6.1) where ω = 2πf is angular frequency, z is pos... | Electromagnetics_Vol2_Page_108_Chunk2283 |
96 CHAPTER 6. WAVEGUIDES Letting ∆β become vanishingly small, we obtain vg ≜∂ω ∂β (6.5) Note the similarity to the definition of phase velocity in Equation 6.3. Group velocity can be interpreted as the speed at which a disturbance in the wave propagates. Information may be conveyed as meaningful disturbances relative to... | Electromagnetics_Vol2_Page_109_Chunk2284 |
6.2. PARALLEL PLATE WAVEGUIDE: INTRODUCTION 97 • “Phase velocity” on Wikipedia. • “Speed of light” on Wikipedia. 6.2 Parallel Plate Waveguide: Introduction [m0173] A parallel plate waveguide is a device for guiding the propagation of waves between two perfectly-conducting plates. Our primary interest in this structure ... | Electromagnetics_Vol2_Page_110_Chunk2285 |
98 CHAPTER 6. WAVEGUIDES imposed by the perfectly-conducting plates, is sufficient to determine a unique solution. This is most easily done in Cartesian coordinates, as we shall now demonstrate. First we express eE in Cartesian coordinates: eE = ˆx eEx + ˆy eEy + ˆz eEz (6.18) This facilitates the decomposition of Equat... | Electromagnetics_Vol2_Page_111_Chunk2286 |
6.3. PARALLEL PLATE WAVEGUIDE: TE CASE, ELECTRIC FIELD 99 6.3 Parallel Plate Waveguide: TE Case, Electric Field [m0174] In Section 6.2, the parallel plate waveguide was introduced. At the end of that section, we described the decomposition of the problem into its TE and TM components. In this section, we find the electr... | Electromagnetics_Vol2_Page_112_Chunk2287 |
100 CHAPTER 6. WAVEGUIDES This confirms that kx and kz are in fact the components of the propagation vector k ≜βˆk = ˆxkx + ˆyky + ˆzkz (6.39) where ˆk is the unit vector pointing in the direction of propagation, and ky = 0 in this particular problem. The solution has now been reduced to finding the constants A, B, and e... | Electromagnetics_Vol2_Page_113_Chunk2288 |
6.3. PARALLEL PLATE WAVEGUIDE: TE CASE, ELECTRIC FIELD 101 where eE(m) y ≜ ( 0, f < f (m) c E(m) y0 e−jk(m) z z sin k(m) x x, f ≥f (m) c (6.53) where m enumerates modes (m = 1, 2, ...), k(m) z ≜ r β2 − h k(m) x i2 (6.54) and k(m) x ≜mπ/a (6.55) Finally, E(m) y0 is a complex-valued constant that depends on sources or bo... | Electromagnetics_Vol2_Page_114_Chunk2289 |
102 CHAPTER 6. WAVEGUIDES (and only one) propagating TE mode is assured. Solution. Single-mode TE propagation is assured by limiting frequency f to greater than the cutoff frequency for m = 1, but lower than the cutoff frequency for m = 2. (Any frequency higher than the cutoff frequency for m = 2 allows at least 2 mode... | Electromagnetics_Vol2_Page_115_Chunk2290 |
6.4. PARALLEL PLATE WAVEGUIDE: TE CASE, MAGNETIC FIELD 103 and ∂eE(m) y ∂x = ∂ ∂x E(m) y0 e−jk(m) z z sin k(m) x x = E(m) y0 e−jk(m) z z cos k(m) x x +k(m) x (6.68) We may now assemble a solution for the magnetic field as follows: ˆx eHx + ˆz eHz = ˆx ∞ X m=1 eH(m) x + ˆz ∞ X m=1 eH(m) z (6.69) where eH(m) x = −... | Electromagnetics_Vol2_Page_116_Chunk2291 |
104 CHAPTER 6. WAVEGUIDES 6.5 Parallel Plate Waveguide: TM Case, Electric Field [m0177] In Section 6.2, the parallel plate waveguide shown in Figure 6.4 was introduced. At the end of that section, we decomposed the problem into its TE and TM components. In this section, we find the TM component of the fields in the waveg... | Electromagnetics_Vol2_Page_117_Chunk2292 |
6.5. PARALLEL PLATE WAVEGUIDE: TM CASE, ELECTRIC FIELD 105 This is precisely the same constraint identified in the TE case, and confirms that kx and ky are in fact the components of the propagation vector k ≜βˆk = ˆxkx + ˆyky + ˆzkz (6.83) where ˆk is the unit vector pointing in the direction of propagation, and ky = 0 i... | Electromagnetics_Vol2_Page_118_Chunk2293 |
106 CHAPTER 6. WAVEGUIDES where m is an integer. Note that this is precisely the same relationship that we identified in the TE case. There is an important difference, however. In the TE case, m = 0 was not of interest because this yields kx = 0, and the associated field turned out to be identically zero. In the present ... | Electromagnetics_Vol2_Page_119_Chunk2294 |
6.6. PARALLEL PLATE WAVEGUIDE: THE TM0 MODE 107 6.6 Parallel Plate Waveguide: The TM0 Mode [m0220] In Section 6.2, the parallel plate waveguide (also shown in Figure 6.5) was introduced. At the end of that section we decomposed the problem into its constituent TE and TM fields. In Section 6.5, we determined the electric... | Electromagnetics_Vol2_Page_120_Chunk2295 |
108 CHAPTER 6. WAVEGUIDES the only mode that can propagate inside the PCB is TM0. Therefore, the field deep inside the PCB may be interpreted as a single plane wave having the TM0 structure shown in Figure 6.5, propagating away from the source end of the PCB. The phase velocity is simply that of the apparent plane wave:... | Electromagnetics_Vol2_Page_121_Chunk2296 |
6.7. GENERAL RELATIONSHIPS FOR UNIDIRECTIONAL WAVES 109 find: eEx = 1 jωǫ ∂eHz ∂y −∂eHy ∂z ! (6.121) eEy = 1 jωǫ ∂eHx ∂z −∂eHz ∂x ! (6.122) eEz = 1 jωǫ ∂eHy ∂x −∂eHx ∂y ! (6.123) Without loss of generality, we may assume that the single direction in which the wave is traveling is in the +ˆz direction. If this is the cas... | Electromagnetics_Vol2_Page_122_Chunk2297 |
110 CHAPTER 6. WAVEGUIDES role in determining the structure of fields within the waveguide, and this provides additional motivation to identify this quantity explicitly in the field equations. Summarizing: If you know the wave is unidirectional, then knowledge of the components of eE and eH in the direction of propagatio... | Electromagnetics_Vol2_Page_123_Chunk2298 |
6.8. RECTANGULAR WAVEGUIDE: TM MODES 111 Equation 6.141 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 demonstr... | Electromagnetics_Vol2_Page_124_Chunk2299 |
112 CHAPTER 6. WAVEGUIDES Substituting this expression into Equation 6.158, we obtain: Y ∂2 ∂x2 X + X ∂2 ∂y2 Y + k2 ρXY = 0 (6.160) Next dividing through by XY , we obtain: 1 X ∂2 ∂x2 X + 1 Y ∂2 ∂y2 Y + k2 ρ = 0 (6.161) Note that the first term depends only on x, the second term depends only on y, and the remaining term... | Electromagnetics_Vol2_Page_125_Chunk2300 |
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