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n1 sin ψi = n2 sin ψt (5.214) When both media are non-magnetic, Equation 5.212 simplifies to ψt = arcsin (√ ǫr1 ǫr2 sin ψi ) (5.215) When ǫr2 > ǫr1, we observe that ψt <ψi. In other words, the transmitted wave travels in a direction that is closer to the surface normal than the angle of incidence. This scenario is demon...
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82 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION c⃝ Z. 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 observ ed to be equal to ψi. The angle of refraction ψt is observed to be 35◦. What is the relative permit...
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the surface normal. Figure 5.16 shows an example from common experience. In non-magnetic media, when ǫr1 < ǫr2, ψt < ψi (refraction toward the surface normal). When ǫr1 > ǫr2, ψt > ψi (refraction away from the surface normal). c⃝ G. Saini CC BY -SA 4.0 Figure 5.16: Refraction accounts for the apparent dis- placement of...
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all we can say is that when ǫr2 <ǫr1, ψt is able to reach π/2 radians, which corresponds to propagation parallel to the boundary . Beyond that threshold, we must account for the unique physical considerations associated with total internal reflection. W e conclude this section with a description of the common waveguidin...
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frequency . Thus, each frequency is refracted by a different amount. Conversely , a prism comprised of a material whose permittivity exhibits negligible variation with frequency will not separate incident
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5.9. TE REFLECTION IN NON-MAGNETIC MEDIA 83 c⃝ D-K uru 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. • “Snel...
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respectively . Many materials of practical interest are non-magnetic; that is, they have permeability that is not significantly different from the permeability of free space. In this section, we consider the behavior of the reflection coefficient for this class of materials. T o begin, recall the general form of Snell’s l...
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84 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION Since permittivity ǫcan be expressed as ǫ0 times the relative permittivity ǫr, we may reduce further to: β1 β2 = √ ǫr1 ǫr2 (5.220) No w Equation 5.218 reduces to: sin ψt = √ ǫr1 ǫr2 sin ψi (5.221) Ne xt, note that for any value ψ, one may write cosine in terms of sine as...
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Finally , by substituting Equation 5.223, we obtain: Γ TE = cos ψi − √ ǫr2/ǫr1 − s in 2 ψi cos ψi + √ ǫr2/ǫr1 − s in 2 ψi (5.228) This expression has the advantage that it is now entirely in terms of ψi, with no need to first calculate ψt. Using Equation 5.228, we can see how different combinations of material affect th...
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ǫr2/ǫr1 − sin2 ψi to be negative, which makes Γ TE complex-valued. This results in total internal reflection, and is addressed elsewhere in another section. When ǫr1 <ǫr2 (e.g., wave traveling in air toward glass), we see that ǫr2/ǫr1 − sin2 ψi is always positive, so Γ TE is always real-valued. Let us continue with the ...
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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: Γ TM = −η1 cos ψi + η2 cos ψt +η1 cos ψi + η2 cos ψt (5.22...
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In non-magnetic media, the permeabilities µ1 and µ2 are assumed equal to µ0. Thus: β1 β2 = ω√µ1ǫ1 ω√µ2ǫ2 = √ ǫ1 ǫ2 (5.231) c⃝ C. W ang CC BY -SA 4.0 Figure 5.21: A transverse magnetic uniform plane wave obliquely incident on the planar boundary be- tween two semi-infinite material regions. Since permittivity ǫcan be exp...
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Γ TM = − ( η0/√ǫr1 ) cos ψi + ( η0/√ǫr2 ) cos ψt + ( η0/√ǫr1 ) cos ψi + ( η0/√ǫr2 ) cos ψt (5.238) Multiplying numerator and denominator by √ǫr2/η0, we obtain: Γ TM = − √ ǫr2/ǫr1 cos ψi + cos ψt + √ ǫr2/ǫr1 cos ψi + cos ψt (5.239) Substituting Equation 5.235, we obtain: Γ TM = − √ ǫr2/ǫr1 cos ψi + √ 1 − (ǫr1/ǫr2) sin2 ...
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86 CHAPTER 5. W A VE 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 Γ TM as a func- tion of angle of incidence ψi for various media combi- nations, parameterized as ǫr2/ǫr1. Using Equati...
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Γ TM is always real-valued. Let us continue with the ǫr1 <ǫr2 condition. Figure 5.22 shows Γ TM plotted for various combinations of media over all possible angles of incidence from 0 (normal incidence) to π/2 (grazing incidence). W e observe: In non-magnetic media with ǫr1 < ǫr2, Γ T M is real-valued and increases from...
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would obtain for a perfect conductor in Region 2. Also note that when ǫr1 <ǫr2, Γ TM changes sign from negative to positive as angle of incidence increases from 0 to π/2. This behavior is quite different from that of the TE component, which is always negative for ǫr1 <ǫr2. The angle of incidence at which Γ TM = 0 is re...
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Applying the principle of superposition, we may consider these components separately . The TE component of the incident wave will scatter as reflected and transmitted waves which are also TE. However, Γ TM = 0 when ψi = ψi B, so the TM
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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 o...
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B = R− sin2 ψi B (5.244) Now employing a trigonometric identity on the left side of the equation, we obtain: R2 ( 1 − sin2 ψi B ) = R− sin2 ψi B (5.245) R2 − R2 sin2 ψi B = R− sin2 ψi B (5.246) ( 1 − R2) sin2 ψi B = R− R2 (5.247) and finally sin ψi B = √ R− R2 1 − R2 (5.248) Although this equation gets the job done, it ...
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ster’s angle for non-magnetic media. Thus, we have found tan ψi B = √ ǫr2 ǫr1 (5.253) Example 5.10. Polarizing angle for an air-to-glass interface. A plane wave is incident from air onto the planar boundary with a glass region. The glass exhibits relative permittivity of 2.1. The incident wave contains both TE and TM c...
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88 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION 5.11 T otal Internal Reflection [m0169] T otal 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...
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ψr = ψi (5.255) i.e., angle of reflection equals angle of incidence. Also, from Snell’s law: √ µr1ǫr1 sin ψi = √µr2ǫr2 sin ψt (5.256) where “r” in the subscripts indicates the relative (unitless) quantities. The associated formula for ψt c⃝ C. W ang CC BY -SA 4.0 (modified) Figure 5.25: A uniform plane wave obliquely inc...
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increased? When calculating ψt using Equation 5.257, one finds that the argument of the arcsine function becomes greater than 1. Since the possible values of the sine function are between −1 and +1, the arcsine function is undefined. Clearly our analysis is inadequate in this situation. T o make sense of this, let us beg...
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TE component and non-magnetic materials is Γ TE = cos ψi − √ ǫr2/ǫr1 − s in 2 ψi cos ψi + √ ǫr2/ǫr1 − s in 2 ψi (5.261) From Equation 5.260, we see that sin2 ψi c = ǫr2 ǫr1 (5.262)
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5.11. TOT AL INTERNAL REFLECTION 89 So, when ψi >ψi c, we see that ǫr2/ǫr1 − sin2 ψi <0 (ψ i >ψi c) (5.263) and therefore√ ǫr2/ǫr1 − s in 2 ψi = jB (ψi >ψi c) (5.264) where Bis a positive real-valued number. Now we may write Equation 5.261 as follows: Γ TE = A− jB A+ jB ( ψi >ψi c) (5.265) where A≜ cos ψi is also a pos...
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component, the identical conclusion is obtained for TM component as well. This is left as an exercise for the student. Example 5.11. T otal internal reflection in glass. Figure 5.15 (Section 5.8) shows a demonstration of refraction of a beam of light incident from air onto a planar boundary with glass. Analysis of that ...
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Figure 5.26: T otal internal reflection of a light wave incident on a planar boundary between glass and air. c⃝ Ti mwether CC BY -SA 3.0 Figure 5.27: Laser light in a dielectric rod exhibiting the Goos-H ¨anchen effect. The angle of incidence in Figure 5.26 is seen to be about ∼= 50◦, which is greater than the critical ...
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90 CHAPTER 5. W A VE 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...
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• “Goos-H ¨anchen effect” on Wikipedia. • “Snell’s law” on Wikipedia. • “T otal internal reflection” on Wikipedia. 5.12 Evanescent W aves [m0170] Consider the situation shown in Figure 5.28: A uniform plane wave obliquely incident on the planar boundary between two semi-infinite material regions, and total internal reflec...
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the boundary if no power is transmitted across the boundary? There must be a field on the opposite side of the boundary , but – somehow – it must have zero power. T o make sense of this, let us attempt to find a solution for the transmitted field. c⃝ C. W ang CC BY -SA 4.0 Figure 5.28: A uniform plane wave obliquely inci-...
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5.12. EV ANESCENT W A VES 91 W e 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 ( ejθ − e−j θ) (5.268) co...
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real part of ψtc remains fixed (parallel to the boundary) and that an imaginary component jψ′′ emerges to satisfy the boundary conditions. For clarity , let us assign the variable ψ′ to represent the real part of ψtc in Equations 5.270–5.272. Then we may refer to all three cases using a single expression as follows: ψtc...
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sin jψ′′ = 1 j2 ( ej(jψ′ ′) − e−j(jψ′′)) (5.279) = 1 j2 ( e−ψ′′ − e+ψ′ ′ ) (5.280) = j1 2 ( e+ψ′′ − e−ψ′ ′ ) (5.281) = jsinh ψ′′ (5.282) In other words, the sine of jψ′′ is jtimes hyperbolic sine (“sinh ”) of ψ′′. Now note that sinh of a real-valued argument is real-valued, so sin jψ′′ is imaginary-valued. Using these ...
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When total internal reflection is in effect, ψi >ψi c, so ψ′ = π/2. In this case, Equations 5.283 and 5.284 yield sin ψtc = cosh ψ′′ (5.287) cos ψtc = −jsinh ψ′′ (5.288) Let us now consider what this means for the field in Region 2. According to the formalism adopted in previous sections, the propagation of wave
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92 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION components in this region is described by the factor e−jkt·r where kt = β2 ˆkt = β2 (ˆx sin ψtc + ˆz cos ψtc) (5.289) and r = ˆxx+ ˆyy+ ˆzz (5.290) so kt · r = ( β2 sin ψtc) x+ ( β2 cos ψtc) z = ( β2 cosh ψ′′) x+ (−jβ2 sinh ψ′′) z (5.291) Therefore, the wave in Region 2 ...
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unlike the transmitted wave in the ψi <ψi c case (also a uniform plane wave). The transmitted wave that we have derived in the ψi >ψi c case gives the impression of being somehow attached to the boundary , and so may be described as a surface wave. However, in this case we have a particular kind of surface wave, known ...
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propagation and attenuation constants for the evanescent wave. It suffices to say that the magnitude of the evanescent field becomes negligible beyond a few wavelengths of the boundary . Finally , we return to the strangest characteristic of this field: It acts like a wave, but conveys no power. This field exists solely to...
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and continues to propagate to infinity , even after light is no longer incident on the boundary . In contrast, the evanescent wave vanishes at the moment laser light ceases to illuminate the boundary . In other words, the evanescent field does not continue to propagate along the boundary to infinity . The reason for this ...
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5.12. EV ANESCENT W A VES 93 Image Credits Fig. 5.1: c⃝ Sevenchw (C. W ang), 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: c⃝ Sevenchw (C. W ang), https://commons.wikimedia.org/wiki/File:Upw incident on a slab.svg...
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CC BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.6: c⃝ Sevenchw (C. W ang), https://commons.wikimedia.org/wiki/File:Plane wave in another rotation coord.svg, CC BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.7: c⃝ Sevenchw (C. W ang), https://commons.wikimedia.org/wiki/File:...
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CC BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.11: c⃝ Sevenchw (C. W ang), https://commons.wikimedia.org/wiki/File:TE-polarized upw incident from air to glass.svg, CC BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.12: c⃝ Sevenchw (C. W ang), https://commons.wikimedia.org/w...
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94 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION Fig. 5.15: c⃝ Z. 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: c⃝ G. Saini, https://kids.kiddle.co/Image:Refractionn.jpg, CC BY -SA 4.0 (https://creativecommons....
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Fig. 5.21: c⃝ Sevenchw (C. W ang), https://commons.wikimedia.org/wiki/File:TM-polarized upw incident on planar boundary .svg, CC BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.23: c⃝ Sevenchw (C. W ang), https://commons.wikimedia.org/wiki/File:Reflection of plane wave incidence angle equals polariz...
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https://en.wikipedia.org/wiki/File:T eljes f%C3%A9nyvisszaver%C5%91d%C3%A9s.jpg, CC BY -SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/). Fig. 5.27: c⃝ Timwether, https://en.wikipedia.org/wiki/File:Laser in fibre.jpg, CC BY -SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/). Fig. 5.28: c⃝ Sevenchw (C. W a...
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Chapter 6 W a veguides 6.1 Phase and Group V elocity [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 . T o see this, consider the wave: Acos (ωt− βz+ ψ) (6.1) where ω= 2πf is angular frequency , zis ...
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1 Formally , “velocity” is a vector which indicates both the di- rection and rate of motion. It is common practice to use the terms “phase velocity” and “group velocity” even though we are actually referring merely to rate of motion. The direction is, of course, in the direction of propagation. Central to the concept o...
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phase modulation). In other words, information can be transmitted only by making the wave non-uniform in some respect. Furthermore, some materials and structures can cause changes in ψor other combinations of parameters which vary with position or time. Examples include dispersion and propagation within waveguides. Reg...
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Group velocity, vg, is the ratio of the apparent change in frequency ωto the associated change in the phase propagation constant β; i.e., ∆ω/∆β . Electromagnetics V ol. 2. c⃝ 2020 S.W . Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics- vol- 2
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96 CHAPTER 6. W A VEGUIDES 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...
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and dispersion (frequency-dependent constitutive parameters) are examples in which vg is not necessarily equal to vp. Here’s an example involving dispersion: Example 6.1. Phase and group velocity for a material exhibiting square-law dispersion. A broad class of non-magnetic dispersive media exhibit relative permittivit...
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(6.12) No w simplifying using Equation 6.8: vg = ( 2 β ω )−1 (6.13) = 1 2 ω β (6.14) = 1 2vp (6.15) Thus, we see that in this case the group velocity is always half the phase velocity . Another commonly-encountered example for which vg is not necessarily equal to vp is the propagation of guided waves; e.g., waves withi...
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6.2. P ARALLEL PLA TE W A VEGUIDE: INTRODUCTION 97 • “Phase velocity” on Wikipedia. • “Speed of light” on Wikipedia. 6.2 Parallel Plate W aveguide: Introduction [m0173] A parallel plate waveguide is a device for guiding the propag ation of waves between two perfectly-conducting plates. Our primary interest in this stru...
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exhibiting real-valued permeability µand real-valued permittivity ǫ. Let us limit our attention to a region within the waveguide which is free of sources. Expressed in phasor form, the electric field intensity is governed by the wave equation ∇2 ˜E + β2 ˜E = 0 (6.16) where β = ω√ µǫ (6.17) Equation 6.16 is a partial dif...
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98 CHAPTER 6. W A VEGUIDES 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 ˜E in Cartesian coordinates: ˜E = ˆx ˜Ex + ˆy ˜Ey + ˆz ˜Ez (6.18) This facilitates the decomposition of Equ...
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∂y2 ˜Ez + ∂2 ∂z2 ˜Ez = −β2 ˜Ez (6.25) Let us restrict our attention to scenarios that can be completely described in two dimensions; namely x and z, and for which there is no variation in y. This is not necessarily required, however, it is representative of a broad class of relevant problems, and allows Equations 6.23–...
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z= d; namely , that the tangent component of ˜E is zero at these surfaces. At this point, the problem has been reduced to a routine exercise in the solution of partial differential equations. However, a somewhat more useful approach is to first decompose the total electric field into transverse electric (TE) and transver...
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components. The TE and TM solutions are presented in Sections 6.3 (Electric component of the TE solution), 6.4 (Magnetic component of the TE solution), and 6.5 (Electric component of the TM solution). The magnetic component of the TM solution can be determined via a straightforward variation of the preceding three case...
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6.3. P ARALLEL PLA TE W A VEGUIDE: TE CASE, ELECTRIC FIELD 99 6.3 Parallel Plate W aveguide: 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 e...
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and kx and kz are real-valued constants. W e have assigned variable names to these constants with advance knowledge of their physical interpretation; however, at this moment they remain simply unknown constants whose values must be determined by enforcement of boundary conditions. c⃝ C. W ang CC BY -SA 4.0 Figure 6.2: ...
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be no wave components propagating in the −ˆz direction. In this case, C = D= 0 and Equation 6.30 simplifies to: ˜Ey = e−jkz z[ Ae−jkxx + Be+jkxx] (6.31) Before proceeding, let’s make sure that Equation 6.31 is actually a solution to Equation 6.29. As we shall see in a moment, performing this check will reveal some addit...
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100 CHAPTER 6. W A VEGUIDES 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, an...
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This expression is simplified using a trigonometric identity: 1 2j [ e+jkxx − e−j kxx] = sin kxx (6.42) Let us now make the definition Ey0 ≜ j2B. Then: ˜Ey = Ey0e−jkz zsin kxx (6.43) Now applying the boundary condition at x= a: Ey0e−jkz zsin kxa= 0 (6.44) The factor e−jkz z cannot be zero, and Ey0 = 0 yields only trivial...
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given by Equation 6.43 and Equation 6.46 with m= 1,2,.... Each solution associated with a particular value of mis referred to as a mode, which (via Equation 6.46) has a particular value of kx. The value of kz for mode mis obtained using Equation 6.38 as follows: kz = √ β2 − k2x = √ β2 − (mπ a )2 (6.47) Since kz is spec...
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component in order to propagate; therefore, these modes do not propagate and may be ignored. Let us now summarize the solution. For the scenario depicted in Figure 6.2, the electric field component of the TE solution is given by: ˆy ˜Ey = ˆy ∞∑ m=1 ˜E(m) y (6.52)
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6.3. P ARALLEL PLA TE W A VEGUIDE: TE CASE, ELECTRIC FIELD 101 where ˜E(m) y ≜ { 0, f <f(m) c E(m) y0 e−jk(m) z zsin k(m) x x, f ≥ f(m) c (6.53) where menumerates modes (m = 1,2,...), k(m) z ≜ √ β2 − [ k(m) x ] 2 (6.54) and k(m) x ≜ mπ /a (6.55) Finally , E(m) y0 is a complex-valued constant that depends on sources or ...
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f >1/2a√µǫ. Also k( 1) x = π/a, so k(1) z = √ β2 − (π a )2 (6.56) Subsequently , ˜E(1) y = E(1) y0 e−jk(1) z zsin πx a (6.57) Note that this mode has the form of a plane wave. The plane wave propagates in the +ˆz direction with phase propagation constant k(1) z . Also, we observe that the apparent plane wave is non-uni...
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Right: m= 2. f >1/a√µǫ. This frequency is higher than f( 1) c , so the m= 1 mode can exist at any frequency at which the m= 2 mode exists. Also k(2) x = 2π/a, so k(2) z = √ β2 − (2π a )2 (6.58) Subsequently , ˜E(2) y = E(2) y0 e−jk(2) z zsin 2πx a (6.59) In this case, the apparent plane wave propagates in the +ˆz direc...
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increasing integer number of sinusoidal half-periods in magnitude. Example 6.2. Single-mode TE propagation in a parallel plate waveguide. Consider an air-filled parallel plate waveguide consisting of plates separated by 1 cm. Determine the frequency range for which one
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102 CHAPTER 6. W A VEGUIDES (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 modes...
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1/√µ0ǫ0 = c. However, the phase velocity indicated by Equation 6.62 is greater than 1/√µǫ; e.g., faster than light would travel in the same material (presuming it were transparent). At first glance, this may seem to be impossible. However, recall that information travels at the group velocity vg, and not necessarily the...
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dispersion, and sometimes specifically as mode dispersion or modal dispersion. 6.4 Parallel Plate W aveguide: TE Case, Magnetic Field [m0175] In Section 6.2, the parallel plate waveguide was introduced. In Section 6.3, we determined the TE component of the electric field. In this section, we determine the TE component of...
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of ˜E are zero. The two remaining terms are −ˆx∂˜Ey/∂z and +ˆz∂˜Ey/∂x. Thus: ˜H = j ωµ ( −ˆx∂˜Ey ∂z + ˆz∂˜Ey ∂x ) (6.66) Recall that ˜Ey is the sum of modes, as indicated in Equation 6.63. Since differentiation (i.e., ∂/∂z and ∂/∂x) is a linear operator, we may evaluate Equation 6.66 for modes one at a time, and then s...
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6.4. P ARALLEL PLA TE W A VEGUIDE: TE CASE, MAGNETIC FIELD 103 and ∂˜E(m) y ∂x = ∂ ∂x E(m) y0 e−jk(m) z zs in k(m) x x = ( E(m) y0 e−jk(m) z zcos k(m) x x )( +k(m) x ) (6.68) W e may now assemble a solution for the magnetic field as follows: ˆx ˜Hx + ˆz ˜Hz = ˆx ∞∑ m=1 ˜H(m) x + ˆz ∞∑ m=1 ˜H(m) z (6.69) where ˜H(m) x = ...
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surface at x= 0, we see ˜H(m) x (x= 0) = 0 ˜H(m) z (x= 0) = +j k(m) x ωµ E(m) y0 e−jk(m) z z (6.72) Similarly , on the PEC surface at x= a, we see ˜H(m) x (x= a) = 0 ˜H(m) z (x= a) = −jk(m) x ωµ E(m) y0 e−jk(m) z z (6.73) Thus, we see the magnetic field vector at the PEC surfaces is non-zero and parallel to the PEC surf...
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on the inner and outer conductors, or the electromagnetic fields between the conductors. The parallel plate waveguide is only slightly more complicated because the field in a properly-designed coaxial cable is a single transverse electromagnetic (TEM) mode, whereas the fields in a parallel plate waveguide are combinations...
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104 CHAPTER 6. W A VEGUIDES 6.5 Parallel Plate W aveguide: 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 wa...
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+ e+jkz z[ Ce−jkxx + De+jkxx] (6.75) where A, B, C, and Dare complex-valued constants; and kx and kz are real-valued constants. W e have assigned variable names to these constants with advance knowledge of their physical interpretation; however, at this moment they remain simply unknown constants whose values must be d...
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on the right (z > 0) side of Figure 6.4, then there can be no wave components propagating in the −ˆz direction. In this case, C = D= 0 and Equation 6.75 simplifies to: ˜Hy = e−jkz z[ Ae−jkxx + Be+jkxx] (6.76) Before proceeding, let’s make sure that Equation 6.76 is actually a solution to Equation 6.74. As in the TE case...
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6.5. P ARALLEL PLA TE W A VEGUIDE: 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 ...
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−ˆx∂˜Hy/∂z and +ˆz∂˜Hy/∂x. Thus: ˜E = 1 jωǫ ( −ˆx∂˜Hy ∂z + ˆz∂˜Hy ∂x ) (6.86) W e may further develop this expression using Equations 6.77 and 6.79. W e find the ˆx component of ˜E is: ˜Ex = kz ωǫ e−jkz z[ Ae−j kxx + Be+jkxx] (6.87) and the ˆz component of ˜E is: ˜Ez = kx ωǫ e−jkz z[ −Ae−j kxx + Be+jkxx] (6.88) The solu...
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this is unnecessarily restrictive. Instead, we require A= Band we may rewrite Equation 6.88 as follows: ˜Ez = Bkx ωǫ e−jkz z[ e+j kxx − e−jkxx] (6.90) This expression is simplified using a trigonometric identity: sin kxa= 1 2j [ e+jkxa − e−j kxa] (6.91) Thus: ˜Ez = j2Bkx ωǫ e−jkz zs in kxx (6.92) Now following up with ˜...
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106 CHAPTER 6. W A VEGUIDES where mis 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 ...
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Continuing: The value of kz for mode mis obtained using Equation 6.82 as follows: kz = √ β2 − k2x = √ β2 − (mπ a )2 (6.99) Since kz is specified to be real-valued, we require: β2 − (mπ a )2 >0 (6.100) This constrains β; specifically: β >mπ a (6.101) Recall that β = ω√µǫand ω= 2πf where f is frequency . Solving for f, we ...
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which m= 0 is not available. Let us now summarize the solution. With respect to Figure 6.4, we find that the electric field component of the TM field is given by: ˜E = ∞∑ m=0 [ ˆx ˜E(m) x + ˆz ˜E(m) z ] (6.104) where ˜E(m) x ≜ { 0 , f <f(m) c E(m) x0 e−jk(m) z zcos k(m) x x, f ≥ f(m) c (6.105) and ˜E(m) z ≜ { 0 , f <f(m) ...
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left of the region of interest, with no additional sources or boundary conditions to the right of the region of interest. The m= 0 mode, commonly referred to as the “TM 0” mode, is of particular importance in the analysis of microstrip transmission line, and is addressed in Section 6.6.
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6.6. P ARALLEL PLA TE W A VEGUIDE: THE TM 0 MODE 107 6.6 Parallel Plate W aveguide: The TM 0 Mode [m0220] In Section 6.2, the parallel plate waveguide (also sho wn 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 ...
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TE case. W e also noted that the cutoff frequency for this mode is zero, so it may exist at any frequency , and within any non-zero plate separation a. For this mode k(0) x = 0, k(0) z = β, and we find ˜E = ˆxE(0) x0 e−jβz (TM0 mode) (6.109) Remarkably , we find that this mode has the form of a uniform plane wave which p...
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(6.111) = ˆy E(0) x0 η e−jβ z (TM0 mode) (6.112) Example 6.3. Guided waves in a printed circuit board (PCB). A very common form of PCB consists of a 1.575 mm-thick slab of low-loss dielectric having relative permittivity ≈ 4.5 sandwiched between two copper planes. Characterize the electromagnetic field in a long strip o...
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of TE and TM modes. The active (non-zero) modes depend on the source (a mode must be “stimulated” by the source in order to propagate) and modal cutoff frequencies. The cutoff frequency for mode mis f(m) c = m 2a√µǫ (6.113) In this case, a= 1.575 mm, µ≈ µ0, and ǫ≈ 4.5ǫ0. Therefore: f(m) c ≈ (44.9 GHz) m (6.114) Since t...
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108 CHAPTER 6. W A VEGUIDES 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 TM 0 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 wa...
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Within a straight waveguide, waves can travel either “forward” or “backward. ” The principle of superposition allows one to consider these two unidirectional cases separately , and then to simply sum the results. In this section, the equations that relate the various components of a unidirectional wave are derived. The...
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as follows: ˜E = ˆx ˜Ex + ˆy ˜Ey + ˆz ˜Ez (6.119) ˜H = ˆx ˜Hx + ˆy ˜Hy + ˆz ˜Hz (6.120) Now applying the equation for curl in Cartesian coordinates (Equation B.16 in Appendix B.2), we
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6.7. GENERAL RELA TIONSHIPS FOR UNIDIRECTIONAL W A VES 109 find: ˜Ex = 1 jωǫ ( ∂˜Hz ∂y − ∂˜Hy ∂z ) (6.121) ˜Ey = 1 jωǫ ( ∂˜Hx ∂z − ∂˜Hz ∂x ) (6.122) ˜Ez = 1 jωǫ ( ∂˜Hy ∂x − ∂˜Hx ∂y ) (6.123) W ithout loss of generality , we may assume that the single direction in which the wave is traveling is in the +ˆz direction. If t...
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to zreduce to algebraic operations; i.e.: ∂˜Hx ∂z = −jkz˜hx( x,y)e−jkz z = −jkz ˜Hx (6.127) ∂˜Hy ∂z = −jkz˜hy( x,y)e−jkz z = −jkz ˜Hy (6.128) ∂˜Hz ∂z = −jkz˜hz( x,y)e−jkz z = −jkz ˜Hz (6.129) W e now substitute Equations 6.124–6.126 into Equations 6.121–6.123 and then use Equations 6.127–6.129 to eliminate partial deri...
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(specifically , a +ˆz-traveling) wave. With just a little bit of algebraic manipulation of these equations, it is possible to obtain expressions for the ˆx and ˆy components of ˜E and ˜H which depend only on the ˆz components of ˜E and ˜H. Here they are: 3 ˜Ex = −j k2ρ ( +kz ∂˜Ez ∂x + ωµ∂˜Hz ∂y ) (6.136) ˜Ey = +j k2ρ ( ...
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propagation in the ˆz direction in the waveguide, kρ must be associated with variation in fields in directions perpendicular to ˆz. In the cylindrical coordinate system, this is the ˆρdirection, hence the subscript “ρ. ” W e shall see later that kρ plays a special 3 Students are encouraged to derive these for themselves...
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110 CHAPTER 6. W A VEGUIDES 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 ˜E and ˜H in the direction of propagat...
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– either ˜Ez or ˜Hz must be non-zero and kρ will be non-zero. In this case, Equations 6.136–6.139 are both usable and useful since they allow determination of all field components given just the z components. In fact, we may further exploit this simplicity by taking one additional step: Decomposition of the unidirection...
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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 TM modes. Figure 6.6 shows the geometr...
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6.8. RECT ANGULAR W A VEGUIDE: 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 demon...
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