Transmission Lines, Waveguides, and Resonators

Transcription

1 Chapter 7 Transmission Lines, Waveguides, and Resonators General Properties of Guided Waves 7.. TM, TE, and TEM Modes 7.3. Coaxial Lines 7.4. Two-Wire Lines 7.5. Parallel-Plate Waveguides 7.6. Rectangular Waveguides 7.7. Cavity Resonators Transmission of power or signal through the free space is convenient. However, the drawbacks are that the radiated power is limited, the propagation path may be disturbed, the signal may be interfered, and frequency can not be reused. For point-to-point transmission of power or signal, such as in power distribution, telephone, CATV, and network, some man-made media are helpful for efficient transmission. The common transmission media are two-wire line, parallel-plate line, coaxial line, optical fiber, and waveguide. All these five media have been used in network. Two-wire line and coaxial line are used in local ethernet; parallel-plate lines are used in IC and printed-circuit boards; optical fiber is used in the main bone of network as FDDI (Fiber Distributed Data Interface); and waveguides are used in the feeding of the dish antenna for overseas transmission through satellite. Except optical fiber where total internal reflection is used, the guidance is made possible by utilizing the reflection from metal surfaces. At microwave frequencies, the hollow metallic pipes are the preferred media for the guidance of electromagnetic wave. This is because that the radiation loss, the metallic loss (associated with the increase of surface resistance due to the skin depth effect), and the dielectric loss can be avoided or reduced. For wave propagating in a waveguide, the frequency should be high enough such that the size of the cross section is at least larger than one-half wavelength General Properties of Guided Wave Consider a time-harmonic wave propagating in a source-free medium whose permeability µ is a constant. As in the preceding Chapter, one has E(r) = jωµ H(r), by taking curl to both sides of Faraday s law. Then, from Ampere s law and a vector identity, one has the Helmholtz equation E(r) E(r) + ω µεe(r) = 0. Further, suppose that the permittivity is also a constant. Thereby, the Helmholtz equation becomes a simpler form of E(r) + k E(r) = 0, where k = ω µε.

2 em7 Guided modes and cutoff frequency Suppose the medium is invariant along a particular, say z, direction. Then the field can be decomposed into spatial harmonics in the axial z direction. That is, the variation of wave along the axial z direction is space-harmonic with a propagation constant k z as E(x, y, z, t) = Re{E(x, y)e jkzz e jωt }. Then, in a source-free and homogeneous region, the Helmholtz equation of field E becomes t E(x, y) + k t E(x, y) = 0, where the transverse propagation constant k t is given by k t = k k z and the two-dimensional transverse Laplacian t = / x + / x. guiding structures and guided modes The guiding structures considered in this Chapter are composed of metallic conductors of arbitrary cross sectional shape invariant in the z direction. The conductor can be single or multiple. For a single conductor forming a tube, the field propagates within the tube. On the other hand, for multiple conductors, the field propagates between the tubes. The material within the metallic tube or between the tubes is supposed to be homogeneous. Subjecting to the boundary conditions, it is found that only some particular discrete values of the transverse propagation constant k t can lead to solutions of the Helmholtz equations, in the absence of any incident field for the two-dimensional problem. And these discrete values of k t are known as the eigenvalues. Each eigenvalue corresponds to an eigenfunction having a particular distribution of electromagnetic field called a mode. The fields corresponding to a pair of eigenvalue and eigenfunction interfere constructively in the guiding structure and will survive in the absence of incident field without changing its field pattern during the propagation. The eigenvalue k t is independent of ω. The propagation constant k z of a guided wave can be pure imaginary and hence the wave becomes evanescent in the axial z direction, if the angular frequency ω is lower than a critical value ω c, where ω c = k t µɛ. Thus, for a given mode with an eigenvalue k t, the lower bound for the angular frequency is ω ω c. The corresponding frequency f c is then known as the cutoff frequency of that mode. Ordinarily, different modes have different cutoff frequency and field distributions. In terms of the cutoff frequency, the two-dimensional Helmholtz equation becomes t E(x, y) + ω c µɛe(x, y) = 0. For different guiding structures, the solutions for ω c and the corresponding modal field tend to be different. For some structures it is possible to get a solution for the situation of zero cutoff frequency ω c = 0. Then the Helmholtz equation reduces to the Poisson equation t E(x, y) = 0.

3 em7 3 It will be shown that the corresponding mode is known as the TEM mode. Dispersion of guided waves phase constant and guided wavelength For a lossless waveguide, if the operating frequency is higher than the cutoff frequency, then propagation constant k z is real. Let the real k z be designated as β. The phase constant is related to frequency as β = (ω ω c )µɛ = k k t. If the cutoff frequency ω c is not equal to zero, the relation between the phase constant β and the frequency ω becomes dispersive and the ω-β curve is known as the dispersion diagram. The phase constant can be rewritten as β = n g k, where ( ) kt n g = 1 = 1 k ( ) ωc ω is called the waveguide obliquity factor (or waveguide factor, for short) which depends on the ratio of the cutoff frequency to the operating frequency. For the TEM mode with ω c = 0, n g = 1 and β = k. Otherwise, the waveguide factor increases from zero toward unity as the frequency is increasing from the cutoff frequency. That is, 0 n g < 1 when 0 < ω c ω. At the cutoff frequency f = f c, the waveguide factor n g = 0. (It is noticed that the waveguide factor n g looks similar to the refractive index of a plasma.) The corresponding guided wavelength λ g is λ g = π β = λ n g λ, where λ = π/k is the wavelength in the unbound space of the same µ and ɛ. For the TEM mode with ω c = 0, λ = λ g. If ω c > 0, then λ g > λ. In terms of wavelength, the relation k = β + k t can be rewritten as where the cutoff wavelength λ c = π/k t. 1 λ = 1 λ g + 1, λ c phase and group velocities Note that the phase speed in a guiding structure is given by v p = ω β = 1 µɛng 1 µɛ. The phase speed is always faster than that in the unbounded space of the same µ and ɛ, if ω c > 0. Note also that the phase speed decreases with increasing frequency. The group speed is given by v g = dω dβ = 1 µɛ n g 1 µɛ, where βdβ = µɛωdω has been made used of, The group speed is always slower than that in the unbounded space of the same µ and ɛ, if ω c > 0. The group speed is always slower than the phase

4 em7 4 speed. Thus a guiding structure is normally dispersive medium. It is seen that the phase speed decreases and the group speed increases with increasing frequency. Note that v g v p = 1 µɛ. Then the group speed is related to the phase constant as v g = 1 µɛω β. From the dispersion relation and the preceding relation, we have (ω ω c )µɛ = (ωµɛv g ). Thus, in terms of the group speed v g, the frequency is related to a nonzero cutoff frequency as ω = ω c 1 v g µɛ. For the case with µ = µ 0 and ɛ = ɛ 0, this relation becomes ω = ω c 1 v g /c. It is noted that by replacing ω c with ω p, this relation between angular frequency and group speed reduces to that in plasma. 7.. TM, TE, and TEM Modes Depending on the axial field components, a guided mode in a homogeneous medium can be categorized into TM, TE, and TEM modes. In a TM (transverse magnetic to the propagation ẑ direction) or TE (transverse electric) mode, the axial component of the magnetic or electric field is zero (H z = 0 or E z = 0), respectively. The TEM (transverse electric and magnetic) mode is a special case of the TM mode with both E z = 0 and H z = 0. Properties of TM modes Consider the guiding structure in a medium of constant constitutive parameters µ and ɛ. In this homogeneous medium the potentials are given by and Φ(r) + k Φ(r) = 1 ɛ ρ f(r) A(r) + k A(r) = µj f (r), where the charge and current induced in the homogeneous medium are excluded. Consider the case where all the currents are time-harmonic and flow only in the axial direction. Then the vector potential is z-directed, that is, A = ẑa z. For the space-harmonic case, the potential becomes A z (x, y, z) = A z (x, y)e jβz. From the Lorenz gauge A = jωµɛφ, one has Φ(x, y) = β ωµɛ A z(x, y).

5 em7 5 This relationship states that in TM modes, the potential Φ is equal to the potential A z, aside from a multiplying constant. fields of TM modes The associated electromagnetic field can be determined solely from the z-directed vector potential. Explicitly, the fields are given by E(r) = jω [ A(r) + A(r) k ] = jω B(r) = A(r) = ẑ t A z (r), [ ẑa z (r) + jβ ] k ( t jẑβ)a z (r) where the del operator = t jẑβ for an axially space-harmonic field, and A = ẑ A z as (fg) = f g + f g. Evidently, for the present case, the axial component of the magnetic field is zero (H z = 0). Thus the field due to axial currents is TM. Thus the space-harmonic field can be expressed in terms of the potentials. The nonvanishing axial component is E z (x, y) = jω { 1 β k } A z (x, y) = k t jωµɛ A z(x, y) = k t Φ(x, y) jβ Note that the axial component of electric field is linearly proportional to the potential A z or Φ. And the transverse components are E t (x, y) = β ωµɛ ta z (x, y) = t Φ(x, y) H t (x, y) = 1 µẑ ta z (x, y). Note that the relation E t = t Φ is owing to the situation that the vector potential is in the axial z direction only. Thus the transverse component of electric field is determined solely by the scalar potential Φ and is originated from charges. Note that the transverse electric and magnetic fields are simply related by H t (x, y) = ωɛ β ẑ E t(x, y). Thus the transverse electric and magnetic fields are perpendicular to each other. In Cartesian coordinates, the transverse fields become E t (x, y) = β ( ˆx ωµɛ x + ŷ ) A z (x, y) y H t (x, y) = 1 µ ( ˆx y ŷ x ) A z (x, y). In a guided mode, the ratio of one transverse electric field to the perpendicular transverse magnetic field is known as the wave impedance. The wave impedance Z T M in TM modes is given as Z T M = E x = E y = β H y H x ωɛ = 1 v p ɛ = ηn g η. Note that the wave impedance of TM modes tends to be smaller than the intrinsic impedance in the unbounded space filled with the same material of the guiding structure.

6 em7 6 Note that the potentials satisfy the two-dimensional Helmholtz equation [ t + ω c µɛ ] ψ(x, y) = 0. where ψ = A z or Φ. The boundary conditions are that the tangential components of the electric field vanishes on the conductors surfaces. As shown in a previous Chapter, this boundary condition is due to the fact that the electric field vanishes inside a conductor and the tangential component of electric field is continuous. As the tangential component of E t along the conductors vanishes, the relation E t = t Φ leads to that the potential difference between any two points on a normal cross section of a conductor is zero. In other words, the potential Φ is constant on each normal cross section of each separate conductor s surface. Thereby, the boundary condition on field is equivalent to the constant-potential condition. Properties of TE modes Note that in the TM mode H z = 0 and E z 0. A dual situation is that H z 0 and E z = 0. This kind of guided modes will be shown to be possible and are known as the TE modes. From Faraday s law, it can be shown that for axially space-harmonic TE fields, the relationship between E x and H y and that between E y and H x become as simple as { βey = ωµh x βe x = ωµh y, where E z = 0 has been made use of. Similar to the TM modes, the transverse electric and magnetic fields are simply related by H t (x, y) = β ωµẑ E t(x, y). Thus the transverse electric and magnetic fields are perpendicular to each other both in TM and TE modes. Then, from Ampere s law, it can be shown that the transverse fields are given by H x = j β kt H z x H y = j β k t H z y. Note that unlike in the TM case, the transverse electric and magnetic fields of the TE case are given in terms of the axial field component, instead of a potential. The wave impedance Z T E in TE modes is given as Z T E = E x H y = E y H x = ωµ β = v pµ = η n g > η. Note that the wave impedance of TE modes is larger than the intrinsic impedance in the unbounded space filled with the same material of the guiding structure. Further, it is noted that Z T M Z T E = η. Properties of TEM Modes

7 em7 7 Consider a special case of TM modes with zero cutoff frequency ω c = 0. Thereby, β = k and the potentials are then related as Φ(x, y) = 1 µɛ A z (x, y). Thus the scalar potential Φ satisfies the two-dimensional Laplace equation t Φ(x, y) = 0. Note that this equation is similar to that in the source-free electrostatic case. Recall the relation E z = (k t /jβ)φ in the TM case. Then it is seen that the axial field E z as well as H z becomes zero and the corresponding field is called TEM. The electromagnetic field of the TEM mode can be given in terms of the scalar potential Φ by E t (x, y) = t Φ(x, y) H t (x, y) = 1 η ẑ tφ(x, y). It is seen that the transverse electric and magnetic fields are related as H t (x, y) = 1 η ẑ E t(x, y). Thus the wave impedance of TEM modes is given as Z T EM = E x = E y µ = H y H x ɛ = η, which is identical to the intrinsic impedance of the unbounded medium. guiding structures for TEM mode As in a TM mode, the potential Φ in a TEM mode is a constant on a normal cross section of a single conductor. A transmission line that supports a TEM wave must consist of at least two isolated conductors over which the electric scalar potential Φ can be made to be of different values. Otherwise, a trivial solution Φ(x, y) = constant is obtained for the Laplace equation. Thus a single-conductor structure can not support the TEM mode. Further, to support a truly TEM mode, the conductors should be perfect. Since, if the conductors are not perfectly conducting, a longitudinal component of electric field is needed to drift an electric current. Thus the mode is not a true TEM mode. The resistance of the transmission lines is approximately proportional to the inverse of the penetration depth, which in turn is inversely proportional to the square root of frequency. Thus a true TEM mode becomes more difficult at higher frequencies. voltage and current in TEM mode Since the potential Φ in a TEM mode is a constant on a normal cross section of a single conductor, the voltage across two separated conductors of a transmission line can be unambiguously defined as a function of the axial position z. Remark that the condition of constant-φ on a normal cross section is a general property of TM modes. However, it does not hold in TE mode. This is because that the boundary condition

8 em7 8 E ˆl = 0 implies that ( l Φ + jωa) ˆl = 0, where ˆl denotes a unit vector tangential to conductor s surface on a transverse plane. As the transverse component A ˆl is not necessarily equal to zero in TE mode, l Φ 0 in general. Remark that the current density on the surface of a perfect conductor is given by the boundary condition J s = ˆn H. In a TM mode, the total current I flowing over a normal cross section enclosing all the conductors is then given by I = J s ẑdl = [ˆn H t ] ẑdl, C where C denotes the circumference of the conductors on a normal cross section. By expressing field H t in terms of the potential A z and by using vector identities, the divergence theorem, and the Helmholtz equation of the vector potential, it can be shown that the current is proportional to the cutoff frequency. Thus, the total current is zero for the TEM mode, while it is not zero for a TM mode in general. For a TE mode, the current itself even can not be defined. Thus, for a two-conductor transmission line of arbitrary shape, the currents on each conductor are of the same magnitude but of opposite signs. That is, the current distribution in the two-wire TEM mode is simply (1 1). Thus the current flows into one conductor will be equal to that flows out from the other conductor. Consequently, the current flowing in each of the two conductors can be unambiguously defined with a single value at a particular axial position z. Since both voltage and current can be unambiguously defined, the impedance can be unambiguously defined with a single value at a particular axial position. power flux in TEM mode Since both voltage and current can be defined in a TEM mode, the power P flowing along the axial z direction over a transverse cross section can be determined in these quantities. Starting from the complex Poynting vector, the power P flowing along the axial z direction in a TEM mode is given by P = 1 E H ẑds, S where the integration area S is the transverse plane excluding the conductors and their surfaces. As E z = 0 and A t = 0, E = t Φ. Then, by expressing the electric field in terms of the scalar potential Φ and by using a vector identity and Stokes s theorem, it can be shown that for a TEM mode on multiple conductors, the power flux P is given as C P = 1 Φ i Ii. i For a two-wire line, the power flux can be simplified to be P = 1 (Φ 1 Φ )I 1 = 1 V I. where I is the longitudinal current on the conductor at the higher potential. Circuit-model of TEM modes For a TEM mode, both current and voltage can be defined with a single value at each axial position z. Thus a circuit model can be used. The equivalent circuit of a differential segment of a lossless transmission line is composed of a serial inductance and a shunt capacitance. The

9 em7 9 inductance is due to the currents of opposite signs flowing on the two wires. The inductance in turn consists of a self inductance of a single straight segment of conducing wire and a mutual inductance due to a corresponding segment of wire on the other line. The capacitance is due to the charges of opposite signs accumulated on the two wires. The distributed parameters can denoted as L and C which are the inductance and capacitance per unit length, respectively. A circuit analysis in time domain yields that Let z 0, one has i(z, t) v(z + z, t) + v(z, t) = L z t i(z + z, t) + i(z, t) = C z v(z, t) z i(z, t) = L t i(z, t) v(z, t) = C, z t which are known as the transmission-line equations. In time-harmonic, the equations become dv (z) dz di(z) dz = jωli(z) = jωcv (z). v(z, t). t These two coupled first-order differential equations can be combined into two uncoupled secondorder differential equations. That is, where d V (z) dz + k V (z) = 0 d I(z) dz + k I(z) = 0, k = ω LC. It is seen that voltage v and current i propagate along the wire with a spatial variation similar to the one of a plane wave propagating in a uniform medium. From the first-order differential equation, it is easy to show that Z 0 = V (z) I(z) = ωl k = k L ωc = C, where Z 0 is known as the characteristic impedance of a lossless transmission line. Note that the electric field associated with a spatial variation of the voltage is perpendicular to the direction of the current in TEM mode. Thus the characteristic impedance has nothing to do with power dissipation. This situation is different from that in a resistor, where the electric field is parallel to the current Coaxial Lines

10 em7 10 TEM modes in coaxial lines Consider coaxial lines consisting of two concentric cylinders of conductors of radii a and b (a < b), the space between them is filled with a homogeneous dielectric. The outer conductor is grounded and the inner conductor is connected to a voltage source with output voltage V. Consider the TEM mode in coaxial lines. Suppose the induced current and field are rotationally symmetric and hence do not depend on azimuthal angle ϕ. The two-dimensional Laplace equation of the potential, t ψ(x, y) = 0, then further reduces to the one-dimensional ordinary differential equation. That is, ( 1 d ρ d ) ψ(ρ) = 0. ρ dρ dρ where ψ = A z /µ = Φ/η. The solution for ψ is given by ψ(ρ, ϕ, z) = (c 0 + c 1 ln ρ)e jkz, where c 0 and c 1 are arbitrary constants. The constant c 1 should not be zero; otherwise, the solution is trivial. Putting c 1 = I 0 /π and c 0 = 0, the function Φ = (I 0 /π)η ln ρ e jkz. It is seen that the solution satifies the constant-potential condition on the conductors surfaces. Without the assumption of azimuthal invariance, the Laplace equation is expected to have no solution of TEM mode. The corresponding field distributions are then E ρ (ρ, ϕ, z) = d dρ Φ(ρ, ϕ, z) = η I 0 H ϕ (ρ, ϕ, z) = 1 η πρ e jkz d dρ Φ(ρ, ϕ, z) = I 0 πρ e jkz. Such a field distribution is permissible if ρ 0; that is, there must exist a nonvanishing inner cylinder. Again, it is seen that two separated conductors are necessary to support such a TEM wave. Note that the electric field is perpendicular to the surface everywhere and that the electric field is orthogonal to the magnetic field. Such an electric field is due to the effect of charge only; the contribution of current has been cancelled out. Note that the transverse fields are related to each other in the simple form of E ρ (ρ, ϕ, z) H ϕ (ρ, ϕ, z) = η, which is just the wave impedance of the TEM mode obtained previously. current and charge densities It is of conceptual importance to note that prior to knowing the distribution of the longitudinal current, one can determine the distribution of the potential A z and hence the field distributions by finding out the possible solution of the associated differential equation subjecting to the boundary conditions. This is a great success of mathematics in dealing with the eigenvalue problems of which the solutions for the potential or field must be some particular distributions corresponding to constructive interference in the guiding structure. The associated current which generates the vector potential can appear only on the conductors surfaces. The surface current density in turn can be determined from the permissible magnetic field

11 em7 11 distribution by the boundary condition. That is, J sf = ˆn H on conductors surfaces ẑ I 0 πb e jkz ρ = b = ẑ I 0 πa e jkz, ρ = a where ˆn denotes a unit vector outward normal to the respective conductor s surfaces. Note that the current density is in the axial z direction, which is self-consistent with the earlier assumption of A = ẑµψ. The total current I(z) on each conductor is equal to ±I 0 e jkz. Unlike the current in a lumped circuit, the current varies along the conductors with the space harmonic e jkz. The corresponding surface charge density in turn can be determined from the permissible electric field distribution by the boundary condition. That is, ρ sf = ɛˆn E on conductors surfaces µɛ I 0 πb e jkz ρ = b = I 0 µɛ πa e jkz. ρ = a Note that the surface current and charge densities are related to each other in the simple form of J sf (ϕ, z) ρ sf (ϕ, z) = 1. µɛ This ratio is just the propagation speed of electromagnetic wave in the homogeneous medium. voltage The scalar potential Φ is given as Φ(ρ, ϕ, z) = (I 0 /π)η ln ρ e jkz, which does not depend on azimuthal angle ϕ. Thus the voltage V, defined as the potential difference Φ between the inner and outer conductors, can be given unambiguously and is then meaningful. That is, V (z) = Φ(a,z) Φ(b,z) = η I ( 0 ln b ) e jkz. π a This result agrees with the following path integral of the electric field a V (z) = E ρ (ρ, ϕ, z)dρ. b The variation of voltage V (z) along the axial z direction, as well as that of current I(z), is the space harmonic e jkz. characteristic impedance The ratio V (z)/i(z) is simply a constant Z 0, which is known as the characteristic impedance of the coaxial lines and is given as Z 0 = V (z) I(z) = η π ln b a.

12 em7 1 Note that this impedance is independent of the operating frequency. For example, consider the coaxial lines with µ = µ 0, ɛ = ɛ 0, a = 0.1 in, and b = 0.36 in. The characteristic impedance is given as Z 0 = 50.1 Ω. So, this is called a 50-Ω line. Coaxial lines of a much higher characteristic impedance will be difficult. For example, for coaxial lines of characteristic impedance as high as Z 0 = 300 Ω, the ratio b/a becomes as large as (3.6) 6 = 100. power flux Since the electric field is orthogonal to the current density, it is obviously that E J = 0. Thus the TEM mode in coaxial lines does not absorb power. The complex Poynting vector is in the axial direction given by S(ρ, ϕ, z) = ẑ 1 E ρ(ρ, ϕ, z)h ϕ(ρ, ϕ, z) = ẑ η 8π I 0 1 ρ, which is independent of ϕ and z. The total power flux over a cross section is P(z) = ẑ S(ρ, ϕ, z)ρdρdϕ = η 4π I 0 ln b a. It is noted that the power flux is independent of z, which reflects zero absorption power. In terms of the characteristic impedance Z 0, the power flux can be written as P(z) = 1 I 0 Z 0. It is seen that the power flux is proportional to the characteristic impedance. In other words, the characteristic impedance corresponds to the power flux, rather than to the power absorption Two-Wire Lines TEM mode in two-wire transmission lines The two-wire transmission line consists of two parallel conducting cylinders separated by a distance. Again, we consider the TEM case. Suppose the space between these two conductors is filled with a homogeneous medium, where the potential Φ satisfies the two-dimensional Laplace equation t Φ(x, y) = 0. The boundary condition is that the tangential component of the electric field vanishes on the conductors surfaces, which equivalent to the potential Φ should be constant on the respective surfaces. Consider a special case where the conducting cylinders are circular with identical radius a and these cylinder separated by a distance h (center to center). It is not easy to obtain the solution for Φ without a trick, even for this special case. Here, we just write down the result which reads Φ(x, y, z) = I 0η 4π ln (x + d) + y (x d) + y e jkz, where d = h a. It is easy to verify that this potential satisfies the two-dimensional Laplace equation. Then it is also easy to show that on the circle given by (x h) + y = a, (x + d) + y (x d) + y = (x + d) (x h) + a (x d) (x h) + a = h + d h d,

13 em7 13 which is a constant. Thus the potential is a constant over a transverse cross section of the corresponding conductor s surface, as required by the constant-potential condition. Similarly, over a cross section of the other conductor s surface given by the circle of (x + h) + y = a, the potential Φ is also a constant associated with ln(h d)/(h + d). Thus the voltage, defined as the difference in the scalar potential between these two conductors, is given by V (z) = I 0η π ln h + d h d e jkz = I 0η π ln h + d a e jkz. A formal procedure to evaluate the current on the conductor is to evaluate the magnetic field from the potential and then to evaluate the surface current density from the magnetic field. It is may be a tedious work to show that the total current on the respective conductors is equal to ±I 0 e jkz. A trick for evaluating the current is given in what follows. It is noted that the potential can be rewritten as Φ(x, y, z) = I 0η π { } ln (x d) + y + ln (x + d) + y e jkz, which is just the potential due to two equivalent line currents ±I 0 e jkz placed at (d, 0) and ( d, 0), respectively. From Ampere law the axial current is given by a closed path integral of H l, as E z = 0. Since the potential and hence the magnetic field are identical to those due to the image currents, the total current on the respective conductors is then equal to the image current ±I 0 e jkz. Thus the characteristic impedance of a two-wire line is then given by the ratio Z 0 = V (z) I(z) = η π ln h + d a, which is a constant independent of z and ω. For example, consider a two-wire line with µ = µ 0, ɛ = ɛ 0, a = 0.8 mm, and h = 1 cm. The characteristic impedance is then given as Z 0 = 300 Ω. Further, we consider the potential due to the pair of the image line currents. It is seen that the potential is a constant value given by Φ(x, y) = (I 0 η/π) ln K at all those points whose coordinates (x, y) satisfy the relation (x + d) + y (x d) + y = K, where K is an arbitrary positive constant. It can be shown that for a given K, the solutions form a circle of radius a and centered at (h, 0), where a = d K K 1 h = d K + 1 K 1. It is easy to show that d = h a, regardless of the K value. If K > 1 (Φ > 0), both a and h are positive and decrease with increasing K value. Further, as K, then a 0 and h d. On the other hand, as K 1 + (Φ 0 + ), then a, h, which is practically the y axis. Conversely, the parameter K can be determined from the radius a and center position h. It is straightforward to shown that K = (x + d) + y (x d) + y = (x + d) (x h ) + a (x d) (x h ) + a = h + d h d = h + d a,

14 em7 14 It is noted that these two solutions for K is reciprocal to each other. The values of K correspond to the circles with (h, 0) and ( h, 0) are reciprocal to each other. Thus there is a change in sign in the potential Φ. That is, if the potential is equal to a positive constant (I 0 η/π) ln K over the circle of radius a and centered at x = h, then the potential is equal to the negative constant (I 0 η/π) ln K over the circle of identical radius but centered at x = h. Remark that the transverse fields are given by E t (x, y) = t Φ(x, y) H t (x, y) = 1 η ẑ tφ(x, y). The electric field E is orthogonal to these equipotential circles. The magnetic field H coincides with the circles, since field H is orthogonal to field E. The surface current density is given by the boundary condition as J sf (x, y, z) = ˆn ˆlH l = ẑ ψ(x, y) e jkz, n where H l is the transverse component tangential to the conducting cylinder surface and ψ = A z /µ = Φ/η. The total current on one conductor is given by the closed contour integral I(z) = e jkz C ψ(x, y)dl, n where C is the circumference of the conductor on a normal cross section. It seems not easy to carry out this integration directly Parallel-Plate Waveguide Consider a parallel-plate structure composed of two conducting plates separated by a vertical distance h (in the y direction). The space between the plates is filled with air or a lossless homogeneous medium with constant parameters µ and ɛ. Suppose the plate is sufficiently wide (in the x direction) and hence the field has no variation in the x direction. That is, all the spatial variation on the horizontal plane is attributed to the axial z direction. The spatial variation of the field may be assumed to be ψ(x, y, z, t) = ψ(y)e jβz e jωt. TEM mode We first consider the TEM mode where β = k. The governing equation is the one-dimensional ODE d ψ(y) = 0, dy where ψ = A z /µ = Φ/η. Obviously, a general solution of the above equation is ψ(y) = c 0 + c 1 y. where c 0 and c 1 are arbitrary constants. The constant c 1 should not be zero; otherwise, the solution is trivial. Due to symmetry, the potential should be zero at the plane (y = 0) midway between the two plates. Then putting c 1 = J s0 and c 0 = 0, the potential is given by Φ(x, y, z) = J s0 ηy e jkz.

15 em7 15 The corresponding field distributions are E y (x, y, z) H x (x, y, z) = d Φ(x, y, z) dy = J s0ηe jkz = 1 d η dy Φ(x, y, z) = J s0e jkz. It is seen that the ratio of E y (x, y, z) to H x (x, y, z) is the wave impedance η. The associated current which generates the vector potential can appear only on the conductor s surfaces. The surface current density in turn can be determined from the permissible magnetic field distribution by the boundary condition. That is, J sf = ˆn H on conductors surfaces ẑj s0 e jkz top = ẑj s0 e jkz, bottom where ˆn denotes a unit vector outward normal to the respective conductor s surfaces. The currents on the top and the bottom plates are then given as I(z) = ±J s0 we jkz. The corresponding surface charge density in turn can be determined from the permissible electric field distribution by the boundary condition. That is, ρ sf = ɛˆn E on conductors surface µɛjs0 e jkz top = µɛj s0 e jkz. bottom Again, it is noted that the surface current and charge densities are related by J sf (x, z) = ρ sf (x, z)/ µɛ. The voltage, defined as the difference of the scalar potential between the top and the bottom plates separated by a vertical distance h, is given by V (z) = J s0 ηh e jkz. This result agrees with the path integral of the electric field. The characteristic impedance of the parallel plates of width w and separation distance h is then given by Z 0 = V (z) I(z) = η h w, which is independent of the operating frequency. As expected intuitively, the characteristic impedance decreases with increasing width. In a printed-circuit board (PCB), the whole bottom of the substrate is coated with a metal film as the ground plane. Meanwhile, metal strips were formed on the top of the substrate and can be used for transmission of power or signal. By using the image current, the structure of a strip and a ground can be replaced with the structure of two parallel strips of vertical separation d, where d is the thickness of the substrate. The characteristic impedance of the grounded strip of width w is given as Z 0 = η d w. The factor of cancels, since the voltage here is given by the potential at the top plate with respect to the ground, which is just one half of the plate-to-plate voltage. For example, consider the case

16 em7 16 where the material of the substrate has parameters µ = µ 0 and ɛ = ɛ 0 and the thickness d of the substrate is 1 mm. To make the metal strip matched to 50-Ω coaxial lines, the required strip width is 5.33 mm. TM and TE modes For the parallel-plate structure, TEM mode is not the only candidate for wave propagation. Both TM and TE modes are possible. For these two kinds of modes, the governing equation is [ ] d dy + k t ψ(y) = 0. where k t = k β and ψ = E z and H z for TM and TE modes, respectively. Remark that E z = (k t /jβ)φ = (k t /jωµɛ)a z for TM modes. A general solution of the above equation for both TM and TE modes is ψ(y) = a n sin k t y + b n cos k t y. Let the two plates be positioned at y = 0 and b, respectively. Thus the vertical distance h is replaced with b. TM mode For TM modes, ψ corresponds to E z, which is a tangential component to the plates. Thus, in TM modes, the boundary conditions are that ψ = 0 at y = 0, b. This is possible if the coefficient b n = 0 and the eigenvalue k t is given by k t = nπ b. n = 1,, Note that the index n can not be zero; otherwise, it corresponds to a trivial solution. Note also that the eigenvalue or the cutoff frequency depends on the size of the guiding structure. The corresponding eigenfunction for TM modes is ( ) nπ E z (y) = a n sin b y, where a n is an arbitrary constant which can not be determined by the eigenvalue problem and are determined physically by the actual excitation. A larger eigenvalue corresponds to a stronger spatial variation in the field distribution of a higher-order mode. The other nonvanishing field components can be determined from E z. That is, E y (y) = j β k t d dy E z(y) = j βb nπ a n cos nπ b y H x (y) = j ωɛ d kt dy E z(y) = j ωɛb nπ a n cos nπ b y. It is noted that the transverse field components have a phase difference of ±90 with respect to the axial component. Thus the peaks of axial and transverse components of fields will appear at different positions. The wave impedance of TM mode is Z T M = E y /H x = β/ωɛ = ηn g, which agrees with the result obtained previously. The complex power flux density is given as S(y, z) = ẑ 1 E y(y, z)h x(y, z) + ŷ 1 E z(y, z)h x(y, z) = ẑ 1 βωɛ a kt n cos k t y ŷj 1 ωɛ a n sin k t y cos k t y, k t

17 em7 17 which is independent of z. Note that due to the 90 phase difference between the axial and the transverse components of fields, the transverse component of the Poynting vector is pure imaginary. Thereby, the time-average power flux along the transverse direction vanishes. Physically, this is due to the total reflection from the perfect conducting plates. TE modes For TE modes, the only possible tangential component of electric field is E x. Then the boundary condition that E x vanishes on the conductor surfaces leads to dψ(y)/dy = 0 at y = 0, b. This is possible if the coefficient a n = 0 and the eigenvalue k t is identical to that of the corresponding TM mode. The corresponding eigenfunction for TE mode is H z (y) = b n cos nπ b y, where b n is an arbitrary constant determined by the actual excitation. The other nonvanishing field components can be determined from H z. That is, H y (y) = j β k t E x (y) = j ωµ k t d dy H z(y) d dy H z(y) = jb n βb nπ sin nπ b y = jb n ωµb nπ sinnπ b y. In TE modes the transverse field components also have a phase difference of ±90 with respect to the axial component. The wave impedance of TE mode is Z T E = E x /H y = ωµ/β = η/n g, which agrees with the result obtained previously. The complex power flux density is given as S(y, z) = ẑ 1 E x(y, z)h y(y, z) ŷ 1 E x(y, z)h z (y, z) = ẑ 1 βωµ b kt n sin k t y + ŷj 1 ωµ b n sin k t y cos k t y, k t which is independent of z. Like the TM modes, the transverse component of the Poynting vector is pure imaginary and the time-average power flux along the transverse direction vanishes. phase constant and cutoff frequency The phase constant of both TM n and TE n modes is ( ) nπ. β = k kt = ω µɛ b If the frequency is not high enough such that the condition ω µɛ > nπ b, is not met, the propagation constant β will be pure imaginary. Thereby, the field will decay exponentially in the axial direction. Thus the minimum frequency at which electromagnetic wave can propagate along the waveguide via the TM n or TE n mode is given by f cn = n b µɛ.

18 em7 18 The modes having the lowest cutoff frequency tend to be easier to be excited and are known as the fundamental modes. In parallel plates, the fundamental modes are the TM 1 and TE 1 modes. The wavelength of the wave to be guided via the TM or TE modes should be short enough that λ < λ c = 1/f c µɛ. The cutoff wavelength of both the TMn and TE n modes is given as λ cn = 1 = b f cn µɛ n. Thus, for a wave to be guided in parallel plates, its wavelength in the corresponding unbounded medium should be at least shorter than b. Note that the cutoff wavelength is independent of the constitutive parameters. Consider the example where the distance b is 5 mm and the space between the plates is supposed to be air. Then the minimum frequency would be as high as 30 GHz. By using a thicker b or by using a material of higher permittivity or permeability, the required frequency can be lowered. The phase constant β determines the propagation property in the axial z direction. In terms of the waveguide factor n g, β is given as β = kn g. The waveguide factor n g is turn is determined by the eigenvalue k t as n g = 1 k t k = 1 (nπ/b) ω µɛ. As β < k, the guided wavelength λ g is longer than the wavelength in the corresponding unbounded medium by a factor of 1/n g. Decomposition of modal wave in parallel-plate waveguide In what follows, a physical interpretation of the variation in the phase constant, the guided wavelength, the phase speed, the group speed, and the wave impedance in TM and TE modes is given. Recall that the electric field distribution of TM n mode in a parallel-plate waveguide with the two plates separated by a vertical distance b is known as E z (x, y, z) = a n sin k t y e jβz E y (x, y.z) = jβ k t a n cos k t y e jβz, where k t = nπ/b and β = k kt = k 1 fc /f. These field components are standing waves in the y direction and can be decomposed as E z (y, z) = ja n E y (y, z) = ja n { e jk ty e jkty} e jβz β k t { e jk t y + e jk ty } e jβz. Apparently, the modal electric field can be represented by two waves, each of them traveling both in y and z directions as E(y, z) = ja n 1 k t { (ŷβ + ẑkt )e jkty e jβz + (ŷβ ẑk t )e jkty e jβz}.

19 em7 19 These two waves propagate at the direction of ŷk t ẑβ and ŷk t + ẑβ, respectively, which are perpendicular to the respective polarizations. It is seen that each individual wave does not actually propagate along the axial z direction, but along a direction with a tilt angle θ g from the z axis, where the waveguide obliquity angle θ g is given as θ g = cot 1 β k t = cos 1 β k = sin 1 k t k = sin 1 f c f. Note that the waveguide obliquity factor n g is related to obliquity angle θ g and can be given in several ways as ( ) ( ) fc n g = cos θ g = 1 λ = 1 = f λ c Thus, just owing to the obliquity in propagation, it is seen that 1 ( kt k ) = β k. β = k cos θ g, λ g = λ/ cos θ g, v p = v/ cos θ g, v g = v cos θ g, and α d = α/ cos θ g. In words, β is merely the z component of k. And, for the same reason, the wave impedance Z T M for TM modes becomes Z T M = E y(y, z) H x (y, z) = E cos θ g H = η cos θ g = η β k. And the power flow and power loss will also depend on the waveguide factor. Similar situations due to obliquity hold for TE n modes of which the electric field is H z (x, y, z) = b n cos k t y e jβz H y (x, y, z) The wave impedance Z T E for TE modes becomes It is noted that Z T E = E x(y, z) H y (y, z) = = jb n βb nπ sin k ty e jβz E H cos θ g = Z T M Z T E = η. This result has been obtained earlier in this Chapter. η cos θ g Rectangular Waveguides Compared to the parallel-plate waveguide, a tube structure is better in field confinement. Consider a waveguide composed of a metallic tube of rectangular shape of size a b. The space within the tube is filled with air or a homogeneous medium with uniform µ and ɛ. As all the metallic walls are connected, no TEM mode can propagate along a waveguide. The spatial variation of the field may be assumed as ψ(x, y, z, t) = ψ(x, y)e jkz e jωt, where ψ = E z and H z for TM and TE modes, respectively. Thus, for these two kinds of modes, the governing equation is the two-dimensional Helmholtz equation as { x + y + k t } ψ(x, y) = 0.

20 em7 0 By using the method of separation of variables, a general solution of the above equation can be given as ψ(x, y) = (a n sin k x x + b n cos k x x)(c n sin k y y + d n cos k y y), where k x + k y = k t. Thus the eigenvalue k t or the cutoff frequency f c is determined by the field distribution over the transverse x-y plane. TM modes For TM modes, the boundary conditions are that ψ = 0 on all the four metallic walls at x = 0, a and y = 0, b. This is possible if the coefficients b n = d n = 0 and if the eigenvalues k x and k y are given as k x = mπ a m = 1,, k y = nπ. n = 1,, b Thus ( ) mπ ( ) nπ kt = +. a b The corresponding eigenfunction for TM mn mode is E z (x, y) = E 0 sin mπ a x sin nπ b y, mn 0 where E 0 is an arbitrary constant determined by the actual excitation. Note that neither m nor n can be zero. The other nonvanishing field components are β mπ mπ E x (x, y) = je 0 cos kt a a x sin nπ b y β nπ E y (x, y) = je 0 kt b nπ H x (x, y) = je 0 ωɛ k t H y (x, y) = je 0 ωɛ k t b mπ a sin mπ a x cos nπ b y mπ sin a x cos nπ b y cos mπ a x sin nπ b y. It is noted that the transverse field components have a phase difference of ±90 with respect to the axial component. TE modes For TE modes, the boundary conditions are the tangential component E y vanishes on the two metallic walls at x = 0, a and the other tangential component E x vanishes on the other two metallic walls at y = 0, b. These boundary conditions lead to that ψ(x, y)/ x = 0 at x = 0, a and ψ(x, y)/ y = 0 at y = 0, b. This is possible if the coefficients a n = c n = 0 and if the eigenvalues k x and k y are identical to those of the corresponding TM modes. The corresponding eigenfunction for TE mn mode is H z (x, y) = H 0 cos mπ a x cos nπ b y, m + n 0

21 em7 1 where H 0 is an arbitrary constant determined by the actual excitation. The other nonvanishing field components are ωµ nπ E x (x, y) = jh 0 kt b E y (x, y) = jh 0 ωµ k t mπ a mπ cos a x sin nπ b y sin mπ a x cos nπ b y β mπ mπ H x (x, y) = jh 0 sin kt a a x cos nπ b y β nπ mπ H y (x, y) = jh 0 cos kt b a x sin nπ b y The transverse field components also have a phase difference of ±90 with respect to the axial component. It is noted that except for the coefficients the functional distributions of each transverse field component in TM and TE modes are identical. cutoff frequency and wavelength The propagation constant β of both TM mn and TE mn modes is determined from the relation β = ω µɛ k t = ω µɛ ( ) mπ ( nπ a b The minimum frequency at which an electromagnetic wave can be guided via the TM mn or TE mn mode is given by f cmn = k (mπ ) t π µɛ = 1 ( ) nπ. π + µɛ a b Thus the waveguide acts as a high-pass filter. In other words, the wavelength of the wave that can be guided should be short enough that λ < λ c = 1/f c µɛ. The cutoff wavelength of both TMmn or TE mn mode is given as ). λ cmn = π k t = (m a ) + ( n b ) = a b m n ( ) ( ). a b + m n The cutoff wavelength can be rewritten as ( ) ( ) m ( ) n = +. a b λ cmn Note that the cutoff wavelength λ c is independent of µ and ɛ, while the cutoff frequency f c decreases with increasing µ and ɛ. For the TE 10 mode the cutoff wavelength λ c = a. That is, for supporting an electromagnetic wave with wavelength λ, the waveguide dimension a (> b) must be greater than λ/, where λ the wavelength in the corresponding unbounded homogeneous medium. To be operated in a single mode, the frequency f or wavelength λ should be chosen in such a way that f c10 < f < f c0, f c01

22 em7 or λ c10 > λ > λ c0, λ c01. In term of waveguide dimension a and b, the preceding relations become a > λ > a, b Thus the bandwidth for the single-moded operation is f c0 f c10 when b < a/ or is f c01 f c10 when b > a/. dispersion relation When the frequency is above the cutoff frequency, the propagation constant β for both TM mn and TE mn nodes is given as β mn = (ω ωc )µɛ = kn g. This relation between β and ω is known as the dispersion relation of waveguide. The corresponding guided wavelength of both TM mn and TE mn modes is given by λ mm = π β mn = λ n g. The guided wavelength is longer than that in the corresponding unbounded homogeneous medium. TM or TE modes with different indices m and n tend to have different guided wavelengths. The phase speed is given as v pmn = ω β mn = ω kn g = 1 > 1 µɛng µɛ and the group speed Note that v gmn = dω dβ mn = β mn ωµɛ = v g v p = 1 µɛ. n g < 1. µɛ µɛ TM or TE modes with different indices m and n tend to have different phase and group speeds. The dominant mode in rectangular waveguide Note that for TM mode m 0 and n 0, but for TE modes either m or n can be zero. The TE 10 mode (assuming a > b) has the lowest cutoff frequency and is known as the dominant (fundamental) mode of the rectangular waveguide. For this mode, ( ) π, β = ω µɛ λc = a, and f c = a 1 a µɛ. The nonzero field components of the TE 10 mode are given as functions of space and time as ωµa E y (x, y, z, t) = H 0 π sin π x sin(ωt βz) a βa H x (x, y, z, t) = H 0 π sin π x sin(ωt βz) a H z (x, y, z, t) = H 0 cos π x cos(ωt βz). a

23 em7 3 By using the boundary condition J s = ˆn H, where ˆn is the outward unit normal vector of the metallic walls, the surface current density induced on the metallic walls in the TE 10 mode are given as J s (0, y, z, t) = ŷh 0 cos(ωt βz) J s (a, y, z, t) = J s (0, y, z, t) βa J s (x, 0, z, t) = ẑh 0 π sin π a x sin(ωt βz) + ˆxH 0 cos π x cos(ωt βz) a J s (x, b, z, t) = J s (x, 0, z, t). It is noted that unlike TM modes, transverse components of current density appear. A graph is helpful in understanding the current distribution. The surface charge density can be given from the boundary condition ρ s = ɛˆn E. It is of conceptual importance to note that the sources are determined after the fields are solved from the eigenvalue problem. This is because that the presence of metal provides a rather simple boundary conditions, from which the tangential component of electric field can be determined a priori without knowing the sources on the metal. The knowledge of field distribution is helpful in determining the current and charge distributions and the propagated power. The knowledge of the current distribution in turn is helpful in determining the metallic attenuation and in designing the excitation method and the cutting slots for external coupling. power flux The axial power flux over a cross section of waveguide is given by the integral P = 1 b a 0 0 (E x H y E y H x)dxdy = 1 Z b 0 a 0 H t (x, y) dxdy, where Z = Z T M (= ηn g ) or Z T E (= η/n g ) for TM or TE mode, respectively. For a lossless waveguide, the power flux is independent of z. It can be shown that the power flux for TM and TE modes for with m 0, n 0 are given by η n f g E f 0 ab TM c P = 1 8 ηn f g H 0 ab, TE f c where E 0 and H 0 are the amplitudes of axial component in the respective mode. Meanwhile, for TE m0 modes the power flux is slightly different. That is, P = 1 4 ηn g f fc H 0 ab, TE m0 The power flux increases with increasing waveguide size. It is noted that both in TM and TE modes, the power flux is proportional to the waveguide factor n g, which in turn depends on frequency as n g = 1 (f c /f). Right at the cutoff frequency, n g = 0 and hence the power flux vanishes. Physically, this is due to the wave propagates transversely only. For the fundamental TE 10 mode, the transverse fields are E y = E y0 sin πx and H a x = E y0 (n g /η) sin πx. a Then the power flux is P = 1 b a E y H xdxdy = E y0 ab 0 0 4η n g.

24 em7 4 Again, the power flux is proportional to the waveguide factor n g. power dissipation in lossy waveguides The loss in a rectangular waveguide consists of dielectric loss and metallic loss. Metallic walls of finite conductivity result in power dissipation. The dissipation power P a on the metallic walls per unit length in z is given as P a = 1 S E(x, y) J (x, y)ds = 1 J(x, y) ds. σ S From the propagation of uniform plane wave in a lossy medium, the field and current density are confined in a small region associated with the skip depth δ and the dissipation power can be given in terms of the surface current density. In a waveguide it is supposed that P a = 1 R s C J s (x, y) dl, where R s = 1/σδ = ωµ/σ is the surface resistance (sheet resistance) and the contour C is the metallic boundary on the cross section S. Note that the dissipation power increases with the square root of frequency and decreases with that of conductivity. This is owing to the skin-depth effect. The dissipation power can be given in terms of magnetic field as P a = 1 R s C { Hl (x, y) + H z (x, y) } dl. For TM modes, the current flows in the axial direction. Thus the dissipation power can be given in terms of the transverse magnetic field as P a = 1 R s C H l (x, y) dl. Note that the magnitude of field H l increases with f and the dissipation power increases with f 5/ for a given amplitude E 0 of the axial component of electric field in TM modes. In a lossy waveguide, since both electric and magnetic fields decay as e α cz, the power flux and the dissipation power decay as P(z) = P(0)e α cz. The dissipation power can be given in terms of rate of decrease of the power flux as P a (z) = dp(z) dz = α c P(z). Thus the attenuation constant due to the metallic loss can be given as α c = P a(z) P(z). It is expected that α c is independent of z. However, the attenuation constant depends on frequency. Right at cutoff, P(z) = 0 and hence the attenuation constant α c increases drastically near the cutoff. On the other hand, when the frequency is high, the dissipation power increases with f 5/, while the power flux is nearly proportional to f. Thus the attenuation constant α c increase approximately with f 1/. The frequency dependence of TE modes are more complicated.