Zhang K., Li D. Electromagnetic Theory for Microwaves and Optoelectronics

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3.7 Impedance Transducers 171

Figure 3.25: Reﬂection of wave at two reﬂection points.

characteristic impedance of each section is determined by (3.179), (3.177),

and (3.180). The length of each section is obtained from (3.191) or (3.192):

l =

max

arccos p

2π

, or l =

max

(π − arccos p)

2π

. (3.197)

Note that the wavelength in the above equations is the wavelength in the

medium of the given layer or the guided wavelength in the waveguide.

The calculation in the design of a multi-section transducer or a multi-layer

anti-reﬂection coating is quite complicated, and it can be done by means of

design tables [36, 118], or computers.

3.7.4 The Small-Reﬂection Approach

If the diﬀerence between the wave impedances of the two adjacent media

or two adjacent waveguides is small enough, the reﬂection at each reﬂection

point must be small too. In this case, a small-reﬂection approach is developed

to simplify the design of a multi-section transducer.

We begin with two reﬂection planes, as shown in Fig. 3.25. This means

that between the input and output waveguides there is only one section of

intermediate waveguide.

Suppose that the reﬂection coeﬃcient at the ﬁrst reﬂection plane is Γ

and the transmission coeﬃcient is T

, the reﬂection coeﬃcient and the trans-

mission coeﬃcient at the ﬁrst plane in the reverse direction are Γ

and T

respectively, and the reﬂection coeﬃcient at the second reﬂection plane is Γ

The expressions for them are as follows:

− Z

+ Z

, T

+ Z

= 1 + Γ

− Z

+ Z

= −Γ

, T

+ Z

= 1 + Γ

= 1 − Γ

− Z

+ Z

172 3. Transmission-Line and Network Theory for Electromagnetic Waves

Figure 3.26: Reﬂection of wave at a number of reﬂection points.

where Z

, Z

and Z

denote the characteristic impedances of the input

section, the output section, and the intermediate section.

The total reﬂection at the ﬁrst plane is given by

Γ = Γ

+ T

−j2θ

+ T

−j4θ

+ T

−j6θ

+ ···

= Γ

+ T

−j2θ

1 + Γ

−j2θ

+ Γ

−j4θ

+ Γ

−j6θ

+ ···

= Γ

+ T

−j2θ

1 − Γ

−j2θ

Substituting the above relations among the reﬂection coeﬃcients and the

transmission coeﬃcients into this expression yields

Γ = Γ

1 − Γ

−j2θ

1 + Γ

−j2θ

. (3.198)

Under the small-reﬂection condition, |Γ

| ¿ 1 and |Γ

| ¿ 1, the terms of

the product of the reﬂection coeﬃcients may be neglected, then the above

expression becomes

Γ ≈ Γ

+ Γ

−j2θ

. (3.199)

For an N-section transducer, there are N intermediate waveguides with

characteristic impedance Z

to Z

, and with the same electrical lengths

θ = β

l. There are N + 1 reﬂection planes between the input and output

waveguides. The reﬂection coeﬃcients are denoted by Γ

to Γ

. See Fig. 3.26.

If each reﬂection plane satisﬁes the small-reﬂection condition, the expression

for the total reﬂection at the ﬁrst reﬂection plane is given by

Γ =Γ

+Γ

−j2θ

+Γ

−j4θ

+ ···Γ

N−1

−j2(N−1)θ

+Γ

−j2Nθ

n=0

−j2nθ

(3.200)

3.7 Impedance Transducers 173

where Γ

denotes the reﬂection coeﬃcient at the nth reﬂection plane,

n+1

− Z

n+1

+ Z

, n 6= 0, n 6= N, (3.201)

− Z

+ Z

, Γ

− Z

+ Z

. (3.202)

If the reﬂection coeﬃcients are symmetric with respect to the center of

the transducer, i.e., Γ

= Γ

, Γ

= Γ

N−1

, Γ

= Γ

N−2

, ······, then (3.200)

becomes

Γ = e

−jNθ

jNθ

+ e

−jNθ

+ Γ

j(N−2)θ

+ e

−j(N−2)θ

+ ···

··· + Γ

(N−1)/2

jθ

+ e

−jθ

, n odd, (3.203)

Γ = e

−jNθ

jNθ

+ e

−jNθ

+ Γ

j(N−2)θ

+ e

−j(N−2)θ

+ ···

··· + Γ

N/2

, n even. (3.204)

By applying the Eulerian formula, the above two expressions become:

Γ = 2e

−jNθ

[Γ

cos N θ + Γ

cos(N − 2)θ + ··· + Γ

cos(N − 2n)θ + ···

(N−1)/2

cos θ], n odd,

N/2

], n even.

(3.205)

This is the expression of the reﬂection coeﬃcient of the multi-section trans-

ducer in the small-reﬂection approach. For the ﬁtting of the frequency re-

sponse of the transducer, there are two design approaches, the Chebyshev

approach of an equal ripple response and the binomial approach of a ﬂatness

response.

(1) The Binomial Transducer, The Flattest Response

Take the following binomial as the ﬁtting function of an N-section transducer

Γ = A

1 + e

−j2θ

, i.e., Γ = A2

−jNθ

cos

θ. (3.206)

At the center of the band, cos θ = 0, θ = π/2, we have

∂

|Γ |

∂θ

θ=π/2

= 2

(N − n − 1)!

sin θ cos

(N−n)

θ=π/2

= 0, n = 0 to N − 1.

The reﬂection coeﬃcient of the transducer at the center of the band will be

zero and the response will be most ﬂat.

174 3. Transmission-Line and Network Theory for Electromagnetic Waves

When θ = 0, the intermediate sections do not exist and Γ will be the

reﬂection coeﬃcient of the connector of a waveguide with Z

and a waveguide

with Z

, so the constant A can be determined

Γ |

θ=0

= A2

− Z

+ Z

, A = 2

−N

− Z

+ Z

. (3.207)

Substituting it into (3.206) yields

Γ = 2

−N

− Z

+ Z

1 + e

−j2θ

. (3.208)

The expansion of the binomial

1 + e

−j2θ

is as follows:

1 + e

−j2θ

= C

N,0

+ C

N,1

−j2θ

+ C

N,2

−j4θ

+ ··· + C

N,n

−j2nθ

+ ···

N,N−1

−j2(N−1)θ

+ C

N,N

−j2Nθ

where

N,n

n(N − 1) ···(N − n + 1)

(N − n)!n!

and C

N,n

= C

N,N−n

Then, we have

1 + e

−j2θ

= 2e

−jNθ

N,0

cos N θ + C

N,1

cos(N − 2)θ + ···

N,n

cos(N − 2n)θ + ···

N,(N−1)/2

cos θ], N odd,

N,N/2

], N even.

(3.209)

Substituting (3.209) into (3.208), and ﬁtting the expression for Γ in the

small-reﬂection approach (3.205), we have

Γ = 2e

−jNθ

[Γ

cos N θ + Γ

cos(N − 2) θ + ··· + Γ

cos(N − 2n) θ + ···

(N−1)/2

cos θ ], n odd

N/2

], n even

− Z

+ Z

−N+1

−jNθ

N,0

cos N θ + C

N,1

cos(N − 2)θ + ···

N,n

cos(N − 2n)θ + ··· +

N,(N−1)/2

cos θ], N odd,

N,N/2

], N even.

(3.210)

Comparing the coeﬃcients of the above equation, we have the reﬂection co-

eﬃcient of each reﬂection plane Γ

− Z

+ Z

−N

N,n

− Z

+ Z

−N

(N − n)!n!

. (3.211)

Then the characteristic impedance of each section can be determined by

means of (3.201)–(3.202).

The frequency responses of binomial transducers with ﬂat responses are

shown in Fig. 3.27(a). The bandwidth becomes wider when the number of

sections increases, and the ﬂatness of the response remains unchanged.

3.7 Impedance Transducers 175

0.0 .2 .4 .6 .8 1.0 1.2 1.4 1.6 1.8 2.0

1.0

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2.0

N = 1

(VSWR)

(a)

C1C2

0.0 .2 .4 .6 .8 1.0 1.2 1.4 1.6 1.8 2.0

1.0

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2.0

(VSWR)

N = 1

C1C2

(b)

Figure 3.27: Frequency responses of binomial (a) and Chebyshev (b) trans-

ducers.

(2) The Chebyshev Transducer, The Equal-Ripple Response

For a Chebyshev design, the ﬁtting function is given by

cos θ

where p is given in (3.194) and may be expressed as

p = cos φ, φ =

1 + q

. (3.212)

Then the ﬁtting function becomes

cos θ

cos φ

= hT

(secφ cos θ).

Chebyshev polynomials include the terms in cos

θ, which can be expanded

into a series of cos nθ by means of the expansion given in (3.209):

cos

θ = 2

−n

−jnθ

(1 + e

j2θ

)

= 2

−n+1

cos nθ + c

cos(n − 2) θ + ··· + c

cos(n − m) θ + ···

n,(n−1)/2

cos θ], n odd,

n,n/2

], n even.

(3.213)

Therefor, the Chebyshev polynomials T

(secφ cos θ) can also be expanded

into a series of cos nθ, for example

(sec φ cos θ) = sec φ cos θ,

(sec φ cos θ) = 2 (sec φ cos θ)

− 1 = sec

φ cos 2 θ + tan

φ,

176 3. Transmission-Line and Network Theory for Electromagnetic Waves

(sec φ cos θ) = 4 (sec φ cos θ)

− 3 sec φ cos θ

= sec

φ cos 3 θ + 3 sec φ tan

φ cos θ,

(sec φ cos θ) = 8 (sec φ cos θ)

− 8 (sec φ cos θ)

+ 1

= sec

φ cos 4 θ + 4 sec

φ tan

φ cos 2 θ + tan

φ(3 sec

φ − 1),

······.

These expressions are used as the ﬁtting functions of the reﬂection coeﬃcient

expressed in (3.205). Let h = Ae

−jNθ

. This yields

Γ = Ae

−jNθ

(sec φ cos θ). (3.214)

When θ = 0, the intermediate sections do not exist and Γ will be the reﬂection

coeﬃcient of the connector of a waveguide with Z

and a waveguide with

, so the constant A can be determined by

Γ |

θ=0

= AT

(sec φ) =

− Z

+ Z

, A =

− Z

+ Z

(sec φ)

. (3.215)

Substituting (3.205) into (3.214) and considering (3.215), we have the ﬁtting

equation

Γ = 2e

−jNθ

[Γ

cos N θ + Γ

cos(N − 2) θ + ··· + Γ

cos(N − 2n) θ + ···

(N−1)/2

cos θ ], n odd

N/2

], n even

− Z

+ Z

(sec φ)

−jNθ

(sec φ cos θ). (3.216)

Comparing the coeﬃcients of the above equation, we can have Γ

, the reﬂec-

tion coeﬃcients of each reﬂection plane, then the characteristic impedance of

each section can be determined by means of (3.201)–(3.202).

The reﬂection coeﬃcient reaches a maximum when T

(sec φ cos θ) = 1:

|Γ |

max

− Z

+ Z

(sec φ)

, i.e., T

(sec φ) =

− Z

+ Z

|Γ |

max

(3.217)

According to the deﬁnition of the Chebyshev polynomial (3.183), we have

(sec φ) = cos [N arccos(sec φ)].

Since sec φ ≥ 1, this becomes

(sec φ) = cosh[N arccosh(sec φ)], i.e., sec φ = cosh

arccoshT

(sec φ)

Substituting (3.217) into it, yields

sec φ = cosh

arccosh

− Z

+ Z

|Γ |

max

¶¸

Problems 177

Figure 3.28: Problem 3.3. T and Π networks.

Hence

N =

arccosh

− Z

+ Z

|Γ |

max

arccosh(sec φ)

. (3.218)

The relation among the bandwidth ratio q = (π − φ)/φ, the maximum re-

ﬂection coeﬃcient |Γ |

max

, and the number of sections or layers N is given

in (3.218). One of them is determined by (3.218) when the other two are

given. The frequency responses of Chebyshev transducers with equal-ripple

response are shown in Fig. 3.27(b).

Problems

3.1 For a lossy transmission line, the voltage and current along the line are

given in (3.8), (3.10) and (3.11). Show that the expression for the

impedance transformation is given by

Z(z

) = Z

Z(z

) + Z

tanh γl

+ Z(z

) tanh γl

3.2 Show that the magnitude of the reﬂection coeﬃcient for a lossy trans-

mission line changes along the line as |Γ (z)| = |Γ (0)|e

−2αz

, where α

denotes the attenuation coeﬃcient of the line.

3.3 Find the (a) and (S) parameters of the T and Π networks given in

Fig. 3.28.

3.4 A metallic insulator for a coaxial line is shown in Fig. 3.29. The short-

circuit branch line does not inﬂuence the wave propagation through the

main line when the length of the branch line is just λ/4, and its input

impedance is ∞. Find the relative bandwidth of this structure if the

maximum allowed VSWR is 1.2.

178 3. Transmission-Line and Network Theory for Electromagnetic Waves

Figure 3.29: Problem 3.4. Metallic insulator.

Figure 3.30: Problem 3.7. Backward-wave structure.

3.5 Two equal admittances jB are parallel connected on a transmission line

with a characteristic impedance Z

and phase factor β = 2π/λ. The

distance between the two admittances is l. Show that the condition of

zero reﬂection is given by cot(2πl/λ) = BZ

3.6 A parallel resonant circuit of L, C, G is connected at the end of a

transmission line.

(1) Plot the loci of the input impedance and admittance at a ﬁxed plane

on the line versus frequency on the Smith chart.

(2) Show how the loci change when the input plane moves.

3.7 Find the propagation coeﬃcient β of a lossless transmission line which

consists of distributed series capacitances and shunt inductances as

shown in Fig. 3.30. Show that, in this structure, β decreases versus

frequency. This means that the wave in this structure is a backward

wave.

3.8 As exercises, try to do the basic applications of the Smith chart given in

Section 3.3.4 by setting up appropriate data,

Chapter 4

Time-Varying

Boundary-Value Problems

Wave equations and Helmholtz’s equations are three-dimensional vector par-

tial diﬀerential equations. In Chapter 2, the uniform plane waves are dis-

cussed, where the three-dimensional vector Helmholtz equations are reduced

directly to one-dimensional scalar diﬀerential equations. The telegraph equa-

tions given in Chapter 3 are also one-dimensional scalar diﬀerential equations.

The solutions of both equations are one-dimensional traveling-waves.

In general, There are three classes of electromagnetic ﬁeld and wave prob-

lems: mixed problems, i.e., problems with given both boundary values and

initial values; initial-value problems, i.e., problems without boundary val-

ues; and boundary-value problems, i.e., problems without initial values. The

initial-value problem is the problem with boundary far enough from the in-

terested region of the problem, and the inﬂuence of the boundary value can

be neglected. The boundary-value problem is the problem with initial time

far enough before the interested time period of the problem, and the inﬂu-

ence of the initial value is damped out and can be neglected. The problems

of steady-state sinusoidal electromagnetic oscillation and wave propagation

in bounded regions, which will be discussed in the next three chapters, are

boundary-value problems of Helmholtz’s equations.

In this chapter, the general solutions of the three-dimensional boundary-

value problems of time-varying ﬁelds are formulated. The three-dimensional

vector partial diﬀerential equations are to be reduced to three-dimensional

scalar partial diﬀerential equations and then, in an appropriate coordinate

system, reduced to three one-dimensional ordinary diﬀerential equations by

means of separation of variables. The ordinary diﬀerential equations with

given boundary conditions are ﬁnally solved. The problems become eigen-

value problems or Sturm–Liouville problems, and the general solution may

be expressed in terms of a set of orthogonal eigenfunctions.

180 4. Time-Varying Boundary-Value Problems

4.1 Uniqueness Theorem for

Time-Varying-Field Problems

For the solution of a boundary-value problem of Helmholtz’s equations,

the question is that what are the boundary conditions appropriate for the

Helmholtz’s equation so that a unique solution exists inside a bounded vol-

ume. If excessive boundary conditions are given, no solution can entirely

ﬁt the conditions, i.e., the solution does not exist. If insuﬃcient boundary

conditions are given, more than one set of solutions can ﬁt the conditions,

i.e., the solution is not unique.

4.1.1 Uniqueness Theorem for the Boundary-Value

Problems of Helmholtz’s Equations

Theorem

In the steady sinusoidal time-varying state, in a volume of interest, V , sur-

rounded by a closed surface, S, refer to Fig. 4.1(a), if the following conditions

are satisﬁed, the solution of the complex Maxwell equations or Helmholtz’s

equations is unique.

1. The sources, namely the complex amplitude of the electric current den-

sity J and the equivalent magnetic current density J

are given every-

where in the given volume V , including the source-free problems, J = 0

and/or J

= 0.

2. The complex amplitude of the tangential component of the electric ﬁeld

n ×E|

or the tangential component of the magnetic ﬁeld n × H|

given everywhere over the boundary S of the given volume.

Proof

Suppose that two sets of complex vector functions, E

, H

, and E

, H

both

are solutions of the given boundary-value problem in a volume V bounded

by a closed surface S, and

∆E = E

− E

, ∆H = H

− H

denotes the diﬀerence functions.

The diﬀerence functions ∆E and ∆H must satisfy the complex Maxwell

equations, because both E

, H

and E

, H

satisfy Maxwell’s equations and

the Maxwell’s equations are linear equations. Then the diﬀerence functions

∆E, ∆H are sure to satisfy the complex Poynting theorem, which is derived