Appendix 2D: Nonuniform Transmission Lines

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2D.1Nonuniform Transmission Lines

A nonuniform transmission line can represent many physical phenomena in an inhomogeneous medium whose medium properties are functions of position. A one-dimensional equivalent ideal transmission line will have the per unit values of series inductance L′ and the shunt capacitance C′ as functions of one spatial coordinate.

Let

(2D.1a)

L' = L' (z),

$L^{'} = L^{'} (z),$

(2D.1b)

C' = C' (z) .

$C^{'} = C^{'} (z) .$

If ε and μ are homogenous in the transverse plane, then

(2D.2)

1 L ' C ' - - - - \sqrt = 1 ε μ - - \sqrt = u p,

$\frac{1}{\sqrt{L^{'} C^{'}}} = \frac{1}{\sqrt{ε μ}} = u_{p},$

where u_p is the phase velocity.

The first-order coupled differential equations are

(2D.3a)

- \partial V \partial z = L' (z) \partial I \partial t,

$- \frac{\partial V}{\partial z} = L^{'} (z) \frac{\partial I}{\partial t},$

(2D.3b)

- \partial I \partial z = C' (z) \partial V \partial t .

$- \frac{\partial I}{\partial z} = C^{'} (z) \frac{\partial V}{\partial t} .$

In the phasor form,

(2D.4a)

- \partial V ˜ \partial z = L' (z) j ω I ˜,

$- \frac{\partial \tilde{V}}{\partial z} = L^{'} (z) j ω \tilde{I},$

(2D.4b)

- \partial I ˜ \partial z = C' (z) j ω V ˜ .

$- \frac{\partial \tilde{I}}{\partial z} = C^{'} (z) j ω \tilde{V} .$

From Equations 2D.3a and 2D.3b, we can obtain

(2D.5a)

\partial 2 V \partial z 2 - 1 L ' ( z ) \partial L ' \partial z \partial V \partial z - L' (z) C' (z) \partial 2 V \partial t 2 = 0.

$\frac{\partial^{2} V}{\partial z^{2}} - \frac{1}{L^{'} (z)} \frac{\partial L^{'}}{\partial z} \frac{\partial V}{\partial z} - L^{'} (z) C^{'} (z) \frac{\partial^{2} V}{\partial t^{2}} = 0.$

The equation for I can be obtained similarly as

(2D.5b)

\partial 2 I \partial z 2 - 1 C ' ( z ) \partial C ' \partial z \partial I \partial z - L' (z) C' (z) \partial 2 I \partial t 2 = 0.

$\frac{\partial^{2} I}{\partial z^{2}} - \frac{1}{C^{'} (z)} \frac{\partial C^{'}}{\partial z} \frac{\partial I}{\partial z} - L^{'} (z) C^{'} (z) \frac{\partial^{2} I}{\partial t^{2}} = 0.$

The phasor form of Equations 2D.5a and 2D.5b are

(2D.6a)

\partial 2 V ˜ \partial z 2 - 1 L ' ( z ) \partial L ' \partial z \partial V ˜ \partial z + ω 2 L' (z) C' (z) V ˜ = 0,

$\frac{\partial^{2} \tilde{V}}{\partial z^{2}} - \frac{1}{L^{'} (z)} \frac{\partial L^{'}}{\partial z} \frac{\partial \tilde{V}}{\partial z} + ω^{2} L^{'} (z) C^{'} (z) \tilde{V} = 0,$

(2D.6b)

\partial 2 I ˜ \partial z 2 - 1 C ' ( z ) \partial C ' \partial z \partial I ˜ \partial z + ω 2 L' (z) C' (z) I ˜ = 0.

$\frac{\partial^{2} \tilde{I}}{\partial z^{2}} - \frac{1}{C^{'} (z)} \frac{\partial C^{'}}{\partial z} \frac{\partial \tilde{I}}{\partial z} + ω^{2} L^{'} (z) C^{'} (z) \tilde{I} = 0.$

Analytical solutions can be obtained for an exponential transmission line and this aspect is discussed in the following section.

2D.2Exponential Transmission Line

In this section, we define an exponential transmission line where both the inductance and capacitance are exponential functions as shown below.

Let

(2D.7a)

L' = L' 0 e q z,

$L^{'} = L_{0}^{'} e^{q z},$

(2D.7b)

C' = C' 0 e - q z .

$C^{'} = C_{0}^{'} e^{- q z} .$

Note that

(2D.7c)

L' C' = L' 0 C' 0 = ε μ .

$L^{'} C^{'} = L_{0}^{'} C_{0}^{'} = ε μ .$

Equations 2D.5a and 2D.5b assume the form of

(2D.8)

\partial 2 V \partial z 2 - q \partial V \partial z - L' 0 C' 0 \partial 2 V \partial t 2 = 0,

$\frac{\partial^{2} V}{\partial z^{2}} - q \frac{\partial V}{\partial z} - L_{0}^{'} C_{0}^{'} \frac{\partial^{2} V}{\partial t^{2}} = 0,$

(2D.9)

\partial 2 I \partial z 2 + q \partial I \partial z - L' 0 C' 0 \partial 2 I \partial t 2 = 0.

$\frac{\partial^{2} I}{\partial z^{2}} + q \frac{\partial I}{\partial z} - L_{0}^{'} C_{0}^{'} \frac{\partial^{2} I}{\partial t^{2}} = 0.$

For harmonic variation in space and time, let

(2D.10)

V (z, t) = V 0 e j (ω t - k v z) .

$V (z, t) = V_{0} e^{j (ω t - k_{v} z)} .$

From Equations 2D.8 and 2D.10, we obtain

(2D.11)

(- j k v) 2 - q (- j k v) - L' 0 C' 0 (j ω) 2 = 0, - k 2 v + j k v q + ω 2 L' 0 C' 0 = 0, ω 2 L' 0 C' 0 = k 2 v - j k v q .

$\begin{matrix} {(- j k_{v})}^{2} - q (- j k_{v}) - L_{0}^{'} C_{0}^{'} {(j ω)}^{2} = 0, \\ - k_{v}^{2} + j k_{v} q + ω^{2} L_{0}^{'} C_{0}^{'} = 0, \\ ω^{2} L_{0}^{'} C_{0}^{'} = k_{v}^{2} - j k_{v} q . \end{matrix}$

For a real ω, k_v will be complex, so

(2D.12)

k v = β v - j α v .

$k_{v} = β_{v} - j α_{v} .$

By substituting Equation 2D.12 into Equation 2D.11, we obtain

(2D.13a)

ω 2 L' 0 C' 0 ω 2 L' 0 C' 0 = = (β v - j α v) 2 - j (β v - j α v) q, β 2 v - α 2 v - α v q,

$\begin{matrix} ω^{2} L_{0}^{'} C_{0}^{'} & = & {(β_{v} - j α_{v})}^{2} - j (β_{v} - j α_{v}) q, \\ ω^{2} L_{0}^{'} C_{0}^{'} & = & β_{v}^{2} - α_{v}^{2} - α_{v} q, \end{matrix}$

(2D.13b)

- β v (2 α v + q) = 0.

$- β_{v} (2 α_{v} + q) = 0.$

Thus, we have

(2D.14)

α v = - q 2 .

$α_{v} = \frac{- q}{2} .$

Substituting Equation 2D.14 into Equation 2D.13a, we obtain

(2D.15a)

β 2 v - q 2 4 + q 2 2 β v = = ω 2 L' 0 C' 0, \pm β,

$\begin{matrix} β_{v}^{2} - \frac{q^{2}}{4} + \frac{q^{2}}{2} & = & ω^{2} L_{0}^{'} C_{0}^{'}, \\ β_{v} & = & \pm β, \end{matrix}$

where

(2D.15b)

β = ω 2 L' 0 C' 0 - q 2 4 - - - - - - - - - - - \sqrt .

$β = \sqrt{ω^{2} L_{0}^{'} C_{0}^{'} - \frac{q^{2}}{4}} .$

Thus k_v = ±β − jα_v = ±β + j(q/2),

(2D.16)

e - j k v z = e \pm j β z e q 2 z .

$e^{- j k_{v} z} = e^{\pm j β z} e^{\frac{q}{2} z} .$

Let

(2D.17a)

ω c = q 2 L ' 0 C ' 0 - - - - \sqrt,

$ω_{c} = \frac{q}{2 \sqrt{L_{0}^{'} C_{0}^{'}}},$

(2D.17b)

β 0 = ω L' 0 C' 0 - - - - \sqrt,

$β_{0} = ω \sqrt{L_{0}^{'} C_{0}^{'}},$

then

(2D.18)

β v = β 0 1 - ω 2 c ω 2 - - - - - - \sqrt,

$β_{v} = β_{0} \sqrt{1 - \frac{ω_{c}^{2}}{ω^{2}}},$

and

(2D.19)

e - j k v z = e [\pm j β 0 1 - ω 2 c ω 2 \sqrt + q 2] z .

$e^{- j k_{v} z} = e^{[\pm j β_{0} \sqrt{1 - \frac{ω_{c}^{2}}{ω^{2}}} + \frac{q}{2}] z} .$

Therefore, the time-harmonic solution for an exponential transmission line described by Equation 2D.8 is given as

(2D.20)

V ˜ (z) = V + 0 e (q / 2) z e - j β z + V - 0 e (q / 2) z e + j β z .

$\tilde{V} (z) = V_{0}^{+} e^{(q / 2) z} e^{- j β z} + V_{0}^{-} e^{(q / 2) z} e^{+ j β z} .$

The first term on the RHS is the positive traveling wave and the second term is the negative traveling wave given by

(2D.21)

V ˜ + (z) = V + 0 e (q / 2) z e - j β z (positive traveling wave),

${\tilde{V}}^{+} (z) = V_{0}^{+} e^{(q / 2) z} e^{- j β z} (positive traveling wave),$

(2D.22)

V ˜ - (z) = V - 0 e (q / 2) z e + j β z (negative traveling wave) .

${\tilde{V}}^{-} (z) = V_{0}^{-} e^{(q / 2) z} e^{+ j β z} (negative traveling wave) .$

The solution for the currents can be obtained in a similar fashion by solving Equation 2D.9 and it can be shown that

(2D.23a)

I ˜ (z) = I + 0 e - (q / 2) z e - j β z + I - 0 e - (q / 2) z e + j β z,

$\tilde{I} (z) = I_{0}^{+} e^{- (q / 2) z} e^{- j β z} + I_{0}^{-} e^{- (q / 2) z} e^{+ j β z},$

(2D.23b)

${\tilde{I}}^{+} (z) = I_{0}^{+} e^{- (q / 2) z} e^{- j β z} (positive traveling wave),$

(2D.23c)

${\tilde{I}}^{-} (z) = I_{0}^{-} e^{- (q / 2) z} e^{+ j β z} (negative traveling wave) .$

The relationship between ${\tilde{V}}_{0}^{+}$ and ${\tilde{I}}_{0}^{+}$ , and ${\tilde{V}}_{0}^{-}$ and ${\tilde{I}}_{0}^{-}$ can be obtained from Equations 2D.4a, 2D.21, and 2D.7a.

(2D.24)

$\begin{matrix} {\tilde{I}}^{+} (z) = - \frac{1}{j ω L^{'} (z)} \frac{\partial {\tilde{V}}^{+} (z)}{\partial z}, \\ {\tilde{I}}^{+} (z) = - \frac{1}{j ω L^{'} (z)} (\frac{q}{2} - j β) V_{0}^{+} e^{(q / 2) z} e^{- j β z}, \\ {\tilde{I}}^{+} (z) = - \frac{q / 2 - j β}{j ω L_{0}^{'} e^{q z}} V_{0}^{+} e^{(q / 2) z} e^{- j β z}, \\ {\tilde{I}}^{+} (z) = - \frac{V_{0}^{+} (q / 2 - j β)}{j ω L_{0}^{'}} e^{- (q / 2) z} e^{- j β z}, \\ {\tilde{I}}^{+} (z) = {\tilde{I}}_{0}^{+} e^{- (q / 2) z} e^{- j β z}, \\ Z_{0}^{+} = \frac{V_{0}^{+}}{I_{0}^{+}} = \frac{- j ω L_{0}^{'}}{q / 2 - j β} = \frac{ω L_{0}^{'}}{β + j (q/ 2)} . \end{matrix}$

Similarly,

(2D.25)

$Z_{0}^{-} = - \frac{V_{0}^{-}}{I_{0}^{-}} = \frac{ω L_{0}^{'}}{β - j (q / 2)} .$

One should note that in a nonuniform transmission line, the characteristic impedances for the oppositely traveling waves are not the same:

(2D.26)

$Z_{0}^{+} \neq Z_{0}^{-} .$

Furthermore,

(2D.27)

$\begin{matrix} Z^{+} (z) = \frac{V^{+} (z)}{I^{+} (z)} = \frac{V_{0}^{+} e^{(q / 2) z} e^{- j β z}}{I_{0}^{+} e^{- (q / 2) z} e^{- j β z}}, \\ Z^{+} (z) = Z_{0}^{+} e^{q z} . \end{matrix}$

Similarly,

(2D.28)

$Z^{-} (z) = Z_{0}^{-} e^{q z} .$

Substituting for β from Equation 2D.18 and Equation 2D.24 can be written as

(2D.29a)

$\begin{matrix} Z_{0}^{+} & = & \frac{ω L_{0}^{'}}{β + j (q / 2)} = \frac{ω L_{0}^{'}}{β_{0} [1 - ω_{c}^{2} / ω^{2}] + j ω_{c} \sqrt{L_{0}^{'} C_{0}^{'}}}, \\ Z_{0}^{+} & = & \frac{ω L_{0}^{'}}{β_{0} {{[1 - ω_{c}^{2} / ω^{2}]}^{1 / 2} + j (ω_{c} / ω)}}, \\ Z_{0}^{+} & = & \sqrt{\frac{L_{0}^{'}}{C_{0}^{'}}} \frac{[{(1 - ω_{c}^{2} / ω^{2})}^{1 / 2} - j (ω_{c} / ω)]}{1 - ω_{c}^{2} / ω^{2} + ω_{c}^{2} / ω^{2}}, \\ Z_{0}^{+} & = & Z_{0} [{(1 - \frac{ω_{c}^{2}}{ω^{2}})}^{1 / 2} - j \frac{ω_{c}}{ω}] . \end{matrix}$

Similarly, we can show that

(2D.29b)

$Z_{0}^{-} = Z_{0} [{(1 - \frac{ω_{c}^{2}}{ω^{2}})}^{1 / 2} + j \frac{ω_{c}}{ω}],$

where Z₀ is the nominal characteristic impedance given by

(2D.30)

$Z_{0} = \sqrt{\frac{L_{0}^{'}}{C_{0}^{'}}} .$

For a uniform line q = ω_c = 0, all the formulas of a uniform line can be obtained from

(2D.31)

$Z_{0}^{+} = Z_{0}^{-} = \sqrt{\frac{L^{'}}{C^{'}}} (uniform line ω_{c} = 0) .$

2D.3The Input Impedance

(2D.32)

$\begin{matrix} Z (z) & = & \frac{{\tilde{V}}_{0}^{+} e^{(q / 2) z} e^{- j β z} [1 + Γ_{0 V} e^{j 2 β z}]}{{\tilde{I}}_{0}^{+} e^{- (q / 2) z} e^{- j β z} [1 + Γ_{0 I} e^{j 2 β z}]}, \\ Z (z) & = & Z_{0}^{+} e^{q z} \frac{[1 + Γ_{0 V} e^{j 2 β z}]}{[1 + Γ_{0 I} e^{j 2 β z}]}, \end{matrix}$

where

(2D.33)

$Γ_{0 V} = \frac{V_{0}^{-}}{V_{0}^{+}},$

(2D.34)

$Γ_{0 I} = \frac{I_{0}^{-}}{I_{0}^{+}} .$

Note that

(2D.35)

$\begin{matrix} Γ_{0 V} & = & \frac{V_{0}^{-}}{V_{0}^{+}} = - \frac{Z_{0}^{-} I_{0}^{-}}{Z_{0}^{+} I_{0}^{+}}, \\ Γ_{0 V} & = & - \frac{Z_{0}^{-}}{Z_{0}^{+}} Γ_{0 I}, \\ Γ_{0 I} & = & - \frac{Z_{0}^{+}}{Z_{0}^{-}} Γ_{0 V} . \end{matrix}$

2D.4Arbitrary Load at z = 0

(2D.36)

$\begin{matrix} Z_{L} & = & \frac{V_{L}}{I_{L}} = \frac{V_{0}^{+} + V_{0}^{-}}{I_{0}^{+} + I_{0}^{-}}, \\ Z_{L} & = & \frac{V_{0}^{+}}{I_{0}^{+}} (\frac{1 + Γ_{0 V}}{1 + Γ_{0 I}}), \\ Z_{L} & = & Z_{0}^{+} (\frac{1 + Γ_{0 V}}{1 + Γ_{0 I}}), \\ Z_{L} & = & Z_{0}^{+} [\frac{1 + Γ_{0 V}}{1 + (Z_{0}^{+} / Z_{0}^{-}) Γ_{0 V}}], \\ Z_{L} & = & Z_{0}^{+} Z_{0}^{-} [\frac{1 + Γ_{0 V}}{Z_{0}^{-} - Z_{0}^{+} Γ_{0 V}}], \\ Z_{L} (Z_{0}^{-} - Z_{0}^{+} Γ_{0 V}) & = & Z_{0}^{+} Z_{0}^{-} (1 + Γ_{0 V}), \\ Γ_{0 V} [Z_{0}^{+} Z_{0}^{+} + Z_{L} Z_{0}^{+}] & = & Z_{L} Z_{0}^{-} - Z_{0}^{+} Z_{0}^{-}, \\ Γ_{0 V} & = & \frac{Z_{0}^{-} [Z_{L} - Z_{0}^{+}]}{Z_{0}^{+} [Z_{L} - Z_{0}^{-}]} . \end{matrix}$

Similarly,

(2D.37)

$Γ_{0 I} = - \frac{[Z_{L} - Z_{0}^{+}]}{[Z_{L} - Z_{0}^{-}]} .$

From Equation 2D.32, the input impedance at z = −h is given by

(2D.38)

$\begin{matrix} Z (h) = Z_{0}^{+} e^{- q h} [\frac{1 + Γ_{0 V} e^{- j 2 β h}}{1 + Γ_{0 I} e^{- j 2 β h}}], \\ Z (h) = Z_{0}^{+} e^{- q h} [\frac{e^{j β h} + Γ_{0 V} e^{- j β h}}{e^{j β h} + Γ_{0 I} e^{- j β h}}], \\ Z (h) = \frac{Z_{0}^{+} e^{- q h} (Z_{L} + Z_{0}^{-})}{Z_{0}^{+} (Z_{L} + Z_{0}^{-})} \frac{Z_{0}^{+} (Z_{L} + Z_{0}^{-}) e^{j β h} + Z_{0}^{-} (Z_{L} + Z_{0}^{+}) e^{- j β h}}{(Z_{L} + Z_{0}^{-}) e^{j β h} - (Z_{L} + Z_{0}^{+}) e^{- j 2 β h}} . \end{matrix}$

The above can be written as

(2D.39)

$\begin{matrix} Z (h) = e^{- q h} \frac{N}{D}, \\ N = \cos β h [Z_{0}^{+} (Z_{L} + Z_{0}^{-}) + Z_{0}^{-} (Z_{L} - Z_{0}^{+})] + j \sin β h [Z_{0}^{+} (Z_{L} + Z_{0}^{-}) - Z_{0}^{-} (Z_{L} - Z_{0}^{+})], \\ N = \cos β h [Z_{L} (Z_{0}^{+} + Z_{0}^{-})] + j \sin β h [Z_{0}^{+} Z_{L} - Z_{0}^{-} Z_{L} + 2 Z_{0}^{+} Z_{0}^{-}], \\ D = \cos β h [Z_{L} + Z_{0}^{-} + Z_{L} + Z_{0}^{+}] + j \sin β h [Z_{L} + Z_{0}^{-} + Z_{L} - Z_{0}^{+}], \\ D = \cos β h [Z_{0}^{+} + Z_{0}^{-}] + jsin β h [2 Z_{L} + Z_{0}^{-} - Z_{0}^{+}], \\ Z (h) = e^{- q h} \frac{Z_{L} (Z_{0}^{+} + Z_{0}^{-})}{(Z_{0}^{+} + Z_{0}^{-})} {\frac{1 + j \tan β h [(Z_{0}^{+} Z_{L} - Z_{0}^{-} Z_{L} + 2 Z_{0}^{+} Z_{0}^{-}) / Z_{L} (Z_{0}^{+} + Z_{0}^{-})]}{1 + jtan β h [(2 Z_{L} + Z_{0}^{-} - Z_{0}^{+}) / (Z_{0}^{+} + Z_{0}^{-})]}}, \\ Z (h) = e^{- q h} \frac{(Z_{0}^{+} + Z_{0}^{-})}{2} {\frac{Z_{L} + j \tan β h [(Z_{0}^{+} Z_{L} - Z_{0}^{-} Z_{L} + 2 Z_{0}^{+} Z_{0}^{-}) / (Z_{0}^{+} + Z_{0}^{-})]}{(Z_{0}^{+} + Z_{0}^{-}) / 2 + jtan β h [(2 Z_{L} + Z_{0}^{-} - Z_{0}^{+}) / 2]}} . \end{matrix}$

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Table of Contents for Appendix 2D: Nonuniform Transmission Lines

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2D.1Nonuniform Transmission Lines

2D.2Exponential Transmission Line

2D.3The Input Impedance

2D.4Arbitrary Load at z = 0

Table of Contents for
Appendix 2D: Nonuniform Transmission Lines