# 5.7: Terminated Coupled Lines


A pair of coupled lines supports even and odd modes and these can be coupled by the terminations at the end of a line. Thus at a termination, such as that shown in Figure 5.6.6(a), where the termination is general and represented by a port-based admittance matrix $$\mathbf{Y}$$, the reflected even and odd modes will have contributions from both incident even and odd modes:

$\label{eq:1}V_{e}^{-}=\Gamma_{Le}V_{e}^{+}+C_{eo}V_{o}^{+}\quad\text{and}\quad V_{o}^{-}=\Gamma_{Lo}V_{o}^{+}+C_{oe}V_{e}^{+}$

where $$\Gamma_{Le}$$ and $$\Gamma_{Lo}$$ are even- and odd-mode reflection coefficients and $$C_{oe}$$ and $$C_{eo}$$ describe coupling first from the odd mode to the even mode and then from the even mode to the odd mode. The derivation of the reflection coefficients and coupling coefficients is left for an example at the end of this section. The summary results are

\begin{align}\label{eq:2}\Gamma_{Le}&=\frac{2 + Z_{0o}Y_{\Delta} − Z_{0e}Y_{\Sigma} − 2Z_{0e}Z_{0o}Y_{D}}{2 + Z_{0o}Y_{\Delta} + Z_{0e}Y_{\Sigma} + 2Z_{0e}Z_{0o}Y_{D}}\\ \label{eq:3}C_{eo}&=\frac{2Z_{oe}Y_{E}}{2 + Z_{0o}Y_{\Delta} + Z_{0e}Y_{\Sigma} + 2Z_{0e}Z_{0o}Y_{D}}=-C_{oe} \\ \label{eq:4}\Gamma_{Lo}&=\frac{1 + Z_{0o}(y_{12} + y_{21}) − Z_{0e}Z_{0o}Y_{D}}{1 + Z_{0o}Y_{\Delta} + Z_{0e}Y_{\Sigma} + Z_{0e}Z_{0o}Y_{D}}\end{align}

where

$\label{eq:5}\left.\begin{array}{ll}{Y_{\Sigma} = (y_{11} + y_{12} + y_{21} + y_{22})}&{Y_{\Delta} = (y_{11} − y_{12} − y_{21} + y_{22})}\\{Y_{D} = (y_{11}y_{22} − y_{12}y_{21})\quad\text{and}}&{Y_{E} = (y_{11} − y_{12} + y_{21} − y_{22})}\end{array}\right\}$

Controlling the coupling can be used in filter design based on parallel coupled lines.

## 5.7.1 Reflectionless Condition

In design of many coupled line sections it is important to minimize reflections of coupled lines. One quasi-reflectionless condition is to terminate individual lines of the coupled line in the system reference impedance $$Z_{0S} = \sqrt{Z_{0e}Z_{0o}}$$. Then $$y_{11} = 1/Z_{0S} = y_{22}$$, and $$y_{12} =0= y_{21}$$ and referring to Equation $$\eqref{eq:5}$$, $$Y_{\Sigma} = Y_{\Delta} = 2/Z_{0S} = 2/\sqrt{Z_{0e}Z_{0o}}$$, $$Y_{E} = 0$$, and $$Y_{D} = 1/Z_{0e}Z_{0o}$$ and Equation $$\eqref{eq:2}$$ becomes

$\label{eq:6}\Gamma_{Le}=\frac{2+2\sqrt{Z_{0o}Z_{0e}} − 2\sqrt{Z_{0e}Z_{0o}} − 2}{2+2\sqrt{Z_{0o}Z_{0e}} + 2\sqrt{Z_{0e}Z_{0o}} + 2}=\frac{\sqrt{Z_{0o}Z_{0e}} − 2\sqrt{Z_{0e}Z_{0o}}}{2 + \sqrt{Z_{0o}Z_{0e}} +\sqrt{Z_{0e}Z_{0o}}}$

which is small and

$\label{eq:7}\Gamma_{Lo}=\frac{1+0-1}{D}=0$

Also note that the mode couplings $$C_{oe}$$ and $$C_{oe}$$ are both zero. So terminating the individual lines of a pair of coupled lines in $$Z_{0S}$$ results in the odd-mode reflection coefficient $$\Gamma_{Lo}$$ being zero and the odd-mode reflection coefficient $$\Gamma_{Lo}$$ being small.

The even and odd-mode reflection coefficients can both be set to zero by simultaneously setting the numerators of Equations $$\eqref{eq:2}$$ and $$\eqref{eq:2}$$ to zero while maintaining symmetry, i.e. $$y_{11} = y_{22}$$, and because of reciprocity $$y_{12} = y_{21}$$ thus ensuring that $$Y_{E} = 0$$ and coupling is zero.

##### Example $$\PageIndex{1}$$: Reflection at the End of a Pair of Coupled Lines

A pair of coupled lines with even- and odd-mode characteristic impedances $$Z_{0e}$$ and $$Z_{0o}$$ respectively is loaded as shown in Figure 5.6.6(a) where the load has the two-port $$y$$-parameter matrix, $$\mathbf{Y}$$. Find the even and odd-mode reflections.

Solution

The analysis begins by writing out the expressions relating the traveling-wave and total voltages and currents at the termination.

$\begin{array}{lllllll}{V_{e} = \frac{1}{2}(V_{1} + V_{2})}&{}&{V_{o} =\frac{1}{2} (V_{1} − V_{2})}&{}&{I_{e} =\frac{1}{2} (I_{1} + I_{2})}&{}&{I_{o} = \frac{1}{2} (I_{1} − I_{2})}\\{V_{1} = (V_{1}^{+} + V_{1}^{−})}&{}&{V_{2} = (V_{2}^{+} + V_{2}^{−})}&{}&{I_{1} = (I_{1}^{+} + I_{1}^{−})}&{}&{I_{2} = (I_{2}^{+} + I_{2}^{−})}\\{V_{1}^{+} = (V_{e}^{+} + V_{o}^{+})}&{}&{V_{2}^{+}= (V_{e}^{+} − V_{o}^{+})}&{}&{I_{1}^{+} = (I_{e}^{+} + I_{o}^{+})}&{}&{I_{2}^{+} = (I_{e}^{−} − I_{o}^{+})}\\{V_{1}^{−} = (V_{e}^{−} + V_{o}^{−})}&{}&{V_{2}^{−} = (V_{e}^{−} − V_{o}^{−})}&{}&{I_{1}^{−} = (I_{e}^{−} + I_{o}^{−})}&{}&{I_{2}^{−} = (I_{e}^{−} − I_{o}^{−})}\\{I_{e}^{+} = V_{e}^{+}/Z_{0e}}&{}&{I_{e}^{−} = −V_{e}^{−} /Z_{0e}}&{}&{I_{o}^{+} = V_{o}^{+}/Z_{0e}}&{}&{I_{o}^{−} = −V_{o}^{−}/Z_{0o}}\end{array}\nonumber$

At the termination

\begin{align} \left[\begin{array}{c}{I_{1}}\\{I_{2}}\end{array}\right] = \left[\begin{array}{c}{I_{e}^{+} + I_{o}^{+} + I_{e}^{−} + I_{o}^{−}}\\{I_{e}^{+} − I_{o}^{+} + I_{e}^{−} − I_{o}^{−}}\end{array}\right]&=\mathbf{Y}\left[\begin{array}{c}{V_{1}}\\{V_{2}}\end{array}\right] =\left[\begin{array}{cc}{y_{11}}&{y_{12}}\\{y_{21}}&{y_{22}}\end{array}\right]\left[\begin{array}{c}{V_{1}}\\{V_{2}}\end{array}\right] \nonumber \\ \label{eq:8}\left[\begin{array}{c}{(V_{e}^{+} − V_{e}^{−})/Z_{0e} + (V_{o}^{+} − V_{o}^{−})/Z_{0o}}\\{(V_{e}^{+} − V_{e}^{−})/Z_{0e} − (V_{o}^{+}− V_{o}^{−})/Z_{0o}}\end{array}\right]&=\left[\begin{array}{cc}{y_{11}}&{y_{12}}\\{y_{21}}&{y_{22}}\end{array}\right]\left[\begin{array}{c}{V_{e}^{+} + V_{o}^{+} + V_{e}^{−} + V_{o}^{−}}\\{V_{e}^{+} − V_{o}^{+} + V_{e}^{−} − V_{o}^{−}}\end{array}\right]\end{align}

The reflected even and odd modes have contributions from incident even and odd modes:

$\label{eq:9}V_{e}^{−}= \Gamma_{Le}V_{e}^{+} + C_{eo}V_{o}^{+}\quad\text{and}\quad V_{o}^{−} = \Gamma_{Lo}V_{o}^{+} + C_{oe}V_{e}^{+}$

where $$\Gamma_{Le}$$ and $$\Gamma_{Lo}$$ are even- and odd-mode reflection coefficients, $$C_{oe}$$ and $$C_{eo}$$ describe coupling, and

$\label{eq:10}\Gamma_{Le}=\left.\frac{V_{e}^{-}}{V_{e}^{+}}\right|_{V_{0}^{+}=0},\quad\Gamma_{Lo}=\left.\frac{V_{o}^{-}}{V_{o}^{+}}\right|_{V_{e}^{+}=0},\quad C_{eo}=\left.\frac{V_{e}^{-}}{V_{o}^{+}}\right|_{V_{e}^{+}=0}\quad\text{and}\quad C_{oe}=\left.\frac{V_{o}^{-}}{V_{e}^{+}}\right|_{V_{o}^{+}=0}$

$$\Gamma_{Le}$$ and $$C_{eo}$$ are obtained by expanding Equation $$\eqref{eq:8}$$ with $$V_{o}^{+} = 0$$:

\begin{align} (V_{e}^{+} − V_{e}^{−})/Z_{0e} − V_{o}^{−}/Z_{0o} &= y_{11}\left( V_{e}^{+} + V_{e}^{−} + V_{o}^{−}\right) + y_{12}\left( V_{e}^{+} + V_{e}^{−} − V_{o}^{−}\right)\nonumber \\ \label{eq:11}\text{i.e. }\frac{V_{e}^{-}}{Z_{0e}}(1 + y_{11}Z_{0e} + y_{12}Z_{0e}) &= \frac{V_{e}^{+}}{Z_{0e}} (1 − y_{11}Z_{0e} − y_{12}Z_{0e}) − \frac{V_{o}^{-}}{Z_{0o}} (1 + y_{11}Z_{0o} − y_{12}Z_{0o}) \\ (V_{e}^{+} − V_{e}^{-} )/Z_{0e} + V_{o}^{−})/Z_{0o} &= y_{21}\left(V_{e}^{+} + V_{e}^{−} + V_{o}^{−}\right) + y_{22}\left( V_{e}^{+} + V_{e}^{−} − V_{o}^{−}\right) \\ \label{eq:12} \text{i.e. }\frac{V_{o}^{−}}{Z_{0o}} (−1 + y_{21}Z_{0o} − y_{22}Z_{0o}) &= \frac{V_{e}^{+}}{Z_{0e}}(1 − y_{21}Z_{0e} − y_{22}Z_{0e}) +\frac{V_{e}^{−}}{Z_{0e}} (−1 − y_{21}Z_{0e} − y_{22}Z_{0e})\end{align}

$$\Gamma_{Le}$$ is obtained by eliminating $$V_{o}^{−}$$ by multiplying Equation $$\eqref{eq:12}$$ by $$(1 + y_{11}Z_{0o} − y_{12}Z_{0o})$$ and subtracting it from Equation $$\eqref{eq:11}$$ multiplied by $$(−1 + y_{21}Z_{0o} − y_{22}Z_{0o})$$:

$\label{eq:13}V_{e}^{-}\frac{a}{Z_{0e}}=V_{e}^{+}\left.\frac{b}{Z_{0e}}\right|_{V_{o}^{+}=0}\quad\text{Thus}\quad\Gamma_{Le}=\frac{b}{a}$

where

\begin{align} a &= (1 + y_{11}Z_{0e} + y_{12}Z_{0e})(−1 + y_{21}Z_{0o} − y_{22}Z_{0o})−(1 + y_{21}Z_{0e} + y_{22}Z_{0e})(1 + y_{11}Z_{0o} − y_{12}Z_{0o})\nonumber \\ \label{eq:14}&\quad = −2 − Z_{0o}Y_{\Delta} − Z_{0e}Y_{\Sigma} − 2Z_{0e}Z_{0o}Y_{D} \\ b &= (1 − y_{11}Z_{0e} − y_{12}Z_{0e})(−1 + y_{21}Z_{0o} − y_{22}Z_{0o})−(1 − y_{21}Z_{0e} − y_{22}Z_{0e})(1 + y_{11}Z_{0o} − y_{12}Z_{0o}) \\ \label{eq:15}&\quad = −2 − Z_{0o}Y_{\Delta} + Z_{0e}Y_{\Sigma} + 2Z_{0e}Z_{0o}Y_{D}\end{align}

\begin{align}\label{eq:16}Y_{\Sigma} &= (y_{11} + y_{12} + y_{21} + y_{22}),\: Y_{\Delta} = (y_{11} − y_{12} − y_{21} + y_{22}),\text{ and }Y_{D} = (y_{11}y_{22} − y_{12}y_{21})\\ \label{eq:17} \Gamma_{Le}&=\frac{b}{a}=\frac{2 + Z_{0o}Y_{\Delta} − Z_{0e}Y_{\Sigma} − 2Z_{0e}Z_{0o}Y_{D}}{2 + Z_{0o}Y_{\Delta} + Z_{0e}Y_{\Sigma} + 2Z_{0e}Z_{0o}Y_{D}}\end{align}

As a sanity check, if the load is symmetrical $$y_{11} = y_{22} = Y_{L} = 1/Z_{L}$$, the differential load impedance is infinite $$y_{12} =0= y_{21}$$, if the lines are far apart $$Z_{0e} = Z_{0o} = Z_{0}$$, then $$Y_{\Sigma} = 2Y_{L},\: Y_{\Delta} = 2Y_{L},$$ and $$Y_{D} = Y_{L}^{2}$$ and Equation $$\eqref{eq:17}$$ becomes

\begin{align}\Gamma_{Le}&=\frac{2+2Z_{0}Y_{L} − 2Z_{0}Y_{L} − 2Z_{0}^{2}Y_{L}^{2}}{2+2Z_{0}Y_{L} + 2Z_{0}Y_{L} + 2Z_{0}^{2}Y_{L}^{2}}=\frac{1 − Z_{0}^{2}Y_{L}^{2}}{1+2Z_{0}Y_{L} + Z_{0}^{2}Y_{L}^{2}}=\frac{(1 − Z_{0}Y_{L})(1 + Z_{0}Y_{L})}{(1 + Z_{0}Y_{L})(1 + Z_{0}Y_{L})}\nonumber \\ \label{eq:18}&=\frac{1 − Z_{0}Y_{L}}{1 + Z_{0}Y_{L}}=\frac{Z_{L}-Z_{0}}{Z_{L}+Z_{0}}\end{align}

This is the reflection coefficient of a transmission line of characteristic impedance $$Z_{0}$$ terminated in a load $$Z_{L}$$ as expected.

Find $$C_{eo}$$ by expanding Equation $$\eqref{eq:8}$$ with $$V_{e}^{+} = 0$$:

\begin{align} −V_{e}^{−}/Z_{0e} + (V_{o}^{+} − V_{o}^{−})/Z_{0o} &= y_{11}\left( V_{o}^{+} + V_{e}^{−} + V_{o}^{−}\right) + y_{12}\left( −V_{o}^{+} + V_{e}^{−} − V_{o}^{−}\right)\nonumber \\ \label{eq:19} \text{i.e. }\frac{V_{e}^{−}}{Z_{0e}} (1 + y_{11}Z_{0e} + y_{12}Z_{0e}) &=\frac{V_{o}^{+}}{Z_{0o}}(1 − y_{11}Z_{0o} + y_{12}Z_{0o}) −\frac{V_{o}^{−}}{Z_{0o}}(1 + y_{11}Z_{0o} − y_{12}Z_{0o})\\ −V_{e}^{−}/Z_{0e} − (V_{o}^{+} − V_{o}^{−})/Z_{0o} &= y_{21}\left( +V_{o}^{+} + V_{e}^{−} + V_{o}^{−}\right) + y_{22}\left( −V_{o}^{+} + V_{e}^{−} − V_{o}^{−}\right) \\ \label{eq:20} \text{i.e. }\frac{V_{o}^{−}}{Z_{0o}} (−1 + y_{21}Z_{0o} − y_{22}Z_{0o}) &= −\frac{V_{o}^{+}}{Z_{0o}} (1 + y_{21}Z_{0o} − y_{22}Z_{0o}) −\frac{V_{e}^{−}}{Z_{0e}} (1 + y_{21}Z_{0e} + y_{22}Z_{0e})\end{align}

To eliminate $$V_{o}^{−}$$ multiply Equation $$\eqref{eq:20}$$ by $$(1 + y_{11}Z_{0o} − y_{12}Z_{0o})$$ and subtract it from Equation $$\eqref{eq:19}$$ multiplied by $$(−1 + y_{21}Z_{0o} − y_{22}Z_{0o})$$:

$\label{eq:21}\frac{V_{e}^{-}}{Z_{0e}}a=\frac{V_{o}^{+}}{Z_{0o}}c\quad\text{such that}\quad C_{eo}=\frac{c}{a}\frac{Z_{0e}}{Z_{0o}}$

where $$a$$ is as in Equation $$\eqref{eq:14}$$ and

\begin{align} c &= (1 − y_{11}Z_{0o} + y_{12}Z_{0o})(−1 + y_{21}Z_{0o} − y_{22}Z_{0o})+(1 + y_{21}Z_{0o} − y_{22}Z_{0o})(1 + y_{11}Z_{0o} − y_{12}Z_{0o})\nonumber \\ \label{eq:22} &= −2Z_{0o}Y_{E}\quad\text{where}\quad Y_{E} = (y_{11} − y_{12} + y_{21} − y_{22}) \\ \label{eq:23} C_{eo}&=\frac{2Z_{0o}Y_{E}}{2 + Z_{0o}Y_{\Delta} + Z_{0e}Y_{\Sigma} + 2Z_{0e}Z_{0o}Y_{D}}\end{align}

Now for a sanity check. If the load is symmetrical $$y_{11} = y_{22}$$ and $$y_{12} = y_{21}$$, then $$Y_{E} = 0$$ and $$C_{eo} = 0$$ indicating that a signal is not converted from odd mode to even mode with reflection from a symmetrical load, as expected.

The analysis is now repeated to find $$\Gamma_{Lo}$$ and $$C_{oe}$$. $$\Gamma_{Lo}$$ is found by eliminating $$V_{e}^{−}$$ from Equations $$\eqref{eq:19}$$ and $$\eqref{eq:20}$$ by multiplying Equation $$\eqref{eq:20}$$ by $$(1 + y_{11}Z_{0e} + y_{12}Z_{0e})$$ and subtracting it from Equation $$\eqref{eq:19}$$ multiplied by $$(1 + y_{21}Z_{0e} + y_{22}Z_{0e})$$:

$\label{eq:24}\frac{V_{o}^{+}}{Z_{0o}}d=\left.\frac{V_{o}^{-}}{Z_{0o}}e\right|_{V_{e}^{+}=0}\quad\text{and}\quad \Gamma_{Lo}=\left.\frac{V_{o}^{-}}{V_{o}^{+}}\right|_{V_{e}^{+}=0}=\frac{d}{e}$

\begin{align} d &= (1 − y_{11}Z_{0o} + y_{12}Z_{0o}) (1 + y_{21}Z_{0e} + y_{22}Z_{0e})+(1 + y_{21}Z_{0o} − y_{22}Z_{0o}) (1 + y_{11}Z_{0e} + y_{12}Z_{0e}) \nonumber \\ \label{eq:25} &=2+2Z_{0o}(y_{12} + y_{21}) − 2Z_{0e}Z_{0o}Y_{D} \\ e&= (1 + y_{11}Z_{0o} − y_{12}Z_{0o}) (1 + y_{21}Z_{0e} + y_{22}Z_{0e})−(−1 + y_{21}Z_{0o} − y_{22}Z_{0o}) (1 + y_{11}Z_{0e} + y_{12}Z_{0e}) \\ \label{eq:26} &=2+2Z_{0o}Y_{\Delta} + 2Z_{0e}Y_{\Sigma} + 2Z_{0e}Z_{0o}Y_{D} \\ \label{eq:27}\Gamma_{Lo}&=\frac{1 + Z_{0o}(y_{12} + y_{21}) − Z_{0e}Z_{0o}Y_{D}}{1 + Z_{0o}Y_{\Delta} + Z_{0e}Y_{\Sigma} + Z_{0e}Z_{0o}Y_{D}}\end{align}

$$C_{oe}$$ is found by eliminating $$V_{e}^{−}$$ achieved by multiplying Equation $$\eqref{eq:20}$$ by $$(1 +y_{11}Z_{0e} + y_{12}Z_{0e})$$ and subtracting it from Equation $$\eqref{eq:19}$$ multiplied by $$(1 + y_{21}Z_{0e} + y_{22}Z_{0e})$$:

$\label{eq:28}V_{o}^{-}\frac{f}{Z_{0o}}=V_{e}^{+}\left.\frac{g}{Z_{0e}}\right|_{V_{o}^{+}=0}\quad\text{Thus}\quad C_{oe}=\frac{g}{f}\frac{Z_{0o}}{Z_{0e}}\quad\text{where}$

\begin{align} g &= −(1 − y_{11}Z_{0e} − y_{12}Z_{0e})(1 + y_{21}Z_{0e} + y_{22}Z_{0e})+(1 − y_{21}Z_{0e} − y_{22}Z_{0e})(1 + y_{11}Z_{0e} + y_{12}Z_{0e})\nonumber \\ \label{eq:29} &= 2Z_{0e}(y_{11} + y_{12} − y_{21} − y_{22})=2Z_{0e}Y_{E} \\ f &= −(1 + y_{11}Z_{0o} − y_{12}Z_{0o})(1 + y_{21}Z_{0e} + y_{22}Z_{0e})+(−1 + y_{21}Z_{0e} − y_{22}Z_{0e})(1 + y_{11}Z_{0e} + y_{12}Z_{0e}) \\ \label{eq:30} &= −2 − Z_{0o}Y_{\Delta} − Z_{0e}Y_{\Sigma} − 2Z_{0e}Z_{0o}Y_{D} \\ \label{eq:31} C_{oe}&=\frac{−2Z_{0o}Y_{E}}{2 + Z_{0o}Y_{\Delta} − Z_{0e}Y_{\Sigma} + 2Z_{0e}Z_{0o}Y_{D}}=-C_{eo}\end{align}

##### Example $$\PageIndex{2}$$: Even and Odd Mode Loads

A pair of coupled lines is loaded as shown in Figure 5.6.6(a). The even-mode characteristic impedance is $$Z_{0e} = 90\:\Omega$$ and the odd-mode characteristic impedance is $$Z_{0o} = 45\:\Omega$$.

1. What is the even-mode reflection coefficient?
To simplify calculations let the forward-traveling even-mode voltage at the load $$V_{e}^{+}=1\text{ V}$$, and there is no forward-traveling odd-mode voltage. $$V_{e}^{−}$$ and $$V_{o}^{−}$$ are the backward-traveling even- and odd-mode voltages and $$I_{e}^{+},\: I_{e}^{−}$$ and $$I_{o}^{−}$$ are the corresponding currents. At Port $$\mathsf{1}$$
\begin{aligned} I_{1} &= \frac{1}{60} V_{1} = V_{e}^{+} + V_{o}^{+} + V_{e}^{−} + V_{o}^{−} = \frac{1}{60} (1 +0+ V_{e}^{−} + V_{o}^{−}) \\ I_{1} &= I_{e}^{+} + I_{o}^{+} + I_{e}^{−} + I_{o}^{−} = \frac{V_{e}^{+}}{Z_{0e}} + 0 −\frac{V_{e}^{−}}{Z_{0e}} −\frac{V_{o}^{−}}{Z_{0o}} = \frac{1}{90} − \frac{V_{e}^{−}}{90} −\frac{V_{o}^{−}}{45}\end{aligned}\nonumber
Equating these and multiplying by $$180$$:
$\label{eq:32} 1+5V_{e}^{−} + 7V_{o}^{−} = 0$
At Port $$\mathsf{2}$$
\begin{aligned}I_{2} &= \frac{1}{100}V_{1} = \frac{1}{100} (V_{e}^{−} V_{o}^{+} + V_{e}^{−} − V_{o}^{−}) = \frac{1}{100} (1 +0+ V_{e}^{−} − V_{o}^{−}) \\ I_{2} &= I_{e}^{+} − I_{o}^{+} + I_{e}^{−} − I_{o}^{−} = \frac{V_{e}^{+}}{Z_{0e}} + 0 −\frac{V_{e}^{−}}{Z_{0e}} + \frac{V_{o}^{−}}{Z_{0o}} = \frac{1}{90} − \frac{V_{e}^{-}}{90} + \frac{V_{o}^{−}}{45} \end{aligned}\nonumber
Equating these and multiplying by $$900$$:
$\label{eq:33}1-19V_{e}^{-}+29V_{o}^{-}=0$
Multiplying Equation $$\eqref{eq:33}$$ by $$7$$ and subtracting it from Equation $$\eqref{eq:32}$$ times $$29$$:
$(29 − 7) + (145 + 133)V_{e}^{−} + (203 − 203)V_{o}^{−} = 0 ⇒ V_{e}^{−} = −0.07914\text{ V} \nonumber$
Thus
$\label{eq:34}\Gamma_{Le} = V_{e}^{−} /V_{e}^{+} = V_{e}^{−}/1 = −0.07914$
Alternatively the general formula, Equation $$\eqref{eq:2}$$, can be used noting that $$Z_{0e} = 90\:\Omega,\: Z_{0e} = 45\:\Omega,\: y_{11} = 1/60,\: y_{22} = 1/100,$$ and $$y_{12} =0= y_{21}$$. Then $$y_{\Delta} = 0.026667\text{ S},\: y_{\Sigma} = 0.026667\text{ S},\: y_{D} = 0.00016667\text{ S}$$:
$\label{eq:35}\Gamma_{Le}=\frac{2 + 45 \cdot 0.026667 − 90\cdot 0.026667 − 2\cdot 45\cdot 90\cdot 0.00016667}{2 + 45\cdot 0.026667 + 90\cdot 0.026667 + 2\cdot 45\cdot 90\cdot 0.00016667}=-0.07914$
The two evaluations of $$\Gamma_{Le}$$ are in agreement.
2. What is the effective even-mode load resistance $$R_{Le}$$?
$$R_{Le}$$ is the resistance that yields $$\Gamma_{Le}$$ when referred to $$Z_{0e}$$. Thus
$\label{eq:36}R_{Le}=Z_{0e}\frac{1+\Gamma_{Le}}{1-\Gamma_{Le}}=90\frac{1+(−0.07914)}{1 − (−0.07914)}=76.80\:\Omega$
3. What is the odd-mode reflection coefficient?
Calculating this directly using Equation $$\eqref{eq:4}$$:
$\label{eq:37}\Gamma_{Lo}=\frac{1+0 − 45\cdot 90\cdot 0.00016667}{1 + 45\cdot 0.026667 + 90\cdot 0.026667 + 45\cdot 90\cdot 0.00016667}=0.06180$
4. What is the effective odd-mode load resistance $$R_{Lo}$$?
$\label{eq:38}R_{Lo}=Z_{0o}\frac{1+\Gamma_{Lo}}{1-\Gamma_{Lo}}=45\frac{1+0.06180}{1− 0.06180}=50.92\:\Omega$

This page titled 5.7: Terminated Coupled Lines is shared under a not declared license and was authored, remixed, and/or curated by Michael Steer.