# 5.5: Symmetric Coupled Transmission Lines

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In this section, even and odd modes are considered as defining independent transmission lines. The development is restricted to a symmetrical pair of coupled lines; that is, each transmission line of the pair is identical. Thus the strips have the same self-inductance, $$L_{s} = L_{11} = L_{22}$$, and self-capacitance, $$C_{s} = C_{11} = C_{22}$$, where the subscript $$s$$ stands for “self.” $$L_{m} = L_{12} = L_{21}$$ and $$C_{m} = C_{12} = C_{21}$$ are the mutual inductance and capacitance of the lines, and the subscript $$m$$ stands for “mutual.” Equations (5.3.1)–(5.3.4) can thus be written as

$\label{eq:1}\frac{dV_{1}(x)}{dx}=-\jmath\omega L_{s}I_{1}(x)-\jmath\omega L_{m}I_{2}(x)$

$\label{eq:2}\frac{dV_{2}(x)}{dx}=-\jmath\omega L_{m}I_{1}(x)-\jmath\omega L_{s}I_{2}(x)$

$\label{eq:3}\frac{dI_{1}(x)}{dx}=-\jmath\omega C_{s}V_{1}(x)-\jmath\omega C_{m}V_{2}(x)$

$\label{eq:4}\frac{dI_{2}(x)}{dx}=-\jmath\omega C_{m}V_{1}(x)-\jmath\omega C_{s}V_{2}(x)$

The even mode is defined as the mode corresponding to both conductors being at the same potential and carrying the same currents:$$^{1}$$

$\label{eq:5}V_{1}=V_{2}=V_{e}\quad\text{and}\quad I_{1}=I_{2}=I_{e}$

The odd mode is defined as the mode corresponding to the conductors being at opposite potentials relative to the reference conductor and carrying currents of equal amplitude but of opposite sign:$$^{2}$$

$\label{eq:6}V_{1}=-V_{2}=V_{o}\quad\text{and}\quad I_{1}=-I_{2}=I_{o}$

The characteristics of the two possible modes of the coupled transmission lines are now described. For the even mode, from Equations $$\eqref{eq:1}$$ and $$\eqref{eq:2}$$,

$\label{eq:7}\frac{d}{dx}\left[V_{1}(x)+V_{2}(x)\right] =-\jmath\omega\left[ L_{m}+L_{s}\right]\left[ I_{1}(x)+I_{2}(x)\right]$

which becomes

$\label{eq:8}\frac{dV_{e}(x)}{dx}=-\jmath\omega (L_{s}+L_{m})I_{e}(x)$

Similarly, using Equations $$\eqref{eq:3}$$ and $$\eqref{eq:4}$$,

$\label{eq:9}\frac{d}{dx}\left[I_{1}(x)+I_{2}(x)\right]=-\jmath\omega (C_{s}+C_{m})\left[V_{1}(x)+V_{2}(x)\right]$

which in turn becomes

$\label{eq:10}\frac{dI_{e}(x)}{dx}=-\jmath\omega (C_{s}+C_{m})V_{e}(x)$

Defining the even-mode inductance and capacitance, $$L_{e}$$ and $$C_{e}$$, respectively, as

$\label{eq:11}L_{e}=L_{s}+L_{m}=L_{11}+L_{12}\quad\text{and}\quad C_{e}=C_{s}+C_{m}=C_{11}+C_{12}$

leads to the even-mode telegrapher’s equations:

$\label{eq:12}\frac{dV_{e}(x)}{dx}=-\jmath\omega L_{e}I_{e}(x)$

and

$\label{eq:13}\frac{dI_{e}(x)}{dx}=-\jmath\omega C_{e}V_{e}(x)$

From these, the even-mode characteristic impedance can be found,

$\label{eq:14}Z_{0e}=\sqrt{\frac{L_{e}}{C_{e}}}=\sqrt{\frac{L_{s}+L_{m}}{C_{s}+C_{m}}}$

and also the even-mode phase velocity,

$\label{eq:15}v_{pe}=\frac{1}{\sqrt{L_{e}C_{e}}}$

The characteristics of the odd-mode operation of the coupled transmission line can be determined in a similar procedure to that used for the even mode. Using Equations $$\eqref{eq:1}$$–$$\eqref{eq:4}$$, the odd-mode telegrapher’s equations become

$\label{eq:16}\frac{dV_{o}(x)}{dx}=-\jmath\omega (L_{s}-L_{m})I_{o}(x)$

and

$\label{eq:17}\frac{dI_{o}(x)}{dx}=-\jmath\omega (C_{s}-C_{m})V_{o}(x)$

Defining $$L_{o}$$ and $$C_{o}$$ for the odd mode such that

$\label{eq:18}L_{o}=L_{s}-L_{m}=L_{11}-L_{12}\quad\text{and}\quad C_{o}=C_{s}-C_{m}=C_{11}-C_{12}$

then the odd-mode characteristic impedance is

$\label{eq:19}Z_{0o}=\sqrt{\frac{L_{o}}{C_{o}}}=\sqrt{\frac{L_{s}-L_{m}}{C_{s}-C_{m}}}$

and the odd-mode phase velocity is

$\label{eq:20}v_{po}=\frac{1}{\sqrt{L_{o}C_{o}}}$

Now for a sanity check. If the individual strips are widely separated, $$L_{m}$$ and $$C_{m}$$ will become very small and $$Z_{0e}$$ and $$Z_{0o}$$ will be almost equal. As the strips become closer, $$L_{m}$$ and $$C_{m}$$ will become larger and $$Z_{0e}$$ and $$Z_{0o}$$ will diverge. This is as expected.

## 5.5.1 Odd-Mode and Even-Mode Capacitances

The previous section used the even- and odd-mode capacitances for two coupled microstrip lines with the strips having equal cross section so that each strip had the same self-capacitances. In this section this restriction is removed and the results apply to any pair of coupled lines in any technology and of any cross section. The results enable the capacitances of the capacitance matrix to be determined from calculations of the charges on the lines. Repeating Equations (5.3.18) and (5.3.19),

$\label{eq:21}Q_{1}=C_{11}V_{1}+C_{12}V_{2}\quad\text{and}\quad Q_{2}=C_{21}V_{1}+C_{22}V_{2}$

the capacitance matrix is

$\label{eq:22}\mathbf{C}=\left[\begin{array}{cc}{C_{11}}&{C_{12}}\\{C_{21}}&{C_{22}}\end{array}\right]$

In the even mode $$V_{1} = V_{2} = V_{e}$$ and so the even-mode charges on strip $$\mathsf{1}$$ and strip $$\mathsf{2}$$ are

$\label{eq:23}Q_{1e} = (C_{11} + C_{12}) V_{e}\quad\text{and}\quad Q_{2e} = (C_{21} + C_{22}) V_{e}$

respectively. Defining the even-mode charge as

$\label{eq:24}Q_{e} = (Q_{1e} + Q_{2e}) /2$

then the even-mode charge becomes

$\label{eq:25}Q_{e} = V_{e}(C_{11} + C_{22} + C_{12} + C_{21})/2$

This leads to the even-mode per unit length capacitance,

$\label{eq:26}C_{e} = Q_{e}/V_{e} = (C_{11} + C_{22} + C_{12} + C_{21})/2$

Similarly, in the odd mode, $$V_{o} = V_{1} = −V_{2}$$, and the odd-mode charges on strip $$\mathsf{1}$$ and strip $$\mathsf{2}$$ are

$\label{eq:27}Q_{1o} = (C_{11} − C_{12}) V_{o}\quad\text{and}\quad Q_{2o} = (C_{21} − C_{22}) V_{o}$

respectively. The odd-mode charge is then

$\label{eq:28}Q_{o} = (Q_{1o} − Q_{2o}) /2=(C_{11} + C_{22} − C_{12} − C_{21}) V_{o}/2$

The odd-mode capacitance is

$\label{eq:29}C_{o} = Q_{o}/V_{o} = (C_{11} + C_{22} − C_{12} − C_{21})/2$

##### Example $$\PageIndex{1}$$: Parallel Line Capacitance

EM software can be used to determine the even- and odd-mode parameters of a coupled line. This is usually done by setting the phasor voltages on the coupled line and evaluating the phasor charges. Consider a pair of coupled microstrip lines as in Figure 5.2.4. The voltage applied to the left strip is designated as $$V_{1}$$ and the voltage applied to the right strip is $$V_{2}$$. The phasor charge on the strips are $$Q_{1}$$ and $$Q_{2}$$, respectively. The analysis is repeated, but this time with the substrate removed, and so establishing the free-space situation. In this case the charges are denoted by $$Q_{01}$$ and $$Q_{02}$$. The matrix of (computer-based) measurements is as follows:

Case $$V_{1}\:(\text{V})$$ $$V_{2}\:(\text{V})$$ $$Q_{1}\:(\text{pC/m})$$ $$Q_{2}\:(\text{pC/m})$$ $$Q_{01}\:(\text{pC/m})$$ $$Q_{02}\:(\text{pC/m})$$
A $$1$$ $$-1$$ $$70$$ $$-80$$ $$22.2$$ $$-24.7$$
B $$1$$ $$1$$ $$30$$ $$40$$ $$2.82$$ $$5.32$$

Table $$\PageIndex{1}$$

1. What is the two-port capacitance matrix?
2. What is the even-mode capacitance?
3. What is the odd-mode capacitance?
4. What is the free-space (no dielectric) two-port capacitance matrix?
5. What is the free-space even-mode capacitance?
6. What is the free-space odd-mode capacitance?
7. What is the even-mode effective relative permittivity?
8. What is the odd-mode effective relative permittivity?
9. Note that both the odd mode and even mode are TEM modes, so the phase velocity in the free-space situation is $$c$$. Determine the odd-mode and even-mode inductances per unit length in free space.
10. Note that the inductances do not change when the dielectric is replaced. What is the even-mode impedance?
11. What is the odd-mode impedance?
12. What is the even-mode phase velocity?
13. What is the odd-mode phase velocity?

Solution

1. Begin by considering the cross section of a coupled line shown to the right and use the basic equations relating the charges on the line to the voltages on them:

Figure $$\PageIndex{1}$$
$Q_{1} = C_{11}V_{1} + C_{12}V_{2}\quad\text{and}\quad Q_{2} = C_{21}V_{1} + C_{22}V_{2}\nonumber$
and consider two sets of voltage conditions.
Case A: This is the odd excitation, $$V_{1} = 1\text{ V}$$ and $$V_{2} = −1\text{ V}$$ and the charges are
$\label{eq:30}Q_{1A} = C_{11} − C_{12}$
and
$\label{eq:31}Q_{2A} = C_{21} − C_{22}$
Case B: This is the even excitation, $$V_{1} = V_{2} = 1\text{ V}$$ and the charges are
$\label{eq:32}Q_{1B}=C_{11}+C_{12}$
and
$\label{eq:33}Q_{2B}=C_{21}+C_{22}$
Adding Equations $$\eqref{eq:30}$$ and $$\eqref{eq:32}$$ results in
$\label{eq:34}Q_{1A} + Q_{1B} = 2C_{11},\quad\text{thus}\quad C_{11} =\frac{1}{2}(Q_{1A} + Q_{1B}) = (70 + 30)/2\text{ pF/m} = 50\text{ pF}$
Subtracting Equation $$\eqref{eq:30}$$ from Equation $$\eqref{eq:32}$$ yields
$\label{eq:35}Q_{1B} − Q_{1A} = 2C_{12},\quad\text{thus}\quad C_{12} =\frac{1}{2}(Q_{1B} − Q_{1A}) =\frac{1}{2}(30 − 70)\text{ pF/m} = −20\text{ pF/m}$
Because of reciprocity, $$C_{21} = C_{12} = −20\text{ pF/m}$$.
Subtracting Equation $$\eqref{eq:31}$$ from Equation $$\eqref{eq:33}$$ results in
$\label{eq:36}Q_{2B} − Q_{2A} = 2C_{22},\quad\text{thus}\quad C_{22} = \frac{1}{2} (Q_{2B} − Q_{2A}) =\frac{1}{2} (40 + 80)\text{ pF/m} = 60\text{ pF/m}$
Thus the per unit length capacitance matrix of the coupled line is
$\label{eq:37}\mathbf{C}=\left[\begin{array}{cc}{C_{11}}&{C_{21}}\\{C_{12}}&{C_{22}}\end{array}\right]=\left[\begin{array}{cc}{50}&{-20}\\{-20}&{60}\end{array}\right]\text{ pF/m}$
2. Even-mode capacitance, $$C_{e}$$:
The even mode has $$V_{1} = V_{2}$$ and the even-mode voltage is $$V_{e} = (V_{1} + V_{2})/2$$.
The even-mode charge is $$Q_{e} = (Q_{1} + Q_{2})/2 = (30 + 40)/2\text{ pC/m} = 35\text{ pC/m}$$, so
$\label{eq:38}C_{e}=Q_{e}/V_{e}=35\text{ pF/m}$
3. Odd-mode capacitance, $$C_{o}$$:
The odd-mode voltage is $$V_{o} = (V_{1} − V_{2})/2$$ and the odd-mode charge is $$Q_{o} = (Q_{1} − Q_{2})/2$$. With $$V_{1} = +1\text{ V}$$ and $$V_{2} = −1\text{ V},\: V_{0} = 1$$,
$\label{eq:39}Q_{o} = \frac{1}{2} [70 − (−80)]\text{ pC/m} = 75\text{ pC/m},\quad\text{thus}\quad C_{o} = Q_{0}/V_{0} = 75\text{ pF/m}$
4. Using a similar procedure to that in (a), but now using the free-space charge calculations, $$Q_{01}$$ and $$Q_{02}$$ results in the unit capacitance matrix:
$\label{eq:40}\mathbf{C}_{0}=\left[\begin{array}{cc}{12.5}&{-9.69} \\ {-9.69}&{15.0}\end{array}\right]\text{ pF/m}$
5. $$C_{e0} = 4.07\text{ pF/m}$$
6. $$C_{o0} = 23.5\text{ pF/m}$$
7. $$\varepsilon_{re} = C_{e}/C_{e0} = 35/4.07 = 8.6$$
8. $$\varepsilon_{ro} = C_{o}/C_{o0} = 75/23.5=3.2$$
9. Phase velocity, $$v_{p} = 1/\sqrt{LC}$$. With no dielectric, the phase velocity is $$c$$:
$v_{p} = c = 1/\sqrt{L_{0}C_{0}}\to L_{0} = 1/(c^{2}C_{0})\nonumber$
and the odd-mode free-space inductance is
$\label{eq:41}L_{o0} = 1/(c^{2} C_{o0}) = 1/\left[(3\cdot 10^{8})^{2}\: 23.5\cdot 10^{−12}\right]\text{ H/m} = 473\text{ nH/m}$
The free-space even-mode inductance is
$\label{eq:42}L_{e0} = 1/(c^{2} C_{e0})= 1/\left[(3\cdot 10^{8})^{2}\:\cdot 4.07\cdot 10^{−12}\right]\text{ H/m} = 2.73\:\mu\text{H/m}$
10. $$Z_{0} =\sqrt{L_{0}/C_{0}};\: L_{o} = L_{o0}\quad Z_{0} =\sqrt{473\cdot 10^{−9}/ (75\cdot 10^{−12})}\:\Omega = 79.4\:\Omega$$
11. $$Z_{e} = \sqrt{L_{e}/C_{e}} ⇒ Z_{e} =\sqrt{2.73\cdot 10^{−6}/ (35\cdot 10^{−12})}\:\Omega = 279\:\Omega$$
12. $$v_{pe} = 1/\sqrt{L_{e}C_{e}} = (2.73\cdot 10^{−6}\cdot 35\cdot 10^{−12})^{-\frac{1}{2}} = 1.023\cdot 10^{8}\text{ m/s}$$
13. $$v_{po} = 1/\sqrt{L_{o}C_{o}} = (473\cdot 10^{−9}\cdot 75\cdot 10^{−12})^{-\frac{1}{2}} = 1.68\cdot 10^{8}\text{ m/s}$$

## Footnotes

[1] Here $$I_{e} = (I_{1} + I_{2})/2$$ and $$V_{e} = (V_{1} + V_{2})/2$$. The reason for the supposedly equally valid definition $$I_{e} = I_{1} + I_{2}$$ not being used is that the adopted definition results in the desirable form of the even-mode characteristic impedance.

[2] Here $$I_{o} = (I_{1} − I_{2})/2$$ and $$V_{o} = (V_{1} − V_{2})/2$$.

This page titled 5.5: Symmetric Coupled Transmission Lines is shared under a not declared license and was authored, remixed, and/or curated by Michael Steer.