# Why don't these even/odd mode transmission line results math up?

In Pozar's Microwave Engineering, he analyzes the Wilkinson power divider using even-odd mode analysis. His results match the expected values and power conservation is satisfied. Similar derivations can be found online. Here's where I'm losing the analysis.

Consider a line of 1 ohm feeding a quarter-wave transformer of impedance sqrt(2) ohms, leading to a load of 2 ohms. This results in a matched input, $$\Gamma_{in} = 0$$ The transmission coefficient for the transmission line, confirmed by simple algebra as well as simulation, is $$T_{load} = -j$$ Where the subscript load indicates that this is the output transmission coefficient, S21. Using this result, we would expect to be able to write $$V_{load} = TV_{inc}$$ Or, for a coordinate system in x from the input port x=-lambda/2 to the load at x=0, $$V(0) = TV(-\lambda/4) = -jV_0$$ for amplitude V_0 i.e. the output voltage is a phase-shift input voltage.

This result in incorrect, however. The transmission coefficient is correct, but the output voltage is not found by it. The output voltage, V(0), can be shown to be $$V(0) = -j V_0 \sqrt 2$$ Which is scaled up slightly from the incident voltage. The reason, in short, is that by assuming: $$V(x) = V^+ (e^{-j\beta x}+e^{j\beta x}\Gamma)$$ We use V(lambda/4), $$V(\lambda/4) = j(1 - \Gamma) = V_0$$ And V(0), $$V(0) = V^+ (1 + \Gamma)$$ And we find $$\frac{V(0)}{V(-\lambda/4)} = -j \frac{1+\Gamma}{1-\Gamma}$$ Where Gamma is the reflection coefficient at the load. Then we arrive at $$\frac{V(0)}{V(-\lambda/4)} = -j \sqrt 2$$

So what happened? Where is the problem with the first derivation, and why in the second are we allowed to assume a matched input but that the load voltage will depend on the load reflection coefficient, when any analysis of a quarter-wave transformer is in some way based on multiple reflections, which makes this a strange idea?

• You can have a matched source and open or short load with any wavelength – Tony Stewart EE75 Mar 6 '20 at 5:55

In a two port network (e.g. your quarter-wave transmission line) $$\S_{21}\$$ is the transmission coefficient only if port 2 is matched. This is an important distinction that trips up a lot of people.

Let's define the reference impedance of port 1 and port 2 of the transmission line as $$\\sqrt{2}\Omega\$$, the same impedance as the transmission line. Then the $$\S_{21}\$$ of the two port network (ie the quarter-wave transmission line with characteristic impedance $$\\sqrt2\$$) is $$\-j\$$ like you calculated. Now say you connect a $$\\sqrt{2}\Omega\$$ source to port 1 and a $$\\sqrt{2}\Omega\$$ load to port 2. The fact that we matched port 2 to its reference impedance now implies that $$T_{load}=S_{21}=-j$$

But now suppose we replace the $$\\sqrt{2}\Omega\$$ load at port 2 with a $$\2\Omega\$$ load and drive port 1 with a $$\1\Omega\$$ source. Port 2 (and 1) is no longer matched! This means that we can no longer equate $$\T_{load}\$$ and $$\S_{21}\$$. Note that the $$\S_{21}\$$ of the network hasn't changed though, only the transmission coefficient. So in other words, we previously defined the reference impedance of port 2 as $$\\sqrt{2}\Omega\$$, so port 2 "expects" to see $$\\sqrt{2}\Omega\$$. Instead it sees $$\2\Omega\$$, meaning the reflection coefficient at port 2 has changed from the matched condition. (Note that matched does not mean zero reflection coefficient in this case, only that the load impedance is equal to $$\\sqrt{2}\Omega\$$).

So how can we solve this problem? One way is to redefine the reference impedances for port 1 and 2 as $$\1\Omega\$$ and $$\2\Omega\$$, respectively. Then you would have to recalculate $$\S_{21}\$$, essentially how you've done using your second (correct) method. In this case, $$\S_{21}\$$ would correspond to the transmission coefficient.

Another way of looking at this problem is in terms of power conservation. The transmission line is assumed to be lossless, thus the only potential spots for power dissipation are in the source impedance and in the load impedance. But we already know that the source sees no reflected wave, so we can ignore any loss in the source. Similar to standard electromagnetic waves incident on a boundary, the power of an incident voltage wave must equal the total power of the transmitted and reflected voltage waves.

So let's consider port 1: there is an incident voltage wave, $$\V_0\$$. There is no reflected wave, as you have calculated. The transmitted wave's power must all be dissipated in the $$\2\Omega\$$ load. The incident power is $$P_i=\frac{|V_0|^2}{Z_0}=\frac{|V_0|^2}{1\Omega}=|V_0|^2$$ The reflected power is $$P_r=0$$ The transmitted power is $$P_t=\frac{|V_{load}|^2}{2Z_0}=\frac{|V_{load}|^2}{2\Omega}$$ Now performing the power conservation step, $$P_i=P_r+P_t => |V_0|^2 = \frac{|V_{load}|^2}{2\Omega}$$ From which we find $$|V_{load}|=\sqrt{2}|V_0|$$ like you found in your correct derivation. Now of course, this method won't give you phase information, but power conservation is a good way to verify a result found from a different method.

A way of interpreting this result is that voltage amplitude will change to whatever it needs to be in order to maintain power conservation, similar to the electric field incident on a boundary. The amount it changes depends on the square root of the ratio between the load and the source (in this case $$\\sqrt{2}\$$), which follows from the ratio for power.