In one of my previous posts I mentioned a result of a RF gain-versus-frequency measurement that looked like a sine wave. I said the sine wave probably means that there is some impedance mismatch in the measured signal path. I wasn't completely sure back then so I did a quick refresh on transmission line theory. Indeed, if you have a transmission line that is driven by a mismatched source *and* also has a mismatched load on the other end at the same time, the amplitude on the load will vary periodically in relation to the signal frequency.

Consider the following circuit. You have a sine wave source *u*_{g} with a source impedance that has a reflection coefficient Γ_{g}. The source is connected through a transmission line to a load with its own reflection coefficient Γ_{l}. The transmission line has a characteristic impedance *Z*_{0}, length *d* and propagation speed *c*. I've also marked the forward voltage wave in the transmission line *u*_{f} and the reverse voltage wave *u*_{r}.

We keep the amplitude of the signal source *u*_{g} constant and vary the frequency. If we measure the amplitude of the signal on the load *u*_{l} the amplitude will vary with applied frequency in a sinusoid like on the following graph. Note that the horizontal axis is frequency, not time. The vertical axis shows amplitude, not instantaneous voltage:

The amplitude on the load *u*_{l} will show peaks every Δ*f*. At the peaks, the signal will have the amplitude *u*_{max} and in the valleys it will have the amplitude *u*_{min}.

It turns out that Δ*f* has the following relation to transmission line length and propagation speed:

\Delta f = \frac{c}{2 d}

This equation can be useful in debugging since it can point to the part of the signal path where the mismatch is happening. For example, if you know the propagation speed it allows you to calculate the length of the mismatched segment.

The ratio of the maximum and minimum amplitude (in linear scale) on the load depends on the reflection coefficients at the source and the load ends of the transmission line:

\frac{u_{max}}{u_{min}} = \frac{1 + |\Gamma_g\Gamma_l|}{1 - |\Gamma_g\Gamma_l|}

This last equation is interesting. At the first glance it makes sense. If either of the ends is perfectly matched (Γ_{g} = 0 or Γ_{l} = 0), then the ratio is 1 and there is no dependency on frequency. This is the expected result. A transmission line that is mismatched on only one end still has a non-optimal power transfer but the amplitude on the load is constant and doesn't depend on the signal frequency.

A signal flow diagram, similar to the one mentioned in this article, can also be used to verify its correctness.

The equation for the ratio of amplitudes on the load is very similar to the equation for the voltage standing wave ratio:

\mathrm{VSWR} = \frac{1 + |\Gamma_l|}{1 - |\Gamma_l|}

In a VSWR calculation you only have a mismatched load. In the case where you have two mismatched ends, it then kind of makes sense that you multiply both reflection coefficients together.

However if you think about it, it's quite weird that it turns out this way. In the VSWR case you are calculating the ratio of the sum and difference of the forward wave (*u*_{f} = 1) and the reflected wave (reflected once off the load, hence *u*_{r} = Γ_{l}). Since the source is perfectly matched in that case, the reflected wave doesn't reflect back the second time. It's obvious that the formula for the ratio is this simple.

On the other hand in the case where both the source and load are mismatched, you get an infinite number of reflections. When the source first gets switched on, the first forward wave driven by the source reflects off the load with Γ_{l}. When the reflected wave gets to the source it reflects off the mismatched source impedance with Γ_{g} and travels back to the load superimposed on the original forward wave. This combination of the signal driven by the source and the first reflection again reflects off the load and so on to infinity. The steady-state amplitude of the signal on the load is the sum of all those infinite reflections.

If you think about it this way, it's surprising that the ratio ends up being this simple equation that is the same as if the signal only reflected once from the load and once from the source.