 ## Monday, 25 December 2017

### SWR, Transmission Lines and Terminations

Well,

Following a discussion at a local radio club, I have been trying to put together a simple demonstration to illustrate standing waves on transmission lines. Now, this can be quite a complex topic, but here I will try and simplify the problem so we can see what's actually going on. There is a great explanation on Wiki here. W2AEW also has an excellent YouTube channel which covers some of this topic very well.

I am going to assume some prior knowledge about transmission lines, but here are the two basic questions that I often hear:
1. Why do I need to match my antenna to my coax and transmitter?
2. What is a good SWR and why does it matter?
Well, in the world of ham radio we match an antenna to a transmission line and a transmitter to minimise reflected waves, let's try and see why reflected waves are a bad thing.

My test setup here comprises a 50 ohm signal generator putting out a 10MHz square wave. We have a short length of 50 ohm coax to a BNC T-Piece and then another length of 50 ohm coax to a BNC connector.

The BNC connector on the end is connected to channel one of my scope (yellow waveforms) and the T-piece is connected to channel 2 of the scope (blue waveforms).

Now, with the 'scope set so that the end of the coax is seeing a 50 ohm load here are the signals that we see on the scope:

The signal at the end of the transmission line (yellow) is much the same as the signal part way along the transmission line (blue). We can even measure the delay from the t-piece part way along my transmission line to the end:

It is clear from the above scope screen that it has taken about 11ns for the signal to move from the T-Piece part way along my transmission line to the end.

Everything looks exactly as expected because the line is terminated at the design impedence of 50 ohms.

Now, if I change the line termination impedance to be 1M ohm, this is the scope screen now:

Here, two interesting things have happened. Firstly the amplitude of the signal at the end of the line (yellow) has doubled, secondly the signal at the T-piece part way alone the line is very distorted.

Because there is a mismatch (in this example a very bad mismatch), the change in impedance will generate a voltage spike which in turn will create a current flowing in the opposite direction, and that is exactly what we can see.

In the case of the yellow waveform, the signal ariving at the end of the transmission line has started to reflect 100% back down the line, but because we are "looking" right at the end of the line, there is no delay so the two signals simply sum together to double the waveform amplitude.

I've annotated the blue waveform to explain whats happening here:

So here we have at point A the start of the signal traveling from the signal generator to the load, 11 ns later (we measured the time earlier and can see its the same here) we reach point B, at this point the reflected waveform is arriving in the opposite direction on the transmission line and the two sum to make the total amplitude of the waveform. Point C is the end of the outgoing waveform and point D the end of the reflected waveform. This is why we see the two signals adding but slightly out of phase with each other - each step in the waveform is the 11ns we know it takes for the signal to travel from the t-piece in my transmission line to the end (or the same distance but in the other direction).

I can set up the scope to illustrate this kind of thing in another way, and the resultant view looks like this video here:

So in this video you need to imagine that the yellow waveform is travelling from left to right from the transmitter to the antenna. Similarly the blue waveform is the reflected waveform travelling from the antenna back to the transmitter. The purple waveform in the middle is the resultant signal on the transmission line - hopefully it's clear that this wave is "standing" and not moving - hence the name. The standing wave peaks at twice the amplitude of each of the individual signals because it is the sum of the two.

So in this mock-up example all the power from the transmitter is reaching the end of the transmission line and reflecting backwards to go back from whence it came - that's very bad for the RF source.

If we think about the calculation of SWR:

Hopefully, we can see that if the reflected power is the same as the forward power then the SWR is infinite.

If we take a second example whereby the forward waveform amplitude is 1 Vpp and the reflected 0.5 Vpp (50% of the forward voltage), then the forward power is 2.5 mW and the reflected 0.625 mW, then the maths tells us:

that the SWR is 3 : 1.

So before the SWR reaches a value considered "acceptable" we need to have a reflected power equal to 4% or less of the transmitted power as that will deliver an SWR of 1.5:1 or less.

I also hope you can see for the SWR to be 1:1 the reflected power needs to be zero and chasing this ultimate aim is rather pointless.

Good, egh?