# Phase Delay Changes with Physical Length Changes

1. Feb 12, 2015

### jasonleroy

Hopefully I can word this correctly. Consider viewing S11 of a short on a smith chart. It starts at the left and moves clockwise around the edge. If you then add a small length of line between the reference plane and the short, the phase angle wraps further around the outside in the clockwise direction.

Question: Why does the phase angle of S11 get smaller with more physical length added to the line? I'm aware of the formulas to predict this, but what is the "physical" explanation?

2. Feb 15, 2015

### Baluncore

Such a seemingly simple question. There must be a really simple answer.

The first point to make is that port1, at any particular frequency, has a reflection coefficient called S11. The reflected signal S11 is measured relative to the incident “unit vector” wave on port1. The incident wave and the reflected wave are directional and so are treated as independent for the scattering matrix.

The Smith Chart on the other hand sums the voltages and currents to show what lumped model a line of that length and termination will appear to be at that frequency. The Smith Chart will be showing you the sum of the incident unit vector, Cos(0) = 1, and S11. That is (1 + S11).

So now the answer to your question. There is a double transit time delay in the reflected signal path, so S11 is an older sample of the port1 incident wave. That means it will have an earlier phase and so for short lines, (less than ¼ of a wavelength), the longer the line the more negative is the phase of S11.

There may have been a confusing complexity in that S11 is measured as an outward travelling wave, while a short circuited line reverses the voltage but the current continues to flow in the same physical direction. I believe those two inversions cancel.

3. Feb 15, 2015

### jasonleroy

Thanks for the response. What seems to be counter-intuitive to me is that a line of longer physical length shows up as having a smaller phase angle than the original. In my mind, longer length = larger phase angle. However, I must be thinking of it from the wrong perspective. The phase angle is a result of the entire system, not a measure of length from input to termination.

4. Feb 15, 2015

### Baluncore

It also depends on what you mean by the word "smaller". Is that closer to zero, or more negative?
Can you specify the system and show the equation that demonstrates your observation, over what range of lengths?

5. Feb 15, 2015

### jasonleroy

I suppose the simplest example would be measuring a perfect short. If there's no distance between you and the short, S11 is 1∠180°. Now, if you add a length equivalent to 30 degrees for example, you have a new S11 of 1∠120°. In my simplistic way of thinking about it, why isn't it 1∠240°? You are, after all, adding length and the signal is traveling further. There is obviously a fundamental flaw in the way I'm conceptualizing what's happening (the professor's answer was the reflected wave has a negative sign, so that is why you subtract the phase length Γin = ΓL*e-j2Φ ). I willing to accept that that's the way the math is set-up, but why does that happen?