# Standing waves on a transmission line and time

1. Sep 27, 2012

### FrankJ777

Hi, I'm hoping somebody can help me understand something. I'm studying transmission lines and I'm confused about SWR and standing waves on a transmission line. According to my book the voltage on the transmision line is the super position of an incident and reflected voltage wave given by:
V$_{(z)}$ = V$^{+}_{0}$ ( e$^{-jBz}$ + $\Gamma$ e$^{jBz}$)

I can see how this produces a standing wave with voltage minimums and maximums at fixed points every λ/2. What I'm confused about is why the voltage is not represented as a function of time. From what I thought I understood about the phasor representation of the voltage we dropped the factor e$^{jωt}$ in our notation, but that it's understood that it's still there and the wave is still a funtion of time and frequency (jωt). So my questions are these. Are standing waves also a function of time and frequency, and if so would they be traveling waves? It seems that according to my text the min and max points are at fixed distances, so it seems that the voltage wave doesn't travel. So what happened to time and frequency?

Thanks to anyone who can set me straight.

2. Sep 28, 2012

### es1

3. Sep 28, 2012

### yungman

Standing wave is not time dependent, like a guitar string that the fundamental has the peak in the middle. If you go through the the wave traveling on the a transmission from the source at one end and termination on the other end. If the source and termination is not match, you will develop a standing wave pattern. It is the voltage traveling forward and reflect back that cause the standing wave pattern. If you solve the equation you posted with z, you'll see the standing wave pattern.

SWR only tells you the Vmax/Vmin or something like that, it does not tell you what standing wave pattern looks like, but from a standing wave pattern, you can find the SWR by just measuring it.

For a cosine wave,

$$\cos (\omega t-\beta z) = Re[e^{j(\omega t-\beta z)}]\;=\; Re[e^{j\omega t}e^{-\beta z}]\;=\;Re [ e^{\omega t}\tilde A ]$$

Phazor is defined as $\tilde A = e^{-\beta z}$

Phasor do not have time information, but it is understood that phasor is ONLY part of the function with time function ignored for simplification.

4. Sep 28, 2012

### FrankJ777

Thanks guys.
I found this video and it seems to to describe it pretty well visually.