1
$\begingroup$

let's consider a transmission line of length L closed on a mismatched load. So, there will be a travelling wave and a reverse travelling wave:

V(z) = V+(z) + V-(z)

My question is: will there always be a standing wave, or does it depend on if it is true that L = n * Lambda/2, with n integer?

Sometimes I read that to have a standing wave it is sufficient to have two waves of same frequency going in opposite direction, while sometimes I read that it is necessary that the length of the medium respects the previous formula.

$\endgroup$
1
$\begingroup$

There will be a standing wave whenever you terminate a transmission line with a mismatched load.

The line is resonant when the standing wave has a minimum or maximum of amplitude at the input (or feed point) of the line. If the termination is short, open, or has purely real impedance, then this occurs when the line length is $L = n\frac{\lambda}{4}$ for some integer $n$. If the termination is capacitive or inductive, resonance will occur for lengths that are offset from $n\frac{\lambda}{4}$, but still spaced by quarter wavelengths.

$\endgroup$
0
$\begingroup$

A pure standing-wave occurs only when no active power is delivered to the load, i.e., when the load is purely reactive, or a perfect short- or open-circuit. For example, if the load is a short-circuit, the voltage wave will have a node at the load (and at periodic intervals from the load). The length of the line makes no difference in his regard. It only determines what impedance will be seen by the source.

Now, say you add a small resistance to the load so that it is no longer a short, but the impedance mismatch is still large, then you will not get a standing wave, but the nodes of the wave will move slowly, i.e., you will have almost a standing wave. The standing-wave ratio is a measure of how close the wave is to being a standing wave.

$\endgroup$
  • $\begingroup$ So saying that mismatch => standing wave is not exactly true? $\endgroup$ – Kinka-Byo Oct 15 '19 at 4:14
  • $\begingroup$ I think so. You can always describe the total wave as having a standing wave component, which will be large when the mismatch is large. But if there is active power delivered to the load, the wave cannot be a pure standing wave because a standing wave carries no active power. $\endgroup$ – Amit Hochman Oct 15 '19 at 6:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.