0
$\begingroup$

I've come upon a (seemingly) very elementary problem in my research, and can't quite get an answer that satisfies me among my peers, so here I am. It's about the reflection and transmission coefficients of a plane EM wave encountering a boundary between two different media, perpendicularly. I'll present the situation in terms of transmission lines and impedances, as this is the language we use in my work. I'm asking it here because I'm (nominally) a physicist doing physics research, but I might send it over to E.E. if you can't help.

Imagine a transmission line with a discontinuity separating region $A$ with impedance $Z_A$ (before the discontinuity) from region $B$ with impedance $Z_B$ (after the discontinuity). The reflection coefficient $\Gamma$ is given by

$$ \Gamma = \frac{Z_B-Z_A}{Z_B+Z_A}. $$

Let an incoming wave have amplitude 1. As it reaches the discontinuity, a wave of (complex) amplitude $\Gamma$ is reflected back in region $A$, and wave of (complex) amplitude $T=1+\Gamma$ is transmitted into region $B$.

I'm interested in the scenario where $Z_B$ is some multiple of $Z_A$, and to facilitate intuition, let $Z_B=3Z_A$, with both $Z_A$ and $Z_B$ real. In this case,

$$\Gamma = \frac{3-1}{3+1} = \frac{1}{2},\quad\text{and}\quad T=\frac{1}{2} + 1 = \frac{3}{2}.$$

Immediately, we see that the transmitted wave has a larger amplitude than the incident wave ($3/2>1$). This doesn't violate energy conservation, as the power delivered by a transmission line is proportional to $|V|^2/Z$, and we still have that $1^2/Z_A=|\Gamma|^2/Z_A + |T|^2/Z_B$. However, and this is my question, what's the intuition for the fact that a transmitted wave has an increased amplitude compared to the incident wave, as the boundary goes from a region of lower to higher impedance? Is anything wrong in my math, or my physical reasoning? This seems counter to the intuition of the analogy with a wave travelling along a string: when it goes from thinner/lighter string (lower impedance) to thicker/heavier string (higher impedance), we'd expect the amplitude to decrease, not increase!

Let me know if my question is not clear.

$\endgroup$
0

1 Answer 1

0
$\begingroup$

Here is one way to imagine this, using a tall sea wall being struck with plane waves. This wall is perfect in the sense that it moves almost not at all upon being struck by a wave and as such represents a very severe high-impedance mismatch.

We observe waves striking the wall and notice that the peak amplitude of the incident waves as they splash against the wall is much higher than their amplitude before striking the wall.

Thus we see that when a wave hits a higher impedance medium, its amplitude grows and the accompanying displacement becomes very small.

Now we imagine a sea wall that is low and has a deep hole behind it. Any wave that hits this wall encounters an extremely low impedance and as soon as its height exceeds that of the wall, the wave sloshes right over it and its amplitude almost vanishes.

This low-impedance load accepts the incoming wave but transforms its height-to-displacement ratio into something with almost no amplitude, but with significant displacement.

The state of affairs described here is completely analogous to the electrical case with which you are dealing.

$\endgroup$
2
  • $\begingroup$ Thanks for the reply, I'm partly convinced but still have some misunderstanding. In your analogy, I am not sure what the "displacement" refers to--is it the speed of the wave? In other word, would it be correct to summarize your point as that fact that when the wave gets into a higher impedance medium, the voltage increases but the speed of propagation diminishes? That might actually be a convincing answer for me, I had not thought of it that way. $\endgroup$
    – vinc.mack
    Dec 8, 2020 at 16:40
  • $\begingroup$ @vinc.mack, by displacement I meant the quantity of the flow variable which, when multiplied by the effort variable, yields power. propagation speed depends on the characteristic impedance of the medium and the impedance mismatch can be calculated on the basis of the propagation speed difference between the two media. $\endgroup$ Dec 8, 2020 at 17:40

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.