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edit: I just found a site almost perfectly describe the issue, https://ophysics.com/w3.html

I found this graph on Google but couldn't find an explanation of it, it is similar to the one presented in my high-frequency engineering class.

  1. Why aren't the curves touching 0.00(except when reflection coefficient = 1)?

-> Shouldn't all cases of standing waves have points where it is always 0.00(nodes), so the amplitude is 0.00 too?

  1. Why aren't their minima as "sharp/pointy" as when reflection coefficient = 1?

I appreciate it if you can explain the problem in a more "graphical/less math" sense. Thanks!

Standing wave patterns on a line for various reflection coefficients showing maxima and minima

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  • \$\begingroup\$ When the reflection coefficient isn't 1, it's a standing wave plus a normal wave. \$\endgroup\$
    – user253751
    Jun 23, 2020 at 17:00
  • \$\begingroup\$ Are other cases therefore not standing waves then? Uh my lecturer really confused me with his vague wordings! \$\endgroup\$
    – Ming Pang
    Jun 23, 2020 at 17:12

2 Answers 2

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The voltage represented by your graph is the superposition (sum) of two AC signals, one propagating in the forward direction along the transmission line, and one propagating back.

We can represent an AC signal of a given frequency as a vector that represents its amplitude and phase. When we add the two signals together, we add their vectors, and the value in your graph is the length of the sum vector (from the origin).

It should be clear that you can only get total cancellation if the two signals have the same amplitude (i.e., total reflection of the signal). Otherwise, the sum can never hit the origin directly.

The "pointiness" is a little harder to explain. As you move along the transmission line, the two vectors that we're adding are rotating in opposite directions. If the sum passes through zero, it does so very quickly, creating a "pointy" graph. If it does not, the distance to the origin changes much more gradually and smoothly.

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  • \$\begingroup\$ Using vectors does explain things elegantly! Also I'd like to add that the amp. does not reach 0 when reflection is not 1, but the instantaneous value does. Apparently I mixed up both. \$\endgroup\$
    – Ming Pang
    Jun 23, 2020 at 18:20
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When reflection coefficient \$\rho=0\$, there are: -

  • no reflected waves
  • no standing waves and,
  • minima cannot exist or fall to zero

If, in the real world we had a reflection coefficient that wasn't quite 0 would anyone expect the reflected wave to have significant amplitude? Clearly not, and this does not cause the standing wave amplitude to drop to zero because the forward wave is hardly cancelled at any point by the reflected wave.

If \$\rho\$ grew a bit then the reflected wave will still not have the same amplitude hence it can never produce a standing wave with zero points.

Only when \$\rho=1\$ can a reflected wave have the same amplitude as the forward wave and only then can there be total cancellation at various points along the transmission line.

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  • \$\begingroup\$ I mixed up amplitude plot and the actual waveform. The waves do cross 0, but the amplitudes do not. The link I posted shows that they can act like "moving wave with changing amplitude". I like your deduction from when it's 0 to when it "grew a bit", thanks! \$\endgroup\$
    – Ming Pang
    Jun 23, 2020 at 18:04

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