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Reflections in a transmission line is causing ringing in square wave because certain frequencies are getting enhanced, but when frequency of the square wave changes won't the amplitude of those enhanced frequency decrease or increase because the amplitude of higher harmonics decrease or increase when we change the frequency of square wave according to Fourier series expansion? Still we don't see any change in amplitude of ringing in square wave. Why is that so?

I understand why the frequency won't change but I don't understand why the amplitude of ringing won't change with the change in frequency of square wave. Please help me understand.

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  • \$\begingroup\$ The highest frequency components of a square wave are dependent on the rise and fall times. This is what is leading to the ringing not the frequency of the square wave it self so why do you think the amplitude will change? \$\endgroup\$
    – Warren Hill
    Mar 7, 2020 at 14:22
  • \$\begingroup\$ @Warren Hill But what about the Fourier series expansion? Is it not contradicting? Won't the amplitude of higher frequency decrease when we reduce frequency of square wave? \$\endgroup\$
    – Trilok Girish Kamagond
    Mar 7, 2020 at 14:33
  • \$\begingroup\$ The highest frequency component \$ \omega_n \approx \dfrac{1.8}{\text{rise time}} \$ If you double the frequency this component is still there but it is now say the 33rd harmonic instead of the 66th. \$\endgroup\$
    – Warren Hill
    Mar 7, 2020 at 14:48
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    \$\begingroup\$ it's not inherent to a square wave, it's a response of a system to a step change... \$\endgroup\$
    – vicatcu
    Mar 7, 2020 at 15:19
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    \$\begingroup\$ @vicatcu Thank you. I am getting sense of it. \$\endgroup\$
    – Trilok Girish Kamagond
    Mar 7, 2020 at 15:28

4 Answers 4

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That is because the edges of the square wave is identical regardless of the rate of when the edges happen. So what happens at the edge does not depend on the frequency of the square wave.

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  • \$\begingroup\$ But what about the Fourier series expansion of square wave? Is it not contradicting? Won't the amplitude of higher frequency decrease when we reduce frequency of square wave? \$\endgroup\$
    – Trilok Girish Kamagond
    Mar 7, 2020 at 14:33
  • \$\begingroup\$ Can you please relate how your explanation is relating to Fourier series expansion? Please. \$\endgroup\$
    – Trilok Girish Kamagond
    Mar 7, 2020 at 14:47
  • \$\begingroup\$ In practice 1 Hz and 1 kHz square wave do not differ that much on the whole, if they both have infinitely fast edges so their bandwidth extends to infinity. You basically have two things superimposed and mixing together to consider; the slope of each edge, and the repetition rate of edges. The repetition rate of edges is just so much less than the frequencies of the edge. \$\endgroup\$
    – Justme
    Mar 7, 2020 at 15:16
  • \$\begingroup\$ so in this case, I should consider my system response to slope of edge rather than the square wave right? \$\endgroup\$
    – Trilok Girish Kamagond
    Mar 7, 2020 at 15:40
  • \$\begingroup\$ Can you also recommend any reference where I can read more on this ringing? \$\endgroup\$
    – Trilok Girish Kamagond
    Mar 7, 2020 at 15:44
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Here is the logic of solving math problems using the frequency domain:

  • This differential equation stuff is hard. In fact, it's practically impossible, and there's all this housekeeping mixed in. I wish there were a better way!
  • Oh look, if I'm analyzing a linear, time-invariant system, I can transform my entire problem into an entirely different domain, and the math gets easier!
  • I do need to always remember that if the system is LTI, the two domains are equivalent, just a different way of seeing the same thing. Then I can go back and forth at will.
  • Gosh it's hard to think about things in the frequency domain, but look! I can actually do the math. I guess I'll stick with it because I can find solutions.

So you're faced with a problem that's easy to think about in the time domain, and even more difficult than usual to think about in the frequency domain. Then remember that the two domains are equivalent, and if an answer is valid in one, it's valid in the other.

Then think it through in the time domain, where you'll see that as long as your square waves take longer to switch state than the settling time of your cable assembly, you can reason everything out in the time domain without getting wrapped up in the frequency domain stuff.

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  • \$\begingroup\$ But what about the Fourier series expansion of square wave? Is it not contradicting? Won't the amplitude of higher frequency decrease when we reduce frequency of square wave? \$\endgroup\$
    – Trilok Girish Kamagond
    Mar 7, 2020 at 14:35
  • \$\begingroup\$ Is it not contradicting what? It can't -- the Fourier transform is complete and exact; it loses no information. Why don't you edit your question with a drawing or sketch of the circuit you're asking about, and possibly a graph of what you think is going on in the frequency domain. \$\endgroup\$
    – TimWescott
    Mar 7, 2020 at 16:07
  • \$\begingroup\$ Thank you. I am getting the sense of my mistake. \$\endgroup\$
    – Trilok Girish Kamagond
    Mar 8, 2020 at 4:46
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To expand on the other answers a bit -- yes, it's true that any LTI system will not have any frequencies in its output that are not present in its input.

However, the "ringing" you see in a mismatched transmission line is the result of multiple reflections of the fast edges in the driving signal, and the timing of those reflections is independent of the fundamental frequency of that signal. And as Tim says, this is easier to think about in the time domain than in the frequency domain.

In the frequency domain, the source signal (and each of its reflections) contains a rich array of harmonics. When you consider the delayed reflections, the components at each harmonic frequency add up in complicated ways because of the varying phase angles associated with each reflection. But the end result gives you the same time-domain waveform, which appears to have frequencies in it that are not present in the original signal. This is an illusion.

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  • \$\begingroup\$ can you please elaborate on "timing of reflections independent of fundamental frequency" ? Please!! I am actually mind blown by the answer. \$\endgroup\$
    – Trilok Girish Kamagond
    Mar 7, 2020 at 15:35
  • \$\begingroup\$ Can you also please tell me where I can read on this? I am unable to find any reference regarding the same.. \$\endgroup\$
    – Trilok Girish Kamagond
    Mar 7, 2020 at 15:37
  • \$\begingroup\$ And also the "delayed reflections, the components at each harmonics add in complicated way" what do you mean by components at each harmonic? \$\endgroup\$
    – Trilok Girish Kamagond
    Mar 7, 2020 at 15:51
  • \$\begingroup\$ The timing of the reflections is strictly a property of the transmission line (and its terminations), and not at all dependent on the signal you feed into it. As for the phase relationships, consider that a reflection of the signal has a delay of time \$t\$. The square wave contains frequency components at 1x, 3x, 5x, etc. of the fundamental. The phase angle between the original signal and its reflection(s) for any given frequency component depends on \$t\$, and it is in general different for each component. \$\endgroup\$
    – Dave Tweed
    Mar 7, 2020 at 18:50
  • \$\begingroup\$ I still don't get what is meant by timing of reflections? Is it the delay or what is it? Please explain. \$\endgroup\$
    – Trilok Girish Kamagond
    Mar 8, 2020 at 5:12
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Ringing is due to the speed of light (or electromagnetic propagation thru a region of given permittivity and permeability) and the length of the wire. That controls how long it takes for a signal to "bounce" back and forth. (given some impedance mismatch).

Note the length of a wire is completely independent of the frequency of any signal. (and thus any harmonics of the signal).

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