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This question might be suited for the Electrical Engineering forum, but I think I might get more scientific explanation/discussion (rather than anecdotal) on this forum.

Is the reflection phenomenon common in RF work with unmatched source/load pairs also present at low frequencies? Or are the wavelengths so much longer than the physical conductor lengths that reflections do not matter? Ie, are reflections strictly a function of wavelength relative to the conductors lengths?

Say for example I have a square wave fed into a pi-filter whose purpose is to extract a 100kHz fundamental (1/4 $\lambda$ = about half a mile). The source and the pi-filter (and load) are unmatched. Does the power reflect back to the source? the wavelength of the signal is much, much greater than the length of the conductors (about 10,000 times, for example, for 3" traces on a PCB)so I'm not sure how that would interfere with the source signal. Perhaps the wavelengths of the higher harmonics would eventually be of "meaningful" lengths, but they should be at much lower power levels....

Would I see a distorted signal at the source due to a reflection from the pi-filter? Could the reflection damage the source? If the power is not reflected, is it absorbed by the filter? where does it go?

Since we can often "get away" without terminating signals in low frequency designs, I hardly ever worry about reflections. I've typically only worried about them at at high MHz signals, and when interfacing signals across backplanes, or long board interconnects. But, it occurs to me that there should be reflections all the time. How is it that they don't typically matter (in low frequency designs)? How is it that sources driving unmatched loads are not "blown up" more often due to these reflections?

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  • $\begingroup$ This question has also been posted on EE.SE: electronics.stackexchange.com/q/533007/6334. $\endgroup$
    – The Photon
    Commented Nov 18, 2020 at 17:37
  • $\begingroup$ yes...I just did....different audiences.... $\endgroup$
    – jrive
    Commented Nov 18, 2020 at 17:37
  • $\begingroup$ Not really. Cross-posting is discouraged unless you've given the question a couple of days on the first place you posted it and not gotten a good answer. $\endgroup$
    – The Photon
    Commented Nov 18, 2020 at 17:38

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You always have a reflection from your termination but the real question is if it matters. Terminating a power amplifier with a sufficiently small impedance usually leads to sparks and smoke. If your connection to the termination is short, say less than $\lambda/10$, then the input impedance is very close to that of the termination so blowing up the generator is then up to the selection of the load but if the transmission line is long then the input impedance will strongly depend on its length and unpleasant surprises may abound, see the input impedance section in https://en.wikipedia.org/wiki/Transmission_line $$Z_{in}(\ell) = Z_0 \frac{Z_L + Z_0\mathfrak{j}\rm{tan}(\beta \ell)} {Z_0+Z_L \mathfrak{j}\rm{tan}(\beta \ell)}\\=Z_L\frac{1+(Z_0/Z_L)\mathfrak{j}\rm{tan}(\beta \ell)} {1+(Z_L/Z_0) \mathfrak{j}\rm{tan}(\beta \ell)}$$ Now let $\beta \ell <0.1$ you have $|\rm{tan}(\beta \ell)|<0.1$ and assume also that your load impedance is, say, $Z_0/10 <|Z_L| < 10Z_L$, then you will also have that $|Z_{in}| \gg 0$ and no worries; the shorter the length $\ell$ is the closer $Z_{in}$ is to $Z_L$ irrespective of the wave impedance $Z_0$. enter image description here

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  • $\begingroup$ Thank you for your response. Not sure I follow. When you say,"terminating a power amplifier with a sufficiently small impedance usually leads to sparks and smoke..." -- is that because of reflections or because excessive current draw? Also, if I understand you, so Zin is close to ZL due to long wavelengths and short conductor lengths, so the source sees practically ZL. What does that do to the reflected voltage ultimately reaching the source? $\endgroup$
    – jrive
    Commented Nov 18, 2020 at 17:52
  • $\begingroup$ a sufficiently small impedance is almost a short circuit, and shorting the terminals of a power amplifier will draw a large current, burn the output stage, etc. $\endgroup$
    – hyportnex
    Commented Nov 18, 2020 at 18:06
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    $\begingroup$ then the result of the reflection is that the load sees a voltage drop of the open circuit voltage from the source impedance on the load: $V_{in} \approx V_{source} Z_{in}/(Z_{source}+Z_{in})$ $\endgroup$
    – hyportnex
    Commented Nov 18, 2020 at 18:17
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    $\begingroup$ (1) you always get a reflection at the source if $Z_s \ne Z_0$ because the wave reflected from the load is still travelling on $Z_0$ line and it sees a load of $Z_s$ for the voltage has impedance $0$. (2) think of an LC circuit driven by the source $V_s, Z_s$, can the voltage on the LC at its resonance be larger than $|V_s|$ ? $\endgroup$
    – hyportnex
    Commented Apr 9, 2021 at 17:59
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    $\begingroup$ no problem, am glad to help $\endgroup$
    – hyportnex
    Commented Apr 9, 2021 at 20:24

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