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? I.e, 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 λ = 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 higher), 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?

  • \$\begingroup\$ The reflections settle out before the source voltage has changed. So they have minimal effect. \$\endgroup\$
    – user57037
    Commented Nov 18, 2020 at 16:30
  • \$\begingroup\$ @mkeith, ok, but why/how? \$\endgroup\$
    – jrive
    Commented Apr 8, 2021 at 14:49

2 Answers 2


The equation shown below describes the input impedance to a transmission line in terms of length, terminating impedance and characteristic impedance: -

enter image description here

So, if \$\ell\$ is zero then: -

$$Z_{IN} = Z_L$$

Do the math, plug some numbers in.

And of course for very long lines, any reflections are more highly attenuated by the return journey to the source so it's not usually a big deal except in power AC analysis.

So, as the line extends from zero length it starts to alter the impedance presented to the source but, for short lines and long wavelength this is pretty much all that it manifests as; it just looks like a slightly odd-ball impedance.

  • \$\begingroup\$ Thanks.....but, my question was about short lines and long wavelengths. And, are you saying this works for all conductors, not just transmissions lines (ie, when conductors are long relative to the wavelength? How do I figure the characteristic impedance of a conductor (not a 50ohm or other Z0-coax)? \$\endgroup\$
    – jrive
    Commented Nov 18, 2020 at 17:28
  • \$\begingroup\$ sorry, I accidentally voted you down. edit the question so I can bump you back up \$\endgroup\$
    – jrive
    Commented Nov 18, 2020 at 17:32
  • \$\begingroup\$ @jrive OK edited.... it applies to all conductor types because they are still transmission lines. All lines have inductance and capacitance per unit length so, if you know those values then characteristic impedance is $$\sqrt{\dfrac{R+j\omega L}{G+j\omega C}}$$ where G is conductance path between conductors per unit length. All lines are t-lines. \$\endgroup\$
    – Andy aka
    Commented Nov 18, 2020 at 17:39

Correct, it happens too fast to see.

Transmission Line theory says you get maximal power transferred at matched impedance yet of course when the load = supply impedance only half the voltage of no load.

So with typical analog and logic signals, the source impedance is far less than the load so you get twice the matched Z voltage from the reflected wave but since the prop. delay is so short for the presumed long wavelength, it happens instantly.

Yet we consider open circuit voltage as unity then consider losses in regulators from load effects unlike RF where we consider s21 the transmission loss relative to the ha,f the open circuit voltage or unity matched impedance voltage.

However you can see the effect in a DSO with a long ground clip lead as the coax impedance may be 100-150 Ohms for low pF/m loading effects, the added mismatch effects on the 10:1 RC divider network causes reflections in a 1m cable >20 MHz. As the reflections decay on each cycle, it merely appears as pulse ringing from an edge where the rise time and DSO capture bandwidth is greater than the standing wave frequency of the cable and ground clip length.

  • if you examine many photos on this site, you will find there are solutions to false ringing and evidence of poor test engineering test methods with lots of ringing in the 30MHz region. This is due to the propagation delay in the ground clip with L being 5 ~ 10 nH /cm and the C of the cable capacitance not the tip capacitance which is 10% of that. The LC filter effectively rings due an inherent group delay , impedance mismatch and resulting reflections seen as ringing.
    • to avoid the ringing arrange 2 test pins or vias to allow use of the 10:1 probe without a ground and tip clips, using a spring probe or carefully using just the pin and coaxial barrel of the probe. Or simply apply the 20 MHz LPF on the DSO, but otherwise, this improves the capture BW accuracy to approx 200 MHz. High BW is possible but requires more of the same with better connections and impedance matching.
  • \$\begingroup\$ thank you. ...so, what happens when you say "so you get twice the matched Z voltage from the reflected wave but since the prop. delay is so short for the presumed long wavelength...", shouldn't this then make my source output twice as large? Is the source absorbing this, and can potentially be damaged if levels are too high? \$\endgroup\$
    – jrive
    Commented Nov 18, 2020 at 17:36
  • \$\begingroup\$ No . A 50 ohm gen into 50 ohm load is 1V and no load is 2V steady state, yet the initial wave is 1V. CMOS is rarely terminated with matched impedance, but almost for RS485. Damage is not a concern when these properties are understood, but there are some power transmitters that need to be protected from opens or shorts . \$\endgroup\$ Commented Nov 18, 2020 at 19:10

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