How do rise and fall times affect reflection in a transmission line?

Reflection will occur with shorter lengths as rise times becomes shorter, does this mean risetime effects the wavelength of a signal?

  • \$\begingroup\$ Consider how changing the rise and fall times affects the frequency components of the signal. \$\endgroup\$ – Hearth Jun 15 '19 at 16:56
  • \$\begingroup\$ The spectrum of a square wave would have higher amplitudes at higher frequencies. Then it makes sense for risetime to affect the wavelength of a signal. But how can I then define the wavelength since the fourier series extends to infinity? \$\endgroup\$ – N. Berg Jun 15 '19 at 17:02
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    \$\begingroup\$ There are losses in a transmission line. The higher frequencies are attenuated more and more severely. \$\endgroup\$ – analogsystemsrf Jun 15 '19 at 17:04
  • \$\begingroup\$ What do you mean by reflections will occur "sooner"? What do you mean by "sooner"? Do you mean reflections will occur at SHORTER transmission line lengths when rise times are short? It's not quite right to say that risetimes affect the wavelength of a signal (it's a bit ambiguous when you say a signal has a wavelength to begin with. A wave has a well-defined wavelengths and a signal contains many waves.). It's more correct to say that shorter risetimes introduce higher frequency components (that have a shorter wavelength) into the signal. \$\endgroup\$ – DKNguyen Jun 15 '19 at 18:16
  • \$\begingroup\$ I do mean on shorter line lengths. So I guess its more accurate to talk about a bandwidth. Since rise and fall times are related to amount frequency components in the square wave signal. I found online that the bandwidth required is about 0.35/Trise. Somehow it does not feel right talking about it this way and I feel like I'm missing something or confusing it with something else. \$\endgroup\$ – N. Berg Jun 15 '19 at 19:10

Reflections will occur no matter how fast the edges are.

The reflected wave will become superimposed with the incoming wave itself. If the rise/fall times are short enough the edges of the two waves will be distinguishable from each other. On the other side, if the edges are slow compared to the propagation times involved then the transmission line effects can be neglected for practical purposes. This is different than stating that no reflection is taking place.

Wavelength is a concept associated with periodic signals, it’s a frequency domain concept, while rise/fall time is a time domain concept. One can still associate a frequency domain spectrum to a non-periodic signal, like a single voltage step. It will be a continuous spectrum with no discrete frequency or wavelength.

Having said all that I have two answer for your question:

1 - No, the reflection does not change the wavelength of a signal, or in a more appropriate language, it does not change the frequency spectrum of a signal as long as (wait for it) the reflection characteristics (characterized by the reflection coefficient, if you want to dig more into it) is not frequency depend, what is the case for a pure short, pure open, or pure resistive load.

2 - Yes, the reflection can change the frequency spectrum of a signal if the reflection characteristics change with frequency. For instance you can have a matching circuit that produces no reflection at all at a particular frequency.

Going back to your wavelength... if you have a period signal... let’s think about a sine wave for simplicity... no linear phenomenon can change its frequency. In the case of a reflection, as long as there isn’t any non-linearity involved, the frequency of the reflected signal will be the same (amplitude and phase may change). Assuming there is no change in the medium, the propagation speed will be the same, and consequently the wavelength will be the same (remember that speed=wavelength*frequency).

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  • \$\begingroup\$ Thank you for the comment. I did not know propagation speed was dependent on frequency. Would this mean that even if a load was completely matched with a transmission line, distortion of the signal would still occur? When talking about a square-wave signal of course. \$\endgroup\$ – N. Berg Jun 21 '19 at 11:02
  • \$\begingroup\$ For most practical purposes you can assume that the propagation speed only depends on the medium and is frequency independent. For long transmission lines this is not totally true and differences in propagation times result in delay distortion of signals, but this is a second order effect - don't let it cloud your understanding of waves for now. \$\endgroup\$ – joribama Jun 21 '19 at 19:23
  • \$\begingroup\$ As far as matching, some differentiation is needed: there are narrowband matchings, typically used for RF circuits where you are only interested in a narrow frequency window, and there matchings that are broadband in nature, typically used in logic circuits where we deal with pulses with a wide spectral content. In broadband matching we don't expect to see much distortion on the reflection because it should treat all the frequency components of the signal the same way. Keep in mind that in a perfect matching we don't expect to see any reflection at all... \$\endgroup\$ – joribama Jun 21 '19 at 19:29
  • \$\begingroup\$ ... This is not the case for narrowband matching. Frequency components outside the bandwidth for which the matching was designed will be treated differently, what typically result in more reflection. In this case the waveform of the reflected signal can be significantly different than the waveform of the incident signal because it will have a different spectral content. In fact this is how an RF filter works: providing selective matching for the frequency band of interest. \$\endgroup\$ – joribama Jun 21 '19 at 19:36

There is a formula for a 1st order approximation of risetime to bandwidth, BW.

for \$t_r\$ (10 to 90%) = \$\dfrac{0.35}{ BW_{(-3dB)}}\$

Reflections depend on this risetime and not the repetition pulse rate.

There is a formula for propagation time which depends on the speed of light,c reduced by the relative medium constants, permittivity = \$\epsilon _r\$ and distance, x, for non-magnetic, μr=1,

\$v_p= \dfrac{c}{\sqrt{μ_rϵ_r}}\$

For coax. a common value for propagation speed is 2/3 c = 2e8 m/s= 20 cm/ns

If the rise time > 20x the prop. time to include reflection time then there is no significant transmission line effect. This is the same as 1/20 wavelength. Some applications neglect < 1/10λ track lengths for the effect of a single delay.

Thus 20x/(20cm/ns) means the Rule of Thumb is you can neglect (in most cases ) the Transmission Line effects using 1ns risetime per cm. ( microwave filters and grid phase power corrections would be examples of exceptions where less sensitivity to error is desirable.)


The most common mistake in computing prop. delay in FR4 is computing the effective reduced dielectric constant, Er(eff) for thin tracks small (w/h and t/h ratios) . For a ground plane, it is accurate but as the conductor changes from a plane wave to a thin stripline the inductance/mm increases and the relative ϵ reduces so using a good calculator such as Saturn avoids these errors or get electrical testing paid for on Striplines designs.

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