Why connection of longer stub lengths cause more intense ringing in a Communication Network like CAN?

As stub lengths increase won't the attenuation increase and lead to less ringing? Also the time taken for the multiple reflections will increase so won't the ringing be less?

Please explain in terms of reflections.

One more doubt regarding multiple reflections I have is that if there is no attenuation will these multiple reflection cause continuous increase in amplitude of ringing?

  • \$\begingroup\$ The longer the stub is, the longer it disturbs the main transmission line by causing an impedance mismatch while the signal enters into the stub, and the longer it takes for the reflection to happen back from the open end of the stub back to main transmission line. But it all depends on bus length, bit rate, driver slew rate, distance between devices, etc. \$\endgroup\$
    – Justme
    Mar 8, 2020 at 16:29
  • \$\begingroup\$ It's not ringing; it's reflections superimposing on themselves. \$\endgroup\$
    – Andy aka
    Mar 9, 2020 at 8:10
  • \$\begingroup\$ @Andy aka, the reflection forms standing waves and the multiple reflections keep on adding to the standing waves making ringing greater right? Is this correct? \$\endgroup\$ Mar 10, 2020 at 11:56
  • \$\begingroup\$ It's not ringing; blah blah...... \$\endgroup\$
    – Andy aka
    Mar 10, 2020 at 12:02
  • \$\begingroup\$ @Justme can you please elaborate on what you mentioned. I am unable to understand how the longer stub line disturbs the transmission line? \$\endgroup\$ Mar 10, 2020 at 12:08

4 Answers 4



  1. The signal isn't sinewave, it covers a wide frequency range.
  2. A stub works as a resonator. Multiple reflections ring a long time at frequencies which have wavelength 4x the length of the stub or that divided by an integer (NOTE the wavelength must be calculated by using the signal velocity in the cable.)

So, let the lowest resonant frequency of a stub be Fo. The longer is the stub the lower is Fo. Also 2Fo, 3Fo, 4Fo.... are proper resonant frequencies. The longer the stub the more resonant frequencies fit into the frequency range of the signal, the higher is the total resonant reactive power.

  • \$\begingroup\$ If the multiple harmonics add on, I think the shape of resulting waveform will be different, I agree that the peak amplitude would be higher, but I was asking that as stub length increases, a lower frequency can resonate on the stub but it will have a higher amplitude compared the resonant frequency for lower length? \$\endgroup\$ Mar 10, 2020 at 12:06

Depends on the stub length. If the time it takes an EM wave to propagate down the stub and back is somewhat less than the rise time of the signal, then the size of any reflections due to impedance mismatches will be hidden inside the shape of the rise time waveform, e.g. the rise may be slightly faster or slower, but won't "bounce" or have any reverse dip, so might be hard to even notice with an oscilloscope. Make the stub longer, and the reflection voltage wave, if any, will arrive back after the signal has already completed risen, and thus will be quite visible.

And if a low loss wire stub is driven continuously at one of its resonant frequencies by the right impedance source, then it can act like an end-fed antenna, which is capable of generating some very high voltages, such as at the tips of a half wave dipole, where kilovolts are possible before the energy is radiated off and dissipated into EM waves.


"As stub lengths increase won't the attenuation increase and lead to less ringing?"

Yes, although the effects are very minor due to the relatively low resistance of the interconnects (copper in most cases). Sometimes you can help this situation by putting 10 ohm or 20 ohm resistors in series with the stubs that serve to dampen the reflections. But you really have to model your entire interface to see how much this gains you.

As to the effect of stub length on ringing, I thought I had posted some Hyperlynx simulations previously that showed this. I will try to locate the post or re-post those results here.


As stubs lengths get longer, at some point, depending on the frequencies of concern, they no longer act like stubs but rather act like transmission line segments. When this happens, at the junction of the main transmission line and the long stub, the signal "sees" an impedance that's half that of each transmission line segment. This is cause by having effectively two transmission lines in parallel. So you're going to get a significant reflection at this junction, back to the source.

EDIT2 - To answer Trilok's question

The following discussion of stubs assumes a single frequency signal, because it makes the discussions a bit easier. But keep in mind that for a square wave or even an edge, many frequencies are involved, and so the explanation is not quite as straightforward. Also, this applies to all frequencies, the only difference being the physical lengths at which they become apparent.

If the stub is short compared to the frequency of interest, then it just acts like a lumped-circuit. For a stub length less than ~1/20 of the wavelength, reflections come back to the main bus within the rise/fall time of the signal, and all you see is a little perturbation of the rising or falling edge.

With a stub length of ¼ wavelength, the reflections come back to the main bus 180 deg out of phase. This cancels out the signal on the bus, causing the classic “suck out” at that frequency. In an ideal case, the receiver sees no signal at this frequency.

Conversely, if the stub length is ½ the wavelength, the reflected signal comes back shifted 360 deg, so is in phase with the signal on the main bus, and adds to the amplitude.

You can extend this analysis to stubs that are multiples of ¼ wavelength or ½ wavelength. With longer length stubs, the resistive losses start to come into play. Note that resistive losses depend on frequency because high speed signals want to travel on the surface of a conductor and surface roughness of a copper trace can increase that resistance apart from what you might expect just from the resistivity of the conductor.

Now what if the stub is infinitely long? You get no reflection from the end (since the signal never reaches the end), but there is that impedance discontinuity I mentioned at the junction of the main bus and the stub. If the main bus is designed for 50 ohm impedance (close to 100 ohm differential impedance), as is the infinitely long stub, then at that junction the signal “sees” an impedance of 25 ohms (single ended), ~50 ohms diff, and so there is a partial (1/3) reflection at that junction back to the source.

Finally. The following picture, from one of Dr. Eric Bogatin’s Rule-of-Thumb papers, illustrates the suck-out effect of a stub (in this case an 0.2 in via) as a function of frequency.

Effect of 0.2 inch Stub, from Dr. Eric Bogatin

  • \$\begingroup\$ can you explain why stubs starts behaving like segments based on frequencies? And how long stubs make the impedance half? Won't the impedance seen at the junction be parallel combination of impedance seen through each stub or transmission line? \$\endgroup\$ Mar 10, 2020 at 12:11
  • \$\begingroup\$ a small clarification - I understand how the impedance is decreasing at the junction due to infinite stubs but the finite length stubs have capacitive or inductive impedance right? How do we analyse that case? \$\endgroup\$ Mar 11, 2020 at 4:04

copper does not have much loss, below 1,000 MHz.

most of your CAN energy is well below 100MHz.

  • 1
    \$\begingroup\$ Please elaborate your answer, describing what losses there are and motivate how these are negligable for CAN. \$\endgroup\$
    – Huisman
    Mar 9, 2020 at 8:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.