If you put a wave packet through the passband of a 1st-order low-pass filter, it will be delayed by the group delay of the filter, and remain the same amplitude, right?

If you put the same wave packet through a complementary 1st-order highpass filter with the same cutoff frequency, the group delay curve is the same, so the delay of the packet will be the same, but the gain is much lower, so it will be both delayed and attenuated to negligibility.

Since the output of the highpass filter is very small, if you sum the outputs of these two filters (as in an audio crossover), I would expect it to be negligibly different from the output of the lowpass filter: Large delayed signal + very small delayed signal = large delayed signal.

Yet if you sum the filter responses, the amplitude is 0 dB everywhere, and the phase is 0 everywhere, and therefore the group delay becomes 0, which would mean that the wave packet comes out with no delay and no changes. I don't understand how this can be possible. Don't filters always incur delay? How can a filter (which also has positive group delay) undo the delay caused by the other channel, especially when this is happening in the stopband?

Which part am I misunderstanding here?

The best-known crossover types with linear phase are first-order non-inverted crossovers, ... The first-order crossover is minimum phase when its outputs are summed normally; it has a flat phase plot at 0°. - The Design of Active Crossovers


Here the result of summing the outputs together produces 0° phase shift, which is to say that the summed amplitude and phase shift of a 1st-order crossover is equivalent to a piece of wire. - Linkwitz-Riley Crossovers: A Primer: 1st-Order Crossover Networks

First-order crossover frequency response

Testing on actual pulses shows how the lowpass (blue) delays the pulse, as expected, and how the highpass (green) can combine with it to produce the original (red) pulse, but how is the highpass pulse occurring before the original if the highpass filter is causal and has positive group delay? Intuition is failing me.

enter image description here

It does show that the highpass output is not as negligible as I imagined, and the delay is more negligible than I imagined, and as you move the carrier frequency around, these two properties change in a proportional way (smaller delay requires lower amplitude highpass output to correct it). But I still don't really understand it.

  • $\begingroup$ So you're implying that the two filters are matched such that their transfer functions sum to unity (i.e. $H_{lp}(z) + H_{hp}(z) = 1$)? That would also imply that the sum of their impulse responses is just an impulse at $n=0$, which would agree with your observation of zero group delay. I think your assumption that the phase of the two filters sum to zero is probably faulty. $\endgroup$ – Jason R Feb 25 '13 at 3:24
  • $\begingroup$ @JasonR: Yes, 1st-order filters, highpass and lowpass, with the same fc. en.wikipedia.org/wiki/Audio_crossover#First_order $\endgroup$ – endolith Feb 25 '13 at 5:09
  • 2
    $\begingroup$ @Jason: endolith is indeed correct. First order hi/lo pass reconstructs perfectly in parallel. There are other cases which do this as well $\endgroup$ – Hilmar Feb 25 '13 at 13:05
  • $\begingroup$ Sorry guys; I was thinking of series cascades only. Disregard. $\endgroup$ – Jason R Feb 25 '13 at 14:00

There are a couple of interesting aspects of "reconstruction to unity". First, there are two ways of combining two filters: parallel and in series. For a parallel topology it is ALWAYS possible to find a complimentary filter so that the pairs add to unity. It's easy enough, actually. Simply do $\tilde{H}(\omega) = 1-H(\omega)$. In the time domain that means that the impulse response of the complimentary filter is simply the negative of the original impulse response with 1 added to the first sample. So all the "ringy" stuff cancels out. Now the shape of this complimentary filter is not always what one would expect. For a 1st order low pass it actually is a first order high pass but for higher order filters it tends to have over/under swings in the cutoff region. However, it always exists as a stable causal filter.

Series (or cascade) "reconstruction to unity" is bit more complicated. Obviously the filters would have to be the inverse of each others, i.e. $\tilde{H}(\omega) = \frac{1}{H(\omega)}$. In general this can be done for any minimum phase filter. The inverse of a minimum phase filter is minimum phase as well and both are causal and stable.

So this leaves us with the question of how to interpret group delay in these cases. The cascade case is actually the more interesting. Since the filters are inverse of each other, the phase, and hence the group delay, of one is the negative of the other. So at frequencies were one filter has positive group delay, the other has negative group delay. An easy example would be a low shelf with +6dB of gain and a low shelf with 6dB of cut. So negative group delays are very real and certainly not a violation of causality. In practice, these show up in areas of the filter that are fairly "non-flat" so the traditional interpretation of "delay of the envelope" doesn't quite apply since there is a fair amount of amplitude distortion as well.

If you Google "negative group delay", you can find a few IEEE articles that have tackled the subject.

  • $\begingroup$ Ok, but the part that's confusing is that both filters have positive group delay, yet combine to produce an output with zero group delay. $\endgroup$ – endolith Feb 26 '13 at 0:03
  • 2
    $\begingroup$ Remember that group delay is the (negative) derivative of phase. For a parallel cascade, the phases of the two systems do not add, as they would in a series connection. Therefore, we shouldn't expect that the group delays of the two systems would add either. $\endgroup$ – Jason R Feb 26 '13 at 3:27
  • 1
    $\begingroup$ Here is another way to think about. The group delay is the same, but the delayed parts are out of phase so they cancel each other out. $\endgroup$ – Hilmar Feb 26 '13 at 12:25

I thought this was a rather interesting question so I will try to answer it, albeit 5 years late.

I think you have discovered a way to misapply one of the ways to measure group delay, which is to say, calculating it as the negative derivative of the phase. In this situation, this method is not appropriate.

In this situation a more appropriate way to measure the group delay is to use a sine wave input and and measure the delay between the input and summed output. Of course, in order to get a complete picture you will need to do a frequency sweep, which is a hassle, but accurate.

If you do this, I think we can all agree you will measure a nonzero group delay.


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.