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, 2013 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, 2013 at 5:09
  • 3
    $\begingroup$ @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, 2013 at 13:05
  • $\begingroup$ Sorry guys; I was thinking of series cascades only. Disregard. $\endgroup$
    – Jason R
    Feb 25, 2013 at 14:00

4 Answers 4


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, 2013 at 0:03
  • 3
    $\begingroup$ 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, 2013 at 3:27
  • 3
    $\begingroup$ 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, 2013 at 12:25

There is no misapplication of group delay nor a violation of physics or causality in this problem. The definition of group delay as the negative derivative of phase with respect to frequency still holds, in that each filter on its own has a positive time delay that is not constant over frequency. The details are revealed in what happens when filters are connected in parallel or series.

For this example of a cross-over filter the two filters are obviously in parallel to achieve the result shown, which is then very intuitive how the result could have 0 group delay: the two filters are a complimentary low pass and high pass; and when connected in parallel act as if neither of the filters are present (all-pass, with 0 delay). If these filters were connected in series the result would be a bandpass at the cross-over with an expected delay; the highpass would attenuate the low frequencies, and the lowpass would attenuate the high frequencies, and at the cross-over both signals will pass -3 dB of the signal, resulting in a magnitude of 0.5 and phase = 0° at the cross-over: $\frac{1}{\sqrt{2}}e^{j\pi/2}\frac{1}{\sqrt{2}}e^{-j\pi/2}$

Consider the generic case of two linear systems in parallel and in frequency as shown in the block diagram below with their frequency responses. Note that the coefficients and the exponents are functions of frequency which I omitted to keep the expressions simple and clear; $A_1 e^{j\phi_1}$ represents $A_1(\omega)e^{j\phi_1(\omega)}$ and both expressions are the Fourier Transform of the impulse response of those systems (frequency response), such as the plots shown by the OP for the high pass and low pass systems.

Parallel and Series systems

Consider the first case in light of the OP's question. At the cross over each filter has a magnitude and phase given as:

Highpass at cross-over: $\frac{1}{\sqrt{2}}e^{j\pi/2}$

Lowpass at cross-over: $\frac{1}{\sqrt{2}}e^{-j\pi/2}$

In parallel the result would be: $\frac{1}{\sqrt{2}}e^{j\pi/2} + \frac{1}{\sqrt{2}}e^{-j\pi/2}$ which is equal to 1 with angle 0. This case is easiest to see graphically as the addition of the two vectors:

addition at cross-over

In series the result would be $\frac{1}{\sqrt{2}}e^{j\pi/2}\frac{1}{\sqrt{2}}e^{-j\pi/2}$. When you multiply vectors you multiply the magnitudes and add the phases (exponents) so this result is simply 0.5 with phase 0.

And at the highest frequency each filter has a magnitude and phase given as:

Highpass as f $\rightarrow \infty$: $1e^{j0}$

Lowpass as f $\rightarrow \infty$: $0e^{-j\pi}$

Note that the result for the parallel case is still 1 with angle 0, but in the series case it approaches 0 (with angle $-\pi$) as the frequency approaches $\infty$. This makes sense, the high pass passes the signal (with no delay - the filter is transparant at the highest frequencies) but the lowpass completely blocks it, so nothing gets through. Further we see how the phase is changing in a negative direction as we pass through the cross-over, and the there is a delay in the bandpass result of the summed filters as given by the negative of the slope of the net phase shift versus frequency.

What happens in between requires a special mathematical relationship between the two filters in order for the parallel combination to sum to a zero phase (and therefore zero group delay, essentially making the parallel combination also transparent). Consider the OP's example where we can clearly see there is a quadrature relationship in the phase of the the two filters. Thus we have:

$$A_1e^{j\phi_1} + A_2e^{j\phi_2}$$ $$=A_1e^{j\phi_1} + A_2e^{j(\phi_1 - \pi/2})$$ $$=A_1e^{j\phi_1} + A_2e^{-j\pi/2}e^{j\phi_1}$$ $$=A_1e^{j\phi_1} - A_2je^{j\phi_1}$$ $$=e^{j\phi_1}(A_1-jA_2)$$

In order for this result to always have zero phase for all frequencies, the following equality must hold:

$$A_1-jA_2 = e^{-j\phi_1}$$

Or alternatively described as:

$$A_1+jA_2 = e^{j\phi_1}$$

Which is simply the real and imaginary component of the unit circle as we sweep $\phi_1$ over all possible values. Therefore when $A_1 = cos(\phi_1)$ and $A_2 = sin(\phi_1)$, the vector summation of the two filters will result in zero phase for all $\phi_1$ and therefore all frequencies.

relationship of A1 and A2

As for a possible intuition with the final plot that the OP showed and his question, consider that the derivative is a high pass function-- if you took the derivative of the red pulse you would get the green pulse as a result. You could not start to get this result until the red pulse is present, so there is no violation of causality.


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.

  • 2
    $\begingroup$ Sorry, that's not correct. Group delay is defined as the negative derivative of the phase vs frequency. That's the definition and as such can't be "misapplied". What you describe would actually measure the phase delay, not the group delay. In the case of a cascaded first order lowpass and highpass filter the results would be the same. Both group delay and phase delay are zero at all frequencies. $\endgroup$
    – Hilmar
    Oct 26, 2019 at 13:55
  • $\begingroup$ @Hilmar It believe it is the parallel combination of high pass and low pass filters (see my answer) not a cascade, agree? Also the measurement is indeed group delay at that frequency if we are measuring time delay. We can convert the time delay measurement to phase by multiplying the measured time delay by $2\pi / f$. $\endgroup$ Oct 26, 2019 at 15:15
  • 3
    $\begingroup$ You can't really measure time delay directly, if the time delay is frequency dependent. Hence the definition as either phase delay or group delay. Phase delay is $f/\omega$ , group delay is $\partial f/ \partial \omega$ Since the group delay is a derivative you can't determine it with a single measurement, you need a few measurements around the frequency of interest. $\endgroup$
    – Hilmar
    Oct 26, 2019 at 18:29
  • $\begingroup$ $f/\omega$ is $1/(2\pi)$? $\endgroup$ Oct 26, 2019 at 21:39
  • $\begingroup$ Yes. $\omega = 2 \cdot \pi \cdot f$, if that's what you are asking $\endgroup$
    – Hilmar
    Oct 27, 2019 at 17:20

Group delay is related to the group i.e. modulated signal thus the measurement of group delay should be done using the group (modulated signal). The group entering the filter should be the same with respect to its shape at the output of the filter. The shape means e.g. the spectrum of the group. Measurements done at a single frequency bear no information about the group delay.

  • 1
    $\begingroup$ I don't think this is accurate. Group delay is the measure of the slope of the phase response at any given frequency. We compute the group delay at each frequency, and over a bandwidth we use "group delay variation" to specify how much the group delay will vary over a bandwidth of interest. We need a range of frequencies of course in which to compute the derivative of the phase, but my understanding is that the computed delay based on taking the derivative of phase with respect to frequency is indeed the time delay that you would measure for single sine waves at each of those frequencies. $\endgroup$ Oct 25, 2019 at 3:15
  • 1
    $\begingroup$ Group delay is DEFINED as the negative derivative of the phase vs frequency. As long as you measure that, it doesn't matter how exactly you measure it and the results will be the same. Group delay can be INTERPRETED as the envelope delay of narrow band modulated signals, but the validity of the interpretation depends a lot on the exact circumstances. $\endgroup$
    – Hilmar
    Oct 26, 2019 at 14:02

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