I keep reading that a constant group delay over a frequency band is desirable for a filter. But a constant group delay means a linear decrease in the phase delay, which would distort the signal.

Let's assume that I have a group delay of 5 [time units] over a frequency band from 100 Hz to 110 Hz. That means that the phase delay is $-5\omega + c$.


  • the 100 Hz component will be delayed by 500 + c rad
  • the 101 Hz component will be delayed by 505 + c rad
  • the 102 Hz component will be delayed by 510 + c rad
  • etc

Considering that different frequencies components have different phase delays, the signal will be distorted.

So why is a flat group delay response desirable for a filter?


1 Answer 1


This is actually pretty simple. Take a pure-tone sine wave, say

$$s(t) = \cos\left(\omega t + \phi \right).$$

Now, fix a time $t_0$ and find the phase $\psi(t_0)$ of $s(t_0)$. We get

$$\psi(t_0) = \omega t_0 + \phi$$

Now let's vary the frequency $\omega$ and keep the time fixed. The above expression becomes a linear function in $\omega$! Even though we haven't actually "delayed" the wave.

Imagine it visually. You have a point in time (not 0, that will always stay the same) and a sine wave which is, say, shrinking (frequency increases). As it shrinks, different phases of this sine wave will cross your fixed time point for different frequencies.

This is why phase response of a simple delay system is a sloped line, not a horizontal line. Its derivative, however, will be horizontal and proportional to the amount of delay we introduced in the system, i.e

$$\frac{\partial}{\partial \omega} \left( \omega t_0 + \phi \right) = t_0,$$

Which is precisely the system delay.

  • 1
    $\begingroup$ sorry to rain on the parade, but this answer is about "phase delay" and not about "group delay". they have a relationship (like they are the same value, for a linear-phase filter), but they are not the same. $\endgroup$ Commented May 21, 2014 at 19:44
  • $\begingroup$ @robertbristow-johnson Ah, thanks for pointing that out. I've misread the question. Will edit. $\endgroup$
    – Phonon
    Commented May 22, 2014 at 9:04

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