Linear phase filters delay all frequencies by the same amount. Why aren't they called Constant phase filters instead of Linear phase?

As I understand, if there is an input signal with two components $f_1$ and $f_2$, and it is passed through a linear phase system which introduces a delay $t$, $f_1$ is delayed by $t$, and $f_2$ is also delayed by $t$.

To what property does the phase have a linear relationship with? Could someone provide some insight into this? Some Math would be helpful.


1 Answer 1


Your premise is wrong. Linear phase filters offer linear phase.

Phase is not the same as time delay – at $f_1$, time delay $t$ will lead to a phase of $\varphi_1=-\omega_1\cdot t$, whereas at $f_2$, the phase will be $\varphi_2=-\omega_2\cdot t$ ($\omega$ is $2\pi f$); as you can immediately see, phase is a linear function of frequency.

(It's always good to remind oneself that frequency is just the derivative of phase over time, $\omega=\frac{d\varphi}{dt}$, and thus phase is just an integral of frequency over time.)

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    $\begingroup$ It can be added that the group delay is minus the derivative of phase with respect to frequency : $\tau = -\frac{d\phi(f)}{df}$, so a linear phase response (i.e. $\phi(f) = \alpha f$) will result in a constant group delay, as said in the title of the question. $\endgroup$
    – Albits
    Commented Dec 5, 2017 at 11:29
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    $\begingroup$ Just a minor correction: the phase corresponding to a delay of $\tau$ is given by $\phi(\omega)=-\omega\tau=-2\pi f\tau$. So there should be a negative sign and a factor of $2\pi$. $\endgroup$
    – Matt L.
    Commented Dec 5, 2017 at 13:12
  • $\begingroup$ @MattL. true; I probably wanted to go all $\omega$ on that notation, but frankly, it was written left-handedly on my bike ride to the office… $\endgroup$ Commented Dec 5, 2017 at 19:43

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