I know that for a linear-phase filter with frequency response given by $$H(e^{j\omega}) = |H(e^{j\omega})|e^{j\phi(\omega)} $$ if the input of the system is $$x[n] = s[n]\cos[\omega_0 n] $$ where $s[n] $ is a narrowband signal with bandwidth $W\ll\omega_0$, then the output of the system is approximately $$y[n] \approx |H(e^{j\omega_0})|s[n-\tau_g]\cos[\omega_0 (n -\tau_p)] $$ (if $\tau_g$ and $\tau_p$ are integers) where $$\tau_g =\left.-\frac{d\phi(\omega)}{d\omega}\right|_{\omega=\omega_o} $$ $$\tau_p =\left.-\frac{\phi(\omega)}{\omega}\right|_{\omega=\omega_o} $$ A broadband signal can be thought as a superposition of narrowband signals, this means that if we have a filter with linear phase (i.e. $\phi(\omega)= \tau\omega$) both group delay and phase delay are constant and equal, therefore all the narrowband signal packets will be shifted in the same amount. But what about generalized linear phase filters? In this case $$\phi(\omega) = n_d \omega + \phi_o $$ so the output of a narrowband input centered at $\omega_0$ will now be $$y[n]= |H(e^{j\omega_0})|s[n-n_d]\cos\left[\omega_0\left(n -n_d-\frac{\phi_0}{\omega_0}\right)\right] $$ and the term $\frac{\phi_0}{\omega_0} $ will differ from one signal packet to another as the center frequency ($\omega_0$) of each packet changes. So why are generalized linear phase filters as important as linear phase ones? Isn't the term $\frac{\phi_0}{\omega_0} $ important?

The only related question I could find here did not have a complete answer to my question.


1 Answer 1


Note that with the definition of generalized linear phase $\phi(\omega)$ according to




the restriction that the impulse response $h[n]=\text{IDTFT}\{H(e^{j\omega})\}$ be real-valued only allows two possible values for $\beta$:


This is due to the required conjugate symmetry of $H(e^{j\omega})$: $H(e^{j\omega})=H^*(e^{-j\omega})$. The value $\beta=0$ obviously results in a system without phase distortion, and the value $\beta=\pi$ results in a sign reversal without any other phase distortions.

However, if you define generalized linear phase according to


with $\phi(\omega)$ given by $(2)$, and with a real-valued but possibly bipolar amplitude function $A(\omega)$, then for real-valued filters the constant $\beta$ can take on four different values:


For $\beta=0$ and $\beta=\pi$ the amplitude function $A(\omega)$ is even, and we get no phase distortion apart from a sign reversal for $\beta=\pi$. However, for $\beta=\pi/2$ and $\beta=3\pi/2$, $A(\omega)$ is odd, and there will be phase distortion, i.e., the group delay does not equal the phase delay. So for narrow-band signals, the envelope experiences a different delay than the carrier. Basically, - apart from the delay - the carrier is just shifted by $\pm 90$ degrees. This is what is desired when implementing differentiators or Hilbert transformers. Consequently, systems with generalized linear phase and with $\beta=\pi/2$ or $\beta=3\pi/2$ are important for implementing causal (FIR) approximations to differentiators and Hilbert transformers, where the phase shift is implemented exactly, apart from a delay necessary to make the filter causal.

In sum, systems with generalized linear phase generally distort the phase (unless $\beta=0$ or $\beta=\pi$), but these distortions are desired when implementing differentiators and Hilbert transformers, and - apart from a delay - they implement the desired phase exactly.

As a final remark, note that the approximation of the response of an LTI system to a narrow-band input signal of bandwidth $W$ using phase delay and group delay is only valid if the magnitude of the system's frequency response is (approximately) constant in the band $[\omega_0-W,\omega_0+W]$.

  • 1
    $\begingroup$ I know it may be beyond the scope of my original question, but for example for audio, if my filter has GLP with $\beta = \pi/2$ how will this distortion affect the way my signal sounds? If I want a FIR filter with absolutely no distortion, my two only possibilities are type I and II ? $\endgroup$
    – diegobatt
    Apr 6, 2017 at 2:11
  • 1
    $\begingroup$ @diegobatt: You can best try this yourself using Matlab or some similar tool. You will probably only notice the amplitude distortion, not so much the phase distortion. If you want zero (phase) distortion in the filter's pass band, you indeed have to use type I or II FIR filters. $\endgroup$
    – Matt L.
    Apr 6, 2017 at 7:14

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.