I am new to the world of digital filters and am educating myself with the book Introduction to Digital Filters by J.O Smith III,. The author derives the frequency response of a very simplistic filter:

$$y(n) = x(n) + x(n-1)$$ $$H\left(e^{j\omega T}\right) = 1+e^{-j\omega T}$$

where the phase response is $\Theta(\omega)=-\omega T/2$, which varies linearly with $\omega$. He then claims this phase response gives rise to a constant time delay irrespective of the signal frequency.

Could someone kindly shed light on why time delay is constant? Isn't the time delay the same as phase shift as reflected by the phase response? It looks like the time delay is the derivative of the phase response but I don't know why.

  • 2
    $\begingroup$ Group delay describes the time delay of the amplitude envelope of the sinusoidal components and is defined as the negative derivative of the phase response with respect to angular frequency $\tau_g(\omega)=-d\Theta(\omega)/d\omega$. en.wikipedia.org/wiki/Group_delay_and_phase_delay $\endgroup$
    – ZR Han
    Dec 30, 2021 at 1:39

2 Answers 2


To help your intuition, consider a sinusoidal signal with frequency $\omega_0$ and some arbitrary but constant phase $\phi$:


Delaying the signal $x[n]$ by $n_0$ samples gives


with the additional phase term


Consequently, for the delay $n_0$ to be independent of the sinusoids frequency, the additional phase must be a linear function of frequency. Note that a linear time-invariant (LTI) system introduces exactly such an additional phase term to a sinusoidal input signal.

For general signals and general LTI systems, very little can be said about the time delay introduced by the system. There are a few special cases, however, for which something useful can be said:

1. Systems with linear phase: $H(j\omega)=A(\omega)e^{-jn_0\omega}$

Apart from amplitude scaling by $A(\omega)$, each frequency component of the input signal is delayed by $n_0$ samples (assuming that $n_0$ is an integer).

2. Sinusoidal input signals are delayed by the system's phase delay evaluated at the input frequency: $$\tau=-\frac{\phi(\omega_0)}{\omega_0}$$

3. For narrow-band input signals, the delay of the signal's envelope is approximately given by the group delay of the system at the input signal's center frequency: $$\tau=-\frac{d\phi(\omega_0)}{d\omega}$$


One can show this mathematically by factoring and using Euler's formula

\begin{align} H(\exp(j\omega T)) &= 1 + \exp(-j\omega T) \\ &= (\exp(j\omega T/2) + \exp(-j\omega T/2)) \exp(-j\omega T/2) \\ &= (\cos(\omega T/2) + j\sin(\omega T/2) + \cos(\omega T/2) - j\sin(\omega T/2)) \exp(-j\omega T/2) \\ &= 2 \cos(\omega T/2) \exp(-j\omega T/2) \end{align}

One could argue that when $ \cos(\omega T/2)$ changes sign that it still has a contribution to the phase/delay, but at least for all other frequencies only $\exp(-j\omega T/2)$ contributes to the phase/delay.


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