Phase Delay

The phase response $\Theta(\omega)$ of an LTI filter gives the radian phase shift added to the phase of each sinusoidal component of the input signal. It is often more intuitive to consider instead the phase delay, defined as

$$\zbox {P(\omega) \isdefs - \frac{\Theta(\omega)}{\omega}.} \qquad\hbox{(Phase Delay)}$$

The phase delay gives the time delay in seconds experienced by each sinusoidal component of the input signal. For example, in the simplest lowpass filter of Chapter 1, we found that the phase response was $\Theta(\omega) = -\omega T/2$ , which corresponds to a phase delay $P(\omega) = T/2$ , or one-half sample. Thus, we can say precisely that the filter $y(n) = x(n) + x(n - 1)$ exhibits half a sample of time delay at every frequency. (Regarding the discussion in §1.3.2, it is now obvious how we should define the filter phase response at frequencies 0 and $f_s/2$ .)

From a sinewave-analysis point of view, if the input to a filter with frequency response $H(e^{j\omega T})= G(\omega)e^{j\Theta(\omega)}$ is

$$x(n) = \cos(\omega nT)$$

then the output is

$$\begin{eqnarray*} y(n) &=& G(\omega) \cos[\omega nT + \Theta(\omega)]\\ &=& G(\omega) \cos\{\omega[nT - P(\omega)]\} \end{eqnarray*}$$

and it can be clearly seen in this form that the phase delay expresses the phase response as a time delay in seconds.