Phase Response

Now we may isolate the filter phase response $\Theta(\omega)$ by taking a ratio of the $a(\omega)$ and $b(\omega)$ in Eq.(1.5):

$$\begin{eqnarray*} \frac{b(\omega)}{a(\omega)} &=& -\frac{G(\omega) \sin\left[\Theta(\omega)\right]}{G(\omega) \cos\left[\Theta(\omega)\right]}\\ &=& -\frac{\sin\left[\Theta(\omega)\right]}{\cos\left[\Theta(\omega)\right]}\\ &\isdef & - \tan[\Theta(\omega)] \end{eqnarray*}$$

Substituting the expansions of $a(\omega)$ and $b(\omega)$ yields

$$\begin{eqnarray*} \tan[\Theta(\omega)] &=& - \frac{b(\omega)}{a(\omega)} \\ &=& - \frac{\sin(\omega T)}{1 + \cos(\omega T)}\\ &=& - \frac{2\sin(\omega T/2)\cos(\omega T/2)}{1 + [\cos^2(\omega T/2) - \sin^2(\omega T/2)]}\\ &=& - \frac{2\sin(\omega T/2)\cos(\omega T/2)}{2\cos^2(\omega T/2)} = - \frac{\sin(\omega T/2)}{\cos(\omega T/2)} = \tan\left(-\omega T/2\right). \end{eqnarray*}$$

Thus, the phase response of the simple lowpass filter $y(n) = x(n) + x(n - 1)$ is

$$\zbox {\Theta(\omega) = -\omega T/2.} \protect$$ (2.7)

We have completely solved for the frequency response of the simplest low-pass filter given in Eq.(1.1) using only trigonometric identities. We found that an input sinusoid of the form

$$x(n) = A \cos(2\pi fnT + \phi)$$

produces the output

$$y(n) = 2A \cos(\pi f T) \cos(2\pi fnT + \phi - \pi fT).$$

Thus, the gain versus frequency is $2\cos(\pi fT)$ and the change in phase at each frequency is given by $-\pi fT$ radians. These functions are shown in Fig.1.7. With these functions at our disposal, we can predict the filter output for any sinusoidal input. Since, by Fourier theory [84], every signal can be represented as a sum of sinusoids, we've also solved the more general problem of predicting the output given any input signal.