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Introduction to Digital Filters
    The Simplest Lowpass Filter
       Finding the Frequency Response
          Phase Response

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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):

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

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

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

Thus, the phase response of the simple lowpass filter is

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

(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

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

produces the output

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

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.

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written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.

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