Frequency Response

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Frequency Response

Given the transfer function , the frequency response is obtained by evaluating it on the unit circle in the complex plane, i.e., by setting $z=e^{j\omega T}$ , where is the sampling interval in seconds, and $\omega$ is radian frequency:^4.3

$\displaystyle H(e^{j\omega T}) \isdef \frac{1 + g_1 e^{-jM_1\omega T}}{1 + g_2 ... ... T}}, \qquad -\pi \le \omega T < \pi \qquad\hbox{(Frequency Response)} \protect$

(4.4)

In the special case

, we obtain

$\begin{eqnarray*} H(e^{j\omega T}) &=& \frac{1 + e^{-jM_1\omega T}}{1 + e^{-jM_2... ...\cos\left(M_1\omega T/2\right)}{\cos\left(M_2\omega T/2\right)}. \end{eqnarray*}$

When $M_1\neq M_2$ , the frequency response is a ratio of cosines in $\omega$ times a linear phase term $e^{j(M_2-M_1)\omega T/2}$ (which corresponds to a pure delay of samples). This special case gives insight into the behavior of the filter as its coefficients and approach 1.

When $M_1=M_2\isdef M$ , the filter degenerates to which corresponds to ; in this case, the delayed input and output signals cancel each other out. As a check, let's verify this in the time domain:

$\begin{eqnarray*} y(n) &=& x(n) + x(n-M) - y(n-M)\\ &=& x(n) + x(n-M) - [x(n-M... ...) - y(n-3M)]\\ &=& x(n) + y(n-3M)\\ &=& \cdots\\ &=& x(n). \end{eqnarray*}$

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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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Introduction to Digital Filters
Analysis of a Digital Comb Filter
Frequency Response