Frequency Response

Given the transfer function $ H(z)$, 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 $ T$ 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 $ g_1=g_2=1$, 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 $ M_1-M_2$ samples). This special case gives insight into the behavior of the filter as its coefficients $ g_1$ and $ g_2$ approach 1.

When $ M_1=M_2\isdef M$, the filter degenerates to $ H(z)=1$ which corresponds to $ y(n)=x(n)$; 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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Transfer Function