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Determine the frequency response and impulse response, $$ y(n)-\frac{1}{6} y(n-1)-\frac{1}{6} y(n-2)=x(n) $$
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Solution:

$ y(n)-\frac{1}{6} y(n-1)-\frac{1}{6} y(n-2)=x(n)\ $

Applying DTFT,

$ Y\left(e^{j \omega}\right)-\frac{1}{6} e^{-j \omega} Y\left(e^{j \omega}\right)-\frac{1}{6} e^{-2 j \omega} Y\left(e^{j \omega}\right)=X\left(e^{j \omega}\right)\ $

Frequency response,

$ H\left(e^{j \omega}\right)=\frac{Y\left(e^{j \omega}\right)}{X\left(e^{j \omega}\right)}=\frac{1}{1-\frac{1}{6} e^{-j \omega}-\frac{1}{6} e^{-2 j \omega}}\\ $

$ =\frac{e^{2 j \omega}}{e^{2 j \omega}-\frac{1}{6} e^{j \omega}-\frac{1}{6}}\\ $

$ \begin{gathered}\\ \frac{H\left(e^{j \omega}\right)}{e^{j \omega}}=\frac{e^{j \omega}}{e^{2 j \omega}-\frac{1}{6} e^{j \omega}-\frac{1}{6}}=\frac{A}{e^{j \omega}-\frac{1}{2}}+\frac{B}{e^{j \omega}+\frac{1}{3}} \\\\ e^{j \omega}=A\left(e^{j \omega}+\frac{1}{3}\right)+B\left(e^{j \omega}-\frac{1}{2}\right)\\ \end{gathered} $

$ \begin{aligned} & \text { At } e^{j \omega}=\frac{1}{2} \\\\ & \frac{1}{2}=A\left(\frac{1}{2}+\frac{1}{3}\right)\\ \end{aligned}\\ $

$ \therefore A=\frac{3}{5}\\ $

$ H\left(e^{j \omega}\right)=\frac{\frac{3}{5} e^{j \omega}}{e^{j \omega}-\frac{1}{2}}+\frac{\frac{2}{5} e^{j \omega}}{e^{j \omega}+\frac{1}{3}}\\ $

Applying inverse DTFT,

$ h(n)=\frac{3}{5}\left(\frac{1}{2}\right)^n u(n)+\frac{2}{5}\left(-\frac{1}{3}\right)^n u(n)\\ $

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