3.9.1
Given \(y\left ( k+2\right ) +\frac {1}{2}y\left ( k\right ) =\frac {1}{4}u\left ( k+2\right ) -\frac {1}{4}u\left ( k\right ) \) find the frequency transfer function \(H\left ( e^{j\omega }\right ) \)
Answer
I will use the Z transform as it is a little faster. Let \(Y\left ( z\right ) \) be the Z transform of \(y\left ( k\right ) \) and let \(U\left ( z\right ) \) be the Z
transform of \(u\left ( k\right ) \) , we obtain from the above
\[ z^{2}Y\left ( z\right ) +\frac {1}{2}Y\left ( z\right ) =\frac {1}{4}z^{2}U\left ( z\right ) -\frac {1}{4}U\left ( z\right ) \]
Hence
\[ H\left ( z\right ) =\frac {Y\left ( z\right ) }{U\left ( z\right ) }=\frac {\frac {1}{4}z^{2}-\frac {1}{4}}{z^{2}+\frac {1}{2}}=\frac {1}{4}\frac {-1+z^{2}}{\frac {1}{2}+z^{2}}\]
Since the DTFT \(H\left ( z\right ) \) at the unit circle, then let \(z=e^{j\omega }\) in the
above we obtain
\begin{align*} H\left ( e^{j\omega }\right ) & =\frac {1}{4}\left ( \frac {-1+e^{2j\omega }}{\frac {1}{2}+e^{2j\omega }}\right ) \\ & =\frac {1}{4}\left ( \frac {-1+e^{2j\omega }}{\frac {1}{2}+e^{2j\omega }}\right ) \left ( \frac {\frac {1}{2}+e^{-2j\omega }}{\frac {1}{2}+e^{-2j\omega }}\right ) \\ & =\frac {1}{4}\left ( \frac {-\frac {1}{2}-e^{-2j\omega }+\frac {1}{2}e^{2j\omega }+1}{\frac {1}{4}+\frac {1}{2}e^{-2j\omega }+\frac {1}{2}e^{2j\omega }+1}\right ) \\ & =\frac {1}{4}\left ( \frac {\frac {1}{2}-\left ( \cos 2\omega -j\sin 2\omega \right ) +\frac {1}{2}\left ( \cos 2\omega +j\sin 2\omega \right ) }{\frac {5}{4}+\cos 2\omega }\right ) \\ & =\frac {1}{4}\left ( \frac {\frac {1}{2}-\frac {1}{2}\cos 2\omega +\frac {3}{2}j\sin 2\omega }{\frac {5}{4}+\cos 2\omega }\right ) \end{align*}
Hence
\[ H\left ( e^{j}\omega \right ) =\frac {\left ( \frac {1}{2}-\frac {1}{2}\cos 2\omega \right ) +j\left ( \frac {3}{2}\sin 2\omega \right ) }{5+4\cos 2\omega }\]
Hence
\begin{align*} \left \vert H\left ( e^{j\omega }\right ) \right \vert ^{2} & =\frac {\left ( \frac {1}{2}-\frac {1}{2}\cos 2\omega \right ) ^{2}+\left ( \frac {3}{2}\sin 2\omega \right ) ^{2}}{\left ( 5+4\cos 2\omega \right ) ^{2}}\\ & =\frac {\left ( \frac {1}{4}+\frac {1}{4}\cos ^{2}2\omega -\frac {1}{4}\cos 2\omega \right ) +\left ( \frac {9}{4}\sin ^{2}2\omega \right ) }{\left ( 5+4\cos 2\omega \right ) ^{2}}\\ & =\frac {\frac {1}{4}+\frac {1}{4}\cos ^{2}2\omega -\frac {1}{4}\cos 2\omega +\frac {9}{4}\sin ^{2}2\omega }{\left ( 5+4\cos 2\omega \right ) ^{2}}\\ & =\frac {\sin ^{2}\omega }{5+4\cos 2\omega }\end{align*}
And
\[ \arg \left ( H\left ( e^{j\omega }\right ) \right ) =\arctan \left ( \frac {3}{\tan \left ( \omega \right ) }\right ) \]
Please note, for the final 2 lines calculation above, I wanted to obtain the most simple result,
so I used Mathematica to simplify.
Here is a plot of the magnitude and phase frequency response from Mathematica: (this is a
bandpass filter).