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).
