I have the following transfer function.
$$ H(j\omega) = \frac{1+0.5 e^{-j\omega}}{1-1.8 \cos(\frac{\pi}{16}) e^{-j\omega}+0.81 e^{-j2\omega}}$$
I'm trying to find the impulse response of the system. However, I couldn't separate the expression above and I couldn't figure out how I can find the impulse response. Can anybody help me how to solve this equation? Any help would be appreciated.
One can see that the given expression can be decomposed as $$ \begin{align} H(j\omega) &= \frac{1+0.5 e^{-j\omega}}{1-1.8 \cos(\frac{\pi}{16}) e^{-j\omega}+0.81 e^{-j2\omega}} \\ &= \frac{A }{1-0.9 e^{j\frac{\pi}{16}} e^{-j\omega}} + \frac{B}{1-0.9 e^{-j\frac{\pi}{16}} e^{-j\omega}} \end{align}$$
where $A = B^* = 0.5 - j 3.9375 = 8/16 - j 63/16 $.
Then the impulse respons will be:
$$h[n] = A (0.9 e^{j\frac{\pi}{16}})^n u[n] + A^* (0.9 e^{-j\frac{\pi}{16}})^n u[n]$$
which can be simplified as:
$$h[n] = 2 \cdot 0.9^n |A| \cos( \frac{\pi}{16}n + \angle{A}) u[n]$$
where $\angle{A}$ is the phase angle of $A$. Following is the resulting sequence plotted, from $n=0$ to $n=35$, using MATLAB/OCTAVE.
