Finding the impulse response of a system
One can see that the given expression can be decomposed as
$$
\begin{align}
H(j\omega) &amp;= \frac{1+0.5 e^{-j\omega}}{1-1.8 \cos(\frac{\pi}{16}) e^{-j\omega}+0.81 e^{-j2\omega}} \\
&amp;= \frac{A }{1-0.9 e^{j\frac{\pi}{16}} e^{-j\omega}} + \frac{B}{1-0.9 e^{-j\frac{\pi}{16}} e^{-j\omega}} 
\end{align}$$

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.