# Finding the impulse response of a system

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.

• Hi! Is that 1/8 or 1/9 in the denominator? – Juancho Oct 16 '19 at 21:44
• I corrected it sir, it should be 1.8. – Jason Oct 16 '19 at 21:47
• get the roots of the denominator. do a partial fraction expansion, and you can get the impulse response by adding the terms from the expansion – user28715 Oct 17 '19 at 0:58
• Sir, I couldn’t find the roots of the denominator since the roots are complex. Can you please help me? – Jason Oct 17 '19 at 8:56
• Hi Jason. Is this a homework problem? – Fat32 Oct 17 '19 at 9:08

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.