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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.

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  • $\begingroup$ Hi! Is that 1/8 or 1/9 in the denominator? $\endgroup$ – Juancho Oct 16 at 21:44
  • $\begingroup$ I corrected it sir, it should be 1.8. $\endgroup$ – Jason Oct 16 at 21:47
  • $\begingroup$ 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 $\endgroup$ – Stanley Pawlukiewicz Oct 17 at 0:58
  • $\begingroup$ Sir, I couldn’t find the roots of the denominator since the roots are complex. Can you please help me? $\endgroup$ – Jason Oct 17 at 8:56
  • $\begingroup$ Hi Jason. Is this a homework problem? $\endgroup$ – Fat32 Oct 17 at 9:08
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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.

enter image description here

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