Please help me to find the impulse response $h(n)$ of a Transfer function $H(z) = \dfrac{1+\frac{1}{6}z^{-1}}{1-\frac{1}{4}z^{-1}}$ given $h(n)$ is left sided.

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    $\begingroup$ The solution for such a prblem has 3 steps: 1-) perform long division of H(z). 2-) perform partial fraction expansion on the remaning fractional part. 3-) perform inverse Z-transform considering the left-sided inverses... Which step are you stuck in ? $\endgroup$ – Fat32 Jan 2 '20 at 15:56
  • $\begingroup$ I'd be happy if I get (2) Partial fraction expansion and (3) Inverse Z transform part $\endgroup$ – Shehin Jan 2 '20 at 16:11
  • $\begingroup$ If $h(n)$ was right sided, the answer would be $h(n) = \left(\dfrac{1}{4}\right)^nu[n]+\dfrac{1}{6}\left(\dfrac{1}{4}\right)^{n-1}u[n-1]$... I think.. But I am confused how to arrive at the answer when $h(n)$ is left sided $\endgroup$ – Shehin Jan 2 '20 at 16:28

Note that the order of your numerator equals the denominator, first perform a long division to simplify, before performing partial fraction expansion:

$$ H(z) = \frac{ 1 + 1/6 z^{-1}}{1 - 0.25 z^{-1}} = -\frac{2}{3} + \frac{5/3}{1 - 0.25 z^{-1}} $$

(Note: long divison already simplified the expression that it does not require a further partial fraction expansion to be applied, hence skip step-2 now)

Now apply the inzerve Z-transform and note that for left sided inverse sequence wil be $-a^n u[-n-1]$ for a term like $1/(1-az^{-1})$.

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    $\begingroup$ Got the long division correct. Now on taking the inverse transform, $$h(n) = \dfrac{-2}{3}\delta (n)-\dfrac{5}{3}\left(\dfrac{1}{4}\right)^nu(-n-1)$$ \\ Is this correct? $\endgroup$ – Shehin Jan 2 '20 at 17:53
  • $\begingroup$ yes it does seem so ;-) $\endgroup$ – Fat32 Jan 2 '20 at 18:16
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    $\begingroup$ Now that you got your solution right, can you figure out the answer without long division (just by breaking the initial ratio into two and taking inverse Z on both)? And if yes, can you verify that this answer is identical with the one you've already got? In other words, can you start from $$h[n] = (1/4)^n u[-n-1] - (1/4)^{n-1}u[-n]$$ and reach the same result? I think it's a good exercise. :) $\endgroup$ – GKH Jan 2 '20 at 21:14
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    $\begingroup$ @GKH good point! one should also be able to see the solution from other perspectives. Although its correct application should yield : $$h[n] = -(0.25)^n u[-n-1] - \frac{1}{6}(0.25)^{n-1} u[-n] $$ $\endgroup$ – Fat32 Jan 2 '20 at 21:38
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    $\begingroup$ @Shehin, you should play around with the second term. Try to extract a delta function at $n=0$ out of it and see what is left. $\endgroup$ – GKH Jan 4 '20 at 14:44

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