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How can I perform the inverse z-transform on the following $H(z)$ to be able to calculate a real-valued impulse response $h[n]$? $$ H(z)=\frac{z^2}{z^2+0.8\sqrt{2}z+0.64} $$ My idea was to find an inverse z-transform on my list of transformations, which basically does what I want with a similar transfer function (nothing too special about that idea). Well, I found the following: $$ H(z)=\frac{p z \sin(\alpha)}{z^2-2p z \cos(\alpha)+p^2} $$ I then tried to adjust the transfer function. First I did that for the denominator and found it had a pair of complex conjugate poles $0.8e^{+j\frac{3\pi}{4}}$, and $0.8e^{-j\frac{3\pi}{4}}$. That seems great, my desired form of the transfer function has similar poles $pe^{+j\alpha}$, and $pe^{-j\alpha}$. Therefore I would substitute $p=0.8$ and $\alpha=\frac{3\pi}{4}$. But now I don't know how to go further, what do I do with the numerator? Is this approach valid at all?

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  • $\begingroup$ That seems fine. You just need to include another gain term on the numerator, because you may no get $p\sin(\alpha)$. $\endgroup$ – Peter K. Sep 24 '15 at 14:47
  • $\begingroup$ You mean I add a term in the numerator of the first $H(z)$, like $az^2$ and $a=psin(\alpha)$? Is it ok to do that? $\endgroup$ – Daiz Sep 24 '15 at 15:02
  • $\begingroup$ Yes, the thing to do is to decouple the constant in the numerator from $p$ and $\alpha$ in the denominator. $\endgroup$ – Peter K. Sep 24 '15 at 15:05
  • $\begingroup$ The result I got: $$ H(z)=\frac{az^2}{z^2+0.8\sqrt{2}z+0.64}=\frac{0.8 z \sin(\frac{3\pi}{4})}{z^2-1.6 z \cos(\frac{3\pi}{4})+0.64}z=0.8^nsin(\frac{3\pi}{4}n)u[n+1] $$ Is it correct like this? Still not sure why I can add the gain term $\endgroup$ – Daiz Sep 24 '15 at 15:17
  • $\begingroup$ It may be that the gain term is already $p \sin(\alpha)$. In general, it won't be. $\endgroup$ – Peter K. Sep 24 '15 at 15:19

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