Z-transform a transfer function

Question

Could someone help me invers Z-transform of this transfer function. $H_k(z) = \frac{Y_k(z)}{X(z)} = \frac{1}{1-cos(\frac{2·\pi ·k}{N})·z^{-1}+z{^-2}}$

Answer 1

I think this is what you had in mind. The polynomial $p(z)=z^{2}-\cos(2\pi k/N)z+1$ has roots $$ \frac{\cos(2\pi k/N)\pm j\sqrt{4-\cos^{2}(2\pi k/N)}}{2},\;\;\; j = \sqrt{-1}. $$ Letting $\alpha$ be the root with $+$, then $\alpha^{\star}$ is the root with $-$, where the $\star$ is complex conjugation. So the original Z-transform is $$ H_{k}(z)=\frac{1}{(1-\frac{\alpha}{z})(1-\frac{\alpha^{\star}}{z})}=\left[\frac{1}{1-\frac{\alpha}{z}}-\frac{1}{1-\frac{\alpha^{\star}}{z}}\right]=\frac{z}{\alpha-\alpha^{\star}}\sum_{n=1}^{\infty}\frac{\alpha^{n}-(\alpha^{\star})^{n}}{z^{n}} = \sum_{n=0}^{\infty}\frac{\alpha^{n+1}-(\alpha^{n+1})^{\star}}{\alpha-\alpha^{\star}}\frac{1}{z^{n}}. $$ You can read off the coefficients. This is with forward delay where the region of convergence is $|z| > |\alpha|=1$. Things switch in $|z|<1$. So your function is the $Z$ transform of $h[n]=\frac{\alpha^{n+1}-(\alpha^{n+1})^{\star}}{\alpha-\alpha^{\star}}$.

NOTE: If there had been a 2 in front of the $cos$ term in your original problem, then $$ \frac{1}{1-2\cos(2\pi k/N)z^{-1}+z^{-2}}=\sum_{n=0}^{\infty}\frac{\sin(2\pi (n+1)k/N)}{\sin(2\pi k/N)}\frac{1}{z^{n}} $$ This is because the above analysis still holds with $$ \alpha = \frac{2\cos(2\pi k/N)\pm j\sqrt{4-4\cos^{2}(2\pi k/N)}}{2}=e^{\pm j(2\pi k/N)}. $$ I had suspected you might have missed the 2, and that's why I did the analysis the way I did. The other works out the same way because $\alpha=e^{j\theta}$ for some $\theta$ due to the fact that $|\alpha|^{2}=1$ anyway. But with the 2 there, $\theta=2\pi k/N$.