IDFT of H(z) sampled in N values
The fact that sampling in one domain corresponds to aliasing in the other domain is an important property of the Fourier transform. Of course, this can also be shown formally. If $X[k]$ is the length $N$ DFT of $x[n]$, we have

$$x[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}\tag{1}$$

Furthermore, $H(z)$ is related to the impulse response by

$$H(z)=\sum_{m=-\infty}^{\infty}h[m]z^{-m}\tag{2}$$

If we define

$$X[k]=H(e^{j2\pi k/N})=\sum_{m=-\infty}^{\infty}h[m]e^{-j2\pi mk/N}\tag{3}$$

we can determine $x[n]$ as follows. From $(1)$ and $(3)$ we have

$$\begin{align}x[n]&amp;=\frac{1}{N}\sum_{k=0}^{N-1}\sum_{m=-\infty}^{\infty}h[m]e^{-j2\pi mk/N}e^{j2\pi nk/N}\\&amp;=\frac{1}{N}\sum_{m=-\infty}^{\infty}h[m]\sum_{k=0}^{N-1}e^{j2\pi(n-m)k/N}\tag{4}\end{align}$$

The last sum in $(4)$ equals $N$ for $n-m=lN$ for integer $l$, and zero otherwise. Consequently, this last sum can be rewritten as

$$\sum_{k=0}^{N-1}e^{j2\pi(n-m)k/N}=N\sum_{l=-\infty}^{\infty}\delta[n-m-lN]$$

and $(4)$ becomes

$$\begin{align}x[n]&amp;=\sum_{m=-\infty}^{\infty}h[m]\sum_{l=-\infty}^{\infty}\delta[n-m-lN]\\&amp;=\sum_{l=-\infty}^{\infty}\sum_{m=-\infty}^{\infty}h[m]\delta[n-m-lN]\\&amp;=\sum_{l=-\infty}^{\infty}h[n-lN]\tag{5}\end{align}$$

where I've used the fact that the sum over $m$ is the convolution of $h[n]$ with $\delta[n-lN]$, and consequently equals $h[n-lN]$.

Eq. $(5)$ establishes the result that the IDFT of an equidistantly sampled discrete-time Fourier transform (DTFT) equals an aliased version of the sequence corresponding to the given DTFT.