IDFT of H(z) sampled in N values

If a have a causal IIR filter described by $H(z)$ and I sample it in $N$ equispaced values around the unit circle, I get a DFT of $N$ points. That DFT corresponds to $h[n]$ truncated in $n=N-1$ or to the DFT of $h[n]$ periodized with period $N$ (i.e. with copies of itself overlapped every $N$ points)?

• The second option is the correct one. Sampling in one domain always corresponds to aliasing in the other domain. – Matt L. Feb 9 '16 at 7:45

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]&=\frac{1}{N}\sum_{k=0}^{N-1}\sum_{m=-\infty}^{\infty}h[m]e^{-j2\pi mk/N}e^{j2\pi nk/N}\\&=\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]&=\sum_{m=-\infty}^{\infty}h[m]\sum_{l=-\infty}^{\infty}\delta[n-m-lN]\\&=\sum_{l=-\infty}^{\infty}\sum_{m=-\infty}^{\infty}h[m]\delta[n-m-lN]\\&=\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.

• I got very confused when you rewrote the last sum in eq. $(4)$ as $N\sum_{l=-\infty}^{\infty}\delta[n-m-lN]$. I see why it is equal to $N$ when $n-m=lN$, but why is it zero otherwise? – Tendero Feb 9 '16 at 15:29
• @M.S.: It's a geometric series, just use the formula and see what happens. – Matt L. Feb 9 '16 at 15:33
• I get that $\sum_{k=0}^{N-1}e^{j2\pi(n-m)k/N}=\sum_{k=0}^{N-1}(e^{j2\pi(n-m)/N})^k=\frac{1-e^{j2\pi(n-m)}}{1-e^{j2\pi(n-m)/N}}$, and that would be zero for any $n-m$, right? So I don't get it, what am I doing wrong? – Tendero Feb 9 '16 at 15:40
• Now I just realized that if $n-m$ is $lN$, then you would have an indetermination $0/0$ which, I suppuse, tends to $N$ if you take the limit. And it would be zero for any other values of $n-m$ then. – Tendero Feb 9 '16 at 15:43
• @M.S. No need to take limits. The formula is only valid for $n-m\neq lN$, because otherwise you would divide by zero. For $n-m=lN$ you can directly evaluate the sum because each term equals $1$ and you have $N$ terms. – Matt L. Feb 9 '16 at 15:48