I struggle with a part of the solution of an exercise and would be grateful for your help.
"Given a time discrete LTI (Linear time-invariant) system H, and an input signal $x[n]$, we have the following relation:
$(Hx)[n]=1$"
(i.e. $(h*x)[n]=1$ where $*$ denotes the convolution)
"In the frequency domain, this implies :
$\hat{h}(\theta)\hat{x}(\theta)=\sum_{k=-\infty}^{\infty}\delta(\theta-k)$"
My question is here:
The DFT of $\delta[n]$ is equal to $1$, but why is the DFT of $1$ equal to $\sum_{k=-\infty}^{\infty}\delta(\theta-k)$ ?
I mean, by defintion, the DFT $\hat{z}[n]=\sum_{0}^{N-1}z[n]e^{-2\pi i k n/N}$ where $0 \leq n \leq N-1$. So if $z[n]=1$, we would have $\sum_{0}^{N-1}1 \cdot e^{-2\pi i k n/N}$, not $\sum_{k=-\infty}^{\infty}\delta(\theta-k)$.
I googled several times but could not find an answer. Or is their solution wrong ?
Thanks for your help !
Let $x = e^{- 2 \pi i n k /N}$. Note that $x = 1$ if and only if $k$ is a multiple of $N$. So, in the case that $k$ is such a multiple, we have $$ \hat z[k] = \frac 1N\sum_{n=0}^{N-1}1 \cdot e^{-2\pi i k n/N} = \frac 1N\sum_{n=0}^{N-1} x^n = \frac 1N\sum_{n=0}^{N-1} 1 = 1. $$ Otherwise, we have $$ \hat z[k] = \frac 1N\sum_{n=0}^{N-1}1 \cdot e^{-2\pi i k n/N} = \frac 1N\sum_{n=0}^{N-1} x^n = \frac 1N\frac{1 - x^{N}}{1 - x} = \frac{0}{1 - x} = 0. $$ This corresponds to the sum $$ \hat z[k] = \sum_{j=-\infty}^\infty \delta[k - Nj]. $$