Circular Convolution Formula Deduction from DFT

Question

I am trying to get to the circular convolution formula from the product of the DFT of two input signals. I came across this webpage, which contains the following.

Could someone help me to solve that equation?

I have something like this

$$ x_3(n) = \frac{1}{N} \sum^{N-1}_{k=0} \exp{\left [ \frac{j 2 \pi k n}{N} \right ]} \left [ \sum^{N-1}_{m=0} x_1(m) \exp{\left [ \frac{-j 2 \pi k m}{N} \right ]} \right ] \left [ \sum^{N-1}_{m=0} x_2(m) \exp{\left [ \frac{-j 2 \pi k m}{N} \right ]} \right ] $$

I am not sure what is next. The following doesn't seem to be right.

$$ \begin{aligned} x_3(n) &= \frac{1}{N} \sum^{N-1}_{k=0} \left [ \sum^{N-1}_{m=0} x_1(m) \exp{\left [ \frac{-j 2 \pi k m}{N} \right ] \exp{\left [ \frac{j 2 \pi k n}{N} \right ]}} \right ] \left [ \sum^{N-1}_{m=0} x_2(m) \exp{\left [ \frac{-j 2 \pi k m}{N} \right ]} \right ]\\ \\ &= \frac{1}{N} \sum^{N-1}_{k=0} \left [ \sum^{N-1}_{m=0} x_1(m) \exp{\left [ \frac{-j 2 \pi k (m-n)}{N} \right ]} \right ] \left [ \sum^{N-1}_{m=0} x_2(m) \exp{\left [ \frac{-j 2 \pi k m}{N} \right ]} \right ]\\ \end{aligned} $$

Answer 1

Let $x$ and $y$ be signals of $N$ samples each, numbered as $x(0),\ldots,x(N-1)$. Then their DFTs are $X$ and $Y$, which also have $N$ entries each: \begin{eqnarray} X(k) &=& \sum_{n=0}^{N-1}x(n)e^{-2\pi i kn/N},\\ Y(k) &=& \sum_{m=0}^{N-1}y(m)e^{-2\pi i k m/N}, \end{eqnarray} where the indices run from $0$ to $N-1$.

The $k^{\textrm{th}}$ entry of the entry-by-entry product of $X$ and $Y$ is \begin{equation} \begin{split} X(k)Y(k) ~=& \left(\sum_{n=0}^{N-1}x(n)e^{-2\pi i kn/N}\right)\left(\sum_{m=0}^{N-1}y(m)e^{-2\pi i k m/N}\right)\\ ~=& \sum_{n=0}^{N-1}\sum_{m=0}^{N-1}x(n)y(m)e^{-2\pi i k(n+m)/N} \end{split} \end{equation}

Now we consider the $\ell^{\textrm{th}}$ entry of the IDFT of this entry-by-entry product: \begin{equation} \begin{split} \mathsf{IDFT}(XY)(\ell) ~=& \frac{1}{N}\sum_{k=0}^{N-1}X(k)Y(k)e^{2\pi i\ell k/N }\\ ~=& \frac{1}{N}\sum_{k=0}^{N-1}\left[\sum_{n=0}^{N-1}\sum_{m=0}^{N-1}x(n)y(m)e^{-2\pi i k(n+m)/N}\right]e^{2\pi i\ell k/N }\\ ~=& \frac{1}{N}\sum_{n=0}^{N-1}\sum_{m=0}^{N-1}x(n)y(m)\sum_{k=0}^{N-1}e^{-2\pi i k(n+m-\ell)/N} \end{split} \end{equation} The sum over $k$ is equal to 0 unless \begin{equation} n+m-\ell=0~\textrm{mod}~N~~~(m=\ell-n~\textrm{mod}~N), \end{equation} in which case it is equal to $N$: \begin{equation} \sum_{k=0}^{N-1}e^{-2\pi i k(n+m-\ell)/N} = N\delta_{m,\ell-n~\textrm{mod}~N}, \end{equation} where $\delta_{a,b}$ is the Kronecker delta. One intuitive way to see this is to see that

$\bullet$ if the exponent is not 0, we are adding all $N$ of the $N^{\textrm{th}}$ roots of unity (view them in the complex plane), which will cancel one another out when added; As noted by the OP, my struck-through claim is true for all $n$, $m$, and $\ell$ only if $N$ is prime. The correct proof is to note that if $e^{-2\pi i(n+m-\ell)/N}\neq 1$, then \begin{equation} \begin{split} \sum_{k=0}^{N-1}e^{-2\pi i k(n+m-\ell)/N} &=~ \sum_{k=0}^{N-1}\left(e^{-2\pi i (n+m-\ell)/N}\right)^k\\ &=~\frac{ 1- \left(e^{-2\pi i (n+m-\ell)/N}\right)^N}{1 - e^{-2\pi i (n+m-\ell)/N}}\\ &=~\frac{ 1- e^{N\times(-2\pi i (n+m-\ell)/N)}}{1 - e^{-2\pi i (n+m-\ell)/N}}\\ &=~\frac{ 1- e^{-2\pi i (n+m-\ell)}}{1 - e^{-2\pi i (n+m-\ell)/N}}\\ &=~\frac{ 1- 1}{1 - e^{-2\pi i (n+m-\ell)/N}} ~~=~~ 0. \end{split} \end{equation}

$\bullet$ if the exponent is 0, then we are adding 1 $N$ times.

(Another not as intuitive away can be found in this video.)

So far, we have \begin{equation} \begin{split} \mathsf{IDFT}(XY)(\ell) ~=& \frac{1}{N}\sum_{n=0}^{N-1}\sum_{m=0}^{N-1}x(n)y(m)N\delta_{m,\ell-n~\textrm{mod}~N}\\ ~=& \sum_{n=0}^{N-1}\sum_{m=0}^{N-1}x(n)y(m)\delta_{m,\ell-n~\textrm{mod}~N} \end{split} \end{equation} When we perform the sum over $m$, the only nonzero term is the one for which $m=\ell-n~\textrm{mod}~N$, so \begin{equation} \begin{split} \mathsf{IDFT}(XY)(\ell) ~=& \sum_{n=0}^{N-1}x(n)y(\ell -n~\textrm{mod}~N). \end{split} \end{equation} The expression on the right-hand side is the $\ell^{\textrm{th}}$ entry of the circular convolution of $x$ and $y$. \begin{equation} \mathsf{IDFT}(XY)(\ell) = (x\circledast y)(\ell), \end{equation} so $\mathsf{IDFT}(XY) = x\circledast y$.