Convolution property of DTFS

Statement

Let $x[n]$ and $y[n]$ be periodic with the same period $N$, with $x[n] \overset{FS}{\longleftrightarrow} a_k$ and $y[n] \overset{FS}{\longleftrightarrow} b_k$. Their periodic convolution

$$z[n] = \sum_{r=\langle N\rangle}x[r]\,y[n - r]$$

is periodic with period $N$, and

$$z[n] \overset{FS}{\longleftrightarrow} c_k = N a_k b_k$$

Convolution in time becomes multiplication of coefficients.

Proof

With $\omega_0 = 2\pi/N$:

$$\begin{aligned} c_k &= \frac{1}{N}\sum_{n=\langle N\rangle}z[n]e^{-jk\omega_0 n} = \frac{1}{N}\sum_{n=\langle N\rangle}\sum_{r=\langle N\rangle}x[r]\,y[n-r]\,e^{-jk\omega_0 n} \\ &= \frac{1}{N}\sum_{r=\langle N\rangle}x[r]\,e^{-jk\omega_0 r}\sum_{n=\langle N\rangle}y[n-r]\,e^{-jk\omega_0(n-r)} \end{aligned}$$

Put $m = n - r$; since $y$ is periodic, $m$ still runs over one full period, so the inner sum is $\sum_{m=\langle N\rangle}y[m]e^{-jk\omega_0 m} = N b_k$. Then

$$c_k = \frac{1}{N}\sum_{r=\langle N\rangle}x[r]e^{-jk\omega_0 r}\cdot N b_k = N\,a_k\,b_k$$

Hence $\sum_{r=\langle N\rangle}x[r]y[n-r] \overset{FS}{\longleftrightarrow} N a_k b_k$. (The dual property: multiplication $x[n]y[n]$ gives the periodic convolution of coefficients $\sum_{l=\langle N\rangle}a_l b_{k-l}$.)