Convolution property of CTFS

Statement

Let $x(t)$ and $y(t)$ be periodic with the same period $T$ ($\omega_0 = 2\pi/T$), with $x(t) \xleftrightarrow{FS} a_k$ and $y(t) \xleftrightarrow{FS} b_k$. The periodic convolution of the two signals has Fourier series coefficients $T a_k b_k$:

$$z(t) = \int_T x(\tau)\,y(t-\tau)\,d\tau \;\xleftrightarrow{FS}\; c_k = T a_k b_k$$

So convolution in time becomes multiplication of the coefficients.

Proof

$z(t)$ is periodic with period $T$, because $y(t - \tau)$ is periodic in $t$. Its coefficients are

$$c_k = \frac{1}{T}\int_T z(t)e^{-jk\omega_0 t}dt = \frac{1}{T}\int_T\left[\int_T x(\tau)y(t-\tau)d\tau\right]e^{-jk\omega_0 t}dt$$

Swap the order of integration, and write $e^{-jk\omega_0 t} = e^{-jk\omega_0 \tau}e^{-jk\omega_0 (t-\tau)}$:

$$c_k = \frac{1}{T}\int_T x(\tau)e^{-jk\omega_0 \tau}\left[\int_T y(t-\tau)e^{-jk\omega_0 (t-\tau)}dt\right]d\tau$$

Put $\sigma = t - \tau$. The integrand is periodic with period $T$, so integrating over any interval of length $T$ gives the same result:

$$\int_T y(\sigma)e^{-jk\omega_0 \sigma}d\sigma = T b_k$$

Therefore

$$c_k = \frac{1}{T}\cdot T b_k\int_T x(\tau)e^{-jk\omega_0 \tau}d\tau = T b_k\cdot a_k$$ $$\boxed{c_k = T a_k b_k}$$

Use: the output of an LTI system for a periodic input can be found by multiplying spectra instead of convolving. The dual result is the multiplication property: $x(t)y(t) \xleftrightarrow{FS} \sum_{l} a_l b_{k-l}$.