Complex exponential Fourier series coefficient of periodic convolution

Question

Let the complex exponential Fourier series coefficients of two periodic signals $x_1(t)$ and $x_2(t)$ be $C_{1n}$ and $C_{2n}$, respectively, with $T_0$ being the fundamental time period of both the signals.

Let the signal $x(t)$ be equal to $x_1(t) * x_2(t)$, where $*$ denotes the periodic convolution operation, and its complex exponential Fourier series coefficient be $C_n$.

The aim is to find a relationship between $C_n$, $C_{1n}$ and $C_{2n}$.

$$C_n = \frac1{T_0} \int_{T_0}\left(x(t)\cdot e^{-jnw_0t}\right)dt$$ $$= \frac{1}{T_0} \int_0^{T_0}\left((x_1(t) * x_2(t)) \cdot e^{-jnw_0t}\right)dt $$$$= \frac{1}{T_0} \int_0^{T_0}\left(\left(\int_0^{T_0}\left(x_1(\tau) \cdot x_2(t - \tau)\right)d\tau\right) \cdot e^{-jnw_0t}\right)dt$$

Since $e^{-jnw_0t}$ is constant with respect to the integral $\int_0^{T_0}\left(x_1(\tau) \cdot x_2(t - \tau)\right)d\tau$, therefore we can write

$$C_n = \frac{1}{T_0} \int_0^{T_0}\left(\int_0^{T_0}\left(e^{-jnw_0t} \cdot x_1(\tau) \cdot x_2(t - \tau)\right)d\tau\right)dt $$$$= \frac{1}{T_0} \int_0^{T_0}\left(\int_0^{T_0}\left(\left(x_1(\tau) \cdot e^{-jnw_0\tau}\right) \cdot \left(x_2(t-\tau) \cdot e^{-jnw_0(t-\tau)}\right)\right)d\tau\right)dt$$

Now, I would like to break this double integral into a product of two integrals, but since $x_2(t-\tau) \cdot e^{-jnw_0(t-\tau)}$ involves both $t$ and $\tau$, therefore I don't know how to proceed further.

Could someone help me with this?

Answer 1

It looks like you're almost there. Note that for any $T$-periodic function $f(t)$ you have

$$\int_{0}^Tf(t)dt=\int_{0}^Tf(t+\tau)dt$$

because as long as you integrate over one whole period (or an integer number of periods), it doesn't matter where you start integrating.

Hence you can split the last expression in your question into the product of two integrals, the first over $\tau$ and the second over $t$. This will give you the desired solution.