Proof of the convolution property of Fourier Series in continuous time

Question

I am facing problem in understanding the proof of Convolution property of Fourier Series (FS) in continuous time CT; that is: $$\mathrm{FS} \big\{x_1(t)\star x_2(t)\big\}=T\sum_{n=-\infty}^{\infty}c_n d_ne^{j \Omega_0 n t}$$ where $x_1(t)$ and $x_2(t)$ are periodic with $T$, and $c_n$ and $d_n$ are the Fourier series coefficient of $x_1(t)$ and $x_2(t)$.

Below I am showing the proof and my confusion,

• PROOF:
  \begin{align} x_1(t)\star x_2(t) &= \int_T x_1(\tau)x_2(t-\tau)d\tau\\ \therefore \mathrm{FS}\big\{x_1(t)\star x_2(t)\big\}&=\int_T FS[x_1(\tau)] FS[x_2(t-\tau)]d\tau\\ &=\int_T \sum_{n=-\infty}^{\infty}c_n e^{j \Omega_0 n \tau} \sum_{p=-\infty}^{\infty}d_p e^{j \Omega_0 p (t- \tau)} d\tau\\ &=\sum_{n=-\infty}^{\infty} \sum_{p=-\infty}^{\infty} c_n d_p \int_T e^{j \Omega_0 (n-p) \tau} e^{j \Omega_0 p t} d\tau\\ &=\sum_{n=-\infty}^{\infty} c_n d_n T e^{j \Omega_0 n t} \end{align}

• CONFUSION:
  Now I am confused how we are getting
  $$ \sum_{n=-\infty}^{\infty} c_n d_n T e^{j \Omega_0 n t} \quad\text{from} \quad\sum_{n=-\infty}^{\infty} \sum_{p=-\infty}^{\infty} c_n d_p \int_T e^{j \Omega_0 (n-p) \tau} e^{j \Omega_0 p t} d\tau \quad ?$$

Please explain.

Answer 1

HINT:

Show that with $\Omega_0=2\pi/T$

$$\int_T e^{j\Omega_0(n-p)\tau} d\tau=\begin{cases}T,&\quad n=p\\0,&\quad\text{otherwise}\end{cases}$$

So the sum over $p$ reduces to a single element $Td_ne^{j\Omega_0 nt}$, and the result follows.