Proving Parseval's Theorem

Question

My goal is to prove Parseval's Thoerem, that is:

$$\frac{1}{T_0} \int \limits_{-T_0/2}^{T_0/2} |x(t)|^2~dt = \sum \limits_{n=-\infty}^{\infty} |c_n|^2$$

To do this, let's say I have a periodic signal $x(t)$ that is the product of two periodic signals both with period $T_0$, that is:

$$x(t) = x_1(t) ~x_2(t)$$

The Complex Fourier Series expression for $x_1(t)$ and $x_2(t)$ is:

$$\begin{matrix}x_1(t) = \sum \limits_{k=-\infty}^{\infty} d_k ~e^{jk\omega_0 t}&&x_2(t) = \sum \limits_{n=-\infty}^{\infty} b_n ~e^{jn\omega_0 t}\\ d_k = \frac{1}{T_0}\int \limits_{-T_0/2}^{T_0/2}x_1(t) ~e^{-jk\omega_0 t} && b_n = \frac{1}{T_0}\int \limits_{-T_0/2}^{T_0/2}x_2(t) ~e^{-jn\omega_0 t} \end{matrix}$$

The Complex Fourier Series expression for x(t) is:

$$\begin{matrix}x(t)=\sum \limits_{m=-\infty}^{\infty} c_m~e^{jm\omega_0 t} \\ c_m = \frac{1}{T_0} \int \limits_{-T_0/2}^{T_0/2} x(t)~e^{-jm\omega_0 t} \end{matrix}$$

---

starting with coefficients for x(t):

$$c_m = \frac{1}{T_0} \int \limits_{-T_0/2}^{T_0/2} x(t)~e^{-jm\omega_0 t}$$

$$c_m = \frac{1}{T_0} \int \limits_{-T_0/2}^{T_0/2} x_1(t)~x_2(t)~e^{-jm\omega_0 t}$$

$$c_m = \frac{1}{T_0} \int \limits_{-T_0/2}^{T_0/2} \Big(\sum \limits_{k=-\infty}^{\infty} d_k~e^{jk\omega_0 t} \Big)~x_2(t)~e^{-jm\omega_0 t}~dt$$

$$c_m = \sum \limits_{k=-\infty}^{\infty} \Bigg[ d_k \frac{1}{T_0} \int \limits_{-T_0/2}^{T_0/2} x_2(t)~e^{-j(m-k)\omega_0 t}~dt \Bigg]$$

$$c_m = \sum \limits_{k=-\infty}^{\infty} d_k~b_{m-k}$$

equating $c_m$ from earlier step:

$$\frac{1}{T_0} \int \limits_{-T_0/2}^{T_0/2} x_1(t)~x_2(t)~e^{-jm\omega_0 t}~dt = \sum \limits_{k=-\infty}^{\infty} d_k~b_{m-k}$$

now if i set m = 0:

$$\boxed{\frac{1}{T_0} \int \limits_{-T_0/2}^{T_0/2} x_1(t)~x_2(t)~ dt = \sum \limits_{k=-\infty}^{\infty} d_k~b_{-k}}$$

This step is similar to parseval's theorem but not quite all the way there. what I need is for this equation to become:

$$\frac{1}{T_0} \int \limits_{T_0/2}^{-T_0/2} |x(t)|^2~dt = \sum \limits_{n=-\infty}^{\infty} |c_n|^2$$

I could set $$y(t) = x_1(t) = x_2(t)$$, which, would make $b_n = d_n$, but i'm uncertain about the negative index on the Right hand side $d_k~b_{-k}$. Does that mean the coefficients are always symmetric about n=0? that is $b_n = b_{-n}$? or maybe the absolute value of the coefficients is always even symmetry for complex fourier series? that is: $|b_n| = |b_{-n}|$???

Answer 1

$$ \frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2dt = \sum_{n=-\infty}^{\infty}|c_n|^2 $$ Notice that $$ |x(t)|^2 = x(t)\overline{x(t)} $$ where the $\bar{}$ indicates the complex conjugation. We have also $$ \overline{x(t)}=\overline{\sum_{n=-\infty}^{\infty}c_n \exp(in\omega t)}=\sum_{n=-\infty}^{\infty}\overline{c_n} \exp(-in\omega t) $$ $$ |x(t)|^2 = (\sum_{m=-\infty}^{\infty} c_m \exp(im\omega t))(\sum_{n=-\infty}^{\infty}\overline{c_n} \exp(-in\omega t))\\ = \sum_{m,n=-\infty}^{\infty} c_m \overline{c_n} \exp(i(m-n)\omega t) $$ Now integrate $|x(t)|^2$ $$ \frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2dt = \frac{1}{T}\int_{-T/2}^{T/2}\sum_{m,n=-\infty}^{\infty} c_m \overline{c_n} \exp(i(m-n)\omega t)dt \\ = \frac{1}{T}\sum_{m,n=-\infty}^{\infty} c_m \overline{c_n} \int_{-T/2}^{T/2} \exp(i(m-n)\omega t)dt \\ $$ Now the integrals $$ \frac{1}{T}\int_{-T/2}^{T/2} \exp(i(m-n)\omega t)dt = \delta_{mn} $$ So we get $$ \frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2dt = \frac{1}{T}\sum_{m,n=-\infty}^{\infty} c_m \overline{c_n} \delta_{mn} \ dt \\ = \sum_{m=-\infty}^{\infty} c_m \overline{c_m} \\ = \sum_{m=-\infty}^{\infty} |c_m|^2 $$

Answer 2

Somewhere you need convergence of the Fourier series of $x(t)$ to $x$ in some sense. If you use a normalized inner product on $[-T_0/2,T_0/2]$ given by $$ \langle f,g\rangle = \frac{1}{T_0}\int_{-T_0/2}^{T_0/2}f(t)\overline{g(t)}dt, $$ then the Fourier series for $f$ becomes $\sum_{n=-\infty}^{\infty}\langle f,e^{ik\omega t}\rangle e^{ik\omega t}$. Furthermore, $$ \|f\|^2= \left\|f-\sum_{n=-N}^{N}\langle f,e^{ik\omega t}\rangle e^{ik\omega t}\right\|^2+\sum_{n=-N}^{N}|\langle f,e^{ik\omega t}\rangle|^2 $$ This shows you that you end up with identity you want iff the Fourier series for $f$ converges in $L^2[-T_0/2,T_0/2]$ to $f$. These two things are equivalent.