Parseval's Theorem for discrete series

Question

I need to use Parseval's Theorem to find \begin{equation} S\:=\:\sum _{n=-\infty }^{\infty }\:\left[\left(\frac{\sin\left(\frac{\pi }{4}n\right)}{2\pi n}\right)\left(\frac{\sin\left(\frac{\pi \:}{6}n\right)}{5\pi \:n}\right)\right] \end{equation} I used the fact that \begin{equation} x_1\left[n\right]\:=\:\frac{\sin\left(\frac{\pi }{4}n\right)}{2\pi n}\:=\:\frac{1}{8}\operatorname{sinc}\left(\frac{n}{4}\right) \end{equation} and \begin{equation} x_2\left[n\right]\:=\:\frac{\sin\left(\frac{\pi }{6}n\right)}{5\pi n}\:=\:\frac{1}{30}\operatorname{sinc}\left(\frac{n}{6}\right) \end{equation} Therefore, the Fourier Transform of each would be \begin{equation} X_1\left(e^{j\omega }\right)\:=\:1\:\:\:\:0<\omega <\frac{\pi }{4} \end{equation} \begin{equation} X_2\left(e^{j\omega }\right)\:=\:1\:\:\:\:0<\omega <\frac{\pi }{6} \end{equation} so using Parseval's theorem \begin{equation} S\:=\frac{1}{2\pi }\int _0^{2\pi }X_1\left(e^{j\omega }\right)X_2\left(e^{j\omega }\right)d\omega \:=\:\frac{1}{2\pi }\int _0^{\frac{\pi }{6}\:}\left(1\right)\left(1\right)d\omega \:\:=\:\frac{1}{12} \end{equation} Is this correct? am kinda doubting I did a mistake in the theory somewhere. Thanks in advance.

Answer 1

You are right, but you have mistaken the DTFTs...

Considering the Parseval's relation for real discrete-time sequences to be:

$$ \sum_{n} x[n]y[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} X(e^{j\omega})Y(e^{j\omega}) d\omega $$

We shall define $x_1[n]$, $x_2[n]$ and find their corresponding DTFT as:

$$ x_1[n] = \frac{\sin(\frac{1}{4} \pi n)}{2\pi n} \Leftrightarrow X_1(e^{j \omega}) = \begin{cases} \frac{1}{2} &,& \text{ for } |w| < \frac{\pi}{4} \\ 0 &,& \text{otherwise} \end{cases} $$

$$ x_2[n] = \frac{\sin(\frac{1}{6} \pi n)}{5\pi n} \Leftrightarrow X_1(e^{j \omega}) = \begin{cases} \frac{1}{5} &,& \text{ for } |w| < \frac{\pi}{6} \\ 0 &,& \text{otherwise} \end{cases} $$

Where we have used the DTFT pair; $$ x[n] = A\frac{\sin(\omega_c n)}{\pi n} \Leftrightarrow X(e^{j \omega}) = \begin{cases} A &,& \text{ for } |w| < \omega_c \\ 0 &,& \text{otherwise} \end{cases} $$

Hence applying Parseval's theorem we see that:

$$ S = \sum_{n} \frac{\sin(\frac{1}{4} \pi n)}{2\pi n} \frac{\sin(\frac{1}{6} \pi n)}{5\pi n} = \frac{1}{2\pi} \int_{-\pi/6}^{\pi/6} \frac{1}{2}\frac{1}{5} d\omega = \frac{1}{2\pi} \frac{2\pi}{6} \frac{1}{10} = \frac{1}{60} = 0.01667$$

To check this result you can use the following OCTAVE/matlab code for getting the sum in the time domain:

L = 128;    % select large enough L for better approximation
n = -L:L;
x =  sin(pi*n/4).*sin(pi*n/6)./(2*5*pi*pi.*n.*n);
x(L+1)= (1/4)*(1/6)/(2*5);
S = sum(x)
>> ans = 0.01667

which verifies that our calculation was correct.