Verify Parseval’s theorem of the sequence $x(n) = \frac{1^n}{4}u(n)$
Solution − $\displaystyle\sum\limits_{-\infty}^\infty|x_1(n)|^2 = \frac{1}{2\pi}\int_{-\pi}^{\pi}|X_1(e^{j\omega})|^2d\omega$
L.H.S $\displaystyle\sum\limits_{-\infty}^\infty|x_1(n)|^2$
$= \displaystyle\sum\limits_{-\infty}^{\infty}x(n)x^*(n)$
$= \displaystyle\sum\limits_{-\infty}^\infty(\frac{1}{4})^{2n}u(n) = \frac{1}{1-\frac{1}{16}} = \frac{16}{15}$
R.H.S. $X(e^{j\omega}) = \frac{1}{1-\frac{1}{4}e-j\omega} = \frac{1}{1-0.25\cos \omega+j0.25\sin \omega}$
$\Longleftrightarrow X^*(e^{j\omega}) = \frac{1}{1-0.25\cos \omega-j0.25\sin \omega}$
Calculating, $X(e^{j\omega}).X^*(e^{j\omega})$
$= \frac{1}{(1-0.25\cos \omega)^2+(0.25\sin \omega)^2} = \frac{1}{1.0625-0.5\cos \omega}$
$\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{1}{1.0625-0.5\cos \omega}d\omega$
$\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{1}{1.0625-0.5\cos \omega}d\omega = 16/15$
We can see that, LHS = RHS.(Hence Proved)
Compute the N-point DFT of $x(n) = 3\delta (n)$
Solution − We know that,
$X(K) = \displaystyle\sum\limits_{n = 0}^{N-1}x(n)e^{\frac{j2\Pi kn}{N}}$
$= \displaystyle\sum\limits_{n = 0}^{N-1}3\delta(n)e^{\frac{j2\Pi kn}{N}}$
$ = 3\delta (0)\times e^0 = 1$
So,$x(k) = 3,0\leq k\leq N-1$… Ans.
Compute the N-point DFT of $x(n) = 7(n-n_0)$
Solution − We know that,
$X(K) = \displaystyle\sum\limits_{n = 0}^{N-1}x(n)e^{\frac{j2\Pi kn}{N}}$
Substituting the value of x(n),
$\displaystyle\sum\limits_{n = 0}^{N-1}7\delta (n-n_0)e^{-\frac{j2\Pi kn}{N}}$
$= e^{-kj14\Pi kn_0/N}$… Ans