Example 1

Verify Parseval’s theorem of the sequence $x(n) = frac{1^n}{4}u(n)$

Solution − $displaystylesumlimits_{-infty}^infty|x_1(n)|^2 = frac{1}{2pi}int_{-pi}^{pi}|X_1(e^{jomega})|^2domega$

L.H.S $displaystylesumlimits_{-infty}^infty|x_1(n)|^2$

$= displaystylesumlimits_{-infty}^{infty}x(n)x^*(n)$

$= displaystylesumlimits_{-infty}^infty(frac{1}{4})^{2n}u(n) = frac{1}{1-frac{1}{16}} = frac{16}{15}$

R.H.S. $X(e^{jomega}) = frac{1}{1-frac{1}{4}e-jomega} = frac{1}{1-0.25cos omega+j0.25sin omega}$

$Longleftrightarrow X^*(e^{jomega}) = frac{1}{1-0.25cos omega-j0.25sin omega}$

Calculating, $X(e^{jomega}).X^*(e^{jomega})$

$= frac{1}{(1-0.25cos omega)^2+(0.25sin omega)^2} = frac{1}{1.0625-0.5cos omega}$

$frac{1}{2pi}int_{-pi}^{pi}frac{1}{1.0625-0.5cos omega}domega$

$frac{1}{2pi}int_{-pi}^{pi}frac{1}{1.0625-0.5cos omega}domega = 16/15$

We can see that, LHS = RHS.(Hence Proved)

Example 2

Compute the N-point DFT of $x(n) = 3delta (n)$

Solution − We know that,

$X(K) = displaystylesumlimits_{n = 0}^{N-1}x(n)e^{frac{j2Pi kn}{N}}$

$= displaystylesumlimits_{n = 0}^{N-1}3delta(n)e^{frac{j2Pi kn}{N}}$

$ = 3delta (0)times e^0 = 1$

So,$x(k) = 3,0leq kleq N-1$… Ans.

Example 3

Compute the N-point DFT of $x(n) = 7(n-n_0)$

Solution − We know that,

$X(K) = displaystylesumlimits_{n = 0}^{N-1}x(n)e^{frac{j2Pi kn}{N}}$

Substituting the value of x(n),

$displaystylesumlimits_{n = 0}^{N-1}7delta (n-n_0)e^{-frac{j2Pi kn}{N}}$

$= e^{-kj14Pi kn_0/N}$… Ans