Signals and Systems - Electrical Engineering

474 C H A P T E R 8: Discrete-Time Signals and Systems

Solution

Using the unit-sample signal generic representation, we have

r[n]=

∑∞

k=−∞

(ku[k])δ[n−k]=

∑∞

k= 0

kδ[n−k]= 0 δ[n]+ 1 δ[n−1]+ 2 δ[n−2]+···

Lettingm=n−k, we write the above equation as

r[n]=

∑n

m=−∞

(n−m)δ[m]=

∑n

m=−∞

(n−m)(u[m]−u[m−1])

=

∑n

m= 0

(n−m)−

∑n

m= 1

(n−m)=n+

∑n

m= 1

(n−m)−

∑n

m= 1

(n−m)=n n≥ 0

Forn<0,r[n]=0. n

nExample 8.16

Consider a discrete pulse

x[n]=

{

1 0≤n≤N− 1

0 otherwise

Obtain representations ofx[n] using unit-sample and unit-step signals.

Solution

The signalx[n] can be represented as

x[n]=

N∑− 1

k= 0

δ[n−k]

and usingδ[n]=u[n]−u[n−1], we obtain a representation of the discrete pulse in terms of unit-

step signals,

x[n]=

N∑− 1

k= 0

(u[n−k]−u[n−k−1])=(u[n]−u[n−1])+(u[n−1]−u[n−2])

+···−u[n−N]

=u[n]−u[n−N]

because of the cancellation of consecutive terms. n