Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10M

Year : DEC 2015

Given: x[n] = $\frac{1}{3}^n$ u[n]

h[n] = $\frac{1}{2}^n$ u[n]

Therefore, y[n] = x[n]*h[n] ------------ 1

y[n] = $\sum_{ k = - {\infty}}^{\infty}$ x[k] h[n-k] -------------- 2

Consider the first term i.e. x[k]

Since x[n] = $\frac{1}{3}^n$ u[n], x[k] = $\frac{1}{3}^k$ u[k]

u[k] = 0 for k < 0.

Hence, x[k] …