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] = $\frac{1}{3}^k$ u[k] for k ≥ 0 -----------3

Consider the second term in convolution of equation 2 i.e. h[n -k].

It is given that h[n] = $\frac{1}{2}^n$ u[n]

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

u[n - k] = 0 for n < k.

Hence, h[n - k] = $\frac{1}{2}^{n-k}$ u[n-k] for n ≥ k ------------ 4

Putting the values of equation 3 and equation 4 in equation 2, we get

y[n] = $\sum_{ k = - {\infty}}^{\infty} \frac{1}{3}^{k} u[k] \frac{1}{2}^{n-k} u[n-k] $ ------------- 5

Since k ≥ 0, the lower limit of summation in above equation will be k = 0.

From equation 4, n ≥ k always

Hence upper limit of summation in above equation will be k = n.

y[n] = $\sum_{k = 0}^{n} (\frac{1}{3})^{k} (\frac{1}{2})^{n-k}$

u[k] and u[n - k] will be 1 within the limits k = 0 to k = n

y[n] = $\sum_{k = 0}^{n} (\frac{1}{3})^{k} (\frac{1}{2})^{n} (\frac{1}{2})^{-k}$

= $(\frac{1}{2})^{n} \sum_{k = 0}^{n} (\frac{1}{3})^{k} (\frac{1}{2})^{-k}$

= $(\frac{1}{2})^{n} \sum_{k = 0}^{n} (\frac{1}{3^k}) (\frac{1}{2^{-k}})$

= $(\frac{1}{2})^{n} \sum_{k = 0}^{n} (\frac{1}{3^k}) {2^{k}}$

= $(\frac{1}{2})^{n} \sum_{k = 0}^{n} (\frac{2}{3})^k $

Using the formula,

$ \sum_{k = 0}^{N} a^k = (\frac{1 - a^{N-1}}{1-a}) $

$y[n] = (\frac{1}{2})^n [\frac{1- (\frac{2}{3})^{n+1}}{1- \frac{2}{3}}]$

$y[n] = 3 (\frac{1}{2})^n [1- (\frac{2}{3})^{n+1}]$