Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : DEC 2014

By using convolution sum formula i.e.,

$x[n]= (\dfrac {1^n}3)u(n) ∴x[k]= (\dfrac 13)^k u(k). \\ \text {Similarly } \space h[n]= (\dfrac 12)^n u(n) ∴h[n-k]=(\dfrac 12)^{n-k} u(n-k). \\ ∴y[n]= ∑\limits_{-∞}^∞ x[k] × h[n-k]. \\ y[n] = ∑\limits_{k=-∞}^∞.(\dfrac 13)^k \space u(k) (\dfrac 12)^{n-k} u(n-k).$

Since k>0, the lower limit of summation in above equation will be k=0. From above equation we know that n>k always. Hence upper limit of summation in above equation will be k=n.

This means if n tends to infinity then k also tends to infinity, but k will have maximum value equal to n.

Thus we can write equation as,

$y[n] = ∑\limits_{k=0}^n.(\dfrac 13)^k (\dfrac 12)^{n-k}. $

In above equation, we have not written u(k) and u(n-k) since the limits of summation are modified accordingly.

$y[n] = ∑\limits_{k=0}^n.(\dfrac 13)^k (\dfrac 12)^n (\dfrac 12)^{-k} \\ =(\dfrac12)^n ∑\limits_{k=0}^n .(\dfrac13)^k (\dfrac12)^{-k} \\ =(\dfrac12)^n ∑\limits_{k=0}^n.(\dfrac23)^k $

By using the formula,$ ∑\limits_{k=0}^n\space a^k =\dfrac{a^{n+1}-1}{a-1}\\ y[n] = (\dfrac12)^n \dfrac { (\dfrac23)^{n+1}-1}{(\dfrac23)-1} \\ y[n] = 3 (\dfrac12)^n \dfrac{1-(\dfrac23)^{n+1}}1$

Verification:

By z-transform,

$y (z) = \dfrac z{z-1⁄3}\dfrac z{z-1⁄2} \\ \dfrac {y [z]}z= \dfrac z{(z-1⁄3)(z-1⁄2)}$

By using partial fraction,

$\dfrac {y [z]}z= \dfrac A{(z-1⁄3)} +\dfrac B{(z-1⁄2)} $

$(A=-2) $ and $(B=3)$

$Y (z) = \dfrac {-2z}{(z-1⁄3)} + \dfrac {3z}{(z-1⁄2)} $

By Inverse z-transform,

$y [n] =-2 (\dfrac13)^n u(n)+3(\dfrac12)^n u(n). $