Question

Determine the output of the LTI system for the input x[n) and impulse response of the
system h[n]
x[n] = {2,7,1,5)
h[n] = {6,2,5,9,1)
(a) (12,46,30,85,78,42,47,5)
() {12,46,30,85,80,40,47,5
(c) (12,46,30,85,78,47,42,5}
(d) {12,46,30,85,80, 41, 46,5
Answer
A
A
B​

Answer 1

Answer:

h(n)=nu(n)

x(n) = \delta(n)-2\delta(n-5)+\delta(n-10)x(n)=δ(n)−2δ(n−5)+δ(n−10)

Output y(n)y(n) is convolution of h(n)h(n) and x(n)x(n) .i.e

y(n) = x(n) * h(n)y(n)=x(n)∗h(n) (here '*∗' implies convolution)

i.e. y(n) = \displaystyle\sum_{k=-\infin}^\infiny(n)=

k=−∞

∑

∞

x(n)h(n-k)x(n)h(n−k)

i.e. y(n) = \displaystyle\sum_{k=-\infin}^\infiny(n)=

k=−∞

∑

∞

[\delta(k)-2\delta(k-5)+\delta(k-10)](n-k)u(n-k)[δ(k)−2δ(k−5)+δ(k−10)](n−k)u(n−k)

y(n) = \displaystyle\sum_{k=-\infin}^\infin [(n-k)u(n-k)\delta(k) - 2(n-k)u(n-k)\delta(k-5)+\\(n-k)u(n-k)\delta(k-10)]y(n)=

k=−∞

∑

∞

[(n−k)u(n−k)δ(k)−2(n−k)u(n−k)δ(k−5)+

(n−k)u(n−k)δ(k−10)]

This equation on doing summation we get,

y(n) = nu(n)-2(n-5)u(n-5)+(n-10)u(n-10)y(n)=nu(n)−2(n−5)u(n−5)+(n−10)u(n−10)

Substituting values for n we get ,

y(n) ={0,1,2,3,4,5,6,4,3,2,1}y(n)=0,1,2,3,4,5,6,4,3,2,1

i.e y(0) = 0, y(1)=1,y(2) = 2. \space so\space on \space and\space y(10) = 1y(0)=0,y(1)=1,y(2)=2. so on and y(10)=1

i.e for all other values of n other than 0\le n \le 100≤n≤10 we have y(n)=0y(n)=0