Question

Consider the cascade interconnection of three causal LTI system shown below;
and the overall impulse response is $h\left(n\right)where{h}_2\left(n\right)=u\left(n\right)-u(n-2)$
$y_1\left(n\right)$ is response of first system for the input. Then


$x\left(n\right)=\delta \left(n\right)-\delta (n-1).$

${\sum }_{n=-\infty }^{\infty }y\left(n\right)=$ ____

• A. 0

Answer 1

The correct option is A 0

$h\left(n\right)=h_1\left(n\right)\ast \left[h_2\right(n)\ast h_2(n\left)\right]$
No.of non zero samples in h(n) = No. of samples in $h_1\left(n\right)$ + No. of samples in $\left(h_2\right(n)\ast h_2(n\left)\right)-1$

7 = No, of samples in $h_1\left(n\right)+3-1$
No. of samples in $h_1\left(n\right)=7-2$
= 5
Let , $h_2\left(n\right)=a,b,c,d,e$
$h_1\left(n\right)=abcde$

$h_2\left(n\right)\ast h_2\left(n\right)=\frac{1,21}{a,b,c,d,e}$

2a 2b 2c 2d 2e

$h\left(n\right)=\frac{a,b,c,d,e}{[a,b+2a,c+2b+a,d+2c+b,c+2d+c,2e+d,e]}$

Compare with h(n),

$h\left(n\right)=[1,5,10,11,8,4,1]$
a = 1
b + 2a = 5
b + 2 = 5
b = 3
c + 2b + a = 10
c + 2(3) + (1) = 10
c + 6 + 1 = 10
c = 3
d + 2(3) + 3= 11
d + 9 = 11
d = 2
e + 2d + c = 8
e + 4 + 3 = 8
e = 1

$\therefore h_1\left(n\right)=1,3,3,2,1$
$y_1\left(n\right)=x\left(n\right)\ast h_1\left(n\right)$

$x\left(n\right)\frac{1-1}{1,3,3,2,1}$

$y_1\left(n\right)=\frac{-1,-3,-3,-2,-1}{1,2,0,-1,-1,-1}$

${\sum }_{n=-\infty }^{\infty }{y}_1\left(n\right)=1+2+0-1-1-1=0$