Question

x[n] = 0; n < -1, n>0 , x[-1] = -1, x[0] = 2 is the input and
y[n] = 0; n < -1, n > 2, y[-1] = -1 =y[1],
y[0] = 3, y[2] = -2 is the output of a discrete-time LTI system
The system impulse response h(n) will be

• A.
  h[n] = 0; n < -2, n > 1, h[-2] = h[1] = h[-1] = -h[0] = 3
• B.
  h[n] = 0; n<0, n>2, h[0] =1, h[1] = h[2] = -1
• C.
  h[n] = 0; n < 0, n > 3, h[0] = -1, h[1] = 2, h[2] =1
• D.
  h[n] = 0; n<-1, n>1, h[-1] = 1, h[0] = h[1] =2

Answer 1

The correct option is B

h[n] = 0; n<0, n>2, h[0] =1, h[1] = h[2] = -1
For finite duration convolution,
x[n] have M terms
h[n] have N terms
then, y[n] should have terms (M + N - 1)
Here, x[n] = {-1, 2}
$\uparrow$
y[n] = {-1, 3, -1, -2}
$\uparrow$
So, h[n] should have only 3 terms and h[n] have value starting from origin or [n=0] only because y[n] have value start from [n = -1]
$\therefore$ Consider option (I)



y[n]={-1, 3, -1, 2}
$\uparrow$
So option (a) is correct.