Why must the sum of all y[n] be 0 here?

Because the sum of h (-2-1+0+1+2) is zero, and sum y = (sum x)(sum h)

Because h is odd so y is always zero

Because the output has an even number of samples

Asked 2/30
2073 Magh · 7 marks
2082 Bhadra (new course) · 5 marks

Output of h[n] = n (−2 ≤ n ≤ 2)

Short answer

Aligning x[0]=-0.25 and h[0]=0, convolution by shifting and summing gives y[n]={-2,-0.5,0.25,3,1.75,0,-0.5,-1.5,0.5,-1} with y[0]=3, for -3≤n≤6.

A discrete time LTI system is defined by the impulse response h[n] = n for −2 ≤ n ≤ 2; 0 otherwise, i.e. h[n] = {−2, −1, 0, 1, 2}. If the input to the given system is x[n] = {1, −0.25, 0, −1, 0.5, −0.5} (↑ at −0.25, i.e. x[0] = −0.25), find and plot the output of the system.

Given:

$h[n] = \{-2, -1, \underset{\uparrow}{0}, 1, 2\}$ for $-2 \le n \le 2$ .
$x[n] = \{1, \underset{\uparrow}{-0.25}, 0, -1, 0.5, -0.5\}$ for $-1 \le n \le 4$ .

y[n] = \sum_k x[k]\,h[n-k]

Output range: start $= -1 + (-2) = -3$ ; length $= 6 + 5 - 1 = 10$ ; so $-3 \le n \le 6$ .

Tabular method

Each row is $x[k]$ times $h$ shifted to start at $n = k - 2$ :

Row	$n=-3$	$n=-2$	$n=-1$	$n=0$	$n=1$	$n=2$	$n=3$	$n=4$	$n=5$	$n=6$
$x[-1]\,h[n+1]$	-2	-1	0	1	2
$x[0]\,h[n]$		0.5	0.25	0	-0.25	-0.5
$x[1]\,h[n-1]$			0	0	0	0	0
$x[2]\,h[n-2]$				2	1	0	-1	-2
$x[3]\,h[n-3]$					-1	-0.5	0	0.5	1
$x[4]\,h[n-4]$						1	0.5	0	-0.5	-1
$y[n]$	-2	-0.5	0.25	3	1.75	0	-0.5	-1.5	0.5	-1

Sample working:

$y[-3] = x[-1]h[-2] = 1(-2) = -2$
$y[0] = x[-1]h[1] + x[0]h[0] + x[1]h[-1] + x[2]h[-2] = 1 + 0 + 0 + 2 = 3$
$y[2] = x[0]h[2] + x[1]h[1] + x[2]h[0] + x[3]h[-1] + x[4]h[-2] = -0.5 + 0 + 0 - 0.5 + 1 = 0$

Sum check: $(\sum x)(\sum h) = (-0.25)(0) = 0$ , and $-2 - 0.5 + 0.25 + 3 + 1.75 + 0 - 0.5 - 1.5 + 0.5 - 1 = 0$ .

Answer:

y[n] = \{-2,\ -0.5,\ 0.25,\ \underset{\uparrow}{3},\ 1.75,\ 0,\ -0.5,\ -1.5,\ 0.5,\ -1\}

for $-3 \le n \le 6$ .

Plot

Stem plot drawn sideways (each # = 0.25; left of | is negative):

  n   y[n]
 -3  -2.00  ########|
 -2  -0.50        ##|
 -1   0.25          |#
  0   3.00          |############
  1   1.75          |#######
  2   0.00          |
  3  -0.50        ##|
  4  -1.50    ######|
  5   0.50          |##
  6  -1.00      ####|

$n$	−3	−2	−1	0	1	2	3	4	5	6
$y[n]$	−2	−0.5	0.25	3	1.75	0	−0.5	−1.5	0.5	−1

✓ Answer checked by Claude Opus 5.5 (high effort)

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Output of h[n] = n (−2 ≤ n ≤ 2)

Tabular method

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