--- title: "Output of h[n] = n (−2 ≤ n ≤ 2)" subject: "Signals and Systems" chapter: 6 key: "semester_4/signals_and_systems:convolution-ramp-input" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-ramp-input site: BCT Engineering --- # Output of h[n] = n (−2 ≤ n ≤ 2) Signals and Systems · Chapter 6: Discrete-Time Systems - Asked in: 2073 Magh, 7 marks; 2082 Bhadra (new course), 5 marks (asked 2 times) - Video: [The impulse response of a linear time invariant system is h[𝒏]. find the response of the system.](https://www.youtube.com/watch?v=wjrDLSzHxMs) — Engg-Course-Made-Easy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-ramp-input - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6 ## Question 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. ## 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. ## Answer **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): ```text 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 | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. x is nonzero for -1 <= n <= 4 (6 samples) and h for -2 <= n <= 2 (5 samples). The output range is: A. -3 to 6 B. -1 to 4 C. -3 to 4 D. -2 to 6 Answer: A. Start = -1 + (-2) = -3; length = 6 + 5 - 1 = 10, so the end is 6. (Key point: Output range) 2. In the tabular method, the row for x[k] is x[k] times h placed starting at: A. n = k B. n = k - 2 C. n = k + 2 D. n = 0 Answer: B. h starts at -2, so the shifted copy begins at k + (-2). (Key point: Row alignment) 3. Why must the sum of all y[n] be 0 here? A. Because the sum of x is zero B. Because the sum of h (-2-1+0+1+2) is zero, and sum y = (sum x)(sum h) C. Because h is odd so y is always zero D. Because the output has an even number of samples Answer: B. sum h = 0, so the product of sums is zero, regardless of sum x = -0.25. (Key point: Sum check) 4. What is y[0]? A. 1 B. 0 C. 3 D. -0.25 Answer: C. y[0] = x[-1]h[1] + x[0]h[0] + x[1]h[-1] + x[2]h[-2] = 1 + 0 + 0 + 2 = 3. 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