--- title: "Output for x = δ[n] + 2δ[n−1] − δ[n−3]" subject: "Signals and Systems" chapter: 6 key: "semester_4/signals_and_systems:convolution-delta-sums" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-delta-sums site: BCT Engineering --- # Output for x = δ[n] + 2δ[n−1] − δ[n−3] Signals and Systems · Chapter 6: Discrete-Time Systems - Asked in: 2080 Chaitra, 8 marks; 2069 Bhadra, 7 marks (asked 2 times) - Video: [(E)SS(O) Q 2.1(a,b,c) || Discrete Time Convolution by Convolution Sum Method || Compute and Plot](https://www.youtube.com/watch?v=3AXXI8Jeq6A) — Electrical Engineering Academy - 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-delta-sums - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6 ## Question Find the output of an LTI system given by: x[n] = δ[n] + 2δ[n−1] − δ[n−3] and h[n] = 2δ[n+1] + 2δ[n−1]. ## Short answer Using distributivity and shift property, y[n]=2x[n+1]+2x[n-1] gives y[n]={2,4,2,2,0,-2} with y[0]=4. ## Answer **Given:** - $x[n] = \delta[n] + 2\delta[n-1] - \delta[n-3] = \{\underset{\uparrow}{1}, 2, 0, -1\}$, for $0 \le n \le 3$. - $h[n] = 2\delta[n+1] + 2\delta[n-1] = \{2, \underset{\uparrow}{0}, 2\}$, for $-1 \le n \le 1$. #### Method 1: Using $x[n] * \delta[n-k] = x[n-k]$ Convolution with a shifted impulse just shifts the signal, and convolution is distributive: $$ \begin{aligned} y[n] &= x[n] * \left(2\delta[n+1] + 2\delta[n-1]\right) = 2x[n+1] + 2x[n-1] \end{aligned} $$ $$ \begin{aligned} 2x[n+1] &= 2\delta[n+1] + 4\delta[n] - 2\delta[n-2] \\ 2x[n-1] &= 2\delta[n-1] + 4\delta[n-2] - 2\delta[n-4] \end{aligned} $$ Adding: $$ y[n] = 2\delta[n+1] + 4\delta[n] + 2\delta[n-1] + 2\delta[n-2] - 2\delta[n-4] $$ #### Method 2: Tabular check Output starts at $0 + (-1) = -1$; length $4 + 3 - 1 = 6$, so $-1 \le n \le 4$. | Row | $n=-1$ | $n=0$ | $n=1$ | $n=2$ | $n=3$ | $n=4$ | |---|---|---|---|---|---|---| | $x[0]\,h[n]$ | 2 | 0 | 2 | | | | | $x[1]\,h[n-1]$ | | 4 | 0 | 4 | | | | $x[2]\,h[n-2]$ | | | 0 | 0 | 0 | | | $x[3]\,h[n-3]$ | | | | -2 | 0 | -2 | | $y[n]$ | **2** | **4** | **2** | **2** | **0** | **-2** | Both methods agree. Sum check: $(\sum x)(\sum h) = 2 \times 4 = 8 = 2 + 4 + 2 + 2 + 0 - 2$. **Answer:** $$ y[n] = \{2,\ \underset{\uparrow}{4},\ 2,\ 2,\ 0,\ -2\} $$ with $y[-1] = 2$, $y[0] = 4$, $y[1] = 2$, $y[2] = 2$, $y[3] = 0$, $y[4] = -2$. ```text y[n] 4 | o | | 2 | o | o o | | | | | 0 +---+---+---+---+---o---+--> n | -1 0 1 2 3 4 -2 | o ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does x[n] * delta[n-k] equal? A. x[n+k] B. x[n-k] C. k x[n] D. x[k] Answer: B. Convolution with a shifted impulse just delays the signal by k: x[n-k]. (Key point: Shifting property) 2. With h[n] = 2delta[n+1] + 2delta[n-1], the output is: A. 2x[n+1] + 2x[n-1] B. 2x[n-1] + 2x[n-1] C. x[n+1] + x[n-1] D. 4x[n] Answer: A. Convolution distributes over the sum and each delta shifts x, with a factor of 2 in each term. (Key point: Using distributivity) 3. Where does the output start, given x begins at n = 0 and h begins at n = -1? A. n = 0 B. n = -1 C. n = 1 D. n = -2 Answer: B. Start of y = start of x + start of h = 0 + (-1) = -1; it runs to n = 4. (Key point: Starting index) 4. What is y[0]? A. 2 B. 0 C. 4 D. 6 Answer: C. y[0] = 2x[1] + 2x[-1] = 2(2) + 0 = 4. 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