--- title: "Output for h[n] = u[n], x[n] = 2ⁿu[−n]" subject: "Signals and Systems" chapter: 6 key: "semester_4/signals_and_systems:convolution-step-anticausal" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-step-anticausal site: BCT Engineering --- # Output for h[n] = u[n], x[n] = 2ⁿu[−n] Signals and Systems · Chapter 6: Discrete-Time Systems - Asked in: 2078 Baisakh, 6 marks - 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-step-anticausal - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6 ## Question If the impulse response of a discrete LTI system is h[n] = u[n] and input to the system is x[n] = 2ⁿu[−n], determine the output y[n] of the system. ## Short answer Evaluate convolution sum with limits: y[n] = 2^{n+1} for n ≤ 0, and y[n] = 2 for n > 0. ## Answer **Given:** $h[n] = u[n]$ (accumulator) and $x[n] = 2^n u[-n]$, which is non-zero only for $n \le 0$: $x[n] = \{\dots, \tfrac{1}{8}, \tfrac{1}{4}, \tfrac{1}{2}, \underset{\uparrow}{1}\}$. $$ y[n] = \sum_{k=-\infty}^{\infty} x[k]\,h[n-k] = \sum_{k=-\infty}^{\infty} 2^k u[-k]\,u[n-k] $$ **Limits:** $u[-k] = 1$ for $k \le 0$; $u[n-k] = 1$ for $k \le n$. So $k$ runs from $-\infty$ to $\min(n, 0)$. #### Case 1: $n \le 0$ (here $\min = n$) $$ \begin{aligned} y[n] &= \sum_{k=-\infty}^{n} 2^k \quad \text{(put } m = n - k \text{, } m = 0 \text{ to } \infty\text{)} \\ &= 2^n\sum_{m=0}^{\infty}\left(\tfrac{1}{2}\right)^m = 2^n\cdot\frac{1}{1 - \frac{1}{2}} \\ &= 2^{n+1} \end{aligned} $$ #### Case 2: $n > 0$ (here $\min = 0$) $$ y[n] = \sum_{k=-\infty}^{0} 2^k = 1 + \tfrac{1}{2} + \tfrac{1}{4} + \dots = \frac{1}{1 - \frac{1}{2}} = 2 $$ **Answer:** $$ y[n] = \begin{cases} 2^{n+1}, & n \le 0 \\ 2, & n > 0 \end{cases} \quad\text{i.e.}\quad y[n] = 2^{n+1}u[-n-1] + 2\,u[n] $$ (At $n = 0$ both forms give 2, so the pieces join.) | $n$ | −3 | −2 | −1 | 0 | 1 | 2 | |---|---|---|---|---|---|---| | $y[n]$ | 0.25 | 0.5 | 1 | 2 | 2 | 2 | ```text y[n] 2 | o o o o ... 1 | o | | | | 0.5 | o | | | | | | o | | | | | | 0 +--+--+--+--+--+--+--+--> n -3 -2 -1 0 1 2 3 ``` This makes sense: $h[n] = u[n]$ is an accumulator, so $y[n] = \sum_{k=-\infty}^{n} x[k]$, the running sum of the input. The sum keeps growing until $n = 0$, then stays at the total $\sum x[k] = 2$ because the input is zero for $n > 0$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For x[n] = 2^n u[-n] and h[n] = u[n], the sum over k runs from -infinity to: A. n B. 0 C. min(n, 0) D. max(n, 0) Answer: C. u[-k] needs k <= 0 and u[n-k] needs k <= n, so both hold for k <= min(n, 0). (Key point: Summation limits) 2. For n <= 0, y[n] = sum of 2^k for k <= n equals: A. 2^n B. 2^(n+1) C. 2 D. 1/2^n Answer: B. 2^n times the geometric series 1 + 1/2 + 1/4 + ... = 2^n x 2 = 2^(n+1). (Key point: Case n <= 0) 3. For n > 0, y[n] equals: A. 2^(n+1) B. 2 (the total sum of x) C. 1 D. infinity Answer: B. The sum stops at k = 0 and gives 1 + 1/2 + 1/4 + ... = 2. (Key point: Case n > 0) 4. Why does y[n] stop growing after n = 0? A. h[n] = u[n] is a running-sum (accumulator) and x is zero for n > 0 B. Because h[n] is zero for n > 0 C. Because the system is memoryless D. Because convolution is commutative Answer: A. y[n] is the running sum of x, which reaches its total 2 at n = 0 and stays there. 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