--- title: "LTI output, h[n]=(1/2)^n window" subject: "Digital Signal Analysis and Processing" chapter: 1 key: "semester_7/digital_signal_analysis_and_processing:lti-output-exp-window-1" source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/lti-output-exp-window-1 site: BCT Engineering --- # LTI output, h[n]=(1/2)^n window Digital Signal Analysis and Processing · Chapter 1: Discrete Time Signals and Systems - Asked in: 2074 Asoj, 3+2 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/lti-output-exp-window-1 - Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-1 ## Question Find the output of LTI system having impulse response h[n] = (1/2)^n {u[n+2] − u[n−2]} and input signal x[n] = {2, 1, 0.5, −1}. Also check the answer. ## Short answer Convolving h[n]={(1/2)^n for -2≤n≤1} with x[n]={2,1,0.5,-1} from n=0 yields y[n]={8,8,6,-1,-1,-0.75,-0.5} for n=-2 to 4. ## Answer The output is y[n] = x[n] * h[n] = Σ x[k] h[n − k]. **Impulse response:** u[n+2] − u[n−2] is 1 for −2 ≤ n ≤ 1, so h[n] = (½)ⁿ for n = −2, −1, 0, 1: h[−2] = 4, h[−1] = 2, h[0] = 1, h[1] = ½. **Input:** x[n] = {2, 1, 0.5, −1}. No origin is marked, so I take the first sample at n = 0: x[0] = 2, x[1] = 1, x[2] = 0.5, x[3] = −1. Output starts at n = −2 + 0 = −2, ends at n = 1 + 3 = 4 (length 4 + 4 − 1 = 7). Using y[n] = Σ x[k] h[n − k]: ```text y[−2] = x0·h[−2] = 2·4 = 8 y[−1] = x0·h[−1] + x1·h[−2] = 4 + 4 = 8 y[ 0] = x0·h[0] + x1·h[−1] + x2·h[−2] = 2 + 2 + 2 = 6 y[ 1] = x0·h[1] + x1·h[0] + x2·h[−1] + x3·h[−2] = 1 + 1 + 1 − 4 = −1 y[ 2] = x1·h[1] + x2·h[0] + x3·h[−1] = 0.5 + 0.5 − 2 = −1 y[ 3] = x2·h[1] + x3·h[0] = 0.25 − 1 = −0.75 y[ 4] = x3·h[1] = −0.5 ``` **Answer:** y[n] = {8, 8, 6, −1, −1, −0.75, −0.5} for n = −2 … 4. #### Check 1. **Sum rule:** Σy = (Σx)(Σh). Σx = 2.5, Σh = 7.5, product = 18.75. Σy = 8 + 8 + 6 − 1 − 1 − 0.75 − 0.5 = 18.75. ✓ 2. **Polynomial (z-transform) method:** X(z) = 2 + z⁻¹ + 0.5z⁻² − z⁻³, H(z) = 4z² + 2z + 1 + 0.5z⁻¹. Multiplying gives 8z² + 8z + 6 − z⁻¹ − z⁻² − 0.75z⁻³ − 0.5z⁻⁴, the same coefficients. ✓ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. u[n+2] - u[n-2] equals 1 for A. n = -2, -1, 0, 1, 2 B. n = 0 ... 2 C. n = -2, -1, 0, 1 D. n = -1 ... 2 Answer: C. u[n+2] starts at n = -2 and u[n-2] starts at n = 2, so the window is -2 ... 1. (Key point: Window of h[n]) 2. h[-2] and h[-1] for h[n] = (1/2)^n in the window are A. 1/4 and 1/2 B. 4 and 2 C. -4 and -2 D. 2 and 1 Answer: B. (1/2)^(-2) = 4 and (1/2)^(-1) = 2. (Key point: Values of h) 3. With x of 4 samples starting at n = 0 and h of 4 samples starting at n = -2, the output A. has 7 samples from n = -2 to 4 B. has 8 samples from n = -2 to 5 C. has 4 samples D. has 7 samples from n = 0 to 6 Answer: A. Length 4+4-1 = 7; start = -2 + 0, end = 1 + 3 = 4. (Key point: Output length and start) 4. Which relation checks the convolution result? A. sum y = sum x + sum h B. sum y = sum x / sum h C. sum y = (sum x)^2 D. sum y = (sum x)(sum h) Answer: D. 2.5 x 7.5 = 18.75 equals the sum of y. (Key point: Sum check) ## Close variants set in other exams - [LTI output with sparse samples, verify](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Alti-output-sparse-samples) — Digital Signal Analysis and Processing, chapter 1, asked 4× - [LTI output, h[n]=(1/3)^n window](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Alti-output-exp-window-2) — Digital Signal Analysis and Processing, chapter 1, asked 4× - [LTI output for h=2u[n]−2u[n−4]](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Alti-output-step-difference) — Digital Signal Analysis and Processing, chapter 1, asked 4× ## Related in Digital Signal Analysis and Processing - [LTI output for 2δ(n) − δ(n−1) input](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Alti-output-impulse-input) — Digital Signal Analysis and Processing, chapter 1 - [Output for 5e^(jπn/3) input](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Alti-output-exp-pi3) — Digital Signal Analysis and Processing, chapter 3, asked 2× - [Output for 5e^(jπn/2) input](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Alti-output-exp-pi2) — Digital Signal Analysis and Processing, chapter 3, asked 2× - [Output for e^(jπn) input, h=(1/2)^n u[n]](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Alti-output-complex-exp-half) — Digital Signal Analysis and Processing, chapter 3 - [Output for sinusoid mix input, h=(1/2)^n u[n]](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Alti-output-sinusoid-mix) — Digital Signal Analysis and Processing, chapter 3 - [Low-pass FIR window design, αs = 41 dB](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Afir-lowpass-window-0-2-0-5-41) — Digital Signal Analysis and Processing, chapter 5 ## Related in other subjects - [Define LTI system and impulse response](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Alti-impulse-response-define) — Data Communication, chapter 2, asked 2× - [LTI output for h(t)=e^(-2t), rectangular input](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Alti-output-rect-input) — Data Communication, chapter 2, asked 3× - [Parseval's theorem; output PSD of LTI system](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aparseval-power-lti-psd) — Communication Systems, chapter 1 - [LTI output of a WSS input is WSS](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Alti-wssp-output) — Communication Systems, chapter 9