--- title: "Graphical convolution of two rectangles" subject: "Digital Signal Analysis and Processing" chapter: 1 key: "semester_7/digital_signal_analysis_and_processing:convolution-graphical-rect" source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/convolution-graphical-rect site: BCT Engineering --- # Graphical convolution of two rectangles Digital Signal Analysis and Processing · Chapter 1: Discrete Time Signals and Systems - Asked in: 2071 Shrawan, 5 marks - Video: [Convolution in 5 Easy Steps](https://www.youtube.com/watch?v=aMaw4EumwyE) — Iain Explains Signals, Systems, and Digital Comms - 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/convolution-graphical-rect - Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-1 ## Question A discrete time LTI system has input signal x[n] = 1 for −1 ≤ n ≤ 1 (0 elsewhere) and impulse response h[n] = 1 for −1 ≤ n ≤ 1 (0 elsewhere). Find the output of the system using the graphical method. ## Short answer Graphical convolution of two length-3 pulses (from −1 to 1) gives a triangular output: y[n] = {1,2,3,2,1} for n=−2,…,2. The sum of products of overlapped samples yields maximum 3 at n=0. ## Answer The output is the convolution sum y[n] = Σ x[k] h[n − k]. **Signals:** x[k] = 1 for k = −1, 0, 1; h[k] = 1 for k = −1, 0, 1. **Graphical method:** 1. Plot x[k] and h[k] against k. 2. Fold h[k] about k = 0 to get h[−k] (here it is symmetric, so it is unchanged). 3. Shift it by n to get h[n − k]. 4. Multiply x[k] · h[n − k] point by point and add all the products. This sum is y[n]. 5. Slide (change n) and repeat. h[n − k] is non-zero for k = n − 1, n, n + 1. Overlap with x[k] (k = −1 … 1): | n | Non-zero h[n−k] at k | Overlap with x | y[n] | |---|---|---|---| | ≤ −3 | n−1 … n+1 | none | 0 | | −2 | −3, −2, −1 | k = −1 | 1 | | −1 | −2, −1, 0 | k = −1, 0 | 2 | | 0 | −1, 0, 1 | k = −1, 0, 1 | 3 | | 1 | 0, 1, 2 | k = 0, 1 | 2 | | 2 | 1, 2, 3 | k = 1 | 1 | | ≥ 3 | – | none | 0 | ```text y[n] 3 | o 2 | o | o 1 | o | | | o 0 +-o--+--+--+--+--+--o-- n -3 -2 -1 0 1 2 3 ``` **Answer:** y[n] = {1, 2, **3**, 2, 1} for n = −2 … 2 (the bold centre sample 3 is at n = 0). **Check:** Σy = 9 = (Σx)(Σh) = 3 × 3. The output length is 3 + 3 − 1 = 5. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The output of an LTI system is A. y[n] = x[n] h[n] B. y[n] = x[n] + h[n] C. y[n] = sum of x[k] h[n-k] D. y[n] = sum of x[k] h[k] Answer: C. The convolution sum folds and shifts h, multiplies by x and adds. (Key point: Convolution sum) 2. The correct order of the graphical method is A. Shift h, multiply, fold, add B. Fold h[k] to get h[-k], shift by n, multiply with x[k], add C. Add x and h, then fold D. Multiply x and h, then shift the result Answer: B. Fold, shift by n, multiply point by point, sum, then slide. (Key point: Graphical steps) 3. What is y[0] for two 3-sample rectangles of ones? A. 3 B. 1 C. 2 D. 9 Answer: A. At n = 0 all three samples overlap, giving 1+1+1 = 3. (Key point: Peak value) 4. The output y[n] is A. {1, 1, 1} for n = -1 ... 1 B. {1, 2, 3, 3, 2, 1} for n = -2 ... 3 C. {3, 3, 3} for n = -1 ... 1 D. {1, 2, 3, 2, 1} for n = -2 ... 2 Answer: D. Length 3+3-1 = 5, symmetric triangle centred at n = 0, sum 9 = 3 x 3. 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