--- title: "Graphical auto-correlation, cross-correlation and convolution" subject: "Communication Systems" chapter: 1 key: "semester_6/communication_systems:graphical-correlation-convolution" source: https://www.bctengineer.com/q/semester_6/communication_systems/graphical-correlation-convolution site: BCT Engineering --- # Graphical auto-correlation, cross-correlation and convolution Communication Systems · Chapter 1: Introduction - Asked in: 2081 Chaitra, 3+3+3 marks - Video: [Convolution|| Auto Correlation|| Cross Correlation](https://www.youtube.com/watch?v=koZum7mCm8k) — Engineering Basics - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/graphical-correlation-convolution - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-1 ## Question Derive auto-correlation, cross-correlation and convolution of the following two signals graphically: [Figure: (a) a rectangular pulse of amplitude 2 from t = 1 to t = 2; (b) a right-angled triangular pulse that jumps to 2 at t = 1 and falls linearly to 0 at t = 2] ## Short answer Graphical overlap yields: rectangle auto-correlation 4(1−|τ|); triangle auto-correlation 4/3−2|τ|+(2/3)|τ|³; cross-correlation R12=2(1−τ²) for −1≤τ≤0, 2(1−τ)² for 0≤τ≤1; convolution y(t)=2(t−2)(4−t) for t∈[2,3], 2(t−4)² for t∈[3,4]. ## Answer Signals (from the figure): - $x_1(t) = 2$ for $1 \le t \le 2$, zero elsewhere (rectangular pulse). - $x_2(t) = 2(2 - t)$ for $1 \le t \le 2$, zero elsewhere (jumps to 2 at $t=1$, falls linearly to 0 at $t=2$). Definitions used (real signals): $R_{11}(\tau) = \int x_1(t)x_1(t+\tau)\,dt$, $R_{12}(\tau) = \int x_1(t)x_2(t+\tau)\,dt$, $y(t) = \int x_1(\lambda)x_2(t-\lambda)\,d\lambda$. #### Auto-correlation Graphically, keep $x(t)$ fixed, slide a copy by $\tau$ (left for $\tau > 0$), multiply and find the area of the overlap. Overlap exists only for $|\tau| < 1$. **Rectangle $x_1$:** for $0 \le \tau \le 1$ the overlap is $1 \le t \le 2-\tau$, of width $1-\tau$, height $2\times2 = 4$: $$ R_{11}(\tau) = \int_{1}^{2-\tau} 4\,dt = 4(1-|\tau|), \quad |\tau| \le 1 $$ It is a triangle of peak $R_{11}(0) = 4$ = energy of $x_1$, base $-1$ to $1$. **Triangle $x_2$:** for $0 \le \tau \le 1$, $$ \begin{aligned} R_{22}(\tau) &= \int_{1}^{2-\tau} 2(2-t)\cdot 2(2-t-\tau)\,dt \\ &= \frac{4}{3} - 2|\tau| + \frac{2}{3}|\tau|^3, \quad |\tau| \le 1 \end{aligned} $$ Peak $R_{22}(0) = 4/3$ = energy of $x_2$; it is even, smooth and reaches 0 at $\tau = \pm1$ (e.g. $R_{22}(0.5) = 5/12$). #### Cross-correlation $R_{12}(\tau)$ Slide $x_2$ relative to $x_1$; overlap again only for $|\tau| < 1$. - $0 \le \tau \le 1$ ($x_2$ moved left, its low tail overlaps): $R_{12}(\tau) = \int_{1}^{2-\tau} 2\cdot 2(2-t-\tau)\,dt = 2(1-\tau)^2$ - $-1 \le \tau \le 0$ (its tall front edge overlaps): $R_{12}(\tau) = \int_{1-\tau}^{2} 2\cdot 2(2-t-\tau)\,dt = 2(1-\tau^2)$ Values: $R_{12}(-1) = 0$, $R_{12}(-0.5) = 1.5$, $R_{12}(0) = 2$, $R_{12}(0.5) = 0.5$, $R_{12}(1) = 0$. The curve is not symmetric (cross-correlation is not even); its peak is at $\tau = 0$ with value 2. #### Convolution $y(t) = x_1(t) * x_2(t)$ Fold $x_2(\lambda)$ to $x_2(t-\lambda)$ (now non-zero for $t-2 \le \lambda \le t-1$, rising from 0 to 2) and slide it over $x_1(\lambda)$ on $[1,2]$. Output exists from $1+1 = 2$ to $2+2 = 4$. - $t < 2$: no overlap, $y = 0$. - $2 \le t \le 3$ (partial overlap $1 \le \lambda \le t-1$): $$ y(t) = \int_{1}^{t-1} 2\cdot 2(2-t+\lambda)\,d\lambda = -2t^2 + 12t - 16 = 2(t-2)(4-t) $$ - $3 \le t \le 4$ (overlap $t-2 \le \lambda \le 2$): $$ y(t) = \int_{t-2}^{2} 2\cdot 2(2-t+\lambda)\,d\lambda = 2(t-4)^2 $$ - $t > 4$: $y = 0$. Check: $y(2) = 0$, $y(3) = 2$ from both pieces (continuous), $y(4) = 0$; area $= 2 \times 1 = 2$ = (area of $x_1$) $\times$ (area of $x_2$) = $2 \times 1$. ```text y(t) 2 | * | * * 1 | * * | * * 0 +--*------------------*----> t 2 3 4 ``` **Answer:** $R_{11} = 4(1-|\tau|)$; $R_{22} = \frac43 - 2|\tau| + \frac23|\tau|^3$; $R_{12} = 2(1-\tau^2)$ for $-1\le\tau\le0$ and $2(1-\tau)^2$ for $0\le\tau\le1$; $y(t) = 2(t-2)(4-t)$ for $2\le t\le3$ and $2(t-4)^2$ for $3\le t\le4$ (all zero elsewhere). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the rectangle x1(t) = 2 on [1,2], what is the auto-correlation R11(tau)? A. 2(1 - |tau|) for |tau| <= 1 B. 4(1 - |tau|) for |tau| <= 1, a triangle with peak 4 C. 4 for |tau| <= 1, a rectangle D. 4(1 - tau^2) for |tau| <= 1 Answer: B. The overlap width is 1 - |tau| and the product height is 2 x 2 = 4, so the triangle peaks at 4, the energy of x1. (Key point: Auto-correlation of the rectangle) 2. What does the value of the auto-correlation at tau = 0 equal? A. The area of the signal B. The energy of the signal C. The peak amplitude of the signal D. Zero for any pulse Answer: B. R(0) is the integral of x^2, so R11(0) = 4 and R22(0) = 4/3 are the energies of x1 and x2. (Key point: Auto-correlation peak) 3. In graphical convolution of x1 (on [1,2]) and x2 (on [1,2]), over what range of t is the output non-zero? A. From 0 to 2 B. From 1 to 2 C. From 2 to 4 D. From -1 to 1 Answer: C. Fold and slide x2; overlap begins when the edges meet at t = 1 + 1 = 2 and ends at t = 2 + 2 = 4. (Key point: Convolution procedure and support) 4. How is the convolution result checked, given x1 has area 2 and x2 has area 1? A. Its peak must equal 4 B. Its area must equal 2 x 1 = 2 C. Its area must equal 2 + 1 = 3 D. Its energy must equal 4/3 Answer: B. The area of a convolution equals the product of the areas of the two signals; y(t) gives 2, and is 2 at t = 3, continuous from both pieces. 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