--- title: "Output for h = u(t)−u(t−3), x = u(t+4)−u(t)" subject: "Signals and Systems" chapter: 5 key: "semester_4/signals_and_systems:convolution-two-step-pulses" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-two-step-pulses site: BCT Engineering --- # Output for h = u(t)−u(t−3), x = u(t+4)−u(t) Signals and Systems · Chapter 5: Continuous-Time Systems - Asked in: 2069 Bhadra, 5 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-two-step-pulses - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-5 ## Question If the impulse response of continuous time linear time invariant system is h(t) = u(t) − u(t−3) and input to the system is x(t) = u(t+4) − u(t), determine the output y(t) of the system. ## Short answer Convolving rectangles of widths 3 and 4 gives a trapezoid: y(t)=0 for t<-4, t+4 for -4<=t<-1, 3 for -1<=t<0, 3-t for 0<=t<3, and 0 for t>=3. ## Answer The output is the convolution $y(t) = \int_{-\infty}^{\infty} h(\tau)\,x(t-\tau)\,d\tau$. **Given:** - $h(t) = u(t) - u(t-3)$: unit pulse on $0 \le t < 3$ (width 3). - $x(t) = u(t+4) - u(t)$: unit pulse on $-4 \le t < 0$ (width 4). The output lasts from $0 + (-4) = -4$ to $3 + 0 = 3$, and convolving two unequal rectangles gives a **trapezoid** with flat top height = smaller width = 3. **Limits:** $x(t-\tau) = 1$ when $-4 \le t-\tau < 0$, i.e. $t < \tau \le t+4$. Combine with $0 \le \tau < 3$ from $h$. **Case 1: $t < -4$** ($t+4 < 0$): no overlap, $y(t) = 0$. **Case 2: $-4 \le t < -1$** (partial overlap, $0 \le \tau \le t+4$): $$ y(t) = \int_0^{t+4} d\tau = t + 4 $$ **Case 3: $-1 \le t < 0$** (window $(t,\ t+4)$ covers all of $[0,3]$): $$ y(t) = \int_0^{3} d\tau = 3 $$ **Case 4: $0 \le t < 3$** (partial overlap, $t \le \tau \le 3$): $$ y(t) = \int_t^{3} d\tau = 3 - t $$ **Case 5: $t \ge 3$:** no overlap, $y(t) = 0$. **Answer:** $$ y(t) = \begin{cases} 0, & t < -4 \\ t + 4, & -4 \le t < -1 \\ 3, & -1 \le t < 0 \\ 3 - t, & 0 \le t < 3 \\ 0, & t \ge 3 \end{cases} $$ Equivalently $y(t) = r(t+4) - r(t+1) - r(t) + r(t-3)$, where $r(t) = t\,u(t)$ is the ramp. **Check:** area of $y$ = (area of $h$) × (area of $x$) = $3 \times 4 = 12$; trapezoid area $= \frac{1}{2}(7 + 1)\times 3 = 12$. ```text y(t) 3 | +----+ | /| |\ | / | | \ | / | | \ 0 +----+---+----+---+----> t -4 -1 0 3 ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. h(t) is a unit pulse of width 3 on [0,3) and x(t) a unit pulse of width 4 on [-4,0). What is the shape and extent of y(t)? A. Triangle from -4 to 0 B. Trapezoid from -4 to 3 C. Rectangle from -4 to 3 D. Trapezoid from 0 to 7 Answer: B. Support runs from 0+(-4) = -4 to 3+0 = 3; unequal rectangles give a trapezoid. (Key point: Output duration and shape) 2. What is the height of the flat top? A. 4 B. 7 C. 3 (the smaller width) D. 12 Answer: C. At full overlap the integral covers the whole shorter pulse, 3 units. (Key point: Flat top height) 3. What is y(t) for -4 <= t < -1? A. t + 4 B. 3 C. 3 - t D. t + 1 Answer: A. The overlap runs from tau = 0 to t+4, giving t + 4. (Key point: Rising part) 4. Which check is consistent with the result? A. Area of y equals area of h plus area of x B. Area of y = 3 x 4 = 12, matching the trapezoid area C. Peak of y equals 3 + 4 D. Area of y = 7 Answer: B. Area of convolution = product of areas; trapezoid area = 0.5(7 + 1)(3) = 12. 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