--- title: "Cascade of two CT LTI systems" subject: "Signals and Systems" chapter: 5 key: "semester_4/signals_and_systems:cascade-impulse-response" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/cascade-impulse-response site: BCT Engineering --- # Cascade of two CT LTI systems Signals and Systems · Chapter 5: Continuous-Time Systems - Asked in: 2070 Magh, 8+2 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/cascade-impulse-response - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-5 ## Question The impulse responses of two LTI systems are given by h₁(t) = e^(−t/2)u(t) and h₂(t) = u(t) − u(t−5). Determine the equivalent impulse response if these two systems are connected in cascade. Also sketch the graph of equivalent impulse response. ## Short answer The equivalent impulse response is h(t) = h1(t) * h2(t). Convolving e^(-t/2)u(t) with a rectangular pulse of width 5 yields: h(t)=0 for t<0; 2(1-e^(-t/2)) for 0≤t<5; 2(e^2.5 -1)e^(-t/2) for t≥5. It rises to a peak of about 1.84 at t=5 and then decays. ## Answer For two LTI systems in **cascade**, the equivalent impulse response is the convolution of the individual impulse responses (associative property): $$ h(t) = h_1(t) * h_2(t) = \int_{-\infty}^{\infty} h_1(\tau)\,h_2(t-\tau)\,d\tau $$ ```text x(t) -->[ h1(t) ]-->[ h2(t) ]--> y(t) == x(t) -->[ h1(t)*h2(t) ]--> y(t) ``` **Given:** $h_1(t) = e^{-t/2}u(t)$; $h_2(t) = u(t) - u(t-5)$, a unit pulse on $0 \le t < 5$. $h_2(t-\tau) = 1$ when $0 \le t-\tau < 5$, i.e. $t-5 < \tau \le t$. Also $h_1(\tau) \ne 0$ only for $\tau \ge 0$. **Useful integral:** $\int_a^b e^{-\tau/2}d\tau = 2\left(e^{-a/2} - e^{-b/2}\right)$. **Case 1: $t < 0$** – no overlap: $h(t) = 0$. **Case 2: $0 \le t < 5$** – limits $0$ to $t$: $$ h(t) = \int_0^{t} e^{-\tau/2}d\tau = 2\left(1 - e^{-t/2}\right) $$ **Case 3: $t \ge 5$** – limits $t-5$ to $t$: $$ \begin{aligned} h(t) &= 2\left(e^{-(t-5)/2} - e^{-t/2}\right) \\ &= 2\left(e^{2.5} - 1\right)e^{-t/2} \approx 22.365\,e^{-t/2} \end{aligned} $$ **Answer:** $$ h(t) = \begin{cases} 0, & t < 0 \\ 2\left(1 - e^{-t/2}\right), & 0 \le t < 5 \\ 2\left(e^{2.5} - 1\right)e^{-t/2}, & t \ge 5 \end{cases} $$ Compactly: $h(t) = 2\left(1 - e^{-t/2}\right)u(t) - 2\left(1 - e^{-(t-5)/2}\right)u(t-5)$. **Values for the sketch:** | $t$ | 0 | 1 | 2 | 5 | 7 | 10 | |---|---|---|---|---|---|---| | $h(t)$ | 0 | 0.787 | 1.264 | 1.836 | 0.675 | 0.151 | **Peak:** $h(5) = 2(1 - e^{-2.5}) = 1.836$, which is continuous from both sides. #### Sketch ```text h(t) 1.84| * | .-' \ | .-' \ | .' `. | .' `-._ | / `--.__ 0 +-------------+---------------> t 0 5 10 ``` It rises like a charging capacitor ($2(1-e^{-t/2})$) from 0 to 1.836 over $0 \le t \le 5$, then decays exponentially to zero with time constant 2 s for $t > 5$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the equivalent impulse response of two LTI systems in cascade? A. h1(t) + h2(t) B. h1(t) h2(t) C. h1(t) * h2(t), their convolution D. h1(t) / h2(t) Answer: C. Cascade uses the associative property: the equivalent h is the convolution of the two. (Key point: Cascade rule) 2. For 0 <= t < 5, what are the limits of integration and the result? A. 0 to t, giving 2(1 - e^(-t/2)) B. t-5 to t, giving 2(e^(-(t-5)/2) - e^(-t/2)) C. 0 to 5, giving 2(1 - e^(-2.5)) D. 0 to t, giving e^(-t/2) Answer: A. h1(tau) starts at 0 and h2(t - tau) = 1 for tau <= t, so the integral of e^(-tau/2) from 0 to t is 2(1 - e^(-t/2)). (Key point: Overlap limits for 0 <= t < 5) 3. For t >= 5 the window is fully inside the support of h1. What is h(t)? A. 2(1 - e^(-t/2)) B. 2(e^2.5 - 1) e^(-t/2) C. 2 e^(-t/2) D. 0 Answer: B. Limits t-5 to t give 2(e^(-(t-5)/2) - e^(-t/2)) = 2(e^2.5 - 1) e^(-t/2). (Key point: Result for t >= 5) 4. Which describes the shape of h(t)? A. Rises to a peak of about 1.84 at t = 5, then decays exponentially with time constant 2 s B. Constant 1 for 0 to 5 then zero C. Decays from 2 at t = 0 to zero at t = 5 D. Rises linearly to t = 5 and then falls linearly Answer: A. h(5) = 2(1 - e^-2.5) = 1.836 is the peak; afterwards the response decays as e^(-t/2). 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