--- title: "What is convolution; convolution integral" subject: "Signals and Systems" chapter: 5 key: "semester_4/signals_and_systems:convolution-integral-what-is" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-integral-what-is site: BCT Engineering --- # What is convolution; convolution integral Signals and Systems · Chapter 5: Continuous-Time Systems - Asked in: 2072 Asoj, 2+6 marks - Video: [The Convolution of Two Functions | Definition & Properties](https://www.youtube.com/watch?v=AgKQQtEc9dk) — Dr. Trefor Bazett - 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-integral-what-is - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-5 ## Question What is convolution? Obtain the expression for convolution integral. ## Short answer Convolution is a mathematical operation that gives an LTI system's output as the superposition of weighted, shifted impulse responses. Its integral expression is y(t) = ∫ x(τ) h(t-τ) dτ, with integration limits 0 to t for causal signals and system. ## Answer #### Convolution **Convolution** is a mathematical operation that combines two signals to produce a third. In signals and systems it gives the output $y(t)$ of an LTI system as the input $x(t)$ "convolved" with the impulse response $h(t)$: $$ y(t) = x(t) * h(t) $$ It works as a weighted, shifted sum: each past value of the input is weighted by the system's impulse response and all contributions are added. #### Expression for the convolution integral **Step 1 – Input as a sum of impulses.** Approximate $x(t)$ by narrow pulses of width $\Delta$: $$ x(t) \approx \sum_{k=-\infty}^{\infty} x(k\Delta)\,\delta_\Delta(t-k\Delta)\,\Delta $$ where $\delta_\Delta(t)$ is a pulse of width $\Delta$ and height $1/\Delta$ (unit area). **Step 2 – Response to one pulse.** Let $h_\Delta(t)$ be the response to $\delta_\Delta(t)$. - **Time invariance:** $\delta_\Delta(t-k\Delta) \rightarrow h_\Delta(t-k\Delta)$. - **Homogeneity:** $x(k\Delta)\Delta\,\delta_\Delta(t-k\Delta) \rightarrow x(k\Delta)\Delta\,h_\Delta(t-k\Delta)$. - **Additivity:** $$ y(t) \approx \sum_{k=-\infty}^{\infty} x(k\Delta)\,h_\Delta(t-k\Delta)\,\Delta $$ **Step 3 – Limit $\Delta \to 0$.** Then $\delta_\Delta \to \delta(t)$, $h_\Delta \to h(t)$, $k\Delta \to \tau$, $\Delta \to d\tau$ and the sum becomes an integral: $$ y(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau $$ This is the **convolution integral**. Substituting $\lambda = t-\tau$ gives the equivalent form $y(t) = \int_{-\infty}^{\infty} h(\tau)\,x(t-\tau)\,d\tau$. **For causal system and causal input** ($h(t) = 0$ and $x(t) = 0$ for $t < 0$): $$ y(t) = \int_{0}^{t} x(\tau)\,h(t-\tau)\,d\tau $$ **Procedure (graphical):** fold $h(\tau)$ to $h(-\tau)$, shift it by $t$, multiply by $x(\tau)$, and find the area of the product; repeat for each $t$. **Properties:** commutative ($x*h = h*x$), associative ($(x*h_1)*h_2 = x*(h_1*h_2)$, used for cascade), distributive ($x*(h_1+h_2) = x*h_1 + x*h_2$, used for parallel connection), and $x(t)*\delta(t-t_0) = x(t-t_0)$. **Example:** $e^{-at}u(t) * u(t) = \int_0^t e^{-a\tau}d\tau = \frac{1}{a}(1-e^{-at})u(t)$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In an LTI system, how is the output y(t) related to the input x(t) and impulse response h(t)? A. y(t) = x(t) h(t), a point-by-point product B. y(t) = x(t) * h(t), the input convolved with the impulse response C. y(t) = x(t) + h(t) D. y(t) = dx/dt * h(t) Answer: B. Convolution gives the LTI output as a weighted, shifted sum of the input weighted by h(t). (Key point: Meaning of convolution) 2. Why can the response to a sum of narrow pulses be written as a sum of shifted copies of h? A. Because the system is causal and stable B. Because the pulses are orthogonal C. Because of linearity (homogeneity, additivity) and time invariance D. Because the Fourier transform of a pulse is constant Answer: C. Time invariance shifts the response, homogeneity scales it by x(k*Delta)*Delta, and additivity adds all of them. (Key point: Derivation approach) 3. What happens in the limit Delta -> 0 in the derivation? A. The sum becomes an integral: y(t) = integral of x(tau) h(t - tau) d tau B. The sum becomes zero C. The pulse delta_Delta becomes a unit step D. h(t) becomes constant Answer: A. k*Delta becomes tau, Delta becomes d tau and the sum becomes the convolution integral. (Key point: Limit step) 4. What is the correct graphical procedure for convolution? A. Shift x only and add the areas B. Differentiate h and multiply by x C. Fold h(tau) to h(-tau), shift by t, multiply by x(tau) and find the area D. Multiply x and h and take the peak value Answer: C. Fold, shift, multiply, integrate (area), repeated for each t. 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