Short notes

Convolution Short Notes

Pass a rectangular pulse of width 2 ms through an RC low-pass filter and the output is not another rectangle but a smoothed, exponentially rising and falling waveform — that output is the convolution of the input with the filter's impulse response h(t) = (1/RC)e^(−t/RC)u(t). Computing that integral analytically, keeping track of the overlapping region as you slide one function over the other, is the graphical convolution procedure that fills half the marks in signals exam questions.

EEE, ECE, EI

How it works

Continuous-time convolution is y(t) = x(t)*h(t) = ∫x(τ)h(t−τ)dτ over all τ. The graphical procedure: flip h(τ) to get h(−τ), slide it by t to get h(t−τ), multiply with x(τ), and integrate the overlap. Limits of integration change as t moves through different regions — tracking these region boundaries is where most errors occur. For discrete-time convolution y[n] = Σ x[k]h[n−k], the same flip-and-shift logic applies but the integral becomes a sum and the result is computed for each integer n separately.

Key points to remember

Convolution is commutative: x*h = h*x. It is also associative and distributive, making it valid to reorder cascaded LTI systems freely. If x[n] has length M and h[n] has length N, their convolution y[n] has length M + N − 1 — a fact directly tested in numerical convolution questions. For LTI systems, the output is entirely determined by the input and impulse response; no other system description is needed. Graphical convolution of two rectangular pulses produces a triangular waveform — the classic exam example. Circular convolution (used in DFT-based filtering) differs from linear convolution and must not be confused with it.

Exam tip

The examiner always asks you to perform graphical convolution of two rectangular pulses and sketch the output — define each region of overlap with precise time limits and write the corresponding integral; losing one region loses 3–4 marks.

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