Impulse response and convolution

Why it matters

For a linear time-invariant (LTI) system, one measured curve — the impulse response — tells you the output for any input. This is how a thermometer's lag, an accelerometer's ringing or a digital moving-average filter is characterised in practice, and convolution is the operation that turns that curve into a prediction.

Key ideas

Impulse response

• The unit impulse δ(t) is zero for t ≠ 0 and has unit area: ∫ δ(t) dt = 1. Its key property is sifting: ∫ x(τ) δ(t − τ) dτ = x(t). The DT impulse δ[n] is simply 1 at n = 0 and 0 elsewhere.
• The impulse response h(t) (or h[n]) is the output of an LTI system, initially at rest, when the input is δ(t) (or δ[n]).
• The step response s(t) is the output for u(t). Because δ(t) = du/dt, h(t) = ds/dt and s(t) = ∫ from −∞ to t of h(τ) dτ. In DT, h[n] = s[n] − s[n − 1].

Why convolution works. Any input can be written as a weighted sum of shifted impulses: x[n] = Σ x[k] δ[n − k]. Time invariance turns each δ[n − k] into h[n − k]; linearity lets you add the weighted responses. The result is the convolution sum. The CT integral follows the same way.

Graphical (flip-and-slide) method

1. Plot x(τ) and h(τ) against the dummy variable τ.
2. Flip h to get h(−τ), then shift by t to get h(t − τ).
3. Multiply x(τ)·h(t − τ) and integrate (or sum) over the overlap.
4. Repeat for every range of t where the overlap changes shape.

Properties

• Commutative x * h = h * x, associative (x * h1) * h2 = x * (h1 * h2) (cascade ⇒ impulse responses convolve), distributive x * (h1 + h2) = x * h1 + x * h2 (parallel ⇒ impulse responses add).
• Identity x * δ = x; delay x(t) * δ(t − t0) = x(t − t0).
• Width: if x lasts T1 and h lasts T2, y lasts T1 + T2. In DT, lengths L and M give L + M − 1 samples, and the start index of y is the sum of the start indices.
• Area: ∫ y = (∫ x)(∫ h) and, in DT, Σ y[n] = (Σ x[n])(Σ h[n]) — a fast check on your answer.
• Causality and stability of an LTI system can be read from h (see the system-properties topic).

Link to transforms. Convolution in time becomes multiplication in frequency: Y(ω) = X(ω)H(ω), Y(s) = X(s)H(s), Y(z) = X(z)H(z). This is why transforms are used for anything longer than a few samples.

Formulas

• y(t) = x(t) * h(t) = ∫ x(τ) h(t − τ) dτ (τ from −∞ to ∞) — CT convolution integral.
• y[n] = x[n] * h[n] = Σ x[k] h[n − k] (k from −∞ to ∞) — DT convolution sum.
• y(t) = ∫ from 0 to t of x(τ) h(t − τ) dτ — when both x and h are zero for negative time (causal input into a causal system).
• s(t) = ∫ from −∞ to t of h(τ) dτ, h(t) = ds(t)/dt — step and impulse responses.
• e^(−at)u(t) * e^(−bt)u(t) = [(e^(−at) − e^(−bt))/(b − a)] u(t), for a ≠ b.
• e^(−at)u(t) * e^(−at)u(t) = t e^(−at) u(t).
• u(t) * u(t) = t·u(t) (the ramp).

Symbols: x input; y output; h impulse response (unit of y per unit of x per second in CT; per sample in DT); τ, k dummy integration/summation variables; t time (s); n sample index; a, b decay rates (s⁻¹); s(t) step response.

Worked examples

Example 1 (standard). An RC low-pass sensor stage has h(t) = 2e^(−2t) u(t) (s⁻¹). Find its step response and its value at t = 1 s.

1. Both x(t) = u(t) and h(t) are causal, so y(t) = ∫ from 0 to t of 1 · 2e^(−2(t − τ)) dτ.
2. Substitute λ = t − τ: y(t) = ∫ from 0 to t of 2e^(−2λ) dλ = [−e^(−2λ)] from 0 to t.
3. y(t) = (1 − e^(−2t)) u(t) — a first-order rise with time constant 0.5 s, ending at the DC gain ∫ h = 1.
4. At t = 1 s: y(1) = 1 − e^(−2) = 1 − 0.1353.
5. y(1) ≈ 0.865 (per unit step).

Example 2 (GATE level). (a) Convolve x[n] = {1, 2, 3} and h[n] = {1, −1, 2}, both starting at n = 0. (b) Find the peak of y(t) = e^(−t)u(t) * e^(−2t)u(t).

1. (a) Length = 3 + 3 − 1 = 5, starting at n = 0.
2. y[0] = 1·1 = 1; y[1] = 1·(−1) + 2·1 = 1; y[2] = 1·2 + 2·(−1) + 3·1 = 3; y[3] = 2·2 + 3·(−1) = 1; y[4] = 3·2 = 6.
3. Check: Σ y = 12 and (Σ x)(Σ h) = 6 × 2 = 12 ✓. y[n] = {1, 1, 3, 1, 6} for n = 0 … 4.
4. (b) Using the exponential-pair formula with a = 1, b = 2: y(t) = (e^(−t) − e^(−2t)) u(t).
5. Set dy/dt = −e^(−t) + 2e^(−2t) = 0 ⇒ e^(−t) = 1/2 ⇒ t = ln 2 = 0.693 s.
6. y(ln 2) = 1/2 − 1/4. Peak = 0.25 at t ≈ 0.693 s.

Common mistakes

• Forgetting to flip h — that computes correlation, not convolution.
• Using the limits 0 to t when the input or the impulse response is not causal (for example a two-sided e^(−|t|)).
• Wrong start index in DT: if x starts at n = −1 and h at n = 0, y starts at n = −1.
• Missing a region of t in the graphical method (partial overlap on the way in and on the way out are separate cases).
• Writing y(t) = 1 − e^(−t) without u(t), which wrongly gives a non-zero output for t < 0.
• Mixing up h(t) and the step response when a question gives "the response to a unit step".

For GATE IN

Common question types: DT convolution of two short sequences (NAT asking for one output sample or the sum of outputs), CT convolution of rectangular pulses or exponentials (value at a given time, or the duration/shape of the result), the impulse response of a cascade or parallel connection, and getting h(t) from a given step response. Practise the flip-and-slide method on pulses and the length/area checks; they catch most slips.

Quick check

1. What is the length of the convolution of a 4-sample and a 6-sample sequence?
2. What is x(t) * δ(t − 3)?
3. The step response of an LTI system is (1 − e^(−5t))u(t). What is h(t)?
4. A rectangular pulse of height 2 and width 1 s is convolved with itself. What is the area of the result?
5. Two systems with h1 and h2 in cascade — overall impulse response? Answers: 1. 4 + 6 − 1 = 9. 2. x(t − 3). 3. 5e^(−5t)u(t). 4. 2 × 2 = 4. 5. h1 * h2.