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Convolution of u(t) with u(t)

Short answer

An LTI system is linear and time-invariant; its output is the convolution of input and impulse response. Convolution of u(t) with u(t) yields the unit ramp r(t)=t u(t). The result grows unbounded, so an integrator with h(t)=u(t) is not BIBO stable.

Define LTI system. Find convolution between x(t) = u(t) and h(t) = u(t) and comment on the result.

LTI system: a system that is linear and time-invariant. Its output is the convolution of the input with the impulse response, y(t) = x(t) * h(t) = ∫ x(τ) h(t - τ) dτ.

Convolution of u(t) with u(t)

y(t) = ∫_{-∞}^{∞} u(τ) u(t - τ) dτ

  • u(τ) = 1 only for τ ≥ 0.
  • u(t - τ) = 1 only for τ ≤ t.

t < 0: there is no overlap, so y(t) = 0.

t ≥ 0: the product is 1 for 0 ≤ τ ≤ t:

y(t) = ∫₀^t 1 dτ = t

y(t) = t·u(t) = r(t), the unit ramp.

y(t)
  |        /
  |      /
  |    /
  |  /
--+--------------> t
  0

Comment

  • Convolving two unit steps gives a ramp, which grows without limit. A bounded input u(t) produces an unbounded output.
  • This shows that a system with h(t) = u(t) (an ideal integrator) is not BIBO stable, since ∫|h(t)| dt = ∞.
  • It is causal, because h(t) = 0 for t < 0.

✓ Answer checked by Claude Opus 5.5 (high effort)

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