Skip to main content

Physical meaning of convolution integral

Short answer

Convolution integrates x(tau) h(t-tau), summing responses to all input impulses weighted by the system memory. For an RC circuit, recent inputs dominate and old ones decay, giving the capacitor voltage. It represents superposition of shifted impulse responses.

Elaborate the physical meaning of convolution integral using suitable example.

The convolution integral

y(t)=∫−∞∞x(τ) h(t−τ) dτy(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau

says that the output of an LTI system at time tt is the sum of the responses to all past (and present) input values, each weighted by how strongly the system still "remembers" it.

Physical meaning:

  • The input is thought of as a chain of very narrow impulses; the impulse at time τ\tau has strength x(τ) dτx(\tau)\,d\tau.
  • Each impulse produces its own response h(t−τ)h(t-\tau), which starts at τ\tau and dies out according to the system's memory.
  • The output is the superposition (linearity) of these shifted copies (time invariance).
  • h(t−τ)h(t - \tau) acts as a weighting function: for a system with short memory, it is small when the age t−τt - \tau of the input is large.

Example – RC circuit charging: h(t)=1RCe−t/RCu(t)h(t) = \frac{1}{RC}e^{-t/RC}u(t). The capacitor voltage now is

y(t)=∫−∞tx(τ)1RCe−(t−τ)/RC dτy(t) = \int_{-\infty}^{t} x(\tau)\frac{1}{RC}e^{-(t-\tau)/RC}\,d\tau

Recent inputs (small t−τt - \tau) count almost fully; inputs older than about 5RC5RC have almost no effect, because that charge has leaked away. Another everyday example is an echo in a hall: what you hear is the present sound plus fading copies of earlier sounds.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Signals and Systems chapter 5 questions →