Structural Dynamics / Impulse response, arbitrary loading and convolution
The convolution integral
Cut any load history into slices, treat each slice as an impulse, add up the free vibrations. That is the whole idea, and the integral is just notation for it.
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Read the lesson register and outstanding review notesWhy this matters
The pulses of the last lesson are idealisations. A real earthquake, a real wind trace, a real machine start-up is none of them.
But every one of them can be cut into slices, and every slice is short. Each slice is an impulse whose response we already know, and the system is linear, so the total response is the sum. That is the convolution integral, and it is worth meeting as a picture before it is met as an equation.
By the end of this lesson you should be able to
- Derive the convolution integral from superposition of impulses
- Explain the role of the shifted argument h(t − τ)
- State exactly what convolution assumes and where it fails
- Say why time-stepping methods exist despite convolution being exact
Build the response from shifted impulses
For a linear time-invariant system initially at rest, divide F(t) into increments F(τ)dτ. Each increment contributes F(τ)h(t−τ)dτ at time t. Summing the contributions gives:
x(t) = ∫₀ᵗ F(τ)h(t−τ)dτ
The elapsed time t−τ matters: an impulse contributes only after it arrives. Add the homogeneous response separately when initial displacement or velocity is nonzero.
Computation and limits
A direct sum over all earlier samples costs O(n²). FFT convolution and recursive exact-step methods can reduce that cost substantially for suitable linear systems, so convolution is a practical computational method as well as a useful derivation.
A fixed shifted kernel cannot describe path-dependent yielding, because later response depends on the changed internal state. Nonlinear time integration instead updates the constitutive model while solving the equation of motion.
From first principles
The convolution integral
We want to show: Derive x(t) = ∫₀ᵗ F(τ)h(t − τ)dτ from superposition of impulses, and be clear about what it assumes.
Every slice of the load history hits the structure once and then lets go. Each slice starts its own free vibration, which carries on independently of what every other slice does. At any moment, the structure's displacement is simply the sum of all those free vibrations, each one evaluated at the time that has passed since its own slice arrived.
Try it
Building an arbitrary response from impulses
Every slice of the force history is an impulse. Each one starts its own free vibration, and the total is the sum.
Force history
- Total response (convolution)
- Same problem, time-stepped
- Contribution of the impulse at t = 0.15 s
- Other individual impulses
Show the numbers behind this plot
- Slice width dt
- 0.0117 s
- Force in this slice
- 60.1 kN
- Impulse of this slice
- 0.702 kN·s
- Its peak contribution
- 4.82 mm
- Peak of the total
- 109 mm
- Convolution vs time-stepping
- 0.67%
Two completely different methods, same answer — which is the check that the implementation is right.
What the picture is telling you
- Each faint curve is one slice of the load acting alone. It is zero until its slice arrives, then it is a decaying free vibration, and it never notices what the other slices do.
- The heavy curve is their sum. Nothing else has been added: no new physics, only addition.
- Slices that arrive when the structure is already moving the same way add to the response; slices that arrive against it subtract. That is the whole mechanism behind resonance, seen at the level of individual impulses.
- Set the damping to zero and look at how long each contribution persists. With no damping, an impulse from the very start of the record is still contributing at the end.
What this shows: The convolution integral is not a new theory. It is linear superposition applied to a load history cut into slices, and the impulse response is the only thing you need to know about the structure.
Worked example
Checking a convolution against a case we already know
Given
- An undamped SDOF system: m = 5 000 kg, k = 1.8 MN/m
- A constant force of 30 kN applied suddenly at t = 0 and held
Find
The response by convolution, and confirmation that it agrees with Module 1.
Assumptions
- Undamped, so the integral can be done in closed form
Practice
In a convolution, an impulse arrives at τ = 1.2 s and the response is being evaluated at t = 3.0 s. At what elapsed time is the impulse response function evaluated, in seconds?
Practice
A convolution is evaluated over 6 000 time steps. Roughly how many multiply–add operations does that require, expressed in millions?
Check yourself
Which single assumption does the convolution integral depend on most fundamentally?
Check yourself
Why does the Duhamel integral apply only to linear systems?
Summary
- Cut the load into slices; each slice is an impulse; add the free vibrations
- x(t) = ∫₀ᵗ F(τ)h(t − τ)dτ — the convolution or Duhamel integral
- The shifted argument (t − τ) is the time elapsed since that impulse arrived
- The upper limit is t because a load cannot act before it arrives
- The shifted-kernel form requires a linear time-invariant system; add initial-condition response separately
- It reproduces the Module 1 step response exactly, which is the check
- Direct summation is O(n²), but faster LTI algorithms exist; a fixed convolution kernel cannot describe yielding