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Queensferry

Module 6 · Lesson 6.3

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.

Why 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

The picture first

Take any force history F(t). Cut it into thin vertical strips of width dτ.

The strip at time τ has area F(τ)dτ — an impulse. From the first lesson of this module, that impulse sets the structure vibrating freely, with a response of magnitude F(τ)dτ times h evaluated at the time ELAPSED SINCE the impulse arrived.

That elapsed time is the important detail. At the observation time t, an impulse that arrived at τ has been running for (t − τ). So its contribution is

F(τ) dτ · h(t − τ)

Every impulse that arrived before t contributes something; impulses that have not arrived yet contribute nothing. Add them all up and you have the response.

Which is the integral

x(t) = ∫₀ᵗ F(τ) h(t − τ) dτ

That is the convolution integral, also called the Duhamel integral. Nothing has been assumed except linearity — which is what permits the addition — and that the structure started at rest.

Why time-stepping exists anyway

Convolution is exact, and it is rarely used in practice. Two reasons.

Cost. Computing the response at n time steps requires n² operations, because every step must sum over every earlier step. For a 40-second earthquake record at 0.005 s that is 8 000 steps and 64 million operations — for one degree of freedom. Newmark's method does it in 8 000.

Linearity. Convolution rests entirely on superposition. The moment a structure yields, superposition fails and convolution fails with it. Time-stepping methods do not care: they solve the equation of motion one step at a time and can update the stiffness whenever they like.

So convolution is the right way to UNDERSTAND arbitrary loading and the wrong way to compute it. Module 8 covers the way it is actually done.

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

s
s
Force against Time. Force reaches a peak magnitude of 80 kN.00.511.522.533.5020406080this sliceTime (s)Force (kN)
Force against Time. Force reaches a peak magnitude of 80 kN.
Displacement against Time. Total response (convolution) reaches a peak magnitude of 109 mm. Same problem, time-stepped reaches a peak magnitude of 109 mm. Contribution of the impulse at t = 0.15 s reaches a peak magnitude of 4.82 mm. Other individual impulses reaches a peak magnitude of 2.46 mm. reaches a peak magnitude of 4.82 mm. reaches a peak magnitude of 6.08 mm. reaches a peak magnitude of 6.4 mm. reaches a peak magnitude of 5.75 mm. reaches a peak magnitude of 3.91 mm. reaches a peak magnitude of 1.66 mm.00.511.522.533.5-80-60-40-20020406080100120Time (s)Displacement (mm)
  • Total response (convolution)
  • Same problem, time-stepped
  • Contribution of the impulse at t = 0.15 s
  • Other individual impulses
Displacement against Time. Total response (convolution) reaches a peak magnitude of 109 mm. Same problem, time-stepped reaches a peak magnitude of 109 mm. Contribution of the impulse at t = 0.15 s reaches a peak magnitude of 4.82 mm. Other individual impulses reaches a peak magnitude of 2.46 mm. reaches a peak magnitude of 4.82 mm. reaches a peak magnitude of 6.08 mm. reaches a peak magnitude of 6.4 mm. reaches a peak magnitude of 5.75 mm. reaches a peak magnitude of 3.91 mm. reaches a peak magnitude of 1.66 mm.
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
    • Linearity is the only real assumption, and it is the one that fails first
    • It reproduces the Module 1 step response exactly, which is the check
    • Cost is O(n²) and it cannot handle yielding — hence Module 8
    Progress is kept in this browser only.

    This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint