Laplace II: Steps, Impulses & Convolution

Study Sheet

Laplace II: Steps, Impulses & Convolution

Piecewise forcing, hammer blows, and the convolution theorem

Discontinuous and Impulsive Forcing

Tip
Unit steps carry delays

The Heaviside step u(t−a)u(t-a) switches on at t=at = a; any piecewise forcing is a sum of stepped pieces. The shift rule both ways: L{g(t−a) u(t−a)}=e−asG(s)\mathcal L\{g(t-a)\,u(t-a)\} = e^{-as}G(s), so a DELAY in time is an EXPONENTIAL factor in ss — and when inverting, every e−ase^{-as} in Y(s)Y(s) announces "the following response, but starting at t=at = a." This is how one algebraic computation handles a force that turns on at t=3t = 3 and off at t=7t = 7.

Tip
The impulse: δ and what it does

The Dirac impulse δ(t−a)\delta(t - a) models a hammer blow: zero except at one instant, total area 11, transform L{δ(t−a)}=e−as\mathcal L\{\delta(t-a)\} = e^{-as}.

Forcing an equation with δ\delta produces an instantaneous JUMP in the highest-derivative-minus-one: y′=δ(t−1)y' = \delta(t-1), y(0)=0y(0)=0 gives the step y=u(t−1)y = u(t-1) — flat, jump by 11, flat. The solution to δ\delta-forcing is the system's IMPULSE RESPONSE, its fingerprint.

Tip
Convolution: multiplication in disguise

The convolution (f∗g)(t)=∫0tf(τ)g(t−τ) dτ(f*g)(t) = \int_0^t f(\tau)g(t-\tau)\,d\tau satisfies L{f∗g}=F(s)G(s)\mathcal L\{f*g\} = F(s)G(s) — products of transforms are convolutions of functions. Consequence: the response to ANY forcing ff is h∗fh * f where hh is the impulse response, because Y=H(s)F(s)Y = H(s)F(s).

One integral formula expresses every solution — the superposition of infinitely many scaled, delayed hammer blows.

Side note
Significance: real inputs are not smooth

Real systems get switched on, struck, and pulsed: a thermostat clicking, a hammer test on a bridge, a defibrillator pulse. Steps and deltas model these honestly, and the convolution integral is how audio engineers apply reverb: recorded impulse response of a cathedral, convolved with your dry vocal track.

Try it
Try it: read a delayed inverse

Invert Y(s)=e−3ss+2Y(s) = \dfrac{e^{-3s}}{s + 2}. Work: without the exponential, 1s+2\tfrac{1}{s+2} inverts to e−2te^{-2t}; the e−3se^{-3s} delays it: y(t)=e−2(t−3)u(t−3)y(t) = e^{-2(t-3)}u(t-3) — silence until t=3t = 3, then the decay begins, shifted intact.

Proofs & Why It Matters

Tip
Proof: the t-shift rule

Directly: L{g(t−a)u(t−a)}=∫a∞g(t−a)e−stdt\mathcal L\{g(t-a)u(t-a)\} = \int_a^\infty g(t-a)e^{-st}dt (the step kills t<at < a). Substitute τ=t−a\tau = t - a: ∫0∞g(τ)e−s(τ+a)dτ=e−as∫0∞g(τ)e−sτdτ=e−asG(s)\int_0^\infty g(\tau)e^{-s(\tau+a)}d\tau = e^{-as}\int_0^\infty g(\tau)e^{-s\tau}d\tau = e^{-as}G(s). ■\blacksquare

Tip
Proof: the convolution theorem

F(s)G(s)=∫0∞ ⁣ ⁣∫0∞f(τ)g(σ)e−s(τ+σ) dσ dτF(s)G(s) = \int_0^\infty\!\!\int_0^\infty f(\tau)g(\sigma)e^{-s(\tau+\sigma)}\,d\sigma\,d\tau — a double integral over the quarter-plane.

Change variables to t=τ+σt = \tau + \sigma (total time) and τ\tau: for each fixed tt, τ\tau ranges over [0,t][0, t], giving ∫0∞e−st[∫0tf(τ)g(t−τ) dτ]dt=L{f∗g}\int_0^\infty e^{-st}\left[\int_0^t f(\tau)g(t-\tau)\,d\tau\right]dt = \mathcal L\{f * g\}. A Fubini swap turns a product into a convolution. ■\blacksquare

Going Deeper: Worked Problems

Example
Worked: a force that switches on, then off

Solve y′′+y=f(t)y'' + y = f(t), y(0)=y′(0)=0y(0) = y'(0) = 0, where f(t)=1f(t) = 1 for 0≤t<π0 \le t < \pi and 00 afterward.

Step 1 — write the forcing with steps: f=1−u(t−π)f = 1 - u(t - \pi), so L{f}=1s−e−πss\mathcal L\{f\} = \dfrac1s - \dfrac{e^{-\pi s}}{s}.

Step 2 — transform: (s2+1)Y=1−e−πss(s^2 + 1)Y = \dfrac{1 - e^{-\pi s}}{s}, so Y=1−e−πss(s2+1)Y = \dfrac{1 - e^{-\pi s}}{s(s^2+1)}.

Step 3 — partial fractions on 1s(s2+1)=1s−ss2+1\dfrac{1}{s(s^2+1)} = \dfrac1s - \dfrac{s}{s^2+1}, which inverts to g(t)=1−cos⁡tg(t) = 1 - \cos t.

Step 4 — the e−πse^{-\pi s} copy is the same response delayed: y=g(t)−g(t−π)u(t−π)y = g(t) - g(t-\pi)u(t-\pi).

Step 5 — read it: for t<πt < \pi, y=1−cos⁡ty = 1 - \cos t (pushed from rest); for t≥πt \ge \pi, y=(1−cos⁡t)−(1−cos⁡(t−π))=−2cos⁡ty = (1 - \cos t) - (1 - \cos(t - \pi)) = -2\cos t — the force released exactly when the oscillator was at its far point, leaving it swinging with amplitude 22 forever. Piecewise forcing, zero case-splitting.

Example
Worked: an impulse response and a convolution

Find the impulse response of y′′+3y′+2y=δ(t)y'' + 3y' + 2y = \delta(t) (zero initial conditions), then use convolution to solve the same equation with forcing f(t)=e−tf(t) = e^{-t}.

Step 1 — transform: (s2+3s+2)H=1(s^2 + 3s + 2)H = 1, so H(s)=1(s+1)(s+2)=1s+1−1s+2H(s) = \dfrac{1}{(s+1)(s+2)} = \dfrac{1}{s+1} - \dfrac{1}{s+2}: impulse response h(t)=e−t−e−2th(t) = e^{-t} - e^{-2t}.

Step 2 — for general forcing, y=h∗fy = h * f: y(t)=∫0t(e−τ−e−2τ)e−(t−τ) dτy(t) = \displaystyle\int_0^t \left(e^{-\tau} - e^{-2\tau}\right)e^{-(t-\tau)}\,d\tau.

Step 3 — simplify: e−t∫0t(1−e−τ)dτ=e−t(t−1+e−t)e^{-t}\displaystyle\int_0^t\left(1 - e^{-\tau}\right)d\tau = e^{-t}\left(t - 1 + e^{-t}\right).

Step 4 — so y=te−t−e−t+e−2ty = te^{-t} - e^{-t} + e^{-2t}; check: y(0)=0y(0) = 0 ✓, and the te−tte^{-t} term is exactly what resonance-style forcing at a homogeneous rate produces. One impulse response, every forcing solved by an integral.