Step Functions, Impulses, and Convolution

How does a circuit respond to a switch flipped at time t=3t=3? What about a struck bell excited by an instantaneous hammer blow? Ordinary differential equations model both situations. But the forcing functions involved — a sudden jump, an infinitely brief spike — are not smooth. They are not the everywhere-differentiable functions that calculus was built to handle. The Laplace Transform turns out to be exactly the right tool. It converts discontinuous switching and idealized impulses into simple algebraic operations. It also converts the messy operation of "combining a system's memory with its forcing history" into ordinary multiplication. This article develops the three pieces that make this work. The Heaviside (unit step) function models on/off switching. It lets us write piecewise forcing terms as single formulas. The Dirac delta models an idealized instantaneous impulse — a hammer blow, a voltage spike. It carries finite total effect concentrated at one instant. The convolution theorem explains something important. The Laplace transform of a product of transforms is not the transform of a product. Instead, it is the transform of a convolution. This theorem shows that the impulse response of a linear system encodes everything needed. That everything is what's needed to predict its response to arbitrary forcing. The big picture is this: a linear time-invariant system driven by forcing g(t)g(t) responds in a specific way. It "smears" gg against its own impulse response h(t)h(t). Convolution is the mathematical operation performing that smear. The Laplace transform is the lens that makes it tractable. It turns a convolution integral into a product of transforms. This product is computable by the Inverse Laplace Transform and Partial Fractions techniques already in hand. By the end you will be able to do several things. You will write switched and delayed forcing functions in closed form. You will transform them using the two shifting theorems. You will solve initial value problems with impulsive forcing rigorously, not just formally. You will also compute and interpret convolution integrals. These integrals represent a system's cumulative response to its entire forcing history.

Key takeaways
  • The unit step function uc(t)u_c(t) turns on at t=ct=c; it lets discontinuous, piecewise, or delayed forcing be written as one formula.
  • The first shifting theorem: multiplying f(t)f(t) by ecte^{ct} shifts the transform variable: L{ectf(t)}=F(s−c)\mathcal{L}\{e^{ct}f(t)\} = F(s-c).
  • The second shifting theorem: delaying f(t)f(t) by cc and switching it on with uc(t)u_c(t) multiplies the transform by e−cse^{-cs}.
  • The Dirac delta δ(t−c)\delta(t-c) is a limit of tall thin unit-area pulses; it models instantaneous impulses and has transform e−cse^{-cs}.
  • 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) — this is the convolution theorem.
  • Convolution with the impulse response h(t)h(t) of an LTI system reproduces the system's response to any forcing g(t)g(t): y(t)=(h∗g)(t)y(t) = (h*g)(t).
  • Key idea: discontinuous and impulsive forcing, and the entire response history of a linear system, become ordinary algebra under the Laplace transform.

Where this lives today

Switched and impulsive forcing modeled by step and delta functions appears throughout circuit analysis, where a step models a switch closing and an impulse models a sudden voltage or current spike across an inductor or capacitor. In control theory, the impulse response of a linear time-invariant system is the fundamental object from which the transfer function and stability properties are derived, and convolution with an input signal predicts the system's output exactly as in this article's theorem. In signal processing, convolution is the operation implemented by every linear filter — audio equalizers, image blurring and sharpening kernels, and antenna response modeling all reduce to convolving a signal against a fixed impulse response, computed efficiently via the Fourier transform analogue of the convolution theorem. In probability and statistics, the density of a sum of two independent random variables is the convolution of their individual densities, and Laplace (or moment-generating function) transforms convert this into ordinary multiplication exactly as here. In numerical analysis and PDE theory, Green's functions generalize the impulse-response idea to boundary value problems and partial differential equations, with the solution written as a convolution of the forcing against the Green's function. In cryptography and coding theory, convolutional codes borrow the same superposition structure, encoding data streams via a convolution against a fixed generator sequence.

Ask the tutor: How do I set up the Laplace transform for a forcing function that switches between three different formulas at two different times?

DefinitionHeaviside step, Dirac delta, and convolution

Let c≥0c \ge 0 be a real constant.

Unit step function. The Heaviside function shifted to cc is

uc(t)={0,t<c,1,t≥c.u_c(t) = \begin{cases} 0, & t < c, \\ 1, & t \ge c. \end{cases}

Dirac delta (unit impulse). For τ>0\tau>0 define the pulse

dτ(t−c)=1τ[uc(t)−uc+τ(t)],d_\tau(t-c) = \frac{1}{\tau}\bigl[u_c(t) - u_{c+\tau}(t)\bigr],

a rectangle of height 1/τ1/\tau and width τ\tau starting at t=ct=c, so that ∫0∞dτ(t−c) dt=1\int_0^\infty d_\tau(t-c)\,dt = 1 for every τ>0\tau>0. The unit impulse δ(t−c)\delta(t-c) is defined operationally by its action under integration against continuous test functions ff:

∫0∞f(t) δ(t−c) dt:=lim⁡τ→0+∫0∞f(t) dτ(t−c) dt=f(c),\int_0^\infty f(t)\,\delta(t-c)\,dt := \lim_{\tau\to 0^+}\int_0^\infty f(t)\,d_\tau(t-c)\,dt = f(c),

for every c≥0c \ge 0 and every ff continuous at cc. It is not a function in the classical sense; it is a generalized object (a distribution) defined by this limiting/sifting property.

Convolution. For piecewise continuous functions f,gf,g on [0,∞)[0,\infty), their convolution is

(f∗g)(t)=∫0tf(τ) g(t−τ) dτ,t≥0.(f*g)(t) = \int_0^t f(\tau)\,g(t-\tau)\,d\tau, \qquad t \ge 0.

Intuition

Think of uc(t)u_c(t) as a light switch: off for all time before cc, on forever after. Multiplying any function g(t)g(t) by uc(t)u_c(t) silences it before cc and lets it through unchanged after cc. Because uc(t)−ud(t)u_c(t) - u_d(t) (for d>cd>c) is 11 exactly on [c,d)[c,d) and 00 elsewhere, sums of steps let us write any piecewise-defined function as one closed formula — the essential trick behind Laplace-transforming switched circuits, forces applied "starting at t=3t=3," or forcing terms that change formula partway through. The Dirac delta pushes this further: instead of a forcing that switches on and stays on, imagine a forcing that switches on, delivers a fixed total impulse (like momentum ∫F dt\int F\,dt), and switches off again — all within a vanishingly short window τ→0\tau \to 0. The rectangle dτ(t−c)d_\tau(t-c) has area 11 regardless of τ\tau; as τ\tau shrinks the rectangle gets taller and narrower but never loses its unit area. In the limit we get an idealized "spike" concentrated at the single instant t=ct = c, carrying total effect 11. This is precisely how a hammer blow, a lightning strike, or a sudden kick to a spring-mass system is modeled: not by any ordinary function, but by this limiting sifting operation.

Figure
Building dτ(t−2)d_\tau(t-2) as τ→0\tau \to 0: constant area, shrinking width, growing height
t height τ=1 τ=0.5 τ=0.25 c = 2 area = 1 always

Now zoom out to the systems view: if a linear time-invariant system's response to a single unit impulse at t=0t=0 is a function h(t)h(t) (its impulse response), then its response to an arbitrary forcing g(t)g(t) is obtained by treating gg as a continuum of tiny impulses g(τ) dτg(\tau)\,d\tau delivered at every instant τ\tau, each producing a shifted, scaled copy g(τ)h(t−τ) dτg(\tau)h(t-\tau)\,d\tau of the impulse response, and summing (integrating) over all τ\tau. That sum is exactly the convolution (h∗g)(t)(h*g)(t). This is why convolution is not an arbitrary algebraic curiosity: it is the precise mathematical expression of "linear superposition of impulse responses," and the convolution theorem is what makes it computable via transforms rather than by evaluating the integral directly.

Figure
Convolution as superposition: forcing sliced into impulses, each producing a shifted response
g(τ): forcing sliced into impulses each slice → shifted, scaled impulse response h(t−τ) y(t) = (h*g)(t) = sum of all responses
TheoremShifting Theorems, Laplace Transform of the Impulse, and the Convolution Theorem

Let f,g:[0,∞)→Rf, g : [0,\infty) \to \mathbb{R} be piecewise continuous on every finite interval [0,A][0,A] and of exponential order: there exist constants M,K,aM, K, a with ∣f(t)∣≤Keat|f(t)| \le K e^{at} for t≥Mt \ge M (and similarly for gg). Let F(s)=L{f}(s)F(s) = \mathcal{L}\{f\}(s) and G(s)=L{g}(s)G(s) = \mathcal{L}\{g\}(s), defined for s>as > a.

(1) First Shifting Theorem. For any real constant cc,

L{ectf(t)}(s)=F(s−c),s>a+c.\mathcal{L}\{e^{ct} f(t)\}(s) = F(s - c), \qquad s > a + c.

(2) Second Shifting Theorem. For any constant c≥0c \ge 0,

L{uc(t) f(t−c)}(s)=e−csF(s),s>a.\mathcal{L}\{u_c(t)\,f(t-c)\}(s) = e^{-cs} F(s), \qquad s > a.

(3) Transform of the Unit Impulse. For c≥0c \ge 0,

L{δ(t−c)}(s)=e−cs,s>0.\mathcal{L}\{\delta(t-c)\}(s) = e^{-cs}, \qquad s > 0.

(4) Convolution Theorem. The convolution f∗gf*g is piecewise continuous, of exponential order, and

L{f∗g}(s)=F(s) G(s),s>a.\mathcal{L}\{f*g\}(s) = F(s)\,G(s), \qquad s > a.

Consequently f∗g=g∗ff*g = g*f, and if yy solves ay′′+by′+cy=g(t)ay'' + by' + cy = g(t) with y(0)=y0,y′(0)=y0′y(0)=y_0, y'(0)=y_0' (constants a≠0,b,ca\ne 0, b, c), then

y(t)=yh(t)+(h∗g)(t),y(t) = y_h(t) + (h * g)(t),

where yhy_h solves the IVP with g≡0g \equiv 0 and the given initial data, and h(t)h(t) is the solution of ah′′+bh′+ch=0ah''+bh'+ch=0 with h(0)=0,h′(0)=1/ah(0)=0, h'(0) = 1/a (the impulse response).

Proof

We prove each part in turn. Throughout, L{f}(s)=∫0∞e−stf(t) dt\mathcal{L}\{f\}(s) = \int_0^\infty e^{-st} f(t)\,dt, and this integral converges absolutely for s>as > a by the exponential order hypothesis (standard existence result for the Laplace transform, assumed as prerequisite background).

(1) First Shifting Theorem. Direct substitution into the definition:

L{ectf(t)}(s)=∫0∞e−stectf(t) dt=∫0∞e−(s−c)tf(t) dt=F(s−c).\mathcal{L}\{e^{ct}f(t)\}(s) = \int_0^\infty e^{-st} e^{ct} f(t)\,dt = \int_0^\infty e^{-(s-c)t} f(t)\,dt = F(s-c).

This integral converges whenever s−c>as - c > a, i.e. s>a+cs > a+c, since FF converges there by hypothesis. No further justification is needed beyond convergence, which follows since ∣ectf(t)∣≤Ke(a+c)t|e^{ct}f(t)| \le Ke^{(a+c)t} for t≥Mt\ge M, giving ectf(t)e^{ct}f(t) exponential order a+ca+c.

(2) Second Shifting Theorem. By definition of ucu_c, the function uc(t)f(t−c)u_c(t) f(t-c) vanishes for t<ct < c and equals f(t−c)f(t-c) for t≥ct \ge c. Hence

L{uc(t)f(t−c)}(s)=∫c∞e−stf(t−c) dt.\mathcal{L}\{u_c(t) f(t-c)\}(s) = \int_c^\infty e^{-st} f(t-c)\,dt.

Substitute τ=t−c\tau = t - c, so t=τ+ct = \tau + c, dt=dτdt = d\tau, and as tt ranges over [c,∞)[c,\infty), τ\tau ranges over [0,∞)[0,\infty):

∫0∞e−s(τ+c)f(τ) dτ=e−sc∫0∞e−sτf(τ) dτ=e−csF(s).\int_0^\infty e^{-s(\tau+c)} f(\tau)\,d\tau = e^{-sc}\int_0^\infty e^{-s\tau} f(\tau)\,d\tau = e^{-cs} F(s).

Convergence for s>as > a is immediate since the integral defining F(s)F(s) already converges there. ■\blacksquare

(3) Transform of the Unit Impulse. First compute the transform of the pulse dτ(t−c)d_\tau(t-c) directly using linearity and part (2) applied to f≡1f\equiv 1 (so F(s)=1/sF(s) = 1/s):

L{dτ(t−c)}(s)=1τ[L{uc(t)}(s)−L{uc+τ(t)}(s)]=1τ[e−css−e−(c+τ)ss]=e−cs⋅1−e−τsτs.\mathcal{L}\{d_\tau(t-c)\}(s) = \frac{1}{\tau}\Bigl[\mathcal{L}\{u_c(t)\}(s) - \mathcal{L}\{u_{c+\tau}(t)\}(s)\Bigr] = \frac{1}{\tau}\left[\frac{e^{-cs}}{s} - \frac{e^{-(c+\tau)s}}{s}\right] = e^{-cs}\cdot\frac{1 - e^{-\tau s}}{\tau s}.

Now let τ→0+\tau \to 0^+ with s>0s>0 fixed. By L'Hôpital's rule (or the standard limit lim⁡x→0(1−e−x)/x=1\lim_{x\to 0}(1-e^{-x})/x = 1 with x=τsx = \tau s),

lim⁡τ→0+1−e−τsτs=1.\lim_{\tau \to 0^+} \frac{1 - e^{-\tau s}}{\tau s} = 1.

Hence

L{δ(t−c)}(s):=lim⁡τ→0+L{dτ(t−c)}(s)=e−cs.\mathcal{L}\{\delta(t-c)\}(s) := \lim_{\tau\to 0^+} \mathcal{L}\{d_\tau(t-c)\}(s) = e^{-cs}.

This matches the defining sifting property in the definition: taking f(t)=e−stf(t) = e^{-st} (continuous at cc) gives ∫0∞e−stδ(t−c) dt=e−cs\int_0^\infty e^{-st}\delta(t-c)\,dt = e^{-cs} directly, consistent with the limit computation above. ■\blacksquare

(4) Convolution Theorem. We compute F(s)G(s)F(s)G(s) as a double integral and reduce it to the transform of f∗gf*g by a change of variables. Since F,GF,G converge absolutely for s>as>a, write

F(s)G(s)=(∫0∞e−sξf(ξ) dξ)(∫0∞e−sηg(η) dη)=∫0∞ ⁣ ⁣∫0∞e−s(ξ+η)f(ξ)g(η) dξ dη,F(s)G(s) = \left(\int_0^\infty e^{-s\xi} f(\xi)\,d\xi\right)\left(\int_0^\infty e^{-s\eta} g(\eta)\,d\eta\right) = \int_0^\infty\!\!\int_0^\infty e^{-s(\xi+\eta)} f(\xi) g(\eta)\,d\xi\,d\eta,

where combining the two integrals into one double integral over the quarter-plane ξ,η≥0\xi,\eta \ge 0 is justified by Fubini's theorem, applicable because the double integral of ∣e−s(ξ+η)f(ξ)g(η)∣|e^{-s(\xi+\eta)}f(\xi)g(\eta)| converges absolutely for s>as>a (each factor is absolutely integrable there, and the product of absolutely convergent one-dimensional integrals is an absolutely convergent double integral). Now introduce new variables t=ξ+ηt = \xi + \eta (so tt ranges over [0,∞)[0,\infty)) and τ=ξ\tau = \xi (so τ\tau ranges over [0,t][0,t] for each fixed tt, and η=t−τ\eta = t - \tau). This is a linear, area-preserving change of variables (Jacobian determinant 11) mapping the quarter-plane {ξ,η≥0}\{\xi,\eta\ge 0\} bijectively (up to the boundary, of measure zero) onto the region {(t,τ):t≥0, 0≤τ≤t}\{(t,\tau): t \ge 0,\ 0\le \tau \le t\}. Hence

F(s)G(s)=∫0∞(∫0te−stf(τ) g(t−τ) dτ)dt=∫0∞e−st(∫0tf(τ)g(t−τ) dτ)dt=∫0∞e−st(f∗g)(t) dt.F(s)G(s) = \int_0^\infty \left(\int_0^t e^{-st} f(\tau)\,g(t-\tau)\,d\tau\right) dt = \int_0^\infty e^{-st}\left(\int_0^t f(\tau)g(t-\tau)\,d\tau\right) dt = \int_0^\infty e^{-st}(f*g)(t)\,dt.

This last expression is exactly L{f∗g}(s)\mathcal{L}\{f*g\}(s), completing the proof that L{f∗g}=FG\mathcal{L}\{f*g\} = FG. Piecewise continuity and exponential order of f∗gf*g: for fixed tt, the integrand τ↦f(τ)g(t−τ)\tau \mapsto f(\tau)g(t-\tau) is piecewise continuous on [0,t][0,t] (finitely many discontinuities, since f,gf,g are piecewise continuous), so the integral defining (f∗g)(t)(f*g)(t) exists; continuity of (f∗g)(t)(f*g)(t) in tt follows from the dominated convergence theorem using the exponential bounds on f,gf,g. For the growth bound, using ∣f(τ)∣≤Keaτ|f(\tau)|\le Ke^{a\tau}, ∣g(t−τ)∣≤Kea(t−τ)|g(t-\tau)|\le Ke^{a(t-\tau)} for τ,t−τ≥M\tau, t-\tau \ge M (and bounding the finitely many remaining pieces by continuity on compacta). We get, for tt sufficiently large,

∣(f∗g)(t)∣≤∫0t∣f(τ)∣∣g(t−τ)∣ dτ≤∫0tK2eaτea(t−τ) dτ=K2eat∫0tdτ=K2t eat≤K2e(a+1)t|(f*g)(t)| \le \int_0^t |f(\tau)||g(t-\tau)|\,d\tau \le \int_0^t K^2 e^{a\tau}e^{a(t-\tau)}\,d\tau = K^2 e^{at}\int_0^t d\tau = K^2 t\, e^{at} \le K^2 e^{(a+1)t}

for tt large enough that t≤ett \le e^t. Thus f∗gf*g has exponential order a+1a+1 (or any order exceeding aa), confirming its transform converges for s>as>a as claimed, and in particular the manipulation above is valid there. Commutativity: the substitution σ=t−τ\sigma = t - \tau in (f∗g)(t)=∫0tf(τ)g(t−τ) dτ(f*g)(t) = \int_0^t f(\tau)g(t-\tau)\,d\tau gives ∫0tf(t−σ)g(σ) dσ=(g∗f)(t)\int_0^t f(t-\sigma)g(\sigma)\,d\sigma = (g*f)(t). Application to the IVP. Consider ay′′+by′+cy=g(t)ay''+by'+cy = g(t), y(0)=y0y(0)=y_0, y′(0)=y0′y'(0)=y_0', with a≠0a\ne 0. Taking Laplace transforms of both sides (using the derivative rules L{y′}=sY−y0\mathcal{L}\{y'\} = sY - y_0, L{y′′}=s2Y−sy0−y0′\mathcal{L}\{y''\} = s^2Y - sy_0 - y_0', standard prerequisite facts) gives

a(s2Y−sy0−y0′)+b(sY−y0)+cY=G(s).a\bigl(s^2 Y - sy_0 - y_0'\bigr) + b\bigl(sY - y_0\bigr) + cY = G(s).

Solve for Y(s)Y(s):

Y(s)=a(sy0+y0′)+by0as2+bs+c+G(s)as2+bs+c.Y(s) = \frac{a(sy_0+y_0') + by_0}{as^2+bs+c} + \frac{G(s)}{as^2+bs+c}.

The first term is exactly L{yh}(s)\mathcal{L}\{y_h\}(s), the transform of the solution of the homogeneous IVP with the same initial data (verified by running the same derivation with g≡0g\equiv 0). For the second term, let h(t)h(t) be the solution of ah′′+bh′+ch=0ah''+bh'+ch=0 with h(0)=0h(0)=0, h′(0)=1/ah'(0)=1/a. Taking its transform: a(s2H−0−1/a)+b(sH−0)+cH=0a(s^2H - 0 - 1/a) + b(sH-0) + cH = 0, i.e. H(s)(as2+bs+c)=1H(s)(as^2+bs+c) = 1, so

H(s)=1as2+bs+c.H(s) = \frac{1}{as^2+bs+c}.

Hence the second term is H(s)G(s)=L{h∗g}(s)H(s)G(s) = \mathcal{L}\{h*g\}(s) by the convolution theorem just proved. Taking inverse transforms (using injectivity of L\mathcal{L} on piecewise-continuous exponential-order functions, a prerequisite fact underlying Inverse Laplace Transform computations) gives

y(t)=yh(t)+(h∗g)(t),y(t) = y_h(t) + (h*g)(t),

as claimed. ■\blacksquare

∎

Worked examples

Example 1 (switched, delayed forcing on a spring-mass system). Solve y′′+4y=g(t)y'' + 4y = g(t), y(0)=1y(0) = 1, y′(0)=0y'(0)=0, where

g(t)={0,0≤t<3,(t−3)2,t≥3.g(t) = \begin{cases} 0, & 0 \le t < 3, \\ (t-3)^2, & t \ge 3. \end{cases}

Note g(t)=u3(t)(t−3)2g(t) = u_3(t)(t-3)^2, a delayed copy of t2t^2 switched on at t=3t=3. Transform both sides. Since L{y′′}=s2Y−s(1)−0=s2Y−s\mathcal{L}\{y''\} = s^2Y - s(1) - 0 = s^2Y - s, and by the second shifting theorem with f(t)=t2f(t)=t^2 (so F(s)=2/s3F(s) = 2/s^3), L{u3(t)(t−3)2}=e−3s⋅2s3\mathcal{L}\{u_3(t)(t-3)^2\} = e^{-3s}\cdot \dfrac{2}{s^3}, we get

s2Y−s+4Y=2e−3ss3  ⟹  Y(s)=ss2+4+2e−3ss3(s2+4).s^2 Y - s + 4Y = \frac{2e^{-3s}}{s^3} \implies Y(s) = \frac{s}{s^2+4} + \frac{2e^{-3s}}{s^3(s^2+4)}.

The first term inverts immediately to cos⁡(2t)\cos(2t). For the second, first find the inverse transform of 2s3(s2+4)\dfrac{2}{s^3(s^2+4)} by Partial Fractions:

2s3(s2+4)=As+Bs2+Cs3+Ds+Es2+4.\frac{2}{s^3(s^2+4)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s^3} + \frac{Ds+E}{s^2+4}.

Multiply through by s3(s2+4)s^3(s^2+4):

2=As2(s2+4)+Bs(s2+4)+C(s2+4)+(Ds+E)s3.2 = As^2(s^2+4) + Bs(s^2+4) + C(s^2+4) + (Ds+E)s^3.

Set s=0s=0: 2=4C⇒C=1/22 = 4C \Rightarrow C = 1/2. Expand and match coefficients of s4,s3,s2,s1,s0s^4, s^3, s^2, s^1, s^0:

2=(A+D)s4+(B+E)s3+(4A+C)s2+4Bs+4C.2 = (A+D)s^4 + (B+E)s^3 + (4A + C)s^2 + 4Bs + 4C.

Matching: s0s^0: 4C=2⇒C=1/24C=2 \Rightarrow C=1/2 (consistent). s1s^1: 4B=0⇒B=04B = 0 \Rightarrow B=0. s2s^2: 4A+C=0⇒4A=−1/2⇒A=−1/84A + C = 0 \Rightarrow 4A = -1/2 \Rightarrow A = -1/8. s4s^4: A+D=0⇒D=1/8A+D=0 \Rightarrow D = 1/8. s3s^3: B+E=0⇒E=0B+E=0 \Rightarrow E=0. So

2s3(s2+4)=−18⋅1s+12⋅1s3+18⋅ss2+4.\frac{2}{s^3(s^2+4)} = -\frac{1}{8}\cdot\frac{1}{s} + \frac{1}{2}\cdot\frac{1}{s^3} + \frac{1}{8}\cdot\frac{s}{s^2+4}.

Inverting term by term (L−1{1/s3}=t2/2\mathcal{L}^{-1}\{1/s^3\} = t^2/2, L−1{s/(s2+4)}=cos⁡2t\mathcal{L}^{-1}\{s/(s^2+4)\} = \cos 2t):

L−1{2s3(s2+4)}=−18+t24+18cos⁡2t=:ϕ(t).\mathcal{L}^{-1}\left\{\frac{2}{s^3(s^2+4)}\right\} = -\frac{1}{8} + \frac{t^2}{4} + \frac{1}{8}\cos 2t =: \phi(t).

By the second shifting theorem in reverse, L−1{e−3s⋅(2/(s3(s2+4)))}=u3(t) ϕ(t−3)\mathcal{L}^{-1}\{e^{-3s}\cdot(2/(s^3(s^2+4)))\} = u_3(t)\,\phi(t-3). Therefore

y(t)=cos⁡2t+u3(t)[−18+(t−3)24+18cos⁡(2(t−3))].y(t) = \cos 2t + u_3(t)\left[-\frac{1}{8} + \frac{(t-3)^2}{4} + \frac{1}{8}\cos\bigl(2(t-3)\bigr)\right].

Check: for t<3t<3, y(t)=cos⁡2ty(t)=\cos 2t, matching y(0)=1y(0)=1, y′(0)=0y'(0)=0 correctly since the bracket contributes nothing there. For t≥3t\ge3, differentiate twice and add 4y4y; the polynomial and cosine pieces are engineered (by construction of H(s)G(s)H(s)G(s)) to satisfy y′′+4y=(t−3)2y''+4y = (t-3)^2, and continuity of y,y′y,y' at t=3t=3 holds because the bracket and its derivative vanish at t=3t=3 (ϕ(0)=−1/8+0+1/8=0\phi(0) = -1/8+0+1/8=0, ϕ′(0)=0+0−14sin⁡0=0\phi'(0) = 0 + 0 -\frac14\sin 0 = 0... let's verify: ϕ′(t)=t/2−14sin⁡2t\phi'(t) = t/2 - \frac14\sin 2t, so ϕ′(0)=0\phi'(0)=0). Both checks pass.

Example 2 (impulsive forcing, convolution form, and consistency check). An underdamped oscillator satisfies y′′+2y′+5y=3δ(t−1)y'' + 2y' + 5y = 3\delta(t-1), y(0)=0y(0)=0, y′(0)=0y'(0)=0, modeling a mass at rest struck by an impulsive force of magnitude 33 at t=1t=1. Transform, using L{δ(t−1)}=e−s\mathcal{L}\{\delta(t-1)\} = e^{-s} from part (3) of the theorem:

s2Y+2sY+5Y=3e−s  ⟹  Y(s)=3e−ss2+2s+5=3e−s⋅1(s+1)2+4.s^2Y + 2sY + 5Y = 3e^{-s} \implies Y(s) = \frac{3e^{-s}}{s^2+2s+5} = 3e^{-s}\cdot \frac{1}{(s+1)^2+4}.

Complete the square: s2+2s+5=(s+1)2+4s^2+2s+5 = (s+1)^2+4. Recall L−1{1/((s+1)2+4)}\mathcal{L}^{-1}\{1/((s+1)^2+4)\}: since L{e−tsin⁡2t}(s)=2/((s+1)2+4)\mathcal{L}\{e^{-t}\sin 2t\}(s) = 2/((s+1)^2+4) by the first shifting theorem applied to L{sin⁡2t}=2/(s2+4)\mathcal{L}\{\sin 2t\}=2/(s^2+4), we get

L−1{1(s+1)2+4}=12e−tsin⁡2t.\mathcal{L}^{-1}\left\{\frac{1}{(s+1)^2+4}\right\} = \frac{1}{2}e^{-t}\sin 2t.

By the second shifting theorem, multiplying by e−se^{-s} delays this by 11 and switches it on with u1(t)u_1(t):

y(t)=3⋅u1(t)⋅12e−(t−1)sin⁡(2(t−1))=32 u1(t) e−(t−1)sin⁡(2(t−1)).y(t) = 3\cdot u_1(t)\cdot \frac{1}{2}e^{-(t-1)}\sin\bigl(2(t-1)\bigr) = \frac{3}{2}\,u_1(t)\,e^{-(t-1)}\sin\bigl(2(t-1)\bigr).

Cross-check via convolution. The impulse response of y′′+2y′+5y=g(t)y''+2y'+5y = g(t) (i.e., solving with unit impulse at t=0t=0) is h(t)=12e−tsin⁡2th(t) = \frac12 e^{-t}\sin 2t (matches H(s)=1/((s+1)2+4)H(s) = 1/((s+1)^2+4), consistent with the theorem's formula H(s)=1/(as2+bs+c)H(s)=1/(as^2+bs+c) with a=1,b=2,c=5a=1,b=2,c=5, and h(0)=0h(0)=0, h′(0)=1h'(0)= 1, matching 1/a=11/a = 1). For forcing g(t)=3δ(t−1)g(t) = 3\delta(t-1), the general solution formula y=h∗gy = h*g gives, using the sifting property directly inside the convolution integral,

y(t)=∫0th(t−τ)⋅3δ(τ−1) dτ=3h(t−1) u1(t)=32u1(t)e−(t−1)sin⁡(2(t−1)),y(t) = \int_0^t h(t-\tau)\cdot 3\delta(\tau - 1)\,d\tau = 3h(t-1)\,u_1(t) = \frac{3}{2}u_1(t)e^{-(t-1)}\sin(2(t-1)),

for t≠1t \ne 1 (using the sifting property, valid whenever t>1t>1 so that τ=1\tau=1 lies in (0,t)(0,t), and giving 00 for t<1t<1). This matches the transform computation exactly, confirming both the impulse-response interpretation and the direct Laplace calculation.

Counterexamplewhy exponential order is needed for the convolution theorem

The convolution theorem requires f,gf,g to be of exponential order so that F(s),G(s)F(s), G(s) converge for some ss and the Fubini-theorem step in the proof applies. Drop this hypothesis and let

f(t)=g(t)=et2.f(t) = g(t) = e^{t^2}.

This function is continuous (hence piecewise continuous) on [0,∞)[0,\infty) but grows faster than any exponential KeatKe^{at}, so it has no exponential order. Consequently ∫0∞e−stet2 dt\int_0^\infty e^{-st}e^{t^2}\,dt diverges for every real ss, since e−st+t2→∞e^{-st+t^2}\to\infty as t→∞t\to\infty regardless of ss. So F(s)F(s) does not exist for any s∈Rs\in\mathbb{R}, and the right-hand side F(s)G(s)F(s)G(s) of the convolution theorem is meaningless: there is no product to equate to L{f∗g}\mathcal{L}\{f*g\}. One might hope L{f∗g}\mathcal{L}\{f*g\} still exists even though F,GF,G individually diverge. But it does not help matters: (f∗g)(t)=∫0teτ2e(t−τ)2 dτ(f*g)(t) = \int_0^t e^{\tau^2}e^{(t-\tau)^2}\,d\tau still grows at least as fast as t et2/2t\,e^{t^2/2} (by evaluating the integrand at τ=t/2\tau = t/2, its minimum-exponent point, and bounding the integral below by the integrand there times a fixed sub-interval width), so f∗gf*g is also not of exponential order, and L{f∗g}(s)\mathcal{L}\{f*g\}(s) diverges for every ss too. Both sides of the claimed identity fail to exist — the theorem's conclusion is not merely false but not even well-posed — which is exactly what happens when its convergence hypothesis is discarded.

Scope, edge cases & common traps

SituationWhat to watch for
Writing piecewise forcing as stepsAlways express uc(t)f(t−c)u_c(t) f(t-c) with the function argument shifted to (t−c)(t-c), matching the second shifting theorem exactly — a common error is transforming uc(t)f(t)u_c(t)f(t) (unshifted argument) as if it were uc(t)f(t−c)u_c(t) f(t-c).
Impulse at t=0t=0If δ(t)\delta(t) acts at the left endpoint of [0,∞)[0,\infty), conventions vary on whether ∫0∞f(t)δ(t)dt\int_0^\infty f(t)\delta(t)dt equals f(0)f(0) or 12f(0)\tfrac12 f(0); state the convention used (this article uses f(0)f(0), consistent with u0(t)≡1u_0(t)\equiv 1 on [0,∞)[0,\infty)).
Convolution vs. productL{fg}≠L{f}L{g}\mathcal{L}\{fg\} \ne \mathcal{L}\{f\}\mathcal{L}\{g\} in general; only the convolution f∗gf*g transforms to the product FGFG.
Existence of transformsBoth shifting theorems and the convolution theorem require exponential order (and piecewise continuity); without it, the transforms may fail to exist altogether (see the counterexample).
Delta "function"δ(t−c)\delta(t-c) is not a function with a pointwise value; manipulations (like differentiating it) require distribution theory, which is beyond ordinary Laplace-transform calculus but the sifting/transform rules used here remain valid.

The mathematician behind it

Origins · 1 minOliver Heaviside and how this was discovered
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Oliver Heaviside
1850–1925 · Victorian telegraphy and mathematical physics, self-taught

Oliver Heaviside was a self-taught English electrical engineer and physicist who developed operational calculus methods — precursors to the modern Laplace transform — to solve the differential equations governing telegraph and transmission lines. Working largely outside academia, he introduced the step function and vector calculus notation still used today, and reformulated Maxwell's twenty original equations into the four compact vector equations now taught worldwide. His methods were denounced by contemporaries as lacking rigor, though they were later fully justified by the operational calculus formalized through the Laplace transform.

History
- 1850: Oliver Heaviside born in Camden Town, London, to a family of modest means.
- 1873: Heaviside begins independent study of Maxwell's Treatise on Electricity and Magnetism, with no university training.
- 1880s: Heaviside develops operational calculus for transmission-line and telegraphy equations, introducing the step function bearing his name.
- 1893–1912: Heaviside publishes Electromagnetic Theory in three volumes, cementing his reformulation of electromagnetism and his operational methods, later rigorized via the Laplace transform.
- 1925: Heaviside dies in Torquay, England; his operational methods are subsequently formalized rigorously within Laplace transform theory.
Why do we need the Dirac delta if it isn't a real function?

Physical situations like a hammer strike or a lightning surge deliver a large, finite total effect (impulse, charge) over a duration so short it is effectively zero. No ordinary function can have zero width, nonzero height at a single point, and finite nonzero integral simultaneously — any function that is zero almost everywhere has integral zero. The Dirac delta resolves this by being defined not as a pointwise function but through its action on other functions (the sifting property), formalized as a distribution. This lets us model idealized instantaneous forcing rigorously, and the resulting Laplace transform e−cse^{-cs} is perfectly well-behaved, making the formalism practically indispensable despite δ\delta not being a classical function.

What's the practical difference between the two shifting theorems?

The first shifting theorem shifts in the ss-domain: multiplying f(t)f(t) by ecte^{ct} in the time domain shifts the transform variable, F(s)↦F(s−c)F(s)\mapsto F(s-c). It is used whenever exponential factors like e−te^{-t} or e3te^{3t} multiply a known transform pair, typically arising from completing the square in a denominator. The second shifting theorem shifts in the tt-domain: delaying a function's start time to cc and switching it on there multiplies the transform by e−cse^{-cs}. It is used whenever forcing changes formula or switches on partway through time, and is the tool for piecewise and delayed forcing.

Why is convolution defined with the integral from 00 to tt, not −∞-\infty to ∞\infty?

This article works with one-sided Laplace transforms, which assume all functions are zero for t<0t<0 (or are only defined on [0,∞)[0,\infty)), matching how initial value problems are posed. Under that convention, f(τ)g(t−τ)f(\tau)g(t-\tau) automatically vanishes outside 0≤τ≤t0\le \tau\le t, since f(τ)=0f(\tau)=0 for τ<0\tau<0 and g(t−τ)=0g(t-\tau)=0 for τ>t\tau>t. So the "natural" convolution integral over all reals collapses to the finite integral from 00 to tt used here. The two-sided convolution (integral over all of R\mathbb{R}) is the more general object used with the Fourier transform, where functions need not vanish for negative arguments.

How do I know which piece of a solution is the "homogeneous part" versus the forced response after using convolution?

In the theorem's IVP formula y(t)=yh(t)+(h∗g)(t)y(t) = y_h(t) + (h*g)(t), the term yh(t)y_h(t) solves the homogeneous equation with the actual initial data y(0)=y0y(0)=y_0, y′(0)=y0′y'(0)=y_0', while h(t)h(t) is a fixed, forcing-independent function (the impulse response) that solves the homogeneous equation with the specific initial data h(0)=0h(0)=0, h′(0)=1/ah'(0)=1/a. The convolution term (h∗g)(t)(h*g)(t) then automatically has zero initial conditions of its own — it contributes nothing to y(0)y(0) or y′(0)y'(0) — so all of the initial data is carried by yhy_h, and all of the forcing's effect is carried by the convolution term. This clean separation is why the formula is useful: hh never needs to be recomputed as gg changes.

Can convolution be computed without Laplace transforms, and when would I do that?

Yes — convolution is defined by an ordinary integral and can always be computed directly, especially when ff or gg is simple (e.g., gg constant or a single exponential). Direct computation is preferable for conceptual understanding and for numerical evaluation when closed-form inverse transforms are hard to find. However, for rational-transform impulse responses (as arise from constant-coefficient linear ODEs), Laplace transforms usually make the calculation far shorter: multiplying two transforms and applying partial fractions is typically easier than evaluating the convolution integral directly, especially when the forcing g(t)g(t) is itself piecewise or delayed.

Does the convolution theorem's proof actually need Fubini's theorem, or is that a technicality that can be skipped?

It is not skippable in a fully rigorous treatment. The proof rewrites the product of two single integrals as one double integral, which requires justifying that the order of integration (or, equivalently, treating it as one combined integral over the quarter-plane) is valid. Fubini's theorem is precisely the tool that licenses this, and it requires absolute convergence of the double integral — which is exactly why the exponential order hypothesis is needed. Without exponential order, as the counterexample shows, the relevant integrals may fail to converge at all, and the entire argument collapses, not merely becomes "less rigorous."

Is δ(t−c)\delta(t-c) the derivative of uc(t)u_c(t)?

Formally yes, in the sense of distributional (generalized) derivatives: uc(t)u_c(t) has a jump of height 11 at t=ct=c and is constant elsewhere, so its derivative is zero everywhere except at the single point t=ct=c, where it must contribute exactly enough to reconstruct the jump — matching the sifting/unit-area behavior that defines δ(t−c)\delta(t-c). This is consistent with the Laplace transform computation: differentiating L{uc(t)}(s)=e−cs/s\mathcal{L}\{u_c(t)\}(s) = e^{-cs}/s using the transform-of-derivative rule (formally, ignoring the discontinuity) gives s⋅(e−cs/s)−uc(0−)=e−css\cdot(e^{-cs}/s) - u_c(0^-) = e^{-cs}, matching L{δ(t−c)}(s)\mathcal{L}\{\delta(t-c)\}(s) computed directly. A fully rigorous justification requires distribution theory, beyond classical calculus, but the relationship is standard and reliable in applications.

Before this