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Unit 7 · Lesson 7-1
The Laplace Transform
1h 40m1300 XP
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The Laplace Transform

From Fourier to Laplace, regions of convergence, the common pairs, the derivative and convolution properties, and the initial- and final-value theorems

1880–1887 — Heaviside's operators
By his own account, Oliver Heaviside knew "only school algebra and trigonometry" when he first opened Maxwell's treatise on electromagnetism. A former telegraph operator, he set out to master it on his own. Between 1880 and 1887 he develops an operational calculus: he writes pp for the operator d/dtd/dt and solves differential equations as if they were algebraic equations (Oliver Heaviside). It gives the right answers for cables and circuits, and it causes a great deal of controversy. Heaviside's methods are not rigorous, and mathematicians do not develop them further; a rigorous justification comes only later, by relating them to a transform Laplace had used more than a century before (Operational calculus). During and after the Second World War, the Laplace transform replaced Heaviside's calculus in engineering (Laplace transform). This lesson is that transform: the tool that handles growth, decay, switching-on and instability, where Fourier's transform cannot.
The Fourier transform needs signals that die away. The unit step, a ramp, a growing exponential or the output of an unstable system has no Fourier transform: the integral does not converge. Multiply such a signal by a decaying exponential e−σte^{-\sigma t}, however, and it often does. The Laplace transform is the Fourier transform of that tamed signal.
Derivation
From Fourier to Laplace
Take the Fourier transform of x(t) e−σtx(t)\,e^{-\sigma t}:
∫−∞∞x(t) e−σt e−jωt dt=∫−∞∞x(t) e−(σ+jω)t dt.\int_{-\infty}^{\infty} x(t)\,e^{-\sigma t}\,e^{-j\omega t}\,dt = \int_{-\infty}^{\infty} x(t)\,e^{-(\sigma + j\omega)t}\,dt .
Write s=σ+jωs = \sigma + j\omega, a point in the complex s-plane: the result is X(s)X(s), the bilateral Laplace transform. On the imaginary axis, σ=0\sigma = 0, it is the Fourier transform again: X(jω)=X(s)∣s=jωX(j\omega) = X(s)\big|_{s = j\omega} — when the integral converges there. For signals that start at t=0t = 0, and for systems switched on with energy already stored, the unilateral transform integrates from 0−0^-, just before the start, so that initial conditions enter the algebra.
X(s)=∫0−∞x(t) e−st dt,s=σ+jωX(s) = \int_{0^-}^{\infty} x(t)\,e^{-st}\,dt, \qquad s = \sigma + j\omega

The unilateral Laplace transform: a Fourier transform with a built-in decay e^{−σt}, evaluated anywhere in the complex s-plane where the integral converges.

💡The region of convergence
The integral converges only for some values of ss, its region of convergence (ROC), which depends on σ=Re⁡s\sigma = \operatorname{Re} s alone:
  • a right-sided signal converges in a right half-plane, Re⁡s>σ0\operatorname{Re} s > \sigma_0: e−2tu(t)↔1/(s+2)e^{-2t}u(t) \leftrightarrow 1/(s + 2) for Re⁡s>−2\operatorname{Re} s > -2;
  • a left-sided signal converges in a left half-plane: −e−2tu(−t)↔1/(s+2)-e^{-2t}u(-t) \leftrightarrow 1/(s + 2) as well — but for Re⁡s<−2\operatorname{Re} s < -2;
  • a two-sided signal converges in a vertical strip, or nowhere: e−2tu(t)+e3tu(−t)e^{-2t}u(t) + e^{3t}u(-t) only for −2<Re⁡s<3-2 < \operatorname{Re} s < 3.
The same formula with different ROCs means different signals. The Fourier transform exists exactly when the ROC contains the imaginary axis.
💡A table worth knowing
| Signal (t≥0t \ge 0) | Transform | ROC | |---|---|---| | δ(t)\delta(t) | 11 | all ss | | u(t)u(t) | 1/s1/s | Re⁡s>0\operatorname{Re} s > 0 | | e−ate^{-at} | 1/(s+a)1/(s + a) | Re⁡s>−a\operatorname{Re} s > -a | | t e−att\,e^{-at} | 1/(s+a)21/(s + a)^2 | Re⁡s>−a\operatorname{Re} s > -a | | sin⁡ω0t\sin\omega_0 t | ω0/(s2+ω02)\omega_0/(s^2 + \omega_0^2) | Re⁡s>0\operatorname{Re} s > 0 | | cos⁡ω0t\cos\omega_0 t | s/(s2+ω02)s/(s^2 + \omega_0^2) | Re⁡s>0\operatorname{Re} s > 0 | | e−atsin⁡ω0te^{-at}\sin\omega_0 t | ω0/((s+a)2+ω02)\omega_0/\big((s + a)^2 + \omega_0^2\big) | Re⁡s>−a\operatorname{Re} s > -a |
Derivation
Differentiation becomes multiplication by s
Integrate by parts:
∫0−∞dxdt e−st dt=[x(t) e−st]0−∞+s∫0−∞x(t) e−st dt=s X(s)−x(0−),\int_{0^-}^{\infty} \frac{dx}{dt}\,e^{-st}\,dt = \Big[x(t)\,e^{-st}\Big]_{0^-}^{\infty} + s\int_{0^-}^{\infty} x(t)\,e^{-st}\,dt = s\,X(s) - x(0^-),
since x(t)e−st→0x(t)e^{-st} \to 0 as t→∞t \to \infty inside the ROC. Applying it twice, x¨↔s2X(s)−s x(0−)−x˙(0−)\ddot x \leftrightarrow s^2X(s) - s\,x(0^-) - \dot x(0^-). Differentiation is multiplication by ss — Heaviside's pp — plus terms that carry the initial conditions. The other workhorses follow as quickly: ∫0−tx dτ↔X(s)/s\int_{0^-}^t x\,d\tau \leftrightarrow X(s)/s; a delay x(t−T)u(t−T)↔e−sTX(s)x(t - T)u(t - T) \leftrightarrow e^{-sT}X(s); e−atx(t)↔X(s+a)e^{-at}x(t) \leftrightarrow X(s + a); and, as for Fourier, convolution ↔\leftrightarrow multiplication: (x∗h)(t)↔X(s)H(s)(x * h)(t) \leftrightarrow X(s)H(s).
💡Initial and final values
The ends of a signal can be read from its transform without inverting it: x(0+)=lim⁡s→∞s X(s),lim⁡t→∞x(t)=lim⁡s→0s X(s).x(0^+) = \lim_{s \to \infty} s\,X(s), \qquad \lim_{t \to \infty} x(t) = \lim_{s \to 0} s\,X(s) . The final-value theorem holds only if the limit exists — every pole of sX(s)sX(s) in the open left half-plane. For Y(s)=10/(s(s2+2s+5))Y(s) = 10/\big(s(s^2 + 2s + 5)\big), the step response of a stable system, it gives 2, and y(t)y(t) does settle at 2. For Z(s)=1/(s(s−1))Z(s) = 1/\big(s(s - 1)\big) it gives −1-1, while z(t)=et−1z(t) = e^t - 1 grows without bound: a pole in the right half-plane voids the theorem.
Rigorous Mathematics is Narrow, Physical Mathematics Bold and Broad.
– Oliver Heaviside, Electromagnetic Theory, Vol. II (1899), quoted on Wikiquote
🕰Laplace before Heaviside
Leonhard Euler studied integrals of this kind from 1744. Laplace's attention turned to them in 1782, and in 1785 he began to use them as transforms, in the sense later made popular; in 1809 he applied his transform to diffusion problems that Fourier series could not handle. Cauchy built an operational calculus on it in 1821. The transform became an everyday engineering tool only during and soon after the Second World War, championed by Gustav Doetsch (Laplace transform).
⚠Same formula, different signal
1/(s+2)1/(s + 2) alone does not name a signal: with Re⁡s>−2\operatorname{Re} s > -2 it is e−2tu(t)e^{-2t}u(t), with Re⁡s<−2\operatorname{Re} s < -2 it is −e−2tu(−t)-e^{-2t}u(-t). Always carry the ROC — or work with the unilateral transform, where every signal starts at t=0t = 0 and the ROC is a right half-plane by construction.
💡Where you will use this
Inverting rational transforms is Lesson 7-2; reading a system's future from its poles is Lesson 7-3; feedback design is Lesson 7-4. The z-transform (Unit 8) does the same job for sequences, and the bilinear transform of Lesson 9-2 maps the s-plane onto the z-plane.
📚Further reading
  • Oppenheim and Willsky, Signals and Systems, sections 9.1–9.5 and 9.9 — the Laplace transform, its ROC, properties and the unilateral transform.
  • Paul J. Nahin, Oliver Heaviside: The Life, Work, and Times of an Electrical Genius of the Victorian Age (Johns Hopkins University Press, 2002).
  • Operational calculus on Wikipedia.
↓But there is a problem
Going from a signal to its transform is a matter of tables. Coming back is the real work: an algebraic answer like (s2+s+20)/(s(s2+s+4))(s^2 + s + 20)/\big(s(s^2 + s + 4)\big) has to become a function of time. The general inversion formula is a contour integral in the complex plane, and the man who found it was the Cambridge mathematician who finally made Heaviside's methods respectable.