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Mathematical Methods

The Laplace transform: properties, inversion and stability

Definition and convergence, table of transforms, properties with proof of the derivative rule, initial and final values, partial fractions and the s-plane.

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The Fourier transform does not exist for growing signals, such as e^{t}, and it does not naturally account for initial conditions. The Laplace transform solves both problems by multiplying the signal by an exponential that damps it. It is the standard tool for circuits in transient conditions and for control engineering: it turns differential equations into algebraic ones and leads directly to the idea of a transfer function.

Definition

For a signal f(t) that is zero for t \lt 0 (as all signals “switched on” at some instant):

    \[ F(s) = \mathcal L\{f\}(s) = \int_0^{\infty}f(t)\,e^{-st}\,dt,\qquad s = \sigma + i\omega \]

With s = i\omega you get back the Fourier transform. The real part \sigma adds a factor e^{-\sigma t} that damps the signal and makes the integral converge even for exponentially growing functions. The integral converges in the half-plane \operatorname{Re}s \gt \sigma_0, where \sigma_0 is the abscissa of convergence.

The basic transforms

Exponential. \mathcal L\{e^{at}\} = \int_0^\infty e^{-(s - a)t}dt = \frac{1}{s - a}, for \operatorname{Re}s \gt a. Every other row of the table follows from this one.

f(t), for t \ge 0F(s)
\delta(t)1
u(t) (step)\dfrac1s
t^n\dfrac{n!}{s^{n+1}}
e^{at}\dfrac{1}{s - a}
\sin\omega t\dfrac{\omega}{s^2 + \omega^2}
\cos\omega t\dfrac{s}{s^2 + \omega^2}
e^{-at}\sin\omega t\dfrac{\omega}{(s + a)^2 + \omega^2}
e^{-at}\cos\omega t\dfrac{s + a}{(s + a)^2 + \omega^2}

Sine and cosine come from \cos\omega t = \frac{e^{i\omega t} + e^{-i\omega t}}{2}: \frac12\left(\frac{1}{s - i\omega} + \frac{1}{s + i\omega}\right) = \frac{s}{s^2 + \omega^2}.

Properties

  • Linearity.
  • Frequency shift: \mathcal L\{e^{-at}f(t)\} = F(s + a). Multiplying by an exponential shifts the transform.
  • Delay: \mathcal L\{f(t - T)u(t - T)\} = e^{-sT}F(s).
  • Derivative: \mathcal L\{f'\} = sF(s) - f(0), and \mathcal L\{f''\} = s^2F(s) - sf(0) - f'(0).
  • Integral: \mathcal L\{\int_0^t f\} = \frac{F(s)}{s}.
  • Convolution: \mathcal L\{f * g\} = F(s)\,G(s).

Proof of the derivative rule

Integrating by parts:

    \[ \int_0^\infty f'(t)e^{-st}dt = \Big[f(t)e^{-st}\Big]_0^\infty + s\int_0^\infty f(t)e^{-st}dt = -f(0) + sF(s) \]

because in the half-plane of convergence f(t)e^{-st}\to 0. It is the key property: differentiation becomes multiplication by s, and the initial conditions enter the equation by themselves. With Fourier you had i\omega; here you have s, plus the f(0) term.

Initial and final value theorems

    \[ f(0^+) = \lim_{s\to\infty}sF(s),\qquad \lim_{t\to\infty}f(t) = \lim_{s\to 0}sF(s) \]

The second holds only if the limit really exists, that is if every pole of sF(s) has a negative real part. They give the start and end of a response without inverting: for example the steady-state output of a system after a step.

Inversion with partial fractions

In practical problems F(s) is almost always a rational function \frac{N(s)}{D(s)}. It is split into partial fractions, and each fraction is inverted with the table.

Example.

    \[ F(s) = \frac{s + 3}{(s + 1)(s + 2)} = \frac{A}{s + 1} + \frac{B}{s + 2} \]

A = \left.\frac{s + 3}{s + 2}\right|_{s = -1} = 2, B = \left.\frac{s + 3}{s + 1}\right|_{s = -2} = -1. So f(t) = 2e^{-t} - e^{-2t}.

Computing A and B is exactly the residue computation of the lesson on the residue theorem: in general the inverse transform is the sum of the residues of F(s)e^{st} at the poles of F. A pair of complex conjugate poles -a\pm i\omega produces a damped oscillation e^{-at}\cos(\omega t + \varphi).

The s-plane and stability

Each pole p = \sigma + i\omega of F(s) contributes to the time response a term e^{pt} = e^{\sigma t}e^{i\omega t}. The position of the pole in the complex plane therefore tells everything about the behaviour of that term.

The s-plane with poles in various positions and the shape of the corresponding time response
Poles in the left half-plane (green) give responses that die away: exponentials if real, damped oscillations if complex. Poles on the imaginary axis give sustained oscillations. Poles in the right half-plane (red) give growing responses: the system is unstable.
  • Negative real part: the term dies away. The further left the pole, the faster.
  • Positive real part: the term grows exponentially. The system is unstable.
  • On the imaginary axis: an oscillation of constant amplitude (or linear growth if the pole is double).
  • The imaginary part gives the frequency of oscillation.

A linear system is stable if and only if every pole of its transfer function has a negative real part. It is the golden rule of control engineering: designing a controller means moving the system’s poles into the right region of the s-plane.

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