The Laplace transform
Fourier leaves out everything that grows. Laplace adds a convergence factor, and with it the step, unstable systems and initial conditions come in. The price is that you always have to say where it converges.
01What for, if we already have Fourier
The Fourier transform requires the signal to be absolutely integrable, and that leaves out such common things as the step or a growing exponential. Laplace’s solution is to multiply by a convergence factor \( e^{-\sigma t} \) before transforming:
That \( \sigma \) is what does the work: if the signal grows, you choose a \( \sigma \) large enough for the product to decay and the integral to converge. The price is that now you have to state for which values of \( \sigma \) it converges, and that is the region of convergence.
| What Laplace adds | Why it matters |
|---|---|
| Transforms growing signals | Makes it possible to analyze unstable systems |
| Handles initial conditions | A charged capacitor enters the calculation |
| Turns the differential equation into an algebraic one | Solving circuits is just solving for the unknown |
| Gives poles and zeros | The whole reading of stability and response |
02The region of convergence, which is not a detail
Here is what is hardest about Laplace and what gets skipped most: \( X(s) \) alone does not determine the signal. The region of convergence is also needed, and with the same algebraic expression and different regions you get completely different signals.
On the left the \( s \)-plane with the poles and the region of convergence shaded; on the right, the corresponding signal in time. Change the region and watch how the signal changes without a single pole moving.
It pays to keep the three rules that summarize all this together:
| If the region… | Then the signal… |
|---|---|
| Is to the right of all the poles | Is causal: it is zero before \( t = 0 \) |
| Is to the left of all of them | Is anticausal: it is zero after \( t = 0 \) |
| Is a strip between poles | Is two-sided: it extends to both sides |
| Contains the \( j\omega \) axis | Has a Fourier transform, and the system is stable |
The last row is the connection between the two transforms, and also the stability criterion: a causal system is stable if and only if all of its poles are to the left of the imaginary axis, because only then does the causal region —which starts at the rightmost pole— manage to cover it.
03The one-sided transform, which is the one used in circuits
For problems with initial conditions, the version that integrates from zero is used:
Its appeal lies in how it transforms the derivative, because that is where the past of the circuit enters:
With that, a circuit is solved as if it were resistive, using generalized impedances:
| Element | Impedance | The initial condition appears as |
|---|---|---|
| Resistor | \( R \) | It has none |
| Capacitor | \( 1/(sC) \) | A voltage source \( v_C(0)/s \) in series |
| Inductor | \( sL \) | A voltage source \( L\,i_L(0) \) in series |
Two shortcuts that avoid inverse transforming when only the endpoints matter: \( x(0^+) = \lim_{s\to\infty} sX(s) \) and \( x(\infty) = \lim_{s\to 0} sX(s) \). The second has a condition that is often forgotten: it holds only if the system is stable. Applied to something that oscillates or grows, it gives a number that means nothing.
04Back to the time domain
For rational functions, which is what comes out of any circuit, the way is partial fractions: you factor the denominator, split into terms of one pole each, and inverse transform each term using the table. And we have already seen that the coefficients of those fractions are the residues, so the two techniques are the same.
| \( X(s) \) | \( x(t) \) for \( t \ge 0 \) |
|---|---|
| \( 1/s \) | \( 1 \), the step |
| \( 1/s^2 \) | \( t \), the ramp |
| \( 1/(s+a) \) | \( e^{-at} \) |
| \( 1/(s+a)^2 \) | \( t\,e^{-at} \) |
| \( \omega/(s^2+\omega^2) \) | \( \operatorname{sin}\omega t \) |
| \( \dfrac{\omega}{(s+a)^2+\omega^2} \) | \( e^{-at}\operatorname{sin}\omega t \) |
| \( e^{-sT}X(s) \) | \( x(t-T) \), the delayed signal |
The fourth row explains where the \( t\,e^{-at} \) of a double pole comes from, and the last one is the one that makes it possible to handle delays: an \( e^{-sT} \) in the transfer function is a dead time, and it is what makes a system with transport delay non-rational.
05In the lab
Set up a series RLC with the capacitor charged to 5 V, write the transformed circuit with the initial-condition sources, solve for \( V_C(s) \) and inverse transform. Compare with the measurement on the oscilloscope.
For \( V(s) = \dfrac{10(s+2)}{s(s^2+3s+2)} \), apply the two theorems and then inverse transform to verify. Repeat with a denominator that has a pole in the right half-plane and explain why the final value theorem fails.
For \( X(s) = \dfrac{1}{(s+1)(s-2)} \), write the three possible signals according to the region of convergence and decide which of the three has a Fourier transform.
06Common mistakes
- Giving \( X(s) \) without the region of convergence. The expression alone does not determine the signal.
- Forgetting the initial condition when transforming a derivative in the one-sided version.
- Applying the final value theorem to an unstable system. It returns a meaningless number.
- Confusing a pole with a zero of the transformed denominator. It is best to always factor.
- Believing that Fourier and Laplace are different things. Fourier is Laplace on the imaginary axis, when that axis is inside the region.
07Self-assessment
What does Laplace add compared with Fourier?
The factor \( e^{-\sigma t} \), which makes it possible to transform growing signals and to handle initial conditions.
Why is the region of convergence needed?
Because the same \( X(s) \) with different regions corresponds to different signals.
When does the Fourier transform exist?
When the region of convergence contains the imaginary axis. For a causal system, that is equivalent to its being stable.
How does a charged capacitor enter the calculation?
As an impedance \( 1/(sC) \) in series with a source \( v_C(0)/s \).
What is \( \mathcal{L}^{-1}\{1/(s+a)^2\} \)?
\( t\,e^{-at} \). That is the term contributed by a double pole.
What does an \( e^{-sT} \) represent in a transfer function?
A pure delay of \( T \) seconds. It is not rational, and that is why it is approximated with a Padé approximant.
08Further reading
- Alan V. Oppenheim and Alan S. Willsky. Signals and Systems. 2nd ed. (Spanish edition, “Señales y sistemas”), Prentice Hall, 1997. Chapter 9, with the careful treatment of the region of convergence that most books omit.
- Charles K. Alexander and Matthew N. O. Sadiku. Fundamentals of Electric Circuits. 7th ed. (Spanish edition, “Fundamentos de circuitos eléctricos”), McGraw-Hill, 2021. Laplace applied to circuits, with the initial-condition sources and many examples.
- Katsuhiko Ogata. Modern Control Engineering. 5th ed. (Spanish edition, “Ingeniería de control moderna”), Prentice Hall, 2010. The table of transforms and the initial and final value theorems, with their conditions of validity.