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4.5 — The Laplace Transform
Chapter 1.5 solved an RC circuit by writing a differential equation, separating variables and integrating. That took half a page for the simplest circuit there is. An RLC circuit needs a second-order equation. A three-stage filter needs a sixth-order one, and nobody solves those by hand.
The Laplace transform turns every one of them into algebra. Differentiation becomes multiplication by s, integration becomes division by s, and a differential equation becomes a polynomial you can rearrange. Initial conditions come along for free, built into the transform rather than fitted afterwards.
The letters this chapter uses, said out loud
s — "ess", the complex frequency, and the one symbol the whole chapter turns on. It is a complex number, s = \sigma + j\omega, holding two things at once: how fast something decays and how fast it oscillates. Fourier's j\omega handles only oscillation, and Laplace's s adds decay to it, which is exactly why Laplace can handle a signal that grows or dies while Fourier cannot.
\sigma — "sigma", the real part of s, in reciprocal seconds. Negative sigma means decay, positive means growth, zero means neither. Section 3's stability test is a statement about the sign of this one number.
\omega — "omega", the imaginary part of s, the ringing frequency in radians per second. Same meaning as in Chapter 4.4.
F(s) or X(s) — the transform of a signal: a function of complex frequency rather than of time. Capital for the transform, lower case for the original, as in Chapter 4.4.
H(s) — the transfer function: the transform of a system's impulse response, and therefore the thing that describes what the system does to any input at all. It is the central object of Part 6.
Poles and zeros — a pole is a value of s where H(s) becomes infinite, because the denominator hits zero; a zero is where H(s) becomes zero. They are marked ✕ and ○ on the s-plane figure below, and between them they determine everything about the system's behaviour.
\mathcal{L}\{\cdot\} — script L, read "the Laplace transform of". The curly brackets hold whatever is being transformed.
Why the letter s? It is generally taken from the German Spektrum or simply as the successor to the earlier operational notation of Oliver Heaviside, who was doing this with an operator p in the 1880s before Laplace's much older integral was recognised as the rigorous version of what he was doing.
1. The definition, and why the extra factor
\boxed{X(s)=\int_0^{\infty}x(t)e^{-st}\,dt}, \qquad s=\sigma+j\omega
Read aloud: "X of s equals the integral from zero to infinity of x of t times e to the minus s t, dee t."
Compare with the Fourier transform, which had e^{-j\omega t}. Laplace has e^{-st} = e^{-\sigma t}e^{-j\omega t} — the same probe, multiplied by a decaying exponential.
That extra factor does two things.
It forces convergence. A step function's Fourier transform needed an impulse to exist at all. Its Laplace transform is simply 1/s, because e^{-\sigma t} pulls the integral down to something finite. Even a growing exponential e^{2t} has a Laplace transform, provided \sigma>2.
It starts at zero. The lower limit is 0, not -\infty, which is exactly right for engineering: something is switched on at t=0 and you want to know what happens next. This is the one-sided or unilateral Laplace transform, and it is the only one used in practice.
The set of s values for which the integral converges is the region of convergence. For a right-sided signal it is always a half-plane \sigma>\sigma_0 to the right of the rightmost pole, and knowing that is enough — the region rarely needs to be computed explicitly in engineering work.
The Fourier transform is the Laplace transform on the imaginary axis, s=j\omega, provided that axis lies inside the region of convergence. So Laplace does not replace Fourier; it contains it.
2. Deriving the essential pairs
Rather than presenting a table, here is where the entries come from.
The step, x(t)=u(t):
X(s)=\int_0^\infty 1\cdot e^{-st}dt=\left[\frac{e^{-st}}{-s}\right]_0^\infty = 0-\left(\frac{1}{-s}\right)=\frac1s
The upper limit vanishes provided \sigma>0, which is the region of convergence.
The impulse, \delta(t):
X(s)=\int_0^\infty\delta(t)e^{-st}dt = e^{-s\cdot0}=1
by the sifting property. Flat in s, exactly as it was flat in \omega — the impulse contains everything.
The exponential, e^{-at}u(t):
X(s)=\int_0^\infty e^{-at}e^{-st}dt=\int_0^\infty e^{-(s+a)t}dt=\frac{1}{s+a}
This is the single most important pair in the subject. Every first-order system produces it, and reading it backwards — seeing \frac{1}{s+a} and thinking "e^{-at}, decaying with time constant 1/a" — is the core skill of the whole chapter.
The ramp, t\,u(t): integrate by parts, or use the general result t^n \leftrightarrow n!/s^{n+1}, giving 1/s^2.
The sinusoid, \cos(\omega_0t)u(t): write it as \tfrac12(e^{j\omega_0t}+e^{-j\omega_0t}) and use the exponential pair twice:
X(s)=\frac12\left(\frac{1}{s-j\omega_0}+\frac{1}{s+j\omega_0}\right)=\frac12\cdot\frac{2s}{s^2+\omega_0^2}=\frac{s}{s^2+\omega_0^2}
Similarly \sin(\omega_0 t) \leftrightarrow \dfrac{\omega_0}{s^2+\omega_0^2}.
The decaying sinusoid follows by shifting: e^{-at}\cos(\omega_0t) \leftrightarrow \dfrac{s+a}{(s+a)^2+\omega_0^2}. Notice the pattern — replacing s by s+a multiplies the time function by e^{-at}. That is the frequency-shift property and it saves an enormous amount of work.
The s-plane, and reading a system off it
Every result in this chapter turns into a picture, and the picture is a single flat plane with a handful of marks on it. Learning to read it replaces most of the algebra.
What the two coordinates mean, since s = \sigma + j\omega has two parts and each controls one thing.
The real part \sigma, measured left–right, is the decay rate. A pole at s = -a contributes a term e^{-at} to the response. Further left means a larger a means a faster die-away. A pole on the vertical axis has \sigma = 0, so e^{0} = 1 and it never decays. A pole to the right has positive \sigma, so the term is e^{+at} and it grows without limit.
That single sentence is the stability test. A system is stable exactly when every pole is strictly in the left half. Nothing else about the poles matters for the yes-or-no question, which is why Chapter 6.4 spends its time proving where the poles are rather than computing them.
The imaginary part \omega, measured up–down, is the ringing frequency. A pole at s = -a + jb contributes e^{-at}\cos(bt) — a decaying oscillation of frequency b. Poles always come in mirror-image pairs above and below the axis when the system is made of real components, because a real system cannot produce a complex response.
Combine the two readings and the picture tells the whole story at a glance: far left and close to the axis is a fast, non-ringing response; near the axis and high up is slow and oscillatory; anywhere on the right is a system that will destroy itself.
3. The properties that do the real work
Differentiation — the whole point
\mathcal{L}\left\{\frac{dx}{dt}\right\}=sX(s)-x(0^-)
Derivation by parts. With u=e^{-st} and dv = x'dt:
\int_0^\infty x'(t)e^{-st}dt = \left[x(t)e^{-st}\right]_0^\infty + s\int_0^\infty x(t)e^{-st}dt = -x(0^-)+sX(s)
Read the result carefully: differentiation became multiplication by s, and the initial condition appeared automatically as a subtracted constant. No separate step is needed to apply initial conditions — they are already in the algebra.
Second derivative:
\mathcal{L}\left\{\frac{d^2x}{dt^2}\right\}=s^2X(s)-sx(0^-)-x'(0^-)
Integration
\mathcal{L}\left\{\int_0^tx(\tau)d\tau\right\}=\frac{X(s)}{s}
Integration divides by s, differentiation multiplies. So s behaves exactly like the operator d/dt, which is not a coincidence — Oliver Heaviside had been using an operator calculus that did precisely this since the 1880s, without proof, to enormous criticism. Laplace's transform, applied later, justified everything Heaviside had been doing.
Convolution
x(t)*h(t) \;\longleftrightarrow\; X(s)H(s)
Same theorem as Fourier's, same proof. This is what makes transfer functions multiply.
The value theorems
x(0^+)=\lim_{s\to\infty}sX(s) \qquad\text{(initial value)}
x(\infty)=\lim_{s\to0}sX(s) \qquad\text{(final value)}
These let you read a system's starting point and settling point straight off the transform without inverting it, which is enormously useful when all you want is the steady-state error of a control loop.
The final value theorem has a trap: it is valid only if the system actually settles. Apply it to an oscillator or an unstable system and it returns a confident, wrong number. Check that all poles are in the left half-plane before using it.
4. Impedances in the s-domain
Chapter 1.6 replaced components with impedances at a single frequency. Laplace does the same thing but for all time behaviour at once.
| Component | Time relation | s-domain impedance |
|---|---|---|
| Resistor | v=iR | R |
| Inductor | v=L\,di/dt | sL |
| Capacitor | i=C\,dv/dt | 1/sC |
Compare with Chapter 1.6's j\omega L and 1/j\omega C — identical with s in place of j\omega. Phasor analysis was Laplace analysis restricted to the imaginary axis all along.
And the initial conditions appear as sources. A capacitor with initial voltage V_0 is 1/sC in series with a voltage source V_0/s. An inductor with initial current I_0 is sL in series with a source LI_0. So you draw the circuit once, in the s-domain, with everything included, and solve it with the ordinary techniques of Chapter 1.3.
Worked example — the RC circuit again
A step of V volts applied at t=0 to an RC series circuit, capacitor initially uncharged.
In the s-domain, it is a voltage divider:
V_C(s)=\frac{V}{s}\cdot\frac{1/sC}{R+1/sC}
Multiply top and bottom by sC:
=\frac{V}{s}\cdot\frac{1}{sRC+1}=\frac{V}{s(sRC+1)}=\frac{V/RC}{s(s+1/RC)}
Now invert. Partial fractions with a=1/RC:
\frac{Va}{s(s+a)}=\frac{A}{s}+\frac{B}{s+a}
Multiply through by s and set s=0: A = Va/a = V. Multiply by (s+a) and set s=-a: B = Va/(-a) = -V.
V_C(s)=\frac{V}{s}-\frac{V}{s+a}
Read each term from the pairs in section 2:
v_C(t)=V-Ve^{-at}=V\left(1-e^{-t/RC}\right)
The same answer as Chapter 1.5, obtained by algebra with no integration. For a first-order circuit that is a fair trade at best. For a fourth-order filter it is the difference between possible and impossible.
5. The transfer function
For a system with zero initial conditions:
\boxed{H(s)=\frac{Y(s)}{X(s)}}
The transfer function is the system, completely. It is also the Laplace transform of the impulse response, since an impulse has X(s)=1.
It always comes out as a ratio of polynomials:
H(s)=\frac{b_ms^m+\cdots+b_1s+b_0}{a_ns^n+\cdots+a_1s+a_0}=K\frac{(s-z_1)(s-z_2)\cdots}{(s-p_1)(s-p_2)\cdots}
- The roots of the numerator are the zeros. At those s values the output is zero — the system blocks that mode entirely.
- The roots of the denominator are the poles. At those s values the output is infinite for a finite input, which means the system produces that mode on its own.
The poles are the system's natural behaviour, and they are the only thing that matters for stability.
6. Reading the pole positions
Each pole p=\sigma+j\omega contributes a term e^{pt}=e^{\sigma t}e^{j\omega t} to the response. So the real part sets growth or decay, and the imaginary part sets oscillation.
| Pole location | Time response |
|---|---|
| Negative real | decaying exponential |
| Positive real | growing exponential |
| Complex pair, left half | decaying oscillation |
| Complex pair, right half | growing oscillation |
| On the imaginary axis | sustained oscillation |
| At the origin | a constant, or a ramp if repeated |
\boxed{\text{A system is stable if and only if every pole has a negative real part.}}
And the geometry is quantitative, not merely qualitative. For a complex pair at \sigma\pm j\omega_d:
- Further left means faster settling. Settling time is roughly 4/|\sigma|, since e^{-|\sigma|t} reaches 2% after four time constants.
- Higher up means faster ringing. The ringing frequency is \omega_d.
- The angle from the negative real axis sets the damping. A pole on the negative real axis is critically damped or slower; a pole close to the imaginary axis rings badly. Specifically \zeta = \cos\theta where \theta is measured from the negative real axis.
This is why control engineers place poles rather than tune gains blindly. Part 6 is built on it.
7. The standard second-order system
H(s)=\frac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}
Poles at
s=-\zeta\omega_n\pm\omega_n\sqrt{\zeta^2-1}
This form appears everywhere — the RLC circuit of Chapter 1.5, a mass on a spring, a motor with inertia, a control loop, the suspension of a car. Learning to recognise it means recognising the behaviour of half the systems in engineering.
| \zeta | Poles | Behaviour |
|---|---|---|
| 0 | on the imaginary axis | rings forever |
| 0\lt\zeta\lt1 | complex, left half | decaying ringing |
| 1 | two equal, real | critically damped, fastest with no overshoot |
| \gt1 | two distinct, real | overdamped, sluggish |
Percentage overshoot depends on \zeta alone:
\%OS = 100\,e^{-\pi\zeta/\sqrt{1-\zeta^2}}
\zeta=0.5 gives 16.3%; \zeta=0.707 gives 4.3%; \zeta=0.8 gives 1.5%.
\zeta = 0.707 is the usual design target because it gives the flattest frequency response with only a few percent of overshoot, and Chapter 6.3 works through why.
8. Inverting by partial fractions
The one procedure you need, in three cases.
Distinct real poles. Write
H(s)=\frac{A}{s-p_1}+\frac{B}{s-p_2}+\cdots
and find each coefficient by the cover-up rule: multiply both sides by (s-p_1) and set s=p_1. Every other term vanishes because it still has (s-p_1) in it.
Worked example.
H(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}=\frac{2}{1}=2, \qquad B=\left.\frac{s+3}{s+1}\right|_{s=-2}=\frac{1}{-1}=-1
h(t)=2e^{-t}-e^{-2t}
Check at t=0: h(0)=2-1=1. And by the initial value theorem, \lim_{s\to\infty}sH(s) = \lim s\cdot\frac{s}{s^2}=1 ✓.
Repeated poles. A pole of multiplicity m needs m terms:
\frac{A_1}{s-p}+\frac{A_2}{(s-p)^2}+\cdots
and (s-p)^{-2} inverts to t\,e^{pt}. The factor of t is the signature of a repeated pole, and it is why a critically damped system's response contains te^{-\omega_nt}.
Complex poles. They always come in conjugate pairs for a real system, and rather than handling them separately, complete the square in the denominator and match against the decaying-sinusoid pairs:
\frac{s+a}{(s+a)^2+\omega^2} \to e^{-at}\cos\omega t, \qquad \frac{\omega}{(s+a)^2+\omega^2}\to e^{-at}\sin\omega t
9. Frequency response from the transfer function
Set s=j\omega and you have the frequency response:
H(j\omega)=|H(j\omega)|\,e^{j\angle H(j\omega)}
Magnitude is the gain at that frequency; angle is the phase shift. This is the bridge back to Chapter 1.6 and forward to Chapter 6.5's Bode plots.
The geometric method is worth knowing because it makes plots readable by eye. For each frequency, mark the point j\omega on the imaginary axis. Then:
|H(j\omega)| = K\cdot\frac{\prod\text{distances from } j\omega \text{ to each zero}}{\prod\text{distances from } j\omega \text{ to each pole}}
\angle H(j\omega)=\sum\text{angles from zeros}-\sum\text{angles from poles}
So as \omega sweeps up the imaginary axis and passes close to a pole, that distance becomes small and the gain peaks. A pole near the axis gives a sharp resonant peak; a pole far to the left gives a gentle roll-off. Passing close to a zero makes the gain dip.
That single picture explains why a notch filter needs a zero on the imaginary axis at the frequency to be removed, and why a high-Q resonator needs poles hugging the axis — which is Chapter 1.6's resonance restated in the language that generalises.
Laplace handles continuous time. Chapter 4.6 does the same job for sampled signals, where a differential equation becomes a difference equation and the s-plane becomes a disc.
Every formula above, built from scratch
None of the results in this chapter are worth memorising, because each one can be rebuilt in under a minute from something simpler. What follows is that rebuilding, one result at a time, so the formula and the reason for it sit on the same page as the explanation that needed them.
Laplace transform
X(s)=\int_0^\infty x(t)e^{-st}dt, \qquad s=\sigma+j\omega
The Fourier transform is the special case s=j\omega, when that axis lies in the region of convergence.
Pairs
| x(t) | X(s) |
|---|---|
| \delta(t) | 1 |
| u(t) | 1/s |
| t\,u(t) | 1/s^2 |
| t^n u(t) | n!/s^{n+1} |
| e^{-at}u(t) | \dfrac{1}{s+a} |
| t e^{-at}u(t) | \dfrac{1}{(s+a)^2} |
| \cos\omega_0t | \dfrac{s}{s^2+\omega_0^2} |
| \sin\omega_0t | \dfrac{\omega_0}{s^2+\omega_0^2} |
| e^{-at}\cos\omega_0t | \dfrac{s+a}{(s+a)^2+\omega_0^2} |
The pattern in the last row: replacing s by s+a multiplies the time function by e^{-at}. That one rule generates half the table.
Properties
\mathcal{L}\{x'\}=sX(s)-x(0^-)
Derivation by parts: \int_0^\infty x'e^{-st}dt = [xe^{-st}]_0^\infty+s\int_0^\infty xe^{-st}dt = -x(0^-)+sX(s).
\mathcal{L}\{x''\}=s^2X-sx(0^-)-x'(0^-)
\mathcal{L}\left\{\int_0^tx\right\}=\frac{X(s)}{s}
x(t-t_0)u(t-t_0)\leftrightarrow e^{-st_0}X(s)
e^{-at}x(t)\leftrightarrow X(s+a)
x*h \leftrightarrow XH
x(0^+)=\lim_{s\to\infty}sX(s), \qquad x(\infty)=\lim_{s\to0}sX(s)
The final value theorem is valid only when all poles are strictly in the left half-plane (or a single pole at the origin).
Circuit impedances
Z_R=R, \qquad Z_L=sL, \qquad Z_C=\frac{1}{sC}
Initial conditions as sources: a capacitor with V_0 becomes 1/sC in series with V_0/s; an inductor with I_0 becomes sL in series with LI_0.
Transfer function and stability
H(s)=\frac{Y(s)}{X(s)}=K\frac{\prod(s-z_i)}{\prod(s-p_i)}
Stable if and only if every \text{Re}(p_i)\lt0.
Second-order standard form:
H(s)=\frac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}, \qquad s=-\zeta\omega_n\pm\omega_n\sqrt{\zeta^2-1}
\%OS = 100\,e^{-\pi\zeta/\sqrt{1-\zeta^2}}, \qquad t_s\approx\frac{4}{\zeta\omega_n}, \qquad \omega_d=\omega_n\sqrt{1-\zeta^2}
Geometric frequency response:
|H(j\omega)|=K\frac{\prod|\text{distance to each zero}|}{\prod|\text{distance to each pole}|}
What the next chapter fixes
Laplace works on continuous signals, and a processor cannot hold one. It holds numbers, taken at intervals. Chapter 4.6 develops the transform built for exactly that case, and shows that the s-plane's left-half stability rule becomes a rule about the inside of a circle instead.