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9.5 — Laplace and Z Transforms
The Fourier transform assumes a signal oscillates and does not blow up. Real engineering signals often do neither: a circuit switched on at t=0, a motor starting from rest, a system whose response grows because it is unstable.
The Laplace transform handles all of these, and it does something more valuable — it turns solving a differential equation into solving an algebraic one. For an engineer, that is the whole point.
\mathcal{L}\{f(t)\} = F(s) = \int_0^\infty f(t)e^{-st}\,dt
Two differences from Fourier. The integral starts at 0, so it deals with signals that begin at a definite moment. And s = \sigma + i\omega is complex, with a real part \sigma that provides a decaying factor e^{-\sigma t} — which is what lets the integral converge even for signals that grow.
Fourier is the special case where \sigma = 0. Laplace is the general tool and Fourier is the slice of it that concerns pure oscillation.
1. The transforms you need
Compute a few from the definition and then use the table forever.
\mathcal{L}\{1\} = \int_0^\infty e^{-st}dt = \frac1s
\mathcal{L}\{e^{at}\} = \int_0^\infty e^{(a-s)t}dt = \frac{1}{s-a}
The rest follow similarly:
| f(t) | F(s) |
|---|---|
| 1 | 1/s |
| t | 1/s^2 |
| t^n | n!/s^{n+1} |
| e^{at} | 1/(s-a) |
| \sin\omega t | \omega/(s^2+\omega^2) |
| \cos\omega t | s/(s^2+\omega^2) |
| \delta(t) | 1 |
That last one, the Dirac delta, is an idealised instantaneous impulse — infinitely tall, infinitely narrow, total area 1. It is not a function in the ordinary sense and is made rigorous as a distribution, but it behaves exactly as a physicist wants an ideal kick to behave: it transforms to 1, meaning it contains every frequency equally, which is why hitting a system with a sharp impulse reveals its whole response at once.
2. The property that matters
\mathcal{L}\{f'(t)\} = sF(s) - f(0)
\mathcal{L}\{f''(t)\} = s^2F(s)-sf(0)-f'(0)
Differentiation becomes multiplication by s, and the initial conditions are folded in automatically rather than being applied at the end.
That single fact is the whole method. A differential equation in t becomes an algebraic equation in s. Solve the algebra, transform back, done.
3. The method, worked
Solve y'' + 3y' + 2y = 0 with y(0)=1 and y'(0)=0.
Transform every term:
\big[s^2Y - s(1) - 0\big] + 3\big[sY - 1\big]+2Y = 0
Collect and solve for Y — this is ordinary algebra:
Y(s^2+3s+2) = s+3 \quad\Longrightarrow\quad Y = \frac{s+3}{s^2+3s+2} = \frac{s+3}{(s+1)(s+2)}
Split by partial fractions (Chapter 2.2):
Y = \frac{2}{s+1}-\frac{1}{s+2}
Read the table backwards:
y(t) = 2e^{-t}-e^{-2t}
Check: y(0) = 2-1 = 1 ✓, and y'(0) = -2+2 = 0 ✓.
No guessing at solution forms, no separate step for the initial conditions, no case analysis on the discriminant. Compare with Chapter 6.2's method, which required all three. This is why every engineering course teaches it, and why partial fractions — which looked like pointless algebra in Chapter 2.2 — is the technique that gets used.
4. Transfer functions
For a system at rest initially, take the transform of the whole input-output relationship and form the ratio:
H(s) = \frac{\text{output}(s)}{\text{input}(s)}
This is the transfer function, and it is the complete description of a linear system. Everything you need to know is in it.
The output for any input is one multiplication: Y(s) = H(s)X(s). In the time domain that would be Chapter 9.4's convolution, which is much harder work. H(s) is the Laplace transform of the impulse response, and the convolution theorem is why this works.
Systems in series multiply their transfer functions, so a whole signal chain collapses into one expression.
Poles and zeros decide the behaviour
Write H(s) as a ratio of polynomials. Its zeros are where the numerator vanishes; its poles are where the denominator vanishes and the response blows up.
Plot the poles in the complex plane and you can read the system's behaviour without solving anything.
Each pole at s = \sigma+i\omega contributes a term e^{\sigma t}\big(\text{oscillation at }\omega\big) to the response. So:
| Pole location | Behaviour |
|---|---|
| Left half-plane (\sigma \lt 0) | decays — stable |
| Right half-plane (\sigma \gt 0) | grows — unstable |
| On the imaginary axis | oscillates forever, marginally stable |
| Further left | decays faster |
| Larger $ | \omega |
The stability rule in one sentence: all poles must be in the left half-plane.
This is the single most-used fact in control engineering. Designing a controller means placing the closed-loop poles where you want them — far enough left to settle quickly, not so lightly damped that it rings, and never on the right.
And it connects back. The poles are the roots of the characteristic equation from Chapter 6.2, which are the eigenvalues from Chapter 4.5. Three chapters, three vocabularies, one set of numbers.
A pole near the imaginary axis is a resonance (Chapter 6.2) — lightly damped and easily excited. A radio tuner deliberately places poles close to the axis at the desired frequency, so the response there is enormous.
5. Feedback and control
A feedback loop with forward gain G(s) and feedback path H(s) has closed-loop transfer function
T(s) = \frac{G(s)}{1+G(s)H(s)}
The poles of the closed-loop system are where 1+G(s)H(s) = 0, which is generally nowhere near the open-loop poles. Feedback moves the poles, and that is exactly why it is useful and exactly why it is dangerous.
Feedback can stabilise something unstable — a modern fighter aircraft is aerodynamically unstable and only flies because a computer corrects it hundreds of times a second. Feedback can also destabilise something stable, if the loop delays the signal enough that the correction arrives out of phase and reinforces the error rather than opposing it.
That is why a microphone near its own speaker howls. The loop gain exceeds 1 at some frequency where the phase has rotated to reinforce, so that frequency grows without bound until something clips.
A PID controller combines three terms: proportional to the error, to its integral, and to its derivative. In the Laplace domain:
C(s) = K_p + \frac{K_i}{s}+K_ds
Each term is one of this Part's operations. The integral term (1/s) removes steady-state error by accumulating it. The derivative term (s) anticipates and damps, because it responds to how fast the error is changing. Volume III, 6.6 covers the tuning.
6. The Z transform: the discrete sibling
For sequences rather than continuous signals:
X(z) = \sum_{n=0}^{\infty}x[n]z^{-n}
The same theory, one step down. Where Laplace has s, the Z transform has z; where a derivative became multiplication by s, a one-step delay becomes multiplication by z^{-1}.
The stability region changes shape. For Laplace it is the left half-plane; for the Z transform it is the inside of the unit circle, |z| \lt 1. The two are related by z = e^{sT} with T the sampling period, and that exponential map sends the left half-plane to the interior of the unit circle.
Every digital filter is designed with this. An FIR filter has only zeros and is always stable. An IIR filter has poles, is far more efficient for the same sharpness, and can be unstable if a pole drifts outside the unit circle — including through rounding error in fixed-point arithmetic, which is a real and famous class of bug in embedded audio code.
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.
The Laplace transform
F(s) = \mathcal{L}\{f(t)\} = \int_0^\infty f(t)e^{-st}\,dt
How it differs from Fourier, and why. It integrates from 0 rather than -\infty, which suits problems with a starting moment. And s is complex, s = \sigma+i\omega, so the e^{-\sigma t} factor damps the integrand — meaning the transform exists for growing functions that have no Fourier transform at all.
The table
| f(t) | F(s) |
|---|---|
| 1 | \frac1s |
| t^n | \frac{n!}{s^{n+1}} |
| e^{at} | \frac{1}{s-a} |
| \sin\omega t | \frac{\omega}{s^2+\omega^2} |
| \cos\omega t | \frac{s}{s^2+\omega^2} |
| e^{at}f(t) | F(s-a) |
| f(t-a)u(t-a) | e^{-as}F(s) |
| \delta(t) | 1 |
Derivation of the first two, so the pattern is clear.
\mathcal{L}\{1\} = \int_0^\infty e^{-st}dt = \left[\frac{e^{-st}}{-s}\right]_0^\infty = 0 - \left(-\frac1s\right) = \frac1s
valid when s>0, so the exponential decays.
\mathcal{L}\{e^{at}\} = \int_0^\infty e^{at}e^{-st}dt = \int_0^\infty e^{-(s-a)t}dt = \frac{1}{s-a}
valid when s>a. The pole at s=a is the transform remembering the growth rate, and reading off pole positions is how engineers judge stability at a glance.
The property that solves differential equations
\mathcal{L}\{f'\} = sF(s)-f(0), \qquad \mathcal{L}\{f''\} = s^2F(s)-sf(0)-f'(0)
Where the first comes from. Integrate by parts:
\int_0^\infty f'(t)e^{-st}dt = \left[f(t)e^{-st}\right]_0^\infty + s\int_0^\infty f(t)e^{-st}dt = -f(0)+sF(s)
What this buys. A differential equation becomes an algebraic equation in s, with the initial conditions already built in — no arbitrary constants to chase at the end. Solve for F(s) by ordinary algebra, split it with partial fractions from 2.2 — polynomials, and read each piece back off the table. The whole method is: transform, do algebra, transform back.
\text{Transfer function}: \quad H(s) = \frac{\text{output transform}}{\text{input transform}}
The system's entire behaviour, in one function of s. A system is stable exactly when every pole of H(s) has a negative real part, because a pole at s=a contributes e^{at}, which grows if a>0.
The Z transform
X(z) = \sum_{n=0}^{\infty}x_nz^{-n}
The discrete counterpart of Laplace: for sequences instead of functions.
| x_n | X(z) |
|---|---|
| \delta_n | 1 |
| 1 (for n\ge0) | \frac{z}{z-1} |
| a^n | \frac{z}{z-a} |
| x_{n-k} | z^{-k}X(z) |
The delay property is the whole point. Shifting a sequence back one step multiplies its transform by z^{-1}, so z^{-1} is the delay operator. A difference equation therefore becomes an algebraic one, exactly as a differential equation did under Laplace.
Stability, in the discrete world, means all poles lie inside the unit circle |z|=1 — because a pole at z=a contributes a^n, which grows if |a|>1. The relationship to Laplace is z = e^{sT} with T the sampling interval, and that mapping takes the left half of the s plane to the inside of the unit circle, which is why the two stability conditions look different and mean the same thing.
7. Where this shows up in your life
Every circuit an engineer analyses. Impedance is defined in the s domain, and an RLC network becomes algebra. Volume III, Part 1.
Every control system. Cruise control, thermostats, drone stabilisation, industrial process control, robot joints, hard-drive head positioning.
Every aircraft. Flight control laws are designed by pole placement.
Every digital audio effect and every software-defined radio. Z-transform filter design.
Every hearing aid and every active noise-cancelling system.
Every discussion of whether a system is stable — in engineering, and by extension in economics and ecology, where the same eigenvalue criterion is used.
Part 9 is finished. The mathematics of this volume is now essentially complete. What remains is the question of what a computer actually does with it, where the answers go wrong, and how a machine finds the best solution when no formula exists.
More places these turn up
Your voice reaches a phone as a waveform and travels as a set of coefficients, because the codec keeps the frequencies you can hear and discards the rest — that is the Fourier transform doing lossy compression. A JPEG does the same thing to an image in two dimensions. Noise-cancelling headphones convolve the outside sound with a filter and add the result back inverted. An MRI machine measures the Fourier transform of your body directly and computes the image by transforming back. Every equaliser slider, every reverb effect, every autotuned vocal, every 4G connection and every seismograph reading is one of the formulas on this page.
Next: 9.P — Worked Problems computes Fourier coefficients from scratch, convolves two signals by hand, and solves a differential equation with Laplace, step by step.