The Laplace Transform and Transfer Functions
The differential equations of Chapter 2 are exact but awkward — solving them by hand is slow, and combining systems is harder still. The Laplace transform turns calculus into algebra: derivatives become multiplications by \(s\), and an entire system collapses into a single ratio of polynomials called the transfer function. This is the language the rest of the course is spoken in.
- The definition of the Laplace transform and the standard transform pairs every controls engineer memorises.
- The properties — linearity, differentiation, integration, time and frequency shift — that turn ODEs into algebra.
- The initial- and final-value theorems, which read off \(t\to0\) and \(t\to\infty\) behaviour without inverting.
- How to recover \(f(t)\) by partial-fraction expansion and the transform table.
- The transfer function \(G(s)=C(s)/R(s)\) — system dynamics captured as one ratio of polynomials.
- Poles, zeros, and the s-plane: how the roots of that ratio foretell a system's behaviour.
Why Transform at All?
A linear system is governed by a differential equation in time. Working in the time domain means solving that equation for every new input, and connecting two systems means substituting one messy solution into another. The Laplace transform sidesteps all of it by moving to a complex frequency variable \(s = \sigma + j\omega\). In this domain, differentiation becomes multiplication, integration becomes division, and convolution becomes ordinary multiplication.
The payoff is enormous: an \(n\)-th order differential equation becomes an algebraic equation in \(s\), initial conditions fold in automatically, and series, parallel, and feedback connections reduce to algebra on ratios of polynomials. We solve in the \(s\)-domain and transform the answer back to time only at the very end.
Transform, solve algebraically, invert. Every technique in the next twenty chapters lives in the middle step, where the hard calculus has already become easy algebra.
Definition & Standard Transform Pairs
The (one-sided) Laplace transform of a function \(f(t)\) defined for \(t \ge 0\) is the integral
In practice nobody evaluates this integral repeatedly — a short table of pairs covers almost every signal in control. These are worth committing to memory:
| \(f(t),\ t\ge 0\) | \(F(s)\) | Notes |
|---|---|---|
| \(\delta(t)\) | \(1\) | Unit impulse |
| \(u(t)\) (unit step) | \(\dfrac{1}{s}\) | Constant input |
| \(t\) | \(\dfrac{1}{s^2}\) | Unit ramp |
| \(\dfrac{t^{n}}{n!}\) | \(\dfrac{1}{s^{n+1}}\) | Polynomial |
| \(e^{-at}\) | \(\dfrac{1}{s+a}\) | Exponential decay |
| \(\sin\omega t\) | \(\dfrac{\omega}{s^2+\omega^2}\) | Oscillation |
| \(\cos\omega t\) | \(\dfrac{s}{s^2+\omega^2}\) | Oscillation |
| \(e^{-at}\sin\omega t\) | \(\dfrac{\omega}{(s+a)^2+\omega^2}\) | Damped oscillation |
| \(e^{-at}\cos\omega t\) | \(\dfrac{s+a}{(s+a)^2+\omega^2}\) | Damped oscillation |
The Properties That Do the Work
The transform's power comes from a handful of properties. Linearity lets us transform term by term. The differentiation property is the one that dissolves an ODE: each time derivative becomes a factor of \(s\), with the initial condition appearing automatically.
| Property | Time domain | s-domain |
|---|---|---|
| Linearity | \(af(t)+bg(t)\) | \(aF(s)+bG(s)\) |
| Integration | \(\int_0^t f\,d\tau\) | \(\dfrac{F(s)}{s}\) |
| Time shift (delay) | \(f(t-T)u(t-T)\) | \(e^{-Ts}F(s)\) |
| Frequency shift | \(e^{-at}f(t)\) | \(F(s+a)\) |
| Scaling | \(f(at)\) | \(\tfrac{1}{a}F\!\left(\tfrac{s}{a}\right)\) |
| Convolution | \(f(t)*g(t)\) | \(F(s)\,G(s)\) |
Initial- & Final-Value Theorems
Often we want only the start or the end of a response, not the whole curve. Two theorems read these off directly from \(F(s)\), no inversion required:
If \(sF(s)\) has poles on the imaginary axis or in the right half-plane, the limit \(t\to\infty\) does not exist (the signal oscillates or grows) and the theorem gives a meaningless number. We will lean on the final-value theorem in Chapter 9 to compute steady-state error — but only after checking stability.
The Inverse Transform by Partial Fractions
To return to the time domain we break \(F(s)\) into a sum of simple terms whose inverses sit in the table. This is partial-fraction expansion. For a ratio with distinct poles \(p_1,p_2,\dots\), write
Each residue is found by covering up its own factor and evaluating at the pole — the cover-up method:
Repeated poles need extra terms \(\dfrac{r}{(s-p)^2}\) and so on; complex-conjugate poles are best grouped into the damped-sine/cosine pairs from the table. Once expanded, the inverse is read off term by term.
The Transfer Function
Apply the differentiation property to a system's ODE with all initial conditions zero, and the equation rearranges into a single ratio: output over input. That ratio is the transfer function.
It depends only on the system, not on the input — a complete fingerprint of the dynamics. The denominator set to zero is the characteristic equation, whose roots decide stability (Part 3). For a physical system the order satisfies \(m \le n\).
For the mass–spring–damper of Chapter 2, transforming \(M\ddot x + B\dot x + Kx = f\) with zero initial conditions gives \((Ms^2+Bs+K)X(s)=F(s)\), hence the transfer function
The system's response to any input is now just \(C(s) = G(s)R(s)\). The impulse response is the special case \(R(s)=1\): the inverse transform of \(G(s)\) itself.
Poles, Zeros & the s-Plane
The roots of the numerator are the zeros; the roots of the denominator are the poles. Plotting them on the complex \(s\)-plane (real axis \(\sigma\), imaginary axis \(j\omega\)) gives an instant read on behaviour. Each pole contributes a term \(e^{p t}\) to the response, so the location of a pole is its destiny:
| Pole location | Term in response | Behaviour |
|---|---|---|
| Negative real (LHP) | \(e^{-at}\) | Decays — stable |
| Positive real (RHP) | \(e^{+at}\) | Grows — unstable |
| Complex pair, LHP | \(e^{-\sigma t}\sin\omega t\) | Damped oscillation |
| On the \(j\omega\)-axis | \(\sin\omega t\) | Sustained oscillation |
| At the origin | constant / ramp | Marginal (integrator) |
Worked Examples
Problem. Find \(\mathcal{L}\{3e^{-2t}\}\).
Solution. By linearity and the exponential pair:
Problem. Solve \(\dot y + 2y = 0\), \(y(0)=5\).
Solution. Transform using the differentiation property:
The initial condition entered the algebra automatically — no constants of integration to chase.
Problem. Find the inverse of \(F(s)=\dfrac{s+3}{(s+1)(s+2)}\).
Solution. Cover-up gives the residues:
Problem. A system's output is \(C(s)=\dfrac{10}{s(s+5)}\). Find its steady-state value.
Solution. The poles of \(sC(s)=10/(s+5)\) lie in the LHP, so the theorem applies:
Problem. Find \(C(s)/R(s)\) for \(\ddot c + 5\dot c + 6c = 2\dot r + r\), zero initial conditions.
Solution. Transform each term and collect:
Poles at \(-2,-3\); a zero at \(-\tfrac12\). All poles in the LHP, so the system is stable.
Problem. Locate the poles and zeros of \(G(s)=\dfrac{4(s+2)}{s^2+2s+5}\) and describe the response shape.
Solution. The numerator gives a zero at \(s=-2\); complete the square in the denominator:
A complex-conjugate pair in the LHP with \(\sigma=1\), \(\omega=2\): the response is a damped oscillation, decaying as \(e^{-t}\) while ringing at 2 rad/s.
Chapter Summary
Transform the ODE to algebra in \(s\), solve, then invert. Calculus becomes arithmetic.
\(\mathcal{L}\{\dot f\}=sF(s)-f(0)\); each derivative is a factor of \(s\), initial conditions included free.
\(f(0^+)=\lim_{s\to\infty}sF\); \(f(\infty)=\lim_{s\to0}sF\) — the latter only if \(sF\) is LHP-stable.
Split \(F(s)\) into table-ready terms; residues by cover-up; invert term by term.
\(G(s)=C/R\) at zero initial conditions — a ratio of polynomials that fingerprints the system.
LHP poles decay, RHP poles grow, complex pairs oscillate. The s-plane shows behaviour at a glance.
Problems
Use the transform table and properties rather than the defining integral. For inverses, expand into partial fractions first. Difficulty rises down the list.
- Find \(\mathcal{L}\{4 + 2t - e^{-3t}\}\).
- Find \(\mathcal{L}\{e^{-t}\cos 3t\}\) using the frequency-shift property.
- Solve \(\dot y + 3y = 6\), \(y(0)=0\), and identify the steady-state value.
- Solve \(\ddot y + 4y = 0\) with \(y(0)=1\), \(\dot y(0)=0\), and describe the motion.
- Invert \(F(s)=\dfrac{6}{s(s+2)(s+3)}\) by partial fractions.
- Invert \(F(s)=\dfrac{s+1}{(s+2)^2}\) (note the repeated pole).
- Use the final-value theorem on \(C(s)=\dfrac{5(s+2)}{s(s^2+3s+2)}\); first verify it is valid.
- Write the transfer function of \(\dddot c + 6\ddot c + 11\dot c + 6c = r\) and find its poles.
- For \(M=1,B=3,K=2\), write the mass–spring–damper transfer function and locate its poles.
- A transfer function is \(G(s)=\dfrac{10}{s^2+2s+10}\). Find \(\sigma\) and \(\omega\) of the pole pair and state the decay rate and ring frequency.
- Explain, using the final-value theorem, why \(C(s)=\dfrac{1}{s^2+1}\) has no steady-state value.
- A delayed step \(r(t)=u(t-2)\) drives \(G(s)=\dfrac{1}{s+1}\). Write \(C(s)\) including the delay term and invert it.