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Controls and Systems · Chapter 2 of 10 · Intermediate

Laplace Transforms and Transfer Functions

The Laplace transform turns a differential equation into algebra. The result is the transfer function, and its poles, a few points in the complex plane, tell you almost everything about how the system behaves.

Readiness check

This chapter trades calculus for algebra. Tick only what you can do closed-notes.

Write a mass-spring-damper differential equation.
Recall the Laplace transform of a derivative.
Do partial-fraction expansion of a rational function.
Factor a quadratic to find its roots.
Plot a point in the complex plane.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsReview the Laplace transform in Math: Laplace.

3 or more weak itemsRevisit differential equations in Math: ODEs.

The core idea

Transforming a differential equation with zero initial conditions turns derivatives into powers of s. The output over input is the transfer function, and its poles set the response.

L{f′(t)} = sF(s) − f(0)G(s) = Y(s)/U(s)poles = roots of the denominator

Under the Laplace transform, each time derivative becomes a multiplication by s, so a linear differential equation becomes a polynomial equation. Solving for output over input gives the transfer function G(s), a ratio of polynomials. Its zeros are where the numerator vanishes and its poles are where the denominator vanishes; the poles, plotted in the complex plane, govern stability and speed. A pole in the left half-plane decays; one in the right half-plane grows. Almost all of control theory is reading and moving those poles.

The skill works when: you transform with zero initial conditions, solve for Y/U, and read the poles.

The skill breaks down when: initial-condition terms are kept in a transfer function, or numerator and denominator roots are confused.

The concept. Poles of a transfer function plotted in the s-plane. Both poles of 1/((s + 1)(s + 2)) lie in the left half-plane, so the system is stable; the more negative pole decays faster.

The skills, taught in order

Five skills cover the transform, turning an equation into a transfer function, reading poles and zeros, and the final value theorem.

2.1 The Laplace transform

The transform maps a function of time to a function of the complex variable s. A short table of pairs covers most control work, and the key property is that differentiation becomes multiplication by s (with the initial value subtracted).

f(t)	F(s)
unit impulse δ(t)	1
unit step u(t)	1/s
ramp t	1/s²
e^−at	1/(s + a)
sin ωt	ω/(s² + ω²)

2.2 From differential equation to transfer function

Transform each term of a linear differential equation with zero initial conditions, collect Y(s) and U(s), and form the ratio G(s) = Y(s)/U(s). The orders of the polynomials match the orders of the derivatives, so a second-order system gives a second-order denominator.

2.3 The transfer function

G(s) is a ratio of polynomials that fully describes a linear time-invariant system's input-output behaviour. It hides the internal details and exposes exactly what matters for response: the locations of its roots.

2.4 Poles and zeros

Zeros are the roots of the numerator; poles are the roots of the denominator. Poles set the natural modes: a real pole at −a gives a decaying e^−at, and a complex pair gives a damped oscillation. Stability is simply whether every pole lies in the left half-plane.

2.5 The final value theorem

If the system is stable, its steady-state output is lim_t→∞ f(t) = lim_s→0 sF(s). This shortcut gives the final value of a response without inverting the whole transform, and it underlies steady-state error in Chapter 5.

Engineering connection: the transfer function is the object every later chapter manipulates, and its poles are what root locus and Bode plots track.

Worked example 1: transfer function of a mass-spring-damper

A mass-spring-damper obeys ẍ + 3ẋ + 2x = f(t) (with m = 1, c = 3, k = 2 in consistent units). Find the transfer function X(s)/F(s) and its poles.

Figure 1. The familiar mechanical second-order system transforms term by term into a second-order transfer function, whose denominator factors into two real poles.

ProblemFind X(s)/F(s) and the poles for the system in Figure 1.
Given / findẍ + 3ẋ + 2x = f(t), zero initial conditions. Find the transfer function and poles.
AssumptionsLinear, time-invariant; zero initial conditions, so derivative terms transform cleanly.
ModelTransform each term (ẍ → s²X, ẋ → sX), solve for X/F, and factor the denominator.
Equations(s² + 3s + 2)X(s) = F(s)X/F = 1/(s² + 3s + 2)
SolveX/F = 1/(s² + 3s + 2) = 1/[(s + 1)(s + 2)], so the poles are at s = −1 and s = −2.
CheckSubstituting s = −1 or s = −2 makes the denominator zero, confirming the poles. Both are in the left half-plane, so the system is stable.
ConclusionThe mechanical equation became algebra, and the two real poles say the response is an overdamped decay. The transfer function is now ready for feedback analysis.

Result. X/F = 1/(s² + 3s + 2), poles at −1 and −2.

Worked example 2: step response by partial fractions

Find the unit-step response of the transfer function G(s) = 2/(s² + 3s + 2), and confirm its steady-state value with the final value theorem.

Figure 2. The two pole modes (e^−t and e^−2t) decay, leaving the steady-state value of 1. The response rises smoothly with no overshoot, as two real poles require.

ProblemFind the step response c(t) of G(s) and its final value.
Given / findG(s) = 2/(s² + 3s + 2), input a unit step (1/s). Find c(t) and c(∞).
AssumptionsStable system, so the final value theorem applies.
ModelForm C(s) = G(s)/s, expand in partial fractions, and inverse-transform each term.
EquationsC(s) = 2/[s(s + 1)(s + 2)] = A/s + B/(s+1) + C/(s+2)c(∞) = lim_s→0 sC(s)
SolveA = 2/[(1)(2)] = 1, B = 2/[(−1)(1)] = −2, C = 2/[(−2)(−1)] = 1. So c(t) = 1 − 2e^−t + e^−2t. The final value is lim_s→0 sC(s) = 2/[(1)(2)] = 1.
CheckAt t = 0, c = 1 − 2 + 1 = 0, the correct start; as t → ∞ the exponentials vanish and c → 1, matching the final value theorem.
ConclusionEach pole contributed one decaying mode, and their sum settles to the DC gain. The final value theorem gave the steady state without the full inversion.

Result. c(t) = 1 − 2e^−t + e^−2t, with c(∞) = 1.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Keeping initial-condition terms	Extra terms in the transfer function	"Are initial conditions zero here?"	A transfer function assumes zero initial conditions.
Confusing poles and zeros	Stability judged from the numerator	"Which polynomial gives the poles?"	Poles are denominator roots; zeros are numerator roots.
Final value theorem on an unstable system	A finite final value for a growing response	"Are all poles in the left half-plane?"	The theorem applies only when the system is stable.
Wrong partial-fraction residues	Step response starts at the wrong value	"Does c(0) match the expansion?"	Recompute residues; check the t = 0 and t → ∞ limits.

Practice ladder

Level 1 · Direct skill

Find the transfer function of ẏ + 5y = 3u(t).

Show answer

(s + 5)Y = 3U, so Y/U = 3/(s + 5). One real pole at −5.

Level 2 · Mixed concept

Find the steady-state step response of G(s) = 6/(s² + 4s + 3) using the final value theorem.

Show answer

c(∞) = lim_s→0 s · [6/(s(s² + 4s + 3))] = 6/3 = 2. The DC gain of G is G(0) = 6/3 = 2, the same value.

Level 3 · Independent problem

Find the poles of G(s) = (s + 1)/(s² + 2s + 5) and say whether the system is stable and oscillatory.

Show answer

Poles: s = −1 ± 2j (from the quadratic formula). Both have negative real part, so it is stable, and the imaginary part means a damped oscillation.

Level 4 · Transfer to real engineering

Take a real second-order system (a suspension, an RLC circuit) and write its transfer function, then locate and interpret its poles.

What good work looks like

A correct second-order transfer function, poles found from the denominator, and an interpretation of whether the response decays, oscillates, or both.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I transformed with zero initial conditions and read the poles correctly."

"Give me four transfer functions; I will classify each pole as stable or not."

"What is the transfer function?" Transforming the equation yourself is the skill.

"Invert this transform." Doing the partial fractions is the point.

Portfolio task

Model a real dynamic system as a differential equation, transform it to a transfer function, find its poles, and predict the step response, then check the steady state with the final value theorem.

Must include: a transfer function, its poles, a predicted response, and a final-value check.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. What does differentiation become under the Laplace transform?

Multiplication by s, minus the initial value: L{f′} = sF(s) − f(0).

2. Define the transfer function.

The ratio of output to input transform, with zero initial conditions: G(s) = Y(s)/U(s).

3. Where are the poles and zeros?

Poles are roots of the denominator; zeros are roots of the numerator.

4. When is a system stable?

When every pole lies in the left half of the s-plane.

5. State the final value theorem.

For a stable system, lim_t→∞ f(t) = lim_s→0 sF(s).

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive a transfer function and step response from a blank page.

+3 daysFind the poles of three new systems.

+7 daysCarry transfer functions into modeling, Chapter 3.

+30 daysReuse poles and the final value theorem in response and error.

Textbook mapping

Item	Mapping
Primary source	Ogata, Modern Control Engineering, Chapter 2 (Mathematical Modeling of Control Systems)
Cross-reference	Nise, Ch. 2 · Math: Laplace
Core topics	2.1 Laplace transform · 2.2 Equation to transfer function · 2.3 Transfer function · 2.4 Poles and zeros · 2.5 Final value theorem
Engineering connection	The transfer function is what every later method manipulates.
Read next	Chapter 3: System Modeling and Block Diagrams.