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System Dynamics · Module 6 of 10

Laplace Transforms and Transfer Functions

The Laplace transform turns a differential equation into algebra. The ratio of output to input becomes a transfer function whose poles, read off by inspection, already tell you how the system will move.

01

Readiness check

This module moves to the frequency domain. Tick only what you can do closed-notes.

  • Recall that the Laplace transform of a derivative brings down an s.
  • Write a mass-spring-damper differential equation.
  • Factor a quadratic to find its roots.
  • Evaluate a function at s = 0.
  • Recall ωn and ζ from a second-order denominator.
0 or 1 weak itemsContinue with this module.
2 weak itemsRefresh the Laplace transform in Mathematics, Module 12.
3 or more weak itemsRevisit the mass-spring-damper in Module 2.
02

The core idea

The Laplace transform replaces d/dt with s, turning a differential equation into algebra. The transfer function G(s) is the ratio of transformed output to input; its poles are the roots of the denominator, and they set the system's response.

d/dt → s, with zero initial conditionsG(s) = Y(s)/U(s)poles = roots of the denominator

The Laplace transform maps a function of time to a function of the complex variable s, and crucially it turns differentiation into multiplication by s (with zero initial conditions). A differential equation becomes an algebraic one, which is solved by ordinary algebra and, if needed, transformed back. The ratio of the transformed output to the transformed input is the transfer function G(s), a complete description of a linear system. Its denominator set to zero is the characteristic equation, and the roots of that equation are the poles. The poles alone determine the form of the response: real poles give exponentials, complex pairs give damped oscillation, and their location fixes the speed and damping. The value G(0) is the DC gain, the steady output per unit steady input.

The skill works when: you transform the equation, form G(s), and read the poles and DC gain by inspection.
The skill breaks down when: initial conditions are kept where the definition assumes zero, or numerator and denominator roles are confused.
The concept. The transfer function's poles, plotted in the complex plane, set the response. Poles in the left half plane decay; the further left, the faster. The value at s = 0 is the DC gain.
03

The skills, taught in order

Five skills move from the transform to the transfer function and the information its poles carry.

6.1 The Laplace transform

The Laplace transform turns a time function into a function of s and, with zero initial conditions, replaces each time derivative with a factor of s. A differential equation in t becomes an algebraic equation in s, far easier to manipulate.

6.2 The transfer function

For a linear system the transfer function G(s) = Y(s)/U(s) is the ratio of transformed output to input. It captures the whole input-output behaviour in one rational function of s and is independent of the particular input.

6.3 Poles and zeros

The roots of the denominator are the poles; the roots of the numerator are the zeros. Poles govern the natural response, its speed and whether it oscillates; zeros shape how strongly each mode appears. Pole location is the single most informative feature of a model.

Pole typeLocationResponse
Real, negativeleft axisdecaying exponential
Complex pair, negative real partleft half planedamped oscillation
On imaginary axisjω axissustained oscillation
Positive real partright half planegrowing, unstable

Pole location maps directly to the shape of the response, which is why poles are read first.

6.4 DC gain and the final value

The DC gain G(0) is the steady output per unit steady input. For a stable system the final value theorem confirms it: the steady response to a unit step is G(0). It is read by setting s = 0 in the transfer function.

6.5 The characteristic equation

Setting the denominator to zero gives the characteristic equation; its roots are the poles. Matching a second-order denominator to s2 + 2ζωns + ωn2 reads off the natural frequency and damping ratio directly from the model.

Engineering connection: transfer functions are the shared language of Control Systems, where the same poles are moved deliberately by feedback.

04

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

A mass-spring-damper has m = 1 kg, c = 3 N·s/m, and k = 2 N/m. Find the transfer function from force to displacement, its poles, and its DC gain.

Figure 1. The denominator factors into two real poles at −1 and −2, so the system is overdamped and decays without oscillation. Setting s = 0 gives the static deflection per unit force.
  1. ProblemFind G(s), its poles, and its DC gain for the system in Figure 1.
  2. Given / findm = 1, c = 3, k = 2. Find G(s) = X(s)/F(s), the poles, and G(0).
  3. AssumptionsZero initial conditions; linear system.
  4. ModelTransform m ẍ + c ẋ + k x = F to (ms2 + cs + k)X = F, so G = 1/(ms2 + cs + k).
  5. EquationsG(s) = 1/(m s2 + c s + k)poles: m s2 + c s + k = 0DC gain = G(0) = 1/k
  6. SolveG(s) = 1/(s2 + 3s + 2) = 1/((s + 1)(s + 2)). Poles at −1 and −2. G(0) = 1/2 = 0.5 m/N.
  7. CheckBoth poles are real and negative, so the response decays without oscillation, consistent with ζ = c/(2√(km)) = 3/(2√2) = 1.06 > 1, overdamped. The DC gain equals the static 1/k.
  8. ConclusionThe transfer function shows at a glance an overdamped system settling to 0.5 m per newton, all read from the poles and G(0).
Result. G(s) = 1/((s+1)(s+2)), poles −1 and −2, DC gain 0.5 m/N.
05

Worked example 2: poles and damping from the characteristic equation

A second-order system has m = 2 kg, c = 12 N·s/m, and k = 10 N/m. Find the characteristic equation, the poles, the natural frequency, and the damping ratio.

Figure 2. Dividing through by the mass gives a monic characteristic equation whose two real roots, −1 and −5, confirm an overdamped system. The natural frequency and damping ratio follow from the coefficients.
  1. ProblemFind the characteristic equation, poles, ωn, and ζ for the system in Figure 2.
  2. Given / findm = 2, c = 12, k = 10. Find the characteristic equation, the poles, ωn, ζ.
  3. AssumptionsLinear second-order system; zero initial conditions.
  4. ModelThe denominator ms2 + cs + k = 0, divided by m, is s2 + 2ζωns + ωn2 = 0.
  5. Equationss2 + (c/m)s + (k/m) = 0ωn = √(k/m), ζ = c/(2√(k m))
  6. SolveDividing by 2: s2 + 6s + 5 = 0, factoring to (s + 1)(s + 5), poles −1 and −5. ωn = √(10/2) = √5 = 2.24 rad/s. ζ = 12/(2√20) = 1.34.
  7. Checkζ > 1, so the poles must be real and distinct, which −1 and −5 are. Their product equals ωn2 = 5 and their sum equals 2ζωn = 6, both consistent.
  8. ConclusionThe characteristic equation reveals an overdamped system; matching it to the standard form gives ωn and ζ without solving any differential equation.
Result. s2 + 6s + 5 = 0, poles −1 and −5, ωn = 2.24 rad/s, ζ = 1.34.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Keeping initial conditionsExtra terms in the transfer function"Did I assume zero initial conditions?"The transfer function is defined with zero initial conditions.
Poles from the numeratorStability judged from zeros"Which polynomial did I set to zero?"Poles are roots of the denominator.
DC gain at the wrong sGain read at s = 1 or s = ∞"Is this G(0)?"DC gain is G(s) evaluated at s = 0.
Not normalising before matchingωn and ζ read off wrong"Is the leading coefficient 1?"Divide by m so s2 stands alone before matching.
07

Practice ladder

Level 1 · Direct skill

Write the transfer function from force to displacement for m = 2, c = 6, k = 4.

Show answer

G(s) = 1/(2s2 + 6s + 4) = 1/(2(s + 1)(s + 2)).

Level 2 · Mixed concept

For that system, find the poles and the DC gain.

Show answer

Poles at −1 and −2. G(0) = 1/4 = 0.25 m/N.

Level 3 · Independent problem

A system has G(s) = 5/(s2 + 2s + 5). Find ωn, ζ, and the steady response to a unit step.

Show answer

ωn = √5 = 2.24 rad/s, ζ = 2/(2√5) = 0.447 (underdamped). Step final value G(0) = 5/5 = 1.

Transfer task | Real engineering

Given only the pole locations of a system, describe what you can already say about its response without solving anything.

What good work looks like

Real negative poles mean a decaying, non-oscillatory response; a complex pair means damped oscillation; the distance from the imaginary axis sets the decay rate, and any right-half-plane pole means instability, all read directly from location.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I took the poles from the denominator, not the numerator."
"Give me three transfer functions; I will classify each from its poles."
"Find the transfer function for me." Transforming the equation is the skill.
"What is the DC gain?" Evaluating G(0) yourself is the point.

Portfolio task

Take a model from an earlier module, write its transfer function, identify the poles and DC gain, and predict the response shape, then check against the time response.

Must include: G(s), the poles, the DC gain, and a predicted response shape.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. What does the Laplace transform do to a derivative?

It replaces d/dt with multiplication by s, given zero initial conditions.

2. Define the transfer function.

G(s) = Y(s)/U(s), the ratio of transformed output to input.

3. Where do the poles come from?

The roots of the denominator, the characteristic equation.

4. What is the DC gain?

G(0), the steady output per unit steady input.

5. What does a right-half-plane pole mean?

A growing response: the system is unstable.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive a transfer function from a differential equation.
+3 daysClassify three systems from their poles.
+7 daysRead the first-order response in detail, Module 7.
+30 daysReuse transfer functions when designing feedback control.
10

Textbook mapping

This module follows Karnopp, Margolis, and Rosenberg, System Dynamics: Modeling, Simulation, and Control of Mechatronic Systems, 5th edition. Use these references to read further.

Topic in this moduleWhere to read more
Laplace transforms and transfer functionsKarnopp, Margolis & Rosenberg, Chapter 8
Poles, zeros, and the characteristic equationKarnopp, Margolis & Rosenberg, Chapter 8
DC gain and the final valueKarnopp, Margolis & Rosenberg, Chapter 8

Chapter numbers refer to the 5th edition. The transform methods are standard, so any recent edition will align closely.