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Math for ME · Chapter 14 of 19 · Advanced

Laplace Transforms and Transfer Functions

The transform turns differential equations into algebra, and systems into transfer functions. Enough is taught here for Controls; no more.

The thread: Solving differential equations by hand is slow. The Laplace transform turns them into algebra, and turns a whole system into the transfer function that control engineers design with.

Readiness check

From ODEs and State-Space Thinking. Tick only what you can do closed-notes.

Solve first and second order linear ODEs.
Perform partial-fraction splits on simple rational expressions.
Work with e^−at and its derivative confidently.
Read pole locations as stability from State-Space Thinking.
Keep algebra organized across several lines.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsRedo the ODEs ladder; the transform assumes ODE fluency.

3 or more weak itemsStep back to ODEs and State-Space Thinking.

The core idea

Differentiation becomes multiplication by s. ODEs become algebra.

L{ẏ} = sY − y(0)G(s) = output/input

Transform the equation, solve for Y(s) by algebra, split into table entries, transform back. The transfer function G(s) is the system's fingerprint: its denominator's roots (poles) are exactly the eigenvalues of State-Space Thinking.

The skill works when: the system is linear with constant coefficients and starts from known initial conditions: the standard controls setting.

The skill breaks down when: coefficients vary or nonlinearity matters; the transform is a linear tool by birth.

The concept. Engineers detour through the s domain because algebra is cheaper than integration. The block diagram is Controls' native drawing.

The skills, taught in order

14.1 The idea: trade calculus for algebra

The Laplace transform weights a time history by e^−st and integrates, producing a function of the complex variable s. Its payoff is one rule: a time derivative becomes multiplication by s, carrying the initial condition along.

L{ẏ} = sY − y(0)L{ÿ} = s²Y − sy(0) − ẏ(0)

So a differential equation in t becomes an algebraic equation in s.

14.2 The short table you actually use

Almost all engineering work needs only these pairs.

f(t)	F(s)
1 (step)	1/s
t	1/s²
e^−at	1/(s + a)
sin ωt	ω/(s² + ω²)
cos ωt	s/(s² + ω²)
e^−at sin ωt	ω/((s + a)² + ω²)

14.3 Solving an ODE: the four steps

The routine never changes: transform the equation (initial conditions enter here), solve the algebra for Y(s), split Y(s) into table-shaped pieces by partial fractions, and invert each piece back to time. The hardest step is usually the partial-fraction split, not any calculus.

14.4 Transfer functions, poles, and zeros

For a linear system at rest, the ratio of the output and input transforms is the transfer function:

G(s) = Y(s)/U(s)

The roots of its denominator are the poles, identical to the eigenvalues of State-Space Thinking, and they set stability and response shape. The roots of its numerator are the zeros, which reshape how strongly each mode appears.

14.5 Test inputs and the final-value shortcut

Two signals probe any system: the step (a sudden constant) and the impulse (a sharp hit, whose response is the system's signature). For a stable system the step's steady value needs no inversion at all, only one substitution:

y(∞) = G(0) × step size

Engineering connection: Control Systems, Vibrations, Mechatronics.

Worked example: a first-order plant hit with a step

A sensor lags its input according to ẏ + 2y = 4u(t), with y(0) = 0 and u a unit step. Solve by Laplace, identify the transfer function, and read the time constant.

Figure 1. Step response y = 2(1 − e^−2t) and the single pole at s = −2 that dictates it.

ProblemSolve ẏ + 2y = 4u(t), y(0) = 0, and extract G(s) and τ.
Given / findFirst-order ODE, unit step input. Find y(t), G(s), τ, and y∞.
AssumptionsLinear, constant coefficients, zero initial state.
ModelTransform both sides; the step becomes 1/s; the derivative becomes sY.
EquationssY + 2Y = 4/s Y = 4/(s(s+2))
SolvePartial fractions: Y = 2/s − 2/(s+2). Inverting from the table: y(t) = 2(1 − e^−2t). The transfer function is G(s) = 4/(s+2) from u to y, with steady gain G(0) = 2 and pole at s = −2, so τ = 1/2 = 0.5 s.
CheckSubstitute back: ẏ = 4e^−2t; ẏ + 2y = 4e^−2t + 4 − 4e^−2t = 4. Limits: y(0) = 0, y(∞) = 2 = G(0), both right. The 63% point sits at 0.5 s on the curve, matching the pole.
ConclusionFive lines of algebra replaced an integration, and the transfer function summarized the device for any future input. Controls will wire blocks like this G(s) into feedback loops and move the pole on purpose.

Result. y(t) = 2(1 − e^−2t); G(s) = 4/(s+2); pole −2; τ = 0.5 s; steady state 2.

04b

Worked example 2: step response of a second-order plant

A plant has transfer function G(s) = 6/(s² + 5s + 6) from input to output. Find its full unit-step response y(t) and its steady-state value.

Given / findG(s) = 6/(s² + 5s + 6), unit step U(s) = 1/s, zero initial conditions. Find y(t) and y(∞).
Factor the poless² + 5s + 6 = (s + 2)(s + 3): two real poles at −2 and −3, so the system is stable and overdamped (no ringing).
Form the outputY(s) = G(s)U(s) = 6/[s(s + 2)(s + 3)].
Partial fractionsY = A/s + B/(s+2) + C/(s+3), with A = 6/(2·3) = 1, B = 6/((−2)(1)) = −3, C = 6/((−3)(−1)) = 2.
Invert term by termy(t) = 1 − 3e^−2t + 2e^−3t.
Steady stateThe final-value shortcut gives y(∞) = G(0) = 6/6 = 1, matching the constant term.
CheckAt t = 0: y = 1 − 3 + 2 = 0, correct for a system starting from rest. As t grows the exponentials vanish, leaving 1. The faster pole at −3 fades first, so the −2 term dominates the tail.
ConclusionFactor, split, invert: a second-order response falls out as cleanly as a first-order one, and the denominator alone announced stability and the overdamped character before any inversion.

Result. y(t) = 1 − 3e^−2t + 2e^−3t; poles −2, −3 (stable, overdamped); y(∞) = 1.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Initial conditions dropped	L{ẏ} written as sY for nonzero y(0)	"What does the system hold at t = 0?"	L{ẏ} = sY − y(0). The transform carries the start state for free; use it.
Inverting without partial fractions	4/(s(s+2)) "inverted" wholesale and wrongly	"Is each term a table entry yet?"	Split first. Only table-shaped pieces invert safely.
Transfer function with initial conditions inside	G(s) different for every start state	"Was G defined at zero initial conditions?"	G(s) is defined for zero ICs; initial-state effects are added separately.
Steady state read from the wrong place	Final value guessed from the numerator	"What is G(0)?"	For stable systems, the step's final value is G(0) times the step size. One substitution.

Practice ladder

Level 1 · Direct skill

From the table: invert Y(s) = 5/(s+3) and Y(s) = 1/s².

Show answer

5e^−3t and t. Two table lookups; speed comes with repetition.

Then write the Laplace transform of f(t) = 3 + 2e^−5t.

Show answer

F(s) = 3/s + 2/(s + 5), using the step and exponential entries term by term.

Level 2 · Mixed concept

Solve ẏ + 4y = 0 with y(0) = 6 by Laplace, and confirm against the ODEs method.

Show answer

sY − 6 + 4Y = 0, so Y = 6/(s+4) and y = 6e^−4t. Separation of variables gives the identical answer; the transform just bookkeeps the IC automatically.

A stable system has G(s) = 10/(s² + 3s + 5). What is its steady response to a step of size 2?

Show answer

y(∞) = G(0) × 2 = (10/5) × 2 = 4. No inversion needed, only G(0) and the step size.

Level 3 · Independent problem

A mass-spring-damper has G(s) = 1/(s² + 4s + 25) from force to displacement. Without inverting, report stability, ωₙ, ζ, and the ringing frequency.

Show answer

Poles: s = −2 ± 4.58i (left half-plane: stable). ωₙ = √25 = 5 rad/s; 2ζωₙ = 4 gives ζ = 0.4; ω_d = 4.58 rad/s. The same plant as the State-Space worked example, recognized from the denominator alone.

Level 4 · Transfer to real engineering

Identify a first-order device you can test (thermometer in hot water, RC low-pass on a breadboard, car cabin heating). Step it, log the response, fit τ and the steady gain, and write its G(s).

What good work looks like

The logged curve, the 63% read of τ, G(s) = K/(τs + 1) with numbers and units, and one sentence on what would move the pole (more thermal mass, different resistor).

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Here is my partial-fraction split. Multiply it back out and tell me if it matches the original; show only yes or no with the failing term."

"Give me four denominators; I will state poles, stability, and response shape before any inversion."

"Invert this transform." Table fluency and splitting are the entire mechanical skill.

"What is the transfer function?" Deriving G from the ODE is one line; owning that line matters in Controls.

Portfolio task

Make your own one-page transform table: the eight entries you actually use (1, t, e^−at, sin, cos, damped sin, step, impulse), each verified once by the definition or by differentiation, plus the two theorems you need (derivative rule, final value).

Must include: one full verification worked in the margin and the final-value theorem applied to the worked example.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Why do engineers transform ODEs at all?

Differentiation becomes multiplication by s, so solving means algebra plus a table lookup instead of integration.

2. State the derivative rule with initial condition.

L{ẏ} = sY(s) − y(0); second derivative: s²Y − sy(0) − ẏ(0).

3. Define the transfer function and its standing assumption.

G(s) = Y(s)/U(s) at zero initial conditions: the system's input-output fingerprint.

4. What do poles and zeros each control?

Poles set the response ingredients (decay rates, ring frequencies, stability); zeros reshape how strongly each ingredient appears.

5. How do you get the steady-state step response in one move?

Final value = G(0) × step size, valid when the system is stable.

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

+1 dayRe-solve the step-response example from a blank page.

+3 daysOne second-order pole reading (Level 3 style), new numbers.

+7 daysMixed set: a transform solve plus a State-Space Thinking pole placement.

+30 daysReuse your fitted G(s) from Level 4 when Controls begins.

Textbook mapping

Item	Mapping
Main sources	Kreyszig, Advanced Engineering Mathematics, Ch 6 (Laplace transforms)
Core topics	14.1 Why transforms · 14.2 The idea · 14.3 Common transforms · 14.4 Inversion · 14.5 Solving ODEs · 14.6 Step input · 14.7 Impulse input · 14.8 Transfer functions · 14.9 Poles, zeros, response
Engineering connection	Control Systems (the native language), Vibrations (frequency response), Mechatronics.
Skip on first pass	Convolution theorem proofs, exotic transform pairs, region-of-convergence theory. Teach enough for controls.
Read next	Fourier Series, Frequency, and Signals.