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Transfer Functions

Laplace turns a differential equation into algebra, cascades into products, and stability into the sign of a pole.

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The transfer function is the Laplace-domain ratio of output to input with zero initial conditions, which converts a differential equation into algebra, makes cascaded systems multiply instead of convolve, and turns stability into a question about which side of the imaginary axis the poles sit on.

The problem it solves

An RC circuit gives RC·dv/dt + v = v_in; a mass on a spring gives m·d²x/dt² + c·dx/dt + kx = F. Same shape, and solving either afresh for every input — step, ramp, sinusoid, disturbance — is intolerable. Worse, it mixes what belongs to the system with what belongs to the input.

Laplace, and what it buys

Under the transform, d/dt becomes × s and ∫dt becomes ÷ s. With zero initial conditions the differential equation becomes algebraic, and rearranging gives G(s) = C(s)/R(s) — a ratio containing no reference to any particular input.

  • Any input can now be run through it by multiplication: C(s) = G(s)·R(s).
  • Series systems multiply, where in the time domain they convolved.
  • Parallel paths add.
  • A closed loop reduces to G/(1 + GH).

Those last three are the whole of block diagram reduction, and they exist only because the transform turned convolution into a product.

Poles and zeros

Factorised, G(s) = K(s+z₁)(s+z₂)/(s+p₁)(s+p₂). Zeros are where the numerator vanishes; poles where the denominator does. Each pole contributes one exponential term to the time response, so poles determine *how* the system behaves while zeros shape *how much* of each mode appears.

A pole at s = −a contributes e^(−at) and decays; one at s = +a contributes e^(+at) and grows. Stability is therefore the question of whether every pole lies in the left half-plane — answered by reading a sign rather than solving anything.

What it assumes

  • Linearity — no saturation, backlash or stiction.
  • Time invariance — parameters do not drift as it runs.
  • Zero initial conditions — or the ratio is not clean.
  • Single input, single output.

So it describes the forced response and says nothing about starting state, which is exactly what state-space methods were developed to handle. The usual workaround for a non-linear plant is to linearise about an operating point, which works within a range.

The numbers you will be asked for

Transfer function

G(s) = C(s) / R(s)

Zero initial conditions.

Series

G₁ · G₂

Convolution in time becomes a product.

Closed loop

T(s) = G / (1 + G·H)

Negative feedback.

Characteristic equation

1 + G(s)H(s) = 0

Its roots are the closed-loop poles.

Advantages and disadvantages

Advantages

  • Turns calculus into algebra, once and for all.
  • Cascades multiply, so block diagrams reduce mechanically.
  • Stability becomes a question about pole position.
  • Independent of the input, so one model serves every test signal.

Disadvantages

  • Requires linearity and time invariance.
  • Assumes zero initial conditions, so it describes only the forced response.
  • SISO only — multivariable systems need state space.
  • A transport delay's e^(−sT) is not a polynomial and must be approximated for some methods.

Watch it work

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Check yourself

question 1 / 4

One question at a time. Pick an answer to see why it is right or wrong, then move on — there is no score to keep and nothing is saved.

What does the Laplace transform actually buy you here?
A system has a pole at s = +3. What does that tell you?
What is the difference between what poles and zeros do?
Which assumption does a transfer function make that real systems most often break?

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4 still unanswered — the dots above jump straight to them.