Laplace Transforms and Transfer Functions
From differential equations to transfer functions, poles and zeros, first- and second-order systems and block algebra.
Why transfer functions
Physical systems are described by differential equations. The Laplace transform turns calculus into algebra: a derivative becomes a multiplication by s, an integral a division by s, and the output is simply the transfer function times the input. Once a system is a transfer function G(s) you can read its speed, damping and stability from the positions of its poles, join systems by multiplying them, and close a loop by one formula.
The Laplace transform
F(s) = ∫₀^∞ f(t) e^(−s t) dt derivative: ℒ{f′} = s F(s) − f(0)
| Time function f(t) | F(s) | Meaning |
|---|---|---|
| unit step u(t) | 1 ÷ s | a switch on |
| unit impulse δ(t) | 1 | a sharp tap |
| t (ramp) | 1 ÷ s² | a steadily increasing setpoint |
| e−a t | 1 ÷ (s + a) | exponential decay with a pole at −a |
| sin ω t | ω ÷ (s² + ω²) | oscillation |
| f(t − T) | e−T s F(s) | a pure delay of T seconds |
Two useful theorems: the final value theorem, limt→∞ y(t) = lims→0 s Y(s) (the steady state, valid if the system is stable), and the initial value theorem, y(0⁺) = lims→∞ s Y(s).
Transfer functions, poles and zeros
The transfer function is G(s) = Y(s) ÷ U(s), a ratio of polynomials in s. The roots of the denominator are the poles; they decide the natural behavior. The roots of the numerator are the zeros; they shape how the input is combined. A pole in the left half of the s-plane (negative real part) is a decaying mode, so the system is stable; a pole on the imaginary axis oscillates for ever; a pole in the right half grows (unstable). The distance of a pole from the imaginary axis sets the speed of decay; its imaginary part sets the oscillation frequency.
First-order systems: the oven
G(s) = K ÷ (τ s + 1) step response y(t) = K U (1 − e−t ÷ τ)
Worked example: the oven without its delay
- G(s) = 2 ÷ (120 s + 1): a pole at s = −1 ÷ 120 = -0.0083 s⁻¹.
- A 10 percent heater step ends at K U = 20 °C (final value theorem: s × K ÷ (τ s + 1) × 10 ÷ s at s = 0). After one time constant it has reached 63 percent: 12.6 °C at t = 120 s, 95 percent at 3 τ = 360 s and 98 percent at 4 τ = 480 s.
- Close the loop with proportional gain Kp = 2: T(s) = Kp G ÷ (1 + Kp G) = 4 ÷ (120 s + 1 + 4) = 0.80 ÷ (24 s + 1). The pole moves from −0.0083 to −0.042: the closed loop is 5 times faster (time constant 24 s) and its DC gain is 0.80: a 20 percent steady-state error, as found in Feedback Control Basics.
Second-order systems: the servo
A mass m on a spring k with a damper c, pushed by a force F, obeys m x″ + c x′ + k x = F. Its transfer function from force to position is G(s) = 1 ÷ (m s² + c s + k) = (1 ÷ k) ωn² ÷ (s² + 2 ζ ωn s + ωn²), with the two parameters that describe every second-order system:
ωn = √(k ÷ m) ζ = c ÷ (2 √(k m))
Worked example: m = 1 kg, k = 100 N/m, c = 10 N·s/m
- ωn = √(100 ÷ 1) = 10 rad/s (1.59 Hz); ζ = 10 ÷ (2 × 10) = 0.50.
- Poles: s = −5 ± j 8.66: a decay rate of 5 s⁻¹ and a damped frequency of 8.66 rad/s.
- The step response overshoots by exp(−π ζ ÷ √(1 − ζ²)) = 16.3 percent (see Time Response and Specifications).
- For ζ ≥ 1 the poles are real and there is no oscillation; ζ = 0 puts them on the imaginary axis (undamped oscillation).
Dead time
Many processes have a dead time θ: a transport delay (a conveyor, a pipe, a thermal lag in a sensor well) during which nothing happens. It is the transfer function e−θ s with a gain of exactly 1 at all frequencies and a phase lag that grows with frequency, φ = −ω θ. The oven’s dead time of 15 s gives −43° at ω = 0.05 rad/s. Dead time is the most damaging element for a feedback loop, because it adds lag without reducing the gain (see Stability: Routh, Root Locus and Margins). For analysis it is approximated by a rational Padé form: e−θ s ≈ (1 − θ s ÷ 2) ÷ (1 + θ s ÷ 2), whose phase at 0.05 rad/s is −41.1° (exact 43.0°): good at low frequency and poor beyond ω θ ≈ 1.
Block diagram algebra
| Connection | Result | Example with G₁ = 3, G₂ = 4 |
|---|---|---|
| Series | G₁ G₂ | 12 |
| Parallel | G₁ + G₂ | 7 |
| Negative feedback | G ÷ (1 + G H) | G₁ with unity feedback: 0.75 |
| Closed loop | T = L ÷ (1 + L), L = C G | the loop gain L decides everything |
| Disturbance to output | S Gd with S = 1 ÷ (1 + L) | a high loop gain makes S small: the disturbance is rejected |
The denominator 1 + L(s) = 0 is the characteristic equation; its roots are the closed-loop poles. Everything about stability and response follows from them (Stability: Routh, Root Locus and Margins).
Related
- Previous: Feedback Control Basics. Next: Time Response and Specifications.
- PID Tuner.
- Structures and Strength of Materials for the mass–spring system.
- Control Systems: the whole guide.
FAQ
What is a pole?
A value of s at which the transfer function becomes infinite: a natural mode of the system. Poles in the left half-plane decay; in the right half-plane they grow.
Do I need to do the Laplace transform by hand?
Rarely. Know the table of the common pairs, the poles-and-zeros picture and the closed-loop formula, and let software do the algebra.
Why do systems have transfer functions only when linear?
The Laplace method applies to linear time-invariant systems. Real systems are nonlinear, but are well approximated near an operating point by linearizing.
What is the difference between a transfer function and a state-space model?
A transfer function relates one input to one output; a state-space model describes all the internal states and handles many inputs and outputs. They carry the same information for a single loop.
References
- Ogata, K., Modern Control Engineering, Pearson (Laplace transform, transfer functions, block diagrams); Franklin, G. F., et al., Feedback Control of Dynamic Systems, Pearson.
- Nise, N. S., Control Systems Engineering, Wiley; Åström, K. J., Murray, R. M., Feedback Systems (free online).
- Oppenheim, A. V., Willsky, A. S., Signals and Systems, Pearson.
Last reviewed: October 2026. The oven and the servo are illustrative models.