ELEN90055 · Control Systems
LTI systems, convolution, Laplace and transfer functions
This chapter is the computational engine of ELEN90055 Control Systems at the University of Melbourne: it shows how a linear time-invariant (LTI) plant turns an input into an output, first in the time domain as a convolution y(t) = (g∗u)(t) with the impulse response g, then in the s-domain where the Laplace transform converts that convolution into a multiplication Y(s) = G(s)U(s).
From a differential-equation model you take Laplace transforms with zero initial conditions and read off the transfer function G(s) = B(s)/A(s) — its denominator roots are the poles, its numerator roots the zeros. It also covers transport delay e-sT and its first-order Padé approximation, and the block-diagram algebra that reduces any single loop to one transfer function. It maps to Part II of the course (loosely following Goodwin, Graebe & Salgado, Chapters 3-4) and underpins every later stability and design question.
What this chapter covers
- 01Describe an LTI system by its unit-step response h(t) and impulse response g(t) = dh/dt
- 02Compute an output as the convolution y(t) = integral of g(t - tau) u(tau) dtau
- 03Use the Laplace transform pairs (step, exponential, t^n/n!, sine, cosine) and rules (derivative, integral, time shift, convolution)
- 04Turn a linear ODE with zero initial conditions into a transfer function G(s) = B(s)/A(s)
- 05Identify poles (roots of A), zeros (roots of B), DC gain G(0) and the relative degree n - m
- 06Read the endpoints of a response with the final-value and initial-value theorems
- 07Approximate a transport delay e^(-sT) by the first-order Pade (1 - Ts/2)/(1 + Ts/2) and spot its right-half-plane zero
- 08Reduce a block diagram with the three rules: series multiply, parallel add, negative feedback divide by 1 + GH
ODE to transfer function to step response
- +1(a) Transfer function. Transform each term with zero initial conditions, using L{y'} = sY: (2s + 1)Y(s) = 3U(s), so G(s) = 3/(2s + 1).
- +1Read off the constants. DC gain G(0) = 3; writing G = 3/(2s + 1) gives time constant tau = 2 s and a single pole at s = -1/2 = -0.5 rad/s (stable, left-half-plane).
- +1(b) Step input. A step of height 4 has U(s) = 4/s, so Y(s) = G(s)U(s) = (3/(2s + 1)) × (4/s) = 12/(s(2s + 1)).
- +2Partial fractions. 12/(s(2s + 1)) = A/s + B/(2s + 1) gives 12 = A(2s + 1) + Bs. At s = 0, A = 12; at s = -1/2, 12 = -B/2 so B = -24. Hence Y = 12/s - 24/(2s + 1) = 12/s - 12/(s + 0.5).
- +1Invert. Using L-inverse{1/s} = 1 and L-inverse{1/(s + 0.5)} = e^(-0.5t): y(t) = 12 - 12 e^(-t/2) for t >= 0.
- +1Final value. As t tends to infinity, e^(-t/2) tends to 0, so y(infinity) = 12 — exactly DC gain × input = 3 × 4 = 12.
- +1(c) FVT check. lim (s to 0) sY(s) = lim (s to 0) 12/(2s + 1) = 12 — it agrees, so the answer is self-consistent.
Key terms
- Impulse response g(t)
- The output of an LTI system to a unit impulse; equivalently the derivative of the unit-step response, g = dh/dt. It is the kernel of the convolution, so it alone determines the whole input-output map, and its Laplace transform is the transfer function G(s).
- Convolution
- The time-domain input-output relation y(t) = integral from 0 to t of g(t - tau) u(tau) dtau: the present output is a weighted running memory of the past input. The lower limit 0 encodes causality. In the s-domain it becomes the multiplication Y(s) = G(s)U(s).
- Laplace transform
- Y(s) = integral from 0 to infinity of y(t) e^(-st) dt, with s = sigma + j*omega the complex frequency (units of 1/s). It converts differentiation into multiplication by s and convolution into multiplication, turning a linear ODE into algebra.
- Transfer function G(s)
- The ratio Y(s)/U(s) = B(s)/A(s) obtained from a linear ODE with zero initial conditions. The roots of A (denominator) are the poles; the roots of B (numerator) are the zeros; G(0) is the DC (steady-state) gain.
- Poles and zeros
- Poles are the denominator roots alpha_k; they set the natural modes and stability (the system is stable exactly when every pole has Re < 0). Zeros are the numerator roots beta_k; they shape the transient, e.g. a right-half-plane zero causes undershoot.
- Relative degree / proper
- The relative degree is n_r = n - m (number of poles minus zeros). A physically realisable transfer function is proper, n_r >= 0; if n_r > 0 it is strictly proper. A negative relative degree is a pure differentiator and is non-causal.
- Final-value theorem (FVT)
- If the limits exist and sY(s) has all poles in the open left-half plane, lim (t to infinity) y(t) = lim (s to 0+) s Y(s). Its partner, the initial-value theorem, gives lim (t to 0+) y(t) = lim (s to infinity) s Y(s).
- Transport delay and Pade
- A dead time of T seconds multiplies the transfer function by e^(-sT) (an all-pass factor). The first-order Pade approximation e^(-sT) approx (1 - Ts/2)/(1 + Ts/2) replaces it by a rational function with a right-half-plane zero at +2/T, which is why delay makes a loop non-minimum-phase.
LTI systems, convolution, Laplace and transfer functions FAQ
Why bother with the Laplace transform instead of solving the differential equation directly?
Because it turns calculus into algebra. In the time domain the output is a convolution integral, which you would have to evaluate for every instant. The Laplace transform maps differentiation to multiplication by s and convolution to an ordinary product Y(s) = G(s)U(s), so a linear constant-coefficient ODE becomes an algebraic equation you solve for Y(s) and then invert with a short table of pairs. Every later tool in the subject — poles, stability tests, frequency response, controller design — operates on the resulting transfer function, so the transform is the gateway to all of it.
What is the difference between a pole and a zero, and why do they matter?
Both come from the transfer function G(s) = B(s)/A(s). Poles are the roots of the denominator A(s); they are the natural frequencies of the system and decide stability — the system is stable exactly when every pole lies strictly in the left-half plane (Re < 0). Zeros are the roots of the numerator B(s); they do not affect stability but shape the transient, for example a right-half-plane zero (as produced by a delay's Pade approximation) causes the step response to undershoot before it rises. Knowing both, plus the DC gain G(0), lets you predict the response before computing it.
Can AI help me with LTI systems, convolution, Laplace and transfer functions in ELEN90055?
Yes, as a study aid. An AI tutor like Sia can explain convolution and the Laplace pairs step by step, walk you through turning an ODE into a transfer function, and check your partial-fraction and inversion algebra or your block-diagram reduction. Use it to understand the method and the sign and units conventions and to practise; it does not sit your open-book exam or guarantee a mark, and you should confirm every formula against the unit notes on Canvas.
Studying with AI? Sia — free AI electrical engineering tutor works through ELEN90055 step by step.
Exam move
Make the ODE-to-transfer-function move automatic, because it is the first step of the typical Q1 and a tool in every other question. Drill three things until they are fast: (1) transforming an ODE term-by-term with zero initial conditions to get G(s) = B(s)/A(s), then reading off poles, zeros, DC gain G(0) and relative degree; (2) partial fractions plus the standard pairs to invert Y(s) = G(s)U(s) back to y(t), always re-checking the endpoint with the final-value theorem; and (3) block-diagram reduction with the three rules — series multiply, parallel add, negative feedback divide by 1 + GH. Keep the sign conventions honest: the derivative rule carries a minus-y(0) term (kept only for non-zero initial conditions), and the closed-loop denominator is 1 + GH with a plus. For delays, remember e^(-sT) and its first-order Pade (1 - Ts/2)/(1 + Ts/2) with its right-half-plane zero at +2/T. The exam is 3 hours, open book, worth 70% with a hurdle, and you answer four questions summing to 50 marks — budget about 3.6 minutes per mark. Since it is open book, marks reward method shown with correct sign and unit conventions, not memorised formulae. Confirm the exact date of the next (Semester 1, ~June 2027) sitting on Canvas.
