ELEN90055 Control Systems

How ELEN90055 is assessed

What ELEN90055 covers

The twelve chapters run one arc — model the system, analyse its stability, respect its limits, then design the loop — mapped to how ELEN90055 Control Systems is examined at the University of Melbourne.

It develops the frequency-domain analysis and design of linear time-invariant feedback loops: modelling and linearising physical systems, deriving transfer functions, testing stability with the Routh, root-locus and Nyquist methods, and designing proportional, lag, lead and PID compensators.

This study guide maps every chapter to how ELEN90055 is actually assessed at the University of Melbourne, so your revision points straight at the 3-hour open-book final that carries a pass hurdle.

Worked example · free

Routh stability: the gain range that keeps a feedback loop stable

Key terms

Transfer function

The Laplace-domain ratio G(s) = B(s)/A(s) of output to input for a linear time-invariant system at zero initial conditions. Its denominator roots are the poles and its numerator roots the zeros, which together fix the response and stability.

Pole

A root of the transfer-function denominator. A system is BIBO stable only when every pole has a strictly negative real part; the pole nearest the imaginary axis is the slow, dominant one that shapes the settling response.

Zero

A root of the transfer-function numerator. Zeros do not affect stability but shape the transient: a right-half-plane (positive) zero causes initial undershoot and extra phase lag (a non-minimum-phase response).

Linearisation

Replacing a nonlinear model by its first-order Taylor approximation about an equilibrium, giving a small-signal incremental model valid only for small deviations. Different operating points give different linear models.

BIBO stability

Bounded-input bounded-output stability: a proper LTI system is stable if and only if all of its poles lie in the open left-half plane (Re < 0). Poles on the imaginary axis give a marginally stable, non-decaying response.

Damping ratio (ζ)

The dimensionless number in the canonical second-order system that sets the overshoot and how oscillatory the response is. With natural frequency ωₙ it places the poles at −ζωₙ ± jωₙ√(1−ζ²).

Sensitivity function (S)

The closed-loop map S = 1/(1 + G₀C) from output disturbance to output (and reference to error). With the complementary sensitivity T it obeys S + T = 1, the identity behind every design trade-off.

Steady-state error

The residual tracking error as t → ∞. For a stable loop with a step input it is k/(1 + A₀(0)); larger DC loop gain gives smaller error, and integral action (a controller pole at s = 0) drives it to zero.

Routh–Hurwitz criterion

An algebraic test that counts right-half-plane roots of the characteristic polynomial from the sign changes in the first column of the Routh array, without factoring it. For a cubic, stability requires a₂a₁ > a₃a₀.

Root locus

The path traced by the closed-loop poles as a single gain varies from 0 to ∞. Branches start at the open-loop poles and end at the zeros or at infinity, governed by the phase condition ∠KF(s) = an odd multiple of π.

Nyquist criterion

A frequency-domain stability test from Cauchy's argument principle: the number of unstable closed-loop poles is Z = N + P, where N counts clockwise encirclements of the −1 point by the loop-gain plot and P is the number of open-loop right-half-plane poles.

Phase margin

How much extra phase lag the loop can tolerate at the gain cross-over frequency before instability: M_f = 180° + ∠A₀(jω_c). A larger phase margin means a better-damped, more robust closed loop.

Internal Model Principle

To track or reject a persistent signal (step, ramp, sinusoid) with zero steady-state error robustly, the controller must contain a model of that signal's generating polynomial as its own poles.

PID controller

A compensator C = K_p + K_i/s + K_d·s that combines proportional action, integral action (which removes steady-state step error) and derivative action (which adds phase lead and damping); a practical realisation rolls off the derivative term to keep it proper.

ELEN90055 FAQ

Questions students ask in each ELEN90055 chapter

Why does feedback keep appearing when feedforward gives perfect tracking?

Feedforward tracks perfectly only when the plant model is exact and there are no disturbances. The moment the real plant differs from the model, or an unmeasured disturbance acts, feedforward passes the full error through (its sensitivity is effectively 1). Feedback measures the output and corrects, attenuating both the plant error and the disturbance by the sensitivity S = 1/(1 + CG). That robustness, not nominal tracking, is why the loop is closed. Chapter 01 · Introduction to feedback control

What is the difference between the time-domain and frequency-domain descriptions of a system?

They describe the same LTI system in two equivalent ways. In the time domain the system is a differential equation, or equivalently the output is the convolution of the input with the impulse response g(t). The Laplace transform turns this into algebra: the transfer function G(s) = Y(s)/U(s), where convolution becomes multiplication Y(s) = G(s)U(s). Transient shape and initial-value behaviour are clearest in the time domain, while interconnection of blocks and steady-state response to sinusoids are trivial in the s-domain, so you switch to whichever makes the question easy. Chapter 02 · System models and block diagrams

Why linearise at all instead of solving the nonlinear ODE directly?

Almost every classical-control tool - transfer functions, poles and zeros, Bode, root locus, Nyquist, and the PID / lead-lag design procedures - is defined only for linear time-invariant systems. Linearising about an operating point produces a small-signal model those tools can act on, and it is accurate for the small perturbations feedback loops are designed to keep the plant near. The price is that the model is local: a different operating point gives different coefficients and can even change the stability verdict, so you must state the point you linearised at. Chapter 03 · ODE models and linearization

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. Chapter 04 · LTI systems, convolution, Laplace and transfer functions

Do zeros affect whether a system is stable?

No. Stability is decided by the poles, the denominator roots: a system is BIBO stable exactly when every pole has a negative real part. Zeros and the gain reshape the transient and the frequency response, and a right-half-plane zero causes undershoot, but moving or adding a zero can never stabilise or destabilise the system. When you are asked about stability, look only at the pole locations. Chapter 05 · Poles, zeros, stability and frequency response

Why can't I just cancel an unstable plant pole with a controller zero?

Because the characteristic polynomial phi = A Q + B P is built from the plant and controller polynomials before any cancellation, so the right-half-plane factor stays in phi and leaves a closed-loop pole in the right-half plane. The reference-to-output map T0 may look stable after the cancellation, but the unstable mode reappears in the input-disturbance sensitivity Si0, and a disturbance at the plant input makes the output grow without bound. The correct approach is to stabilise the unstable pole by feedback (for example a large enough proportional or a designed compensator), which shifts the pole into the left-half plane rather than hiding it. Chapter 06 · Closed-loop sensitivity functions and performance

When should I use Routh, root locus, or Nyquist?

Use Routh-Hurwitz when you have a polynomial characteristic equation and want a yes/no stability answer or a parameter range — it is purely algebraic and needs no plot. Use the root locus when you want to see how the closed-loop poles move as one gain varies, and to choose a gain for a target damping or speed. Use Nyquist when the loop contains something Routh cannot handle as a polynomial, such as a pure time delay e^(-sT), or when the plant is open-loop unstable (non-zero P), because Nyquist counts unstable poles as Z = N + P and also hands you the gain and phase margins directly. Chapter 07 · Stability analysis: Routh, root-locus, Nyquist

Are the phase margin and the gain margin enough to declare a loop robust?

Not on their own. Both margins are measured at a single frequency each: the phase margin at the gain cross-over and the gain margin at the phase cross-over. A Nyquist plot can show a comfortable phase margin and a comfortable gain margin yet still pass close to -1 at some frequency in between, which means a tall sensitivity peak and poor robustness. That is why the peak sensitivity ||S0||_inf, the reciprocal of the closest approach to -1 over all frequencies, is the honest single number, and why the cross-over relation |S0(jwc)| = 1/(2 sin(Mf/2)) links a small phase margin to a large peak. Chapter 08 · Relative stability and robustness

Why can a proportional controller never remove the steady-state error to a step?

With proportional control C = K the step error is e_ss = 1/(1 + K G0(0)), which is finite for any finite gain and only shrinks as K grows, never reaching zero. The interpolation view explains this: zero steady-state error to a step needs S0(0) = 0, which requires a pole at s = 0 in the open loop. A pure gain adds no such pole, so the offset stays. Adding a controller pole at s = 0 (integral action) is exactly the internal model of a step, and it drives the error to zero robustly. Chapter 09 · Fundamental limits, internal model principle, feedforward

How do I decide between a lag/PI and a lead/PD compensator?

Match the compensator to the specification it fixes. A lag or PI raises the loop gain at low frequency, which improves steady-state accuracy (a lag leaves a small residual offset; a PI removes the step offset entirely because its pole sits at the origin), but both add negative phase and so do not help the phase margin. A lead or PD adds positive phase around cross-over, which raises the phase margin and hence the damping, and it lets you push the cross-over higher for a faster response — but it does nothing for low-frequency accuracy and it amplifies high-frequency noise. So if the spec is about steady-state error, reach for lag/PI; if it is about overshoot, phase margin or speed, reach for lead/PD; if you need both at once, cascade them as a lead-lag, which in its limiting three-term form is a PID. Chapter 10 · Design: specifications and P/lag/lead/PID compensation

Why should the loop-gain slope be about -20 dB/dec at cross-over?

By Bode's gain-phase relation, the phase near cross-over is set by the magnitude slope there. A -20 dB/dec slope corresponds to roughly -90 degrees of phase, giving a phase margin near 90 degrees (in practice about 65 to 90 degrees) - healthy. A -40 dB/dec slope corresponds to about -180 degrees, leaving essentially no phase margin, so the loop rings or goes unstable. You can roll off faster well above cross-over for noise, just not through it. Chapter 11 · Classical loop-shaping and Bode gain-phase relation

Why must the plant polynomials A(s) and B(s) be coprime for pole placement to work?

Coprimeness means A and B share no common factor, which is exactly the condition for the Sylvester (resultant) matrix to be invertible. When it is invertible, the Diophantine equation A L + B P = A_cl has a unique solution for the controller coefficients, so any target characteristic polynomial of the right degree is achievable and the closed-loop poles can be placed anywhere. If A and B did share a factor, that factor would divide A L + B P as well, so it would be forced to appear in A_cl: you cannot move a plant pole that is cancelled by a plant zero. In state-space language, coprimeness of A and B is the transfer-function version of the plant being controllable and observable. Chapter 12 · Pole-placement synthesis

How to study for the exam

Treat ELEN90055 as one repeating loop rather than a list of tricks. Because the University of Melbourne final is open book, the marks are not in recall but in method: set up the model, linearise cleanly, derive the transfer function, then read stability and margins off the right test.

Work the recurring exam archetypes by hand — a linearisation to a transfer function, a Routh parameter range, a root-locus or Nyquist stability check, and a Bode-based loop design — until each is automatic, and always finish by re-checking the time-domain step response, because pole dominance is never guaranteed.

Keep up with the weekly problem sets and the LEGO-robot workshops, since the same modelling-and-control skills carry into both the group reports and the exam. Pace your practice at the exam's own rate of about 3.6 minutes per mark so the three-hour paper never feels rushed.

Reserve SWOTVAC — the study week before exams — for final revision under timed conditions, working through past exam papers and practice problems from the subject's Canvas site so that every archetype is fresh going into the paper.

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