ELEN90055 · Control Systems
Introduction to feedback control
This chapter opens ELEN90055 Control Systems at the University of Melbourne by asking the founding question of the subject: given a plant we cannot change, how should we choose its input so its output behaves as we want? It contrasts the two answers — feedforward (open loop) and feedback (closed loop) — and shows why closing the loop, which introduces the factor 1 + CG into every input-output relation, is what delivers disturbance rejection and robustness to plant uncertainty.
It maps to Part I of the course (loosely following Goodwin, Graebe & Salgado, Chapters 1-2) and fixes the block-diagram algebra and the sensitivity language every later chapter reuses.
What this chapter covers
- 01Tell feedforward (open-loop) apart from feedback (closed-loop) control and state what each can and cannot do
- 02Write the negative-feedback error e = r - y and see where the sign convention comes from
- 03Reduce a series/parallel/feedback block diagram to a closed-loop transfer function
- 04Define the sensitivity S = 1/(1+CG) and complementary sensitivity T = CG/(1+CG), and use the identity S + T = 1
- 05Interpret the loop gain L = CG: large at low frequency for tracking, small at high frequency for noise rejection
- 06Show that the relative sensitivity of the closed loop to plant error equals S (the robustness result)
- 07Compute the steady-state step offset e_ss = 1/(1 + K G(0)) for a proportional loop
- 08Recognise the same sense-compare-correct loop in engineered and natural systems
Close a proportional loop on a first-order plant
- +1Loop gain. L(s) = CG = (9 × 3)/(4s + 1) = 27/(4s + 1), so the DC loop gain is L(0) = 27.
- +2Closed-loop transfer function. T = L/(1 + L) = 27/[(4s + 1) + 27] = 27/(4s + 28), i.e. T(s) = (27/28)/((4/28)s + 1) = 0.964/(0.143 s + 1).
- +1DC gain and speed. T(0) = 27/28 ≈ 0.964; the closed-loop time constant is 4/28 ≈ 0.143 s, about 28× faster than the open-loop 4 s.
- +1Steady-state error. For a unit-step reference, e_ss = 1/(1 + L(0)) = 1/28 ≈ 0.0357 = S(0), a 3.57% offset.
- +1Robustness. A 10% rise in plant gain (3 → 3.3) gives L(0) = 29.7, T(0) = 29.7/30.7 ≈ 0.967; the relative change is only about 0.35% ≈ S(0) × 10%. The 10% plant error is shrunk to well under 1%.
Key terms
- Feedforward (open-loop) control
- The controller computes the plant input from the reference alone, using a model of the plant; the output is never measured. Fast and cannot destabilise the loop, but it corrects neither modelling error nor unmeasured disturbances.
- Feedback (closed-loop) control
- The output is measured and subtracted from the reference to form the error e = r - y, which drives the controller. The loop rejects disturbances and tolerates model error, at the cost of possible instability and sensor noise.
- Loop gain L(s)
- The product C(s)G(s) of controller and plant around the loop (the course notes also write A0 = G0 C). Its size versus 1 governs tracking, disturbance rejection and stability margins.
- Sensitivity S(s)
- S = 1/(1 + CG): the map from an output disturbance to the output, and from the reference to the error. It also equals the relative sensitivity of the closed loop to plant error, so it measures robustness. Small where the loop gain is large.
- Complementary sensitivity T(s)
- T = CG/(1 + CG): the reference-to-output transfer function of the closed loop, and the map from sensor noise to the output. Satisfies S + T = 1 at every frequency.
- Steady-state error
- The residual error as time tends to infinity. For a step reference under proportional control C = K, e_ss = 1/(1 + K G(0)) = S(0); it shrinks as the DC loop gain rises but is never zero without integral action.
- Robustness to uncertainty
- The property that the closed-loop behaviour changes little when the true plant differs from the model. Because a relative plant change delta-G/G produces a relative closed-loop change S times as large, feedback with small |S| is robust.
Introduction to feedback control FAQ
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.
What is the difference between the loop gain L and the closed-loop transfer function T?
L = CG is the open-loop gain measured around the loop; it is what you use for stability margins, root locus and Nyquist. T = L/(1 + L) is the closed-loop reference-to-output transfer function. They are different objects: a large L gives a T close to 1 and a small sensitivity S = 1/(1 + L), but L itself is not the closed-loop response.
Can AI help me with introduction to feedback control in ELEN90055?
Yes, as a study aid. An AI tutor like Sia can explain feedforward vs feedback step by step, walk you through reducing a block diagram to a closed-loop transfer function, and check your algebra on sensitivity and steady-state error. Use it to understand method and sign 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
Treat Part I as the vocabulary the rest of the subject speaks. First, be able to draw the standard loop and write e = r - y and T = CG/(1 + CG) from memory, checking the denominator is always 1 + L (the plus sign is negative feedback). Second, drill the pair S = 1/(1 + CG) and T = CG/(1 + CG) and the identity S + T = 1 until they are automatic, because every later 'comment on robustness / disturbance rejection' question is really asking for this in the language of S and T. Third, practise the loop-gain reading: large |L| at low frequency for tracking and rejection, small |L| at high frequency for noise immunity. 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, so always define your symbols and re-check the time-domain response after any frequency-domain step. Confirm the exact date of the next (Semester 1, ~June 2027) sitting on Canvas.
