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ELEN90055 · Control Systems

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Chapter 6 of 12 · ELEN90055

Closed-loop sensitivity functions and performance

This chapter is the analytical core of ELEN90055 Control Systems at the University of Melbourne: once a loop is closed, every input-output relation is one of just four transfer functions — the sensitivity S0, the complementary sensitivity T0, the input-disturbance sensitivity Si0 and the control sensitivity Su0 — all built on the loop gain L = C G0 and tied together by the identity S0 + T0 = 1.

It reads the four functions through the loop gain (tracking where |L| is large, robustness where |L| is small), gets the steady-state error from the final-value theorem, and imposes internal stability via the characteristic polynomial — including the rule that forbids cancelling an unstable plant pole with a controller zero. It corresponds to the frequency-domain analysis part of the course (Part III, loosely following Goodwin, Graebe & Salgado).

In this chapter

What this chapter covers

  • 01Define the four closed-loop sensitivities S0, T0, Si0, Su0 and state which signal each one maps to which output, with units
  • 02Use the three algebraic identities S0 + T0 = 1, Si0 = G0 S0 and Su0 = C S0 = T0/G0
  • 03Write the output equation Y = T0 H R + S0 Do + Si0 Di - T0 Dm and the matching control equation
  • 04Read the loop gain: |L| >> 1 at low frequency for tracking, |L| << 1 at high frequency for robustness, and locate the cross-over where |L| = 1
  • 05Compute the steady-state error e_ss = 1/(1 + L(0)) for a step and 1/(K Ghat(0)) for a ramp using the final-value theorem
  • 06Apply the Internal Model Principle: a step needs a controller pole at s = 0, a sinusoid at omega0 needs the factor s^2 + omega0^2
  • 07Form the characteristic polynomial phi = A Q + B P and test internal stability (all roots in the open left-half plane)
  • 08Explain why cancelling a right-half-plane pole or zero is forbidden, and show the hidden mode survives in Si0
Worked example · free

Build all four sensitivities and read off performance

Q [8 marks]. A stable plant G0(s) = 3/(s + 2) (DC gain 1.5, time constant 0.5 s) is controlled by a proportional C = K = 4 in a unity-feedback loop. (a) Write all four sensitivity functions. (b) Verify the identities. (c) Give the DC values and the steady-state error to a unit-step reference. (d) Find the loop-gain cross-over frequency and confirm internal stability.
  • +1Loop gain and common denominator. L = C G0 = (4 × 3)/(s + 2) = 12/(s + 2), so 1 + L = (s + 2 + 12)/(s + 2) = (s + 14)/(s + 2), and the DC loop gain is L(0) = 12/2 = 6.
  • +2(a) The four functions. S0 = 1/(1 + L) = (s + 2)/(s + 14); T0 = L/(1 + L) = 12/(s + 14); Si0 = G0 S0 = [3/(s + 2)]·[(s + 2)/(s + 14)] = 3/(s + 14); Su0 = C S0 = 4(s + 2)/(s + 14).
  • +1(b) Identities. S0 + T0 = [(s + 2) + 12]/(s + 14) = (s + 14)/(s + 14) = 1 (check). And T0/G0 = [12/(s + 14)] / [3/(s + 2)] = 4(s + 2)/(s + 14) = Su0 (check).
  • +1(c) DC values. S0(0) = 2/14 = 1/7 ≈ 0.143; T0(0) = 12/14 = 6/7 ≈ 0.857 (sum = 1); Si0(0) = 3/14 ≈ 0.214; Su0(0) = 8/14 = 4/7 ≈ 0.571.
  • +2Steady-state error. To a unit-step reference, e_ss = S0(0) = 1/(1 + L(0)) = 1/7 ≈ 0.143, a 14.3% offset (the output settles at 85.7% of the setpoint). A proportional loop cannot remove it — that needs an integrator (Internal Model Principle).
  • +1(d) Cross-over and internal stability. |L(jωc)| = 1 gives 12/√(ωc^2 + 4) = 1, so ωc^2 + 4 = 144, ωc = √140 ≈ 11.8 rad/s. The only closed-loop pole is at s = -14, and all four sensitivities are proper with that single stable pole, so the loop is internally stable.
S0 = (s + 2)/(s + 14), T0 = 12/(s + 14), Si0 = 3/(s + 14), Su0 = 4(s + 2)/(s + 14); DC values 1/7, 6/7, 3/14, 4/7 with S0(0) + T0(0) = 1; step offset e_ss = 1/7 ≈ 14.3%; cross-over ωc = √140 ≈ 11.8 rad/s; the single closed-loop pole is s = -14, so the loop is internally stable.
Sia tip — Always verify S0(0) + T0(0) = 1 as a free sanity check, and apply the final-value theorem for e_ss only after confirming the loop is stable. The same S0(0) that gives the step offset is what attenuates output disturbances at DC.
Glossary

Key terms

Sensitivity S0
S0 = 1/(1 + L) with L = C G0 the loop gain. It maps an output disturbance to the output and the reference to the error; dimensionless. Small where |L| is large, giving tight tracking and disturbance rejection.
Complementary sensitivity T0
T0 = L/(1 + L): the reference-to-output transfer function and the (negated) map from sensor noise to the output. Dimensionless. Satisfies S0 + T0 = 1 at every frequency.
Input-disturbance sensitivity Si0
Si0 = G0/(1 + L) = G0 S0: the map from a disturbance at the plant input to the output. Carries the plant's units. Small where |S0| is small but shaped by the plant G0.
Control sensitivity Su0
Su0 = C/(1 + L) = C S0 = T0/G0: the map from reference or output disturbance to the control signal. Carries the controller's units; it grows where the plant gain |G0| is small, a warning about actuator size.
Loop gain L(s)
L = C(s) G0(s), the product of controller and plant around the loop (the course notes also write A0). Its magnitude relative to 1 sets tracking, disturbance rejection and stability margins; cross-over is where |L(jω)| = 1.
Characteristic polynomial
For plant G0 = B/A and controller C = P/Q (coprime), phi(s) = A(s)Q(s) + B(s)P(s). It is the common denominator of all four sensitivities; the loop is internally stable iff all its roots lie in the open left-half plane.
Internal stability
The property that a bounded signal injected anywhere in the loop produces bounded signals everywhere, i.e. all four sensitivities are proper and stable. Stronger than merely T0 being stable.
Forbidden unstable cancellation
Cancelling a right-half-plane plant pole with a controller zero (or vice versa). The cancelled factor survives in phi = A Q + B P, so the loop is internally unstable even though T0 may look stable; the hidden mode appears in Si0.
FAQ

Closed-loop sensitivity functions and performance FAQ

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.

What is the difference between the closed loop being stable and being internally stable?

A loop can have a stable reference-to-output map T0 while another internal map is unstable. Internal stability is the stronger and correct requirement: all four sensitivities S0, T0, Si0 and Su0 must be proper and stable at once, which is equivalent to the characteristic polynomial phi = A Q + B P having all roots in the open left-half plane. Checking only T0 misses hidden unstable modes created by pole-zero cancellation, so always test phi built from the uncancelled polynomials.

Can AI help me with closed-loop sensitivity functions in ELEN90055?

Yes, as a study aid. An AI tutor like Sia can explain the four sensitivities and the identity S0 + T0 = 1 step by step, walk you through forming the characteristic polynomial and testing internal stability, and check your algebra on steady-state error. Use it to understand the method, the sign conventions and the loop-gain interpretation, 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.

Study strategy

Exam move

Anchor this chapter on the four sensitivities and their identities. First, be able to write S0 = 1/(1 + L), T0 = L/(1 + L), Si0 = G0 S0 and Su0 = C S0 = T0/G0 from the loop gain L = C G0, and use S0 + T0 = 1 as a running sanity check on any answer. Second, drill the loop-gain reading: large |L| at low frequency for tracking and disturbance rejection, small |L| at high frequency for noise immunity and robustness, with cross-over where |L| = 1 — almost every 'comment on performance / robustness' part wants this. Third, get steady-state error automatic: e_ss = 1/(1 + L(0)) for a step, and name the Internal Model Principle pole (an integrator for a step, s^2 + omega0^2 for a sinusoid) whenever zero error is required, applying the final-value theorem only after confirming stability. Fourth, and most heavily marked, master internal stability via the characteristic polynomial phi = A Q + B P and never cancel a right-half-plane pole or zero. 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 signs and units, not memorised formulae. Confirm the exact date of the next (Semester 1, ~June 2027) sitting on Canvas.

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