Questions: Sensitivity and Complementary Sensitivity Functions

Question 1

A control engineer wants to design a feedback controller that simultaneously achieves: (1) excellent disturbance rejection at ALL frequencies (|S(jω)| ≈ 0 for all ω) and (2) perfect tracking at ALL frequencies (|T(jω)| ≈ 1 for all ω). Is this achievable?

• A. Yes — sufficiently high loop gain achieves both goals simultaneously
• B. Yes — a minimum-phase plant with no right-half-plane zeros allows both
• C. No — the identity S + T = 1 makes it impossible to have |S| ≈ 0 and |T| ≈ 1 at different frequencies simultaneously
• D. No — but only because real actuators saturate and cannot provide infinite gain

Answer: (C)

S + T = 1 is an algebraic identity that holds at every frequency. If |T(jω)| ≈ 1 at some frequency, then |S(jω)| ≈ 0 there — which is actually good for tracking. The problem is at high frequencies: making T large there to achieve tracking amplifies sensor noise into the control signal. Conversely, any frequency where |S| is forced small forces |T| ≈ 1 there, which is fine for low frequencies but harmful at high frequencies. The fundamental constraint is not actuator saturation but the mathematical identity S + T = 1 itself.

Question 2

A designer adds integral control to eliminate steady-state tracking error, which forces S(0) = 0 (zero sensitivity at DC). What does Bode's sensitivity integral theorem imply about S(jω) at other frequencies?

• A. The sensitivity function remains near zero at all frequencies due to the integral's persistent correction
• B. The sensitivity function must peak above 1 at some finite frequency to compensate, because the integral of log|S(jω)| over all frequencies must equal zero
• C. The sensitivity function becomes exactly 1 at all frequencies above the crossover frequency
• D. No constraint is imposed — the sensitivity function can be made arbitrarily small at all frequencies with a high-gain integral controller

Answer: (B)

Bode's sensitivity integral states that for a stable, minimum-phase loop, ∫log|S(jω)|dω = 0. If integral action pushes log|S| strongly negative at DC (large negative contribution), the integral constraint forces a compensating positive region — a sensitivity peak — at some finite frequency. This is the 'waterbed effect': pushing S down in one band forces it up in another. The integral controller trades excellent DC rejection for a sensitivity hump at moderate frequencies, which manifests as oscillatory transients in the step response.

Question 3

The identity S(s) + T(s) = 1 means that reducing sensitivity to disturbances at some frequencies necessarily increases sensitivity at other frequencies.

• T. True
• F. False

Answer: True

This is the waterbed effect, formalized by Bode's sensitivity integral. S + T = 1 is an algebraic identity: wherever |S| is small (good disturbance rejection), |T| is close to 1 (good tracking, but also high susceptibility to sensor noise at that frequency). More importantly, Bode's integral theorem shows that suppressing |S| in one frequency band must be compensated by amplification in another. There is no 'free lunch' in feedback design — every improvement in one frequency region comes at a cost somewhere else.

Question 4

Increasing loop gain uniformly across most frequencies reduces the sensitivity function S(jω) everywhere, simultaneously improving disturbance rejection and tracking without any penalty.

• T. True
• F. False

Answer: False

While increasing loop gain does reduce |S| at frequencies where it was already large (improving disturbance rejection and tracking there), the waterbed effect prevents a uniform reduction everywhere. At high frequencies, high loop gain means the feedback loop amplifies sensor noise — large |T| at high frequencies. Additionally, practical plants have phase lag that causes the loop to become unstable at sufficiently high gain. Bode's sensitivity integral formally shows that reducing sensitivity below its open-loop value in one band requires a compensating sensitivity increase elsewhere. Uniform improvement across all frequencies is mathematically impossible for any realizable system.

Question 5

What is the 'waterbed effect' in control systems, and why does it make it impossible to achieve arbitrarily good disturbance rejection across all frequencies simultaneously?

Model answer: The waterbed effect refers to the consequence of Bode's sensitivity integral: for a stable, minimum-phase feedback loop, the area under log|S(jω)| over all frequencies is fixed (equals zero for minimum-phase systems). Suppressing sensitivity in one frequency band — like adding integral action to eliminate DC error — creates a compensating sensitivity peak at another frequency. You cannot push the 'water' down everywhere; reducing it in one region forces it up in another. This is why every feedback design involves tradeoffs: tight performance at low frequencies, where disturbances matter, comes at the cost of reduced robustness at higher frequencies, where unmodeled dynamics and sensor noise dominate.

The waterbed effect is not an engineering limitation that better technology could overcome — it is a mathematical constraint on the class of stable feedback systems. It explains why there is no universally optimal controller: every controller design is a choice about where to push the sensitivity waterbed, trading off low-frequency performance against high-frequency robustness. Understanding this constraint is central to robust control design and explains why bandwidth limitations are fundamental rather than incidental.