--- title: "Tracking, disturbance rejection and pole sensitivity" subject: "Control Systems" chapter: 3 key: "semester_3/control_systems:track-reject-sensitivity-160" source: https://www.bctengineer.com/q/semester_3/control_systems/track-reject-sensitivity-160 site: BCT Engineering --- # Tracking, disturbance rejection and pole sensitivity Control Systems · Chapter 3: System Transfer Function and Responses - Asked in: 2073 Shrawan, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/track-reject-sensitivity-160 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3 ## Question For a closed loop system given by the block diagram below: i) Can the system track a step reference input 'r' with zero steady state error? ii) Can the system reject a step disturbance 'w' with zero steady state error? iii) Compute the sensitivity of closed loop transfer function to change in the plant pole at '−2'. [Figure: R → summing point (+, −) → controller 160(s+4)/(s+30) → summing point (+, + with disturbance w) → plant 1/[s(s+2)] → y; unity negative feedback from y] ## Short answer i) Yes, the system is Type 1, giving zero steady-state step error. ii) No, disturbance enters after the controller; steady error = 30/640 ≈ 0.047w. iii) Sensitivity S = -2s(s+30)/(s³+32s²+220s+640), zero at DC. ## Answer Controller $D(s) = \dfrac{160(s+4)}{s+30}$, plant $P(s) = \dfrac{1}{s(s+2)}$. #### Stability first Characteristic equation $1 + DP = 0$: $$ s(s+2)(s+30) + 160(s+4) = s^3 + 32s^2 + 220s + 640 = 0 $$ Routh array: | Row | Col 1 | Col 2 | |---|---|---| | $s^3$ | 1 | 220 | | $s^2$ | 32 | 640 | | $s^1$ | $(32\cdot 220 - 640)/32 = 200$ | 0 | | $s^0$ | 640 | | No sign change, so the closed loop is stable and the final value theorem can be used. #### i) Tracking a step reference $$ G(s) = DP = \frac{160(s+4)}{s(s+2)(s+30)} $$ There is one pole at the origin, so the system is **Type 1**: $K_p = \infty$. $$ e_{ss} = \frac{1}{1 + K_p} = 0 $$ **Yes**, the system tracks a step reference with zero steady state error. #### ii) Rejecting a step disturbance $w$ With $r = 0$: $$ \frac{Y(s)}{W(s)} = \frac{P}{1 + DP} = \frac{s+30}{s^3 + 32s^2 + 220s + 640} $$ For a unit step $W(s) = 1/s$: $$ y_{ss} = \lim_{s\to 0} s\cdot\frac{1}{s}\cdot\frac{s+30}{s^3+32s^2+220s+640} = \frac{30}{640} = 0.0469 $$ **No**. The disturbance enters after the controller, and the integrator is in the plant, not in the controller. So a steady error of $0.0469w$ (about 4.7% of the disturbance) remains. #### iii) Sensitivity to the plant pole at $-2$ Let the pole be at $-a$, with $a = 2$: $P = \dfrac{1}{s(s+a)}$, $T = \dfrac{DP}{1+DP}$. By the chain rule: $$ S_a^T = S_P^T \cdot S_a^P $$ $$ \begin{aligned} S_P^T &= \frac{1}{1 + DP} = \frac{s(s+2)(s+30)}{s^3 + 32s^2 + 220s + 640} \\ S_a^P &= \frac{\partial P}{\partial a}\cdot\frac{a}{P} = \frac{-1}{s(s+a)^2}\cdot a\,s(s+a) = \frac{-a}{s+a} = \frac{-2}{s+2} \end{aligned} $$ **Answer:** $$ S_a^T = \frac{-2s(s+30)}{s^3 + 32s^2 + 220s + 640} $$ At DC ($s = 0$) the sensitivity is zero, so a change in the pole location has no effect on the steady state output. Its effect grows at higher frequencies. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why check stability before using the final value theorem? A. The theorem needs an unstable system B. Stability determines the type C. It is unnecessary D. The theorem is valid only when the closed loop is stable, confirmed by the Routh array Answer: D. The characteristic equation s^3 + 32s^2 + 220s + 640 has no sign change in the first column. (Key point: Stability first) 2. Can the system track a step reference with zero error? A. Yes, the loop is Type 1 so Kp is infinite B. No, because Kp = 0 C. No, because the controller has no integrator D. Yes, because the disturbance is zero Answer: A. The plant integrator makes G Type 1, so ess = 1/(1 + Kp) = 0. (Key point: Tracking a step) 3. Can it reject a step disturbance w with zero error? A. Yes, because the plant has an integrator B. Yes, because Kp is infinite C. No, the disturbance enters after the controller and the integrator is in the plant, leaving a steady error of 30/640 D. No, because the closed loop is unstable Answer: C. The disturbance enters after the controller and the integrator is in the plant, so Y/W has non-zero DC gain 30/640 = 0.0469. (Key point: Rejecting disturbance) 4. How is the sensitivity of T to the pole at -2 obtained? A. S = 1/(1 + DP) only B. S = S_P^T x S_a^P, giving -2s(s+30)/(s^3 + 32s^2 + 220s + 640) C. S = DP/(1 + DP) D. S = -2/(s+2) only Answer: B. Chain rule with S_P^T = 1/(1+DP) and S_a^P = -a/(s+a); it is zero at DC. 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