--- title: "Compensator for 4/[s(s+2)]: Kv, PM, GM" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:comp-4-s-s2-kv20-pm50" source: https://www.bctengineer.com/q/semester_3/control_systems/comp-4-s-s2-kv20-pm50 site: BCT Engineering --- # Compensator for 4/[s(s+2)]: Kv, PM, GM Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2070 Chaitra (old course), 16 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/comp-4-s-s2-kv20-pm50 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question A system has open loop transfer function Gf(s) = 4/[s(s+2)]. It is desired to design a compensator so that the static velocity error constant Kv is 20 sec⁻¹, phase margin is at least 50° and gain margin is at least 10 dB. ## Short answer Lead compensator Gc(s)=46.1 (s+4.27)/(s+19.66) raises the gain to meet Kv=20 s⁻¹. The compensated system has a phase margin of 52.3° and infinite gain margin, satisfying the specifications. ## Answer The uncompensated system has a low phase margin once the gain is raised for $K_v = 20$, so a **phase-lead compensator** is designed using the Bode plot. Take $G_c(s) = K_c\,\alpha\,\dfrac{Ts+1}{\alpha Ts+1}$, $0 < \alpha < 1$. #### Step 1: Gain for $K_v$ $$ K_v = \lim_{s\to0} s\,G_c(s)G_f(s) = K_c\alpha\cdot\frac{4}{2} = 2K_c\alpha = 20 \;\Rightarrow\; K = K_c\alpha = 10 $$ Gain-adjusted system: $G_1(s) = \dfrac{40}{s(s+2)} = \dfrac{20}{s(0.5s+1)}$. #### Step 2: Phase margin of $G_1$ Gain crossover where $|G_1(j\omega)| = 1$: $$ \begin{aligned} \frac{40}{\omega\sqrt{\omega^2+4}} &= 1 \;\Rightarrow\; \omega^4 + 4\omega^2 - 1600 = 0 \\ \omega^2 &= 38.05,\quad \omega_{gc} = 6.17\ \text{rad/s} \\ \angle G_1 &= -90^\circ - \tan^{-1}(6.17/2) = -162.0^\circ \\ PM &= 180^\circ - 162.0^\circ = 18.0^\circ \end{aligned} $$ The phase never reaches $-180^\circ$ (second-order type-1 system), so $GM = \infty$; the gain margin requirement of 10 dB is automatically met. #### Step 3: Phase lead needed Add about $8^\circ$ for the extra lag caused by the shift of the crossover frequency: $$ \phi_m = 50^\circ - 18^\circ + 8^\circ = 40^\circ $$ $$ \alpha = \frac{1-\sin 40^\circ}{1+\sin 40^\circ} = \frac{0.357}{1.643} = 0.217 $$ #### Step 4: New gain crossover frequency The compensator adds $10\log(1/\alpha) = 6.63$ dB at $\omega_m$. So $\omega_m$ is where $|G_1| = -6.63$ dB, i.e. $|G_1| = \sqrt{\alpha} = 0.466$: $$ \frac{40}{\omega\sqrt{\omega^2+4}} = 0.466 \;\Rightarrow\; \omega_m = 9.16\ \text{rad/s} $$ #### Step 5: Corner frequencies $$ \begin{aligned} T &= \frac{1}{\omega_m\sqrt{\alpha}} = \frac{1}{9.16\times0.466} = 0.234\ \text{s} \\ \text{zero: } \frac{1}{T} &= 4.27\ \text{rad/s}, \qquad \text{pole: } \frac{1}{\alpha T} = 19.66\ \text{rad/s} \\ K_c &= \frac{10}{\alpha} = 46.1 \end{aligned} $$ #### Step 6: Compensator and check $$ G_c(s) = 46.1\,\frac{s+4.27}{s+19.66} = 10\,\frac{0.234s+1}{0.0508s+1} $$ $$ G_c(s)G_f(s) = \frac{40(0.234s+1)}{s(s+2)(0.0508s+1)} $$ | Quantity | Uncompensated ($K=10$) | Compensated | |---|---|---| | Gain crossover | 6.17 rad/s | 9.15 rad/s | | Phase margin | 18.0° | 52.3° | | Gain margin | $\infty$ | $\infty$ | | $K_v$ | 20 s$^{-1}$ | 20 s$^{-1}$ | **Answer:** $G_c(s) = 46.1\dfrac{s+4.27}{s+19.66}$ gives $K_v = 20\ \text{s}^{-1}$, $PM = 52.3^\circ\ (\ge 50^\circ)$ and $GM = \infty\ (\ge 10$ dB$)$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For Gf(s) = 4/[s(s+2)] with compensator Kc*alpha*(Ts+1)/(alpha*Ts+1), how is the required Kv = 20 met? A. Kc*alpha = 10, since Kv = 2*Kc*alpha B. Kc*alpha = 5, since Kv = 4*Kc*alpha C. alpha = 10, since Kv = alpha*2 D. Kc = 20, since Kv = Kc Answer: A. Kv = lim s*Gc*Gf = Kc*alpha*4/2 = 2*Kc*alpha, so Kc*alpha = 10 and the gain-adjusted plant is 40/[s(s+2)]. (Key point: Gain from Kv) 2. Why is a phase-lead compensator chosen here? A. Steady-state error must be reduced without changing the crossover B. The system is type 0 and needs an integrator C. The gain margin of the uncompensated system is negative D. The gain-adjusted system has PM of only about 18 degrees, below the 50 degrees needed Answer: D. After setting K = 10 the phase margin is 18 degrees, so positive phase must be added near crossover. (Key point: Why lead is needed) 3. Why is the 10 dB gain margin requirement automatically satisfied? A. The lead network adds 10 dB at every frequency B. Kv = 20 forces GM to be 10 dB C. The compensator zero cancels the plant pole D. Phase never reaches -180 degrees for this second-order type-1 system, so GM is infinite Answer: D. A second-order type-1 system has phase approaching -180 degrees only asymptotically, so there is no phase crossover and GM is infinite. (Key point: Gain margin) 4. The required phase lead is 50 - 18 + 8 = 40 degrees. How is alpha found, and why is the extra 8 degrees added? A. alpha = 1/sin40; 8 degrees covers the steady-state error B. alpha = (1-sin40)/(1+sin40) = 0.217; 8 degrees covers the plant phase lost as crossover moves up C. alpha = (1+sin40)/(1-sin40); 8 degrees covers gain margin D. alpha = sin40; 8 degrees covers the noise Answer: B. sin(phi_m) = (1-alpha)/(1+alpha) gives alpha = 0.217. The crossover shifts to a higher frequency where the plant phase is lower, so extra lead is allowed for. 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