--- title: "Lead compensator for K/[s(s+10)(s+2)], Kv ≥ 800" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:lead-k-s-s10-s2-kv800" source: https://www.bctengineer.com/q/semester_3/control_systems/lead-k-s-s10-s2-kv800 site: BCT Engineering --- # Lead compensator for K/[s(s+10)(s+2)], Kv ≥ 800 Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2079 Bhadra, 12 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/lead-k-s-s10-s2-kv800 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question The forward path transfer function of unity feedback system is given by G(s) = K/[s(s+10)(s+2)]. Design a suitable location of pole zero pair for a lead compensating network so that phase margin is at least 50° and velocity error constant is maintained at least 800 sec⁻¹. ## Short answer K=16000 gives Kv=800 s^-1. Three identical lead stages, each with zero at -25.1 rad/s and pole at -146.2 rad/s (α=0.1716), provide a phase boost. The compensated system achieves gain crossover 60.6 rad/s and phase margin 56.3°≥50° with Kv=800 sec^-1. ## Answer A lead network $\dfrac{1+Ts}{1+\alpha Ts}$ (unity DC gain) is placed so that its maximum phase lead occurs at the new gain crossover frequency. #### Step 1: Gain for $K_v$ $$ K_v = \lim_{s\to0} s\,\frac{K}{s(s+10)(s+2)} = \frac{K}{20} = 800 \;\Rightarrow\; K = 16000 $$ $$ G(s) = \frac{16000}{s(s+2)(s+10)} = \frac{800}{s(0.5s+1)(0.1s+1)} $$ #### Step 2: Uncompensated margins $$ \begin{aligned} |G(j\omega)| &= 1 \text{ at } \omega_{gc} = 24.5\ \text{rad/s} \\ \angle G &= -90^\circ - \tan^{-1}\frac{24.5}{2} - \tan^{-1}\frac{24.5}{10} = -243.2^\circ \\ PM &= -63.2^\circ,\quad GM = -36.5 \text{ dB at } \omega_{pc} = 4.47\ \text{rad/s} \end{aligned} $$ The system is badly unstable. #### Step 3: Phase lead needed $$ \phi_m = 50^\circ - (-63.2^\circ) + 5^\circ \approx 118^\circ $$ One lead network gives at most about $60^\circ$ in practice, so **three identical lead stages** are used. A first trial of $3 \times 40^\circ$ gives only $PM \approx 43^\circ$ because the crossover shifts, so $3 \times 45^\circ$ is chosen: $$ \alpha = \frac{1-\sin45^\circ}{1+\sin45^\circ} = 0.1716 \text{ per stage} $$ #### Step 4: New gain crossover Three stages add $3\times10\log(1/\alpha) = 22.96$ dB at $\omega_m$, so $\omega_m$ is where $|G| = \alpha^{1.5} = 0.0711$: $$ \frac{16000}{\omega\sqrt{\omega^2+4}\sqrt{\omega^2+100}} = 0.0711 \;\Rightarrow\; \omega_m = 60.6\ \text{rad/s} $$ #### Step 5: Pole–zero locations $$ \begin{aligned} T &= \frac{1}{\omega_m\sqrt{\alpha}} = \frac{1}{60.55\times0.4142} = 0.0399\ \text{s} \\ \text{zero: } s &= -\frac{1}{T} = -25.1 \\ \text{pole: } s &= -\frac{1}{\alpha T} = -146.2 \end{aligned} $$ $$ G_c(s) = \left(\frac{1+0.0399s}{1+0.00684s}\right)^3 = 198.0\left(\frac{s+25.1}{s+146.2}\right)^3 $$ #### Step 6: Check $$ G_cG = \frac{16000(1+0.0399s)^3}{s(s+2)(s+10)(1+0.00684s)^3} $$ At $\omega = 60.6$: plant phase $= -90^\circ - 88.1^\circ - 80.6^\circ = -258.7^\circ$; lead $= 3\times45^\circ = 135^\circ$; $PM = 180^\circ - 258.7^\circ + 135^\circ = 56.3^\circ$. | Quantity | Uncompensated | Compensated | |---|---|---| | Gain crossover | 24.5 rad/s | 60.6 rad/s | | Phase margin | $-63.2^\circ$ | $56.3^\circ$ | | Gain margin | $-36.5$ dB | 13.6 dB | | $K_v$ | 800 s$^{-1}$ | 800 s$^{-1}$ | **Answer:** each stage has its zero at $s = -25.1$ and pole at $s = -146.2$ ($\alpha = 0.1716$); three such stages give $PM = 56.3^\circ \ge 50^\circ$ with $K_v = 800\ \text{s}^{-1}$. (Such a large $K_v$ needs a large lead; in practice a lag–lead compensator would give a lower bandwidth and less noise.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For G = K/[s(s+10)(s+2)], what K gives Kv = 800? A. K = 16000 B. K = 160 C. K = 8000 D. K = 800 Answer: A. Kv = K/(10*2) = K/20 = 800, so K = 16000. (Key point: Gain from Kv) 2. Why are three identical lead stages used? A. The plant has three poles B. The needed lead (about 118 degrees) far exceeds the roughly 60 degrees one network can give C. Each stage adds 800 to Kv D. Three stages reduce alpha to 1 Answer: B. The plant PM is -63.2 degrees so about 118 degrees of lead is needed, beyond one practical lead network. (Key point: Why three stages) 3. Three stages add 22.96 dB at omega_m. How is the new crossover found? A. Where |G| = alpha B. Where plant phase is -180 degrees C. Where |G| = 1 D. Where |G| = alpha^1.5 = 0.0711 (-22.96 dB) Answer: D. The plant must be 22.96 dB below 0 dB at omega_m so that the compensated loop crosses 0 dB there. (Key point: Crossover) 4. With T = 1/(omega_m*sqrt(alpha)), where are each stage's zero and pole? A. Zero at -25.1, pole at -146.2 B. Zero at -146.2, pole at -25.1 C. Zero at -2, pole at -10 D. Zero at -60.6, pole at -60.6 Answer: A. Zero = -1/T = -25.1 and pole = -1/(alpha T) = -146.2; the zero is lower in frequency, giving phase lead. 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