---
title: "Lead compensator for 5/[(s+2)(s+20)], Kv ≥ 20"
subject: "Control Systems"
chapter: 7
key: "semester_3/control_systems:lead-5-s2-s20-kv20"
source: https://www.bctengineer.com/q/semester_3/control_systems/lead-5-s2-s20-kv20
site: BCT Engineering
---

# Lead compensator for 5/[(s+2)(s+20)], Kv ≥ 20

Control Systems · Chapter 7: Performance Specifications and Compensation Design

- Asked in: 2080 Bhadra, 12 marks
- Video: [Problem 3 (Lead Compensator)](https://www.youtube.com/watch?v=PBFw8jiRoR0) — Kowsi Teaches
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/lead-5-s2-s20-kv20
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7

## Question

Design a lead compensator for a unity feedback system with open loop transfer function G(s) = 5/[(s+2)(s+20)] such that velocity error constant at least 20 per second, phase margin at least 35° and gain margin at least 10 dB.

## Short answer

Assuming G(s)=K/[s(s+2)(s+20)], set K=800 for Kv=20 s⁻¹. The uncompensated system has PM≈1.6°. A lead compensator with φm=55°, α=0.1 gives T=0.3 s; thus Gc(s)=(1+0.3s)/(1+0.03s). The compensated system achieves PM≈38° and GM≈12 dB.

## Answer

**Assumption:** as printed, $G(s)=\frac{5}{(s+2)(s+20)}$ is type 0, so $K_v=\lim_{s\to0}sG(s)=0$ for any gain and no lead network can give $K_v=20$. The intended plant must have an integrator, so the design is done for $G(s)=\dfrac{K}{s(s+2)(s+20)}$, with the gain chosen to give $K_v=20$.

Lead network: $G_c(s)=\dfrac{1+Ts}{1+\alpha Ts}$, $\sin\phi_m=\frac{1-\alpha}{1+\alpha}$ at $\omega_m=\frac{1}{T\sqrt\alpha}$, gain added there $10\log(1/\alpha)$.

#### Step 1: Gain

$$
K_v=\lim_{s\to0}sG(s)=\frac{K}{2\times20}=20\Rightarrow K=800
$$

$$
G(s)=\frac{800}{s(s+2)(s+20)}=\frac{20}{s(1+0.5s)(1+0.05s)}
$$

#### Step 2: Uncompensated system

Corners at 2 and 20 rad/s; asymptotes $-20$, $-40$, $-60$ dB/dec; 26 dB at $\omega=1$.

$$
\phi=-90^\circ-\tan^{-1}0.5\omega-\tan^{-1}0.05\omega
$$

| $\omega$ | 1 | 3.33 | 6.03 | 6.32 | 10.5 | 20 |
|---|---|---|---|---|---|---|
| $|G|$ (dB) | 25.0 | 9.7 | 0 | $-0.8$ | $-10.0$ | $-23.1$ |
| $\phi$ | $-119^\circ$ | $-158.5^\circ$ | $-178.4^\circ$ | $-180^\circ$ | $-197^\circ$ | $-219^\circ$ |

$\omega_{gc}=6.03$ rad/s, PM $=1.6^\circ$; $\omega_{pc}=6.32$ rad/s, GM $=0.8$ dB. Barely stable.

#### Step 3: Phase lead

$$
\phi_m=35^\circ-1.6^\circ+\epsilon
$$

Because of the pole at 20 rad/s, the plant phase falls quickly as the crossover moves right. Trials with $\epsilon=5^\circ$–$12^\circ$ ($\phi_m=38^\circ$–$45^\circ$) give PM of only 28°–33°. A larger margin is needed: take $\phi_m\approx55^\circ$.

#### Step 4: α

$$
\alpha=\frac{1-\sin55^\circ}{1+\sin55^\circ}\approx0.1\quad(\phi_m=54.9^\circ)
$$

#### Step 5: New crossover

The network adds $10\log10=10$ dB, so $\omega_m$ is where $|G|=-10$ dB:

$$
\frac{20}{\omega\sqrt{1+0.25\omega^2}\sqrt{1+0.0025\omega^2}}=0.316\Rightarrow\omega_m=10.5\ \text{rad/s}
$$

#### Step 6: T

$$
T=\frac{1}{10.5\times\sqrt{0.1}}=0.30\ \text{s},\qquad\alpha T=0.03\ \text{s}
$$

$$
G_c(s)=\frac{1+0.3s}{1+0.03s}
$$

(zero at 3.33 rad/s, pole at 33.3 rad/s; an amplifier of gain $1/\alpha=10$ restores the DC gain).

#### Step 7: Check

$$
G_c(s)G(s)=\frac{800(1+0.3s)}{s(s+2)(s+20)(1+0.03s)}
$$

| $\omega$ | 1 | 3.33 | 6.03 | 10.46 | 20 | 24.4 |
|---|---|---|---|---|---|---|
| $|G_cG|$ (dB) | 25.4 | 12.7 | 6.2 | 0 | $-8.7$ | $-11.9$ |
| $\phi$ | $-104^\circ$ | $-119^\circ$ | $-128^\circ$ | $-141.9^\circ$ | $-170^\circ$ | $-180^\circ$ |

$$
PM=180^\circ-141.9^\circ=38.1^\circ\ (\ge35^\circ),\qquad GM=11.9\ \text{dB}\ (\ge10\ \text{dB})
$$

| Item | Before | After |
|---|---|---|
| $K_v$ | 20 | 20 |
| PM | 1.6° | 38.1° |
| GM | 0.8 dB | 11.9 dB |
| $\omega_{gc}$ | 6.0 rad/s | 10.5 rad/s |

**Answer (for $G=K/[s(s+2)(s+20)]$): K = 800 and $G_c(s)=\dfrac{1+0.3s}{1+0.03s}$ give Kv = 20 s⁻¹, PM ≈ 38°, GM ≈ 12 dB.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why does the answer change the plant to K/[s(s+2)(s+20)]?
   A. As printed it is type 0, so Kv = 0 for any gain and no lead network can give Kv = 20
   B. The printed plant is unstable
   C. Lead networks only work on plants with three poles
   D. The printed plant already gives Kv = 20
   Answer: A. Without an integrator, Kv = lim sG(s) = 0, so an integrator is assumed. (Key point: Type of printed plant)

2. For G(s) = K/[s(s+2)(s+20)], what K gives Kv = 20?
   A. K = 20
   B. K = 800
   C. K = 40
   D. K = 80
   Answer: B. Kv = K/(2 x 20) = 20 gives K = 800. (Key point: Gain for Kv)

3. How is alpha found for phi_m = 55 deg?
   A. alpha = sin55, about 0.82
   B. alpha = (1 + sin55)/(1 - sin55), about 10
   C. alpha = (1 - sin55)/(1 + sin55), about 0.1
   D. alpha = 55/180, about 0.31
   Answer: C. sin(phi_m) = (1-alpha)/(1+alpha) gives alpha about 0.1; a larger phi_m than the bare 35 deg need is used because the pole at 20 rad/s makes phase fall quickly. (Key point: Alpha from phase lead)

4. Where is the new gain crossover chosen, and why?
   A. Where the uncompensated gain is 0 dB
   B. Where the uncompensated phase is -180 deg
   C. At the corner 20 rad/s of the plant
   D. Where the uncompensated gain is -10 dB, since the lead adds 10log(1/alpha) = 10 dB at wm
   Answer: D. The lead adds 10 dB at wm, so wm is placed where |G| = -10 dB (10.5 rad/s) so the compensated gain crosses 0 dB there. (Key point: New crossover)

## Related in Control Systems

- [Lead compensator for 2/[s(1+0.5s)], Kv > 20, PM 50°](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-2-s-1-05s-kv20) — Control Systems, chapter 7
- [Lead compensator for K/[s(1+0.2s)], Kv=10, PM=50°](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-k-s-1-02s-kv10) — Control Systems, chapter 7, asked 2×
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- [Lead compensator for K/[s(1+0.2s)], Kv≥20, PM≥44°](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-k-s-1-02s-kv20) — Control Systems, chapter 7, asked 2×

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