---
title: "Lead compensator for K/[s(1+0.2s)], Kv≥20, PM≥44°"
subject: "Control Systems"
chapter: 7
key: "semester_3/control_systems:lead-k-s-1-02s-kv20"
source: https://www.bctengineer.com/q/semester_3/control_systems/lead-k-s-1-02s-kv20
site: BCT Engineering
---

# Lead compensator for K/[s(1+0.2s)], Kv≥20, PM≥44°

Control Systems · Chapter 7: Performance Specifications and Compensation Design

- Asked in: 2073 Shrawan, 10 marks; 2067 Asar (old course), 16 marks (asked 2 times)
- 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-k-s-1-02s-kv20
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7

## Question

The open loop transfer function of a unity feedback system is G(s) = K/[s(1+0.2s)]. It is required that Kv ≥ 20 sec⁻¹ and phase margin (φm) = 44°. Design a lead compensating network to satisfy the required specifications.

## Short answer

Lead compensator: Gc(s) = (1+0.126s)/(1+0.06s) with α = 0.473, K=20. Provides Kv = 20 s⁻¹ and phase margin ≈ 44°.

## Answer

Lead network: $G_c(s)=\dfrac{1+Ts}{1+\alpha Ts}$, $\alpha<1$; maximum lead $\sin\phi_m=\frac{1-\alpha}{1+\alpha}$ at $\omega_m=\frac{1}{T\sqrt\alpha}$, where it adds $10\log(1/\alpha)$ dB (after the $1/\alpha$ attenuation is made up by an amplifier).

#### Step 1: K

$$
K_v=\lim_{s\to0}sG(s)=K=20\ \text{s}^{-1}\Rightarrow G(s)=\frac{20}{s(1+0.2s)}
$$

#### Step 2: Uncompensated system

Asymptotes: $-20$ dB/dec through 26 dB at $\omega=1$; corner at 5 rad/s; then $-40$ dB/dec.

$$
\frac{20}{\omega\sqrt{1+0.04\omega^2}}=1\Rightarrow0.04\omega^4+\omega^2-400=0\Rightarrow\omega^2=88.3,\ \omega_{gc}=9.40\ \text{rad/s}
$$

$$
PM=90^\circ-\tan^{-1}(0.2\times9.40)=90^\circ-62.0^\circ=28.0^\circ
$$

#### Step 3: Phase lead needed

$$
\phi_m=44^\circ-28.0^\circ+5^\circ=21.0^\circ
$$

#### Step 4: α

$$
\alpha=\frac{1-\sin21^\circ}{1+\sin21^\circ}=\frac{1-0.358}{1+0.358}=0.473
$$

#### Step 5: New gain crossover

The network adds $10\log(1/0.473)=3.25$ dB at $\omega_m$. Find where $|G|=-3.25$ dB ($=\sqrt\alpha=0.688$):

$$
\frac{20}{\omega\sqrt{1+0.04\omega^2}}=0.688\Rightarrow\omega_m=11.55\ \text{rad/s}
$$

#### Step 6: T

$$
T=\frac{1}{\omega_m\sqrt\alpha}=\frac{1}{11.55\times0.688}=0.126\ \text{s},\qquad\alpha T=0.0595\approx0.06\ \text{s}
$$

Zero at $1/T=7.94$ rad/s, pole at $1/\alpha T=16.8$ rad/s.

$$
G_c(s)=\frac{1+0.126s}{1+0.06s}
$$

Passive R-C network: $\dfrac{E_o}{E_i}=\alpha\dfrac{1+Ts}{1+\alpha Ts}$ with $T=R_1C$, $\alpha=\dfrac{R_2}{R_1+R_2}$; an amplifier of gain $1/\alpha\approx2.1$ restores $K_v$. E.g. $C=1\ \mu$F, $R_1=126\ \text{k}\Omega$, $R_2=113\ \text{k}\Omega$.

#### Step 7: Check

$$
G_c(s)G(s)=\frac{20(1+0.126s)}{s(1+0.2s)(1+0.06s)}
$$

| $\omega$ | 1 | 5 | 7.94 | 9.4 | 11.54 | 16.8 |
|---|---|---|---|---|---|---|
| $|G|$ (dB) | 25.9 | 9.0 | 2.6 | 0 | $-3.2$ | $-9.4$ |
| $|G_cG|$ (dB) | 25.9 | 10.1 | 4.7 | 2.6 | 0 | $-5.0$ |
| $\phi_{comp}$ | $-97.6^\circ$ | $-119.5^\circ$ | $-128.3^\circ$ | $-131.6^\circ$ | $-135.8^\circ$ | $-143.9^\circ$ |

$$
PM=180^\circ-135.8^\circ=44.2^\circ\approx44^\circ
$$

GM = ∞ (phase stays above $-180^\circ$).

| Item | Before | After |
|---|---|---|
| $K_v$ | 20 | 20 |
| $\omega_{gc}$ | 9.40 rad/s | 11.54 rad/s |
| PM | 28° | 44.2° |

**Answer: $G_c(s)=\dfrac{1+0.126s}{1+0.06s}$ (α = 0.473) with K = 20 gives Kv = 20 s⁻¹ and PM ≈ 44°.**

## Check yourself

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

1. For G(s) = K/[s(1+0.2s)], Kv = 20 requires
   A. K = 4
   B. K = 20
   C. K = 100
   D. K = 0.2
   Answer: B. Kv = K = 20. (Key point: K for Kv)

2. With gain crossover 9.40 rad/s, the uncompensated PM is
   A. 62 degrees
   B. 44 degrees
   C. 28 degrees
   D. 90 degrees
   Answer: C. PM = 90 - tan^-1(0.2 x 9.40) = 90 - 62 = 28 degrees. (Key point: Uncompensated PM)

3. Required phi_m = 44 - 28 + 5 = 21 degrees. Alpha is
   A. (1+sin21)/(1-sin21) = 2.1
   B. (1-sin21)/(1+sin21) = 0.473
   C. sin 21 = 0.358
   D. 1/0.473 = 2.1
   Answer: B. For a lead, alpha < 1: alpha = (1 - 0.358)/(1 + 0.358) = 0.473. (Key point: Alpha)

4. Lead adds 3.25 dB at w_m. How are w_m and T obtained?
   A. w_m where plant gain is -3.25 dB (11.55 rad/s); T = 1/(w_m sqrt(alpha)) = 0.126 s
   B. w_m = 9.4 rad/s; T = 0.2 s
   C. w_m where plant gain is +3.25 dB; T = 0.06 s
   D. w_m = 5 rad/s; T = 0.126 s
   Answer: A. The new crossover is at -3.25 dB (11.55 rad/s), then T = 1/(11.55 x 0.688) = 0.126 s and alpha T = 0.06 s. (Key point: New crossover and T)

## Related in Control Systems

- [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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