---
title: "Lag compensator for k/[s(1+0.1s)(1+0.2s)], Kv = 30"
subject: "Control Systems"
chapter: 7
key: "semester_3/control_systems:lag-k-s-1-01s-1"
source: https://www.bctengineer.com/q/semester_3/control_systems/lag-k-s-1-01s-1
site: BCT Engineering
---

# Lag compensator for k/[s(1+0.1s)(1+0.2s)], Kv = 30

Control Systems · Chapter 7: Performance Specifications and Compensation Design

- Asked in: 2074 Chaitra, 12 marks; 2078 Kartik, 12 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/lag-k-s-1-01s-1
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7

## Question

Design a suitable phase lag compensating network for G(s) = K/[s(1+0.1s)(1+0.2s)] to meet the following specifications: Kv = 30 sec⁻¹, P.M ≥ 40°.

## Short answer

Lag compensator: Gc(s) = (1+4s)/(1+42.5s) with β ≈ 10.6, T = 4 s, and K=30. Achieves Kv = 30 s⁻¹ and phase margin ≈ 44.6°.

## Answer

A **phase-lag network** $G_c(s)=\dfrac{1+Ts}{1+\beta Ts}$ ($\beta>1$) has unity DC gain and attenuates high frequencies by $20\log\beta$ dB. It lowers the gain crossover frequency to where the plant phase gives the required PM, while $K_v$ is kept by the gain $K$.

#### Step 1: K from $K_v$

$$
K_v=\lim_{s\to0}sG(s)=K=30\Rightarrow G(s)=\frac{30}{s(1+0.1s)(1+0.2s)}
$$

#### Step 2: Uncompensated system

Corners 5 and 10 rad/s. Asymptotes: $-20$ dB/dec (29.5 dB at $\omega=1$) to 5, $-40$ to 10, $-60$ after.

$$
\phi(\omega)=-90^\circ-\tan^{-1}0.1\omega-\tan^{-1}0.2\omega
$$

| $\omega$ | 0.1 | 0.25 | 1 | 2.46 | 5 | 7.07 | 9.77 |
|---|---|---|---|---|---|---|---|
| $|G|$ (dB) | 49.5 | 41.6 | 29.3 | 20.5 | 11.6 | 6.0 | 0 |
| $\phi$ | $-91.7^\circ$ | $-94.3^\circ$ | $-107^\circ$ | $-130^\circ$ | $-161.6^\circ$ | $-180^\circ$ | $-197^\circ$ |

$\omega_{gc}=9.77$ rad/s, PM $=-17.2^\circ$; $\omega_{pc}=7.07$ rad/s, GM $=-6$ dB. The system is **unstable**.

#### Step 3: New gain crossover

Required phase: $-180^\circ+40^\circ+10^\circ=-130^\circ$ ($10^\circ$ allowance for the lag network's own phase lag).

$$
\tan^{-1}0.1\omega+\tan^{-1}0.2\omega=40^\circ\Rightarrow\omega_{gc}'=2.46\ \text{rad/s}
$$

#### Step 4: β

At 2.46 rad/s, $|G|=20.53$ dB $=10.63$:

$$
20\log\beta=20.53\Rightarrow\beta\approx10.6
$$

#### Step 5: T

Zero about one decade below $\omega_{gc}'$:

$$
\frac1T\approx\frac{2.46}{10}\approx0.25\ \text{rad/s}\Rightarrow T=4\ \text{s},\quad\beta T=42.5\ \text{s}
$$

Pole at $1/\beta T=0.0235$ rad/s.

$$
G_c(s)=\frac{1+4s}{1+42.5s}
$$

Network realisation (passive R-C lag): $T=R_2C$, $\beta=\frac{R_1+R_2}{R_2}=10.6$; e.g. $C=10\ \mu$F, $R_2=400\ \text{k}\Omega$, $R_1=3.84\ \text{M}\Omega$.

#### Step 6: Compensated system

$$
G_c(s)G(s)=\frac{30(1+4s)}{s(1+0.1s)(1+0.2s)(1+42.5s)}
$$

| $\omega$ | 0.0235 | 0.25 | 1 | 2.47 | 6.83 |
|---|---|---|---|---|---|
| $|G_cG|$ (dB) | 59.2 | 24.0 | 9.1 | 0 | $-13.9$ |
| $\phi$ | $-130^\circ$ | $-134^\circ$ | $-120^\circ$ | $-135.4^\circ$ | $-180^\circ$ |

$$
PM=180^\circ-135.4^\circ=44.6^\circ\ (\ge40^\circ),\qquad GM=13.9\ \text{dB}
$$

| Item | Uncompensated | Compensated |
|---|---|---|
| $K_v$ | 30 | 30 |
| $\omega_{gc}$ | 9.77 rad/s | 2.47 rad/s |
| PM | $-17.2^\circ$ | $44.6^\circ$ |
| GM | $-6$ dB | 13.9 dB |

**Answer: $G_c(s)=\dfrac{1+4s}{1+42.5s}$ (β ≈ 10.6, T = 4 s) with K = 30 gives Kv = 30 s⁻¹ and PM ≈ 44.6°.**

## Check yourself

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

1. In the lag design, how is Kv = 30 met?
   A. By the lag network's high-frequency gain
   B. K = 30, since the lag has unity DC gain and does not change Kv
   C. By choosing beta = 30
   D. By making T = 30 s
   Answer: B. Gc has unity DC gain, so Kv = K = 30. (Key point: K from Kv)

2. Required phase at the new crossover is -130 degrees. Why -130 and not -140?
   A. It includes a 10 degree allowance for the lag network's own phase lag
   B. The 10 degrees accounts for gain margin
   C. The 10 degrees accounts for Kv
   D. -130 is the plant phase at crossover
   Answer: A. -180 + 40 + 10 = -130 degrees; the extra 10 degrees covers lag from the compensator. (Key point: New crossover)

3. At w = 2.46 rad/s, |G| = 20.53 dB. Then beta is
   A. beta = 20.53
   B. 20 log beta = 20.53 dB, beta about 10.6
   C. beta = 1/10.6
   D. beta = 2.46
   Answer: B. The lag must attenuate 20.53 dB so that the new crossover has 0 dB: beta = 10.63. (Key point: Finding beta)

4. With the zero one decade below 2.46 rad/s, what is Gc and the result?
   A. Gc = (1+4s)/(1+42.5s), PM about 44.6 degrees
   B. Gc = (1+42.5s)/(1+4s), PM about 44.6 degrees
   C. Gc = (1+0.4s)/(1+4.25s), PM = 40 degrees
   D. Gc = (1+4s)/(1+42.5s), PM about -17 degrees
   Answer: A. 1/T = 0.25 gives T = 4 s and beta T = 42.5 s; the check gives PM = 44.6 degrees (above 40). (Key point: Compensator)

## Close variants set in other exams

- [Lead vs lag; lead design for K/[s²(1+0.25s)]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-k-s2-025s-ka10) — Control Systems, chapter 7, asked 3×
- [Lag compensator for k/[s(s+2)(s+20)], Kv = 20](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alag-k-s-s2-s20-kv20) — Control Systems, chapter 7, asked 4×

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