---
title: "Lead compensator for k/[s²(1+0.25s)], Ka = 10, PM 35°"
subject: "Control Systems"
chapter: 7
key: "semester_3/control_systems:lead-k-s2-025s-ka10-pm35"
source: https://www.bctengineer.com/q/semester_3/control_systems/lead-k-s2-025s-ka10-pm35
site: BCT Engineering
---

# Lead compensator for k/[s²(1+0.25s)], Ka = 10, PM 35°

Control Systems · Chapter 7: Performance Specifications and Compensation Design

- Asked in: 2076 Asoj, 2+14 marks; 2071 Shrawan, 16 marks; 2078 Bhadra, 12 marks; 2076 Chaitra, 16 marks (asked 4 times)
- 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-s2-025s-ka10-pm35
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7

## Question

The open loop transfer function of type-II system with unity feedback is given by G(s) = K/[s²(1+0.25s)]. Design a lead compensator to meet the following specifications: (i) Acceleration error constant (Ka) = 10/sec² (ii) PM at least 35°.

## Short answer

A single lead stage cannot meet the specification; two identical lead networks in cascade are used. With α = 0.13, T = 0.434 s, the compensator is [(1+0.434s)/(1+0.0564s)]² plus amplifier gain 59.2, giving PM ≈ 42.7° and Ka = 10 s⁻².

## Answer

A **lead compensator** $G_c(s)=\dfrac{1+Ts}{1+\alpha Ts}$ ($\alpha<1$) adds positive phase near the new gain crossover frequency, which raises the phase margin. Its maximum phase lead $\phi_m$ occurs at $\omega_m=\dfrac{1}{T\sqrt\alpha}$, with $\sin\phi_m=\dfrac{1-\alpha}{1+\alpha}$, and there it adds a gain of $10\log(1/\alpha)$ dB. (The design below uses the Bode plot method.)

#### Step 1: Gain K from the error constant

$$
K_a=\lim_{s\to0}s^2G(s)=\lim_{s\to0}\frac{s^2K}{s^2(1+0.25s)}=K=10
$$

$$
G(s)=\frac{10}{s^2(1+0.25s)}
$$

#### Step 2: Uncompensated system

$$
|G(j\omega)|=\frac{10}{\omega^2\sqrt{1+0.0625\omega^2}},\qquad \phi=-180^\circ-\tan^{-1}0.25\omega
$$

Asymptotic Bode magnitude: $-40$ dB/dec through $20\log10=20$ dB at $\omega=1$; corner at $\omega=4$, then $-60$ dB/dec.

| $\omega$ | 0.5 | 1 | 2.3 | 2.85 | 4 | 6.39 | 10 |
|---|---|---|---|---|---|---|---|
| $|G|$ (dB) | 32.0 | 19.7 | 4.3 | 0 | $-7.1$ | $-17.7$ | $-28.6$ |
| $\phi$ | $-187^\circ$ | $-194^\circ$ | $-210^\circ$ | $-215.5^\circ$ | $-225^\circ$ | $-238^\circ$ | $-248^\circ$ |

Solving $|G|=1$: $\omega_{gc}=2.85$ rad/s and

$$
PM=180^\circ+(-180^\circ-\tan^{-1}0.713)=-35.5^\circ
$$

The uncompensated system is **unstable** (phase is always below $-180^\circ$).

#### Step 3: Phase lead required

$$
\phi_{req}=PM_d-PM_{unc}+\epsilon = 35^\circ-(-35.5^\circ)+\epsilon\approx 75^\circ\text{–}85^\circ
$$

A single lead stage cannot do this: as the gain crossover moves right, the plant phase falls further (the $(1+0.25s)$ lag). A check shows that one lead section gives a PM of only about $8^\circ$ with $\alpha=0.1$, and stays below $25^\circ$ even as $\alpha\to0$. So we use **two identical lead sections in cascade**, each giving about half the lead (a double-lead compensator):

$$
G_c(s)=\left(\frac{1+Ts}{1+\alpha Ts}\right)^2
$$

#### Step 4: Choose α

Trial 1: $\phi_m=45^\circ$ per stage ($\alpha=0.172$) gives PM ≈ 34.7° (just short).
Trial 2: $\phi_m\approx50^\circ$ per stage. Take $\alpha=0.13$:

$$
\sin\phi_m=\frac{1-0.13}{1+0.13}=0.770\Rightarrow\phi_m=50.3^\circ\ \text{per stage}\ (100.7^\circ\ \text{total})
$$

#### Step 5: New gain crossover frequency

Two stages add $2\times10\log(1/\alpha)=20\log(1/0.13)=17.72$ dB at $\omega_m$. So the new crossover is where the uncompensated magnitude is $-17.72$ dB:

$$
\frac{10}{\omega^2\sqrt{1+0.0625\omega^2}}=0.13\Rightarrow\omega_m=\omega_{gc}'=6.39\ \text{rad/s}
$$

#### Step 6: Time constants

$$
\begin{aligned}
T&=\frac{1}{\omega_m\sqrt\alpha}=\frac{1}{6.39\times\sqrt{0.13}}=0.434\ \text{s}\\
\alpha T&=0.13\times0.434=0.0564\ \text{s}
\end{aligned}
$$

Corner frequencies: zero at $1/T=2.30$ rad/s, pole at $1/\alpha T=17.7$ rad/s.

$$
G_c(s)=\left(\frac{1+0.434s}{1+0.0564s}\right)^2
$$

The attenuation $\alpha^2$ of the passive lead networks is made up by an amplifier of gain $1/\alpha^2=59.2$, so the DC gain, and hence $K_a=10$, is unchanged.

#### Step 7: Compensated system and check

$$
G_c(s)G(s)=\frac{10(1+0.434s)^2}{s^2(1+0.25s)(1+0.0564s)^2}
$$

At $\omega=6.39$ rad/s:

$$
\begin{aligned}
\phi&=-180^\circ-\tan^{-1}(1.60)+2\times50.3^\circ\\
&=-180^\circ-58.0^\circ+100.7^\circ=-137.3^\circ\\
PM&=180^\circ-137.3^\circ=42.7^\circ\ (\ge35^\circ)
\end{aligned}
$$

GM of the compensated system ≈ 12.9 dB (phase crossover at 17.1 rad/s). All closed-loop poles are in the LHP ($-27.0$, $-4.08\pm j7.90$, $-2.14\pm j1.14$).

```text
R(s)->( )->[Amp 59.2]->[(s+2.3)/(s+17.7)]^2 ->[G(s)]-+->C
       ^-                                            |
       |_____________________________________________|
```

| Item | Uncompensated | Compensated |
|---|---|---|
| $K_a$ | 10 | 10 |
| $\omega_{gc}$ | 2.85 rad/s | 6.39 rad/s |
| PM | $-35.5^\circ$ | $42.7^\circ$ |
| Stability | unstable | stable |

**Answer: $G_c(s)=\left(\dfrac{1+0.434s}{1+0.0564s}\right)^2$ (two lead stages, α = 0.13 each) with K = 10 gives Ka = 10 s⁻² and PM ≈ 43° (> 35°).**

## Check yourself

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

1. For G(s) = K/[s^2(1+0.25s)], how is K fixed for Ka = 10?
   A. Ka = lim s G(s) = K, so K = 10 sec^-2 needs K=100
   B. Ka = lim s^2 G(s) = K, so K = 10
   C. Ka = lim G(s), so K = infinity
   D. Ka = lim s^2 G(s) = 4K, so K = 2.5
   Answer: B. s^2 cancels the double pole at the origin, leaving Ka = K = 10. (Key point: Gain from Ka)

2. Why does the design use two cascaded lead sections instead of one?
   A. The plant is type 1 and needs extra type
   B. The required phase lead (about 75-85 degrees) is too large for one stage, as the plant phase also falls when crossover moves right
   C. Two stages are needed to raise Ka
   D. A single lead stage would reduce the DC gain
   Answer: B. One lead gives at most about 18.6 degrees of PM here, so the required lead is split over two identical stages. (Key point: Why a single lead fails)

3. With alpha = 0.13 per stage, what is the maximum phase lead per stage?
   A. sin(phi) = (1+0.13)/(1-0.13), phi = 90 degrees
   B. sin(phi) = (1-0.13)/(1+0.13) = 0.770, phi = 50.3 degrees
   C. sin(phi) = 0.13, phi = 7.5 degrees
   D. sin(phi) = (1-0.13)/(1+0.13) = 0.770, phi = 25 degrees
   Answer: B. sin(phi_m) = (1-alpha)/(1+alpha) = 0.770 gives 50.3 degrees per stage, 100.7 degrees total. (Key point: Choosing alpha and phase per stage)

4. Two stages add 17.72 dB at w_m. How is the new gain crossover found, and what is the final PM?
   A. Where the uncompensated gain is +17.72 dB; PM = 35.5 degrees
   B. Where the uncompensated gain is -17.72 dB (w = 6.39 rad/s); PM = 42.7 degrees
   C. Where the phase is -180 degrees; PM = 0 degrees
   D. At the corner 4 rad/s; PM = 35 degrees
   Answer: B. The lead raises gain by 17.72 dB at w_m, so w_m is placed where the plant gain is -17.72 dB, giving 6.39 rad/s and PM about 42.7 degrees, above 35 degrees. (Key point: New crossover and result)

## Close variants set in other exams

- [Lead compensator for K/[s²(0.2s+1)], Ka = 10, PM 35°](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-k-s2-02s-ka10-pm35) — Control Systems, chapter 7, asked 5×

## Related in Control Systems

- [Lead compensator for k/[s²(s+5)], Ka = 5, PM 30°](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-k-s2-s5-ka5-pm30) — 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×
- [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≥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×
- [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×
- [Lead compensator for 5/[(s+2)(s+20)], Kv ≥ 20](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-5-s2-s20-kv20) — Control Systems, chapter 7

## Related in other subjects

- [Bode design: PM 55°, GM 10 dB, Kv = 5](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Abode-design-phase-margin) — Digital Control System, chapter 4