---
title: "Lead vs lag; lead design for K/[s²(1+0.25s)]"
subject: "Control Systems"
chapter: 7
key: "semester_3/control_systems:lead-k-s2-025s-ka10"
source: https://www.bctengineer.com/q/semester_3/control_systems/lead-k-s2-025s-ka10
site: BCT Engineering
---

# Lead vs lag; lead design for K/[s²(1+0.25s)]

Control Systems · Chapter 7: Performance Specifications and Compensation Design

- Asked in: 2081 Baisakh, 2+14 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-s2-025s-ka10
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7

## Question

Differentiate between lead and lag compensator. Design a suitable lead compensating network for G(s) = K/[s²(1+0.25s)] to meet the following specifications: Ka = 10 sec⁻² and P.M ≥ 40°.

## Short answer

Lead compensator adds phase and improves transient response; lag improves steady-state accuracy. For Ka=10 s⁻², set K=10. Since a single lead stage is insufficient, use two identical stages with α=0.13, T=0.434 s. The compensator is Gc(s)=[(1+0.434s)/(1+0.0564s)]² with gain 59.2, giving PM≈42.7°.

## Answer

#### Lead vs lag compensator

| Point | Lead compensator | Lag compensator |
|---|---|---|
| Transfer function | $\frac{1+Ts}{1+\alpha Ts}$, $\alpha<1$ | $\frac{1+Ts}{1+\beta Ts}$, $\beta>1$ |
| Pole–zero | zero nearer origin than pole | pole nearer origin than zero |
| Phase | positive (lead) | negative (lag) |
| Filter type | high-pass | low-pass |
| Main use | improves PM, speed (transient) | improves steady-state accuracy |
| Bandwidth | increases | decreases |
| Noise | more sensitive | less sensitive |

#### Design

**Step 1: K.** $K_a=\lim_{s\to0}s^2G(s)=K=10$, so $G(s)=\dfrac{10}{s^2(1+0.25s)}$.

**Step 2: Uncompensated system.**

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

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

$\omega_{gc}=2.85$ rad/s, $PM=-35.5^\circ$: **unstable** (phase always below $-180^\circ$).

**Step 3: Lead required.** $\phi=40^\circ-(-35.5^\circ)+\epsilon\approx80^\circ$ or more. One lead stage usefully gives at most about 60°, and for this plant a single stage gives a PM of only about 8° with $\alpha=0.1$ (and under 25° even as $\alpha\to0$). So **two identical lead stages** are used:

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

**Step 4: α per stage.** Take $\phi_m\approx50^\circ$ per stage: $\alpha=0.13$, $\sin\phi_m=\frac{0.87}{1.13}\Rightarrow\phi_m=50.3^\circ$ (total 100.7°).

**Step 5: New crossover.** Two stages add $20\log(1/0.13)=17.7$ dB at $\omega_m$, so

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

**Step 6: T.**

$$
T=\frac{1}{6.39\sqrt{0.13}}=0.434\ \text{s},\qquad\alpha T=0.0564\ \text{s}
$$

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

with an amplifier of gain $1/\alpha^2=59.2$ to cancel the networks' attenuation (so $K_a$ stays 10).

**Step 7: Check at 6.39 rad/s.**

$$
\begin{aligned}
\phi&=-180^\circ-\tan^{-1}1.60+2(50.3^\circ)=-137.3^\circ\\
PM&=42.7^\circ\ (\ge40^\circ)
\end{aligned}
$$

GM ≈ 12.9 dB; all closed-loop poles are in the LHP.

**Answer: $G_c(s)=\left(\dfrac{1+0.434s}{1+0.0564s}\right)^2$ with K = 10: Ka = 10 s⁻², PM ≈ 42.7°.**

## 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)] with Ka = 10, what is K?
   A. K = 10, since Ka = lim s^2 G(s) = K
   B. K = 2.5
   C. K = 40
   D. K = 1, since Ka = 1/K
   Answer: A. The plant is type 2, so Ka = lim s^2 G(s) = K, giving K = 10. (Key point: Finding K)

2. Why does the answer use two identical lead stages instead of one?
   A. The plant has no phase lag, so one stage would over-compensate
   B. The uncompensated PM is -35.5 deg, so about 80 deg or more is needed, more than one stage can usefully give
   C. Two stages are needed to raise Ka
   D. A single lead stage cannot be built with passive components
   Answer: B. Required lead is about 80 deg or more, beyond what a single stage usefully gives (it reaches PM of only about 18.6 deg). (Key point: Why two stages)

3. With alpha = 0.13 per stage, what maximum phase does each stage give?
   A. About 13 deg, equal to alpha
   B. About 87 deg, from 1 - alpha
   C. About 50.3 deg, from sin(phi_m) = (1-alpha)/(1+alpha)
   D. About 25 deg, half of the 50 deg needed
   Answer: C. sin(phi_m) = 0.87/1.13 gives phi_m = 50.3 deg per stage, 100.7 deg for two. (Key point: Choosing alpha)

4. Which statement correctly separates a lead compensator from a lag compensator?
   A. Lead is a low-pass filter that improves steady-state accuracy; lag is a high-pass filter
   B. Lead lowers bandwidth and noise sensitivity; lag raises both
   C. Lead has alpha > 1 and lag has beta < 1, with the same pole-zero order
   D. Lead has its zero nearer the origin, raises bandwidth and improves transient; lag has its pole nearer the origin and improves steady-state accuracy
   Answer: D. Lead (alpha<1) is high-pass, increases bandwidth and noise sensitivity; lag (beta>1) is low-pass and improves steady-state error. (Key point: Lead vs lag)

## Close variants set in other exams

- [Lag compensator for k/[s(1+0.1s)(1+0.2s)], Kv = 30](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alag-k-s-1-01s-1) — Control Systems, chapter 7, asked 4×

## Related in Control Systems

- [Lead compensator for k/[s²(1+0.25s)], Ka = 10, PM 35°](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-k-s2-025s-ka10-pm35) — Control Systems, chapter 7, asked 5×
- [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×
- [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
- [(1+aTs)/(1+Ts) as lead or lag compensator](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-or-lag-1-ats) — Control Systems, chapter 7
- [Compare lag and lead compensator applications](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alag-lead-compare) — Control Systems, chapter 7
- [Short note: Lead-lag compensator](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Ashort-lead-lag) — Control Systems, chapter 7

## Related in other subjects

- [Short note: Phase-lead/phase-lag compensator](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ashort-note-lead-lag-compensator) — Digital Control System, chapter 4