---
title: "Lead condition and maximum phase lead frequency"
subject: "Control Systems"
chapter: 7
key: "semester_3/control_systems:lead-condition-s-tau"
source: https://www.bctengineer.com/q/semester_3/control_systems/lead-condition-s-tau
site: BCT Engineering
---

# Lead condition and maximum phase lead frequency

Control Systems · Chapter 7: Performance Specifications and Compensation Design

- Asked in: 2071 Chaitra, 4 marks
- Video: [Compensator Example for Maximum Phase in Control Systems](https://www.youtube.com/watch?v=oSHVpFsgrQc) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/lead-condition-s-tau
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7

## Question

For a compensator transfer function given by Gc(s) = (s + τ)/(s + aτ), give the condition of lead compensator. For the given value of 'a' what is the frequency that leads to maximum phase angle lead?

## Short answer

The condition for lead is a > 1, meaning the zero (–τ) is closer to the origin than the pole (–aτ). The maximum phase lead occurs at the geometric mean frequency ωm = τ√a rad/s.

## Answer

For $G_c(s) = \dfrac{s+\tau}{s+a\tau}$ the zero is at $s = -\tau$ and the pole is at $s = -a\tau$.

#### Condition for a lead compensator
The network gives phase lead when the zero lies closer to the origin than the pole:

$$
a\tau > \tau \quad \Rightarrow \quad a > 1
$$

Then the phase angle is

$$
\phi(\omega) = \tan^{-1}\frac{\omega}{\tau} - \tan^{-1}\frac{\omega}{a\tau} > 0 \quad \text{for all } \omega > 0
$$

(If $a < 1$ the pole is nearer the origin and the network becomes a lag compensator.)

#### Frequency of maximum phase lead
Differentiate $\phi$ with respect to $\omega$ and set it to zero:

$$
\begin{aligned}
\frac{d\phi}{d\omega} &= \frac{1/\tau}{1+\omega^2/\tau^2} - \frac{1/(a\tau)}{1+\omega^2/(a\tau)^2} = 0 \\
\frac{\tau}{\tau^2+\omega^2} &= \frac{a\tau}{a^2\tau^2+\omega^2} \\
a^2\tau^2 + \omega^2 &= a\tau^2 + a\omega^2 \\
\omega^2 (a-1) &= a\tau^2 (a-1) \\
\omega_m &= \tau\sqrt{a}
\end{aligned}
$$

So the maximum phase lead occurs at $\omega_m = \sqrt{\tau \times a\tau} = \tau\sqrt{a}$, the **geometric mean** of the zero and pole frequencies (the midpoint on the log-frequency scale of the Bode plot).

The maximum lead is

$$
\phi_m = \sin^{-1}\frac{a-1}{a+1}
$$

Example: $\tau = 2$, $a = 4$ gives $\omega_m = 2\sqrt{4} = 4$ rad/s and $\phi_m = \sin^{-1}(3/5) = 36.9^\circ$.

## Check yourself

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

1. For Gc = (s+tau)/(s+a tau), what is the condition for a lead compensator?
   A. a > 1 (zero nearer the origin than the pole)
   B. a < 1
   C. a = 0
   D. a > tau
   Answer: A. a*tau > tau means the pole is further from the origin. (Key point: Lead condition)

2. At what frequency does the maximum phase lead occur?
   A. wm = tau/sqrt(a)
   B. wm = tau*sqrt(a)
   C. wm = a*tau
   D. wm = tau*a^2
   Answer: B. It is the geometric mean sqrt(tau x a tau). (Key point: Frequency of max lead)

3. What is the maximum phase lead?
   A. phi_m = asin((a+1)/(a-1))
   B. phi_m = atan(a)
   C. phi_m = asin((a-1)/(a+1))
   D. phi_m = 90 deg for all a
   Answer: C. Evaluating the phase at wm gives sin(phi_m) = (a-1)/(a+1). (Key point: Maximum lead)

4. For tau = 2, a = 4, what are wm and phi_m?
   A. 8 rad/s and 36.9 deg
   B. 4 rad/s and 53.1 deg
   C. 1 rad/s and 36.9 deg
   D. 4 rad/s and 36.9 deg
   Answer: D. wm = 2 x 2 = 4; sin(phi_m) = 3/5 gives 36.9 deg. (Key point: Example)

## Related in Control Systems

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