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The transfer function of a compensator is given as $G_c(s)=\frac{s+a}{s+b}$. $G_c(s)$ is a lead compensator if

1. a = 1, b = 2
2. a = 3, b = 2
3. a = -3, b = -1
4. a = 3, b = 1

Correct answer: a = 1, b = 2

Solution

A lead compensator has a zero nearer to the origin than its pole, i.e. the zero frequency is lower than the pole frequency. For $G_c(s)=\frac{s+a}{s+b}$ with positive $a,b$, this means $a<b$. Among the options, only $a=1, b=2$ satisfies this.

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