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The open-loop transfer function of a plant is given by $G(s)=\frac{1}{s^2-1}$. If the plant is operated in a unity-feedback configuration, then the lead compensator that can stabilize this control system is

1. $10\frac{s-1}{s+2}$
2. $10\frac{s+4}{s+2}$
3. $10\frac{s+2}{s+10}$
4. $2\frac{s+2}{s+10}$

Correct answer: $10\frac{s+4}{s+2}$

Solution

The plant has unstable poles at $s=\pm 1$, so unity feedback alone is unstable. A lead compensator with a zero nearer the origin than its pole can provide phase advance and move the closed-loop poles into the left half-plane; among the options, $10\frac{s+4}{s+2}$ is the stabilizing choice.

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