Question

The figure shows a closed-loop system with a plant \(G(s) = \frac{1}{s^2}\) and a lead compensator \(C(s)\). The compensator is designed to place the dominant closed-loop poles at \(-1.5 \pm j\frac{\sqrt{27}}{2}\). From the following options, choose the phase lead that the compensator needs to contribute.

• A. $30^\circ$
• B. $60^\circ$
• C. $90^\circ$
• D. $120^\circ$

Hint:

The angle condition for a point $s$ to be on the root locus is $\angle G(s)H(s) = (2k+1)180^\circ$. In a unity feedback system, $H(s) = 1$, so $\angle G(s) = (2k+1)180^\circ$. The phase lead required by the compensator is the angle that $C(s)$ must provide at the desired pole location to satisfy this condition.

Answer 1

The Correct Option is B

Step 1: Determine the open-loop transfer function without the compensator.
The open-loop transfer function is \(G(s) = \frac{1}{s^2}\).
Step 2: Evaluate the angle of the plant at the desired closed-loop pole.
Consider the pole \(s_p = -1.5 + j\frac{\sqrt{27}}{2}\).
The angle of \(s_p\) is \(\angle s_p = \arctan\left(\frac{\sqrt{27}/2}{-1.5}\right) = \arctan(-\sqrt{3}) = 120^\circ\).
The angle of \(G(s_p)\) is \(\angle G(s_p) = \angle \left(\frac{1}{s_p^2}\right) = -2 \angle s_p = -2(120^\circ) = -240^\circ\).
We can also write this as \(-240^\circ + 360^\circ = 120^\circ\).
Step 3: Apply the angle condition for closed-loop poles.
For \(s_p\) to be a closed-loop pole, the angle of the open-loop transfer function \(C(s)G(s)\) at \(s_p\) must be \(-180^\circ + k \cdot 360^\circ\), where \(k\) is an integer.
\(\angle C(s_p) + \angle G(s_p) = -180^\circ \pmod{360^\circ}\)
Let \(\phi_{lead} = \angle C(s_p)\) be the phase lead contributed by the compensator.
\(\phi_{lead} + (-240^\circ) = -180^\circ\)
\(\phi_{lead} = -180^\circ + 240^\circ = 60^\circ\).
Alternatively using the positive equivalent angle for \(G(s_p)\):
\(\phi_{lead} + 120^\circ = -180^\circ \pmod{360^\circ}\)
\(\phi_{lead} = -180^\circ - 120^\circ \pmod{360^\circ}\)
\(\phi_{lead} = -300^\circ \pmod{360^\circ}\)
\(\phi_{lead} = -300^\circ + 360^\circ = 60^\circ\).
Final Answer: (B)