Question

The characteristic equation of a control system is given by \( s(s+1)(s^2+2s+1)+k(s+2)=0 \). The angles of asymptotes of the root loci are

• A. \( 60^\circ, 180^\circ, 300^\circ \)
• B. \( 30^\circ, 60^\circ, 90^\circ \)
• C. \( 0^\circ, 18^\circ, 45^\circ \)
• D. \( 10^\circ, 10^\circ, 30^\circ \)

Hint:

For root locus, identify open-loop poles (P) and zeros (Z).
Number of asymptotes = \(|P-Z|\) (if \(P \neq Z\)).
Angles of asymptotes: \( \phi_a = \frac{(2q+1)180^\circ}{P-Z} \) for \(k>0\), where \(q = 0, 1, \dots, |P-Z|-1\).
Centroid of asymptotes: \( \sigma_a = \frac{\sum (\text{real parts of poles}) - \sum (\text{real parts of zeros})}{P-Z} \).

Answer 1

The Correct Option is A

The characteristic equation is \(1 + G(s)H(s) = 0\), where \(G(s)H(s) = \frac{k(s+2)}{s(s+1)(s^2+2s+1)}\). Note that \(s^2+2s+1 = (s+1)^2\). So, the open-loop transfer function is \(G(s)H(s) = \frac{k(s+2)}{s(s+1)(s+1)^2} = \frac{k(s+2)}{s(s+1)^3}\). Number of open-loop poles (P) = 4 (at \(s=0, s=-1, s=-1, s=-1\)). Number of open-loop zeros (Z) = 1 (at \(s=-2\)). The number of asymptotes is \(P-Z\) if \(P>Z\). Number of asymptotes = \(4 - 1 = 3\). The angles of the asymptotes are given by the formula: \[ \phi_a = \frac{(2q+1)180^\circ}{P-Z} \] where \(q = 0, 1, 2, \dots, (P-Z-1)\). Here \(P-Z = 3\), so \(q = 0, 1, 2\). For \(q=0\): \(\phi_a = \frac{(2(0)+1)180^\circ}{3} = \frac{180^\circ}{3} = 60^\circ\). For \(q=1\): \(\phi_a = \frac{(2(1)+1)180^\circ}{3} = \frac{3 \times 180^\circ}{3} = 180^\circ\). For \(q=2\): \(\phi_a = \frac{(2(2)+1)180^\circ}{3} = \frac{5 \times 180^\circ}{3} = 5 \times 60^\circ = 300^\circ\). (Or \(300^\circ \equiv -60^\circ\)). The angles of the asymptotes are \(60^\circ, 180^\circ, 300^\circ\). This matches option (a). \[ \boxed{60^\circ, 180^\circ, 300^\circ} \]