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

To find the angles of asymptotes of the root loci for the given characteristic equation \( s(s+1)(s^2+2s+1)+k(s+2)=0 \), we first express it in the standard form. The characteristic equation can be rewritten as:

\( s(s+1)(s^2+2s+1)+k(s+2)=0 \Rightarrow s^4+3s^3+3s^2+s+k(s+2)=0 \).

For the analysis of root loci, we consider the open-loop transfer function which is typically expressed as \(1+KG(s)H(s)=0\) for root locus analysis:

\(G(s)H(s) = \frac{(s+2)}{s(s+1)(s^2+2s+1)}\).

The angles of asymptotes in the root locus are calculated using the formula:

\[\theta = \frac{(2q+1)180^\circ}{n-m} \]

where \(n\) is the number of poles, \(m\) is the number of zeros, and \(q\) is an integer taking values \(0,1,2,...,n-m-1\).

For the given function, the poles of \(G(s)H(s)\) are at \(s=0,\,-1,\,-1,-1\) and the zero is at \(s=-2\). Thus, we have:

• Number of poles, \(n = 4\)
• Number of zeros, \(m = 1\)
• Difference \(n-m = 3\)

Substituting these into the formula for angles of asymptotes:

\(\theta_0 = \frac{(2\cdot0+1)180^\circ}{3} = 60^\circ\)

\(\theta_1 = \frac{(2\cdot1+1)180^\circ}{3} = 180^\circ\)

\(\theta_2 = \frac{(2\cdot2+1)180^\circ}{3} = 300^\circ\)

This gives the angles of the asymptotes as \(60^\circ\), \(180^\circ\), and \(300^\circ\).

Therefore, the correct answer is:

\(60^\circ, 180^\circ, 300^\circ\)