Question

For the following system, the angles of asymptotes are \[ G(s)H(s) = \frac{K}{s(s^2+2s+1)} \]

• A. \(45^\circ, 135^\circ, 225^\circ\)
• B. \(60^\circ, 180^\circ, 300^\circ\)
• C. \(90^\circ, 270^\circ\)
• D. \(30^\circ, 150^\circ, 270^\circ\)

Hint:

Asymptote angles are always symmetric with respect to the real axis. For any system where $P-Z = 3$, the angles will always be $60^\circ, 180^\circ, 300^\circ$ (or $\pm 60^\circ, 180^\circ$).

Answer 1

The Correct Option is B

Concept: When a control system has open-loop branches that approach infinity, the directions of these branches are guided by asymptotes. The angles ($\theta_q$) that these asymptotes make with the positive real axis are determined by the equation: \[ \theta_q = \frac{(2q + 1) \times 180^\circ}{P - Z} \] where $P$ is the number of open-loop poles, $Z$ is the number of open-loop zeros, and $q$ is an integer index running from $0$ up to $(P - Z - 1)$.

Step 1: Determine the number of poles (\(P\)) and zeros (\(Z\)).
Given the open-loop transfer function: \[ G(s)H(s) = \frac{K}{s(s^2+2s+1)} = \frac{K}{s(s+1)^2} \] * Setting the denominator to zero gives the poles: $s = 0, s = -1, s = -1$. Thus, $P = 3$. * The numerator contains no $s$ terms, meaning there are no finite zeros. Thus, $Z = 0$. The number of asymptotes required is: \[ P - Z = 3 - 0 = 3 \] The index variable $q$ will take the values $q = 0, 1, 2$.

Step 2: Calculate each asymptote angle.
* For \(q = 0\): \[ \theta_0 = \frac{(2(0) + 1) \times 180^\circ}{3} = \frac{180^\circ}{3} = 60^\circ \] * For \(q = 1\): \[ \theta_1 = \frac{(2(1) + 1) \times 180^\circ}{3} = \frac{3 \times 180^\circ}{3} = 180^\circ \] * For \(q = 2\): \[ \theta_2 = \frac{(2(2) + 1) \times 180^\circ}{3} = \frac{5 \times 180^\circ}{3} = 300^\circ \] The angles of the asymptotes are $60^\circ, 180^\circ$, and $300^\circ$, which matches Option (B).