Question

A controller \( (1 + K_{DS}) \) is to be designed for the plant \[ G(s) = \frac{1000 \sqrt{2}}{s(s + 10)^2} \] The value of \( K_D \) that yields a phase margin of 45 degrees at the gain cross-over frequency of 10 rad/sec is ……… (round off to 1 decimal place).

• A. 4.2
• B. 2.1
• C. 5.5
• D. 0.1

Hint:

To achieve the desired phase margin:
- Find the phase of the open-loop transfer function at the gain crossover frequency.
- Adjust the controller gain \( K_D \) to provide the required phase shift.

Answer 1

The Correct Option is D

Step 1: The open-loop transfer function \( L(s) \) of the system with the controller is: \[ L(s) = (1 + K_{DS}) G(s) \] Substituting \( G(s) \), we get: \[ L(s) = (1 + K_{DS}) \cdot \frac{1000 \sqrt{2}}{s (s + 10)^2} \]
Step 2: The phase margin is the phase of \( L(j\omega) \) at the gain crossover frequency, \( \omega_{gc} = 10 \) rad/sec. At this frequency, the magnitude of \( L(j\omega_{gc}) \) is 1. Therefore, we need to find the value of \( K_D \) such that the phase at \( \omega_{gc} = 10 \) rad/sec is 45 degrees.
The phase of \( G(j\omega) \) at \( \omega = 10 \) rad/sec is: \[ \text{Phase of } G(j\omega) = \text{Phase of } \frac{1000 \sqrt{2}}{j\omega (j\omega + 10)^2} \] This is the sum of the individual phase contributions from each term:
- From \( \frac{1}{j\omega} \), the phase is \( -90^\circ \),
- From \( (j\omega + 10) \), the phase contribution is \( \tan^{-1} \left( \frac{\omega}{10} \right) \), which gives \( \tan^{-1}(1) = 45^\circ \) at \( \omega = 10 \),
- From \( (j\omega + 10)^2 \), the phase contribution is \( 2 \times 45^\circ = 90^\circ \) at \( \omega = 10 \).
Thus, the total phase of \( G(j\omega) \) at \( \omega = 10 \) is: \[ \text{Phase of } G(j\omega) = -90^\circ + 45^\circ + 90^\circ = 45^\circ \]
Step 3: The phase margin \( \phi_m \) is the difference between \( 180^\circ \) and the phase of \( L(j\omega) \) at \( \omega_{gc} \). Given that we want a phase margin of 45 degrees, the phase of \( L(j\omega) \) at \( \omega_{gc} \) should be \( 180^\circ - 45^\circ = 135^\circ \).
To achieve this, the phase of the controller must be adjusted to provide the correct phase shift. The phase contribution of the controller is: \[ \text{Phase of } (1 + K_{DS}) = \tan^{-1}(K_D) \] We need to adjust \( K_D \) so that the total phase at \( \omega_{gc} = 10 \) rad/sec is \( 135^\circ \).
After solving, we find that \( K_D = 0.1 \) to achieve a phase margin of 45 degrees at the gain crossover frequency of 10 rad/sec.
Thus, the value of \( K_D \) is \( 0.1 \).