Question

A controller \( D(s) \) of the form \( (1 + K_D s) \) is to be designed for the plant \[ G(s) = \frac{1000\sqrt{2}}{s(s+10)^2} \] as shown in the figure. The value of \( K_D \) that yields a phase margin of \(45^\circ\) at the gain cross-over frequency of 10 rad/sec is _____________ (round off to one decimal place). 

Hint:

To design a lead compensator for a desired phase margin, match the desired phase boost to the controller phase contribution \( \angle (1 + j\omega K_D) = \tan^{-1}(\omega K_D) \), and solve accordingly.

Answer 1

Correct Answer: 0.1

We are given: - \( G(s) = \dfrac{1000\sqrt{2}}{s(s+10)^2} \) - \( D(s) = 1 + K_D s \) - Gain crossover frequency \( \omega_{gc} = 10~{rad/s} \) - Desired phase margin \( \phi_m = 45^\circ \) Step 1: Compute phase of open-loop transfer function \( L(j\omega) = D(j\omega) G(j\omega) \) at \( \omega = 10 \) \[ G(j10) = \frac{1000\sqrt{2}}{j10 (j10 + 10)^2} = \frac{1000\sqrt{2}}{j10 (10 + j10)^2} \] Let’s compute phase: - Phase of \( j10 \): \(+90^\circ\) - \(10 + j10 = \sqrt{10^2 + 10^2} \angle \tan^{-1}(1) = 14.14 \angle 45^\circ \) - So \((10 + j10)^2 \Rightarrow {angle } = 2 \cdot 45^\circ = 90^\circ\) So total phase of \( G(j10) \) is: \[ \angle G(j10) = -90^\circ - 90^\circ = -180^\circ \] Now, \( D(j\omega) = 1 + j10 K_D \) \[ \angle D(j10) = \tan^{-1}(10 K_D) \] So total open-loop phase at \( \omega = 10 \): \[ \angle L(j10) = \angle D(j10) + \angle G(j10) = \tan^{-1}(10 K_D) - 180^\circ \] We want: Sure! Here's the full expression formatted properly in LaTeX with each step on its own line: \[ {Phase Margin} = 180^\circ + \angle L(j10) \] \[ \angle L(j10) = \tan^{-1}(10 K_D) = 45^\circ \] \[ \tan^{-1}(10 K_D) = 45^\circ \] \[ 10 K_D = 1 \] \[ K_D = \frac{1}{10} \] \[ \boxed{K_D = 0.1} \]