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

Given:

• \( G(s) = \dfrac{1000\sqrt{2}}{s(s+10)^2} \)
• \( D(s) = 1 + K_D s \)
• Gain crossover frequency \( \omega_{gc} = 10 \,\text{rad/s} \)
• Desired phase margin \( \phi_m = 45^\circ \)

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Step 1: Phase of the open-loop transfer function at \( \omega = 10 \)

The open-loop transfer function is:

\[ L(j\omega) = D(j\omega)G(j\omega) \]

Evaluate \( G(j10) \):

\[ G(j10) = \frac{1000\sqrt{2}}{j10\,(j10+10)^2} \]

Determine the phase contributions:

• Phase of \( j10 \) is \( +90^\circ \)
• \( 10 + j10 = 14.14 \angle 45^\circ \)
• \( (10 + j10)^2 \Rightarrow \) phase \( = 2 \times 45^\circ = 90^\circ \)

Hence, the phase of \( G(j10) \) is:

\[ \angle G(j10) = -90^\circ - 90^\circ = -180^\circ \]

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Step 2: Phase contribution of the PD controller

For the controller:

\[ D(j10) = 1 + j10K_D \]

Its phase is:

\[ \angle D(j10) = \tan^{-1}(10K_D) \]

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Step 3: Apply phase margin condition

The total open-loop phase at gain crossover frequency is:

\[ \angle L(j10) = \tan^{-1}(10K_D) - 180^\circ \]

Phase margin is given by:

\[ \text{Phase Margin} = 180^\circ + \angle L(j10) \]

Substitute the desired phase margin:

\[ 45^\circ = 180^\circ + \tan^{-1}(10K_D) - 180^\circ \]

\[ \tan^{-1}(10K_D) = 45^\circ \]

\[ 10K_D = 1 \]

\[ K_D = \frac{1}{10} \]

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Final Answer:

\( K_D = 0.1 \)