EBK SYSTEM DYNAMICS
EBK SYSTEM DYNAMICS
3rd Edition
ISBN: 8220100254963
Author: Palm
Publisher: MCG
Question
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Chapter 12, Problem 12.1P
To determine

The design of PD compensator of the given plant.

Expert Solution & Answer
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Answer to Problem 12.1P

The closed loop transfer function of the plant with PD compensator is 7.25+3.535s5s2+3.535s+1.25withKP=7.25andKD=3.535.

Explanation of Solution

Given information:

The transfer function of the plant is Gp(s)=Θ(s)Φ(s)=15s26. For a closed loop system, the damping ratio (ζ) is 0.707 and the angular frequency (ωn) is 0.5.

The structure of the system with PD compensation is shown below.

EBK SYSTEM DYNAMICS, Chapter 12, Problem 12.1P

Figure-(1)

Write the closed loop transfer function of the system.

C(s)R(s)=(KP+KDs)GP(s)1+(KP+KDs)GP(s)........ (I)

Here, the plant function is GP(s), the constants of PD controller are KP and KD.

Substitute 15s26 for GP(s) in Equation (I).

C(s)R(s)=(KP+KDs)×15s261+(KP+KDs)×15s26=(KP+KDs)5s26+(KP+KDs)=KP+KDs5(s2+KD5s+KP65)........ (II)

Write the characteristic equation of the above system.

s2+KD5s+KP65........ (III)

Compare with the standard characteristic equation of the second degree closed loop system.

s2+2ζωns+ωn2=0........ (IV)

Compare the equation (III) and (IV).

ωn2=KP65........ (V)

2ζωn=KD5........ (VI)

Substitute 0.5 for ωn in the Equation (V).

(0.5)2=KP650.25=KP651.25=KP6KP=7.25

Substitute 0.5 for ωn and 0.707 for ζ in the Equation (VI).

2×0.707×0.5=KD5KD=3.535

Substitute 7.25 for KP and 3.535 for KD in the Equation (II).

C(s)R(s)=7.25+3.535s5(s2+3.5355s+7.2565)=7.25+3.535s5s2+3.535s+1.25.

Conclusion:

The closed loop transfer function of the plant with PD compensator is 7.25+3.535s5s2+3.535s+1.25withKP=7.25andKD=3.535.

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