System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 6, Problem 6.36P
To determine

The modified model of the system shown in figure 1 for the gear pair attached between the motor shaft and the load. Also, compute the transfer functions ΩL(s)Vf(s) and ΩL(s)TL(s).

Expert Solution & Answer
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Answer to Problem 6.36P

The system’s modified model comprises the following equations as shown:

vf=Rfif+LfdifdtN(Im+ILN2)dωLdt=KTif(cm+cLN2)NωLTL

Also, the transfer functions are as follows:

ΩL(s)Vf(s)=KTN1(Rf+Lfs)((Im+ILN2)s+(cm+cLN2))

ΩL(s)TL(s)=1N((Im+ILN2)s+(cm+cLN2)).

Explanation of Solution

Given:

The speed control system for the field-controlled motor has been given as shown in figure 1:

System Dynamics, Chapter 6, Problem 6.36P

Figure 1

This speed-control system is equipped with a gear train of ratio N such that:

ωmωL=N

Concept Used:

  1. The system model equations for this speed-control system are as follows:
  2. For field-circuit, vf=Rfif+Lfdifdt

    For rotational coupling, Idωdt=TcωTL=KTifcωTL

    Here, I and c are the effective inertia and effective damping of the system respectively.

  3. The effective inertia and the effective damping for the rotational part equipped with a gear train of ratio N are as follows:
  4. Ieff=Im+ILN2

    ceff=cm+cLN2

Calculation:

The model equations for the system are:

vf=Rfif+Lfdifdt (1)

For the inertia I,

Idωdt=TcωTL=KTifcωTL (2)

Also, on considering the effect of gearbox in a rotational system, we get

Effective inertia, Ieff=Im+ILN2 (3)

Effective damping, ceff=cm+cLN2 (4)

Thus, on using equations (2), (3) and (4), we have for effective inertia Ieff

Idωdt=TcωTL=KTifcωTL

Ieffdωmdt=KTifceffωmTL ωmωL=NN(Im+ILN2)dωLdt=KTif(cm+cLN2)NωLTL (5)

Taking Laplace transform of equation (1) and (5) while keeping zero initial conditions, we have

vf=Rfif+LfdifdtVf(s)=RfIf(s)+LfsIf(s)If(s)=1(Rf+Lfs)Vf(s) (6)

N(Im+ILN2)dωLdt=KTif(cm+cLN2)NωLTLN(Im+ILN2)sΩL(s)=KTIf(s)(cm+cLN2)NΩL(s)TL(s)((Im+ILN2)s+(cm+cLN2))NΩL(s)=KTIf(s)TL(s) (7)

From equations (6) and (7)

((Im+ILN2)s+(cm+cLN2))NΩL(s)=KTIf(s)TL(s)((Im+ILN2)s+(cm+cLN2))NΩL(s)=KT1(Rf+Lfs)Vf(s)TL(s) (8)

In case, for the multiple inputs system, for finding the transfer function for one system other inputs are temporarily set to zero that is,

For ΩL(s)Vf(s) setting TL(s)=0, therefore from equation (8)

((Im+ILN2)s+(cm+cLN2))NΩL(s)=KT1(Rf+Lfs)Vf(s)TL(s)((Im+ILN2)s+(cm+cLN2))NΩL(s)=KT1(Rf+Lfs)Vf(s)ΩL(s)Vf(s)=KTN1(Rf+Lfs)((Im+ILN2)s+(cm+cLN2))

Similarly, for ΩL(s)TL(s) setting Vf(s)=0, therefore from equation (8)

((Im+ILN2)s+(cm+cLN2))NΩL(s)=KT1(Rf+Lfs)Vf(s)TL(s)((Im+ILN2)s+(cm+cLN2))NΩL(s)=TL(s)ΩL(s)TL(s)=1N((Im+ILN2)s+(cm+cLN2)).

Conclusion:

The system’s modified model comprises the following equations as shown:

vf=Rfif+LfdifdtN(Im+ILN2)dωLdt=KTif(cm+cLN2)NωLTL

Also, the transfer functions are as follows:

ΩL(s)Vf(s)=KTN1(Rf+Lfs)((Im+ILN2)s+(cm+cLN2))

ΩL(s)TL(s)=1N((Im+ILN2)s+(cm+cLN2)).

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