System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 6, Problem 6.47P
To determine

(a)

The transfer function X(s)V(s) for the speaker model for the parameter values L and c equal to zero. Consequently, also compute the expressions for its characteristic roots.

Expert Solution
Check Mark

Answer to Problem 6.47P

The transfer function X(s)V(s) is X(s)V(s)=KfmRs2+KfKbs+kR and the characteristic roots are s=KfKb2mR±12Kf2Kb2m2R24km.

Explanation of Solution

Given:

  1. The electrical and mechanical sub-system of the speaker system is shown in figure below:
  2. System Dynamics, Chapter 6, Problem 6.47P

    Also, the inductance (L) and damping constant (c) are zero.

  3. Characteristic equation is the denominator of the transfer function of the corresponding system. Thus, a transfer function for a system could be represented as follows:
  4. T(s)=Ks2+As+B for the characteristic equation s2+As+B=0.

Concept Used:

The model equations for the speaker system as shown in the figure are:

For mechanical sub-system:

md2xdt2=cdxdtkx+Kfi

For electrical sub-system:

v=Ldidt+Ri+kbdxdt.

Calculation:

Since for the speaker system, we have

md2xdt2=cdxdtkx+Kfi (1)

v=Ldidt+Ri+kbdxdt (2)

On using equations (1) and (2), while keeping initial conditions equal to zero such that

md2xdt2=cdxdtkx+Kfims2X(s)=csX(s)kX(s)+KfI(s)(ms2+cs+k)X(s)=KfI(s)X(s)I(s)=Kf(ms2+cs+k) (3)

Similarly,

v=Ldidt+Ri+KbdxdtV(s)=LsI(s)+RI(s)+KbsX(s)V(s)=(Ls+R)I(s)+KbsX(s)V(s)=(Ls+R)(ms2+cs+k)KfX(s)+KbsX(s)X(s)I(s)=Kf(ms2+cs+k)

X(s)V(s)=Kf(Ls+R)(ms2+cs+k)+KfKbsX(s)V(s)=KfmLs3+s2(cL+mR)+s(Lk+cR+KfKb)+kR (4)

Thus, at L=c=0, we get

X(s)V(s)=KfmLs3+s2(cL+mR)+s(Lk+cR+KfKb)+kRX(s)V(s)=KfmRs2+KfKbs+kR

Therefore, the characteristic equation of this transfer function is as:

mRs2+KfKbs+kR=0s2+KfKbmRs+km=0

On taking out characteristic roots that are,

s2+KfKbmRs+km=0s=KfKb2mR±12Kf2Kb2m2R24km.

Conclusion:

The transfer function X(s)V(s) is,

X(s)V(s)=KfmRs2+KfKbs+kR.

And the characteristic roots are as follows:

s=KfKb2mR±12Kf2Kb2m2R24km.

To determine

(b)

The third order transfer function and the corresponding characteristic roots of the speaker system for the given parameter values. Also, to compare this result with the one obtained in sub-part (a).

Expert Solution
Check Mark

Answer to Problem 6.47P

The transfer function and characteristic roots for third order case of speaker system are:

X(s)V(s)=162×106s3+0.024s2+608s+48×105s=8715.433,1642.28±j16512.92

Similarly, for the second order case of the speaker system, the transfer function and characteristic roots are:

X(s)V(s)=160.024s2+208s+48×105s=4333.33±j13461.88

On comparing these cases, it is concluded that system is stable in both the cases as roots are lying left to the imaginary axis in s-plane. However, the complex roots in the second order case are far into imaginary axis in comparison to the complex roots in third order case of the speaker system.

Explanation of Solution

Given:

Here, the parameter values for the speaker system are as follows:

m=0.002kg,k=4×105N/m,Kf=16N/A, Kb=13Vs/m,R=12ohm,L=103H,c=0.

Concept Used:

  1. Here, the transfer function X(s)V(s) for the speaker system obtained in sub-part (a) is as follows:
  2. X(s)V(s)=KfmLs3+s2(cL+mR)+s(Lk+cR+KfKb)+kRfromequation(4)insub-part(a)

    Also, at L=c=0, we get

    X(s)V(s)=KfmRs2+KfKbs+kR

  3. Characteristic equation is the denominator of the transfer function of the corresponding system. Thus, a transfer function for a system could be represented as follows:
  4. T(s)=Ks2+As+B for the characteristic equation s2+As+B=0.

Calculation:

Since for the speaker system, the transfer function is as follows:

X(s)V(s)=KfmLs3+s2(cL+mR)+s(Lk+cR+KfKb)+kR

On substituting the parameter values in the above expression, we get

X(s)V(s)=KfmLs3+s2(cL+mR)+s(Lk+cR+KfKb)+kR

m=0.002kg,k=4×105N/m,Kf=16N/A, Kb=13Vs/m,R=12ohm, L=103H,c=0X(s)V(s)=160.002×103s3+s2(0+0.002×12)+s(4×105×103+0+16×13)+4×105×12X(s)V(s)=162×106s3+0.024s2+608s+48×105

Therefore, the characteristic equation would be:

2×106s3+0.024s2+608s+48×105=0s3+12×103s2+304×106s+24×1011=0

On taking out characteristic roots that are,

s3+12×103s2+304×106s+24×1011=0s=8715.433,1642.28±j16512.92

Similarly, for the transfer function at L=c=0, we have

X(s)V(s)=KfmRs2+KfKbs+kR

On keeping the parameter values in this expression, we have

X(s)V(s)=KfmRs2+KfKbs+kRm=0.002kg,k=4×105N/m,Kf=16N/A, Kb=13Vs/m,R=12ohmX(s)V(s)=160.002×12s2+16×13s+4×105×12X(s)V(s)=160.024s2+208s+48×105

Therefore, the characteristic equation would be:

0.024s2+208s+48×105=0s2+8666.67s+2×108=0

On taking out characteristic roots that are,

s2+8666.67s+2×108=0s=4333.33±j13461.88.

Conclusion:

The transfer function and characteristic roots for third order case of speaker system are:

X(s)V(s)=162×106s3+0.024s2+608s+48×105s=8715.433,1642.28±j16512.92

Similarly, for the second order case of the speaker system, the transfer function and characteristic roots are:

X(s)V(s)=160.024s2+208s+48×105s=4333.33±j13461.88

On comparing these cases, it is concluded that a system is stable in both the cases as roots are lying left to the imaginary axis in s-plane. However, the complex roots in the second order case are far into imaginary axis in comparison to the complex roots in third order case of the speaker system.

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Chapter 6 Solutions

System Dynamics

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