System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 2, Problem 2.36P
To determine

(a)

The transfer functions X(s)F(s) and X(s)G(s) for the model equations as shown:

x˙=4x+2y+f(t)andy˙=9y5x+g(t).

Expert Solution
Check Mark

Answer to Problem 2.36P

The transfer functions for the model equations are as:

X(s)F(s)=(s+9)(s2+13s+46)andX(s)G(s)=2(s2+13s+46).

Explanation of Solution

Given:

The given model equations are as:

x˙=4x+2y+f(t)andy˙=9y5x+g(t).

Concept Used:

Laplace transform is employed for obtaining the transfer functions for the model conditions keeping the initial conditions zero. In addition, for the system having multiple inputs, the superposition principle is used for finding the impact of one input at a single time by disconnecting the system from other inputs.

Calculation:

Model equations to be solved are as:

x˙=4x+2y+f(t)andy˙=9y5x+g(t)

For X(s)F(s), set G(s) temporarily equal to zero.

On taking Laplace for both the model equations, that is

x˙=4x+2y+f(t)sX(s)=4X(s)+2Y(s)+F(s)X(s)F(s)(s+4)=2Y(s)X(s)+1 (1)

Similarly,

y˙=9y5x+g(t)sY(s)=9Y(s)5X(s)+G(s)sY(s)=9Y(s)5X(s)(s+9)Y(s)F(s)=5X(s)F(s) (2)

Using equations (1) and (2), we get

X(s)F(s)(s+4)=2Y(s)X(s)+1X(s)F(s)(s+4)=10(s+9)X(s)F(s)+1X(s)F(s)((s+4)+10(s+9))=1X(s)F(s)=(s+9)10+(s+9)(s+4)=(s+9)(s2+13s+46)

Similarly, for X(s)G(s), set F(s) temporarily to zero.

On taking Laplace for both the model equations, that is

x˙=4x+2y+f(t)sX(s)=4X(s)+2Y(s)+F(s)sX(s)=4X(s)+2Y(s)X(s)G(s)(s+4)=2Y(s)G(s) (3)

Similarly,

y˙=9y5x+g(t)sY(s)=9Y(s)5X(s)+G(s)(s+9)Y(s)G(s)=5X(s)G(s)+1 (4)

Using equations (3) and (4), we get

Y(s)G(s)(s+9)=5X(s)G(s)+1(s+4)(s+9)2X(s)G(s)+5X(s)G(s)=1X(s)G(s)((s+4)(s+9)2+5)=1X(s)G(s)=210+(s+9)(s+4)=2(s2+13s+46).

Conclusion:

The obtained transfer functions are as:

X(s)F(s)=(s+9)(s2+13s+46)andX(s)G(s)=2(s2+13s+46).

To determine

(b)

The values for τ,ζ,ωdandωn of the model equations as shown:

x˙=4x+2y+f(t)andy˙=9y5x+g(t).

Expert Solution
Check Mark

Answer to Problem 2.36P

The parameter values are:

τ=213,ζ=13246,ωn=46andωd=15215.

Explanation of Solution

Given:

The given system model equations are as:

x˙=4x+2y+f(t)andy˙=9y5x+g(t).

Concept Used:

Characteristic equation is obtained from transfer function obtained in part a, that is

Since,

X(s)F(s)=(s+9)(s2+13s+46)andX(s)G(s)=2(s2+13s+46)

Therefore, the characteristic equation is as

s2+13s+46=0

This equation is compared with ms2+cs+k=0 and the following formulae for the parameters are used:

τ=2mc,

ζ=c2mk,

ωn=km

And ωd=ωn1ζ2.

Calculation:

Characteristic equation of the system is as:

s2+13s+46=0

On comparing this with ms2+cs+k=0, we get

τ=2mc=213,ζ=c2mk=13246,ωn=km=46andωd=ωn1ζ2=461(13246)2=461169184=152=15215.

Conclusion:

The obtained values for the parameters are as shown:

τ=213,ζ=13246,ωn=46andωd=15215.

To determine

(c)

The time taken for the responses x(t)andy(t) to disappear when response f(t)=g(t)=0.

Expert Solution
Check Mark

Answer to Problem 2.36P

At t=4τ=813 seconds, the responses x(t)andy(t) disappear.

Explanation of Solution

Given:

The obtained transfer functions in the first sub-part ‘a’ are:

X(s)F(s)=(s+9)(s2+13s+46)andX(s)G(s)=2(s2+13s+46)

Also, f(t)=g(t)=0.

Concept Used:

Since

f(t)=g(t)=0

Therefore,

x˙=4x+2yandy˙=9y5x

Using these modified model equations, expressions for X(s)andY(s) are evaluated using Laplace transform consequently, the responses for x(t)andy(t) are computed.

Using these responses, the time taken by oscillations to decay is observed.

Calculation:

Since,

x˙=4x+2y+f(t)andy˙=9y5x+g(t)

On taking f(t)=g(t)=0, we get

x˙=4x+2yandy˙=9y5x

On taking Laplace transform, we have

x˙=4x+2ysX(s)x(0)=4X(s)+2Y(s)X(s)(s+4)=2Y(s)+x(0) (1)andy˙=9y5xsY(s)y(0)=9Y(s)5X(s)Y(s)(s+9)=5X(s)+y(0) (2)

Using equations (1) and (2), we have

X(s)(s+4)=2Y(s)+x(0)

X(s)(s+4)=2(5X(s)+y(0)(s+9))+x(0)

X(s)((s+4)+10(s+9))=2y(0)(s+9)+x(0)

X(s)((s+4)(s+9)+10(s+9))=2y(0)(s+9)+x(0)

X(s)((s2+13s+46)(s+9))=2y(0)(s+9)+x(0)

X(s)=2y(0)(s2+13s+46)+x(0)(s+9)(s2+13s+46)

X(s)=2y(0)((s+132)2+(152)2)+x(0)(s+9)((s+132)2+(152)2)

X(s)=215(2y(0)+52x(0))152((s+132)2+(152)2)+x(0)(s+132)((s+132)2+(152)2)

On taking inverse Laplace of this, we get

x(t)=x(0)e132tcos152t+215(2y(0)+52x(0))e132tsin152t

On observing the response x(t), we get

Here, at t=4τ=813, e132t<0.02 that is the oscillations of response x(t) disappear at t=813 seconds.

Similarly, we have

x˙=4x+2yy=x˙+4x2

That is the response y(t) will be of the form as shown

x˙=4x+2yy=x˙+4x2=134e132tA(t)+e132t2dA(t)dt+2e132tA(t)where,A(t)=(x(0)cos152t+215(2y(0)+52x(0))sin152t)y=e132t2dA(t)dt54e132tA(t)y=e132t(54A(t)+12dA(t)dt)

Therefore, at t=4τ=813, e132t<0.02 that is the response y(t) will also disappear at t=813 seconds.

Conclusion:

The obtained responses x(t)andy(t) are disappearing at t=813 seconds.

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Chapter 2 Solutions

System Dynamics

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