System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 5, Problem 5.22P
To determine

The state-variable model in standard form for the transfer function Y(s)F(s)=s+2s2+4s+3 using two methods. Also, relate the initial conditions for the state variables with the given initial values y(0)andy˙(0).

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Answer to Problem 5.22P

The state model for the transfer function Y(s)F(s)=s+2s2+4s+3 using two methods are as shown:

By the first method,

x˙1=4x1+x2+fx˙2=3x1+2fy=x1

By the second method,

x˙1=x1+f2x˙2=3x2f2y=x1x2

And the relation between initial values of state variables and y(0)andy˙(0):

For the state model using the first method,

x1(0)=y(0)x2(0)=y˙(0)+4y(0)f(0)

While for the state model using the second method,

x1(0)=12y˙(0)+32y(0)12f(0)x2(0)=12y˙(0)+y(0)12f(0).

Explanation of Solution

Given:

The transfer function of the system is as shown:

Y(s)F(s)=s+2s2+4s+3

With given initial values y(0)andy˙(0).

Concept Used:

Firstly, the highest order of the denominator of a transfer function is equal to the number of state variable for the model.

Secondly, the methodology for assuming the state variables in case of given transfer function is as follows:

First method: For the transfer function of the form Y(s)F(s)=as+bs+c, try to make a term of ‘1’ in the denominator such that

Y(s)F(s)=a+bs1+csY(s)=aF(s)+bsF(s)csY(s)Y(s)=aF(s)+1s(bF(s)cY(s))

Here, the second term on the right side is the input to an integrator, 1s thus take this as the state variable for the model and solve further to obtain the state model.

Second method: For a transfer function as Y(s)F(s)=s+as+b, transfer the function as shown

Y(s)=(s+a)F(s)s+b

And take F(s)s+b as the state variable for the model and solve further for the state model.

Calculation:

Given transfer function is as shown,

Y(s)F(s)=s+2s2+4s+3

Using the first method,

Divide the numerator and denominator of transfer function by s2 such that

Y(s)F(s)=1s+2s21+4s+3s2Y(s)=4sY(s)3s2Y(s)+1sF(s)+2s2F(s)Y(s)=1s((F(s)4Y(s))+1s(2F(s)3Y(s)))

On taking,

X1(s)=Y(s)=1s((F(s)4Y(s))+1s(2F(s)3Y(s)))andX2(s)=1s(2F(s)3Y(s))

Therefore,

Y(s)=X1(s)y=x1 (1)

Similarly,

X2(s)=1s(2F(s)3Y(s))sX2(s)=2F(s)3X1(s) From(1)x˙2=3x1+2f (2)

X1(s)=Y(s)=1s((F(s)4Y(s))+1s(2F(s)3Y(s)))X1(s)=1s((F(s)4X1(s))+X2(s)) From(1)and(2)sX1(s)=F(s)4X1(s)+X2(s)x˙1=4x1+x2+f (3)

Thus, from equations (1), (2) and (3), we have the state model in standard form as follows

x˙1=4x1+x2+fx˙2=3x1+2fy=x1

From equation (1) at t=0, we get

x1(0)=y(0) (4)

Similarly, from equation (3) at t=0, we have

x˙1=4x1+x2+fx2(0)=x˙1(0)+4x1(0)f(0)x2(0)=y˙(0)+4y(0)f(0) (5) From(4)

Now, using the second method for the transfer function

Y(s)F(s)=s+2s2+4s+3

This could be also written as

Y(s)F(s)=s+2s2+4s+3Y(s)=(s+2)F(s)(s+3)(s+1)Y(s)=(s+2)F(s)2(1(s+1)1(s+3))

On taking the state variables as follows

X1(s)=F(s)2(s+1)andX2(s)=F(s)2(s+3)

We have

X1(s)=F(s)2(s+1)sX1(s)=F(s)2X1(s)x˙1=x1+f2 (6)

Similarly,

X2(s)=F(s)2(s+3)sX2(s)=F(s)23X2(s)x˙2=3x2f2 (7)

Therefore,

Y(s)=(s+2)F(s)2(1(s+1)1(s+3))Y(s)=(s+2)(X1(s)+X2(s))Y(s)=sX1(s)+sX2(s)+2X1(s)+2X2(s)Y(s)=F(s)2X1(s)+F(s)23X2(s)+2X1(s)+2X2(s) From(6)and(7)Y(s)=X1(s)X2(s)y=x1x2 (8) Thus, from equations (6), (7) and (8) the state model for the transfer function is as

x˙1=x1+f2x˙2=3x2f2y=x1x2

On subtracting equations (6) and (7) at t=0, we get

x˙1(0)x˙2(0)=x1(0)+f(0)2+3x2(0)+f(0)2x˙1(0)x˙2(0)=x1(0)+3x2(0)+f(0)y˙(0)=x1(0)+3x2(0)+f(0) y=x1x2y˙=x˙1x˙2y˙(0)=x1(0)+3(x1(0)y(0))+f(0) y=x1x2y˙(0)=2x1(0)3y(0)f(0)x1(0)=12y˙(0)+32y(0)12f(0) (9) Therefore,

x2(0)=x1(0)y(0) y=x1x2x2(0)=12y˙(0)+32y(0)12f(0)y(0) From(9)x2(0)=12y˙(0)+y(0)12f(0) (10).

Conclusion:

The obtained standard form of state-model for the transfer function Y(s)F(s)=s+2s2+4s+3 is as follows:

By the first method,

x˙1=4x1+x2+fx˙2=3x1+2fy=x1

By the second method,

x˙1=x1+f2x˙2=3x2f2y=x1x2

And the relation between initial values of state variables and y(0)andy˙(0):

For the state model using the first method,

x1(0)=y(0)x2(0)=y˙(0)+4y(0)f(0)

While for the state model using the second method,

x1(0)=12y˙(0)+32y(0)12f(0)x2(0)=12y˙(0)+y(0)12f(0)

And the output equation would be

y=x1.

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