Dlscrib Com PDF Solution Manual For Process Control Modeling Design

 

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Chapter 8 Solutions

Using an all-pa all-pass ss facto factorizat rization, ion, we hav havee that   (−3s + 1)(−5s + 1)e g˜ p+ (s) = (3s + 1)(5s + 1)   1.75(3s + 1)(5s + 1) g˜ p (s) = (10s + 1)(4s + 1)2

1.25s

−

8.1

Missing

−

8.2 From the block diagram y (s) =  g p (s)u(s) + gd (s)l(s)

thus, thu s, our strictly proper con controlle trollerr is q (s) =

and





u(s) =  q (s) r (s) −  d˜(s)

b.

but d˜(s) =  y (s) −  ˜ y (s)

  (10s + 1)(4s + 1)2 1.75(3s + 1)(5s + 1)(λs + 1)2

If the model is perfect, then the transfer transfer function is

y˜(s) = g˜ p (s)u(s)

y (s) =  q (s)g p (s)r(s)

  (−3s + 1)(−5s + 1) e 1.25s =   r(s) (3s + 1)(5s + 1)(λs + 1)2 −

then

d˜(s) =  y (s) −  ˜ g p (s)u(s)

and

The sketch of   y (t) should look like figure 8-1. u(s) =  q (s) [r(s) − y (s) + ˜ g p (s)u(s)] Exercise 8−3

[1 − q (s)˜ g p (s)] u(s) =  q (s)r(s) −  q (s)y (s) u(s) =

1.2

q (s)   q (s) r (s) − y (s) 1 − q (s)˜ g p (s) 1 − q (s)˜g p (s)  

1

0.8

then, plugging this into our original expression for y (s), we have y (s) =

0.6

  g p (s)q (s)   g p (s)q (s) r(s)− y (s)+gd (s)l(s) 1 −  q (s)˜ g p (s) 1 − q (s)˜g p (s)

        )         t         (       y

0.4

and we can rearrange this into

0.2

[1 − q (s)˜g p (s)] y (s) =  g p (s)q (s)r(s) − g p (s)q (s)y (s) + [1  − q (s)˜ g p (s)] gd (s)l(s) [1 − q (s)˜g p (s) + g p (s)q (s)] y (s) =

0

−0.2 0

10

20

30

40

50

t

g p (s)q (s)r(s) + [1 − q (s)˜ g p (s)] gd (s)l(s) [1 +  q (s) {g p (s) −  ˜ g p (s)}] y(s) =

Figure 8-1: Plot for 8.3

g p (s)q (s)r(s) + [1 − q (s)˜ g p (s)] gd (s)l(s) y(s) =

g p (s)q (s) r(s) 1 + q (s) {g p (s) −  ˜ g p (s)}

8.4

+

  [1 − q (s)˜g p (s)] gd (s) l(s) 1 +  q (s) {g p (s) − ˜g p (s)}

Missing

 

8.5

8.3 a.

Our transfer function is (with its factorization From the problem statement representation)   2 g ˜ p (s) =   1.75(−3s + 1)(−5s + 1)e 1.25s 5s + 1 g ˜ p (s) = −

(10s + 1)(4s + 1)2 g˜ p (s) = g ˜ p+ (s)˜ g p (s)

g p (s) =   1.5(−s + 1) (s + 1)(4s + 1)

−

8-1

60

70

 

8.7

Then our controller is q (s) =

Missing

  (5s + 1) 2(λs + 1)

8.8

The close closed–loop d–loop relationship relationship is y (s) =

Missing

g p (s)q (s) r(s) 1 + q (s) {g p (s) −  ˜ g p (s)}  

1.5( s+1) (5s+1) (s+1)(4s+1) 2(λs+1) −

= =

s+1) 1 +  2((5λs +1)

  8λs3



  1.5( s+1)   2 (s+1)(4s+1)  − 5s+1 −

 r(s)

1.5(−s + 1)(5s + 1) r(s) + (10λ − 7.5)s2 + (2λ + 6)s + 1.5

we can find the offset using the final value theorem (assuming a unit step) offset = lim s (r(s) −  y (s)) s→0

= 1  −  lim s→

0 8λs3

1.5(−s + 1)(5s + 1) + (10λ − 7.5)s2 + (2λ + 6)s + 1.5

=0 In order to find the smallest value of  λ  that will keep the system stable, we need to use Routh’s stability criterion. Our characteristic equation is 8λs3 + (10λ − 7.5)s2 + (2λ + 6)s + 1.5 = 0 Thus we immediately find that the necessary conditions dictate that λ >  0 λ > 0 .75 λ >  −3

to check the sufficient condition, we need to build the array row 1 8λ   2λ + 6 2 10λ − 7.5 1.5   20λ +33λ 45 3   0 10λ 7.5 4 1.5 2

−

−

thus we need to find the bound for the element in column 3; since we have another condition that makes 10λ − 7.5  >  0, we can just satisfy 20λ2 + 33λ − 45  >  0. We find that   λ <   −2.5369 or   λ >   0.88690 8869068. 68. Thus our most restrictive condition requires λ >  0 .88691

8.6

Missing 8-2

 

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