Solution Manual for Process Control Modeling, Design and Simulation 1st Edition B. Wayne Bequette

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Chapter 8 Solutions

Using an all-pass factorization, we have that (−3s + 1)(−5s + 1)e−1.25s (3s + 1)(5s + 1) 1.75(3s + 1)(5s + 1) g˜p− (s) = (10s + 1)(4s + 1)2

8.1

g˜p+ (s) =

Missing 8.2 From the block diagram

thus, our strictly proper controller is

y(s) = gp (s)u(s) + gd (s)l(s)

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

q(s) =




 ˜ u(s) = q(s) r(s) − d(s)

and

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

but ˜ = y(s) − y˜(s) d(s) y˜(s) = g˜p (s)u(s)

y(s) = q(s)gp (s)r(s)

then

=

˜ = y(s) − g˜p (s)u(s) d(s) and

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

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)˜ gp (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)˜ gp (s) 1 − q(s)˜ gp (s)

1

0.8

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

0.6 y(t)

gp (s)q(s) gp (s)q(s) r(s)− y(s)+gd (s)l(s) y(s) = 1 − q(s)˜ gp (s) 1 − q(s)˜ gp (s)

0.4

and we can rearrange this into

0.2

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

0

−0.2

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

8.3

10

20

30

40

50

t

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

0

Figure 8-1: Plot for 8.3

8.4

Missing 8.5

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 1.5(−s + 1) gp (s) = g˜p (s) = g˜p+ (s)˜ gp− (s) (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 closed–loop relationship is y(s) =

1.5(−s+1) (5s+1) (s+1)(4s+1) 2(λs+1)

= 1+ =

Missing

gp (s)q(s) r(s) 1 + q(s) {gp (s) − g˜p (s)} (5s+1) 2(λs+1)



1.5(−s+1) (s+1)(4s+1)

−

2 5s+1

 r(s)

1.5(−s + 1)(5s + 1) r(s) 8λs3 + (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 20λ2 +33λ−45 3 0 10λ−7.5 4 1.5 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 > 0.8869068. Thus our most restrictive condition requires λ > 0.88691

8.6

Missing 8-2

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