自控第三章作业答案

P3.4 The open-loop transfer function of a unity negative feedback system is

)

1(1)(+=

s s s G

Determine the rise time, peak time, percent overshoot and setting time (using a 5% setting criterion).

Solution: Writing he closed-loop transfer function 2

2

2

2

21

1)(n

n n

s s s s s ωςωωΦ++=

++=

we get 1=n ω, 5.0=ς. Since this is an underdamped second-order system with 5

.0=ς, the

system performance can be estimated as follows.

Rising time

.sec 42.25

.0115

.0arccos 1arccos 2

2

≈-⋅-=

--=

πς

ωςπn r t

Peak time

.sec 62.35

.01112

2

≈-⋅=

-=

π

ς

ωπn p t

Percent overshoot %

3.16% 100% 1002

2

5

.015.01≈⨯=⨯=--πς

πςσ

e

e p

Setting time

.sec 61

5.033

=⨯=

≈

n

s t ςω

(using a 5% setting criterion)

P3.5 A second-order system gives a unit step response shown in Fig. P3.5. Find the open-loop transfer function if the system is a unit negative-feedback system.

Solution: By inspection we have %

30% 1001

13.1=⨯-=

p

σ

Solving the formula for calculating the overshoot,

3.02

1==-ς

πςσ

e

p

, we have

362

.0ln ln 2

2

≈+-=

p

p σ

π

σ

ς

Since .sec 1=p t , solving the formula for calculating the peak time, 2

1ς

ωπ-=

n p t , we get

s e c / 7.33rad n =ω

Hence, the open-loop transfer function is )

4.24(7.1135)

2()(2

+=

+=

s s s s s G n n

ςωω

P3.6 A feedback system is shown in Fig. P3.6(a), and its unit step response curve is shown in Fig. P3.6(b). Determine the values of 1k , 2k , and a .

.1.1

Figure P3.5

Solution: The transfer function between the input and output is given by

2

2

2

1)

()(k as s

k k s R s C ++=

The system is stable and we have, from the response curve,

2

1lim )(lim 12

2

210

==⋅

++⋅

=→∞

→k s

k as s

k k s t c s t

By inspection we have %

9% 10000

.211

.218.2=⨯-=

p

σ

Solving the formula for calculating the overshoot, 09

.02

1==-ς

πςσe

p

, we have

608

.0ln ln 2

2

≈+-=

p

p σ

π

σ

ς

Since .sec 8.0=p t , solving the formula for calculating the peak time,

2

1ς

ωπ-=

n p t , we get

s e c / 95.4rad n =ω

Then, comparing the characteristic polynomial of the system with its standard form, we have

2

2222n n s s k as s ωςω++=++

5.2495.4222===n k ω

02.695.4608.022=⨯⨯==n a ςω

P3.8 For the servomechanism system shown in Fig. P3.8, determine the values of k and a that satisfy the following closed-loop system design requirements. (a) Maximum of 40% overshoot. (b) Peak time of 4s.

Solution: For the closed-loop transfer function we have 22

22

2)(n

n n

s s

k

s k s

k s ω

ςωωαΦ++=++=

hence, by inspection, we get

k n

=2

ω, αςωk n =2, and n

n

k

ως

ςωα22=

=

Taking consideration of %

40% 1002

1=⨯=-ς

πςσ

e

p

results in

280.0=ς

.

In this case, to satisfy the requirement of peak time, 4

12

=-=

ς

ωπn p t , we have

.s e c / 818.0r a d n =ω

.2.2

(a)

(b)

Figure P3.6

Figure P3.8