Sistemas De Control Automatico Benjamin C. Kuo 7ed (Solucionario) PDF

Preview

Summary

http://www.elsolucionario.blogspot.com LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS. Chapter 2 MATHEMATICAL FOUNDATION 2-1 (a) Poles: s = 0, 0, −1, −10...

Description

http://www.ĞůƐŽůƵĐŝŽŶĂƌŝŽ͘ďůŽŐƐƉŽƚ.com

/,%52681,9(5,67$5,26 0

K

or K

< 11 . 36

14.2 s

0

K

Thus, the system is stable for 0 < K < 11.36. When K = 11.36, the system is marginally stable. The auxiliary equation equation is A ( s ) = 14.2 s frequency of oscillation is 0.894 rad/sec.

(b)

s

2

+ 11 . 36 = 0 .

The solution of A(s) = 0 is s

2

= − 0 .8 .

The

+ Ks + 2 s + ( K + 1) s + 10 = 0

4

3

2

Routh Tabulation: 4

s

1

3

s

2

s

2

K

K

− K −1

2K

=

K

K

−1

10

+1

>0

K

10

K

>1

K

−9 K − 1 2

s

1

s

2

−1

K 0

−9 K −1 > 0

10

The conditions for stability are: K > 0, K > 1, and − 9 K − 1 > 0 . Since K is always positive, the last condition cannot be met by any real value of K. Thus, the system is unstable for all values of K. 2

(c)

s

2

+ ( K + 2 ) s + 2 Ks + 10 = 0

3

2

Routh Tabulation: s s

s

3

2

1

K 2K

1

2K

+2

10

2

+ 4 K − 10 K

s

0

K

K

+2

> −2 2

+ 2 K −5 > 0

10

The conditions for stability are: K > −2 and K + 2 K − 5 > 0 or (K +3.4495)(K − 1.4495) > 0, or K > 1.4495. Thus, the condition for stability is K > 1.4495. When K = 1.4495 the system is 2

marginally stable. The auxiliary equation is A ( s ) The frequency of oscillation is 1.7026 rad/sec.

(d)

s

3

+ 20

s

2

+ 5 s + 10

K

= 3.4495

s

=0

Routh Tabulation:

73

2

+ 10 = 0 .

The solution is s

2

= −2 .899

.

s s s

3

1

2

1

5

20

10 K

− 10

100

K

= 5 − 0 .5 K

5

− 0 .5 K > 0

K

>0

or K

< 10

20 s

0

10 K

The conditions for stability are: K > 0 and K < 10. Thus, 0 < K < 10. When K = 10, the system is

= 20

marginally stable. The auxiliary equation is A ( s ) equation is s

(e)

s

= −5 .

2

3

2

+ 100 = 0 .

The solution of the auxiliary

The frequency of oscillation is 2.236 rad/sec.

+ Ks + 5 s + 10 s + 10

4

s

2

K

=0

Routh Tabulation: s s s

4

3

2

1

5

K

10

K

10 K

5K

− 10

5K

10 K

>0 − 10 > 0

or K

>2

K

− 100

50 K s

K 5K

1

− 10

K

2

=

− 10

50 K

− 100 − 10 5K

K

3

5K

− 10

− 10 − K > 0 3

K s

0

β

10 K

γε

K

The conditions for stability are: K > 0, K > 2, and 5 K K

+ 2 . 9055

K

2

− 2 . 9055

K

+ 3 .4419

ϕ

2 and K < is unstable for all values of K.

(f)

s

4

− 10 − K > 0 . 3

>0 The last condition is written as

The second-order term is positive for all values of K.

−2.9055.

Since these are contradictory, the system

+ 12 . 5 s + s + 5 s + K = 0 3

2

Routh Tabulation: s s s

4

3

2

1

1

12 . 5

5

−5

12 . 5

= 0 .6

K

K

12 . 5

s

1

3

− 12 . 5 K

= 5 − 20 .83

K

5

− 20 . 83 K > 0

or K

< 0 .24

0 .6 s K K >0 The condition for stability is 0 < K < 0.24. When K = 0.24 the system is marginally stable. The auxiliary 0

equation is A ( s ) = 0 . 6 s oscillation is 0.632 rad/sec.

2

6-4

The characteristic equation is Ts

3

+ 0 .24 = 0 .

The solution of the auxiliary equation is s

+ ( 2T +1) s + ( 2 + K ) s + 5 K = 0 2

Routh Tabulation:

74

2

= − 0 .4.

The frequency of

s s

s

3

T

2

+1

2T ( 2T

1

0

T

5K

T

+ 1 )( K + 2 ) − 5 KT 2T

s

+2

K

>0 > −1 /

K (1 − 3 T )

+1

5K

T > 0, K > 0, and K

4T

+2

3T

−1

+ 24

Ks

<

+ 4T + 2 > 0

>0

K

The conditions for stability are:

2

. The regions of stability in the

T-versus-K parameter plane is shown below.

6-5 (a)

Characteristic equation: s

5

+ 600

s

4

+ 50000

s

3

+ Ks

2

+ 80

K

=0

Routh Tabulation: s s

s

5

4

1

50000

24 K

600

K

80 K

× 10 − K 7

3

3

2

214080

14320 K K

600 s

3

s

1

0

7

600 00 K

−K

2

80 K

× 10 − K 7

− 7 .2 × 10

+ 3 .113256 × 10

16

600( 214080 s

< 3 × 10

00

11

− 14400

K

−K

K

< 214080

K

00

2

K

2

− 2 .162 × 10

)

80 K

K

Conditions for stability:

75

>0

7

K

+ 5 × 10

12

0

Thus, the final condition for stability is: When K

5

or

+ ( K + 2 ) s + 30 2

Ks

+ 200

K

=0

Routh tabulation: s s

s

3

2

1

1

+2

K

2

30 K

K s

0

30 K

− 140

200 K K

+2

200 K

Stability Condition:

Characteristic equation: s

3

> −2

K

> 4. 6667

K

>0

K > 4.6667

When K = 4.6667, the auxiliary equation is A ( s ) The frequency of oscillation is 11.832 rad/sec.

(c)

K

+ 30

s

+ 200

2

= 6 . 6667

s

2

+ 933

. 333

=0.

The solution is s

2

+K =0

s

Routh tabulation: s s s

3

2

1

1

200

30

K

−K

6000

K

< 6000

K

>0

30 s

0

K

Stabililty Condition:

0

<

< 6000

K

When K = 6000, the auxiliary equation is A ( s ) The frequency of oscillation is 14.142 rad/sec.

(d)

Characteristic equation:

s

3

= 30

s

2

+ 6000 = 0 .

The solution is s

2

= − 200

.

+ 2 s + ( K + 3) s + K + 1 = 0 2

Routh tabulation: s s s

3

2

1

K

+3

1

K

2

K +1

+5

K

> −5

K

> −1

30 s

0

K +1

Stability condition:

K>

−1.

When K =

−1 the zero element occurs in the first element of the

76

= − 140

.

0

= s + ( k 2 − 1 ) s + 20 − 2 k1 − k 2 = 0 2

or k

2

>1

− 2 k1 − k 2 > 0

or

k

< 20 − 2 k 1

2

Parameter plane:

6-7 Characteristic equation of closed-loop system: s −1 0 sI − A + BK = 0 k1

= s + ( k 3 + 3 ) s + ( k 2 + 4 ) s + k1 = 0

s

−1

k2 + 4

s + k3 + 3

3

2

Routh Tabulation:

s

3

s

2

s

1

s

k2 + 4

1 k3 + 3

(k

3

k3 +3>0 or k3 > − 3

k1

+ 3) ( k 2 + 4 ) − k 1

(k

k3 + 3 0

+ 3 )( k + 4) − k > 0 2

1

k >0

k

1

1

Stability Requirements:

k 3 > − 3, 6-8 (a)

3

(k

k 1 > 0,

3

+ 3) ( k 2 + 4 ) − k 1 > 0

Since A is a diagonal matrix with distinct eigenvalues, the states are decoupled from each other. The second row of B is zero; thus, the second state variable, x is uncontrollable. Since the uncontrollable 2

77

state has the eigenvalue at −3 which is stable, and the unstable state x with the eigenvalue at −2 is 3

controllable, the system is stabilizable.

(b) 6-9

Since the uncontrollable state x has an unstable eigenvalue at 1, the system is no stabilizable. 1

The closed-loop transfer function of the sysetm is

Y (s)

=

R (s ) The characteristic equation is:

1000 s + 15.6 s + ( 56 + 100 K t ) s + 1000 3

s

2

3

+ 15 . 6 s + ( 56 + 100 2

K )s t

+ 1000 = 0

Routh Tabulation: s s

s

3

1

2

1

15 . 6 873 . 6

+ 100

56

+ 1560

K

t

1000 Kt

− 1000 1560 K t

15 .6 s

0

.4

>0

1000

Stability Requirements:

6-10

− 126

K

t

> 0 . 081

The closed-loop transfer function is Y (s)

K(s

=

R( s)

s

The characteristic equation:

3

s

+ Ks 3

+ Ks

+ 2 )( s + α )

+ ( 2 K + αK − 1 ) s + 2 αK

2

2

+ ( 2 K + α K − 1) s + 2 αK = 0

Routh Tabulation: s s

s

3

1

2

1

2 αK

K (2

+ αK − 1

2K

K

+ α ) K − K − 2 αK

>0

2

(2 + α )K

K s

0

2 αK

Stability Requirements: α > 0 , K -versus- α Parameter Plane:

α >0 K

> 0,

K

>

1

+ 2α

2

+α

78

.

− 1 − 2α > 0

6-11 (a)

Only the attitude sensor loop is in operation: K

Θ( s) Θr ( s ) If KK

s

If KK

s

t

= 0. The system transfer function is:

G (s)

=

p

1+ K G ( s) s

K

= s

2

p

− α + KK

s

>α,

the characteristic equation roots are on the imaginary axis, and the missible will oscillate.

≤α,

the characteristic equaton roots are at the origin or in the right-half plane, and the system

is unstable. The missile will tumble end over end.

(b)

Both loops are in operation: The system transfer function is

Θ( s) Θr ( s ) For stability: When K

KK

=0

t

G (s) p

1 + K sG t

> 0,

t

and KK

=

KK

>α,

s

(s)

p

s

+ K sG p ( s )

K

= s

2

+ KK

t

s

+ KK s − α

−α > 0 .

the characteristic equation roots are on the imaginary axis, and the missile

will oscillate back and forth. For any KK − α if KK < 0, the characteristic equation roots are in the right-half plane, and the system s

If KK

t

t

is unstable. The missile will tumble end over end. > 0 , and KK < α , the characteristic equation roots are in the right-half plane, and the system is t

unstable. The missile will tumble end over end.

6-12

Let s

1

= s + α,

then when s

= −α,

s

1

= 0.

This transforms the

s = −α axis in the s-plane onto the imaginary

axis of the s -plane. 1

(a)

F (s)

=

Or

s

s 2 1

2

+ 5s +3 = 0

Le t s

= s1 − 1

We get

(s

1

−1) + 5 ( s1 −1) + 3 = 0 2

+ 3 s1 − 1 = 0 2

s1

Routh Tabulation:

s s

−1

1

1

3

1

−1

0 1

Since there is one sign change in the first column of the Routh tabulation, there is one root in the region to the right of s = −1 in the s-plane. The roots are at −3.3028 and 0.3028.

(b)

F (s)

=

s

3

+ 3s + 3s +1 = 0 2

Let s

=

s

1

−1

We get

79

( s1 −1)3

+ 3( s1 −1) + 3 ( s1 −1) + 1 = 0 2

(c)

Or

s

F (s)

=

Or

= 0.

3 1

3

s s

3 1

The three roots in the s -plane are all at s 1

+ 4 s + 3 s + 10 = 0 2

=

Let s

s

1

1

= 0.

Thus, F(s) has three roots at s =

We get

1

−2

1

10

(s

−1

1

−1.

−1) + 4 ( s1 −1) + 3 ( s1 −1) +10 = 0 3

2

+ s 1 − 2 s1 + 10 = 0 2

s s

Routh Tabulation:

3 1 2 1

− 12

1

s1 s

0

10

1

Since there are two sign changes in the first column of the Routh tabulation, F(s) has two roots in the region to the right of s = −1 in the s-plane. The roots are at −3.8897, −0.0552 + j1.605, and −0.0552

(d)

F (s) Or

=

s

+4s +4s +4 = 0

3

s

− j1.6025. 2

3 1

Let s

=

s

1

−1

(s

We get

1

−1) + 4 ( s1 −1) + 4 ( s1 −1) + 4 = 0 3

2

+ s1 − s1 + 3 = 0 2

s s

Routh Tabulation:

3 1 2 1 1

s1 s

1

−1

1

3

−4

0

3

1

Since there are two sign changes in the first column of the Routh tabulation, F(s) has two roots in the region to the right of s = −1 in the s-plane. The roots are at −3.1304, −0.4348 + j1.0434, and −0.4348

−j1.04348.

6-13 (a) Block diagram:

(b) Open-loop transfer function: G ( s) =

H ( s)

=

E (s )

K a K i nK I N

s (R a Js + K i K b ) ( As + K o )

=

16.667 N s( s + 1)( s + 11.767)

Closed-loop transfer function: H (s) R( s)

G (s)

= 1

+G ( s)

16 .667 N

= s

3

+ 12 . 767

(c) Characteristic equation:

80

s

2

+ 11 . 767 s + 16 . 667

N

s

+ 12 . 767

3

s

2

+ 11 . 767

s

+ 16 . 667

=0

N

Routh Tabulation: 3

s

2

s s

1

1

11 . 767

12 .767 150 .22

16 . 667 N

− 16 . 667

N

− 16.667

1 50.22

N

>0

or

N

. 9 − 15 NA

>0

12 . 767 s

0

16 .667N

N

Stability condition:

0

<

N

>0

0

0 . 706 A

+3 > 0

24.92 A

+ 105

N > 0

N < 1.66 + 7.06/A

When A

→∞

N

max

→ 1. 66

For A = 50, the characteristic equation is

3 s + ( 35.3 + 0.06 K o ) s + 0.706 Ko s + 250 N = 0 3

2

Routh tabulation

81

Thus, N

max

= 1.

− 588