自動(dòng)控制原理簡(jiǎn)答題

1_e~Ts

47

48

49

51

52

53

54

n

m

n-m

S

n-m

55

z C(z)

G(z)=

56Nyquist

+ooNyquist (-

1,J0) N, G(s)H s

(NHP) s

57

58 t

59 (Nichocls

1(CD)=201SA(CD)(db)(Nichols

60

61

P7=

Z Z=|P-NI

62

(1)

(3)

(2)

63

PD

64

(wr)=180+

ZG(>JHOg)=-180

(”.)，

|=1

65

66

68

69

70

71

72

73

74

75

76

1

1

2 3

3

4

36.

37.—

(2

3

39.

40.

36.

(3

(2

37.

1

/=T(2.5 f=0

0.632(2.5

(34)

3&

1/5

(2

(2

(1

39.

(1.5

(1.5

W(S)=E(S)H(S),

1 ％)=E(s) (2

40.

S

(2

36.

(2.5

(3

(2.5

37.

(2.5

(2.5

38.0

201gK<>(2

1

-20dB/dec,

?=Ki(1.5

2

40dE/dec,

CDI=KZ(1.5

39.

(3

G(s)=K/(s

(s+l)(s+2))

K

0<K<6o(2

40.1)

s

(1

2)

s

±45

(1

3)

(1

4)

5)

(1

5)

(1

3

36.1)

2) (2

(1

(1

(1

37.

(2.5

(2.5

3

1)

(1.5

2)

(2

3)

(1.5

39.

(3

(2

40.

5

(2.5

(2.5

36.

37.

3

39.

40.

36

(1

(1

(2

(1

37

(2

(3

3&

(3

s

s

(2

39.

(2.5

(2.5

40.

m n,(n-m)

(2.5

2.5

36.

37.

3&

39.

40.

36

(1.5 (1.5 (1 (1

37.

(1

(1

(1

38.

39.

40.

1)

(1

(2

(2.5

(2

(3

(2

(2.5

36.

37.

38.

39.

40.

36.

(1

(1

(1 (2

37.

g=0.707

(3 g=0.707

(2

3&

-aictan0.16?-aictan (2.5

Jl+d/J100+Q，

Q=0 A(0)=1

0)

0

A(s) 0八

(co)=-180(2.5

39.

(2.5

(2.5

40.

(1

(1

(1

(1

(1

36.

37.

0.707

3

G(s)=--

(s+

l)(s+10)

39.

40.

36

(2.5

(2.5

37.

(2 0

(1 1

-90

(1

2

-180

(1

38.

r(r)

= R(s)=

A K“=Inn

=0 (2.5

2

1

2

J=

—=—=oo

111115~G(5)

)(2.5

39.

(2

(3

40.

(1

(1

(1

(1

(1

36.

37.

38.

I

39.

40.

36.

(2

(3

37.

(3

(3

39.

(1.5

(1.5

(2

40.

(3

(2

3

PID

15

3

13 3

4

5

TOC\o"1-5"\h\z

T?1 T?1

PI

H（s）, 1/H（s）,

(3

9

bode （3

w w0-8

0

10

t

nyquist （3

16 5

17

20dB/dec -60dB/dec

-40dB/dec

18

TOC\o"1-5"\h\z

(1) (2) (3)

19

20 3

(1) (2) (3)

21

(2) (3)

(4)

21

S0G(s)

1

（K=l）

(6

3

(3

DDZ-III

(3

PLC

lOOKPa,

(m3)Qmin

0.1%

1) N,2) C,3)

T(2

1g/cm3

54

(TC

(2

2%

4%

(1.5

0.01%

(2

(1.5

(3

)

(3

B,4)

Ts,5)

5

Kp

Kp

Kp

Kp

Kp,

Kp

Kp

(SP186)