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第五章线性系统的频域分析法

5-2 频 率 特 性

C (s) G (s) = R (s)

r(t) = Asinωt
k1 k2 = + + s + jω s jω

Aω R(s) = 2 2 s +ω

Aω C ( s) = 2 G(s) 2 s +ω

n

i =1

ki s si

Aω(s + jω) AG( jω) k1 = C(s)(s + jω) |s = jω = G(s) = (s + jω)(s jω ) s = jω 2j
Aω ( s jω ) k 2 = C ( s )( s jω ) |s = jω = G ( s ) ( s + jω )( s jω ) AG ( jω ) = 2j

s = jω

G( jω) =| G( jω) | e

j∠G( jω)

= A(ω)e

j (ω)

G( jω) = G(s) |s= jω

G( jω) = G(s)|s= jω =| G( jω)| e j∠G( jω) = A(ω)e j(ω)

A A j (ω ) k1 = G( jω ) e 　　　k2 = G( jω ) e j (ω ) 2j 2j

c (t ) = k1e

jω t

+ k2 e

jω t

A G ( jω ) e j (ω t + ) e j (ω t + ) = 2j

= A G ( jω ) sin(ω t + (ω ))

G ( jω )

：反映幅值衰减

(ω ) ：相移

G( jω) = G(s) |s= jω = G( jω) e j(ω)

A(ω ) =| G( jω ) |
(ω ) = ∠G ( jω )

G ( jω ) = P(ω ) + jQ(ω ) 这里 P(ω ) = Re[G ( jω )] 和 Q(ω ) = Im[G( jω )] 分别称为系统的实

K G (s) = Ts + 1

r(t) = sin t

2 c (t ) = sin(t 45o ) 2

K G ( jω ) = 1 + jT ω

A(ω ) =

K 1 + ω 2T 2

ω =1

(ω ) = arctan T ω

ω =0

Q(ω )
A(ω ) (ω ) ω =∞

ω

P(ω )

G(s) =

s +1 s2 + s +1

num=1; den=[1 1]; nyquist(num,den)

K 例 G (s) = Ts + 1

K G ( jω ) = 1 + jT ω

A(ω ) =

K 1 + ω 2T 2

(ω ) = arctan T ω
(ω) = 0

ω =0

A(ω) = K

ω =∞

A(ω) = 0

(ω) = 90

o

σ

Dec Dec Dec Dec

∞...
0

2
0.01

1
0.1

0 1

1 10

2 100

log ω

ω

ω 由于 以对数分度，所以零频率线在 ∞ 处。

dB

L = 20 log A

－40dB/dec －20dB/dec

log ω

20 k = = 20 log1 log0.1 ω
A(ω ) = 1

ω

L = 40 log ω

A(ω ) =

1

ω2

5-3 典型环节频率特性和 开环频率特性

G ( jω ) = K

20 log K K >1 20 log K 20 log K K = 1 log ω K <1

> 0 L(ω ) = 20 lg K = 常数 = = 0 < 0

K >1 K =1 K <1

log ω
0° (ω ) = ∠ K = 180° K ≥0 K <0

(ω )

180°

K ≥0
K <0

180°

K 积分环节的频率特性： ⒉ 积分环节的频率特性：G ( s ) = s π K (ω) = tg1(ω /0) = A(ω) =

K K K π G ( jω ) = =j = e 2 jω ω ω
L(ω) = 20log A(ω) = 20log = 20log K 20log ω, K

ω

2

ω

L(ω ) / dB 40 20
20 40

K = 10

ω

1 10 100
K =1

(ω )
90°

ω
1 10 100

ω= K

A(ω ) = 1 1 + T 2ω 2 ,

G (s) =

1 Ts + 1

G ( jω ) =

1 jT ω + 1

(ω ) = tg 1T ω

1/T
1/ T

L(ω) ≈ 0

L(ω) ≈ 20log Tω = 20logω 20logT

1 Tω = 1, ω o = ，称为转折频率或交接频率。 T

1 20T

1 10T

1 5T

1 2T

1 T

2 T

5 T

10 T

20 T

1 ω = ω o = 处，为 T

max = 20log 1 + T 2ω0 ≈ 3(dB)
1 10T 1 5T 1 2T 1 T 2 T 5 T 10 T

2

(ω ) = tg 1Tω

K Kω n = 2 振荡环节的频率特性： ⒋ 振荡环节的频率特性： G ( s ) = 2 2 T s + 2ζTs + 1 s + 2ζω n s + ω n 2
2

1 G ( jω ) = (1 T 2ω 2 ) + j 2ζωT

A(ω ) =

1

(1 T 2ω 2 ) 2 + ( 2ζωT ) 2 2ζωT (ω ) = tg 1 1 T 2ω 2

L 对数幅频特性为： (ω ) = 20 log A(ω ) = 20 log (1 T 2ω 2 ) 2 + (2ζωT ) 2

2ζωT 相频特性： (ω ) = tg 1 T 2ω 2
1

K = 10,T = 1,ζ = 0.3 10 G( jω) = 2 s + 0.6s +1 1 ωo = T

40dB/ Dec

① 对数相频特性曲线 在半对数坐标系中 对于( ω0, -90°)点 是斜对称的。 ② 对数幅频特性曲线 有峰值。

ω r = ω n 1 2ζ 2

> 1 2

1 = 0.707 时， r = 0 ω 2

M r = A(ω r ) =

<
1

1 2

2ζ 1 ζ 2

1 当ω = ω 0 ， A(ω0 ) = ，(ω 0 ) = 20 lg 2ζ 。 2ζ L

L(ω )(dB )

= 0 .1 = 0 .2 = 0 .3 = 0 .5 = 0 .7 = 1 .0

= 0.1 = 0.2 = 0.3 = 0.4 = 0.5 = 0.6

= 0.7 = 0.8 = 1.0

(ω )(deg)

1 10T

1 5T

1 2T

1 T

2 T

5 T

10 T

= 0.1 = 0.2 = 0.3 = 0.5 = 0.7 = 1 .0

1 10T

1 5T

1 2T

1 T

2 T

5 T

10 T

G ( s ) = 1 + Ts G ( s ) = T 2 s 2 + 2ζTs + 1 频率特性分别为： G ( jω ) = jω

G ( jω ) = 1 + jTω G ( jω ) = 1 T 2ω 2 + j 2ζωT

2

L(ω )(dB)
20dB / dec

1

10 20dB / dec

(ω )(deg)
90°

0 0.1
90°

1 10

A(ω ) = 1 + T 2ω 2 , (ω ) = tg 1Tω L(ω ) = 20 lg 1 + T 2ω 2

ω = 0, (ω ) = 0; 　ω = 1/ T , (ω ) = 450 ; 　ω = ∞, (ω ) = 900

1 T

L(ω )(dB)

30

20

10

0

(ω )(deg)
90°

45°

0° 1 20T

1 10T

1 5T

1 2T

1 T

2 T

5 T

10 T

20 T

1 20T

1 10T

1 5T

1 2T

1 T

2 T

5 T

10 T

20 T

G ( s) = T 2 s 2 + 2ζTs + 1
2 2 2 2 1

2ζωT A(ω ) = (1 T ω ) + ( 2ζωT ) , (ω ) = tg 1 T 2ω 2 L(ω ) = 20 lg (1 T 2ω 2 ) 2 + ( 2ζωT ) 2

1 ，高频段的斜率+40dB/Dec。 T 1 相角： ω = 0时， (ω ) = 0; ω = , (ω ) = π ; ω = +∞, (ω ) = π 当 T 2

( )(deg)

L(ω )(dB)

180°
1 .0

150° 120° 90° 60° 30°

L( )(dB )

0. 7 0. 5 0. 3 0. 2 0. 1

= 0.1 = 0.2 = 0.3 = 0.5 = 0.7 = 1.0

10

1 .0 0 .7

(ω )(deg)

0

0 .5 0 .3 0 .2

-10

0 .1

= 0.1 = 0.2 = 0.3 = 0.5 = 0.7 = 1 .0

-20
1 10T 1 5T 1 2T 1 T 2 T 5 T 10 T

1 10T

1 5T

1 2T

1 T

2 T

5 T

10 T

G 频率特性： ( jω ) = e jτω 幅频特性：A(ω ) = 1

(ω )

1 10τ

1 5τ

1 2τ

1

2

5

10

τ

τ

τ

τ

7. 非最小相位环节 定义：在右半S平面上既无极点也无零点，同时无纯滞后环节 的系统是最小相位系统，相应的传递函数称为最小相位传递函 数；反之，在右半S平面上具有极点或零点，或有纯滞后环节 的系统是非最小相位系统，相应的传递函数称为非最小相位传 递函数。
1 G1 ( s ) = T1 s + 1 1 G2 ( s ) = 1 T1 s

G1 ( jω) = G2 ( jω) =

1 1+ω2T12

1 (ω ) = tan 1 ωT 2 (ω ) = tan 1 ωT

0

T

ω0 = 振荡环节的频率特性—波德图：低频、高频渐进线，斜率 ， 振荡环节的频率特性 波德图：低频、高频渐进线，斜率-40， 波德图 T 转折频率

1

K s (T1 s + 1)(T2 s + 1)

A(ω ) =

K

K jω(1+ jT1ω)(1 + jT2ω)

ω 1 + ω 2T2 2 1 + ω 2T12

(ω ) = 90o tg 1T1ω tg 1T2ω

③ A(0) = ∞ (0) = 90o ⑤
(ωx ) = 180o

A(∞) = 0 (∞) = 270o

ωx

σ

tg 1T1ω x + tg 1T2ω x = 90o
1 ω x T1T2 = 0
2

ωx =

1 T1T 2

ωx

KTT2 A(ωx ) = 1 T1 +T2

：穿越频率

K (τ s + 1) s (T1 s + 1)(T2 s + 1)

② A(ω ) =

K 1 + ω 2τ 2

K jω(1+ jT1ω)(1 + jT2ω)

ω 1 + ω 2T2 2 1 + ω 2T12

(ω ) = 90o tg 1T1ω tg 1T2ω + tg 1τω

③ A(0) = ∞ (0) = 90o ⑤
(ωx ) = 180o

A(∞) = 0 (∞) = 180o

σ

tg 1T1ω x + tg 1T2ω x = 90o + tg 1τω x

ω T1 + ω T2 1 = 2 1 ω x T1T2 τω
(T1T2 T1τ T2τ )ω 2 = 1

TT2 τ< 1 T +T2 1

K (τ s + 1) G ( s) H ( s) = 2 s (Ts + 1)

K (τ 1 s + 1)(τ 2 s 2 + 2ζ 2τ 2 s + 1)… G (s)H (s) = γ s (T1 s + 1)(T2 s 2 + 2ζ 2 T2 s + 1)…

L(ω ) = 20 log A(ω ) = ∑ Li (ω )
i =1

N

(ω ) = ∑ i (ω )
i =1

N

k , T1 > T2 ，试 [例]：开环系统传递函数为：G ( s ) = s (1 + T1s)(1 + T2 s ) 画出该系统的波德图。
[解]：该系统由四个典型环节组成。一个比例环节，一个积分环 节两个惯性环节。手工将它们分别画在一张图上。
20
1 T1 1 T2

(ω )
45o
90o 135o

1 T1

1 T2

20 40 60 80

40

60

180o

270o

ω =1
L(ω ) = 20 log K

K G (s) = ν s
L(ω) = 0

K / ων = 1

ω = Kν

1

４.确定其他特性曲线，在转折频率处斜率发生变化 确定其他特性曲线，

100( s + 2) G (s)H (s) = = s ( s + 2)( s + 20)

ω1 = 1
40 20 -20

ω2 = 2

ω3 = 20

10(0.5 s + 1) s s ( s + 1)( + 1) 20

ωc 2 0 lo g = L1 ω2

-40 1 2 10 -20 -40 20

L1 = 20log K 40log

2 1

ωc = 5

A(ω c ) = 1

10 × 0.5ω c A(ω c ) = =1 ωc × ωc

K G( s) = s( s + 1)( s + 5)

ωc ωc 2 + 12 (0.2ωc )2 + 12 = 2

[例]系统开环特性为： 试画出波德图。
ω 则， 1 =

10 Gk ( s ) = (0.25s + 1)(0.25s 2 + 0.4 s + 1)

[解]：1、该系统是0型系统，所以 ν = 0, k = 10, T1 = 0.25, T2 = 0.5
1 1 = 4, ω 2 = = 2,20 log k = 20dB T1 T2

2、低频渐进线：斜率为 20ν = 0dB，过点（1，20） 3、波德图如下：
A(ω )
20

40
4 60

1

2

10

log ω

[例]已知

10 3 (1 + 100s ) 2 ，试画波德图。 G (s) = 2 s (1 + 10 s )(1 + 0.125s )(1 + 0.05s )
1 1 = 0.01, ω 2 = = 0.1, 100 10

[解]：1、k = 10 3 ,20 log k = 60;ν = 2; ω1 =
1 1 ω3 = = 8, ω 4 = = 20, 0.125 0.05

2、低频渐进线斜率为 20ν = 40dB ，过（1，-60）点。 3、高频渐进线斜率为 ： 20 × (n m) = 60 4、画出波德图如下页：

2
1

2

(1,60)

3

3e j 0.5ω [例]具有延迟环节的开环频率特性为：Gk ( jω ) = ，试画 1 + jω 出波德图。

[解]：Gk ( jω ) =

3 1+ ω 2

e jω e j 0.5ω =

3 1+ ω 2

e j1.5ω

L(ω )
20 log 3

(ω )

1

10
tg1ω

100

ω
1

45o
90
o

10

100

0.5ω (ω)

ω

Ks G(s)H (s) = (T1s +1)(T2 s + 2ζ T2 s +1)
ω1 : ω2 :
ζ:

ω1 20log = 12 ωc1 ωc1 40log = 12 ω2
Mr = 1 2ζ 1 ζ
2

1 ωc1 = K

20 12

ω c1

ω1

ω2

ωc 2

20log Mr = 8