控制工程基础第六章习题答案

6-1 (a)G c (S )=V 0(S )V i (S )

=

Z 1

Z 1+Z 2=

R

R +1

=

RCS

RCS +1=

TS

TS +1

(b)G c (S )=1CS 1

CS

+R =1

RCS +1=1

TS +1

6-2 (a)

V i (S )R 1

=-

V 0(S )R 2

•

R 22R 2

2+R 2

→G c (S )=-

R 2+R 3R 3

(1+

R 3R 2

R 2+R 3

CS )=-K p (1+TdS )

小区中间变量V p (S)得到G c (S )=V 0(S )V i (S )

=-

R 12+R 1R 3RCS +R 3R 2

R 1R 2

(b)V i (S )

R 1

1+R 1C 2S

=-

V 0(S )

R 2+

12∴G c (S )=V 0(S )

V i

(S )=-(R 1C 1S +1)R 2C 2S

R 1C 2S

=-

(T 2S +1)(T 1S +1)

T 2S

(c)

V i (S )R 1

=-

V 0(S )

R 2

2∴G c (S )=

V 0(S )V i (S )

=-

R 21+R 2CS

R 1

=-R 2R 1•

1

1+R 2CS

=-K c •11+TS

K c =R 2R 1

T=R 2C

6-3 G C 1(S )=

S +1S +10

=10•01S +10.1S +1

•1

10

α=10

G C 2(S )=S +1S +20

=

20•0.05S +10.1S +1

•1

20

α=20

G C (S )=

S +10.02S +10

=

50•0.02S +10.02S +1

α=50

αm =sin −1α−1α+1=sin −110−1

10+1=54.90 64.80 73.90 W m =

1√α•T

=1√10•0.1=3.16 1√20•0.05=4.47 1

√50•0.02=7.07

L(W m )=10log α=10log 10=10分贝 13分贝 17分贝 6-4 G C 1 S =S +1

5S +1α=0.2 T=5

幅值5倍 10倍 20倍

φ(w)=-90

-4.550 -2.410

G C 2 S =S +1

10S +1α=0.1 T=10 幅值20log α=−20分贝

相角 -10.150

-5.130 -2.160

G C 3 S =20S +1

20S +1α=0.05 T=20 幅值20log 20+20log 0.05=0分贝

相角φ(w)=-10.60

-5.40 -2.760

6-5 G C (S )=（2.5S +1）（S+1）

(25S +1)(0.1S +1)α=10 T 1=0.1T 2=25

φ m =sin −1

α−1α+1

=550

W m =

1√α•T 1

=3.16

L(W m )=-10log α=-10分贝

6-6 用根轨迹法校正A、B为-1±j√3

未校正导流对AB产生的相角为-193.90所以加上PD调节后的附加相角为β=193.90-1800 =140log14=√3

x

→x=6.95 ∴Z c的值为-7.95

所以G k S=K g S+7.95

S S+1.5K gA=S S+1.5

S+7.95

|s=−1+j

√3

=

√3+6.952

=0.5

所以G c S= K p1+TdS=Td K p(S+1

Td

)

1

Td

=7.95 → Td=0.125 5 Td K p=0.5→K p=0.795

∴G c S=0.795(1+0.126S) 以-0.795为圆心，以R=7.16为半径的圆。

6-7 K g=0.5K=K g•7.95

1.5=

2.65 斜坡下的稳态误差e ss=1

K v

=1

R

=0.377

e ss变为原来的10%，则K变为原来的10倍为26.5

增加一对极点比零点更靠近坐标原点的开环偶极子改善稳态性能，

设S+Z c

S+P c =S+0.5

S+0.05

坐标原点附加分支对瞬态性能基本不影响，达能改善它的稳态性能

6-8 用第一种方法，求K pp和自然振荡周期T

φs=

8K p

S2+5S+4+8K p

特征方程S2+5S+4+8K p=0

S2 1 4+8K p

S1 5 0

S04+8K p

6-9 K v=4（S−1）=K g

10

=4 →K g=40绘制未校正导流的Bode图

20log K=12 dB 转折频率为W1=1 W2=10

G0S=4

S(S+1)(0.1S+1)

∴r=150W c=2∴Kℎ=0.25 Lℎ=12分贝

要求r‘=450选择串联超前校正装置φm=r‘−r+Δ=65.50+150=81.5（比较大）

α=1−sinφm

1+sinφm

=180 选择串联滞后校正装置

r‘+Δ=450+50=500在L0w找出φ0w=130对应的频率作为新的截止频率W c’=0.7 L00.7=34分贝

6-10 e ss≤1

150=1

K v

=1

K

要求K=15 r=450W c≤7.5Lℎ=6分贝

W c=4（S−1）r=120不满足要求

选择超前校正装置W c≤7.5S−1取W c=7.5=W m 则L07.5=-10分贝则10logα=10 ∴α=10

φm=550 T=1

√αW m =0.042 αT=0.42∴G C(S)=0.42S+1

0.042S+1

r=62.50

6-11 要求K a=2校正后应成为Ⅱ型系统8%≤40% h=40+64

40−16

=4.33

取h=5 h=W2

W1

取K=2，绘制未校正系统的Bode图