第四章习题答案4 1已知开环零极点分布如题41图所示试概略绘出相应闭环根

j0j0j0 j0j0j0j0

题4-1

K3j

3j

3j

3j渐近线与实轴交点：

5

(2k dp1 (p1zi)(p1pjp1

p p1

p2ps59s

33s

26skj

26jk53332694512k

32.2)

2

2

k

k7689(舍故与虚轴交点为

0.89，对应此时

35G(sH(s)G(s)

s2(sH(s)=K

001(2k

3 d d2(舍)

s3s2k

j32k

kk4-4 s(s212Ks解：Gk(s

k s1

jj2k

ss22sp1

p21

p31 1j1j p p1p p2

s32s22sk2320,2

k0,kK

G(s)

p2k

p3

p3s(s1)(s2)(s125 a

1

d d dd10.4,d2

k88

k365时系统不稳定5ImaginaryImaginary---- - - G(s)

s(0.1s1)(s试绘制该系统的根轨迹图。KG(s)

p2s(s10)(s

p3 110 1 d

d

0,d

k

G(s)K(ss(s8s118

K=1.5s11

8(s1)s(s ((1j8)(1j8)G(s)KG(s)

s(sK(sK(s(s1j3)s1j3G(s)

s(s1)(s4)(s（1）1

0d

d1d+1+j1d+1+j1d1-j

1d

d

（3）试凑法得到为-0.6和-G(s)K(sG(s)

G(s)

K(ss(s28sG(s)

K(ss2(s29sG(s)

s(s22s1110ImaginaryImaginary-0--1- - - - 20----- - - - - - 35ImaginaryImaginary--1ImaginaryImaginary---- - - - - - 5ImaginaryImaginary--4-10aKKs(s（1）K-0k（2）Gk(s)s(sk

1Gk(s)

1 0s(s0s2ask

1 s2kG(s)k

s2k

90a180,a ddjdjdkd2kdk取dks3ask0代入sk2k0,ka0,0(舍4-11KK(0.25ss(0.5sG(s)1

k(s4)2s(s0 0d

d2

d2d2d(d

dd1 d2以-42.828s11.172代入1G(s)0s26.828代入1G(s0K值范围（23.3，∞）及G(s)

K*(ss(s1)(s求当=0.5K解：

132221 d

d

dd864ImaginaryImaginary0------ - - - - - - （2）

1k*(1

n1s(s1)(sk*(s2)s(s1)(s3)s34s2(3k*)s2k*解得n

k*

sd0.681.18

s3G(s)

解：(s)

D(s)(s210s)(Ts1)=Ts2(s10)s210s1(s210s等效为1 0s2(s1(s210sG(s) s2(sT01/T0P1,2

P3

Z1,25432ImaginaryImaginary0----- - - - - - - - - 1(sa)G(s) s2(sD(s)1s1as3jj0211 0s3s21411 0s(s2s14G(s)

s(4s24s

s(2s1)21 d11

d61 0j(2j14

2

1ImaginaryImaginary---- - - - - G(s)K*(1s(sk*(sG(s)

s(s(s)

k*(ss(s

21 2

2k 33 2331 d

d

，

1

d213d1,213

代入1

k*(ss(s

0

*7.46(d1*

*3)，*

0.54(d1 k*由1

k*(ss(s

0sj代入得k*

221ImaginaryImaginary0----- - - G(s)

s(5s4-16图所示。 s(5s15 s(5s15(0.8s0.8ss(5s115K0K0Root---- - - - - Root110-0--1- - - - -2 - - - -0 4-17 C1KhKs(sKh=0.5K0Kh=0.5，K=10时系统的闭环极点与对应的K=1时，Kh0K=1Kh=0、0.5、4的阶跃响应指标σ%tsKh的大小对系解：(s)

hs2sk(1kh1k(s

G(s) s(s1 d

d2d122d22

221ImaginaryImaginary---- - - - - - - Gk(s)s(s1)(1(s)

s2s(0.5s

s26sD(s)s26s6 366 36

s1,23

11

(s)

khs2sk(1khkk=1k

G(s)

s2sdd132

2

dd13

dd1kh1ImaginaryImaginary0------ - - - - - - - -

0

(s)

ns(sn

(s)

s2s欠阻尼2

2

n

ts

%

kh

(s)

s2s0.5sn s

321,1,%

kh4

D(S

s25s121,

1(1(2n12.5 2.52ts

tsKGK(s)(s1)(sG(s) s(s1)(sG(s) K(s s(s1)(s3)(sG(s) (s1)(s

ImaginaryImaginary0------ - - - - - G(s)

s(s1)(s

0,2,2，

54320------ - - -

G(s) K(s s(s1)(s3)(sRoot5ImaginaryImaginary---- - G(s)

s2(sz1解 G(s)

s2(s（1）

3

2 Root8640----- - - - - K（2）

2

864ImaginaryImaginary0----- - - - - K04-214-21K在0KK(1K(1s(0.2s

C4-21（1）

K(1s(0.2s

5K(s2)2s(s5)1

(s)

15K(s2)2s(s

5K(s2)2s(s

d

1 d

2dd(dd24d10 d2与虚轴交点：sj代入2s(s55K(s2)2210j5Kj10K

2

5K1010,K4320----- - （2）D(s)2s210s5k(s2(j)2(105k)(j)10k2(22)(105k)10k4(105k)4105k5k1042(22)4210k(2)22即以(2,0)为圆心，半径为14根轨迹与虚轴交点为10K2，所以当0K24-224-22NN_Y_1s(sK(s(s4-22f（s设a8K，有几条78a5f(sK7设0a1f(s

G(s) K(s s(s1)(sf(s)s(s1)(s5)K(sa)s36s2(5K)s(2

a87s36s2(5K)sKa(sa)30K7a8，三条根轨迹重合，特征根为-2K=77渐近线与实轴交点

15 2

177

分离点为 2,d 5432ImaginaryImaginary0------ - - - - - 8a 15a3a

取a

1

dd d d5---- - - - - - a

1

K d

d

d

0a1a=0.5，

3a2.75

d

1d

d

dd321ImaginaryImaginary--4-234-23f(s

G(s)s s2(s

0E(KG04-231K31f(sK4f(sK令

(s)s as2(s解：f(ss2s3K(ss33s2Ks11

(s1)3s33s23s1 K3,

s3， 54320------ - -2 - -1 - - （2）1,a

K>0Root8640------3 - - - - - - （3）

34

5ImaginaryImaginary---- - - - - - - - - 5（4）

s s2(s

a0

此时

a2

0KhKhsKs

C(s)4-24图3为使闭环极点为s1 ，试确定增益K和速度反馈系数Kh的数值3KhKKKhK sKKs（1）(s sKKs

s1

1 1s2(1 D(s)(s1

3)(s1

3)s22sK(s

K4,Kh

Gk(s)

K(10.5s)s s

2 d

d4即以(202证：D(s)s20.5KsK (j)20.5K(j)K220.5KK 20.5K0,k2240,(2)224即以(2,0)2210------ - - - - - - - K=4则G(s4(1Khs)

D(s)s24Ks414Khs

G(s)

4Khs

s2

s2以(0,02210------ - - - - - - - 4-254-25K(s(s4)(s4-25K确定闭环特征方程的根的实部均小于-1时KK(s（1）D(s1Gk(s1(s2)(s4)。10----

-- - - - s0时与虚轴相交，此时，K=4，K>4s1K9K9根实部均小于-14-264-26Kg(1s(s4-26a=3Kg0Kg的取值范围及Kg值Kg=3a0的根轨迹，确定系统

a2（1）

Gk(s)

Kg(1s)s(s3)

1Kg(1s)s(s1 d

d1

d1,d

以(1.0221ImaginaryImaginary0------ -

s23sKsK

2(3K)jK3 Kg3,3

3，

3d1

Kg

Kg

0Kg

3时，13(1s)s(s

s2(a3)s31 s23s

Gk(s)

s23sarccos

450sxxj

s2a3)s30x2[1j]2(a3)x[1j]3 a5.45,x21100-0--1-- - - - - 4-274-27Kg1(Kg1(s22s4)(s_4-27