《自动控制原理》黄坚课后习题答案

试建立图所示电路的动态微分方程＋ui1＋ui1R1R-i2i+u-o＋ u1 L +CRu i R1 i1 uCRii oi-解：i=i-i

22 -u=u-u1 2 1 i ou u-u ui=R=i o i=Ro1 1 1 2i=C =Cdu d(i=C =Ci o

u-u u-i o=Ro -

d(u-u)o2

dt dt 1 2(u-u)i=i+i i= i 11 2 1

du duu dui=Ro i=Cdt1

R(u-u2 i

)=R

u-CR10

1R2(dti-dto)21 22Ldu

CR

duo+R

u+R

u=CR

R

i+Ruu-u=R

dto

1 2dt

1o 20

2dt 2i1 o 2u-u u

du ui-uo

Lduo=uo+CCLd2uoi 1=

o

dt

R R RRdt R1 1 12

dt R dt22L1 2duLo

ui=CLd2uo+(C+

L)duo+(1+1)uu=u+Rdt

R R

RR

R R o1 o 2 1 2

12 1 2求下列函数的拉氏变换。(1)f(t)=sin4t+cos4t解siωω coωs2+s2 2+s2L[sin4t+cos4t]= 4

s =s+4s2+16 s2+16s2+16(2)f(t)=t3+e4t

3!+

1=6s+24+s4s4(3)f(t)=tneat解：

s-4n!

s4(s+4)(4)f(t)=(t-1)2e2t

(s-a)n+12(s-2)解： L[(t-1)2e2t]=e-(s-2)(s-2)3s+1求下列函数的拉氏反变换。s+1(1)F(s)=

A A= 1+ 2(s+2)(s+3)

s+2 s+3解：A=(s+2)

s+11 (s+2)(s+3)A=(s+3) s+1 =22F(s)=

(s+2)(s+3)s=-3-1-f(t)=2e-3t-e-2ts+3s

s+2A A A(2)F(s)=(s+1)2(s+2)(s+)2+s+2+s+3解=(s+1)2 s

=-11 (s+1)2(s+2)s=-1A=d[s] =22

s+2

s=-1A=(s+2) s

=-23 (s+1)2(s+2)s=-2f(t)=-2e-2t-te-t+2e-t

(4)F(s)=s+2s(s+1)2(s+3)(4)F(s)=2s2-5s+1As+A A

A A A A1 2+ 3 4(3)F(s)=s(s2+1)=+12+s322

解：=(s+1)2+s+1 s+s+3-1 2 1 -3解：F(s)(s2+1)s=+j=As+A

A=2 A=3

A4=12

A=42s2-5s+1

=A

s=+j

1 2d[s(s+3)]s s=j

2s=j

A= ds-2-5j+1=jA

-5j-1=-A+jA

2 s=-1j 1 2 1 2

=[s(s+3)-(s+2)(2s+3)]

=-3123A=1 A=-5 A=F(s)ss=0=1123

[s(s+3)]2 s=-1 4+1 -5+

f(t)=-te-t-

3e-t+2

+1e-3tF(s)=s

ss2+1

s2+1

f(t)=1+cost-5sint

2 4 3 12(2-4)求解下列微分方程。+5 d2y(t+5

y(0)=·(0)=2dt2 dt 62Y(s)-sy(0)-y'(0)+5sY(s)-5y(0)+6Y(s)=sY(s)= = 1Y(s)= = 1s(s2+5s+6) s

As+2

As+33A=1 A=5 A=-41 2 3y(t)=1+5e-2t-4e-3t2-5 并求传递函数。i+i+1iu-2R1iRur2

解：+U(s)+r _

I(s)Cs1+1 +

I(s)R2

U(s)cc U(s)

I(s)- c 12(1+sC)RU(s) R 2 R+RRsCU(s)= 1 =R+2+12sCc 1+(R+sC)R 1 2 12(2)＋ur-

1 2解：U(s)r_1解：U(s)r_1I(s)I(s)1U(s)cU(s)L1I(s)2U(s)1-R- CsL1Ls1Rc2U(s)1I(s)1213Rc1C R uc2-L=-R/LsL=-/LCs2L=-1/sCR LL=R/LCRs21 2 2 3 1 13 2 1P=R/LCRs2

=1 U(s) R1 2

1 URCLs2+(RR2C+L)s+R+Rc 1 12 1 22-8设有一个初始条件为零的系统，系统的输入、输出曲线如图，求G(s)。δ(t) c(t)K0

解:δ(t) c(t)Kt 0 T tc(t)=

Kt-K(t-T)

C(s)=

K(1-e-T)SC(s)=G(s)T T Ts2若系统在单位阶跃输入作用时，已知初始s条件为零的条件下系统的输出响应，求系统的 r(t)=I(t)c(t)=1-e-2t+e-ts传递函数和脉冲响应。

R(s)=1解

1 1 1 (s2+4s+2)- + s s+2s+1 s(s+1)(s+2)- +

(s2+4s+2)C(s)=(s+1)(s+2)C(s)=G(s)=C(s)/R(s)=

(s2+4s+2)(s+1)(s+2)

c(t)=δ(t)+2e-2t-e-tR(s)-GR(s)-G(s)X(s)G66 3C(s)-GGGG1X(s)2-3X(s)4X(s)C(s)[G(s)-G(s)]C(s)G(s)G57 85-G7G解:X(s)=R(s)G(s)-G(s)[G(s)-G(s)]C(s)1 1 1 7 8={R(s)-C(s)[G(s)-G(s)]}G(s)7 8 1X(s)=G(s)[X(s)-G(s)X(s)]2 2 1 6 3X(s)=G(s)[X(s)-C(s)G(s)] C(s)=G(s)X(s)3 3 2 5 4 3 8R(s) GG

GG

C(s)- 12

326G5

C(s)

GGGGG-G

R(s)1+GGG+GGG+3

GG(G-G)7 8求系统的传递函数

326

345

1234 7 8解:L=-GH Δ=1+GH+GGH1 21 21 122(a)R(s)(a)R(s)G(s)3_G(s)+G(s)C(s)1_2H(s)1H(s)2

P=GG

Δ=1122 1 12 1PΔPΔC(s) Σ

P=GG Δ=12 32 2GG+GGR(s)=k=1k k=1+G2H1+G2G3HΔ 212

122(b)R(s)

G(s)3

C(s)

解:L=-GGHL=-GGHP=GGΔ=1_G(s) + G

1 1

2 1

1 12 11 2 Δ=1+GGH+GGHP=GGΔ=1+GGHG(s) +

14 1

2 32 2 144 C(s)GG+GG+GGGGHH(s)

R(s)=

1GG

23412 14R(s)C(s)_R(s)C(s)_G1G2G+ +3++H1R(s)_GGC(s)12G3H+1+H1C(s)

GG(1–GH)R(s)=1+G1G2+GH3–G1H12 11 31R(s)GR(s)GC(s)1+_G2H

P=G Δ=1解:L=-GH

1 1 11 2C(s)

P=G Δ=12 2 2(e)

R(s)

=(G+G) 11+GH1 21+GH2R(s)GC(s)-R(s)GC(s)-1+G2_G3G

L=GG

L=-GG1 13

2 1

3 23L=GGP=GΔ=1P=GΔ=14 24 1 1 1 2 2 2C(s)

(G+G)R(s)=1+GG+G1

G-GG4 1

23 14 24(f)

解:L=-GG

L=GP=G

Δ=1-GR(s)C(s)_LLR(s)C(s)_LLG11G22+

2 2 1(11 2Δ=1+GG-G

C(s)=

–G)212 2

R(s)1+GG–G12 22-12(a)2-12(a)H(s)D(s)1R(s)__G1+G(s)C(s)+2LLH(s)122H(s)3L=GH L=-GGHP=GGΔ=11 22 2 123 1 12 11C(s)= G2G1R(s)1-GH+GGH22 123P=G Δ=1P=-GGHΔ=11 2 1 2 121 2C(s)

G(1-GH)2 11D(s)1-GH+GGH22 12

解:L=-GG L=-GGHP=GGΔ=11 12 2 12 1 12 1R(s)_ _G1GC(s)2H(b) CR(s)_ _G1GC(s)2HR(s)=1+GG1H2+GGP=GG

12 12Δ=11 n2 1

P=1 Δ=1+GGH2 2 12= n2 1C(s)1+GG+GGHD(= n2 112 12=1C(s)=1C(s)

P=GGΔ=11 2 2

123 1 23 12-13(a)

P=GGGΔ

GG+GGG= 23 123R(s)E(s)_R(s)E(s)_G+1_GGC(s)L23L12

R(s) 1+G+GGGP=-GG

Δ=1 P=1

=12+G1231 23 1 2 2 2C(s) -GG+1+GR(s)=G2 123解:L=-GGL=-GGG

P=GGΔ=11 34 2

235GG1G1E(E(s)1GR(s)-G2-G+GC(s)35G4

P=GG2

Δ=1C(s)=

125 15235 2

R(s)1+GGG+GGP=GG Δ=1 P=1 2315G341 15 1 2 2 34E(s)GG+(1+GG)=15 15R(s)1+GGG+GG235 342-14

G(s)

解:P=GGΔ

=1P=GGΔ=14 1 13

2 23 2R(s)E(s)G(s)

G(s)+C(s) P=G

Δ=1P=GGΔ=1- 1 + 3

3 14

4 24 4G(s)

X(s) L=-G

L=-GG

C(s)(G+G)(G+G)= 1 2 3 42

1 13

23R(s) 1+G(G+G)E(s)= 1

X(s)=G(s)

C(s)=1

3 1 2R(s)1+G(G+G

E(s)

D(s)3 1 2R(s)R(s)1+-GGGC(s)1123H2H1R(s)+2-GGGC(s)2456

解:L=GG L=-GGGHH L=-G1 12 2 14512 3 4Δ=1-GG+GGGHH+G

-GGG12 143512

4 124P=GGG Δ=1+G1 123 1 4C(s)

GGG(1+G)11(s)1+G+GG23H-4G-GGG1 4 14512 12 124C(s)

GGG(1-GG)

C(s) -GGGGGH1R(s)1+G+G456H-12

-GGG

R(s)1+G+GG1

3H4-G5G1-GGG2 4 14512 1

124 2

4 14512 1

124C(s) GGGGHR2(s)=1+G+GG1G4H5H6-G2G-GGG1 4 14512 12 124设温度计需要在一分钟内指示出响应值的并且假设温度计为一阶系统，求时间常数T。如果将温度计放在澡盆内，澡盆的温度以10oC/min的速度线性变化，解:c(t)=c(∞)98% t=4T=1min T=0.25 c(t)=10(t-T+e-t/T)=10(T-e(t)=r(t)-c(t)

=lime(t)=10T=2.5sst→∞R1R1uRr0-+C∞+ucC=2.5μF R=20kΩ R=200kΩ0 1（1）单位阶跃响应,及()=8时的1值．解:)R1s1= K T=RC=0.5 K=R/R=10Ts+1 1 1 01 -t

-2t

8=10(1–e-2tu(t)=K(1–eT)=10(1–e ) )c0.8=1–e-2t e-2t=0.2 t=0.8t1时刻的值．T解: t=0.8 R(s)=1 T1

Ke-t/T=4s2R(s)=1s2

u(t)=K(t-T+Te-t/T)=4cR(s)=1

U(s)=

1=K(

1-T+T2

T2 )s3 c Ts+1s3 s3 s2 s s+1/Tu(t)=10(1t2-0.5t+0.25-0.25e-2t)=1.2c 2G(s)= 4

s(s+5)解:C(s)= 4

R(s)=1 C(s)=

4 =1+1/3-4/3R(s)

s2+5s+4

s s(s+1)(s+4) s s+4 s+1c(t)=1+1e-4t-4e-t3 3G(s)= 1升时间t、峰值时间t、超调量σ%和调整时间t。

s(s+1)解 C(s) 1

ζω

ω=1 ωdn2=0.866:R(s)=

nω2=n

n2ζ=0.5 βζ =60o2t=3.14-3.14/3=2.42 t=π=3.14=3.63rωd 0.866

pωd 0.866o%=π1ζ100%

t= 3=6t= 4=8e-1.8

sζωn sζωn3-6已知系统的单位阶跃响应:c(t)=1+0.2e-60t-1.2e-10t求系统的闭环传递函数。求系统的阻尼比和无阻尼振荡频率。解:C(s)= + 解:C(s)= + - s s+60s+10s(s+60)(s+10)R(s)=1 600

ζωn=70

ω=24.5nns R(s)s2+70s+600 ω2=600n3-7 曲线如图，系统的为单位反

ζ=1.43c(t)1.310 0.1 t解: tπ =0.1ζπ/2=ln3.3=1.19ωn2=3.14=31.4p n2

0.1-ζπ1-ζ=0.3

ζπ)2/1ζ2

ω=33.4e 2ζπ1-=3.3

9.8ζ2=1.42-1.4ζ

n ω

1115.6nne ζ2

ζ=0.35

G(s)=s(s+ω

)=s(s+22.7)3-11 已知闭环系统的特征方程式，试用劳斯判据判断系统的稳定性。s3+20s2+9s+100=0劳斯表如下：s3 1 9s2 20 100s1 4s0100

(3)s4+8s3+18s2+16s+5=0劳斯表如下：s41185s3816s2165s1s116

G(s)=

s0 5 系统稳定。K(0.5s+1)=b解:s4+3s3+4s2+2s+Ks+2K=0=bs4 1 4 s3 3 2+Ks2b 2K31s1bs41

10-2K31 3=bK2+10K-20=b41 10-K(K-1.7)(K+11.7)>0K<10解:G(s)=10(1+解:G(s)=10(1+s)= +s+1τss(+s+1τ)

R(s)--s(s+1)τs10C(s)s3 1 R(s)--s(s+1)τs10C(s)Φb

10(s+1)3+2+1τ=10(1+1τ)-10>0

s1 b31s0 1031 τ>0R(s)-R(s)-τs+1ss(s+1)10C(s)解G(s)=

Φ

10τ(s+1)s3 1

3+2+1τs+10110τ-101s2 1

b= >031s1 b31s0 10

τ>1

r(t)=I(t)+2t+t23-16已知单位反馈系统的开环传递函数，

R(s)=

1 2 2+ s s2 s3+ pK=20p

R1e = 0=1(1)

20

ss11+K 21解： (0.1s+1)(0.2s+)

K=0υ

e =∞ss2K=0ae=∞ss

e =∞ss3K=∞ e =0(2)G(s)=

200 =

10 p

ss1 2 2υυs(s+2)(s+10)

s(0.5s+1)(0.1s+1) K

ess2=K=10K=0ae=∞ss

e =∞ss3(3)G(s)=

10(2s+1)

= (2s+1)

K=∞p

e =0ss1s2(s2+4s+10)

s2(0.1s2+0.4s+1)Kυ

e =0ss2K=1a

e=2ss

e =2ss3(1)单位阶跃输入

R(s)- -K

１C(s)s21t1s

=1.8(5%)确定K1

和值。 τsΦ K K ζωnΦ

e-ζπ1-ζ2=0.2ss解:ss21

s+K1

ω2

1t=ω

=1.8ζ

ω=

n3=3.7 K3

1 sζnτ=0.24nn1.8*0.45

2=13.71 n(2)r(t)=I(t),t

1t22 1

K=∞K 1 R(s)=s

e =0ss11s2解1s2

s=s(

τ1s+1)

υ=1

R(s)=1

K=K υ

=τ=0.24ss2s31 s31

R(s)=1

K=0 a

=∞ss3R(s)- -s(s+2)τsR(s)- -s(s+2)τsKC(s)单位斜坡输入的稳态误差ess=0.25确定KK解:G(s)= K K

K2+τ Φ(s)=和τ值。

s2+2s+τs s( 1 s+1) 2+(2+τ)s+K2+τζω=2+τ=2*0.7K e=2+τ=0.25 K=31.62+τn ss KKω2=KKn

τ=0.25K-2 τ=0.186系统结构如图。

D(s)1

D(s)2r(t)=d(t)=d(t)=I(t)

R(s) E(s)G(s) +F(s)+ C(s)1 2 -r(t)作下的稳态误差．1解

=lims·

s = 1ssr

1+G(s)F(s)1+G(0)F(0)求d(t)和d(t)同时作用下的稳态误差．1 (s)H(s)Ed(s)=1+G(s2)G(s)H(s)·D(s)1e =lims[

2-F(s)

-1 ]1

-[1+F(s)]ssd

s→0

1+G(s)F(s)

1+G(s)F(s)s

1+G(0)F(0)求d(t)作用下的稳态误差．1 K F(s)=1G(s)=Kp+s Js -1)e =lim)

-F(s) 1=lim

Js 1=0ssd

s→01+G(s)F(s)s

s→01+(K+p

K1ssJs4-1 已知系统的零、极点分布如图，大致绘制出系统的根轨迹。解:(1) 0σ

(3)

jω(4)jω(4)jω60000σ(2)0σ(2)0σjω0σ

(6)jω0σ

(7)

1350

jω4500

(8)

1080

jωσ360σ04-2 已知开环传递函数，试用解析法绘制出系统的根轨

G(s)=

K(s+1)rK:Φ(s)=s+1rKrKrK=0

s=-1-K-3+j2-3+j20+j1-2-10σs=-2+j0Kr→∞ Kr

s=0+j1s=-3+j2jωp3jωp3p2pzz01σ21解解(1)G(s)= 1）开环零、极点p=0 p=-1p=-5

4）分离点和会合点A(s)B'(s)=A'(s)B(s)1 2 3z=-1.5 z=-5.521 22）实轴上根轨迹段p~p z~p z~-∞

A(s)=s3+6s2+5sB(s)=s2+7s+8.25A(s)'=3s2+12s+51 2 1 3 2 B(s)'=2s+7根轨迹的渐近线n-m=1 θ=

s=-0.63s=-2.5s1=-3.6 s2=-7.283 4jωp3-1.75pp2jωp3-1.75pp21zσ10开环零、极点p=0 p=-1 p=-41 z2-1.5 31实轴上根轨迹段p~p p~z1 2 3 1n-m=2 θ=2σ=-1-4+1.5=-1.752(3)(3)jω1p2pp-0.6701jω1p2pp-0.6701σ3-1

分离点和会合点A(s)=s3+5s2+4s A(s)'=3s2+10s+4 B(s)'=1s=-0.621 2 3实轴上根轨迹段p~p p~-∞1 2 3根轨迹的渐近线n-m=3θ=+60o,3σ=-1-1=-0.673根轨迹与虚轴的交点s3+2s2+s+K=0

分离点和会合点A(s)=s3+2s2+sB(s)=1A(s)'=3s2+4s+1K=0 rK=2

=0r1=±1

B(s)'=0s=-0.33r 2,3jω6.2jω6.2p4p3 2p pz-5.6701σ1-6.2(4)G(s)=s(s+3)(sr+7)(s+15)开环零、极点p=0 p=-3p=-7p=-15z=-81 2 3 4 1实轴上根轨迹段p~p

p

p~-∞1 2 3 1 4根轨迹的渐近线n-m=3σ=-3-7-15+8=-5.67o θ=+60,+3 oo 根轨迹与虚轴的交点s4+25s3+171s2+323s+8K=0

分离点和会合点A(s)=s4+25s3+171s2+315sA(s)'=4s3+75s2+342s+315B(s)=s+8 B(s)'=2s+7K=0r

=0 K=6381 r

r=±6.22,3

s=-1.4p2 1pz0σ1(1)p2 1pz0σ1((Gs ss1)1解: p=0 p=-1z=-211p~p2z~-∞1 2 1分离点和会合点s2+4s+2=0s=-3.41s=-0.59

)ω=01 2 r闭环特征方程式 4ω2-2(1+Kr

)+2K=0rs2+s+Ks+2K=0

s=-2j

ω=±1.41r r(-2+ω2+(-2j)(1+K)+2K=0

K=3r r r已知系统的开环传递函数，试确定闭环极点ζ=0.5Kr值。jω1.7s1spp332p1jω1.7s1spp332p1-3-10σ-1.7rs(s+1)(s+3)解:p=0p=-1p=-3p~p

p~-81 2

1 2 33σ=-1-3=-1.33

θ=+60o,+180o根轨迹的分离点：A(s)B'(s)=A'(s)B(s)3s2+8s+3=0s=-0.45s=-2.2 舍去1 2与虚轴交点s3+4s2+3s+K=0

ζ=0.5得s=-0.37+j0.81s=-4+0.37×2=-3.26-ω3

r=0 K

ω=0

K3=|s||s+1||s+3|K2=0 Kr=12ω

=±1.7

=3263

2.230.26=1.9r r 2,3 ×(2)

Ks(s+3)(s2+2s+2)r=0p=-3r

=-1±j p~p11 2 3.4 24σ=-3-1-1=-1.25θ=+45o,+135o4根轨迹的出射角θ=π3

-2-4=+π-135º-90º-26.6º=-71.6º与虚轴的交点rs(s+3)(s2+2s+2)+K=0rs4+5s3+8s2+6s+K=0r(ω)4+5(ω)38(ω)+jω+K=0rω2+K=0 K=0 ω1=0

jωpjωp3s11.1-71.613526.6p-2.32-1.2 p0σ190-1.1p44s3+15s2+16s+6=0解得 s=-2.3 ζ=0.5得s=-0.36+j0.751K=|s||s+3||s+1+j||s+1-j|r Kr

=±

r 1 1 1 1=0

=8.16r

2,3

=2.92jω2.8jω2.8s1spp332-4-2p01σ-2.8试绘制出根轨迹图。KG(s)H(s)=s(s+2)(rs+4)解:p=0p=-2p=-4p~p

p~-81 2

1 2 33σ=-2-4=-2θ=+60o,+180o3根轨迹的分离点：A(s)B'(s)=A'(s)B(s)s=-0.85 s=-3.15 舍去1 2

ω3+ω

Kr

ω=01阻尼振荡响应的K值范围

K-6ω2=0

K=48

=±2.8s=-0.85

K=0.85

r

×3.15=3.1 r

2,3s=±j2.8 Kr=48r

(4)ζ=0.5

s=-0.7+j1.21与虚轴交点s3+6s2+8s+K=0r

s=-6+0.7×2=-4.63K=4.6×2.6×0.6=7.2rr(t)=sin(t+30o)，试求系(1)解统Gs)态1出φ(s)= 10

)=

= 10

10=0.905(s+1)

(s+11)

12+ω)112+112211 φ=-tg-1111

c(t)=0.9sin(t+24.8o)s(1)

750 I型系统n-m=3s(s+5)(s+15) ω=0 )=∞φ)=-90oω=∞)=0 φ)=-270o

ω=∞0 Re(3)

10(2s+1)(8s+1)

ω=0(5)

10ω=0ω=0Imω=∞0 Re0型系统n-m=2

ω=∞ω

解：I型系统n-m=20ω=0 )=10φ)=0oω=∞ )=0 φ)=-180o

Reω=0 )=∞φ)=-270o统 n-m=3Imω=0ω=∞0 Reω=∞统 n-m=3Imω=0ω=∞0 Re

10(s+0.2)s2(s+0.1)(s+15)

II型系ω=0

)=∞ φ)=-180oω

)=0

φ(ω)=-270o5-2 G(s)=

750

)dB

(3)G(s)= 10

L(ω)dBs(s+5)(s+15)40-20dB/dec

(2s+1)(8s+1)G(s)=1 101 20

15 解20

-20dB/decs(5

0-20

1 5-40dB/dec

ωω=0.125ω=0.5

0-20

0.125 0.5ω-40dB/dec20lgK=20dBω=5ω=15

φ) -60dB/dec 1 (2

φ(ω)1 2

ωω=0 φω)=0o 0 ωω=0ω

φ(ω)=-90oφ)=-270o

-90-180-270

ω=∞φ(ω)=-180o

-90-180

10s(s-1)40

L(ω)dB-20dB/dec1ω=1 20lgK=20dB2010

-40dB/dec1 ωω=0φ(ω)=-270o-20ω=∞φ)=-180o

φ(ω)0 ω-90-180-270(7)G(s)=

10(s+0.2)

= 1.33(5s+1)L(ω)dB40 L(ω)dB40 -40dB/dec-60dB/dec2000.10.2-20151ω-40dB/decφ)0-90-180-270-60dB/decω解：20lgK=2.5dBω=0.1 ω=0.2

s2(10s+1)(0.67s+1)ω=151 2 3ω=0 φ)=-180oω=∞φ)=-270oL(ω)dB1L(ω)dB1102020dB/decω)dB20lgK10)dB20lgK10-20dB/decωcωK=10 20

(b)

20lgK=-200K=0.10G(s)= 10

G(s)=

0.1s -20(0.1s+1) 0

(0.05s+1)(d) K=251

L(ω)dB4820lgK0

-20dB/dec-40dB/dec1050100ω

(c)

K=100

L(ω)dB-20dB/dec0 0.01

100ωG(s)=

251

-60dB/dec

100

-40dB/dec-60dB/dec(s+1)(0.1s+1)(0.01s+1)(c) K=100

G(s)=s(100s+1)(0.01s+1)G(s)=

100 0L(ω)dB-20dB/decL(ω)dB-20dB/dec1000.01 ω-40dB/dec-60dB/dec)dB)dB-20dB/dec4.58dB100ωω-60dB/dec20lgM=4.58 M=1.7= 1 得：ζζ1ζ2r rζ=±0.94ζ1

=±0.32 ζ=0.3 02ωrωn1-ζ

ωn=50

=100

010002=ω2=ωn

G(s)=s[(0.02s)2+0.01s+1)]5-7 为积分环节个数，试判别系统稳)

(b) p=0p=0-1Imp=0-1Imυ=0ω=00 Re

Imυ=2ω=0ω=0+0 Re系统不稳定 系统稳定(c)p=0Imυ=2(d)ω=0+Imp=0ω=0+-1 0ω=0Re-10υ=3ω=0Re系统不稳定 系统稳定(e) p=0 Imυ=1-1 ω=00 Re