2-14R(s)G(s)-2-14R(s)G(s)-E(s)G(s)41+X(s)G(s)+C(s)3G(s)2D(s)

=GG

Δ=1P

=GG

Δ=11 1P=GG

3 1Δ=1

2 2=GG

3 2Δ=13 1 4 3

4 2 4 4L=-GGL=-GG

C(s) (G+G)(G+G)= 1 2 3 4 1 1 3 2

2 3R(s)

(G+G)E(s)= 1

X(s)

(s)

C(s)=1

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

E(s)

D(s)3 1 2R(s)R(s)1+-GG2G C(s)311H2H1R(s)+2-G4GGC(s)256

解:L=GG

L=-G

GGHH L=-G1 1 2 2 1Δ=1-GG+GGG

4 5 1 2 3 4HH+G-GGG1 2 1 4 35 1 2 4 1 2 4P=GGG Δ=1+G1 1 2 3 1 4C(s)1 =

GGG(1+G)1 2 3 4R(s) 1+G

+GGGHH-GG-GGG1 4 1 4 5 1

1 2 1 2 4C(s)

GG

(1-GG)

C(s) -GGGGGHR2(s)

+G

5

H

2 -GG

R(s)

+GG1

G1-GGG2 4 1 4 5 1 2 1

1 2 4 2

4 1 4 5 1 2 1

1 2 4C(s) GGGGHR2(s)=1+G+GG1G4H5H6-G2G-GGG1 4 1 4 5 1 2 1 2 1 2 4設(shè)溫度計(jì)需要在一分鐘內(nèi)指示出響應(yīng)值的并且假設(shè)溫度計(jì)為一階系統(tǒng)，求時(shí)間常數(shù)T。如果將溫度計(jì)放在澡盆內(nèi)，澡盆的溫度以10oC/min的速度線性變化，解: c(t)=c(∞)98% t=4T=1min T=0.25 r(t)=10tc(t)=10(t-T+e-t/T)=10(T-e-t/T)e(t)=r(t)-c(t)

=lime(t)=10T=2.5sst→∞R1電路如圖，設(shè)系統(tǒng)初始狀態(tài)為零。u R C

=20kΩ

=200kΩ

0 - ∞ u+ c+（1）時(shí)的值．R/R K解:G(s)=R

1

=Ts+

T=RC=0.5 K=R/R=101u(t)=K(1–c

1-Tt)=10(1–e-2t)-T

1 08=10(1–e-2t)0.8=1–e-2t e-2t=0.2 t=0.8(2)求系統(tǒng)的單位脈沖響應(yīng)，單位斜坡響應(yīng),及單位拋物響應(yīng)在t1時(shí)刻的值．解: t1

=0.8 R(s)=1

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

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

U(s)= K 1=K(1

T+T2

T2 )s3 c Ts+1s3 s3 s2 s s+1/T2u(t)=10(2c

1t2-0.5t+0.25-0.25e-2t)=1.2

G(s)= 4

s(s+5)解: C(s) 4 1

4 1

4/3R(s)=R(s)=

C(s)=s(s+1)(s+4)=

+

-s+1c(t)=1+

1e-4t-4e-t 3 3G(s)= 1r p 升時(shí)間t、峰值時(shí)間t、超調(diào)量σ%和調(diào)整時(shí)間tr p

s(s+1)C(s) 1

ω

ω=1 ω

1-ζ2=0.866ζ解: R(s)=ζ

n n d ns2+s+1

ω2=n

=0.5 =t112=60oωt3.14-3.14/3=2.42 ω

=π

3.14

=3.63r d 0.866

p d 0.866ω%=π1100%ω

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

ω ωn nss3-6已知系統(tǒng)的單位階躍響應(yīng):c(t)=1+0.2e-60t-1.2e-10tss系統(tǒng)的閉環(huán)傳遞函數(shù)。系統(tǒng)的阻尼比和無阻尼振蕩頻率。解:C(s)= + 解:C(s)= + - s s+60s+10s(s+60)(s+10)R(s)=1

ω

ω=24.5s R(s)

nω2n

nζ=1.431.31c(t)0 1.31c(t)0 0.1t解:

ωπ=0.1 /12=ln3.3=1.19

1-ζ2=3.14=31.4p n12 π

n 0.1π1ζ

ζ)2/12

ω=33.4e 2π1-

9.82=1.42-1.42

n

1115.6e

ζ=0.35

G(s)=s(s+ωn

)=s(s+22.7)3-11 已知閉環(huán)系統(tǒng)的特征方程式，試用勞斯判據(jù)判斷系統(tǒng)的穩(wěn)定性。s3+20s2+9s+100=0勞斯表如下：s3 1 9s2 20 100s1 4s0100

(3)s4+8s3+18s2+16s+5=0勞斯表如下：s41185s3816s2165s1s116系統(tǒng)穩(wěn)定。

G(s)=

s0 5 系統(tǒng)穩(wěn)定。K(0.5s+1)3-12(+)(.+s+1)=b解:s4+3s3+4s2+2s+Ks+2K=0=bs4 1 4 s3 3 2+Ks2 b 2K31s1 bs41

10-2K31 3=bK2+10K-20=b41 10-K(K-1.7)(K+11.7)>0K<10R(s)-R(s)--s(s+1)τs10C(s)s3110s3110s2(1+1)10s1b31s010解:G(s)= 10(1+s)解:G(s)= +s+1s s(+s+1)b

10(s+1)3=10(1+1)-10>031

τ>0R(s)

τ

10 C(s)3-14已知系統(tǒng)結(jié)構(gòu)如圖，試確定系

- s s(s+1)解:G(s)=1(

Φ(s)=

1(s+1)s2(s+1)s3 1

3+2+1s+10110τ-101s2 1

b = >031s1 b31s0 10

τ>1

r(t)=I(t)+2t+t23-16已知單位反饋系統(tǒng)的開環(huán)傳遞函數(shù)，

R(s)=

1 2 2+ s s2 s3+ =21(1)G(s)==21

R20 R

K=20p

ess1

1+0 1解： (0.1s+1)(0.2s+)

e =∞ss2K=0ae=∞ss

e =∞ss3K=∞ e =0(2)G(s)=

200 =

10 p

ss1 2 2s(s+2)(s+10)

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

ess2=K=10K=0ae=∞ss

e =∞ss3(3)

10(2s+1)=

(2s+1)

K=∞p

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

s2(0.1s2+0.4s+1) K

e =0υ ss2K=1a

e=2ss

e =2ss31已知系統(tǒng)結(jié)構(gòu)如圖(1)單位階躍輸入:R(s) - - K1

１C(s)s2ts

=1.8(5%)確定K1

和值。 K K ω

π

2=0.2解:G(s)= 1

Φ(s)=

1 n 1

ω2=K t

3=1.81 1

1 sω

ω=

=3.7

nn1.8*0.45 1 n(2)r(t)=I(t),t,

1t22 1

K=∞K 1 R(s)=s

e =0ss1解:G(s)=

1 = τ

υ=1

R(s)=1

K=K

=τ=0.24

1s+1)

s2

ss2s31 s31

R(s)=1

K=0 a

=∞ss3R(s)--s(s+2)τR(s)--s(s+2)τsKC(s)

確定K和τ值。解:G(s)=

ssKτ=

KK2+τ (s)=K s2+2s+K

s s( 1 s+1) 2+(2+)s+K2+τω=2+=2*0.7K e=2+=0.25 =31.62+τn ss Kω2=K 0.25K-2 n K系統(tǒng)結(jié)構(gòu)如圖。

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 -作下的穩(wěn)態(tài)誤差．1解

=lims·

s = 1ssr

1+G(s)F(s)1+G(0)F(0)求

(t)和d(t)同時(shí)作用下的穩(wěn)態(tài)誤差．1

(s)H(s)Ed(s)=1+G

(s2)G1

H(s)·D(s)e =lims

-F(s)

-1 ]1

-[1+F(s)]ssd

1+G(s)F(s)

1+G(s)F(s)

1+G(0)F(0)求d(t)作用下的穩(wěn)態(tài)誤差．1 K F(s)=1G(s)=Kp+s Js -1)e =lim)

-F(s) 1=lims Js

1=0ssd

1+G(s)F(s)

1+(Kp

K 1 ss Js4-1 已知系統(tǒng)的零、極點(diǎn)分布如圖，大致繪制出系統(tǒng)的根軌跡。解:(1)

(3)

900

(4)

600σ 0σ 0(2)0σ(2)0σ600

(6)

(7)

1350

450

(8)

1080

3600σ

σ 0σ 0σ4-2 已知開環(huán)傳遞函數(shù)，試用解析法繪制出系統(tǒng)的根軌

G(s)=

K(s+1)解

KKr

s=-1-Kr

r-3+j2r

jω0+j1K=0

s=-2+j0σ0σr→∞

s=0+j1

-2 -1K=r s=-3+j2jωpjωp3p2pzz01σ21解:解:(1)G(s)= jωjω6.2p4p3p2pz-5.6701-6.2(4)G(s)=s(s+3)(sr+7)(s+15)開環(huán)零、極點(diǎn)p=0 1

=-3 p2

=-7p4

=-15z1

=-8實(shí)軸上根軌跡段p~p

p~z

p~-∞1 2 3 1 4根軌跡的漸近線n-m=3σ=-3-7-15+8=-5.67180+60o,+3 o180根軌跡與虛軸的交點(diǎn)

分離點(diǎn)和會(huì)合點(diǎn)A(s)=s4+25s3+171s2+315sA(s)'=4s3+75s2+342s+315s4+25s3+171s2+323s+8K=0K=0ω=0 K=638ωr

B(s)=s+8s=-1.4

B(s)'=2s+7r 1 r 2,34-5 (1)(2)增益Kr復(fù)數(shù)特征根的實(shí)部為-2。pp2p1z10σ(Gs ss1)解: p=0 p=-1 z=-21 2 1p~p z~-∞1 2 1分離點(diǎn)和會(huì)合點(diǎn)s2+4s+2=0s1

=-3.41s2

=-0.59 r

)ω=0閉環(huán)特征方程式 r

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

s=-2j

1.41r r(-2+2+(-2j)(1+K)+2K=0

K=3r r r已知系統(tǒng)的開環(huán)傳遞函數(shù)，試確定閉環(huán)極點(diǎn)ζKr值。jω1.7s1s3p3-3p2jω1.7s1s3p3-3p2-1p1σ0-1.7rs(s+1)(s+3)解:

=0p=-1

=-3p~p p~1 2

1 2 33-1-33

根軌跡的分離點(diǎn)：A(s)B'(s)=A'(s)B(s)3s2+8s+3=0s=-0.45 s1

=-2.2 舍去與虛軸交點(diǎn)

ζ=0.5得s1

=-0.37+j0.8s3+4s2+3s+K=0

s=-4+0.37×2=-3.263

rK

ω=0

=|s

||s

+1||s

+3|KKr=12

=±1.7

=3263

2.230.26=1.9r r 2,3 ×jωp3sjωp3s11.1-71.6p226.6-2.3 -1.2p1901350σ-1.1p4rs(s+3)(s2+2s+2)

=0

=-3

=-1±j p~p11 2 3.4 24-3-1-1=-1.254

+45o,θ=3

- -1 2 4=與虛軸的交點(diǎn)rs(s+3)(s2+2s+2)+K=0rrs4+5s3+8s2+6s+K=0r

分離點(diǎn)和會(huì)合點(diǎn)4s3+15s2+16s+6=0解得s=-2.3 ζ=0.5ω=0

得s=-0.36+j0.7518(j )1rω=0 K=0 1=0

K=|s

||s

+3||s

+1+j||s

+1-j|r Kr

r 1 1 1 1-5ω3+6ω=0

=8.16r

2,3

=2.92jωjω2.8s1spp332-4 -2p0σ1-2.8試?yán)L制出根軌跡圖。K解:

=0p=-2p

=-4p~p p~1 2

1 2 33-2-4=-23根軌跡的分離點(diǎn)：A(s)B'(s)=A'(s)B(s)s=-0.85 s=-3.15 舍去(2) K(2) K阻尼振蕩響應(yīng)的

3+=0

Kr

ω=01r K

K=48ω

=±2.8s=-0.85 KKr=48r

r(4)ζ=0.5

r 2,3s=-0.7+j1.21(3)與虛軸交點(diǎn)s3+6s2+8s+K=0r

s=-6+0.7×2=-4.63K=4.6×2.6×0.6=7.2rr(t)=sin(t+30o)，試求系解統(tǒng)Gs)態(tài)1出(s)= 10

10 = 10

10=0.905(s+1)

(s+11)

112+)112+112211 =-tgω=-tg1111

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

750 I型系統(tǒng)n-m=3s(s+5)(s+15) =0 A)∞)=-9o∞ A)=0 0

ω=∞0 Re解(3)G(s)= 10 Im

ω=0ω=0Im0 Re解ω=0Im0 Re(2s+1)(8s+1)0型系統(tǒng)n-m=2=0 A)=10)=o∞ A)=0 =

ω=∞ω=00

s(s-1)解：I型系統(tǒng)n-m=2=0 A)=∞)=-270o=∞ A)=0 =0

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

II型系統(tǒng)n-m=3 Im=0 A)=∞)=-180o

ω=0

ω=∞0 Re∞ A)=0 )=-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

20

-20dB/dec

0-20

1 5-40dB/dec

ωω=0.125ω=0.5

0-20

0.125 0.5 ω-40dB/dec20lgK=20dBω=5ω=15

-60dB/dec

1 2φ(ω0

φ(ω)1 2

ω

)=0o 0 ω

)=-90o)=-270o

-90-180-270

∞=0

-90-180

10s(s-1)

L(ω)dB40 -20dB/dec1ω=1 20lgK=20dB 2010

-40dB/dec1 ω=0 )=-27o-20∞=

φ(ω)ω-90-180-270(7)G(s)=

10(s+0.2) =

1.33(5s+1)s2(s+0.1)(s+15)解：20lgK=2.5dBω=0.1 ω=0.2

s2(10s+1)(0.67s+1)ω=151 2 3=0 )=-18o∞)=-270odB40dB40200-20-40dB/dec-60dB/dec150.10.2 1ω-40dB/dec0-90-180-270-60dB/decωdB02011020ω20dB/decdB20lgKdB20lgK10-20dB/decωcωK=10 20

(b)

20lgK=-20K=0.1G(s)= 10

G(s)=

s -(0.1s+1) 0

(0.05s+1)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(c) K=100

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

100 0L(ω)dBL(ω)dB-20dB/dec1000.01-40dB/dec-60dB/decωL(ω)dBL(ω)dB-20dB/dec4.58dB100ω-60dB/dec1ζ1220lgM=4.58 M=1.7= 得：dBr r0.94 ζ

01ω

2ω

=1001r n1-2 n 01

1002=ωn

G(s)=s[(0.02s)2+0.01s+1)]5-7 為積分環(huán)節(jié)個(gè)數(shù)，試判別系統(tǒng)穩(wěn))

(b) p=0

Im υ=2p=0Imp=0Imυ=0ω=0-10Re系統(tǒng)不穩(wěn)定 系統(tǒng)穩(wěn)定(c)

p=0-1

Im υ=20

(d) -10

p=0υRe系統(tǒng)不穩(wěn)定

系統(tǒng)穩(wěn)定(e)p=0-1Imυ=1(f)Imp=1υ=00Reω=0-10Re系統(tǒng)穩(wěn)定

ω=0+

系統(tǒng)穩(wěn)定(g)

p=0

Im υ=1

(h)

Im υ=0-1 0

ω=0Re

-1 ω=00 Re系統(tǒng)穩(wěn)定 系統(tǒng)不穩(wěn)定ω=0+L(ω)dB-20dB/dec-40dB/dec0200.1ωL(ω)dB-20dB/dec-40dB/dec0200.1ωc10ω-60dB/dec0-90180ωγ解：K=10 10G(s)=101 ω=1cc=180oω)c=180o-90o-tg-110-tg-10.05=90o-84.3o-2.9o=2.8o -R(s)-2R(s)-20.5s+15s(0.02s+1)1C(s)

1000000000)dB-20dB/dec12ωc50ω-40dB/dec-60dB/dec10 ω=4.5 2cγ c=180o-90o-tg-1-2-tg-10.02×4.5=90o-66o-2.6o=21.4o500K6-1 已知單位負(fù)反饋系統(tǒng)開環(huán)傳遞函數(shù)，G0(s)=s(s+5oK==100vG(s)=

υ=100,4o

L(ω)dB0

L40 L20lgK=40

100dB 0.2 20

0 Lc24 40c°

5 ω

82 ωω=22.4 c

c c-20 L取=5.6oφ- m=45o–12.6o+5.6o=38°1+sinφa=1–sinφm=4.2

900-90

φ0

-40dB/decφcωφmL( )=10lga0 mω=40

-180 γωωm c 12=ma=82T=82=0.01aT1ω=1=23.8aT=0.04aT1G(s)=1+0.04sc 1+0.01sG(s)=G(s)G(s)0 c6-5 已知單位負(fù)反饋系統(tǒng)開環(huán)傳遞函數(shù)，

(s)=0

Ks(0.5s+1)(0.2s+1)dBL02640L20ω0.0050cc0.10.525ω-20LcdBL02640L20ω0.0050cc0.10.525ω-20Lc0-90-180-270φφ0cφωγ10vG(s)=0 s(0.5s+1)(0.2s+1)20lg10=20dB10 ω=4.5ccγ=-18°φ=-180o+γ'+Δ=-180o+50o+10o=-120oc取=0.5cL(')=26dB=-20lβ0 c取ω= 1==0.125 cω=0.0051 2G(s)=1+βTs=10s+1c 1+Ts 200s+1G(s)=G(s)G(s)0 c2-14

G(s)

解:

=GG

Δ=1P

=GG

Δ=14 1 1 3

2 2 3 2R(s) )

已知系統(tǒng)+(()=c(s

PGG

曲線，要求繪制校正G(s)

G(s)

3 1 4

4 2 4 4- 1 + 3

L=-GG

L=-GG

C(s)

(G+G

)(G+G)G(s)

X(s)

1 1 3 2 2

= 1 2 3 4 2

(s)性曲線，并寫出開環(huán)s)

(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+-G1G2G C(s)31H2H1R(s)+2-G4GGC(s)256

解:L=GG

L=-GGGHH L=-G1 1 2 2 1 4 5 1 2 3 4G+GGG HH+G-GGG1 2 1 4 35 1 2

4 1 2 4P=GGG Δ=1+G1C(s)

1 2 3G

1 4GG(1+G)R1(s)=1+G

+G

2

H

G-GGG1 4 1 4 5 1

1 2 1 2 4C(s)

GGG(1-GG)

C(s) -GGGGGH1R2(s)

+G

H

2 -GG

R(s)1+G+GG1 2 H-51-GGG2 4 1 4 5 1 2 1

1 2 4 2

4 1 4 5 1 2 1

1 2 4C(s) GGGGHR2(s)=1+G+GG1G4H5H6-G2G-GGG1 4 1 4 5 1 2 1 2 1 2 4設(shè)溫度計(jì)需要在一分鐘內(nèi)指示出響應(yīng)值的并且假設(shè)溫度計(jì)為一階系統(tǒng)，求時(shí)間常數(shù)T。如果將溫度計(jì)放在澡盆內(nèi)，澡盆的溫度以10oC/min的速度線性變化，解: c(t)=c(∞)98% t=4T=1min T=0.25 r(t)=10tc(t)=10(t-T+e-t/T)=10(T-e-t/T)e(t)=r(t)-c(t)

=lime(t)=10T=2.5sst→∞R1電路如圖，設(shè)系統(tǒng)初始狀態(tài)為零。u R C

=20kΩ

=200kΩ

0 - ∞ u+ c+（1）時(shí)的值．解:

R/RK1 0 =K

T=R

C=0.5

/R=10RCs+1 Ts+1 1 1 01 -t

-2t

8=10(1–e-2t)u(t)=K(1–c

T)=10(1–e )0.8=1–e-2t

e-2t=0.2 t=0.8T時(shí)刻的值．T解: t1

=0.8 R(s)=1

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

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

T+T2- T2 ) 2s3 2

Ts+1

s3 s2 s s+1/Tu(t)=10(c

1t2-0.5t+0.25-0.25e-2t)=1.2

G(s)= 4

s(s+5)解: C(s) 4 1

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

C(s)=s(s+1)(s+4)=s+s+4-s+1c(t)=1+

1e-4t-4e-t 3 3G(s)= 1r p 升時(shí)間t、峰值時(shí)間t、超調(diào)量σ%和調(diào)整時(shí)間tr p

s(s+1)C(s) 1

ω

ω=1 ω

1-ζ2=0.866解: R(s)=

n n d ns2+s+1

ω2=n

=0.5 =t112=60oζpωdt3.14-3.14/3=2.42 t=π=3.14=3.63ζpωdr d 0.866

ω0.866%=π1100%

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

ω ωn nss3-6已知系統(tǒng)的單位階躍響應(yīng):c(t)=1+0.2e-60t-1.2e-10tss系統(tǒng)的閉環(huán)傳遞函數(shù)。系統(tǒng)的阻尼比和無阻尼振蕩頻率。解:C(s)= + - 1 0.2 解:C(s)= + - s s+60s+10s(s+60)(s+10)R(s)=1

ω

ω=24.5s R(s)

nω2n

nζ=1.431.31c(t)0 1.31c(t)0 0.1t解:

ωπ=0.1 /12=ln3.3=1.19

1-ζ2=3.14=31.4p n12 π

n 0.1π1ζ

ζ)2/12

ω=33.4e 2π1-

9.82=1.42-1.42

n

1115.6e

ζ=0.35

G(s)=s(s+ωn

)=s(s+22.7)3-11)s+029+0=3s++821s5穩(wěn)定性。勞斯表如下：s3 1 9s2 20 100s1 4s0100

勞斯表如下：s41185s3816s2165s1s116系統(tǒng)穩(wěn)定。

G(s)=

s0 5 系統(tǒng)穩(wěn)定。K(0.5s+1)3-12(+)(.+s+1)=b解:s4+3s3+4s2+2s+Ks+2K=0=bs4 1 4 s3 3 2+Ks2 b 2K31s1 bs41

10-2K31 3=bK2+10K-20=b41 10-K(K-1.7)(K+11.7)>0K<103-13統(tǒng)穩(wěn)定時(shí)τ值范圍。解:G(s)= 10(1+s解:G(s)= +s+1s s(+s+1)

R(s)-R(s)--s(s+1)τs10C(s)2 (1+1)10b

10(s+1)3=10(1+1)-10>0

s1 b31s0 1031

τ>0R(s)

τ

10 C(s)3-14已知系統(tǒng)結(jié)構(gòu)如圖，試確定系

- s s(s+1)解:G(s)=1(

Φ(s)=

1(s+1)s2(s+1)s3 1

3+2+1s+10110τ-101s2 1

b = >031s1 b31s0 10

τ>1

r(t)=I(t)+2t+t23-16已知單位反饋系統(tǒng)的開環(huán)傳遞函數(shù)，

R(s)=

1 2 2+ s s2 s3+ (1)G(s)=

20

K=20p

e = 0 11+=R21ss11+=R21解： (0.1s+1)(0.2s+)

e =∞ss2K=0ae=∞ss

e =∞ss3K=∞ e =0(2)G(s)=

200 =

10 p

ss1 2 2s(s+2)(s+10)

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

ess2=K=10K=0ae=∞ss

e =∞ss3(3)

10(2s+1) =

(2s+1)

K=∞p

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

s2(0.1s2+0.4s+1) K

e =0υ ss2K=1a

e=2ss

e =2ss31已知系統(tǒng)結(jié)構(gòu)如圖(1)單位階躍輸入:R(s) - - K1

１C(s)s2ts

=1.8(5%) 確定K1

和值。 K K ω

π

2=0.2ω解:G(s)=s2+sω1

1

n 1nω2=K n

=1.8ζ=0.45

ω= 3 =3.7 K

1τ=0.24s nn1.8*0.45

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

1t22 1

K=∞K 1 R(s)=s

e =0ss1解:G(s)=

1 = τ

υ=1

R(s)=1

K=K

=τ=0.24

s(

s+1)

s2

ss2s31 s31

R(s)=1

K=0 a

=∞ss3R(s)--s(s+2)τR(s)--s(s+2)τsKC(s)斜坡輸入的穩(wěn)態(tài)誤差ess

確定K和τ值。解:G(s)= Kτ=ss2+2s+Ks

KK2+τ (s)=s+(2+)s+KK 2+τs( 1 s+1) 2+τω=2+=2*0.7K e=2+=0.25 =31.6n ss KKω2=KKn

0.25K-2 系統(tǒng)結(jié)構(gòu)如圖。

D(s)1

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

R(s) E(s)

+F(s)

+ C(s)1 2 -作下的穩(wěn)態(tài)誤差．1解

=lims·

s = 1ssr

1+G(s)F(s)1+G(0)F(0)求

和d(t)同時(shí)作用下的穩(wěn)態(tài)誤差．1

(s)H(s)Ed(s)=

(2)G1

(s)H(s)·D(s)2e =lims

-F(s)

-1 ]1

-[1+F(s)]ssd

1+G(s)F(s)

1+G(s)F(s)

1+G(0)F(0)求d(t)作用下的穩(wěn)態(tài)誤差．1 K F(s)=1G(s)=Kp+s Js -1)e =lim)

-F(s) 1=lims Js

1=0ssd

1+G(s)F(s)

1+(Kp

K 1 ss Js4-1 已知系統(tǒng)的零、極點(diǎn)分布如圖，大致繪制出系統(tǒng)的根軌跡。解:(1)

(3)

900

(4)

600σ 0σ 0(2)0σ(2)0σ600

(6)

(7)

1350

450

(8)

1080

3600σ

σ 0σ 0σ4-2 已知開環(huán)傳遞函數(shù)，試用解析法繪制出系統(tǒng)的根軌

G(s)=

K(s+1)解

KKr

s=-1-Kr

r-3+j2r

jω0+j1K=0

s=-2+j0σ0σr→∞

s=0+j1

-2 -1K=r s=-3+j2jωpjωp3p2pz2z101σ解:解:(1)G(s)= jω6.2p4p3jω6.2p4p3p2z-5.67p01-6.2(4)G(s)=開環(huán)零、極點(diǎn)p=0 1

=-3 p2

=-7p4

=-15z1

=-8實(shí)軸上根軌跡段p~p

p~z

p~-∞1 2 3 1 4根軌跡的漸近線n-m=3σ=-3-7-15+8=-5.67180+60o,+3 o180根軌跡與虛軸的交點(diǎn)

分離點(diǎn)和會(huì)合點(diǎn)A(s)=s4+25s3+171s2+315sA(s)'=4s3+75s2+342s+315s4+25s3+171s2+323s+8K=0K=0ω=0 K=638ωr

B(s)=s+8s=-1.4

B(s)'=2s+7r 1 r 2,34-5 (1)(2)增益Kr復(fù)數(shù)特征根的實(shí)部為-2。pp2pz10σ1(Gs ss1)解:

=0

=-1

=-21pp2 z~11 2 1分離點(diǎn)和會(huì)合點(diǎn)s2+4s+2=0

=-3.41s

=-0.59

)ω=01閉環(huán)特征方程式

2 r4-ω2-2(1+Kr

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

s=-2j

1.41r r(-2+2+(-2j)(1+K)+2K=0

K=3r r r已知系統(tǒng)的開環(huán)傳遞函數(shù)，試確定閉環(huán)極點(diǎn)ζKr值。jω1.7s1s3p3-3p2jω1.7s1s3p3-3p2p1-10σ-1.7rs(s+1)(s+3)解:

=0p=-1

=-3p~p p~1 2

1 2 33σ=-1-3=-1.3θ=+60o,+180o3根軌跡的分離點(diǎn)：A(s)B'(s)=A'(s)B(s)3s2+8s+3=0s=-0.45 s1

=-2.2 舍去與虛軸交點(diǎn)

ζ=0.5得s1

=-0.37+j0.8s3+4s2+3s+K=0

s=-4+0.37×2=-3.263

rK

ω=0

=|s

||s

+1||s

+3|KKr=12

=±1.7

=3263

2.230.26=1.9r r 2,3 ×(2)

Ks(s+3)(s2+2s+2)rr

=0 1

=-32

3.4

=-1±j p~p1 24-3-1-1=-1.25+45o,4根軌跡的出射角θ=3

- -1 2 4=+π-135o-90o-26.6o=-71.6or與虛軸的交點(diǎn)s(s+3)(s2+2s+2)+K=0s4+5s3+8s2+6s+K=r0rω=0

jωjωp3s11.1-71.6p226.6-2.3 -1.2p1901350σ-1.1p44s3+15s2+16s+6=0解得s=-2.3 ζ=0.5得s=-0.36+j0.75

8(j )=0 K

rω=01

K=|s

1||s

+3||s

+1+j||s

+1-j|r Kr

r 1 1 1 1-5ω3+6ω=0

=8.16r

2,3

=2.92jωjω2.8s1spp332-4 -2p0σ1-2.8試?yán)L制出根軌跡圖。K解:

=0p=-2p

=-4p~p p~1 2

1 2 33-2-4=-23根軌跡的分離點(diǎn)：A(s)B'(s)=A'(s)B(s)s=-0.85 s=-3.15 舍去(2) K(2) K阻尼振蕩響應(yīng)的

3+=0

Kr

ω=01r K

K=48ω

=±2.8s=-0.85 KKr=48r

r(4)ζ=0.5

r 2,3s=-0.7+j1.21(3)與虛軸交點(diǎn)s3+6s2+8s+K=0r

s=-6+0.7×2=-4.63K=4.6×2.6×0.6=7.2r5-2已知單位負(fù)反饋系統(tǒng)開環(huán)傳遞函數(shù)，當(dāng)輸入信號(hào)r(t)=sin(t+30o)，試求系(1)解統(tǒng)Gs)態(tài)1出(s)= 10

10 = 10

10=0.905(s+1)

(s+11)

12+)12+112211 =-tω=-t1111

c(t)=0.9sin(t+24.8o)sI型系統(tǒng)5-2I型系統(tǒng)(1)

750 n-m=3s(s+5)(s+15) Im=0 A)∞)=-9o∞ A)=0 0

ω=∞0 Re解(3)G(s)= 10 Im

ω=0ω=0Im0 Re解ω=0Im0 Re(2s+1)(8s+1)0型系統(tǒng)n-m=2=0 A)=10)=o∞ A)=0 =

ω=∞ω=00

s(s-1)解：I型系統(tǒng)n-m=2=0 A)=∞)=-270o=∞ A)=0 =0

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

II型系統(tǒng)n-m=3 Im=0 A)=∞)=-180o

ω=0

ω=∞0 Re∞ A)=0 )=-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

20

-20dB/dec

0-20

1 5-40dB/dec

ωω=0.125ω=0.5

0-20

0.125 0.5 ω-40dB/dec20lgK=20dBω=5ω=15

-60dB/dec

1 20

φ(ω)1 2

ω

)=0o 0 ω=0 )=-90o∞ 0

-90-180-270

∞=

-90-180

10s(s-1)

L(ω)dB40 -20dB/dec1ω=1 20lgK=20dB 2010

-40dB/dec1 ω=0 )=-27o-20∞=

φ(ω)0 ω-90-180dB40200-20-40dB/dec-60dB/dec150.10.2 1ω-40dB/dec0-90-180-270-60dB/decω(7)G(s)=

10(s+0.2) =

1.33(5s+1)s2(s+0.1)(s+15)解：20lgK=2.5dB

s2(10s+1)(0.67s+1)ω=0.1 ω=0.2 ω=151 2 3=0 )=-18o∞)=-270odBdB11020ω20dB/decdB20lgKdB20lgK-20dB/dec10ωcωK=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/dec1050

(c)

K=100

L(ω)dB-20dB/dec0 0.01

100ωG(s)=

251

-60dB/dec

100

-40dB/dec-60dB/dec(c) K=100

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

100

-20dB