(s)

G(j2)sin(2t+G(j2)

￥

sin(2t+j(j2)習(xí)題5-2某最小相角系統(tǒng)的概略開環(huán)對數(shù)幅頻特性曲線對

-0(0

ww j32j320習(xí)題5-4某最小相角系統(tǒng)K2和K6K<12和0K

4<K<6K<￥和0K--0-w cL(w

-

K=

T=a2

t=a2K(s+-0w1

-

w

20lg|

(jwc(bs+ac(bs+a2s(cs+a22b=2w wa2w2=第五章Ks(s+a)觀測到輸入、輸出的幅值相等，相位差為90Ks(s+a)

M(w

22x wr

wr=w

Mrsinwt

csin(wt+f(w))Ks(s+Ks(s+a)cm

wn=3

,確定w為何值時穩(wěn)態(tài)輸

s+1R(s C(s+11-繪制t0時的開環(huán)幅相曲線，并用奈氏判據(jù)判斷t0wc為4，j2

2·(0-1)G(jw)

(jw)2(2jw 4w4wG'(s)=10(s+1)s2(2s+1)

10(s+1)s2(2s+1)

R(sg=

C(ss+1wcs+1=180 t0wc為4，G'(s)=10(s+1)s2(2s+1)

10(s+1)s2(2s+1)

s+1s+1R(sg=

C(swc=c

=18

t=2

t2

t2

的穩(wěn)態(tài)誤差ess態(tài) 習(xí)題1

1KKwcKwc

KG(scG(sc

0+—j''

c c1800+(-50

cwcwcc

G(scG(sc

1+bTsG(s)1+Ts

c jwc

cc

G(jw''

cc

G(jw''cscs

5 j0.56 G(scG(sc

10 159cbw''c

wc wcGG(scG(scG(scc

40.04

GG(sc習(xí)題2

為原系統(tǒng)的50%(允許誤差–5%)，并計算校正前后系ts； =

1￡

sins wcs

K=2+1.5(M-1)+2.5(M-

1￡Mr習(xí)題2 c求原系統(tǒng)的w' Mr c

2

G(s)?sG(s)?1w'c2cjw'c2c

模模從12 習(xí)題2 c求原系統(tǒng)的w' Mr c

900-2 2

212

singsin Mr

sin

1￡Mrs wcs

K=2+1.5(M-1)+2.5(M-

1￡Mr習(xí)題2

s%-0.16+

900- 2-arctg0.01· garcsin1

aG(s

2

習(xí)題2cc

'

aG(scaG(sc

60+習(xí)題2a

3

aaG(sc

jw習(xí)題2a

3

aaG(scc c

jw習(xí)題2

w

a aaT=awm

aG(aG(scc習(xí)題2

w

a aaT=awm

aaG(sc

習(xí)題2

G(s)?

2s2 G(s)

G(s)?2·0.93s模從1到22s(s(

1)習(xí)題2

G(s)?

2s2 G(s)

cw''=2·0.93=c''

G(s)?2·0.93s模從1到Mr= 習(xí)題2aGaG(sc

w' 4w'c 2+1.5(Mr-1)+2.5(Mr-1 cw''=2·0.93=c''Mr= 習(xí)題2aGaG(sc

w' 4w'cw'c習(xí)題

2siKs(Ts+1)Ks(Ts+1)

C(s 求r(t=1,(=+2t時系統(tǒng)的穩(wěn)態(tài)誤差ess(t(s)=

N(s

s(Ts+1)

Ts2+s+

r(t)=cosT(jw)2+jw+

2sit j2+K-

|j(j2)|=K=T==,(=+2t

N(s

Ks(Ts+1)Ks(Ts+1)G(jw)= jw(jw+1) ww2+

g=

r(t)=cos2t2twc=cf(w)=-900-tg-1c

K=T==,(=+2tG(s)=essr(t)=

N(s

s(Ts+1)s(Ts+1)

essn1(t=,(=+2t0G

N(s

KTs2+s+

N(s

N(sN(s s(Ts+1)=,(=+2t0G

N(s

KTs2+s+

N(s

essn1(t

=limsj(s)N(sN(s s(Ts+1)sfi =,(=+2t G

N(s

KTs2+s+

N(s

N(s

jen

essn2(tN(N(s s(Ts+1)=,(=+2t G

N(s

j(j3KTs2+s+

N(s

(j3)2+j3+

En(s

|j(j3)|= essn2(tjen(j3

—j(j3)=N(N(s s(Ts+1)=,(=+2t j(j3)=- =

N(sN(s Ks(Ts+1)=4(5+j3)=20Ks(Ts+1)

j(j352-(j3

En(s

- (j3)2+j3+|j(j3)|= essn2(tjen(j3

—j(j3)==,(=+2t essn2(t)=

ee(t)=-18sin(3t+310

sin(3t+310

N(s

Ks(Ts+1)Ks(Ts+1)=N(s

sin(3t+310En(

j(j34=-(j3)2+j3+|j(j3)|= essn2(tjen(j3

—j(j3)==,(=+2t 習(xí)題4G(s)

K*s2(s+2)(s+a反饋通道H(s)=1,a為大于0的常數(shù),閉環(huán)根軌跡具有重極點-概略繪制閉環(huán)系統(tǒng)根軌跡(0K*<￥),試分析閉環(huán)系統(tǒng)X(s)為輸出量的傳遞函數(shù)。5(s)C(s1(s)2(s)X(s)3(s4(s)j(s)=X(s)

5(s)C(s1(s)2(s)X(s)3(s4(s)習(xí)題6

K(s+1)s3+as2+2s+試確定反饋系統(tǒng)具有頻率w2弧度/秒的純虛極試確定反饋系統(tǒng)具有頻率w2弧度/1+G(jw)=1+G(j2)=

s3+as2+2s+1(j2)3+a(j2)2+2(j2)+1+K(j2+1)=-8j-4a+4j+1+2Kj+K=0(-8+4+2K)j-4a+1+K=K=

a=4習(xí)題7s%和調(diào)節(jié)時N(sKR(N(sK

C(s

(

-

-r(t)=2sin3t

css

4

+KG 1

G2(s)=s+N(sKN(sK11r(t)=2sin3t

css

4|j(j3)|=

R(sC(sR(sC(sKN(s

1( -0wr(t)=2sin3t

css

4|j(j3)|=

R(sC(sR(sC(sKN(s ()

s(s+1) s2+(1+K)s+2Kr(t)=2sin3t

css

4|j(j3)|=

=| (j3)2+(1+K)(j3)+2K

()

4=

K=s2+(1+K)s+2K=3r(t)=2sin3t

css

4(s)

s2+4s+

G(s)= N(sKR(sN(sK

()

=s2+(1+K)s+2K=3r(t)=2sin3t

css

4(s)

s2+4s+

w2=n2wn=nN(sKN(sK11

3

wn=x=ts

s%=e ·100%=

K設(shè)j

(t)=1wr=，r(t) (s)=

+G s2+as+nw2= K=n

j(jwn)=12wn=

a=

wr|j(jw)|=1=

x=

r(t) wr=wn

wn= (s)=

+G s2+as+w2= K=

1n2wn=

a=

|j(jw)|=1=

x=

r(t) wr=wn

wn= (s)=

() +G s2+as+w2= K=

s2+s+1n2wn=

a=

|j(jw)|=1=

x=

r(t) wr=wn

wn= j(j2)

(jw)2+jw+1(j2)2+j2+1

1j2-

()

s2+s+1|j(j2)

r(t)—j(j2)=-(p-tg-12)=-146.303 j(j2)

(jw)2+jw+1(j2)2+j2+1

1j2-

()

s2+s+1|j(j2)

r(t)—j(j2)=-(p-tg-12)=-146.303css=1+0.554sin(2t-146.30|G(jw)|=

()

s2+s+ jw(jw+1)1

ww2+

wc=g=1800+f(wc

g= g,若在相角 下，使K=10，則此時的a值應(yīng)為多少？|G(jw)|=

()

s2+s+ jw(jw+1)1

ww2+

wc=g=1800+f(wc

g= g,若在相角 下，使K=10，則此時的a值應(yīng)為多少？|G(jw)|=

()

s2+s+ jw(jw+1)1

ww2+

wc=g=1800+f(wc

g= g,若在相角 下，使K=10，則此時的a值應(yīng)為多少？0.781a0.7812a2+

()

s2+s+

a=

G(s)=w 'w w'

a2

w=1 awcg=1800+f(wcwc

'=

g=w'=1800-900-tg-1c=a2.求相 第五章求出r(t 2sin10t時的