自動(dòng)控制原理簡(jiǎn)明教程第四章習(xí)題答案

G(S)

S+S(S+2)(S+a從0fi￥

￡G(S)

S+S(S+2)(S+S+j(S)

G(S 1+G(S

S(S+2)(S+ S+S(S+2)(S+

S+S3+6S2+9S+

S3+6S2+

令G(S)= = S3+6S2+ S(S+a從0fi￥m01 及及0

-

+sa= =n-

：1+ + = d+ d+解得：d1aSS32(-1)·4

2

令S=jw(jw)3+6(jw)2+9(jw)+a= a=- - G(S)

S+

a(S S(S+2)(S+

8S(S+1)(S 8ess

=1=8￡

a16a ￥變化的根軌跡

G開(S

(S+2)(TSS(Sj(S)

G開(S)1G開(S)

=S

(S+2)(TS+2S+2+TS(S+D(SS22S2+TS(S2)1+TS(S+ =S2+2S+G(S)=TS(S+ S2+2S+

TS(S+(S+1+j)(S+1-\開環(huán)零點(diǎn)：m2Z10Z2n12：1+ d+ d+1+ d+1-n mm

=

i=1d-解得：d

i=1d-

q=–180=–

=

T S+S+1+

S+1-

=1T=1(-1,j- -1

0(-1,-j這即為T從0 \0T<1S1S21系統(tǒng)此時(shí)為臨界阻尼T=1。(S+2)(Sj(S)= S(S =1+(S+2)(SS(S

(S+2)(SS(S+1)+(S+2)(S=(S+2)(S+1)=(S+2)(S2(S+1)(S 2(Sk*(S+3)k*(S+3)--k*(S+G(S(S2)2S-13

-

+sa= =n-

=f=(2k+1)p=– n-

+

+ =d- d+ d- d+3333

d2=-2

q=–180=–

= mq=1[(2k+1)p+

P-Z-Pl

j - - -

-

13333

k*-2++33 =133-2++ -2+-3k*3(3-3)31+

=1k*=33k*>330￡k*￡3(3-3)31+0<k*<3k*4k*￡3(3-3)3 1+G(S)

(S+a)S(S+1)2繪制系統(tǒng)的閉環(huán)根軌跡圖（a0~￥當(dāng)r(t)

S3+2S2+S+S+a=S3+2S2+2S+a=

1+S+ =0-S(S-

S3+2S2+\ (S)

S(S2+2S+ S[(S+1)2則開環(huán)極點(diǎn)n310P21j31m

-

-1+j-1- sa= =n-

=-f(2k+1)pp和 n- Sjw(jw)3+2(jw)2+2jw+a=\a-2w2=02w-w3=

2a=2Q存在復(fù)數(shù)極點(diǎn) Pq =(2k+1)p+P-P

=1800-1350-900=-\ =45033r(t=1.2t

1p 3- 3

0

G(S)

S+

a(S=S(S S(S

= =1.2￡0.6a? a?0<a<

2￡a<

+2S2+2S+a=0a=[S-(-1)](S-S2)(S-S3)=S

+2S=S3+S2+S2+2S+1=S2(S+1)+(S+1)2=(S+1)(S2+S3(SS2SS3S3

+S

=-1+323

=-1-

j(S)

(S+

S(S-S1)(S-S2)(S-S3

(S+1)(S--1+2

3)(S--1- 32G(S)

4(S+k)S(S+1)(S+試?yán)L制系統(tǒng)的根軌跡：(k0fi￥)-G(

4(S+k S(S+1)(S+ D(S)=S3+6S2+9S+4k= (S) S(S+1+S(S+m=n=3,=0,=-3,P3=-

-

-3-sa= =n-

f=(2k+1)p=–p n- 1

+ = d+ d+

=–

= (jw)3+6(jw)2+9jw+4k=4k-6w2= 9w-w3=

k- -2

- -3圍為0k13.5，而臨界阻尼時(shí)，也即分離

=-(-1)(-1+3)2+4k=0k=(S-S3)(S+1)2 =S(S+3)2+(S-S3)(S+1)2=(S+4)(S

=S(S

+4(S(SS3S4S34（另外一個(gè)閉環(huán)極點(diǎn)）j(S)

4(S(S+4)(SS SK0~￥變化的根軌跡（要計(jì)算分離--

kG(SS(S2j)(S2根軌跡方程：1+G開(S0或G開(S1均可S1=0,S2=-2-j,S3=-2+j,n=

-

(-2+j)+2- sa= =n-

=-f=(2k+1)p=–p n- 分離點(diǎn)：1 = d+2- d+2+

=-8–6

d1=-

d2=-

q=–180=–

= Sjwk* S(S+2+j)(S+2-S34S25Sk*-jw3-4w2+5jw+k*=\k*-4w2=5w-w3=

5k*=5 =(2k+1)p+

-Zi-Pl- \可畫出根軌跡圖 --1 - -2-k* S(S+2+j)(S+2-d1處的k*在分離點(diǎn)d2=-1.67處，代入特征方程中得d2 的k*=1.85k*

1.85￡k*￡21.85￡k*￡0<k*￡S1k

=-1k*=j(S)=G開(S 1G開(S

kS3+4S2+5S+k(SS1)(SS2SS3

S3+4S2+5S+(S+1)(S-S2)(S-S3)=S

+4S

+5S+則(SS2)(SS3S23S2(S2)(S\S2=- S3=-

S3=- 根之和 = n

(前提nmi s=(-1)n常數(shù)i\s1+s2+s3=-

S3=-kG(S)

S(Sk*0~￥

1f(2k+1)pp 3 sa= =- 1

=

d =– =

S3+2S2+S+k*=-w3+w= -2w2+k*=

或k*=

（舍去k*=1--- -- -1r(t=1(tjj0123

K*(S-(S-2)(S-3)(S+設(shè)特征方程為：(S2)(S3)(SaK*(S-1)令S=0，代入K*\a=10S3+5S2+(K*-44)S+60-K*=-jw3-5w2+(K*-44)jw+60-K*=-w3+(K*-44)w=-5w2+60-K*=