自动控制原理课后答案

1、.第五章 线性系统的频域分析与校正习题与解答 C12Curucuruc2图5-75 R-C网络RK 2R R1RC1R R C1U (s)cU (s)RK ( s2解 211 1T s 111RrsC T1R 12R R112R 112sCU (j ) K j )R j R R Cc21211G (j ) 1 U (j ) R R j R R Car121211R R CU (s)cU (s)s1sC2222T s 11 T (R R CR R r2212sC12U (j )1 j R C1 j c22G (j ) U (j ) 1 j (R R C 1 br122 用时，系统的稳态输出c (

2、t)e (t)sst rt) rt)sin(t 30)2cos(2t 45)1(s) 解 图5-76 系统结构图s212(j ) j242j 2 4.1(j ) 4 2 ( ) arctan()21s1 (s) ,1G(s) s2e12 (j ) ,(j ) arctan arctan(则)2ee42sint 2rt) m12(j arctan( ) 45(j )则2285 (j )e822(j2) arctan 18.46ec r (jt )t )ssme r (jt )0.79sin(2t )ssmee 1rsin(t 30)2cos(2t 45) 当 rt)11 r 2m2251( (

3、) 26.5521 ( ( ) 18.453ee(j c t) r ( t ( r (j t smmt 0.4sin( 3.4) 0.7cos(2 90)t e t) r( t ( r(j t (jsmeemeet 0.63sin( 48.4) 1.58cos(2 26.6)t .h(t) 11.8e 0.8e(t 0)4t9t1 1.80.81C(s) ,R(s) 解s s4 s9 s(s4)(ssC(s)R(s)(s) 则(s4)(s9)(j ) (j j 9) G(s) K/sG(s)G(s) K s/(2)23 K s(3)/K K G(j) ej( )2解j0,G(j G() 0,(

4、) 2KK0,G(j G() 0, ( )3KKG(j )ej()(3)2(j )330,G(j G() 0,.() 2 s(2ss 0.5sG(s)H(s) 2 ( )。0.5 2 和 ) (j )H(j ) 解G j0.5 )j j2 210( ) 21(2 ) ) (0.5 )2220.5 ( ) 90arctan2 arctan1 2 17.8885 (2) 0.3835AA 153.435(2) 327.53 5G(s)(2s8ss)G(s)s25G(j )解 2 ) )2 21G(j )tg 8 tg11 tg 212G(j )00G(j ) G(j )G(j )G(j )1800

5、G(j ) .x 10843210-1-2-3-4-1012Real Axis345-9-8-7-6-5Real Axis-4-3-2-10 x 1014 12 G(j )2G( j ) tg18001G( j ) ,G(j )1800 G( j ) 0,G(j )900 K(T sG(s) K,T ,T0；2sT s1211 G(j ) 180 ， 当 G( j )G(j )。KT s解 G(s)2sT s1KT s(s) GG(j )2sT s01G (s)G(s)0.lim ( )sG s K e1 K 1K 1。K，vs0G( arctanT 90arctanT 18021T Tarc

6、tanT arctanT 121TT1212TT 112 jT jT ) 1TT jT T )T T )G( 0.5121221121T1T1T222122T T 1T T T 12 T 022221222T T T 2T (T 032222222T 2 T 1 T 0.5 K 1，。2121 j(j ) Gj j0.5 ) 10G(s) s(ss 2解 G(j ) 0G(j0 ) 90G(1 ) 135G ) 315.G(j) 0 s 2G(s)G(s)G(s)G(s)G(s)；(2s8s200；s (ss240(ss(s0.2)(s s220(3ss (6ss 4s25)(10s228(s

7、s(s ss 4s222G(s)解 (2s8s .200G(s)s2(ss 40(s100(2sG(s) s(s0.2)(s ss2s(s s20.2 .20(3sG(s)s (6ss 4s25)(10s22sG(s) 4s2s (6s s s2 5 0.8 1s18(s25 0.1G(s)s(s ss 4s2221 4s(s s s s25 25 .KG(s) G (s)sv0G (s)为G(s)01 Kv1 1K证 。svK1 Kvsv1KK120lg 01, 故 K, Kvv(j )v11v1 图 578 511题图K解 G(s)ss( 1220lgK L( ) 40，K .100G(s

8、)11( s s12 sK( 1 G(s)K 2s01Cs ( 22 KsG(s)ss( 231 , K 20lgK011. G (s)G(s)和G (s)123G (s)G (s)2G (s)411G (s)G (s)23L 1K ( ) 20lg1K 1801G (s) K11KsG (s)22s(0.8K21lgK / 20lg0K 120,22L ( )320lgK20lg 0K331 9 , G (s) K s 9sK0.111333GG1G s ( )21G G42318s(0.125sG ,G ,G 代入得：G (s)将1234. 解 题号Z备注PNK102123K23s(Ts

9、T s 0000212KG(s) s2(Ts.K(Ts40002122G(s)0000K(TsT sG(s) 12s35601234K8101K9110121KG(s) s(Ts KG(s) (K,T ；ss2TKK10T K,T 2KK T) jT ) X( )Y( )G(j ) 解 T)2 2j j jT )210 令 Y )X )T1KTX() 11TT1或TK 132TK；2110 0 T；9K,T .10(s22s(s2)(sG(s) 解 G(j )G(j50180G(j)100G(j )210(5 j2 )G(j )(2 j )(0.5 j )222210 )3 j (5.53.5

10、 )222 ) ) 0令G(j ) / 0Re G(j )4.03701Z P2N 12( )22 1s3, H(s) G(s)ss)2(s2.s2 ( ) ( ) G (s) G s H s解 (s400G(j) 01800G (j )G (j )00Z P N 0 2 0 02000P Z 00 Z P2N Z 2N 02( 21 10G(s)s(0.2s 0.8s2解 210100.8 j0.2 )G(j ) 22j j0.2 j ) 0.04 )G(j )G(j 180 270G(j0 )G(j) limReG(j ) 80.T 1G(j ) ( )T 121Z P2N 12( ) 2

11、2 G(s) s2s(s 4G(j )的0解 G(j G(j2 ) 153.4G(j0 )90 333.4G(j2 )G(j) 绘出幅相特性曲线G(j )Z P2N 02( 2. 100G(s)G(s)G(s)s(0.2s50(0.2ss2)(s10s(s25ss100( 2G(s) sss(s 10 20100100G(s)解 s(0.2sss( 5 5 Cg180 G(j ) 180 90 tg 0.2 12.600010C1hG( )g.5050G(s)(0.2ss2)(sss( ( 2s52 6由Bode图得：c50G(j ) 16.3 cc5 2cc1 g5110G(j ) 2 g2

12、gggC5C2001110180 G(j )180 tgtgtg 2 29.4C( ) 1 ( ) 1 (2 ) 1g2g22152gh0.391G( )50g.10s(s25s10G(s)sss( ( 104 410 6.325 410 6.325 00Cgh 1.s100( 2ss1020cg 180 ( ) 24.8ch 0.343 9.3(dB) .as1G(s) s21(a )2G(j )( a )解1021a22A( )1c2c a 1422cc( ) 00c ( ) tg a 180 45 180 1351000cc 1c a0.84。，c 10, H(s) 1K sG(s)s(

13、shK 。h解 K s)G(s)H(s) hs(s10(K j hG(j )H(j ) j (j . 1tg K 180( )c 90 180 tg1000chc1 900tgtg K1chc K chc 1 Kchc1K 20hc1K h2c c 0.1KhK j )h( )( )G(j )H(j ) u j (j K )10(K2u( ) ; v( )hh ( (22v()0 01 KK 令, 则 10(22hh1Kh K )1u()h ( 21K ) (Khh K 1 010K92hh9 1K K ,K 110hhh 10故当K 110hKes G(s)s1.KG(j )ej解1KG(j )121()ej tg ()11KG(j )1即12G 0 ( ) ( ). 1 0.8tg1tg(tg1 ) tg(08 08 . ) tgK1 :1(0.8)2 K 1tg(0.8)2 sec0