《自动控制原理》课程教学资源（PPT课件讲稿）经典控制理论习题答案

经典控制理论习题答案 (联系地址: frasch)

经典控制理论习题答案 （联系地址： xwfr@sohu.com）

习题答案 (仅供参考,不对之处敬请批评指正,谢谢!) (说明:书中有印刷错误的习题,在此无法给出答案,请谅解。28章习题答案联系人王凤如,xwfr@sohu.com) 2-2(a)min(t)=fl[;(t)-x()-/2x() 即:mx0(t)+(f+f2)x0(t)=f1x;(t1) (b)f(k+k2)o(t)+k k2xo(t)=k2fi(t (c)o(t)+(k1+k2)xo()=()+k1x(t 2-3(a)R1R2C1C2i0(t)+(R1C1+R2C2+R1C20(t)+u0(t) =R1R2C1C2i1(t)+(R1C1+R2C2;(t (b)f1f2o(t)+(fikit fik2+)xo(t)+kikyo(t) =ff2x;(1)+(fk2+f2k1)x1(t)+k1k2x2(t) 2-4(a)rr, Cuo(t)+(R/+r2 uo(t)=rr Cui (t)+Rui(t) (b)R'CC2io(t)+(RC2 +2RCuio(t)+uo(t) RCIC2ui (t)+2RCui(t)+u,(t)

习题答案 2-2(a) ( ) [ ( ) ( )] ( ) 1 2 mx t f x t x t f x t o i o o  =  -  -  即： mx (t ) ( f f )x (t ) f x (t ) o 1 2 o 1 i  + +  =  (b) f ( k k )x (t ) k k x (t ) k fx (t ) 1 2 o 1 2 o 2 i +  + =  (c) ( ) ( ) ( ) ( ) ( ) 1 2 1 fx t k k x t fx t k x t o + + o = i + i   2-3(a) R R C C u (t ) ( R C R C R C )u (t ) u (t ) o + + + o + o   1 2 1 2 1 1 2 2 1 2 (b) f f x (t ) ( f k f k f k )x (t ) k k x (t ) 1 2 o + 1 1 + 1 2 + 2 1 o + 1 2 o   2-4(a) R R Cu (t ) ( R R )u (t ) R R Cu (t ) R u (t ) 1 2 o + 1 + 2 o = 1 2 i + 2 i   (b) R C C u (t ) ( RC RC )u (t ) u (t ) o + + o + o   1 2 2 1 2 2 （仅供参考，不对之处敬请批评指正，谢谢！） （2-2题~2-4题） ( ) 2 ( ) ( ) 1 2 1 RC C u t RC u t u t = i + i + i   f f x (t ) ( f k f k )x (t ) k k x (t ) = 1 2 i + 1 2 + 2 1 i + 1 2 i   R R C C u (t ) ( R C R C )u (t ) i i   = 1 2 1 2 + 1 1 + 2 2 （说明：书中有印刷错误的习题，在此无法给出答案，请谅解。2~8章习题答案联系人王凤如，xwfr@sohu.com）

2-5(1)运动模态:et x()=t-2+2e 0.5t (2)运动模态:e" sin 3t x(t)=2e"'sint (3)运动模态(1+t)et x(t)=1-(1+t)e 2-6△Q=△P 27△F=12.II 2-8 Med =-ed (sina(a-do) s2+4s+2 2-9Φ(s)= k(t)= dc(t) δ(t)+2e . 2t e (S+1)(S+2) dt 2-10零初态响应c1(t)=1-2e-+e-2t 2t 零输入响应c2(t)=e1-2 总输出c()=c1()+c2(t)=1-4e+2e2

2-5(1) 运动模态： 0.5 t e − t x t t e 0.5 ( ) = - 2 + 2 (2) e sin t 2 −0.5 t 3 x(t ) e sin t 2 0.5 t 3 3 2 3 − = (3) (1+t)e -t t x(t ) 1 ( 1 t )e − = − + 2-6 Q P 2Qo 2 k  =  2-7 F = 12.11y 2-8 e E (sin )( )  d = − do o  −o 2-9 (s 1)(s 2) s 4s 2 (s) 2 + + + +  = 2t t (t) 2e e dt dc(t) k(t) − − = =  + − 2-10 零初态响应 c (t) 1 2e e 2t t 1 = − + − − 零输入响应 2t t 2 c (t) e 2e − − = − 总输出 t t c t c t c t e e 2 ( ) 1 ( ) 2 ( ) 1 4 2 − − = + = − + （2-5题~2-10题） 运动模态： 运动模态：

C(s)100(4s+1) 2-11 E(s)10(12s2+23s+5) R(s)12s2+23s+25 RO 12s2+23s+25 2-12( (s) R (R1C1S+1)(RCS+10) =R,①.C+D)(b)U() R. CS (b) U(s)R1(R2C2S+1) ;(s)R。(R1+R2)C2S+1 2-13 RR RoR,CCS+rcas+R,R 2-14 Q K I s+1 a(s) TmS+ △u u 2-15 32 s(TmS+1) 11 3126 (s) TmS"+(1+3k3k, km)s+31.26k3k

（2-11题~2-15题） 2-11 12s 23s 25 100(4s 1) R(s) C(s) 2 + + + = 12s 23s 25 10(12s 23s 5) R(s) E(s) 2 2 + + + + = 2-12(a) R C s (R C s 1)(R C s 10) U (s) U (s) o 1 1 1 o o i o + + = − (R C s 1) (b) R R U (s) U (s) o o o 1 i o = − + (b) (R R )C s 1 (R C s 1) R R U (s) U (s) 1 2 2 2 2 o 1 i o + + + = − 2-13 2 1 2 3 o 2 1 1 2 3 o 1 2 i o R R C C s R C s R R R R U (s) U (s) + + = − 2-14 T s 1 K U (s) (s) m 1 a m + =  T s 1 K M (s) (s) m 2 a m + = −  2-15 11 1 3 2 k3 s(T s 1) k m m + 3 k st 11 1 i o ui uo u u1 u2 ua ut 3 t m 3 m 2 i m o T s (1 3k k k )s 31.26k k 31.26 (s) (s) + + + =  

C(s) G+G2 2-17(a)R(s)1+G2G3 (b)=a+2 1111 C(s) (G+G3)G R(S)1+G2H,+G G2H,(d) R(S)I+GH,+G2H2+G3H3+G,HGWs (e) Cs)_Sx1+02H1-G GH, +G,GH2 (f) C(s)_(G,+G3)G2 R(S) R(S) 1+G GrH C(s) G,G2 2-18(a)ws)1+G1G2+G1G2H1 C(s)G3G2-(1+G1G2H1) NS)1+G,G2+G GrHI b) C(s)-(+G1)G2 G4+G3G C(s) G R(s)1+G2G4+G3G4 N(s)1+G2G4+G3G4 2-19与2-17同2-20与2-18同 2-21(a)R=1+,且1+G2H2+6M1B2+6,HG,H2 E(S (1+G3H2)-GG;H2H1 R(S)1+G,H,+GH,+GG,G,H,H2+GHG,H (b)C(s) G1-G2-2G1G2 C(s) 1-G1G2 R(s)1-G1-G2-3G1G2 R(s)1-G1-G2-3G1G2

（2-17题~2-21题） 2-17(a) 2 3 1 2 1 G G G G R(s) C(s) + + = (b) 1 1 1 2 1 2 1 2 1 G H H H G G (1 H H ) R(s) C(s) − + + = (c) 2 1 1 2 2 1 3 2 1 G H G G H (G G )G R(s) C(s) + + + = (d) 1 1 2 2 3 3 1 1 3 3 1 2 3 1 G H G H G H G H G H G G G R(s) C(s) + + + + = (e) 2 1 1 2 1 2 3 2 1 2 3 4 1 G H G G H G G H G G G G R(s) C(s) + − + = + (f) 1 2 1 1 3 2 1 G G H (G G )G R(s) C(s) + + = 2-18(a) 1 2 1 2 1 1 2 1 G G G G H G G R(s) C(s) + + = 1 2 1 2 1 3 2 1 2 1 1 G G G G H G G (1 G G H ) N(s) C(s) + + − + = (b) 2 4 3 4 1 2 4 3 4 1 G G G G (1 G )G G G G R(s) C(s) + + + + = 2 4 3 4 4 1 G G G G G N(s) C(s) + + = 2-19与2-17同 2-20与2-18同 2-21(a) 1 1 3 2 1 2 3 1 2 1 1 3 2 1 2 3 4 3 1 1 1 G H G H G G G H H G H G H G G G G G (1 G H ) R(s) C(s) + + + + + + = 1 1 3 2 1 2 3 1 2 1 1 3 2 3 2 4 3 2 1 1 G H G H G G G H H G H G H ( 1 G H ) G G H H R( s ) E( s ) + + + + + − = (b) 1 2 1 2 1 2 1 2 1 G G 3G G G G 2G G R(s) C(s) − − − − − − = 1 2 1 2 1 2 1 G G 3G G 1 G G R(s) C(s) − − − − =