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 = 12.11y 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) − − − − =
C(s) 2-22(a)R G G2G3GAGs 1+Gah+G,Gh +G,g (b)9个单独回路 -G,HL=-GhL=-GhL=-G, h Ls=-G,G, h 5 -G G3 G4GsG6H5, L7=-GG8G6H5, L8=g,H G8GH5, Lg=GHH 6对两两互不接触回路:L1L2LL3LL3LL2LsL2LL2 三个互不接触回路1组:L1L2L3 4条前向通路及其余子式:P1=G1G2G3G4G3G6,△1=1;P2= GG3G4GSG6,△2=1 P3=-G7HIG8G6, A3=1+GH2; P4=G G8G6, A4=1+G4H2 Cs ∑P△k R(S)1-∑L2+∑L1L-L1L2L3 C(s)590 =15.128 C(s) abcd +ed(I-bg R(s)39 R(s 1-af-bg-ch-ehgf +afc C(s) bcde +ade +(a+ bc(1+eg le(1+cf)-lehbe -leha RI() 1+cf +eg+ bede+cefg +adeh r2(s) 1+cf +eg+ bcdeh+ cefg+ adeh (o) CS)=lahd-1g)+aej+ aegI+(bdh * bdej*+ b degi)+ (cdh t efdej+ ci) g-fg C(s) fdh+fdej+i+fj C(s) h(l-fg)+ej+egi R2(s) 1-f deg-fg R3(s) 1-fdeg-fg
2-22(a) （2-22题） 3 1 3 4 3 2 3 2 1 2 3 4 5 6 1 G H G G H G G H G G G G G G R(s) C(s) + + + = + (b) 9个单独回路： 1 2 1 2 4 2 3 6 3 4 3 4 5 4 L5 G1G2G3G4G5G6H5 L = −G H ,L = −G H ,L = −G H ,L = −G G G H , = − 6 7 3 4 5 6 5 7 1 8 6 5 8 7 1 8 6 5 L9 G8H4H1 L = −G G G G G H ,L = −G G G H ,L = G H G G H , = 6对两两互不接触回路： L1L2 L1L3 L2L3 L7L2 L8L2 L9L2 三个互不接触回路1组： L1L2L3 4条前向通路及其余子式： P1=G1G2G3G4G5G6 ,Δ1=1 ; P2=G7G3G4G5G6 , Δ2=1 ; P3=-G7H1G8G6 ,Δ3=1+G4H2 ; P4=G1G8G6 , Δ4=1+G4H2 ; = = − + − = 9 a 1 6 1 a b c 1 2 3 4 k 1 k k 1 L L L L L L P R(s) C(s) (c) 15.128 39 590 R(s) C(s) = = (d) 1 af bg ch ehgf afch abcd e d(1 bg) R(s) C(s) − − − − + + − = (e) 1 cf e g bcdeh cefg adeh bcde ade (a bc)(1 e g) R (s) C(s) 1 + + + + + + + + + = 1 cf e g bcdeh cefg adeh le(1 cf) lehbc leha R (s) C(s) 2 + + + + + + − − = (f) 1 f deg fg [ah(1 fg) aej aegi] (bdh bdej bdeg i) (cfdh cfdej ci) R (s) C(s) 1 − − − + + + + + + + + = 1 f deg fg fdh fdej i fj R (s) C(s) 2 − − + + + = 1 f deg fg h(1 fg) e j egi R (s) C(s) 3 − − − + + =
3-h()=1T-τ e 3-2(1)k(t)=10h(t)=10t 25 32(2)k(t)=esin4th(t)=1-esin(4t+53.13) 0.0125 3-3(1)@(s)= (2)Φ(s)=+ 5√50(s+4 s+l.25 (3)m=、0,1 s2+16 s(3s+1) 3-42=06 0%=948%tn=196t,=297s 35r=1.00660n.=15=0.5x=2.5B= l.686 t=145tn=3.156t,=601330%=199% 3-65=143an=2453-7k1=144k2=0.31 3-8(a)=0an=1系统临界稳定 s+l s+界+!5=0.5m,=1σ%=29.8%t=7.5ls (c)(s)= s4+s+1 5=0.5an=1σ‰=163%t,=8.08 3-9(1 (b)比(c)多一个零点附加零点有削弱阻尼的作用。 G(s)= k=55 =√10a%=3509%t=3.5s 0.2 S(0.5s+1 (2)G(s)= 10(0.Is+1 k=105 10z=10 B4=.249 S(S+1) 10 y %=37.06%t,=3 0.1
（3-1题~3-9题） 3-1 T t h(t) 1 e T T − − = − 3-2 (1) k(t ) = 10 h(t ) = 10t 3-2 (2) e sin4t 4 25 k(t ) −3t = e sin( 4t 53.13 ) 4 5 h(t ) 1 3t o = − + − 3-3 (1) s 1.25 0.0125 ( s ) + = (2) s 16 50(s 4 ) s 5 (s) 2 2 + + = + (3) s( 3s 1 ) 0.1 ( s ) + = 3-4 0.6 2 % 9.478% t 1.96s t 2.917 s = n = = p = s = 3-5 r 1.0066 1 0.5 z 2.5 1.686 = n = d = = = 2 = − t r = 1.45s t p = 3.156s t s = 6.0133s %= 17.99% 3-6 = 1.43 n = 24.5 3-7 k1 = 1.44 k2 = 0.311 3-8 (a) = 0 n = 1 系统临界稳定 (b) 0.5 1 % 29.8% t 7.51s s s 1 s 1 (s ) 2 = n = = s = + + + = (c) 0.5 1 % 16.3% t 8.08s s s 1 1 (s ) 2 = n = = s = + + = (b)比(c)多一个零点,附加零点有削弱阻尼的作用。 3-9 (1) 10 % 35.09% t 3.5s e 0.2 10 1 k 5 s( 0.5s 1 ) 5 G(s ) = = n = = s = ss = + = (2) 10 z 10 r 1 1.249 10 1 k 10 s(s 1 ) 10( 0.1s 1 ) G(s ) = = n = = = d = + + = % 37.06% t 3s e 0.1 2 = − = s = ss =
3-11劳斯表变号两次,有两个特征根在s右半平面系统不稳定。 3-12(1)有一对纯虚根:S12=±j系统不稳定 (2)S:=土j2S3=土1s=1s=-5系统不稳定 (3)有一对纯虚根:s/2=±j5系统不稳定。 3-1300τ≠0 3-15(1) k=20e=0es=0 (2)k=10e,=0.2e=0 (3)k=0.1en=0en=20 3-16()kn=50k,=0k,=0(2)k,=∞k,=kk.=0 (3) k =oo k=I 3-18(1)es=0(2)esn=0(3)esm?=0 3-20R STS+1) ♀→由题意得:E(s)=R(C(s) B k,< T1+T2 T,s+1 k,(8+T
3-11 劳斯表变号两次，有两个特征根在 （3-11题~3-20 s右半平面 题） ,系统不稳定。 3-12 (1) 有一对纯虚根： s1,2 = j2 系统不稳定。 (2) s1,2 = j 2 s3,4 = 1 s5 = 1 s6 = −5 系统不稳定。 (3) 有一对纯虚根： s1,2 = j 5 系统不稳定。 3-13 0 k 1.7 3-14 0 0 3-15 (1) k = 20 es s = es s = (2) k = 10 es s = 0.2 es s = (3) k = 0.1 ess = 0 ess = 20 3-16 (1) (3) k 50 k 0 k 0 (2) p = v = a = k 0 200 k kp = kv = a = kp = kv = ka = 1 3-18 (1) essr = 0 (2) essn1 = 0 (3) essn2 = 0 3-20 k1 s(Ts 1) k 1 2 + T s 1 1 2 + R u B C 由题意得：E(s)=R(s)-C(s) 2 1 2 1 2 1 2 o 2 k TT T T k k ( T ) 1 + +
(4-4题45题) 4-4(1) k=7 4-4(2) 44(3) 10 1.707 -0.293 -0.9 0.88 Root Locu Root LocuS 4-5(1) 45(2) -4.236 n=士135°
（4-4题~4-5题） -0.88 j 10 k =7 -0.293 -1.707 -0.9 -4.236 o p =± 135 o p =0
「A∞/2 4-8 0=1.036 4-6(1)k=11 k=73.2 ±jy k*=260 k=30 6p=±9273 -2+jy6 199 0,404 Z 663 30 Root Locus 200 70.7 4-9 k=96 150 4-10 k。=150 100 -3.29 100 21.13 s6n=±45°,±1350 k。=9.62 200 300-250 00-150-10050
（4-6题~4-10题） 4-6 (1) k =11 (2) k*=30 6.63 30 199 z = = -0.404 k*=73.2 =1.036 o p =± 92.73 -2 + j 6 -2 ± j 10 k*=260 -3.29 j 21 k*=96 o o p =± 45 ,± 135 kc =150 ko =9.62 -21.13 =70.7