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。 3-130<k<17 3-14τ>0τ≠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