自動控制原理第五版課后答案完整版-2

1、rc rrrc 。rrrd222；d3d23632；dct)t)t；t；2；ct) rr (t)2tcos t c(t)ar(t) 中R u2RR CC s (RC R C RC )s12o212112212Ru2i1212112212R 12C s2R 1C s1即00i1222112212o121221122if ( o11io2f (22oi111221112211221KR1C112C2212 設初始條件均為零，試用拉氏變換法求解下列微分方程式，并概略繪制x(t)曲線，指x(t)2x(t) x(t) (t）。圖 RU U( )U 1iR Ro11oo1UUR2oR2010 和U2,可得

2、：o12323io112o212 所示。已知電位器最大工作角度o ( )/ ( )ss 0i。EK 1100m12KmU (s) T 1am。1t Kt(s)tUUUKsKK1K2Kaiso123tKt1soTss22mti0123m0123m2tt，1s12122111t2tC(s)2112C(s)GG32GH11G1G3121G1N(s)C s( )G3121G1G3121G11213212112121GG12RCGG24G32RCG2342GGG42323124342434423G4243 數(shù) 。4圖 題 1G H G G H G G H312323431123451P G , 262G

3、 12351G H G G H G G H631343232121242363434545123456567345657186587186598416121323728292L L L123P=G G G G G G , =1 ; P =G G G G G , 2=1 ;11234561273456P =-G H G G ,3=1+G H ; P =G G G , 4=1+G H37186424186424kkk1 961 L L L LL Labc1 h(t)1012.5e1.2t t53.1 )0t時p p)pph(t)1012.5e1.2t t53.11012.5e (cos53.1 s

4、in1.6tsin53.1 cos1.6t)000h(t)15e1.2tte1.2tte1.2t sin1.6t001.6t t ( )h t0pp0ppt0.95h() h(t )1.05h()s ps1s(s0.8)K s1t1s1s 25K K )s25K21t1t1s(s2K 1.44,K 0.31nn1t解 44005s 5s 10 04220s310s 0ssss1004222D(s)/輔助方程，3,456G(s) s2(s2 6s100)r(t) 2t和r(t)22tt22s21050s 105000，r(t) 2當當當1p02Kv22te 2K0ss3a，rtess2232s2

5、 1s0 2，220r(t) 2當當當1p22 02Kv22t2e 2K2ss3a0r 2te2d2c t t r 2 2t tT122T22121RukCk12s1B1T s12E(s) R(s)C(s)2 (s)11(s)11e212112212TT s T T )s sK K32121212212sse32212121K K T2 21211k)0令122o22o2KKG(s)s(0.2ss s(s2)(s，K* 10K7(2K ,3333 a，a；111 0 ；2*K*0.5K， ，111d 0.29 d 1.707；12 K (s z)G(s) s2 K 令s jD(s)0ReD(1

6、200 K z 0 ImD( 30 K 0*，K* 30,z 6.63 。G(s) 333a，a；111d d d 12KcD(s) s 150s 5000s5000K32令s D(s)0ReD() 5000K 0 ImD() 5000 023，5000 01,23c將bD(s) s 4sbsb20 s 4s20b(s4)022*2 有2(,4111d 24j d 24j d 4d 8d 402d 1180 arctan290 1350001D(s) s(s10)30(sb) s 40s30b 02bb*2 有240,0 110d d d 20j sh(t)H(s)(s)R(s)H(s)R(s)(s)1則令R() (21G(j)oK 1T 1 ( ) T21T T021e ()E(s) K1T2當100211210(j jj (Ts s s123Ks(Ts s2122K12561234K s（2P 0,N 0Z P2N 020002P 0,N 02P 0,N 01P N 21Z P2N 12 020P 1,N 0Z P2N 120111PN 2122 as1e02)1c2c42即cc ) 00c ) 0a 000ccccK%20%KK；