控制工程基础---作业答案

作业P81-3,1-4。

1-31-4(P72)2-1，2-2。

2-1-a22,u u u u u u i c i c -==+dt du dtdu dtdu c CCCi i c 2-==dtdu dtdu c RCRCR i u i 22-==dt du dtdu i RCu RC=+222-1-b221Kx BBdtdx dtdx =-dtdx dtdx BKx B122=+2-1-c2u u u i c -=2212)()(1212R CCR i i u R u R u dtdu dtdu R c i i -+-=+=i dt du dt du u R C R R u R R C R R i 22121221)(2+=++2-1-d22211121x K x K x K B B dt dx dt dx =-+-1122112)(x K B x K K B dt dx dt dx +=++2-1-e1211R u R u R i i -=⎰⎰⎰-+-=+=dt u dt u u u dt i R i u C R iC R R R i R R R CR 211211212111212i dt du dt du u C R u C R R i +=++22212)((P72)2-1，2-2 2-1-fdtdyBy K X K y K x K x K x K =--=-2222222111,122121)1(x x y K K K K -+=dtdx K BK dt dx K K B x K x K K x K 121221)1()(1122122-+=-+-111212112)(x K BKx K K K B dtdx dt dx +=++2-221212111311)()(dt x d dtdx dtdx dtdx mBB x K t f =----22222122223)(dt x d dtdx dtdx dtdx mBx K B =---22322322221)(x K B B m Bdtdx dt x d dtdx +++=313222212113311)()(dt x d dt x d dt x d dtdx dtt df m B B B K=++--22222223233312222313232222322)(])([])([dt x d dtdx dt x d dt x d B B B dtdx dt x d B K dtt df B KB B m x K B B m ++++-+++-+2223213233321424321)(dt x d B K m dt x d B B B m dt x d B m m ++=+222323424][)(12323121213212312121dt x d dt x d dt x d K m B B B B B B K m B m B m B m B m mm +++++++++dt t df dtdx B x K K B B K B B K )(32213213122)]()([=+++++（P72）2-3:-2)、-4)、-6)；2-4:-2)。

2-3:-2)21)2)(1(321)(++++++==s A s A s s s s G2})1{(1)2)(1(31=+=-=+++s s s s s A1})2{(2)2)(1(32-=+=-=+++s s s s s A2112)(++-=s s s G0,2)(2≥-=--t e e t g t t2-3:-4)41)1()4()1(13022012)(+++++++==s A s A s A s s s G 311)4()1(1201})1{(2=+=-=++s s s s A911)4(11)4()1(1202}{]})1[({22-=-=+=-=+-=++s s s s s ds ds A914)4()1(132)4{(=+=-=++s s s s A41911191)1(1312)(++++-=s s s s G0,)(4919131≥+-=---t ee te t g t t 2-3:-6)14)1)(4(10423212)(+++++++==s A s A s A sA s s s s G 5.2}{0)1)(4(1012===++s s ss sA2})1{(1)1)(4(1042-=+=-=++s s s s s Ajs js s s s A s A s 2322)1)(4(102)(})4{(2==+++=+32)12(2102A j A j j +=+3222A j A j +=--2,3212-=-=A A1122221112425.225.2)(2222221+++++-----=-+=s s s s s s s s ss G 0,22sin 2cos 5.2)(21≥---=-t e t t t g t2-4:-2)使用拉氏变换求解下列微分方程2)0(,1)0(,2cos 5sin 4)()(-==+=-⋅⋅⋅x x t t t x t x41122254)()0()0()(++⋅+=---s s s s X x sx s X s4514222)(2)(+++=-+-s s s s X s s X s4111)4)(1)(1)(1(89610226524321222345)(++++-+++-+++-+--+++==s A s A s A s A s A s A s s s s s s s s s s X 1})1{(1)4)(1)(1_)(1(8961021222345=+=-=+++++-+-s s s s s s s s s s s A 1})1{(1)4)(1)(1)(1(8961022222345=-==++-+++-+-s s s s s s s s s s s A js js s s s s s s s s s A s A s ==++-+++-+-+=+}{})1{(43)4)(1)(1)(1(896102222234543)4)(1)(1(89610222345A jA j j j j j j j j +=+-+++-+-432A jA +=- 2,043-==A Ajs js s s s s s s s s s A s A s 2652)4)(1)(1)(1(8961022}{})4{(222345==++-+++-+-+=+65]1)2)[(12)(12(8)2(9)2(6)2(10)2(2)2(222345A A j j j j j j j j j +=+-+++-+-6522A A j j +=-0,165=-=A A4111111222)(++-+--+=s s s s s s X022≥--+=-t ,t cos t sin e e )t (x tt（P72）2-5:-3)、-4)；2-6；2-9:-a)、b)。

2-5:试求下列传递函数的极点和零点（包括在无穷远处的） -3))1()2(2)(+++=s s s s K s G23213,213,21,0;;2jp p z z ±-==∞=-= -4))2)(1(2)(++-=s s K e ss G2,1;211-=-=∞=p p z2-6: -a)12111211R u R u R u u R i -==-dt du dt du dt u u d C CCCi 212111)(11-==-⎰+++=dt i i R i i u C R CC R )()(11121122+-+-=dtdu dtdu R R R R C R CR u u 211212121221dt CCdtdu dt du R u R u C )(2112112111-+-⎰-+-=2122122121222221dt u d dtdu dtdu dtdu C C R R CR CR CR dtdu dtdu dtu d CR C R u u C C R R 212221111212121-+-+=++++222211121212222)(u C R C R C R C C R R dtdu dt u d 122112121121)(u C R C R C C R R dtdu dt u d +++1)(1)()()(2221112212122112212112+++++++=s C R C R C R s C C R R s C R C R s C C R R s U s U -b)yK BBx x K dty x d dtx x d 1)(1)(2212221)(==+---dtdx K B dtdx K B K K K K x x y 212112121221-+-==-+-dtdx dtdx BBx K x K 212222122221212121212121112121dt x d K B B dt x d K B B dtdx K K B dtdx K K B dtdx B+-+-=++++221121121212222)(x K K K B K B K B B B dtdx dt x d 1211221211212)(x K K K B K B B B dtdx dt x d +++2112112122121122122112)()()()(K K s K B K B K B s B B K K s K B K B s B B s X s X +++++++=2-9: -a)dtdu R u o i Ci 10-==idtdu u C R o -=1s C R s U s U I O 11)()(10-=-b)R u i i =iR R u iR u 011-=-=dt du R R C dtduC i C i 01111-==iR R dtdu R R R C i R R C o u u u R i i u i 0102110221)(---=+--=0210211021211)()(R R R R R R C R R R s R R C s U s U s I O +++--=-=（P74）2-10:a) ；2-11；2-12 ；2-13 。

2-10: -a)1111,R U U R R u u R oi oi I i --==3333,R U R R u R O o I i ==⎰-+-=dt i iR i i u R R CR R o )()(311231)()(311231R R CsR R o I I R I I U -+-=oCsR o CsR i CsR o R R o R R i R R o U U U U U U U 311321212111--+--=--=o i o CsU R R CsU R R CsU R R 323231oo i o U R U R U R CsU R R 13321--+31213231332R R Cs R R Cs R R Cs R R R Cs R R U U io +++++=11313122313+++++=R R Cs R R Cs R Cs R R R R1)(1313122313+++++⋅=Cs R Cs R R R R R RR R2-11 -a)-b)-c)2-122-13bce f ch bg abcd -----=∆1abc P ==∆11,1bce f ch bg abcd abc P P -----∆=∆=1111(p116) 3-1;3-5 3-1（1）惯性环节，阶跃输入)0(1)(≥-=-t e t x Tto指示实际水温98%的数值，所花时间约4T=1分钟 T ＝0.25分钟；（2）惯性环节，10t 速度输入)0)((10)(≥+-=-t Te T t t x Tt o )1(10)()()(1T e T t x t x t e o i --=-= T t e t 10)(lim =∞→稳态误差＝10T ＝2.5°C 3-5 -1))0(2.12.01)(1060≥-+=--t e e t x t t o系统传递函数＝)()(s X s X i O )10)(60(600)(1102.1602.01++-+=++s s ss s s ；-2))60070/(6002++s s ，s rad n n/5.24,6002==ωω，43.1,702==ξξωn(p116) 3-7;3-8 3-7K s K K s Ks k h h s s k s s k s +⋅++++==Φ++)1()1(12)1()1()(，h n n K K K ⋅+==12,2ξωω，s t Mp p1,2.0==456.0,2.01/exp(2==--=ξξξπ解得由p Ms rad ，t n n P /528.31)1/(2==-=ωξωπ解得由s K s rad K h 177.0,/5.1222==s rac t n r 65.0)1/()tan(212=--=-ξωπξξ)02.0(486.24=∆==s t ns ξω3-8？Ks s Ks H s G s G s s Ks s Ks ++++===Φ++5.341)()(1)(2)5.34()5.34()( 当200=K 时，22.1,142.1425.34====nK n ωξω?21===-ξωπωπn dp t当1500=K 时，445.0,73.3825.34====nK n ωξω当5.13=K 时，695.4,674.325.34====nK n ωξω(p117) 3-11;3-12 3-11 -1),)11.0(100lim ∞=+→=s s s K p,100)11.0(1000lim =+→=s s s K s v0)11.0(1002,lim =+→=s s s K s a-2),010011210210a a a K a K a K a ss a vpe ++=++=∞++∞=≠==ss a ss ，e a ，e a 时当时当0,02100213-120,21lim=∞==→ssi sK K s p e K12121110..lim K s s ssn sK K sK s e -==+-→11Kessn essi ess -=+=⏹ (p117) 3-17,,21h v p K K KK K K =∞=∆+∆=+=bh K K KK b K b K a ss K K KK e h V P 211,21,)(2122132121)(hHK K KK S S T T S T T K K KK s ++++=Φ0,0,,02121212121T T K K KK K K KK T T T T h h +⏹ （p168-169）4-2；4-7 4-222)()(dt x d dtx y d mBx y K =+--KyB Kx B mdt dydt dx dt x d +=++22 )()()()()(,)(2ωϕωωj KBs ms K Bs s Y s X eA j G s G ===+++22222)()()()(ωωωωB m K B K A +-+=YY A X t x B m K B K .)()(22222)()()(ωωωω+-+===4-722)()()()(dt t x d o dtt dx o o mt Kx Bt f =--)()()()(22t f t Kx B mo dtt dx dt t x d o o =++)(1)()()()(,)(2ωϕωωj KBs ms s F s X e A j G s G O ===++2222arctan )(,)()()(1ωωωωωϕωm K B B m K A -+--==)2sin()(,2sin 2)(2π-==t t x t t f oss)2(0,1)2(22ϕπ=--=⨯A1=m222)2()2(12222arctan ,12π-=-=⨯-+-k BB K1,4==B k(p171) 4-17、4-18 4-17解1:作图法。