Dap an de Thi CSTD GK1 1213

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  • 7/28/2019 Dap an de Thi CSTD GK1 1213

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    i hc Bch Khoa TPHCM P N KIM TRA GI A K 1. Nm hc 2012-2Khoa in in T Mn: C s t ng

    Bmn KT Ngy thi: 12/10/2012---o0o--- Th i gian lm bi: 60 pht

    Bi 1: (2.5 im) Tm hm truy n t ng ng0)()(

    )()(

    s R N s N

    sY sG ca h thng c s khi Hnh 1.

    Bi gi i:S dng tn hi u t ng ng : c th v nhiu cch (0.5im)

    Cch 1:

    Cch 2:

    ng tin : (0.5im) Vng kn : (0.5im) P1 = G2G3 L1 = G2G4G1 P2 = G 6 L2 = G3G5

    L3 = G2G3G1

    R(s) 1

    N (s)

    G1 G2

    1

    1 G3 Y (s)

    G4 G5

    1 1

    1

    G6

    G1(s) R(s) Y (s)

    Hnh 1

    + G2(s)

    G6(s)

    G4(s)

    _ +

    G5(s)

    + _ G3(s)+

    _

    N (s)

    _ +

    -1 G2 N(s) 1 G3

    -G 5

    Y(s)

    -G 4G1

    G6

    -G1

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    L i gi i:Cu 3.1:+ Cc: 1 2 30; 10 p p p (0.25im) +Zero: 1 1 z ;

    + im tch nh p:3 210

    1

    s s K

    s

    2 3 20 3 20 1 10 0dK s s s s sds

    3 22 13 20 0 s s s

    0 ( )

    2.5 ( )

    4 ( )

    s tach

    s cuc tieu nhap

    s cuc dai tach

    (0.5im)

    + Gc ti m cn: 0180 903 2

    q ;

    + Giao im tim cn v tr c honh:0 0 10 1

    4.53 1OA

    (0.25im)

    Q NS: (0.75im)

    -12 -10 -8 -6 -4 -2 0 2-8

    -6

    -4

    -2

    0

    2

    4

    6

    8Root Locus

    Real Axis

    I m a g

    i n a r y

    A x

    i s

    + T Q NS suy ra h thng n nh v i mi K >0 (0.25im)

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    Cu 3.2:K=12 PTT ca h thng tr thnh: 3 210 12 12 s s s

    Nghim ca PT T: 12,3

    8.79

    0.605

    s

    s i

    (0.25im)

    11

    6.02

    n

    n

    6.01

    6.0

    2

    n (0.25im)

    st n

    qd 67.66.044

    ; (0.25im)

    2

    exp exp 0.6 15.18%1

    POT

    (0.25im)

    Bi 4: (3.5 im) Cho h thng nh Hnh 3.

    4.1. (2.5 im) Cho K C =1. Vbiu Bode bin v pha c a h h trong mi n tn s t 0.01 n 20 rad/seXc nh tn sct bin c v tn s ct pha . nh gi tnh n nh ca hkn.

    4.2. (0.5 im) Cho K C =1. Tnh sai s xc l p ca h thng khi tn hi u vo l hm d c n v. 4.3. (0.5 im) Xc nh mi n gi tr K C sao cho h kn n nh.

    L i gi i:

    4.1Hm truy n vng h :0.1

    2

    100( 0.5)( ) ( )

    ( 10)( 5)

    s

    h C

    s eG s K G s

    s s s

    a hm truy n h v dng:0.1

    2

    (2 1)( )

    (0.1 1)(0.2 1)

    s

    h

    s eG s

    s s s Cc t n s ct: 1 2 30.5( / ), 5( / ), 10( / )rad s rad s rad s (0.25im)

    Xc nh im A: (0.25im) 0

    0

    0.1( / ):

    ( ) 20 log(1) 2* 20 log(0.1) 40

    rad s A

    L dB

    hocdB L

    srad A

    80)01.0log(*2)1log(20)(

    )/(01.0:

    0

    0

    Biu thc pha v b ng: (0.25im) 180

    ( ) 180 arctan(2 ) arctan(0.1 ) arctan(0.2 ) 0.1

    R(s) Y (s)+ _

    Hnh 3

    K C

    )5)(10()5.0(100

    )( 21.0

    s s se s

    sG sG( s)

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    L p bng xc nh mt s im: (0.25im)

    (rad/s) 0.1 0.5 1 5 10 20( ) (deg) -171 -146 -139 -196 -259 -345

    Biu Bode (1.0im), ch :- biu d i y trong mi n 0.1-20rad/s, SV v trong mi n 0.01 n 20 rad/s v n tnh im.- ch c tr n vn 1 im nu ch thch y dc, tn s ct, d tr .

    Tn s ct bin: )/(2 srad C , tn s ct pha: )/(2 srad d tr bin: )(5 dBGM , d tr pha: 030 M T cc k t qu trn bi u Bode, ta suy ra h kn n nh. (0.5im)

    4.2 (0.5im) H svn tc:

    0lim ( )V C s K sK G s

    Suy ra, sai s xc l p:1

    0 xl V

    e K

    4.3 (0.5im)T biu Bode ta th y, n u d ch Bode bin ln 5dB th lc d tr bin s m v h kn s khngnh, do :

    52020 log( ) 5 10 1.8C C K K

    0 1.8C K L u : do d tr bin xc nh g n ng nn gi tr K C c th khc nhau. (tnh chnh xc theo

    Matlab th 0 2.8C K )