FOH - 0.01ms - GW2

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Transcript of FOH - 0.01ms - GW2

  • 1 [L Hi ng 20090710]

    Bi thc hnh s 1: Tm m hnh gin on ng c mt chiu

    ng c c cc tham s:

    * in tr phn ng: = 250

    * in cm phn ng: = 4

    * T thng danh nh : = 0.04

    * Mmen qun tnh: = 0.0122

    * Hng s ng c : = 236.8 = 38.2

    M hnh ng c 1 chiu

    1-2 Tm hm truyn t ca m hnh

    - Hm truyn t vng h:

    = 1

    .

    1

    +1. . .

    1

    2...

    - Hm truyn t vng kn:

    =

    1 .. .

    S dng matlab tnh hm truyn t:

    Tt = 100e-6; La = 4e-3; Ra = 250e-3; Ta = La/Ra; T2 =

    0.01e-3; km = 38.2; ke = 236.8; phi = 0.04; J =

    0.012;

    Gh = 1/Ra*tf([1],[Ta 1])*km*phi*tf([1],[2*pi*J 0])

    Gk = feedback(Gh, ke*phi)

  • 2 [L Hi ng 20090710]

    Thu c cc kt qu nh sau:

    = 81.06

    0.016 2 +

    = 81.06

    0.0162 + + 767.8

    S dng matlab tm hm truyn t trn min nh Z bng cc phng

    php ZOH, FOH v TUSTIN, vi chu k trch mu 1 = 0.1 v 2 = 0.01

    T1 = 0.1e-3; T2 = 0.01e-3;

    Gz1 = c2d(Gk,T1,ZOH);

    Gz2 = c2d(Gk,T2,ZOH);

    Gz3 = c2d(Gk,T1,FOH);

    Gz4 = c2d(Gk,T2,FOH);

    Gz5 = c2d(Gk,T1,TUSTIN);

    Gz6 = c2d(Gk,T2,TUSTIN);

    Ta thu c cc hm truyn t trn min nh Z ng vi mi trng hp,

    sau khi chuyn sang s m m nh sau:

    1 = 2.528 5 . 1 + 2.523 5 . 2

    1 1.993 . 1 + 0.9938 . 2

    2 = 2.533 7 . 1 + 2.532 7 . 2

    1 1.999 . 1 + 0.9994 . 2

    3 = 8.431 6 + 3.367 5 . 1 + 8.404 6 . 2

    1 1.993 . 1 + 0.9938 . 2

    4 = 8.443 8 + 3.377 7 . 1 + 8.44 8 . 2

    1 1.999 . 1 + 0.9994 . 2

    5 = 1.263 5 + 2.525 5 . 1 + 1.263 5 . 2

    1 1.993 . 1 + 0.9938 . 2

    6 = 1.266 7 + 2.532 7 . 1 + 1.266 7 . 2

    1 1.999 . 1 + 0.9994 . 2

  • 3 [L Hi ng 20090710]

    Hm truyn t gin on tnh bng tay theo 2 chu k trch mu:

    Chu k trch mu 1 = 0.1

    7 = 2.524 5 . 1 + 2.521 5 . 2

    1 1.993 1 + 0.9938 2

    Chu k trch mu 2 = 0.01

    8 = 2.531 7 . 1 + 2.533 7 . 2

    1 1.999 1 + 0.9994 2

  • 4 [L Hi ng 20090710]

    3. M phng cc m hnh gin on thu c bng p ng bc nhy

    step(Gk) hoc bng m phng simulink vi s nh sau:

    Vi thi gian trch mu T1 = 0.1ms, ta thu c cc ng th:

  • 5 [L Hi ng 20090710]

    Vi thi gian trch mu T2 = 0.01ms, ta c:

    Nhn xt:

    Vi T1 = 0.1ms, ta thy php bin i Z theo cc phng php ZOH, FOH

    v TUSTIN cho kt qu gn tng ng nhau, v ging vi kt qu bin i

    bng tay

    Nhng vi thi gian trch mu T2 = 0.01ms, ta thy c s khc bit r

    rng gia 4 phng php ny. So vi trng hp T1 th h dao ng nhiu

    hn, v cha i n trng thi n nh ngay, do cc im cc b y ra xa,

    gn bin gii n nh ca ng trn n v. Tuy nhin sai lch ny, v mt gi

    tr l khng ln, v cc h u i ti trng thi n nh sau 1 khong thi gian.

    Kt qu tnh tay gn vi phng php ZOH nht.

    4. Xy dng m hnh trng thi ca CMC trn min thi gian lin tc

    M hnh trng thi ca i tng:

    +1 = . + . = . + .

    = . + . = . + .

    Chng trnh matlab:

  • 6 [L Hi ng 20090710]

    Vi = 81.06

    0.0162++767.8

    T1 = 0.1e-3; T2 = 0.01e-3;

    (A,B,C,D) = tf2ss([81.06],[0.016 1 767.8])

    (phi1,H1) = c2d(A,B,T1)

    (phi2,H2) = c2d(A,B,T2)

    step(A,B,C,D); hold on;

    step(phi1,H1,C,D); hold on;

    step(phi2,H2,C,D)

    ta thu c cc ma trn kt qu nh sau:

    = [62.5 47987.5

    1 0] = [

    10

    ]

    = [0 5066.25] = 0

    p ng bc nhy ca m hnh (A,B,C,D)

  • 7 [L Hi ng 20090710]

    so snh p ng bc nhy ca 3 m hnh:

  • 8 [L Hi ng 20090710]

    Bi thc hnh s 2: Tng hp vng iu chnh dng phn ng

    (iu khin mmen quay).

    S vng iu chnh dng phn ng ca CMC

    Hm truyn t ca m hnh i tng iu khin dng

    = 1

    +1 .

    1

    .

    1

    +1

    -S dng lnh c2d ta tm c hm truyn t trn min nh z ca i tng

    theo phng php FOH (vi chu k trch mu 0,01ms):

    Tt = 100e-6; La = 4e-3; Ra = 250e-3; Ta = La/Ra;

    T2 = 0.01e-3;

    Gi = 1/Ra*tf([1],[Tt 1])*tf(1,[Ta 1])

    c2d(Gi,T2,'foh');

    ta thu c hm truyn ca i tng dng nh sau:

    () = 4.064 5 2 + 0.1585 3 + 3.865 5

    2 1.904 + 0.9043

    Thit k b iu chnh theo phng php cn bng m hnh

    Gi s sau 2 bc, gi tr ca i tng iu khin s ui kp gi tr

    t ca i lng ch o.

    Vi tc p ng ca gi tr thc l 2 chu k T1

    Ta c:

    2() = 1. 1 + 2.

    2

    (vi iu kin 1 + 2 = 1)

    ta chn 1 = 0.6; 2 = 0.4;

  • 9 [L Hi ng 20090710]

    suy ra 2() = 0.6. 1 + 0.4. 2

    Hm truyn t ca b iu chnh l:

    () = 1

    ().

    ()1 ()

    Chng trnh matlab T2 = 0.01e-3;

    Gw = filt([0 0.6 0.4],[1],T2)

    B = filt([4.064e-5 0.1585e-3 3.865e-5],[1],T2)

    A = filt([1 -1.904 0.9043],[1],T2)

    Gz4 = B/A

    Gr = Gw/[Gz4*(1-Gw)]

    Gk = feedback(Gr*Gz4, 1)

    ta thu c cc hm truyn t nh sau:

    4() = 4.064 5 + 0.1585 3 1 + 3.865 5. 2

    1 1.904 1 + 0.9043 2

    = 0.6. 1 0.7424 2 0.219 3 + 0.3617 . 4

    4.064 5 + 0.0001341 . 1 7.271 5 . 2 8.659 5 . 3 1.546 5 . 4

  • 10 [L Hi ng 20090710]

    Kt qu m phng bng p ng bc nhy ca h kn:

    M phng bng simulink vi s nh hnh di cng cho kt qu

    tng t

    Nhn xt:

    T th ta thy ng sau 3 bc i tng iu khin ui kp gi tr t

    ca i lng ch o. Kt thc chu k trch mu u tin u ra t ti gi tr

    1 ca b iu khin ( = 0.6 ). Kt thc chu k trch mu th 2 u ra t ti gi tr 1 + 2 ca b iu khin ( = 1 ) v tin ti xc lp

  • 11 [L Hi ng 20090710]

    Bi thc hnh s 3: tng hp vng iu chnh tc quay

    i tng tc bi ny c tnh bng

    = (2) 1

    2

    1

    vi (2) c tnh xp x thnh 1 khu qun tnh tch phn bc nht

    (2) = 1

    2 . + 1

    vi = 100

    c th, ta tnh c:

    () = 1.528

    1.508 5 . 2 + 0.0754 .

    chuyn sang min nh Z vi s m m, chu k trch mu 2 = 0.01

    () = 1.668 6 + 6.589 6 . 1 + 1.627 6 . 2

    1 1.951. 1 + 0.9512 . 2

    Tng hp b iu khin PI cho tc theo tiu chun tch phn bnh

    phng sai lch.

    B iu khin c dng

    = 0 + 1 .

    1

    1 + 1 . 1

    i tng iu khin

    = 0 + 1 .

    1 + 2 . 2

    0 + 1 . 1 + 2 . 2

    Vi cc h s , tnh theo Gn trn

    ta c sai lch tnh

    =

    1 + .

    chuyn sang min nh Z

    =

    1 +

    sai lch iu chnh vit di dng sai phn:

  • 12 [L Hi ng 20090710]

    = + (1 1). 1 12 (1 1 + 0. 1). 1 (1 + 0. 2 + 1. 1). 2 (0. 3 + 1. 2). 3 1. 3. 4

    y ta ly = 1 0

    chn 0 = 275 vi , xc nh theo . ta cn tm 1 sao cho

    = ()2

    0=

    2=0 t gi tr min

    vi k = 0, ta c 0 = 1

    vi k = 1, ta c 1 = (1 0. 1) = 0.9982

    vi k = 2, ta c 2 = 0.9924 + 1.627 6 . 1

    ..

    khi = 2

    =0

    I t gi tr nh nht th 1 274

    kt hp vi iu kin 0 |1| th ta chn 1 = 274

    M phng bng Simulink

    S simulink nh sau:

    < cc th thu c bng khi scope c chuyn sang th Figure bng

    lnh

    plot(ScopeData.time,ScopeData.signals.values(:));

    thun tin cho vic hin th v in n >

  • 13 [L Hi ng 20090710]

    Cc kt qu thu c:

    p ng u ra ca h thng

    tch phn bnh phng sai lch

  • 14 [L Hi ng 20090710]

    Nhn xt : qua 2 th ta thy, p ng u ra ca h thng vi tn hiu

    bc nhy nhanh chng tin ti gi tr xc lp bng 1. qu iu chnh khng

    vt qu 20%, sai lch tnh nhanh chng tin ti 0

    b iu khin PI t c yu cu ra.

  • 15 [L Hi ng 20090710]

    kim tra tnh bn vng ca h thng

    nhiu gi tr t: ta t thm 1 tn hiu bc nhy tc ng vo

    h thng sau khong thi gian 0.03s

    thu c p ng u ra ca h:

    nhn xt: h p ng tt i vi nhiu gi tr t

  • 16 [L Hi ng 20090710]

    nhiu ph ti: t thm mt tn hiu bc nhy tc ng vo u

    ra ca h thng sau thi im 0.03s

    thu c p ng u ra ca h:

    nhn xt: khi nhn thm tn hiu nhiu ph ti, h thng nhanh

    chng t a v trng thi xc lp ch sau 1 vi chu k trch mu.

    h thng p ng tt i vi nhiu ph ti

  • 17 [L Hi ng 20090710]

    Bi thc hnh s 4: tng hp b C tc quay trn KGTT

    tng hp b C tc quay theo phng php gn im cc,

    sao cho im cc nhn gi tr dng v nm trong ng trn n

    v trn min nh Z v phng php p ng hu hn (dead-beat).

    chn b p1 vi 2 im cc nm trong khong [0,1] (gi tr c

    th-chn sao cho h thng c qu iu chnh khng qu 20%, v

    sai lch tnh tin ti 0), b p2 theo phng php deat-beat vi 2 im

    cc u ti gc. l m hnh ng c xy dng bi 1. Vi chu

    k trch mu 1 = 0.1, 2 = 0.01

    chng trnh matlab

    T1 = 0.1; T2 = 0.01;

    num = [81.06];

    den = [0.016 1 767.8];

    Gk = tf(num,den);

    [A,B,C,D] = tf2ss(num,den);

    [phi1,H1] = c2d(A,B,T1);

    [phi2,H2] = c2d(A,B,T2);

    p2 = [0 0];

  • 18 [L Hi ng 20090710]

    // chu ki trich mau T1 = 0.1s

    p1 = [0.66 0.66];

    K11 = acker(phi1,H1,p1);

    K21 = acker(phi1,H1,p2);

    Gk1 = ss(phi1-H1*K11,H1,C,D,T1);

    Gk3 = ss(phi1-H1*K21,H1,C,D,T1);

    step(Gk1)

    hold on;

    step(Gk3)

    grid on;

    ta c th p ng bc nhy thu c theo 2 phng php

  • 19 [L Hi ng 20090710]

    // voi chu ki trich mau T2 = 0.01s

    p1 = [0.44 0.55];

    K12 = acker(phi2,H2,p1);

    K22 = acker(phi2,H2,p2);

    Gk4 = ss(phi2-H2*K22,H2,C,D,T2);

    Gk2 = ss(phi2-H2*K12,H2,C,D,T2);

    step(Gk2)

    hold on;

    step(Gk4)

    grid on;

    th p ng thu c

    Nhn xt chung:

    Tng hp b iu khin tc quay theo phng php phn hi trng