The Z transform techniques

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Transcription:

h Z trnfor tchniqu h Z trnfor h th rol in dicrt yt tht th Lplc trnfor h in nlyi of continuou yt. h Z trnfor i th principl nlyticl tool for ingl-loop dicrt-ti yt. h Z trnfor h Z trnfor i to dicrt-ti yt wht Lplc trnfor i to continuou- ti yt. Concptully, th yol Z cn ocitd with ti hifting in diffrnc qution in th wy tht S cn ocitd with diffrntition in diffrntil qution. h Z trnfor of i Function h dicrt-ti function f*t cn Lplc trnford -t L{f*t } f*t dt 0 W hll oti u th nottion h function f*t i rgrdd tking vlu ro xcpt t intnt whr tk, whr k tk only intgrl vlu. h Z trnfor of ft i dfind ing qul to th Lplc trnfor of f*t: tht i if Z{ ft } i digntd F thn finlly, if w dfin, L{f*t } F* -t -k f*t dt fk fk -k 0 k0 Z{ft }L{f*t } ENG640 Control Syt Dign : Hn Prchidh, Dr. Brry Hyn Pg k0 F F* fk -k k0 -k thn F fk k0

Expl : find th Z-trnfor of unit-tp function ft ut whr ft 0 whn t < 0 nd ft whn t 0. Solution For dicrt for ft fk uk which i unity for ll vlu of K F fk -k - - -3-4... k0 -k - F - fk - - -3-4... k0 Sutrct fro F - - F > F - - hrfor F - - - Expl : find th Z-trnfor of xponntil function whr ft 0 whn t < 0 nd ft -t whn t 0. olution For dicrt for ft fk -k F fk -k - - - - -3-3 -4-4... k0 F - - - - -3-3 -4-4... Sutrct fro hrfor - - F - - - - -3-3 -4-4... F - - - F > F - - - F - - - - - Prol : Find th Z-trnfor of polynoil function k whr ft 0 whn t < 0 nd ft k whn t 0. Prol : Find th Z-trnfor of unit-rp function whr ft 0 whn t < 0 nd ft t whn t 0. ENG640 Control Syt Dign : Hn Prchidh, Dr. Brry Hyn Pg

h Z trnfor of Lplc Function Anothr thod of driving th Z-trnfor of ti function, i to trnfor ft into G nd thn driv th quivlnt G, uing th following thod :- Au G N/D, whr D h finit nur of ditinct root. hn p Nx G n Xn - n D x n Whr D dd/d nd Xn, n,.,p r th root of qution D 0. Expl: find th Z-trnfor of f /. not : f ut, th tp function. Solution N, D, D, n nd x 0. p Nx G n n D x n Xn - n 0 - - Expl: find th Z-trnfor of f /. not : f L ft -t. olution N, D, D, n nd x -. p Nx G n n D x n Xn - n - - - Prol : Cn you find th Z-trnfor of i f / 3 ii / rp. So Proprti of Z trnfor. Linrity Z[ft ft] Z[ft] Z[ft] F F. Multipliction y contnt Z[ * {ft}] * Z[ft] * F 3. Shift proprty Z[fk n] n F nd Z[fk - n] -n F 4. Initil-vlu thor Li fk li F K 0 5. Finl-vlu thor Li fk li - F K ENG640 Control Syt Dign : Hn Prchidh, Dr. Brry Hyn Pg 3

ENG640 Control Syt Dign : Hn Prchidh, Dr. Brry Hyn Pg 4 Lplc trnfor f i function ft t>0 -trnfor f Modifid -trnfor f, t u t 3 t 3 3 t t t t t t t t int co in co in in cot co co co co co int t co in co in in co co co co co co

Expl: find th Z-trnfor of ft to gnrt in3t t intrvl 0f 0. Sc. Solution Uing Z trnfor tl Z { } Sint Z in Z in3 0.955Z - 0.955Z Z Z co Z Z co3 Z.90673Z.90673Z - Z - Expl: find th Z-trnfor of D givn low with pling ti of 0. Sc. Solution 6 D 3 6 D 5 6 A B A 3A B B A B 3A B 3 [...] [...] thrfor A B 0 nd 3A B 6 A -B nd 3A -A 6 > A 6 nd B -6 6 6 D - 6 6 3 Z { D } - - -3 Uing 0. cond Expl: find th Z-trnfor of th continuou plnt low ing upplid with pl fro digitl controllr through ZOH DAC with pling ti 0. Sc. Ipul Fro th Ipul o th DAC 3 Gp 4 0.4674 -.5595-0.6065 - Continuou Output fro h yt ENG640 Control Syt Dign : Hn Prchidh, Dr. Brry Hyn Pg 5

Solution - - D ------------ Uing prtil frction G - - x[ 3/ 3/8 _ 3/8 ] 4 G Z{ - - - - x[ 3/ 3/8 _ 3/8 ] } Z{ [ 3/ 3/8 _ 3/8 } 4 G 3/ _ [ 3/8 3/8 ] -4 Uing 0. cond U of odifid Z trnfor including ti dly Up to thi point it h n ud tht ngligil ti lp twn pling th rror nd producing th corrponding input voltg to th input of th plnt. For xpl, in hip utopilot pling t 0 cond, k th illicond pnd on coputtion inignificnt. In n ircrft howvr, with n utopilot pling, y, vry 50 illicond, th coputtion ti i ignificnt proportion of th pling intrvl. h dly would thrfor hv to includd in th thticl odl of th yt dd ti or trnport lg. h odifid trnfor i n to contin prtr ; thi rlt to th ount of dly tht ut ccountd for. Lt th dd ti D cond nd p proportionfrction of th pling intrvl, tht i n p D And of odifid tl - p 0.7363-0.478 -.6703-0.6703-4 ] D D t n0 n4 t h frctionl portion of th dly i thn trtd in th nnr dcrid ov nd th intgr portion i ccountd for y ultiplying th rult y -n. ENG640 Control Syt Dign : Hn Prchidh, Dr. Brry Hyn Pg 6

Expl: find th Z-trnfor of th continuou plnt low ing upplid with pl fro digitl controllr through ZOH DAC with pling ti 0.5 Sc nd hving finit dly D of : i 0. cond, nd ii.8 cond. Ipul Fro th Ipul o th DAC G - D Continuou Output fro h yt olution G - - - D x - - - D - D - - G - Z { } - Z { } i 0.5, D 0. thrfor D 0 p > 0. 0 p 0.5 > p 0.4 nd 0.6 > 0.6 - G - [ ] - 0.3 0.5 - [ ] ii 0.5, D.8 thrfor D n p >.8 3 p 0.5 > p 0.3 nd 0.3 > 0.7 - -3 0.5 [ ] G - [ ] - - -3 0.35 Not : -3 which i cu of n3. ENG640 Control Syt Dign : Hn Prchidh, Dr. Brry Hyn Pg 7

Prol: find th Z-trnfor of th continuou plnt low ing upplid with pl fro digitl controllr through ZOH DAC with pling ti Sc nd hving finit dly D of : i 0.3 cond, nd ii 3.5 cond. Ipul Fro th Ipul o th DAC G - D Continuou Output fro h yt Rltionhip twn Z nd diffrnc Eqution Whr continuou yt r dcrid y diffrntil qution, dicrt yt r dcrid y diffrnc qution. h diffrnc qution how th rltionhip twn th input ignl k nd th output ignl uk. Expl: find th diffrnc qution of D givn low: olution u 0.4674 - D.5595-0.6065 - u[ - -.5595 0.6065.5595 u 0.6065 u 0.4674 - u - - ] x 0.4674 - un.5595 un- 0.6065 un- 0.4674 n- un.5595 un- 0.6065 un- 0.4674 n- Prol : find th diffrnc qution of D givn low y finding D : 3 4 ENG640 Control Syt Dign : Hn Prchidh, Dr. Brry Hyn Pg 8