sổ tay cdt chuong 29-xly sign so

8/6/2019 s tay cdt Chuong 29-Xly Sign So

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29X l tn hiu s

trong cc ng dng c in t

Bonnie S. HeckGeorgia Institute of Technology

Tomas R. KurfessGeorgia Institute of Technology

29.1 Gii thiu ............................................................1

29.2 C s v x l tn hiu ........................................1

29.3 nh x tn hiu lin tc sang tn hiu ri rc ..... .4

29.4 Thit k b lc s ................................................829.5 Thit k iu khin s .......................................18

29.1 Gii thiu

a s cc k s lm vic trong lnh vc C in t ch lm v nhng h thng hon ton l ckh hoc in t. C rt nhiu cc h thng in h tr cc h c v ngc li. V d, hu ht ccb vi x l trong my tnh ngy nay u c b tn nhit v qut lm mt my tnh hot ngtrong nhit n nh. Cc h thng in c s dng rng ri theo di v iu khin ton bh thng c kh. Mi n khi cc chp x l s gi thnh r, b lc s v iu khin s cho cc hc mi tr ln ph bin. Cc v d v trng hp ny c th thy trong xe t v hu ht cc thit

b gia nh. V d, cc tn hiu cm bin s dng trong mn hnh v iu khin cc h c i hivi dng ca x l tn hiu. Dng x l tn hiu ny c th xp t dng n gin, tn hiu s dngmt b lc thng thp n cc vic phn tch cao cp hn nh m men v gim st ngun trongng c servo mt chiu. Chng ny trnh by khi qut v cc phng php x l tn hiu sph hp trong cc h c. Trong khun kh chng ny, khng th phn tch chi tit, chi tit hnxem [1,2].

29.2 C s v x l tn hiu

Mt vi khi nim c bn v x l tn hiu c gii thiu trc khi tho lun v b lc hociu khin.

Tn hiu lin tc

Vic phn tch h thng lin tc vi thi gian s dng bin i Laplace, gii quyt p ngh thng, v thit k iu khin. Bin i Laplace l duy nht i vi tn hiu lin tc, x(t),vc cho bi cng thc

1


2/20

S tay C in t

0( ) ( ) stX s x t e dt

=

Mt hm truyn ca h tuyn tnh, H(s), l t l ca bin i Laplace u ra trn bin iLaplace u vo (vi iu kin ban u l 0).

Bin i Fourier s dng xc nh ph tn ca tn hiu. Bin i Fourier ca x(t) c chobi cng thc:

( ) ( ) j tX x t e dt

= (29.1)

Trong c n v l radian trn giy. Lu rng khi ( ) 0x t = m 0t , bin i Laplacetng ng vi bin i Fourier bng vic t s j= . (Nn ch rng cn c thm cc iukin hi t ca bin i Fourier). p ng tn s ca h thng c nh ngha l t l ca bini Fourier u ra trn bin i Forier u vo ca n. Hon ton tng ng, c th nhn c

t hm truyn bng cch thay ( )H bng ( ) ( ) ( ) s jH H j H s == = . n gin k hiu, j

khng a vo danh sch cc tham s, ch ghi k hiu l ( )H miu t p ng tn s. Bng

thng ca mt h thng c nh ngha l p ng ti ( ) 0.707 (0)H H = .

Tn hiu ri rc

Bin iz c s dng gii phng trnh sai phn v phn tch h thng. Bin iz ca tnhiu ri rc, [n]x , c nh ngha l

( ) [ ] nn

X z x n z

=

=

Bin i Fourier ri rc (DTFT discrete-time Fourier transform) s dng xc nh ph tnca tn hiu. DTFT v DTFT ngc c nh ngha

( ) [ ] j nn

X x n e

=

= (29.2)

v

1( ) ( )

2j nx n X e d

= (29.3)

Ch rng DTFT c th nhn c t bin i z bng cch t jz e = . (Mt khc, c mt sgi thit v hi t trong kt qu ny). DTFT theo chu k 2 , di in hnh l [ , ] hoc[0, 2 ] , trong 0 = (tn s thp), = (tn s cao). p ng tn s ca h ri rc l t lgia DTFT ca tn hiu ra chia cho DTFT ca tn hiu vo. Hoc n cng c th nhn c t hm

truyn ( ) ( ) ( ) jj

z eH H e H z

= = . n gin hay dng k hiu ( )H hn ( )jH e . Cng

ging nh trng hp tn hiu lin tc, bng thng c nh ngha l tn s ti ( ) 0.707 (0)H H = .

Trong khi DTFT l lin tc i vi bin i ca tn s , bin i Fourier ri rc (DFT discrete Fourier transform) c cha cc im ri rc i vi tham s k. Xt trong khong gii hn

[ ]x n , vi [ ] 0x n = , vi n < 0 v vi n N . DFT v DTF ngc ca [ ]x n c nh ngha l:

12 /

0

[ ] , 0,1...., 1N

j nk N

k

n

X x n e k N

=

= = (29.4)

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X l tn hiu s trong cc ng dng c in t

12 /

0

1, 0,1,...., 1

N j nk N

n k

k

x X e n N N

=

= =

Ch rng DFT l bin i c ri rc ha ca DTFT vi ( )kX X= trong di t 0 n2 . Vic tnh cng thc tng qut cho DTFT c th ch c lm vi cc tn hiu n gin nh lxung vung hoc xung tam gic. V vy DFT c s dng tng qut nh l mt phng php s tnh DTFT ti cc im ri rc vi tn s thuc ri t 0 n 2 . Trong thc t, t c

mt th ca DTFT, phi v th ca Xk theo k, vi kc qut trn cc im 2 /N . Vi mttn hiu bt k, nh l cc tn hiu o t mt thit b, Vic tnh ton DFT thay v DTFT l mtphng php hon ho hn tm ph tn ca tn hiu. t c nhiu kt qu hn trong vicv mt DTFT t cc im c tnh t DFT, thm cc im khng vo cui ca dy do gi trN c tng ln.

Gi thit l tn hiu trong min thi gian khng b hn ch v mt thi gian quan st, v vykhng c gi tr N lm cho x[n] = 0 vi n > N. thc hin DFT, tn hiu phi b cht. C 2trng hp xy ra: Trng hp x[n] suy gim thnh 0 v trng hp x[n] c cc thnh phn chuk. Trng hp x[n] suy gim v 0 thNphi c chn ln cc gi tr khng ng k c thb qua. Kt qu DFT l mt xp x (khng phi l phin bn c ri rc) ca DTFT. Nu tn hiul chu k, DTFT khng th c tnh bng phng php s v kt qu DTFT c cc p ng xung.Tuy nhin, cc tn s biu din trong tn hiu c th vn c xc nh nu gi tr Ndng ctc chn sao cho tn hiu c ct i bng mt s nguyn ln chu k. Nu khng DFT thu cs b d trong th tn s khi so snh vi tn hiu DTFT thc. V d, xt mt tn hiu x[n] =cos(0.4 n). y l tn hiu tun hon vi chu k n = 5 v c DTFT c cho bi cng thc

( ) [ ( +0.4 )]+ ( -0.4 )]X = vi . Tt c ph tn c t ti 0.4 = v0.4 = . DTFT c chu k 2 , cng c mt xung ti 2 0.4 . DFT c tnh cho 2 im

cht ca tn hiu, mt tiN= 20 (bng 4 ln chu k) v ti mt im khcN= 22. DFT viN= 20c v trn hnh 29.1(a) trong bin c lp k c qut 2 /N cho mi im. th ny xuthin ph tn l cc im 0 ti im 0.4 (1.2566) = v 2 0.4 ( 5.0265) = = , t c v trchnh xc ca xung trong DTFT. Tng t nh vy, DFT vi N = 22 c v nh trn hnh29.1(b), ch kt qu c s r r trong c tnh tn.

Hnh 29.1 DFT ca tn hiu tun hon (a) ct sau 4 chu k (b) ct sau 4.4 chu k

Nu mt tn hiu c ph tun hon, nhng n khng phi l tn hiu tun hon, nh l x[n] =

cos(0.5n) + cos(0.2n), th vic r r l khng th trnh khi khi chn N. Mt cch thay i gimr r l nh th nht ca tn hiu bng khng ti im u v im cui trc khi tnh tonDFT. Qu trnh ny, c hiu l ca s d liu, c thc hin bng vic nhn x[n] vi mt hmca s w[n] v sau thc hin DFT tch s x[n] * w[n]. Ba ca s ch yu l ca s hnh chnht, cht ct nh nhn, ca s Hanning v ca s Hamming [1].

Ca s ch nht:3


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S tay C in t

[ ] 1, 0 1w n n N =

Ca s Hanning:

1 2[ ] 1 cos , 0 1

2 1

nw n n N

N

=

Ca s Hamming:

2[ ] 0.54 0.46cos , 0 1

1

nw n n N

N

=

Nu gi tr ca N l s ln hn 2, c mt phng php tnh DFT c gi l bin iFourier nhanh (FFT fast Fourier transform). Nu gi tr Nkhng ln hn 2, cc im khng cth c thm vo phn cui ca tn hiu theo th t tnh FFT. iu ny khng nh hng n chnh xc ca kt qu, nhng n lm tng phn gii ca th thu c khi DFT (hoc FFT)c s dng tnh DTFT. Trong nhiu trng hp, Biu thc c s dng trong (29.4) tnh DFT. Cng thm sc mnh tnh ton ca vi x l ngy nay gim bt s cn thit v kh nngs ca FFT. Chi tit v thut ton ca FFT nm ngoi gii hn ca cun sch ny. Bn c xem[1] hoc [2] bit chi tit.

29.3 nh x tn hiu lin tc sang tn hiu ri rcTrong khi hu ht cc h vt l hot ng trong thi gian lin tc, cc my tnh hot ng trong

thi gian ri rc. V vy, theo th t s dng my tnh x l cc php o phi chuyn t h lintc sang h ri rc, phi c cch nh x gia th gii lin tc v th gii ri rc.

S ri rc ha

Trc khi mt tn hiu analog c phn tch s dng cc cng ngh s, n phi c ri rcho (ngha l, c chuyn sang tn hiu ri rc). Phng php ch yu ri rc ha l ly mu,khi cc gi tr ca tn hiu c xc nh ti cc im ri rc vi thi gian. Thng thng, tn hiuc ly mu ti mt di c nh c coi nh l chu k ly mu. Di ly mu (n v l Hz) l snghch o ca chu k ly mu. Hnh 29.2 miu t tn hiu 1Hz c ly mu ti 2 di. Cc im

ti c ly mu chu k 15ms, cc im sng hn c ly mu vi chu k 250 ms. T hnh 29.2,s xp x ca dng sng suy bin trn tru nh l tn s ly mu c gim v tin ti tn s tnhiu. Trong thc t, c th c biu din nh l mt tn hiu phi c ly mu ti mt tn s lnhn 2 ln ph tn ln nht ca n. iu ny c bit nh l ly mu Nyquist. V d, nu tnhiu trong hnh 29.2 c ly mu ti 0.5 Hz, th tt c cc gi tr ly mu u c gi tr bng 0,cc im nh l 0, 500, 1000 ms gi tr ca tn hiu l 0.

Biu din sai tn hiu l v mt tn s ly mu qu nh c coi nh l nh thin (aliasing).nh l ly mu Nyquist a ra 2 phng php. Phng php th nht l dng mt tn s ly muln gp 2 ln ph tn cao nht ca tn hiu ang c ly mu. Tn s ny c gi l tn sNyquist. Khng chc chn rng ph tn thc l ca tn hiu l tng, s dng mt b lc thngthp m bo rng mt tn hiu khng b nh hng bi cc tn s cao hn mt mc gii hnnht nh. B lc c gi l b lc chng nh thin (anti-aliasing filter). y l phng phpth 2 v l phng php thc hin nhiu trong thc t, n c s dng tha mn nh l ly

mu Nyquist. V vy, s kt hp ca mt b lc chng nh thin c thit k tt cng vi mttn s ly mu chnh l tn s gii hn trn ca b lc s m bo rng nh l ly mu Nyquistc tha mn.

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Hnh 29.2 Tn hiu 1Hz

Hnh 29.3 Mt tn hiu c ly mu v xy dng li s dng ly mu v gi mu bc khng(ZOH)

C 2 im quan trng cn c ch khi s dng mt b lc chng nh thin. Th nht, blc chng nh thin s dng trc khi tn hiu c ly mu, khi ly mu th gy ra thin nh. Vc bn, iu ny i hi b lc chng nh thin c thc hin s dng mt b lc tng t trckhi tn hiu c s ha. Khi tn hiu b nh thin trong qu trnh ly mu, khng th sa c sdng b lc s. iu th 2 l trong thc t, tn s ct ca b lc chng nh thin nn chn nhhn t 5 n 10 tn s Nyquist. Nn ch rng b lc chng nh thin a thm pha tr vo gi

tr o, n c th lm gim s n nh v vic thc hin vng phn hi tr khi bng thng ca blc chng nh thin ln hn nhiu bng thng ca h kn. Cc thit b dng thc hin ly mul cc b chuyn i tng t sang s (ADCs), v b lc chng nh thin c s dng trcthit b ny.

Ngc li ca vic ly mu l ti to khi mt tn hiu ri rc c chuyn i sang mt tn hiulin tc. Di ly mu Nyquist m bo rng nu mt tn hiu lin tc c ly mu ti mt di mdi t nht gp 2 ln thnh phn tn s cao nht trong tn hiu, th tn hiu lin tc c th cti to li chnh xc t tn hiu ly mu. Tuy nhin, nh l ny gi thit rng mi qu trnh ti tol l tng, iu ny khng thc t. Hu ht cc cch thc hin ti to mt tn hiu l gi mu bckhng (ZOH). ZOH mc nh l gi tr ca tn hiu l hng s gia cc ln ly mu. S xp x nyl kh hp l nu tn hiu c ly mu v c bn khng thay i gia cc mu ring bit. Hnh29.3 l mt v d ca tn hiu v biu din gi mu bc khng ca n. Mu xm, ng trn biuhin tn hiu tng t ban u. Cc im mu en dc theo tn hiu ch ra cc gi tr ly mu catn hiu. Mi im en c ni ti im tip theo bng mt ng nm ngang sau l ngthng ng. ng nm ngang miu t vic gi mu bc khng, gi tr ca tn hiu vn gi lhng s gia cc gi tr ly mu. ng thng ng khng nh rng tn hiu khng gi nguynhng s trong sut chu k ly mu. Khi khong thi gian gia cc im ly mu c tng ln, chnh xc ca vic gi mu bc khng b gim i. Ngc li, khi chu k ly mu gim, chnh

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S tay C in t

xc ca gi mu bc khng ci thin. Cc thit b thc hin ti to l cc b chuyn i DACs,chng thng s dng phng php gi mu bc khng.

nh x t min s sang min z

Mt phng php lin h mt phng s vi mt phng z th xut pht t biu din ton hcca tn hiu lin tc c ly mu x(t) v tnh ton bin i Laplace ca n. Kt qu bin iLaplace ca tn hiu c ly mu lin quan vi bin i z cax[n] bng vic t sTz e= trong Tl chu k ly mu. Biu thc sTz e= thng thng c gi l nh x chnh xc gia mt phng

z v mt phngs. ( bit chi tit hn, xem [1]). V d, Mt biu din s ( )dH z ca mt h lin

tc, ( )H s c th thu c bng nh x ( ) ( ) sTd z eH z H s == . Tuy nhin, nh x ny dn n ( )dH z l mt hm khng hu t. nh x xp x gia mt phng s v mt phngz thng c s dng dn n mt hm hu t cho ( )dH z . Ba nh x l bin i song tuyn, bin i thun v bini ngc.

Bin i song tuyn:

2( 1)

( 1)

zs

T z

=

+

Bin i thun:1

( 1)s zT

=

Bin i ngc:

1( 1)s z

Tz=

Bin i song tuyn (cn c gi l quy tc Tustin hoc quy tc hnh thang) l bin i chnhxc nht trong cc nh x. N nh x ton b min bn tri ca mt phng s sang vng trn n vca mt phng z, v vy n duy tr c tnh n nh. Xt mt v d o hm bc nht ca

( ) 1/( 2)H s s= + . Biu din ri rc ca hm truyn ny l

2 ( 1)

( 1)

( 1)( ) ( )

2 2 2zd s

T z

T zH z H s

z T

=+

+= =

+

Ch rng kt qu ca hm truyn trn l hu t i vi z.

Mt phng php khc ca hm truyn nh x gia min lin tc v min ri rc l nh xp ng u ra.

p ng u ra: Gi sx(t) l u vo ca mt h thngH(s) vi kt qu u ray(t).x[n] vy[n] c ly mu t x(t) v y(t). Sau ,Hd(z) l t s gia cc bin i z,x[n] v y[n]. Hu htcc p ng u ra l p ng step trong x(t) l hm step,x(t) = 1 vi t> 0 vx(t) = 0 vi t< 0.Biu thc ( )dH z c cho bi cng thc sau :

1 ( )

( ) (1 )dH s

H z z Z s

=

Trong [ ( )/ ]Z H s s biu din bin i z p ng step ca h lin tc. Dng cho h bc nhttng qut nh sau:

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1

1

(1 )( ) ( )

1

aT

d aT

a e zH s H z

s a e z

= =

+

Phng php p ng u ra (c bit l u vo step) hu ht c s dng nh x h lintc vi h ri rc khi thit k mt b iu khin s trong min ri rc. Tt c cc b iu khin sc thc hin s dng gi mu bc khng u ra ca b iu khin, vi h a vo mt tnhiu step ging nh l mt tng cc tn hiu step b tr. V vy, phng php p ng u ra stepl cch chnh xc nht nh x mt h c gi mu bc 0 u vo.

nh x trong min tn s

Bin i Fourier lin tc c th c thay th bng DTFT thng qua biu thc:

( ) ( )T

X TX vi

=

=

Trong ( )X c nh ngha trong phng trnh (29.1) v biu din bin i Fourier lin

tc ca x(t), ( )X c nh ngha trong (29.2) v biu din DTFT ca tn hiu c ly mux[n]. nh x ny rt hu dng trong vic tnh ton bin i Fourier ca d liu o. Ni mt cchc th, gi s mt tn hiu lin tc c o bng vic ly mu t ADC v lu tr n nh l mt tnhiu ri rc. Nu tn hiux(t) l gii hn trong mt khong (l hu hn), th DTFT ca x[n] ctnh ti cc im ri rc vi tn s s dng DFT, X k c cho bi phng trnh (29.4). S dng

cc quan h /T = , 2 /k N = , v2 /

( )k k NX X == , trong N l chiu di ca chuix[n] v T l chu k ly mu, c cho bi quan h sau:

2 /( ) , 0 / 2, 0 ( 1) / 2k sk NTX TX k N = =

Trong =2 /Ts l tn s ly mu c n v rad trn giy. tng chnh xc c thgim chu k ly mu T, v tng kt qu trn th bng cch tngNT.

Nu tn hiux(t) khng b gii hn trong mt khong, n phi c ct b s dng phngphp s tnh bin i Fourier lin tc. Nh trnh by trong phn Tn hiu ri rc, nu tn hiuc ly mu gim dn v 0, th chn s im c ly mu N ln sao cho x(t) nh v cth b qua k t gi tr . Nu tn hiu ly mux[n] l chu k, chn chu k ly mu Tv s im

Nsao cho tn hiux[n] bng mt s nguyn ln chu k. V d, xt tn hiu ( ) os( t)x t c = . Nu chuk ly mu c chn l T= 0.4s, th tn hiu c ri rc ha l [n]=x(nT)=cos(0.4 n)x , ltn hiu ging nh tn hiu c phn tch trong mc tn hiu ri rc. ChnN=5, 10, 15,... ns em li kt qu ng vi DFT cc gi tr khc s lm kt qu r r. Nu tn hiu x(t) c ph tnchu k, nhng n khng phi l tn hiu chu k, th s dng mt hm ca s nh trnh by trongmc Tn hiu ri rc gim r r khi tnh ton DFT.

Ch rng DFT cng c th c s dng xc nh cc h s Fourier ca cc tnh hiu chuk. Xt mt chui Fourier c dng

( ) j ktkk

x t c e

=

=

Ly mu tn hiux(t) th nht m bo l tn hiu c ly mu i qua mt s nguyn ln chu

k. Cc h s ck vi 0,...( 1) /2k N= c th c tnh bng ck= Xk/N trong Xk c thnhn c t DFT. Cc h s cn li c tnh t cng thc *k kc c =

p ng tn s ca mt h thng l t s ca bin i Fourier u ra trn bin i Fourier uvo, mt nh x gia mt h thng lin tc, ( )H , v mt h ri rc tng ng, ( )dH , c thc nhn t php nh x trc l

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S tay C in t

( ) ( )d TH H vi ==

nh x ny th rt hu dng trong c vic thit k cc b lc s v cc b iu khin s.

29.4 Thit k b lc s

Chc nng ca p ng tn s ca h ri rc l miu t cch h thng x l cc tn hiu u

vo vi cc tn s khc nhau. Xt mt tn hiu vo 0[n]=Acos( )x n ti mt h thng vi p ngtn s l ( )H trong 0 2 . u ra tng ng c tnh bi

0 0 0[ ] ( ) cos( ( ))y n H n H = +

Vi cc tn hiu khng chu k, c tnh lc ca bin i Fourier c mi quan h:

( ) ( ) ( )Y H X =

V vy, nu ( )H nh qu mt di tn xc nh, th tn hiu u vo vi ph tn trong di tnb suy gim s cho chng i qua h thng.

Thng th rt tin li khi lc mt tn hiu bng mt b lc s, nh hnh 29.4. B bin itng t s ADC ly mu tn hiu lin tc to ra mt chui cc tn hiu ri rc x l bi my

tnh hoc b vi x l tn hiu. Tn hiu c lc c th c lu tr di dng s hoc c giqua mt b bin i s tng t (DAC). B lc s c thc hin bng phn mm bng mtphng trnh quy nhn c t phng trnh sai phn. Xt mt b lc s vi hm truyn:

1 1

1 2 1 1 2 1

1 1

1 2 1 1 2 1

( ) N N N

N N

N N N

N N

b z b z b b b z b zH z

a z a z a a a z a z

+ +

+ +

+ + + + + += =

+ + + + + +L L

L L

Gi tr hin thi ca u ra y[n] c tnh s dng cng thc quy cho bi phng trnh saiphn:

1 2 1 2 1

1

1[ ] ( [ ] [ 1] [ ] [ 1] [ ])N Ny n b x n b x n b x n N a y n a y n N a + +

= + + + L L (29.5)

Ch rng cc gi tr trc ca y v x phi c gi li s dng trong tnh quy.

By gi xt p ng xung ca b lc s, vi y[n] c tnh t mt u vo x[n] l mt xung(v d [n]=1 khi n=0 v cn li [n]=0 ). Php tnh quy trnh by trn em li mt p ngca y[n] l v hn (khng c gi tr no ca M lm cho y[n]=0 vi mi n>M). B lc kiu nyc gi l b lc p ng xung v hn (infinite impulse response IIR).

Hnh 29.4 Cu hnh ca phn cng x l tn hiu

By gi xt trng hp cc h s khuch i ca b lc am = 0 vi m > 1. Biu thc kt qu cay[n] t phng trnh (29.5) khng cn phi tnh quy na khi n ch ph thuc vo gi tr hin tiv cc gi tr trc ca x, v khng ph thuc vo gi tr trc ca y. Kt qu l p ng xungc th c trong khong N. B lc kiu ny c gi l mt b lc p ng xung hu hn (FIRfinite impulse response filter). Cc b lc FIR i khi c chung hn cc b lc IIR v chng cpha tuyn tnh trong p ng tn s. Pha tuyn tnh ngha l gc ca p ng tn s l khi

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l hng s. p ng ny s l mt khong tr trong min thi gian. Cc phng php thit k choc 2 kiu b lc c miu t trong 2 mc tip theo.

Thit k b lc IIR

Hai phng php thit k cc b lc IIR c gi l m phng tng t (thit k gin tip)v thit k trc tip. Thit k gin tit c thc hin bng cch thit k mt b lc analog trcsau s dng mt nh x c miu t trong mc nh x t min s sang minz chuyn

i n thnh mt b lc s. Phng php ny c cc u im l s dng c nhiu cc cng nghthit k cho cc b lc tng t s dng cho thit k b lc s. Phng php thit k trc tip nichung bao gm cc cng ngh s, v n thng c u tin hn thit k gin tip khi chu k lymu khng qu nh. Thit k trc tip l phm vi vt ra ngoi phm vi ca cun sch ny; Thamkho [2] bit chi tit hn.

Thit k b lc tng t bt u bng vic chn mt di thng, mt mu ca b lc v bc cab lc. Thm na, xc nh s lng ca gn sng c th c ci cho php trong di thng(passband) hoc di chn (stopband). Hai mu thit k analog in hnh l b lc Butterworth vb lc Chebyshev.

B lc butterworth: B lc Butterworth c c tnh ha bng vic khng c im khng vc cc im cc nm trn na ng trn bn tri ca mt phng s. Khong cch ca cc im cc khi to rng ca di tn v c biu th bng b . Gc ca cc im cc c xc nh

bng vic cch 2 im cc bng nhau xung quang mt ng trn kp kn vi bn knh b vsau gi cc im cc ch nm trn bn tri mt phng, nh hnh 29.5. B lc bc N th c chobi cng thc:

)

( )(

N

b

k b k

H ss p

=

Trong :

/

( 0.5) /

1 3 1,

2 23 2

,

2 2

jk N

k

j k N

N Ne k n vi N l

pN N

e k n vi N ch n

+

+ == =

Hnh 29.5 Phn b im cc ca b lc Butterworth bc 4

9


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S tay C in t

Hnh 29.6 So snh cc b lc Butterworth tng t

B lc ny l thng thp trong bin ca p ng tn s trong b lc l mt di hp l vgn vi gi tr 1 vi b < v gim pha xa tn s di thng. Bc ca b lc cng ln th gimcng nhanh. So snh 3 b lc trong hnh 29.6. Ch rng cc im chuyn ca b lc c bc caophi tr gi l pha th cng b gim i mt cch t ngt. S nh pha tr ln quan trng trong hthng o thi gian thc nh yu cu ca cc b iu khin phn hi.

B lc Chebyshev: Khng n iu nh b lc Butterworth, b lc Chebyshev cho php mtvi gn sng trong th bin cho di thng hoc di chn. Kiu 1 ca b lc Cheyshev cho phpgn sng trong di thng trong khi b lc Chebyshev kiu 2 cho php gn trong di chn. Viccho php mt gn sng dn n b lc Chebyshev c im chuyn i nhn hn gn bng thnghn b lc Butterworth cng bc. Trong thit k Chebyshev, tn s ct c thng c xc nhtri vi bng thng. Tn s ct l tn s ti im bin ca b lc suy gim xung mt t s ttrc ca gi tr DC. Khi t s ny l 0.707, tn s ct l bng thng. Bnh thng, trong thit k,

t s ny c chn ph hp vi tng s gn sng c php trong di thng. Mt b lc thngthp kiu 1 c nh ngha bng mi quan h:

2 2

1( )

1 ( / )N cH

C

=+

v

1 2( ) 2 ( ) ( ) N N N C x xC x C x =

Biu thc ( )NC x c gi l a thc Chebyshev bc thN , v n c tnh quy bt u vi

0( ) 1C x = v 1( )C x x= . Gi tr tng 0> xc nh tng s gn sng cho php trong di thng;

Ni mt cch c th, gn sng nm gia gi tr 1 v 21/ 1 + . Xt v d, b lc Chebyshev kiu

1 nh trong hnh 29.7; Cc b lc ny c thit k c gn l 1dB trong di thng (0.51= ). Ch rng 1= vi gn 3dB.

Nh cp trn cc b lc mu ny l thng thp. thit k mt kiu b lc khc (thngcao, thng mt di), u tin l thit k b lc thng thp, ( )H s , vi mt tn s ct c (thng

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thng c chn l 1). Sau s dng mt php bin i tn s chuyn i b lc ny sangkiu mong mun. Cc php bin i trong min tn s chun s c trnh by di y.

T thng thp ti thng thp: t c mt b lc thng thp vi tn s ct 1 , thays trong

H(s) ban u bi 1/cs .

Hnh 29.7 S so snh ca cc b lc Chebshev kim analog

T thng thp n thng cao: t c mt b lc thng cao vi mt di thng t 1 n , thay s trong hm truynH(s) bi 1 /c s .

T thng thp ti thng di: t c mt b lc vi di thng t 1 n 2 , thay s tronghm truynH(s) bi

2

2 1

2 1( )

s

s

+

T thng thp n chn mt di: t c mt b lc chn di vi di chn t 1 n 2 ,thays trong hm truynH(s) bi

2 1

2

2 1

( )s

s

+

Thit kt b lc FIR

Mt cch c c b lc FIR l cht p ng xung ca mt b lc IIR l tng. V d, mtb lc thng thp IIR l tng c p ng tn s:

,( )

0,

c cAHtruong hopkhac

=

Trong A l mt hng s v c l tn s ct. p ng xung ca b lc ny nhn c bngvic nghch o phng trnh DTFT (29.3):

11


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S tay C in t

/ 2[ ] sincjnc cA n

h n e c

=

Ch rng phng trnh ny c khong gii hn trong c n < 0 v n > 0. To mt b lc FIRphi cht p ng xung t n N< v n N> . Tuy nhin, b lc IIR ban u v b lc FIR b chtu phi nhn qu; iu ngha l p ng xung l khc 0 vi n < 0. Cc b lc phi nhn qu cncc gi tr tng lai ca u vo tnh gi tr u ra hin ti; V vy, n khng th c thc

hin trong thi gian thc. V l do ny, thit k IIR ch yu s dng cc b lc khng l tng(thng da trn cc mu analog) chng xp x p ng tn s l tng. khi lc mt tn hiu off-line c lu tr, Tnh nhn qu khng cn thm v tt c cc gi tr ca tn hiu l c sn (gm ccc gi tr tng lai).

Th t thc hin thi gian thc ca mt b lc FIR c tng qut bng vic cht mt b lcIIR l tng, b lc ny phi c tr sao cho tt c thng tin tn hiu ca p ng xung xut hint khi n > 0. Vic tr trong min thi gian tng ng vi mt lch pha trong min tn s. Vvy, mt b lc FIR c thit k bng cch trc tin l chn mt b lc IIR l tng (thngthp, thng cao), sau thc hin nghch o DTFT tm p ng xung, v cht p ngxung, v cui cng, lm tr n trong min thi gian. Mt mt phng php tng ng v honho hn l xp xp li cc bc c miu t trn. Th nht, cng thm mt lch pha trongp ng tn s ca b lc IIR l tng. Vic ny c lm bng cch nhn p ng tn s vi

( 1) / 2j Ne . Sau thc hin nghch o DTFT v cht n vi n < 0 v n >N 1. Kt qu l mt b

lc FIR nhn qu vi bc N.Sau y l b lc FIR tng qut bc Nn c tng qut ho s dng phng php miu t

trn. Chn ( 1) / 2m N= .

B lc FIR thng thp vi tn s ct c :

, 0

( )[ ] sin , 0 1

0, -

c

c c

n

n mh n c n N

C ctr ng hpcnl i

=

= <

Trong sinc(x) = sin( x)/ x

B lc FIR thng cao vi di thng t 1 :

1

1 1

1 , 0

( )[ ] sin 0, 0 1

0, -

vi n

n mh n c vi n N

C ctr ng hpcnl i

=

= <

B lc FIR thng mt di vi di thng 1 2n :

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2 1

2 1

2 1

, 0

[ ] sin [ ( ) / ] sin[ ( / ], 0 1

0, -

vi N

h n c n m n m vi n N

C ctr ng hpcnl i

=

= <

thc hin cc b lc ny, cc h s trong phng trnh (29.5) c t [m-1]mb h= ,

11a = , v 0ma = vi 0m > .

Cc b lc FIR c thit k s dng phng php ny c p ng tn s c im chuyn igia di thng v di chn nhn hn (bc cng ln th im chuyn i cng nhn), nhng chnggi cho ng dc thnh mt ng gn gia di thng v di chn. ng gn ny nhn c tvic cht p ng xung ca b lc IIR. Vic ct t t s dng mt hm ca s lm cho p ngtn s tr ln trn tru. Hm ca s c trnh by trong phn nh x t min s sang minzdng thu thp d liu cng c s dng trong thit k b lc FIR, khi cc b lc c thayi thnh [n]w[n]h . Thit FIR s dng cc ca s khc c trnh by chi tit hn trong [1,2].

Thit k b lc s vi tr gip ca my tnh

Mablab l mt gi phn mm phi bin phn tch x l tn hiu v thit k. Toolbox x ltn hiu cha nhiu lnh thit k v m phng cc b lc s. V d, cc lnh butter v cheby1thit k t ng mt b lc tng t cho mt b lc IIR v sau s dng php bin i songtuyn nh x ti cc b lc trong min ri rc. Cc b lc thng thp, thng cao, di chn v dithng c th c thit k s dng cc lnh ny min l cc tn s ct trong min s, c xcnh theo . thit k mt b lc thng thp s da trn b lc Butterworth anolog vi tn sct w1, s dng lnh [b, a]=butter(N,w1*T/pi, 'hight') . Trong Nl s im cc, Tl chu k, vw1*Tl tn s ct. Lnh ny t h s khuch i ca b lc, c nh ngha trong phng trnh(29.5), trong cc vc tb v a theo th t. thit k mt b lc thng cao vi tn s ct analogw1, s dng lnh [b, a]=butter(N,w1*T/pi, 'hight') . thit k mt b lc thng mt di s vi di

thng tng t t w1 n w2, nh ngha 1 2[ , ]w w w= v s dng lnh [b, a]=butter(N,w1*T/pi) .

thit k b lc di chn s vi di chn tw

1 tiw

2, nh nghaw=[w1,w2]

v s dng lnh[b, a]=butter(N,w1*T/pi,'stop') . Thit k mt b lc Chebyshev kiu 1 bc thN c thc hinging nh cc phng php cho butter tr butter c thay bng cheby1.

Toolbox x l tn hiu cng cung cp cc lnh thit k cc b lc FIR. c mt b lc FIRthng thp vi chiu diNv tn s ct analog w1, s dng lnh ir1(N-1, w1*T/pi)h f= . Kt qu vcth cha p ng xung ca FIR trong h(1) l gi tr ca h[0]. Cc gi tr trong vector h cng bngcc h s b trong phng trnh (29.5) vi bc ln hn. ( 1 1a = v 0ma = vi m > 1). Mt b lc FIRthng cao chiu di N vi tn s ct analog w1 c thit k bng cch s dng lnh

1( 1, 1* / , ' ')h fir N w T pi high= . Mt b lc FIR thng mt di vi bng thng t w1 ti w2 cthit k bng lnh 1( 1, 1* / )h fir N w T pi= trong [ 1, 2]w w w= . Mt b lc FIR chn mt divi di chn t w1 n w2 c thit k bng lnh 1( 1, 1* / , ' ')h fir N w T pi stop= trong

[ 1, 2]w w w= . Lnh fir1 mc nh s dng ca s Hamming. s dng ca s khc a thm lachn hanning hoc boxcar ( l nhng ca s hnh thang) vi cc tham s; v d,

1( 1, 1* / , ' ', oxcar(N))h fir N w T pi high b= to b lc FIR thng cao vi tn s ct tng t w1 sdng mt ca s hnh thang.

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S tay C in t

Lnh b lc trong Matlab c s dng tnh u ra ca b lc s thm ch u vo l chui.Mt v d v trng hp ny l er(b, a, x)y fit = trong b v a l cc h s ca b lc v x lu vo chui u vo.

Cc v d v b lc

Thng thng, nhiu vi tn s 60 Hz b ln vo cc gi tr o ca h thng c in do inp tuyn tnh. (Lu , Chu u thng ly nhiu tn s 50 Hz). Vi mc ch chng minh,mt tn hiu vi tn s 60 Hz c chng vo tn hiu c tn s thp hn nh hnh 29.8. gimbi s nh hng khng tt ca nhiu 60 Hz, s dng b lc chn mt di. Thng thng, b lcchn mt di c thit k vi tn s khong 60Hz trnh p ng kiu nh hnh 29.8. Cclnh Matlab sau y c th c dng thit k b lc s chn mt diButterworth bc 8 m cctn s gy l 50 v 70Hz. V vy, b lc ny loi b c nhiu 60Hz.

T = 0.001; % Chu k ly mu

n = 4; % mt na bc ca b lc

low_freq = 50 * (2*pi); %Tn hiu chn gia 50 v 70 Hzhigh_freq = 70 * (2*pi);

w1 = low_freq*(T/pi); % tn s gy

w2 = high_freq*(T/pi);

w = [w1 w2];

[b,a] = butter(n,w,stop); % Cc h s ca b lc

W = -pi:pi/200:pi; % nh ngha mt vc t tn s ri rc(digital frequency)

H = freqz(b,a,W); % Tnh ton p ng tn s

Hnh 29.8 Gi tr o b ln nhiu tn s 60 Hz

Hnh 29.9 b lc chn

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Hnh 29.10 Gi tr o c lc

Hnh 29.9 minh ha bin ca p ng tn s i vi b lc IIR thu c. Ch rng bintn s c v trong di [ - , ] trong tn s DC tng ng vi 0 = v tn s cao nht chophp l = . Trong v d ny, cc tn s gy tng ng vi 1 50(2 ) 0.314T = = v

270(2 ) 0.44T = = Hnh 29.10 l kt qu p dng ca b lc ny vi tn hiu nhiu. Vi cc

mc ch thc tin, nhiu tn s 60Hz c lm suy gim hon ton. Nh ta c th nhn thy tronghnh 29.10, Ch c nhiu trong khong thi gian qu 100ms u ca p ng step. y l s kthp ca b lc Butterworth bc bn v qu h thng vi tn s 60 Hz. Nn ch rng tn sly mu ca 1000kHz th nhanh n thc hin chnh xc tn hiu 60Hz. Nu s dng tn s ly

mu nh hn 120Hz, th tn hiu tn s 60Hz s nh thin, v d lc n mc no vn khng loib c cc hiu ng ca nhiu tn s 60Hz.

Thm mt ng dng khc ca b lc s trong c in t c s dng khi thc hin c lng dch chuyn t vic o gia tc. Mt phng php n gin tnh dch chuyn l tch phn2 ln gia tc. Trong mins, 2 b tch phn ny tng ng vi nhn 21/s . S dng bin i songtuyn chuyn 21/ s sang minz sinh ra hm truyn sau,

2 2 2 1 2

2 ( 1)2 2 1 2

2( 1)( 1)( 1)

1 2 1 1 2( ) ( )

4 42 1 1 2z

s zT z sT z

T z z T z zH z H s

s z z z z

= + =

+

+ + + += = = = + +

Phng trnh sai phn tng ng dng tnh dch chuyn []y t phng trnh dd[]y l2

dd dd4(y[n]-2y[n-1] + y[n-2])=T (y [n] + 2y [n-1] + ydd[n-2]) . Tuy nhin, cc gia tc k thngkhng p ng tt ti cc tn s thp; trong thc t, ngi ta thng chn mt ng dc vo dliu lm gim s sai lnh trong tnh ton dch chuyn. Cng rt nhy cm vi cc chuyn ngngu nhin. Mt phng php khc x l d liu gia tc bng b lc thng mt di trc khi sdng phng trnh sai phn tch hp n v s lng. di bng thng phi cha cc tn s trungtm trong h thng.

Hnh 29.11 S o gia tc

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S tay C in t

Hnh 29.12 dch chuyn thc t

Hnh 29.13 dch chuyn c c lng khng s dng b lc trc

Xt v d, d liu gia tc nh hnh 29.11, vi mt vi tn hiu nhiu ngu nhin. Tn hiu nyc ly mu di 6400Hz, tn s trung tm ca h thng l 50Hz. Hnh 29.12 minh ha dch chuyn thc, cn hnh 29.13 minh ha dch chuyn ti u c tnh bng cch tch hpton b d liu gia tc, s dng phng trnh sai phn trn. c lng ny rt km.

Vi s lu chn khc, mt b lc thng mt di Chebyshev kiu 1 bc 8 analog vi bng thng25 500Hz c thit k v c ri rc ha s dng php bin i song tuyn. D liu gia tcc x l bng b lc ny trc, v sau d liu lc c tch hp v s lng vi kt qu nhhnh 29.14. Ch rng c lng ny tt hn nhiu kt qu t c m khng s dng b lcthng di. Mt b lc thng di FIR bc 500 cng c thit k trong v d ny vi bng thng 25

500 Hz. Sau khi d liu qua b lc FIR, th n c tch hp s lng em li dch chuync c lng nh hnh 29.15. Do c c

tnh pha tuyn tnh ca b lcFIR, nn n c t s mo hn b lc IIR, nhng n c tr ln hn. Bc cng ln, chnh xcca kt qu cng cao, v cng t thng tin c ngha b mt trong qu trnh cht p ng xung ca

b lc thng mt di IIR, nhng tr th ln hn. Bin ca b lc IIR v cc b lc thng mtdi FIR nh hnh 29.16.

Hnh 29.14 dch chuyn c c lng vi b lc IIR

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Hnh 29.15 dch chuyn c c lng vi b lc FIR

Hnh 29.16 Cc b lc thng di s: (a) b lc IIR Chebyshev v (b) b lc FIR

Hai iu c cp trong v d ny l:

1. Vic tnh ton khng lc v cng nhy cm vi dc trong d liu (c giiquyt bng cch nhn 2 b tch phn). V vy, ng dc c d b khi gia tctrc khi x l. C 2 b lc gi b ng cho mt cch hiu qu, do kt quo khng thay i nu nh ng cho vn bnh thng.

2. B lc FIR th hin mt vi tr. C l, nguyn nhn ca tr ny l b lc dng nh vn c mt vi cn tr vi vng di chn nh gn vng ban u. Tng

vng di chn lm gim sai lch. iu ny c th thc hin c bng cch gim tns ly mu hoc bng cch tng tn s bng thng. C 2 bin php ny u gim sailch nhng tng cc sai s khc trong tn hiu. Tng chiu di ca b lc gim sai str khng cn a vo cc sai s khc.

Cc lnh Matlab s dng thit k cc b lc v cc kt qu l:[num, den]=c2dm(1, [1 0 0], T, tustin); % S ha bc 2

y1=filter(num, den, ydd); % 2 b tch phn ca ydd

Wbreak=[2*pi*25*T, 2*pi*500*T]; % Cc tn s gy

[b,a] =heby1(4,1,Wbreak); % Thit k b lc IIR vi 1dB gn

W =pi:pi/200:pi; % nh ngha di tn s ri rc cho th

H =reqz(b,a,W); % kt qu p ng tn splot(W,abs(H)); % th bin ca p ng tn s

yddfilt=filter(b,a,ydd); % Tnh ton u ra ca b lc ydd

y2=ilter(num,den,yddfilt); % 2 b tch phn ca yddfilt

hfir =fir1(500,Wbreak); % thit k b lc FIR vi bc 500

17


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S tay C in t

yddfilt=filter(hfir,1,ydd); % Tnh ton u ra cho b lcFIR

y3 =filter(num,den,yddfilt); % 2 b tch phn ca yddfilt

29.5 Thit k iu khin s

Cng nh trong trng hp thit k b lc s, c 2 phng php ch yu thit k mt b

iu khin s: phng php thit k gin tip c da trn vic ri rc ha mt thit k analog,v phng php trc tip da trn mt m hnh (thng s dng phng php p ng step) vsau thit k b iu khin trc tip trn min ri rc. Hu ht cc k s hc iu khin lin tckinh in, v hc nhiu v thit k b iu khin lin tc hn l b iu khin ri rc hoc thit kiu khin s. Rt may, cc cng c thit k iu khin lin tc thng cng c th c s dngthit k cc h iu khin s. to cc b iu khin thit k trong min lin tc, s dng mtnh x t mins sang minz. Cc nh x trnh by trong chng ny c th c s dng chocc loi b iu khin. Cch xc nh bng nh x l cch tt nht tm tng b iu khin hocb lc. Mc d vic xp x song tuyn phc tp hn xp x thun hoc xp x ngc, nhng n vnc s dng cho hu ht cc h c in t. iu ny l do trong thc t hu ht cc b iu khinhin i u c sc tnh ton qun l phc tp c tng ln cc di thng yu cu cah c in t.

Mt v d v phng php thit k gin tip, xt mt b iu khin t l vi phn PD

(proportional derivative) c s dng lm tng cht lng ca h thng. cc h s vi phn vt l cho b iu khin ln lt lKd vKp. B iu khin PD,K(s), c cho bi cng thc sau:

( ) d pK s K s K = + (29.6)

Phng trnh (29.6) c chuyn sang min s (digital) s dng bt k nh x no t min ssang minz trnh by trong phn trc ca chng ny. Mt v d bin i song tuyn s dngb iu khin s tng qut,K(z).

2 ( 1)

( 1)

(2 ) ( 2 )( ) ( ) d p p d z

sT z

K TK z TK K K z K s

Tz T

=+

+ + = =

+(29.7)

Ngoi cc h s iu khin, ch c 1 tha s cn trong phng trnh (29.7) l T, thi gian trchmu. Nh trnh by trn, thi gian trch mu phi t nht phi ln gp 5-10 ln hng s thigian h thng nhanh nht. Tuy nhin, thi gian trch mu th thng c chn nhanh hn vi trmln hng s thi gian nhanh nht. Mt chin lc thay i cho mt h thng phn hi l chn dithi gian ly mu ti thiu l 20 ln di thng h kn l tng. Thi gian trch mu v c bnnhanh hn h thng thc gim bt s khc bit gia b iu khin c thit k trong min lintc v thc hin trong min ri rc. Nn ch rng tn s ly mu cng ln, cc h s ca b iukhin cng nh. V d, trong phng trnh (29.7), khi thi gian ly mu cng nh, Tcng nh yucu phn gii s tt hn cho cc h s ca b iu khin. Nu Ttr ln nh hn phn gii hs b iu khin, th n c th thc hin sai ti gi tr 0 dn n sai lut iu khin.

V d v iu khin s

Xt mt ng c c nh v tc cao vi ng hc ng c c cho bi phng trnh bcnht

( )( )

( ) 1m

in m

KsG s

V s T s

= =

+

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Trong mK l hng s khuch i ca ng c, mT l hng s thi gian ca ng c, ( )s l

bin i Laplace ca vn tc, v ( )inV s l bin i Laplace ca in p vo ng c. xc nh

gi tr ca mT v mK , s dng p ng step ca vn tc. Hnh 29.17 l p ng ca ng c vi

u vo l step 1V. H s khuch i ng c, mK , l gi tr n nh ca tc ng c v l 5.Kt qu ny cng c th c xc nh s dng cc lp lun nh sau:

0 0

0 0

lim ( ) lim ( ) lim ( ) ( )

1 1lim ( ) lim

1

int s s

mm

s sm

t s s sG s V s

KsG s s K

s T s s

= =

= = =+

Hng s thi gian, mT c th c tnh bng cch xc nh vn tc ng c vi p ng step

ti mt T= nh sau:

/ 1( ) (1 ) (1 ) 0.632mt Tm m m mt T K e K e K = = = =

V thi gian cn ng c t c 63.2% trng thi n nh ca p ng step chnh l hngs thi gian. T hnh 29.18, hng s thi gian ca ng c ny l 0.05s. V vy hm truyn cang c ny l:

( )( ) 5( ) 29.8( ) 1 0.05 1

m

in m

KsG sV s T s s= = =

+ +

Hnh 29.17 p ng step ca vn tc ng c

Hnh 29.18 p ng step ca v tr ng c.

Trong v d ny, ng c c s dng ch iu khin v tr. V tr ng c l tch phn cavn tc, Phng trnh (29.8) c thm vi mt b tch phn to ra hm truyn ca mi quanh in p u vo ng c vi v tr u ra ng c ( )s :

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S tay C in t

( )( )

( ) ( 1)m

p

in m

KsG s

V s s T s

= =

+(29.9)

Mt b iu khin PD c chn s dng trong v d ny lm tng cht lng ca h thng.Thit k p ng nhanh v khng c qu iu chnh, h s vi phn, dK , v h s t l, pK ,c chn ln lt l 0.05 v 1, to ra lut iu khin nh sau:

( ) 0.05 1d pK s K s K s= + = + (29.10)Trn danh ngha, thit k ny b qua im cc tn s cao ca ng hc ng c trong phng

trnh (29.9). Chu k ly mu trong v d ny chn l 1 ms l nhanh hn ng k so vi hng sthi gian ca h thng, v n khng phi l gi tr khng hp l cho cc b iu khin hin i.Nh trnh by trn, s dng tn s trch mu 1kHz (1ms) lm gim bt s khc nhau gia biu khin c thit k trong min lin tc v s thc hin ca n trong min ri rc. S dngphp bin i song tuyn nh trnh by trong mc nh x t mim s sang minz kt qu mtb iu khin s c dng:

101 99( )

1Dz

K zz

=

+

Trong thc t, p ng h kn ca h thng s dng b iu khin s khng d phn bit vi h

thng s dng b iu khin tng t cho bi phng trnh (29.10). p ng v tr h kn ca ngc vi u vo 100 nh hnh 29.18.

Nh cp trong mc V d v b lc, tn hiu nhiu tn s 60 Hz thng xut hin trongo lng cc h c in, v vy b lc chn di thng c s dng lm gim i tn hiunhiu. Trong hot ng ca h kn, b lc s thng di c ni tng vi b iu khin PD s.

Ti liu tham kho

[1] Kamen, E.W., and Heck, B.S., Signals and Systems Using the Web and Matlab,2nd ed., Prentice-Hall, Englewood Cliffs, NJ, 2000.

[2] Britton Rorabaugh, C., Digital Filter Designers Handbook: with C++Algorithms, 2nd ed., McGraw-Hill, New York, 1997.

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sổ tay cdt chuong 29-xly sign so

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