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By Hoang Le Uyen Thuc , Electronics And Telecommunication Department , Đà Nẵng University Of Technologies , Việt Nam
Transcript of DSP Vietnamese
Chng I
Chng
1
GII THIU X L TN HIU SChng ny nu tng qut cc vn lin quan n mn hc. Ni dung chnh chng ny l: Gii thch cc khi nim nh: Tn hiu, Tn hiu s, X l tn hiu, X l tn hiu s... Cc khu c bn trong h thng x l tn hiu s Nu mt s ng dng ca x l tn hiu s So snh x l tng t v x l s Gii thch khi nim Tn s Cc bc c bn chuyn i tn hiu t tng t sang s Cc bc c bn chuyn i tn hiu t s sang tng t
1.1 TN HIU, H THNG v X L TN HIU hiu X l tn hiu l g, ta s tm hiu ngha ca tng t. Tn hiu(signal) dng ch mt i lng vt l mang tin tc. V mt ton hc, ta c th m t tn hiu nh l mt hm theo bin thi gian, khng gian hay cc bin c lp khc. Chng hn nh, hm: x(t ) = 20t 2 m t tn hiu bin thin theo bin thi gian t. Hay mt v d khc, hm:
s ( x, y ) = 3 x + 5 xy + y 2 m t tn hiu l hm theo hai bin c lp x v y, trong x v ybiu din cho hai ta khng gian trong mt phng. Hai tn hiu trong v d trn thuc v lp tn hiu c th c biu din chnh xc bng hm theo bin c lp. Tuy nhin, trong thc t, cc mi quan h gia cc i lng vt l v cc bin c lp thng rt phc tp nn khng th biu din tn hiu nh trong hai v d va nu trn.
Hnh 1.1 V d tn hiu ting ni Ly v d tn hiu ting ni- l s bin thin ca p sut khng kh theo thi gian. Chng hn khi ta pht m t away, dng sng ca t c biu din trn hnh 1.1. Mt v d khc l tn hiu in tm (ECG)- cung cp cho bc s nhng tin tc v tnh trng tim ca bnh nhn, hay l tn hiu in no (EEG) cung cp tin tc v hot ng ca no. Cc tn hiu ting ni, ECG, EEG l cc v d v tn hiu mang tin c th biu din l hm theo bin thi gian. Thc t c nhng tn hiu l hm theo nhiu bin c lp. V d nh tn -1-
Chng I
hiu nh (image)- l s thay i ca cng nh sng theo khng gian, c th xem l hm sng theo hai bin khng gian. Tt c cc tn hiu u do mt ngun no to ra, theo mt cch thc no . V d tn hiu ting ni c to ra bng cch p khng kh i qua dy thanh m. Mt bc nh c c bng cch phi sng mt tm phim chp mt cnh/ i tng no . Qu trnh to ra tn hiu nh vy thng lin quan n mt h thng, h thng ny p ng li mt kch thch no . Trong tn hiu ting ni, h thng l h thng pht m, gm mi, rng, li, dy thanh... Kch thch lin quan n h thng c gi l ngun tn hiu (signal source). Nh vy ta c ngun ting ni, ngun nh v cc ngun tn hiu khc. C th nh ngha h thng (system) l mt thit b vt l thc hin mt tc ng no ln tn hiu. V d, b lc dng gim nhiu trong tn hiu mang tin c gi l mt h thng. Khi ta truyn tn hiu qua mt h thng, nh b lc chng hn, ta ni rng ta x l tn hiu . Trong trng hp ny, x l tn hiu lin quan n lc nhiu ra khi tn hiu mong mun. Nh vy, x l tn hiu (signal processing) l mun ni n mt lot cc cng vic hay cc php ton c thc hin trn tn hiu nhm t mt mc ch no , nh l tch ly tin tc cha bn trong tn hiu hoc l truyn tn hiu mang tin t ni ny n ni khc. y ta cn lu n nh ngha h thng, khng ch n thun l thit b vt l m cn l cc phn mm x l tn hiu hoc l s kt hp gia phn cng v phn mm.V d khi x l s tn hiu bng cc mch logic, h thng x l y l phn cng. Khi x l bng my tnh s, tc ng ln tn hiu bao gm mt lot cc php ton thc hin bi chng trnh phn mm. Khi x l bng cc b vi x l- h thng bao gm kt hp c phn cng v phn mm, mi phn thc hin cc cng vic ring no . 1.2 PHN LOI TN HIU Cc phng php ta s dng trong x l tn hiu ph thuc cht ch vo c im ca tn hiu. C nhng phng php ring p dng cho mt s loi tn hiu no . Do vy, trc tin ta cn xem qua cch phn loi tn hiu lin quan n nhng ng dng c th. 1.2.1 Tn hiu nhiu hng v tn hiu nhiu knh Nh ni trong mc 1.1, tn hiu c th c m t l hm theo mt hoc nhiu bin c lp. Nu tn hiu l hm theo mt bin, ta gi l cc tn hiu mt hng (one-dimention signal), nh tn hiu ting ni, ECG, EEG. Ngc li ta gi l tn hiu nhiu hng (multidimention signal), v d nh tn hiu nh trng en, mi im nh l hm theo 2 bin c lp.y
y1
I(x1,y1)
x x1
Hnh 1.2 V d tn hiu nh mu (2 hng- 3 knh) -2-
Chng I
Trong mt s ng dng, tn hiu c to ra khng phi t mt m l nhiu ngun hay nhiu b cm bin. Cc tn hiu nh vy c gi l tn hiu a knh (multi-channel signal). Bc nh trn hnh 1.2 l mt v d v tn hiu 2 hng, 3 knh. Ta thy sng I(x,y) mi mt im l hm theo 2 bin khng gian c lp, sng ny li ph thuc vo sng ca 3 mu c bn red, green v blue. Mt v d khc, tn hiu nh TV mu l tn hiu 3 hng- 3 knh, c th biu din bi vector sau :
I r (x, y, t) I(x, y, t) = Ig (x, y, t) I (x, y, t) b Trong gio trnh ny, ta tp trung xt tn hiu mt hng- mt knh, bin l bin thi gian (mc d thc t khng phi lc no bin cng l bin thi gian) 1.2.2 Tn hiu lin tc v tn hiu ri rc Tn hiu lin tc (continuous-time signal) hay cn gi l tn hiu tng t l tn hiu c xc nh ti tt c cc gi tr thi gian. V mt ton hc, c th m t tn hiu ny l hm ca mt bin lin tc, v d tn hiu ting ni. Tn hiu ri rc (discrete-time signal) ch c xc nh ti mt s thi im no . Khong cch gia cc thi im ny khng nht thit phi bng nhau, nhng trong thc t thng l ly bng nhau d tnh ton. C th to ra tn hiu ri rc t tn hiu lin tc bng 2 cch. Mt l ly mu tn hiu lin tc, hai l o hay m mt i lng vt l no theo mt chu k nht nh, v d cn em b hng thng, o p sut khng kh theo gi... Tn hiu x(t n ) = e n , n = 0, 1, 2, 3,... l mt v d v tn hiu ri rc. Ta c th dng bin nguyn n thay cho bin thi gian ri rc tn. Lc ny, tn hiu tr thnh mt hm theo bin nguyn, v mt ton ta c th biu din tn hiu ri rc l mt dy s (thc hoc phc). Ta s dng k hiu x(n) thay cho x(tn), ngha l tn = nT vi T l hng s- khong cch gia hai thi im ri rc cnh nhau. Hnh 1.3 l mt v d v tn hiu ting ni ri rc.t
Hnh 1.3 V d tn hiu ri rc 1.2.3 Tn hiu bin lin tc v tn hiu bin ri rc Bin ca c tn hiu lin tc v ri rc u c th lin tc hay ri rc. Nu tn hiu c tt c cc gi tr trong mt di bin no th ta gi l tn hiu bin lin tc (continuous-valued signal). Ngc li, nu tn hiu ch ly mt s gi tr no (cn gi l mc) trong mt di bin th l tn hiu bin ri rc (discrete-valued signal). -3-
Chng I
Khong cch gia cc mc bin ny c th bng nhau hay khng bng nhau. Thng th ta biu din cc mc bin ny bng mt s nguyn, l bi s ca khong cch gia hai mc bin cnh nhau. Tn hiu ri rc theo c thi gian v bin c gi l tn hiu s (digital signal). Hnh 1.4 l mt v d v tn hiu s.
Hnh 1.4 V d tn hiu s vi 6 mc bin khc nhau x l tn hiu, trc ht phi thu ly c tn hiu. V d ta thu ly tn hiu m thanh bng microphone, chuyn i tn hiu m thanh sang tn hiu in. Hay nh tn hiu nh, ta c th thu ly bng my nh. Trong my nh tng t chng hn, tn hiu nh sng iu khin cc phn ng ha hc trn mt tm phim nh. V bn cht, cc tn hiu t nhin u l tng t, c s mc bin v s thi im u l v hn. Do vy, tn hiu tng t khng ph hp x l bng cc h thng s. x l s, tn hiu tng t c ly mu vo cc thi im ri rc, to thnh tn hiu ri rc, sau lng t ha bin ca n thnh mt tp cc mc bin ri rc. Qu trnh lng t ha (quantization) tn hiu, v c bn l mt qu trnh xp x ha. N c th c thc hin d dng bng cch lm trn hay ct gt. V d tn hiu c gi tr l 8.62 c th c xp x ha thnh 8 (nu lng t ha bng cch ct gt) hay l 9 (nu lng t ha bng cch lm trn) 1.2.4 Tn hiu xc nh v tn hiu ngu nhin Qu trnh phn tch ton hc v x l tn hiu yu cu phi m t c tn hiu. S m t ny lin quan n mt m hnh tn hiu. Da vo m hnh tn hiu, ta c mt cch phn loi tn hiu khc. Cc tn hiu c th c m t duy nht bng mt biu din ton hc r rng nh l th, bng d liu... c gi l tn hiu xc nh (deterministic signal). T xc nh mun nhn mnh l ta bit r v chc chn cc gi tr ca tn hiu trong qu kh, hin ti v tng lai. Tuy nhin trong nhiu ng dng thc t, c nhng tn hiu khng th biu din chnh xc bng cc cng thc ton hc hay nhng m t ton nh vy l qu phc tp. Ta khng th on trc s bin thin ca cc gi tr ca loi tn hiu ny. Ta gi y l tn hiu ngu nhin (random signal). V d tn hiu nhiu l tn hiu ngu nhin. Ta cn lu rng vic phn loi tn hiu thc thnh xc nh hay ngu nhin khng phi lc no cng r rng. i khi, xem tn hiu l xc nh hay ngu nhin u dn n nhng kt qu c ngha. Nhng i khi, vic phn loi sai s dn n kt qu b li, bi v c nhng cng c ton ch c th p dng cho tn hiu xc nh, trong khi cc cng c khc li ch p dng cho tn hiu ngu nhin. iu ny s tr nn r rng hn khi ta kim tra cc cng c ton c th. 1.3 H THNG X L TN HIU 1.3.1 Cc khu c bn trong mt h thng x l s tn hiu Nh ni trn, hu ht cc tn hiu bt gp trong khoa hc v k thut u l tng t. C th x l trc tip cc tn hiu bng mt h thng tng t thch hp. Trong trng hp -4-
Chng I
ny, ta ni tn hiu c x l trc tip dng tng t, nh minh ha trn hnh 1.5. C tn hiu vo v ra u l tn hiu tng t.T/h tng t vo B x l tn hiu tng t T/h tng t ra
Hnh 1.5 X l tn hiu tng t X l s l mt phng php khc x l tn hiu tng t, nh minh ha trn hnh 1.6. Tn hiu tng t phi c chuyn i thnh dng s (A/D) trc khi x l. iu khng may l qu trnh chuyn i tng t/ s ny khng bao gi hon ho, ngha l tn hiu s khng phi l biu din chnh xc cho tn hiu tng t ban u. Khi tn hiu tng t c chuyn thnh tn hiu s gn ng nht, qu trnh x l s c thc hin bng mt b x l tn hiu s DSP (Digital Signal Processor), to ra mt tn hiu s mi. Trong hu ht cc ng dng, tn hiu s cn c chuyn i ngc li thnh tn hiu tng t (D/A) cui qu trnh x l. Tuy nhin, cng c nhng ng dng lin quan n phn tch tn hiu, trong khng cn chuyn i D/A. Hnh 1.6 l s khi mt h thng x l tn hiu bng phng php s. B x l tn hiu s DSP c th l mt mch logic, mt my tnh s hoc l mt b vi x l lp trnh c.T/h tng t vo B chuyn i A/D T/h s vo B x l tn hiu s DSP T/h s ra B chuyn i D/A T/h tng t ra
Hnh 1.6 X l s tn hiu 1.3.2 u im ca x l s so vi x l tng t C nhiu nguyn nhn khc nhau khin cho x l s c a chung hn l x l trc tip tn hiu tng t. Trc tin, h thng s c th lp trnh c, to ta tnh mm do trong vic cu hnh li cc hot ng x l bng cch n gin l thay i chng trnh, trong khi cu hnh li h tng t, ta phi thit k li phn cng, ri kim tra v thm nh xem cc hot ng c ng khng. chnh xc cng ng mt vai tr qua trng trong vic la chn b x l tn hiu. sai lch ca cc linh kin tng t khin cho cc nh thit k h thng v cng kh khn trong vic iu khin chnh xc ca h thng tng t. Trong khi , vic iu khin chnh xc ca h thng s li rt d dng, ch cn ta xc nh r yu cu v chnh xc ri quyt nh la chn cc b chuyn i A/D v DSP c di t thch hp, c kiu nh dng du phy tnh hay du phy ng. Tn hiu s d dng lu tr trn cc thit b bng a t m khng b mt mt hay gim cht lng. Nh vy tn hiu s c th truyn i xa v c th c x l t xa. Phng php x l s cng cho php thc hin cc thut ton x l tn hiu tinh vi phc tp hn nhiu so vi x l tng t, nh vic x l c thc hin bng phn mm trn cc my tnh s. Trong mt vi trng hp, x l s r hn x l tng t. Gi thnh thp hn l do cc phn cng s r hn, hoc l do tnh mm do trong x l s. Tuy nhin, x l s cng c mt vi hn ch. Trc tin l s hn ch v tc hot ng ca cc b chuyn i A/D v b x l s DSP. Sau ny ta s thy nhng tn hiu bng thng -5-
Chng I
cc ln yu cu tc ly mu ca b A/D cc nhanh v tc x l ca DSP cng phi cc nhanh. V vy, phng php x l s cha p dng c cho cc tn hiu tng t bng thng ln. Nh s pht trin nhanh chng ca cng ngh my tnh v cng ngh sn xut vi mch m lnh vc x l tn hiu s (DSP) pht trin rt mnh trong vi thp nin gn y. ng dng ca DSP ngy cng nhiu trong khoa hc v cng ngh. DSP ng vai tr quan trng trong s pht trin ca cc lnh vc nh vin thng, a phng tin, y hc, x l nh v tng tc ngi-my... thy r nh hng to ln ca x l tn hiu s, ta xem v d v s pht trin ca my nh, t my nh tng t truyn thng n my nh s ngy nay. My nh truyn thng hot ng da trn cc c im vt l ca thu knh quang hc, trong cht lng bc nh cng p khi h thng thu knh cng to v rng. Khi my nh s mi ra i vi thu knh nh hn th cht lng nh chp thp hn nhiu so vi tng t. Tuy nhin, khi nng lc x l ca cc b vi x l mnh hn v cc thut ton x l tn hiu s tinh vi hn c p dng th cc nhc im v quang hc c khc phc v cht lng nh c ci thin r rt. Hin nay, cc my nh s cho cht lng nh vt tri hn so vi tng t. Hn na, cc my nh s ci trong in thoi di ng hin nay c thu knh rt nh nhng vn c th cho cht lng nh rt tt. Cht lng nh y ph thuc vo nng lc ca DSP ch khng phi ph thuc vo kch thc ca thu knh quang hc. Ni cch khc, cng ngh my nh s s dng nng lc tnh ton ca DSP khc phc cc hn ch v vt l. Tm li, DSP l mt lnh vc da trn nguyn ca ton hc, vt l v khoa hc my tnh v c nhng ng dng rt rng ri trong nhiu lnh vc khc nhau. 1.4 KHI NIM TN S TRONG TN HIU LIN TC V TN HIU RI RC T vt l chng ta bit rng tn s lin quan cht ch vi kiu chuyn ng c chu k gi l dao ng v c m t bng hm sin. Khi nim tn s lin quan trc tip n khi nim thi gian. Thc t th tn s c th nguyn l o ngc ca thi gian. Do vy bn cht ca thi gian (lin tc hoc ri rc) s c nh hng n bn cht ca tn s. 1.4.1 Tn hiu sin lin tc Mt dao ng iu ha n gin c m t ton hc bng hm sin lin tc sau:
x a (t) = Acos(t+ ), - 0, tn hiu ra ph thuc vo tn hiu vo v vo chnh tn hiu ra cc thi im trc 2.4.2 Gii phng trnh sai phn tuyn tnh h s hng V c bn, mc ch ca gii phng trnh l xc nh tn hiu ra y[n], n 0 ca h thng ng vi mt tn hiu vo c th x[n], n 0 v ng vi cc iu kin ban u c th no . Nghim ca phng trnh l tng ca 2 phn: y[n ] = y 0 [n ] + y p [n ] Trong y0[n] l nghim tng qut ca phng trnh thun nht v yp[n] l nghim ring. Nghim tng qut y0[n] l nghim ca phng trnh v phi bng 0, tc l khng c tn hiu vo. Dng tng qut ca y0[n] l: y 0 [n ] = C11 + C 2 2 + ... + C N N Trong i l nghim ca phng trnh c trng:
ak =0
N
k
ni k
v Ci l cc h s trng s, c xc nh da vo iu kin u v tn hiu vo. Nghim ring yp[n] l mt nghim no tha phng trnh sai phn trn vi mt tn hiu vo c th x[n], n 0 . Ni cch khc, yp[n] l mt nghim no ca phng trnh:
a k y[n k ] = b r x[n r], a 0 = 1k =0 r =0
N
M
Ta tm yp[n] c dng ging nh dng ca x[n], chng hn nh: x[n] yp [n]
A A.M n A n .n M A cos 0 n A sin 0 n V d:
K K.M n A n (K 0 n M + K1n M 1 + ... + K M ) K1 cos 0 n + K 2 sin 0 n
Tm nghim tng qut y[n ], n 0 ca phng trnh:- 46 -
Chng II
y[n ] + a 1 y[n 1] = x[n ]
vi x[n] l tn hiu bc nhy v y[-1] l iu kin u. Cho x[n] = 0, nghim tng qut y0[n] lc ny c dng: y 0 [n ] = n Gii ra ta c: = a 1
Do vy, y0[n] l: y 0 [ n ] = C( a 1 ) n Do x[n] l tn hiu bc nhy n v nn chn yp[n] c dng: y p [n ] = Ku[n ] y K l mt h s, c xc nh sao cho phng trnh tha mn. Thay yp[n] vo phng trnh trn ta c:Ku[n ] + a 1Ku[n 1] = u[n ]
xc nh K, ta tnh vi n 1 v trong di khng c s hng no b trit tiu. Vy, K + a 1K = 1 K= Nh vy, nghim ring ca phng trnh l: y p [n ] = Nghim tng qut ca phng trnh trn l: y[n ] = y 0 [n ] + y p [n ] = C(a 1 ) n + C c xc nh sao cho tha mn iu kin ban u. Cho n = 0, t phng trnh ta c:y[0] + a 1 y[1] = 1 y[0] = a 1 y[1] + 1
1 1 + a1
1 u[n ] 1 + a1 1 , n0 1 + a1
Mt khc, kt hp y[0] va tm c vi nghim tng qut ca phng trnh, ta c: y[0] = C + a 1 = a 1 y[1] + 1 C = a 1 y[1] + 1 1 + a1 1 + a1 1 (a 1 ) n +1 , n0 1 + a1
Thay C vo nghim y[n] ta c kt qu cui cng nh sau: y[n ] = (a 1 ) n +1 y[1] + = y zi [n ] + y zs [n ] Ta nhn thy nghim ca phng trnh gm c hai phn: - 47 -
Chng II
1. yzi[n] l p ng u vo 0 (zero-input response) ca h thng. p ng ny ch ph thuc vo bn cht ca h thng v cc iu kin ban u. V vy n cn c tn gi l p ng t do (free response). 2. yzs[n] ph thuc vo bn cht ca h thng v vo tn hiu vo, do n cn c gi l p ng cng bc (forced response). N c xc nh khi khng n iu kin u hay l iu kin u bng 0. Khi iu kin u bng 0, ta c th ni h thng trng thi 0. Do vy, yzs[n] cn c gi l p ng trng thi 0 (zero-state response)
Qua y ta cng thy: C ph thuc vo c iu kin u v tn hiu vo. Nh vy, C nh hng n c p ng u vo 0 v p ng trng thi 0. Ni cch khc, nu ta mun ch c p ng trng thi 0, ta gii tm C vi iu kin u bng 0. Ta cng thy rng c th tm nghim ring ca phng trnh t p ng trng thi 0: y p [n ] = lim y zs [n ]n
V d:
Tm y[n ], n 0 ca h sau:
y[n ] 3y[n 1] 4 y[n 2] = x[n ] + 2 x[n 1]vi x[n] = 4n u[n] v cc iu kin u bng 0.
- 48 -
Chng II
2.4.3 Thc hin h ri rc LTI
T phng trnh m t quan h vo-ra ta thy thc hin h LTI, ta cn cc khu nhn, tr v cng. C nhiu cch khc nhau thc hin h ri rc, y ta xt cch trc tip- l cch thc hin trc tip da vo phng trnh sai phn m khng qua mt php bn i no1. Dng chun tc 1
y[n ] + a 1 y[n 1] + ... + a N y[n N] = b 0 x[n ] + b1 x[n 1]] + ... + b M x[n M ] y[n ] = b 0 x[n ] + b1x[n 1]] + ... + b M x[n M ] + (a 1 ) y[n 1] + ... + (a N ) y[n N]
2. Dng chun tc 2
thy dng chun tc 1, h thng gm 2 h mc ni tip. Theo tnh cht giao hon ca tng chp th th t cc h con mc ni tip c th thay i c. Do vy, ta c th thay i h dng 1 thnh:
- 49 -
Chng III
Chng
3
PHN TCH H RI RC LTI DNG PHP BIN I ZPhp bin i Z l mt cng c quan trng trong vic phn tch h ri rc LTI. Trong chng ny ta s tm hiu v php bin i Z, cc tnh cht v ng dng ca n vo vic phn tch h ri rc LTI. Ni dung chnh chng ny l: Php bin i Z Php bin i Z ngc Cc tnh cht ca php bin i Z Phn tch h ri rc LTI da vo hm truyn t ng dng bin i Z gii phng trnh sai phn
2.1 PHP BIN I Z (Z-Transform) Php bin i Z l bn sao ri rc ha ca php bin i Laplace. Laplace transform : F ( s ) = z -transform : F ( z ) =
f (t )e st dt f [ n] z n
n =
Tht vy, xt tn hiu lin tc f (t ) v ly mu n, ta c:
f s (t ) = f (t ) (t nT ) =n =
n =
f (nT ) (t nT )
Bin i Laplace ca tn hiu ly mu (cn gi l ri rc) l: L[ f s (t )] = f (nT ) (t nT ) e st dt = f (nT ) (t nT )e st dt n = n =
=
n =
f (nT ) sT
(t nT )e st dt =
n =
f (nT )e snT
Cho f [n] = f (nT ) v z = e , ta c:F ( z) =
n =
f [ n] z n
F ( z )|z =esT = f [n]e sTnn =
= f (nT )e snTn =
= L[ f s (t )]
Nh vy, bin i Z vi z = e sT chnh l bin i Laplace ca tn hiu ri rc.3.1.1 nh ngha php bin i Z - 50 -
Chng III Nh va trnh by trn, php bin i Z hai pha (bilateral Z-Transform) ca h[n] l: H ( z ) = Z [ h[n]] =
n =
h[n]z
n
Ta cng c nh ngha php bin i Z mt pha (unilateral Z-transform ) l:
H ( z ) = h[n]z n .n=0
Php bin i Z hai pha c dng cho tt c tn hiu, c nhn qu v khng nhn qu. Theo nh ngha trn ta thy: X(z) l mt chui lu tha v hn nn ch tn ti i vi cc gi tr z m ti X(z) hi t. Tp cc bin z m ti X(z) hi t gi l min hi t ca X(z)k hiu l ROC (Region of Convergence ). Ta s thy c th c nhng tn hiu khc nhau nhng c bin i Z trng nhau. im khc bit y chnh l min hi t. Ta cn lu n hai khi nim lin quan n bin i Z- l im khng (zero) v im cc (pole). im khng l im m ti X(z) = 0 v im cc l im m ti X(z) = . Do ROC l tp cc z m X(z) tn ti nn ROC khng bao gi cha im cc.V d:
Tm bin i Z, v ROC v biu din im cc-khng:
x1[n] = a nu[n] and
x2 [n] = (a n )u[ n 1]
Ta thy hai tn hiu khc nhau trn c bin i Z trng nhau nhng ROC khc nhau.- 51 -
Chng III3.1.2 Min hi t ca php bin i Z 1. x[n] lch phi x[n] = 0, n < n0
X ( z) =
n = n0
x[n]z
n
X ( z) =
n = n0
1 x[n] z
n
Khi n , cn (1/z ) n 0 tng hi t. Nh vy, iu kin hi t s tha vi cc gi tr ca z nm ngoi ng trn i qua im cc xa gc nht, ngha l | z |> rmax .
2. x[n] lch tri x[n] = 0, n > n0X ( z) =n
n =
x[n]z
n0
n
Khi n , cn (1/z ) 0 hay z 0 tng hi t. Vy ROC l min nm trong ng trn i qua im cc gn gc nht, ngha l | z |< rmin
Lu trong trng hp tn hiu x[n] = 0 vi n > n0 > 0 nhng x[n0 ] 0 , ROC khng cha im 0. Chng hn nh vi x[n] = u[n + 1] th X ( z) =
n =
1
z n = z 1 + z nn=0
khng hi t z = 0 nn z = 0 khng nm trong ROC.3. Tn hiu x[n] lch hai pha
ROC c dng:r1 < z < r2 (hnh vnh khn hoc rng)
4. Tn hiu x[n] di hu hn
ROC l ton b mt phng z ngoi tr z = 0 v/hoc z = - 52 -
Chng III
[n 1] z ,| z |> 01
[n + 1] z,| z |< V d:
Tm bin i Z v ROC ca: x[n] = a|n| where | a |< 1 .
V d:
Tm bin i Z v ROC ca: x[n] = 3n u[n 1] + 4n u[n 1].
- 53 -
Chng IIIV d:
Tm bin i Z v ROC ca:
1 2
[n 1] + 3 [n + 1]
V d:
Tm bin i Z ca: h[n] = (.5) n u[n 1] + 3n u[ n 1]. H biu din bng p ng xung nh trn c n nh BIBO khng?
V d: Tm bin i Z ca: x[n] = r n sin(bn)u[n]
- 54 -
Chng III2.2 PHP BIN I Z NGC IZT 2.2.1 Biu thc tnh IZT
Biu thc tnh IZT c xy dng da trn nh l tch phn Cauchy. nh l nh sau:1, n = 0 1 n 1 z dz = 0, n 0 2j C vi C l ng cong kn bao quanh gc ta theo chiu dng v nm trong mt phng z. Nhn 2 v ca biu thc tnh ZT viz l1 ri ly tch phn theo ng cong C, ta c: 2j
1 1 1 X(z)z l1dz = x[n ]z n +l1dz = x[n ] z n +l1dz n 2j C 2j C = 2j n = C
p dng nh l tch phn Cauchy ta rt ra c:1 X(z)z l1dz = x[l] 2j C
Thay l = n, ta c biu thc tnh IZT nh sau:x[n ] = 1 n 1 X(z)z dz 2j C
T y ta thy c th tnh IZT trc tip t cng thc va tm c. Cch tnh l da vo nh l v gi tr thng d (xem sch). Tuy nhin, cch tnh ny kh phc tp nn khng c s dng trong thc t. Sau y ta xt hai phng php tnh IZT c dng trong thc t:2.2.2 Phng php khai trin chui ly tha (Power Series Expansion)
Ta c th tnh IZT bng cch khai trin X(z) thnh chui ly tha:X ( z ) = x[k ]z k = x[0] + x[1]z 1 + x[2]z 2 + Lk =0
x[n] = x[k ] [n k ] = x[0] [n] + x[1] [ n 1] + x[2] [n 2] + Lk =0
Ta c:
[n k ] z kSau ng nht cc h s ca chui lu tha vi x[n].V d:
z
Tm IZT ca:
X ( z ) = 1 + 2 z 1 + 3z 2
- 55 -
Chng IIIV d:
Tm IZT ca:
X(z) =
1 , ROC : z > a 1 az 1
V d:
Tm IZT bit: X ( z) =
8 z 19 , | z |> 3 z 5z + 62
Cch khai trin X(z) thnh chui ly tha nh trn c im khng thun tin l kh/khng th biu din c x[n] dng tng minh.- 56 -
Chng III2.2.3 Phng php khai trin ring phn (Partial Fraction Expansion)
Phng php ny tng t nh tnh bin i Laplace ngc bit. Gi s cn tnh IZT{X(z)}. Ta khai trin X(z) thnh dng sau: X(z) = X p (z) + X i (z)i
Trong Xp (z) c dng a thc, Xi(z) c dng phn thc vi bc ca t s nh hn bc ca mu s. Tu im cc m Xi(z) c th c cc dng nh sau: 1. Nu pi l im cc n: X i (z) = ri vi z pis
ri = (z p i )X(z)
z = pi
2. Nu pi l im cc bi bc s: X i (z) = vi ck =
ck k k =1 ( z p i )
1 d sk s k (z p i ) s X(z) (s k )! dz
[
]
z = pi
Sau khi khai trin X(z) ta s dng bng 3.1 suy ra IZT. ( n ) 1 ( n m ) z m z a n u[n ] za az na n u[n ] (z a ) 2 az(z + a ) n 2 a n u[n ] (z a ) 3 z(z a cos ) a n cos(n )u[n ] 2 z 2z cos + a 2 az sin a n sin(n )u[n ] 2 z 2z cos + a 2 Kz K *z 2 | K | a n cos(n + )u[n ] + , p = ae j & K =| K | e j z p z p*Bng 3.1 Cc cp x[n] X(z) thng dng V d:2 z 2 5z ,| z |> 3 ( z 2)( z 3)
Tm IZT ca: X ( z ) = Ta khai trin
X ( z) 2z 5 = z ( z 2)( z 3) - 57 -
Chng III
V d:
Tm IZT ca: X(z) = 2z , (z 2)(z 1) 2 z >2
V d:
Tm IZT ca: X ( z) = z z 0.5z + 0.252
- 58 -
Chng III2.3 CC TNH CHT CA PHP BIN I Z
Trong phn ny, ta xt nhng tnh cht quan trng nht ca php bin i Z.2.3.1 Tuyn tnhZ
ax[n] + by[n] aX ( z ) + bY ( z )
Min hi t mi ph thuc vo min hi t ca c X ( z ) v Y(z) , l giao ca hai min hi t R x R y . Tuy nhin, nu t hp aX(z) + bY(z) lm kh i mt s im cc ca X(z)hoc Y(z) th min hi t s m rng ra, nn:R Rx Ry
2.3.2 Dch chuyn thi gianx[n n0 ] z n0 X ( z )Z
y min hi t mi ging min hi t Rx , c th thm vo hoc bt i im gc hay im v cng ty n0 dng hay m
V d:
Tm w[n] bit:W ( z) = z 4 ,| z |> 3 z2 2z 3
- 59 -
Chng III Tnh cht tuyn tnh v dch thi gian rt hiu qu i vi cc h thng m t bi phng trnh sai phn tuyn tnh h s hng.2.3.3 Tng chpZ
y[n] = x[n] h[n] X ( z ) H ( z )
y min hi t mi l Ry Rx Rh Tnh cht tng chp ca bin i Z gip ta tnh ton tng chp tuyn tnh ri rc mt cch n gin hn. Tnh cht ny s c s dng rt nhiu.Chng minh:Z
y[n] = x[n] h[n] Thay i th t ly tng, ta c: y[n] = t m = (n k ) , ta c: y[n] = =
n = k =
[ x[k ]h[n k ]]z
n
k =
x[k ] h[n k ]z nn =
k =
x[k ][ h[m]z ( m + k ) ]m =k m
k =
x[k ]z h[m]zm =
= X ( z)H ( z) Min hi t mi ph thuc vo min hi t ca c X ( z ) v H ( z ) , l giao ca hai min hi t Rx Rh . Tuy nhin, nu mt tha s X(z) hoc H(z) c im khng, im khng ny kh im cc ca tha s kia th min hi t s m rng ra, nn Ry Rx Rh
V d:
Cho h[n] = a n u[n] , ( | a |< 1 ) v x[n] = u[n] . Tm y[n] = x[n] h[n]. Nu x[n] = u[n 2] th y[n] thay i nh th no?
- 60 -
Chng III
V d:
Tm u ra y[n] vi u vo x[n] = u[n] v h LTI c p ng xung: h[n] = 3n u[ n 1].
- 61 -
Chng III2.3.4 nh l gi tr u v gi tr cui
nh l gi tr u v gi tr cui thng lin quan n bin i Z mt pha, nhng chng cng ng vi bin i Z hai pha nu tn hiu x[n] = 0 vi n < 0.1. nh l gi tr u(initial value theorem)
Biu din: F ( z ) = f [n]z n = f [0] + f [1]z 1 + f [2]z 2 + ,n=0
Ly gii hn lim F ( z ) , ta s c gi tr u ca f[n]- chnh l f[0]z
2. nh l gi tr cui(final value theorem)
Nu gi tr cui ca f[n] tn ti th: lim f [n] = f [] = lim( z 1) F ( z )n z 1
V d:
Tm gi tr u v gi tr cui ca tn hiu f [n] , bit rng: F ( z) = z z .6
2.4 PHN TCH H RI RC LTI
Ta bit trong min thi gian, c th biu din h ri rc LTI bng s , tng chp, p ng xung, p ng bc v phng trnh sai phn . Sau y ta s xt mt cch khc - rt hiu qu biu din h thng ri rc LTI. l biu din bng hm truyn t (transfer function) hay cn gi l hm h thng (system function)2.4.1 nh ngha hm truyn t
T tnh cht tng chp ca ZT v t quan h gia tn hiu vo x[n], tn hiu ra y[n] vi p ng xung h[n], ta c: Y(z) = X(z).H(z) y X(z) l bin i Z ca x[n], Y(z) l bin i Z ca y[n] v H(z) l bin i Z ca p ng xung h[n]. Da vo p ng xung h[n], ta bit c cc c tnh ca h thng, vy r rng l da vo H(z) ta cng s bit c cc c tnh ca h thng. Ni cch khc, H(z) l biu din ca h thng trong min z. Ta gi H(z) l hm truyn t hay hm h thng. Ta c th xc nh H(z) rt n gin da vo phng trnh sai phn: - 62 -
Chng III
ak =0
N
k
y[n k ] = b r x[n r ]r =0
M
Ly bin i Z hai v, s dng tnh cht tuyn tnh v dch thi gian, ta c:
a k z k Y ( z) = b r z r X (z)k =0 r =0
N
M
Suy ra hm truyn t nh sau: Y(z) H ( z) = = X(z)
b z ak =0 r =0 N r k
M
r
z k
Da vo hm truyn t, ta bit c cc c tnh ca h thng, gm tnh nh, tnh kh o, tnh nhn qu, tnh n nh BIBO.2.4.2 Tnh nh
H khng nh phi c p ng xung c dng:
h[n] = K [n].H(z) = K Vy h c nh c hm truyn t l mt hng s.2.4.3 Tnh kh o
h[n] hi [n] = [n] H ( z ) H i ( z ) = 1 y:hi [n] H i ( z ) l o ca h[n] H ( z ) .z z
V d:
Tm h o hi [n] ca h: h[n] = a nu[n]. Kim tra kt qu bng cch tnh tng chp ca h[n] vi hi [n] .
- 63 -
Chng IIIV d:
Tm h o ca h h[n] nhn qu bit: H ( z) = za . z b
2.4.4 Tnh nhn qu
h[n] = 0, n < 0ROC: | z |> rmax H nhn qu c min hi t ca H(z) nm ngoi ng trn i ngang qua im cc xa gc nht.2.4.5 Tnh n nh BIBO
k =
h[k ] <
H(z) =
n =
h[n]z n | H(z) |
n =
| h[n]z n | = | h[n] || z n |n =
Khi ta tnh trn ng trn n v (tc l |z| = 1) th: | H (z) |
n =
| h[n] |
Nh vy, nu h thng n nh BIBO th ng trn n v nm trong ROC. iu ngc li cng ng. Kt hp vi tnh nhn qu va xt trong 2.4.4 ta c kt lun: H nhn qu s n nh BIBO nu v ch nu tt c cc im cc ca H(z) nm bn trong ng trn n v trong mt phng z:| p k |< 1, k
V d:
H c p ng xung l u[n] c nhn qu khng? C n nh BIBO khng?
- 64 -
Chng IIIV d:
Xt tnh nhn qu v n nh ca h c p ng xung l: h[n ] = (.9) n u[n ]
V d:
Xt tnh nhn qu v n nh BIBO ca h c hm truyn t l: H ( z) = 2z2 5 z 1 2 , 1 . Nu | a |< 1 ?
- 68 -
Chng IVV d:
Cho p[n] = u[n] u[n N ] . Tm P () . Hy chng t rng bin i Fourier ny c pha tuyn tnh (linear phase)
V d:
Tm H () ca h LTI c p ng xung sau h[n] = [n] + 2 [n 1] + 2 [n 2] + [n 3] V chng t rng h c pha tuyn tnh
4.1.4 Quan h gia bin i Z v bin i Fourier
Biu thc tnh ZT l: X(z) =
n =
x[n]z
n
Gi s ROC c cha ng trn n v. Tnh X(z) trn ng trn n v, ta c: X(z)z =ej
=
n =
x[n]e
jn
= X ()
Nh vy, bin i Fourier chnh l bin i Z tnh trn ng trn n v. Da vo y, ta c th pht biu li iu kin tn ti ca DTFT nh sau:
- 69 -
Chng IV Bin i Fourier ca mt tn hiu ch tn ti khi ROC ca bin i Z ca tn hiu c cha ng trn n v.V d:
Lm li cc v d trn- Tm bin i Fourier ca: (a) x[n] = a n u[n] , | a |< 1 . Nu | a |> 1 ?
(b) y[n] = a nu[ n] , | a |> 1 . Nu | a |< 1 ?
(c) p[n] = u[n] u[n N ]
(d) h[n] = [n] + 2 [n 1] + 2 [n 2] + [n 3]
4.2 PHP BIN I FOURIER NGC 4.2.1 Biu thc tnh bin i Fourier ngc
Ta thy X() l mt hm tun hon vi chu k 2 , do e j tun hon vi chu k 2 : e j = e j ( + 2 ) = e j e j 2 = e j . Do di tn s ca tn hiu ri rc l mt di tn bt k rng 2 , thng chn l: (, ) hay (0,2) . Vy ta c th khai trin X() thnh chi Fourier trong khong (, ) hay (0,2) nu iu kin tn ti X() tha mn. Cc h s Fourier l x[n], ta c th tnh c x[n] t X() theo cch sau: Nhn 2 v ca biu thc tnh DTFT vi
1 jl e ri ly tch phn trong khong ( , ) ta c: 2 1 1 1 X()e jl d = x[n ]e jn e jl d = x[n ] e j ( l n ) d = x[l] n 2 2 = n = 2
Thay l = n v thay cn tch phn, khng nht thit phi l ( , ) m ch cn khong cch gia cn trn v di l 2 , ta c biu thc tnh bin i Fourier ngc (IDTFT) nh sau:
- 70 -
Chng IV
x[n] =
1 2
X ( )e2
jn
d
Ta c th tnh IDTFT bng hai cch: mt l tnh trc tip tch phn trn, hai l chuyn v bin i Z ri tnh nh tnh bin i Z ngc. Ty vo tng trng hp c th m ta chn phng php no cho thun tin.4.2.2 Mt s v d tnh bin i Fourier ngc V d:
Tm x[n] nu bit: 1, c X () = 0, c < <
V d:
Tm x[n] nu bit: X() = cos 2
- 71 -
Chng IV4.3 CC TNH CHT CA PHP BIN I FOURIER
Sau y ta s xt mt s tnh cht quan trng ca DTFT, phn cn li xem sch.4.3.1 Tnh tuyn tnh
ax1[n] + bx2 [n] aX 1 () + bX 2 ()
4.3.2 Tnh dch thi gian
x[n] X ()
x[n n0 ] e jn0 X () Qua y ta thy s dch chuyn tn hiu trong min thi gian s khng nh hng n bin ca DTFT, tuy nhin pha c cng thm mt lng.
4.3.3 Tnh dch tn s/ iu ch
x[n] X ()e j0n x[n ] X( 0 ) cos( 0 n ) x[n ] 1 1 X ( 0 ) + X ( + 0 ) 2 2
Nh vy, vic iu ch tn hiu gy ra s dch tn s.- 72 -
Chng IV4.3.4 Tnh chp thi gian
Tng t nh bin i Z, vi bin i Fourier ta cng c:x1[n] x2 [n] X 1 () X 2 ()F
V d:
Cho h[n] = a nu[n],| a |< 1 . Tm h o ca n hi [n] , nhng khng dng bin i Z.
4.3.5 Tnh nhn thi gian
x 1 [n ].x 2 [n ]
1 X 1 ()X 2 ( )d 2 2
4.4 PHN TCH TN S (PH) CHO TN HIU RI RC 4.4.1 ngha ca ph
Trong min tn s, mi tn hiu u c c im ring ca n. V d nh, tn hiu sin ch c duy nht mt tn s n, trong khi nhiu trng cha tt c cc thnh phn tn s. S bin thin chm ca tn hiu l do tn s thp, trong khi s bin thin nhanh v nhng sn nhn l do tn s cao. Nh xung vung chng hn, n cha c tn s thp v c tn s cao. Hnh sau minh ha cho iu . Hnh (a) l mt sng sin tn s thp, cc hnh sau (b)-(c) cng thm dn cc sng sin tn s cao dn. Hnh cui cng (e) l tng ca 7 sng sin. Trong hnh (e) ta thy tng ca 7 sng sin c dng xp x vi dng ca mt xung vung. Ph ca tn hiu l m t chi tit cc thnh phn tn s cha bn trong tn hiu. V d nh vi tn hiu xung vung va ni trn, ph ca n ch ra tt c cc nh nhn ca cc sng sin ring c th kt hp li vi nhau to ra xung vung. Thng tin ny quan trng v nhiu l do. V d nh, thnh phn tn s trong mt mu nhc ch cho ta bit cc c trng ca loa, t khi sn xut li ta c th ci tin cho hay hn. Mt v d khc, micro trong h thng nhn dng ting ni phi c di tn rng c th bt c tt c cc tn s quan trng trong ting ni u vo. d on cc nh hng ca b lc trn tn hiu, cn phi bit khng ch bn cht ca b lc m cn phi bit c ph ca tn hiu na. - 73 -
Chng IV
4.4.2 Ph bin v ph pha
Ph ca tn hiu gm c hai phn: ph bin (magnitude spectrum) v ph pha (phase spectrum). Ph bin ch ra ln ca tng hnh phn tn s. Ph pha ch ra quan h pha gia cc thnh phn tn s khc nhau. Trong phn ny, ta xt tn hiu ri rc khng tun hon. Cng c tnh ph tn hiu ri rc khng tun hon l DTFT. tnh ph tn hiu, ta qua hai bc: mt l tnh DTFT ca tn hiu- l X() , hai l tnh bin v pha ca X() :X ( ) = X ( ) e j ( )
y | X() | l ph bin v () l ph pha. Ta d dng chng minh c rng i vi tn hiu thc, ph bin l mt hm chn theo tn s v ph pha l mt hm l theo .
Do , nu bit ph X() trong khong 0 n , ta c th suy ra ph trong ton di tn s. - 74 -
Chng IV d gii thch ph, tn s s t 0 n thng c chuyn i thnh tn s tng t f t 0 n fS/2 nu tn s ly mu l fS.V d:
Tm ph bin v ph pha ca tn hiu ch nht: x[n] = u[n] - u[n-4]
V d:
Mt mu nguyn m ting ni eee c ly mu tn s 8 kHz. Ph bin ca tn hiu ny nh trn hnh. Hi tn s c bn ca tn hiu ny l bao nhiu?
- 75 -
Chng IV4.4.3 Mt ph nng lng
Nng lng ca tn hiu x[n] c nh ngha l: E= By gi ta biu din nng lng theo ph:
n =
| x[n ] |
2
1 * E = x[n ]x [n ] = x[n ] X ()e jn d n = n = 2 *
Thay i th t ly tng v tch phn, ta c:1 1 2 E= X * () x[n ]e jn d = X() d 2 2 n =
Vy quan h v nng lng gia x[n] v X() l:E=
n =
| x[n ] |2 =2
1 2 X() d (quan h Parseval) 2
i lng S xx () = X() gi l mt ph nng lng.V d:
Xc nh mt ph nng lng ca tn hiu sau: x[n] = an u[n] vi -1 < a < 1
4.4.4 Bng thng
Bng thng (bandwidth) l di tn s tp trung hu ht nng lng (cng sut) ca tn hiu. Gi s 95% nng lng ca tn hiu tp trung trong di tn s F1 F F2 , ta ni bng thng 95% ca tn hiu l F2 F1 . Ta c th nh ngha cc bng thng 75%, bng thng 90%, bng thng 99%... theo kiu tng t nh bng thng 95% ni trn.Da vo bng thng ca tn hiu, ta c th phn loi tn hiu nh sau: Nu nng lng tn hiu tp trung quanh tn s 0 th l tn hiu tn s thp (low-frequency signal). Nu nng lng tn hiu tp trung min tn s cao th l tn hiu cao tn (highfrequency signal).
- 76 -
Chng IV Nu nng lng tn hiu tp trung vo mt di tn s no gia tn s thp v tn s cao th l tn hiu thng di (bandpass signal) Trong trng hp tn hiu thng di, khi nim bng hp (narrowband) c dng ch tn hiu c bng thng F2 F1 rt nh (khong 10% hoc nh hn) so vi tn s trung tm (F1 + F2 ) / 2 . Ngc li, tn hiu c gi l bng rng (wideband). Tn hiu c gi l c bng thng hu hn (bandlimited) nu ph ca n bng 0 ngoi di tn F B . Tn hiu nng lng x[n] c gi l c bng thng hu hn nu:X () = 0, 0 < <
4.5 PHN TCH TN S CHO H THNG RI RC LTI
Trong min tn s, h thng ri rc LTI c m t bng mt hm theo tn s- gi l p ng tn s (frequency response)- l bin i Fourier ca p ng xung h[n]: Quan h gia tn hiu vo- ra v h thng trong min tn s nh sau: y[n ] = x[n ] h[n ] Y() = X().H() p ng tn s hon ton c trng cho h ri rc LTI trong min tn s. N cho php ta: - xc nh cc p ng ca h thng vi cc u vo c dng t hp tuyn tnh ca tn hiu sin hay hm m phc. - xc nh cc c tnh ca h LTI l b lc tn s.4.5.1 Tnh p ng tn s 1. Tnh t p ng xung
Theo nh ngha, p ng tn s l H() c tnh nh sau: H () =
n =
h[n] eM
jn
2. Tnh t phng trnh sai phn tuyn tnh h s hng
a k y[n k] = b r x[n r]k =0 r =0
N
Ly DTFT 2 v, s dng tnh cht tuyn tnh v dch thi gian, ta c:
[a k e jk ]Y( = [b r e jr ]X()k =0 r =0 M
N
M
Y () = H ( ) = X ()V d:
b e ak =0 r =0 N r k
jr
e jk
Tm p ng tn s ca h:
y[[n ]] + 0.1y[[n 1]] + 0.85y[n 2] = x[n ] 0.3x[n 1]- 77 -
Chng IV
3. Tnh t hm truyn t
Theo quan h gia php bin i Z v php bin i Fourier, ta c th tnh c p ng tn s t hm truyn t bng cch thay z = e j (vi iu kin l ROC c cha ng trn n v): H () = H ( z )4.5.2 p ng bin v p ng phaz = e j
Do p ng tn s H() l hm theo bin phc nn c th biu din nh sau:H ( ) = H ( ) e j ( )
| H() | c gi l p ng bin v () c gi l p ng pha.V d:
Cho p ng tn s ca h sau: H () = Tm p ng bin v pha. 1 1 0.4e j
4.5.3 p ng ca h LTI i vi u vo l t hp tuyn tnh ca cc tn hiu dng sin hay hm m phc 1. p ng trng thi 0 i vi u vo dng hm m phc
T chng II, ta bit p ng ca h (iu kin u l 0) l: y[n ] =
k =
h[k] x[n k ]
Gi s tn hiu vo l tn hiu hm m phc sau: - 78 -
Chng IV x[n ] = Ae jn , < n < vi A l bin v l mt tn s trong di tn ( , ) . Thay x[n] vo biu thc y[n] trn, ta c: y[n ] =k =
h[k ](Ae
j ( n k )
)jn
= A h[k ] e jk k = = (Ae jn )H() = x[n ]H()
(
)e
Ta thy p ng ca h c dng ging dng ca u vo, tc l dng hm m phc vi cng tn s, ch khc nhau mt h s nhn l H() . iu ny cng ng trong trng hp tn hiu vo c dng sin/cos.
V d:
Xc nh u ra ca h thng c p ng xung l: h[n ] = (1 / 2) n u[n ] khi u vo c dng: (a) x[n ] = Ae j n 2
1 2 j26.60 , < n < . Cho bit H = = e 1 5 2 1+ j 2
(b) x[n ] = 10 5 sin
n + 20 cos n, < n < 2
- 79 -
Chng IV2. Eigenfunction v eigenvalue
Nu ta c tn hiu vo v tn hiu ra c th phn tch thnh cc hm c s l:
x[n] = akk [n]k
y[n] = ak k [n]k
Cc hm c s ny c cng dng l k [n] , ch khc nhau mt h s nhn (thc/ phc) bk :
k [n] = k [n] h[n] v k [n] = bkk [n]th k [n] c gi l mt eigenfunction ca h ri rc LTI vi eigenvalue l bk . Trong trng hp ny, tn hiu vo c dng hm m phc nh trn l eigenfunction v H() tnh ti cng tn s ca tn hiu vo l eigenvalue tng ng.3. p ng trng thi bn v p ng nht thi
Ta c th phn tch p ng ca h thng thnh hai thnh phn. Thnh phn th nht khng tin ti 0 khi n tin ti v cng, c gi l p ng trng thi bn (steady-sate response) yss[n]. Thnh phn ny tn ti trong cng khong thi gian tn ti ca u vo. Thnh phn kia tin ti 0 khi n tin ti v cng, c gi l p ng nht thi (transient response) ytr[n] Trong nhiu ng dng th p ng nht thi khng quan trng v ch tn ti trong mt khong thi gian ngn v do vy m n thng c b qua.V d:
Cho tn hiu x[n ] = Ae jn , n 0 i vo h thng y[n ] ay[n 1] = x[n ] (|a| < 1) Cho iu kin u l y[-1]. Tm p ng ca h, p ng trng thi bn, p ng nht thi.
Tn hiu ra l: y[n ] = an +1
Aa n +1e j ( n +1) jn A y[1] e + e jn , n 0 j j 1 ae 1 ae - 80 -
Chng IV Ta c p ng trng thi bn l: y ss [n ] = lim y[n ] =n
A = AH()e jn j 1 ae
Hai s hng u ca y[n] gim v 0 khi n tin ti v cng. l p ng nht thi: y tr [n ] = an +1
Aa n +1e j ( n +1) jn y[1] e ,n0 1 ae jn x[n] = k =1 X k zk M
Tng qut, khi tn hiu vo l:
Bng cch xp chng, ta tm c p ng trng thi bn nh sau:n yss [n] = k =1 X k H ( zk ) zk . M
V d: n Cho u vo x[n] = ( 3 ) , v 4
h[n] = (.5) n u[n] Tm p ng trng thi bn. 3 3 yss [n] = H 4 4 n
4.5.4 H LTI l b lc tn s
B lc (filter) l mt h thng x l tn hiu bng cch thay i cc c trng tn s ca tn hiu theo mt iu kin no . Ni cch khc, b lc thay i ph ca tn hiu vo X() theo p ng tn s H() to ra tn hiu ra c ph l: Y() = X()H() . p ng tn s y ng vai tr l mt hm trng s hay mt hm thay i dng ph i vi cc thnh phn tn s khc nhau trong tn hiu vo. Khi xt theo quan im ny th bt k mt h LTI no cng c th c xem l mt b lc tn s, ngay c khi n khng ngn mt vi hay tt c cc thnh phn tn s trong tn hiu vo. Do vy ta c th ng nht hai khi nim b lc tn s v h LTI. Trong mn hc ny, ta dng thut ng b lc l ch cc h LTI thc hin chc nng chn lc tn hiu theo tn s. B lc cho cc thnh phn tn s ca tn hiu trong mt di tn no i qua v ngn khng cho cc thnh phn tn s khc i qua. Di tn s cho qua gi l di thng (passband) v di tn s khng cho qua gi l di chn (stopband/block-band). Tn s gii hn gia di thng v di chn gi l tn s ct (cut-off frequency)
- 81 -
Chng IV Cch m t b lc n gin nht l biu din dng ca n trong min tn s. chnh l p ng tn s, gm p ng bin v p ng pha. Xt b lc c di thng l (1 , 2 ) . Nu y l b lc l tng th p ng tn s c dng nh sau:Ce jn 0 , 1 < < 2 H () = 0,
y C v n0 l hng s. Tn hiu ra b lc l tng c dng: Y() = X()H() = CX ()e jn 0 , 1 < < 2 y[n ] = Cx[n n 0 ] Ta thy tn hiu ra n gin ch l tn hiu vo b thay i mt h s nhn v b tr i mt khong thi gian. S thay i bin v tr ny khng lm mo tn hiu. Vy b lc l tng l b lc c p ng bin c dng ch nht v p ng pha l tuyn tnh trong di thng: | H() |= C, 1 < < 2 () = n 0 , 1 < < 2 C rt nhiu loi b lc khc nhau vi rt nhiu ng dng khc nhau, trong thng dng nht l b lc thng thp, thng cao, thng di v chn di. Hnh sau v cc p ng bin ca 4 loi b lc thng dng.
- 82 -
Chng IV Cc p ng bin trn khng c dng ch nht v y khng phi l b lc l tng. Gia di thng v di chn c mt di chuyn tip (transition band). li (gain) ca b lc ti mt tn s no l gi tr ca p ng bin ti tn s . Tn s ct l tn s ti im m li l 1 / 2 ca gi tr ln nht. B lc cng tin gn n b lc l tng hn khi dc ca b lc cng ln, di chuyn tip cng nh. iu ny yu cu bc ca b lc phi ln. Ta s quay li tm hiu k hn v b lc v thit k b lc sau ny.
- 83 -
Chng V
Chng
5
PHP BIN I FOURIER RI RC V NG DNGT chng trc, ta thy ngha ca vic phn tch tn s cho tn hiu ri rc. Cng vic ny thng c thc hin trn cc b x l tn hiu s DSP. thc hin phn tch tn s, ta phi chuyn tn hiu trong min thi gian thnh biu din tng ng trong min tn s. Ta bit biu din l bin i Fourier X() ca tn hiu x[n]. Tuy nhin, X() l mt hm lin tc theo tn s v do , n khng ph hp cho tnh ton thc t. Hn na, tn hiu a vo tnh DTFT l tn hiu di v hn, trong khi thc t ta ch c tn hiu di hu hn, v d nh mt bc nh, mt on ting ni
Trong chng ny, ta s xt mt php bin i mi khc phc c cc khuyt im trn ca DTFT. l php bin i Fourier ri rc DFT (Discrete Fourier Transform). y l mt cng c tnh ton rt mnh thc hin phn tch tn s cho tn hiu ri rc trong thc t. Ni dung chnh chng ny gm: DTFT ca tn hiu ri rc tun hon. y l php bin i trung gian dn dt n DFT DFT thun v ngc Cc tnh cht ca DFT Mt s ng dng ca DFT Thut ton tnh nhanh DFT, gi l FFT
5.1 PHP BIN I FOURIER CA TN HIU RI RC TUN HON 5.1.1 Khai trin chui Fourier cho tn hiu ri rc tun hon
Nhc li khai trin chui Fourier cho tn hiu lin tc tun hon:x(t ) =
k =
aek
jk0t
synthesis equation
ak =
1 x(t )e jk0t dt T T
analysis equation
Tng t, ta c khai trin chui Fourier cho tn hiu ri rc tun hon (cn c gi l chui Fourier ri rc DFS- Discrete Fourier Serie) nh sau: x[n] = ak = 1 Nk< N >
ak e jk 0 n x[n]e jk 0 n
synthesis equation analysis equation
n< N >
Khc vi khai trin chui Fourier cho tn hiu lin tc tun hon, php ly tch phn by gi c thay bng mt tng. V c im khc quan trng na l tng y l tng hu hn, ly trong mt khong bng mt chu k ca tn hiu. L do l:ejk 0 n
=e
jk
2 n N
=e
jk
2 n N
.e
jk 2 n
=e
j( k + N )
2 n N
= e j( k + N ) 0 n
- 88 -
Chng V5.1.2 Biu thc tnh bin i Fourier ca tn hiu ri rc tun hon
Ta c hai cch xy dng biu thc tnh bin di Fourier ca tn hiu ri rc tun hon nh sau:1. Cch th nht:
Ta bt u t tn hiu lin tc tun hon. Ta c:e j0t 2 ( 0 )F
Nn: x[n ] =
k =
a k e jk0t X() = 2 a k ( k0 )k =
F
Vy, ph ca tn hiu tun hon l ph vch (line spectrum), c v s vch ph vi chiu cao l 2a k nm cch u nhau nhng khong l 0 trn trc tn s By gi chuyn sang tm bin i Fourier ca tn hiu ri rc tun hon: Trc ht, ta tm DTFT ca e j0n . Ta c th on l DTFT ca e j0n cng c dng xung tng t nh DTFT ca e j0 t , nhng khc im DTFT ny tun hon vi chu k 2 : DT : e j0 n 2 ( 0 + 2 l )l = F
Ta c th kim tra li iu ny bng cch ly DTFT ngc: x[n] ==
1 21 2
< 2 >
X ()e jn d 2 ( 0 )e jn d
0 +
0
= e j 0 n Kt hp kt qu DTFT ca e j0 n vi khai trin chui Fourier ca x[n], tng t nh vi tn hiu lin tc, ta c:
x[n] 2
F
k< N > l =
a ( k k k = k 0
0
+ 2 l )
= 2
a ( k ) (do ak tun hon)
- 89 -
Chng V Vi 0 =2 N
, ta c: x[n] periodic with period N 2F
k =
a ( k
2 k ) N
vi ak l h s ca chui Fourier, tng c ly trong mt chu k ca tn hiu.ak = =V d:
1 N
n< N >
x[n]e j 2 nk /N x[n]e j 2 nk /N
1 N
n0 + N 1 n = n0
Tm DTFT ca dy xung ri rc sau: p[n] =
k =
[n kN ].
Cui cng ta c: p[n] =
k =
[n kN ]
2 N
k =
(
2 k ) = P ( ) N 2 v c N
Nh vy, DTFT ca dy xung ri rc l tp v s xung ri rc c chiu cao l khong cch gia hai xung cnh nhau l 2 N- 90 -
Chng V2. Cch th hai:
Ta c th rt ra kt qu DTFT ca tn hiu ri rc tun hon nh trn nhng bng cch khc. Ta xt mt chu k ca tn hiu tun hon x[n] , k hiu l: x0 [n] : x[n], 0 n N 1 x0 [n] = otherwise. 0,
Sau tnh DTFT ca x0 [n] X 0 () =
n =
x0 [n]e jn = x0 [n]e jnn =0
N 1
Vit li x[n] di dng tng ca v s chu k x0 [n] : x[n] =
k =
x0 [n kN ] =
k =
x0 [n] [n kN ] = x0 [n]
k =
[n kN ]
Theo tnh cht chp tuyn tnh ta c:x[n] = x0 [n] p[n] X 0 () P() = X () Thay P () va tm c trong v d trn vo biu thc ny, ta c: 2 X () = X 0 () N = 2 NF
( k
2 k ) N
Xk
0
(
2 k 2 k ) ( ) (t/c nhn vi mt xung) N N
y X 0 ( 2Nk ) c N gi tr phn bit, ngha l k = 0,1,2,..., N 1 .
Biu thc tnh DTFT ngc l: x[n] = = 1 2 1 N
2 k =
X ()e jn d = X0(
1 2
2
0
[
2 N
k =
X0(
2 k 2 k jn ) ( )]e d N N
2 k 2 2 k jn 1 ) ( )e d = 0 N N N
Xk =0
N 1
0
(
2 k j 2N kn )e N
Nu so snh vi cng thc chui Fourier trn, ta c: ak = 1 2k X0 vi k = 0,1,2,..., N 1 N N - 91 -
Chng V Tm li, ta c: x[n] = x0 [n]
k =
[n kN ]
X 0 () = x0 [n]e jnn =0
N 1
X () =
2 N
k =
1 N
X0(
2 k 2 k ) ( ) N N0
x[n] =
Xk =0
N 1
(
2 k j 2N kn )e N
ak =
1 2 k X0( ) N N
Vy, tnh DTFT X () ca tn hiu x[n] ri rc tun hon vi chu k N , ta tin hnh theo cc bc sau y:1. Bt u vi mt chu k x0 [n] ca tn hiu x[n] , lu x0 [n] khng tun hon 2. Tm DTFT ca tn hiu khng tun hon trn:X 0 () = n = x0 [n]e jn
3. Tnh X 0 () ti cc gi tr =
2 k N
, k = 0,1,, N 1 2 N
4. T y c DTFT ca tn hiu tun hon theo nh cng thc va tm:
X () =V d:
k =
X0(
2 k 2 k ) ( ) N N
Cho x[n] = 1 . Tm X ()
- 92 -
Chng VV d:
Cho x0 [n] = [n] + [n 1] + 2 [n 3] . Gi s N = 4 . Tm X 0 () v X () v xc nh 4 gi tr phn bit ca X 0 ( 2Nk ) .
V d:
Cho tn hiu tun hon x[n] vi chu k N = 3 v mt chu k l: x0 [n] = [n] + 2 [n 2] . Tm X 0 () v X () . Kim tra kt qu bng cch tnh DTFT ngc khi phc li x[n] .
- 93 -
Chng VV d:
Cho tn hiu tun hon y[n] vi chu k N = 3 v mt chu k l: y0 [n] = [n] + 2 [n 1] + 3 [n 2] . Tm Y0 () v Y () . Kim tra kt qu bng cch tnh DTFT ngc khi phc li y[n] .
5.2 PHP BIN I FOURIER CA TN HIU RI RC DI HU HN 5.2.1 Biu thc tnh bin i Fourier ri rc thun ca tn hiu ri rc tun hon
Trong mc trn, ta xt mt chu k x0 [n] ca tn hiu tun hon x[n] . Ta c th xem phn chu k ny c c bng cch ly ca s (windowing) tn hiu di v hn x[n] : x0 [n] = x[n]wR [n] Vi wR [ n] l ca s ch nht ( y n cn c gi l ca s DFT): 1, n = 0,1,L, N 1 wR [n] = otherwise 0, x0 [n] = x[n]wR [n] ch l cc mu ca x[n] nm gia n = 0 v n = N 1. (khng quan tm n cc mu nm ngoi ca s). Ta c th tnh DTFT ca x0 [n] nh sau: X 0 () = DTFT( x0 [n]) = Vy, X 0 () = x[n]e jn = x0 [n]e jnn=0 n=0 N 1 N 1
n =
x0 [n]e jn =
n =
x[n]wR [n]e jn = x[n]e jnn=0
N 1
By gi ta tin hnh ly mu X 0 () lu tr trn my tnh. Do X 0 () lin tc v tun hon vi chu k 2 nn ch cn cc mu trong di tn s c bn. thun tin, ta ly N mu- 94 -
cch u nhau trong on [0, 2 ) :
Chng V
0, 2 / N, 4 / N, K, ( N 1)2 / N
Ni cch khc, cc im l: =2 k N
, k = 0,1,, N 1
Ta nh ngha php bin i Fourier ri rc DFT (Discrete Fourier Transform) nh sau: X [k ] = X 0 ( 2 k ) vi k = 0, 1, K, N 1 N
X[k] c gi l ph ri rc (discrete spectrum) ca tn hiu ri rc.Lu 1:
X[k] l hm phc theo bin nguyn, c th c biu din di dng: X[k ] =| X[k ] | e j[ k ] y |X[k]| l ph bin v [k ] ph pha.Lu 2: phn gii (resolution) ca ph ri rc l 2N v ta ly mu ph lin tc ti cc im cch nhau 2N trong min tn s, ngha l: = 2N .
Ta cng c th biu din phn gii theo tn s tng t f. Ta nh li quan h: F= Do :f = fs N
f fs
Lu 3: Nu ta xem xt cc mu ca X 0 () l 2Nk vi k = n th ta s thy DFT chnh l mt chu k ca DFS, nhng DFT hiu qu hn nhiu so vi DFS bi v s mu ca DFT l hu hn:
- 95 -
Chng VX [ k ] = X 0 ( ) | =N 1 n=0
2 k , k = 0,1,L, N 1 NN
= x[n]e jn |= 2 k ,k =0,1,L, N 1 = x[n]e jn=0 N 12 kn N
,k =0,1,L, N 1
cho gn, ta k hiu:
WN = e
j
2 N
Khi khng cn n N, ta c th vit n gin W thay cho WN Vy,X [k ] = x[n]WNkn , k = 0,1,L , N 1n=0 N 1
l DFT ca dy x0 [n]. ly ca s t x[n]V d:
Tnh DFT ca x[n ] = u[n ] u[n N]
(en=0
N 1
j 2 k N
) n = W knn =0
N 1
Suy ra DFT ca x[n] = 1, n = 0,1,L, 7.
V d:
n=0 1, . Tm X [k ], k = 0,1,, 7 Cho x[n] = 0, n = 1,, 7- 96 -
Chng V
- 97 -
Chng VV d:
Cho y[n] = [n 2] v N = 8 . Tm Y [k ]
V d: Cho x[n] = cWN pn , n = 0,1,, N 1 , vi p l mt s nguyn p [0,1,, N 1] v WN = e Tm DFT ca x[n] . j 2N
5.2.2 Biu thc tnh bin i Fourier ri rc ngc
Trong mc ny, ta s i thit lp cng thc khi phc x[n] t X [k ] . S khi phc ny c gi l tng hp hay DFT ngc (IDFT) T biu thc tnh DTFT ngc c thit lp trong mc 5.2.1 v do tnh tng h gia min thi gian v tn s, ta c th suy ra biu thc tnh IDFT nh sau: x[n] = 1 N
X [k ]Wk =0
N 1
kn N
, n = 0,1,, N 1
- 98 -
Chng V Sau y ta s chng minh iu ny ng:x[n] =
1 N
x[l ]Wk =0 l =0
N 1 N 1
kl N
WN kn
=
N 1 1 x[l ] WN (l n ) k =0 k N l =0
N 1
Ta c
Wk =0
N 1
k (l n ) N
N, l = n = 0, l n
Thay kt qu ny vo x[n] ta c c biu thc tnh IDFT trn l ng x[n] =N 1 1 N 1 1 x[l ] WNk (l n ) = k =0 N l =0 N 1 = ( Nx[n]) = x[n] N
x[l ]N [n l ]l =0
N 1
V d:
Tm IDFT ca X [k ] = 1, k = 0,1,, 7 .
V d:
Cho x[n] = [n] + 2 [n 1] + 3 [n 2] + [n 3] v N = 4 , tm X [k ] .
- 99 -
Chng VV d:
Cho X [k ] = 2 [k ] + 2 [k 2] v N = 4 , tm x[n] .
5.2.3 Chn s mu tn s N
Qua mc 5.2.1 ta thy biu thc tnh DFT c thnh lp t vic ly mu DTFT vi s mu l N. S mu N ny cng chnh l s mu ca tn hiu ri rc trong min thi gian hay l di ca ca s DFT, ni ngn gn l s mu tn s bng s mu thi gian.V d:
Cho tn hiu x[n] nh hnh bn. Tnh ri v hai loi ph bin | X() | v |X[k]| trn th. Xem th ta thy r rng rng: cc mu |X[k]| bng vi | X() | ti cng tn s.
- 100 -
Chng V Vic chn N nh hng n phn gii ca ph ri rc. Chn N cng ln, phn gii cng tt, ngha l khong cch gia hai vch ph cnh nhau X[k] v X[k+1] cng nh, ngha l ng bao ca ph ri rc X[k] cng gn vi hnh nh ca ph lin tc | X() | . vic tng N khng lm nh hng n kt qu, ta ko di tn hiu trong min thi gian ra bng cch chn thm cc mu bng 0 (zero-padding) vo pha cui ca tn hiu.V d:
Cho x[n] = u[n] u[n 5] . Tm X[k] vi N nh sau: (a) N = 5.
- 101 -
Chng V (b) N = 10
5.2.4 Cc tnh cht ca bin i Fourier ri rc
Hu ht cc tnh cht ca DFT tng t nh cc tnh cht ca DTFT, nhng c vi im khc nhau. im khc nhau l do DFT chnh l mt chu k trch ra t dy DFS tun hon vi chu k N.% By gi ta thay i k hiu, k hiu x[n] l dy tun hon chu k N, x[n] l mt chu k trch % ra t x[n] : % x[n] = x[n] k =
[n kN ]
=
k =
x[n kN]- 102 -
Chng V1. Dch vng
Nu x[n] X [k ] th x[n m] W km X[k ] vi WN = eV d:DFT j 2N
DFT
Dch vng i m mu s cho kt qu trng vi dich vng i (m mod N) mu.2. Tng chp vngx1[n] x2 [n] X 1[k ] X 2 [k ]DFT , N
y:y[n] = x1[n] x2 [n] = x1[ p]x2 [n p]mod Np =0 N 1
Du l k hiu tng chp vng. Nhc li cng thc tng chp tuyn tnh:y[n] = x1[n] x2 [n] =
p =
x1[ p ]x2 [n p]
- 103 -
Chng V Thot nhn, ta thy biu thc tnh tng chp vng rt ging tng chp tuyn tnh. Tuy nhin, hai php chp khc nhau nhng im sau y: Php chp vng ch p dng cho hai dy di hu hn v bng nhau, kt qu cng l mt dy cng chiu di, ngha l x1[n] , x2 [n] , and y[n] u c chiu di l N. Trong khi , php chp tuyn tnh p dng cho hai dy c chiu di bt k: nu x1[n] di N x1 , x2 [n] di N x1 th y[n] di Php dch trong tng chp vng l php dch vng, khc vi php dch trong tng chp tuyn tnh l php dch tuyn tnh.
-
V nhng im khc nhau trn nn kt qu ca tng chp vng v tng chp tuyn tnh ca cng hai dy c th khng trng nhau. Tuy nhin, ta c cch lm cho hai kt qu trng nhau nh sau: Chuyn tng chp tuyn tnh sang min tn s:Y() = X 1 ().X 2 ()
-
Ly mu Y() vi s mu l N N y = N x1 + N x 2 1 , ta c: Y[k ] = X[k ].H[k ] Tnh DFT ngc, ta c: y[n] = x[n] * h[n]
-
y chiu di ca y[n] , x[n] v h[n] l:
N N y = N x1 + N x 2 1 Nh vy, bng cch ko di cc tn hiu x1[n] v x2[n] ra n chiu di N N y = N x1 + N x 2 1 ri ly chp vng, ta c hai kt qu ca tng chp vng v chp tuyn tnh l trng nhau:y[n ] = x 1 [n ] x 2 [n ] = x 1 [n ] x 2 [n ]
V d:
Tm x1[n] x2 [n] = z[n] , vi x1[n] = [1, 2, 0, 0] , x2 [n] = [1,1, 0, 0] v N = 4. Kt qu ny c trng vi tng chp tuyn tnh khng?
- 104 -
Chng VV d:
Tm y[n] = x[n] x[n] , vi x[n] = [1, 0,1,1] trong hai trng hp: (a) N = 4 (b) N = 8 N bng bao nhiu l tng chp vng trng vi tng chp tuyn tnh?
5.3 MT S NG DNG CA DFT
Phn ny s gii thiu s lc v mt s ng dng ca DFT trong thc t5.3.1 Phn tch ph tn hiu
Trong chng trc, ta bit c ngha ca ph trong vic phn tch tn hiu, t ph ca tn hiu ta bit c mt s thng tin cn thit. tm ph ca tn hiu (c lin tc v ri rc), ta cn phi bit gi tr ca tn hiu ti tt c cc thi im. Tuy nhin trong thc t, do ta ch quan st c tn hiu trong mt khong thi gian hu hn nn ph tnh c ch l xp x ca ph chnh xc. DFT c ng dng rt hiu qu trong vic tnh ton ph xp x ny. Trong thc t, nu tn hiu cn phn tch l tn hiu lin tc, trc ht ta cho tn hiu i qua mt b lc chng chng ph ri ly mu vi tn s Fs 2B , vi B l bng thng ca tn hiu sau khi lc. Nh vy, tn s cao nht cha trong tn hiu ri rc l Fs/2. Sau , ta phi gii hn chiu di ca tn hiu trong khong thi gian T0 = LT, vi L l s mu v T l khong cch gia hai mu. Cui cng, ta tnh DFT ca tn hiu ri rc L mu. Nh trnh by trn, mun tng phn gii ca ph ri rc, ta tng chiu di ca DFT bng cch b thm s 0 vo cui tn hiu ri rc trc khi tnh DFT. V d sau y minh ha mt ng dng ca DFT trong vic phn tch ph tn hiu in tm (ECG): Hnh v (a) l th ca 11 nhp tim ca mt bnh nhn. 11 nhp tim ny xut hin trong khong thi gian 9 giy, tng ng vi 11/9 = 1.22 nhp trong mt giy, hay 73 nhp trong mt pht. Hnh (b) l chi tit na u ca nhp tim th t. Hnh (c) l mt on ph bin DFT c c sau khi ly mu on 11 nhp tim (a) vi tn s ly mu l 8 kHz. Nhn (c) ta thy c hai im bin cao nht xut hin tn s 88 Hz- 105 -
Chng V v 235 Hz. tm hiu ph k hn, ta tnh DFT ca tn hiu hnh (b)- ph ny th hin hnh (d), y ta thy r hai im bin cao nht tn s 88 Hz v 235 Hz bn trong mi nhp tim. Tuy nhin, ta khng thy tn s lp li nhp tim l 1.22 Hz trong DFT hnh (c). Hnh (e) gii thch r hn iu ny. N l phin bn m rng ca cc nh nhn trong di tn t 60 Hz n 100 Hz. Trong khi tn s 1.22 Hz qu nh nn khng thy r trong hnh (c) th trong hnh (e) ny, ta thy r cc hi ca tn s 1.22 Hz v thy r khong cch gia hai nh nhn l 1.22 Hz.
5.3.2 Tnh tn hiu ra h thng ri rc LTI
Tn hiu ra h thng ri rc LTI c tnh bng cch chp tn hiu vo vi p ng xung ca h thng:
y[n ] = x[n ] h[n ]Ta c hai cch tnh tng chp ny: mt l tnh trc tip, hai l tnh thng qua tng chp vng nh phn tch trong mc 5.2.4. Cch tnh qua tng chp vng s c li hn v mt thi gian. L do l tng chp vng c th tnh thng qua DFT, m DFT c th c tnh nhanh nh thut ton tnh nhanh FFT. tnh y[n], ta thc hin theo cc bc sau y: Ko di x[n] n di N = Nx + Nh - 1- 106 -
Chng V Ko di h[n] n di N = Nx + Nh - 1 Tnh DFT ca x[n] N mu, ta c X[k] Tnh DFT ca h[n] N mu, ta c H[k] Nhn X[k] vi H[k], ta c Y[k]: Y[k] = X[k].H[k] Tnh DFT ngc ca Y[k], ta c y[n] Vic tnh DFT v DFT ngc c thc hin nh mt thut ton tnh nhanh DFT, gi l FFT (Fast Fourier Transform). Phn sau s trnh by v thut ton FFT.5.4 TNH NHANH DFT BNG THUT TON FFT
DFT c ng dng rng ri trong x l tn hiu ri rc/ s nn nhiu nh ton hc, k s rt quan tm n vic rt ngn thi gian tnh ton. Nm 1965, Cooley v Tukey tm ra thut ton tnh DFT mt cch hiu qu gi l thut ton FFT. Cn lu FFT khng phi l mt php bin i m l mt thut ton tnh DFT nhanh v gn hn. nh gi hiu qu ca thut ton, ta s dng s php tnh nhn v cng phc. S php nhn v cng phc lin quan trc tip n tc tnh ton khi thut ton c thc hin trn cc my tnh hay l cc b x l chuyn dng.5.4.1 Hiu qu tnh ton ca FFT
Cng thc tnh DFT ca dy di N:X [k ] = x[n]W knn =0 N 1
Qua y ta thy tnh mi gi tr DFT ta cn N php nhn v cng phc. tnh ton b DFT ta cn N 2 php nhn v cng phc. Tuy nhin, nu tnh DFT nh thut ton FFT th s php nhn v cng phc gim xung ch cn N log 2 N . 2 V d nh N = 210 = 1024 th nu tnh trc tip DFT cn N 2 = 220 = 106 php nhn v cng phc, trong khi tnh qua FFT th s php nhn v cng phc gim xung ch cn N log 2 N = 2 5120. S php tnh gim i gn 200 ln! Hnh sau cho thy r hiu qu ca thut ton FFT:10000 Number of Operations
8000
6000
4000
2000
0 0
20
40 60 N, Size of DFT or FFT
80
100
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Chng V C nhiu thut ton FFT khc nhau bao gm FFT phn chia theo thi gian v FFT phn chia theo tn s. Trong phn ny ta tp trung vo thut ton FFT c s 2 ( N = 2i where i is an integer ) phn chia theo thi gian.5.4.2 Nguyn tc ca FFT
Nguyn tc c bn m cc thut ton FFT u da vo l phn chia DFT N mu thnh cc DFT nh hn mt cch lin tc: Vi N = 2i, u tin ta phn chia DFT N mu thnh cc DFT N mu, sau phn chia DFT 2 N N 2 mu thnh DFT 4 mu v c tip tc nh th cho n khi c cc DFT di N = 2. Vic tnh DFT nh hn r rng s cn t php tnh nhn v cng phc hn. Trc tin, chia x[n] thnh cc dy con chn v l:X [k ] =neven
x[n]W2 mk
kn
+ x[n]W knnodd
t n = 2m vi n chn v n = 2m + 1 vi n l:X [k ] = x[2m]Wm=0N 1 2
+ x[2m + 1]W k (2 m +1) =m=0
N 1 2
m=0
x[2m](W )
N 1 2
2 mk
+W
k
m =0
x[2m + 1](W
N 2
1
2 mk
)
=
X [k ] = X e [k ] + W k X o [k ] = G[k ] + W k H [k ] X e [k ] v X o [k ] l DFTTip theo chia dy conN 2 N 2
mu.
mu l x[2m] lm i bng cch t m = 2 p :N 1 4
X e [k ] = x[4 p](W ) + W4 kp p =0
2k
x[4 p + 2](Wp =0
N 4
1
4 kp
) =
Thc hin tng t nh vy cho dy con x[2m+1]V d: N = 8
Qu trnh phn chia DFT 8 mu thnh cc DFT nh hn c minh ha trn lu . u tin, chia x[n] thnh 2 dy con, dy th nht l dy chn x[0], x[2], x[4], x[6] v dy th hai l dy l x[1], x[3], x[5], x[7]. Tip theo, chia dy chn thnh 2 dy con, dy th nht l x[0], x[4] v dy th hai l x[2], x[6]. Tng t, dy l c chia thnh 2 dy con, l dy x[1], x[5] v dy x[3], x[7]. Cc DFT 2 mu c tnh n gin nh sau: G[k ] = g[n ]W , 0 k 1, W = enk n =0 1 2 2
j
= 1 (ch cn php cng v tr)
G[0] = g[0]W 0.0 + g[1]W 1.0 = g[0] + g[1] G[1] = g[0]W0.1
+ g[1]W
1.1
= g[0] g[1]- 108 -
Chng V
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Chng VFFT c s: A Butterfly
0 WNr
WN(r + N/2) Lu : WN(r + N/2) gin nh sau: = WN N/2 WNr = -1 WNr = - WNr , do c th v li lu FFT n
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Chng V
Ph lc 1Summary: The Common Types of Fourier TransformsContinuous in Time x(t ) = Aperiodic in Frequency Fourier Series (FS): Periodic in Time, = Discrete in Frequencyak =1 jk t T x(t )e 0 dt Tk =
Discrete in Time x[n] = Periodic in Frequency Discrete Fourier Series (DFS) and Discrete Fourier Transform (DFT):X [k ] = x[n]WNkn , 0 k N 1n=0 N 1
x(t ) =
aek
jk0t
x[n] =
1 N
X [k ]Wk =0 j 2N
N 1
kn N
, 0 n N 1
where WN = e
.
Aperiodic in Time, = Continuous in Frequency
Fourier Transform (FT):X ( ) = x(t )e jt dt
Discrete-Time Fourier Transform (DTFT):X () =n =
x(t ) = X ( )e
jt
dt
x[n]e2
jn
1 x[n] = 2
X ()e
jn
d
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Chng V
Ph lc 2Some Fourier RelationshipsThe Fourier transform is the Laplace transform evaluated on the j axis. X ( ) = x(t )e jt dt = X ( s ) s = j = x(t )e st dt s = j
The discrete-time Fourier transform is the z-transform evaluated around the unit circle. X () = x[n]e jn = X ( z ) z =e j = x[n]x n n = n = z = e j
Discrete-time periodic signals can also be described by a Fourier Series expansion: x[n] = and ak = 1 Nn< N > k< N >
ak e jk 0 n
synthesis equation
x[n]e jk 0 n
analysis equation
then using the DTFT of the impulse train, P () that we previously found, the DTFT of an arbitrary discrete-time periodic signal can be found from X 0 () the DTFT of one period x0 [n] 2 X () = X 0 () N = 2 N
( k
2 k ) N
Xk
0
(
2 k 2 k ) ( ) N N
The DFT is simply a scaled version of the terms of one period of the discrete time Fourier transform for a periodic sequence: X [k ] = X 0 ( for =2 k NN 1 2 k kn ) = x[n]WN , 0 k N 1 N n=0
, k = 0,1,, N 1 , i.e. only look at the N distinct sampled frequencies of X 0 () .N 1 n=0
Also important, the orthogonality of exponentials:
Wwhere WN = e j 2N
kn N
= N [k ]
.
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