Chng 1: C s l thuyt v x l tn hiu s1.1 Tn hiu v h thng theo thi gian1.1.1 Cc loi tn hiu v h thng1.1.1.1 Tn hiu nh ngha tn hiui lng vt l bin thin theo thi gian, theo khng gian, theo mt hoc nhiu bin c lp khc a chn: chn ng a l ~ thi gian m thanh, ting ni: dao ng sng ~ thi gian (t) Hnh nh: cng nh sng ~ khng gian (x,y,z)Biu din ton hc: hm theo bin c lp u(t) = 2t2 5 f(x,y) = x2 2xy 6y2 Cc t/h t nhin thng khng biu din c bi mt hm s cpHm xp x cho cc t/h t nhin Phn loi tn hiuTn hiu c phn loi da vo nhiu c s khc nhau v tng ng c cc cch phn loi khc nhau. y, ta da vo s lin tc hay ri rc ca thi gian v bin phn loi. C 4 loi tn hiu nh sau:- Tn hiu tng t (Analog signal): thi gian lin tc v bin cng lin tc.- Tn hiu ri rc (Discrete signal): thi gian ri rc v bin lin tc. Ta c th thu c mt tn hiu ri rc bng cch ly mu mt tn hiu lin tc. V vy tn hiu ri rc cn c gi l tn hiu ly mu (sampled signal).- Tn hiu lng t ha (Quantified signal): thi gian lin tc v bin ri rc. y l tn hiu tng t c bin c ri rc ha.- Tn hiu s (Digital signal): thi gian ri rc v bin cng ri rc. y l tn hiu ri rc c bin c lng t ha. Cc loi tn hiu trn c minh ha trong hnh 1.1.

tx(t)x(n)x(t)nx(n)H1.1 Tn hiu tng tH1.2 Tn hiu ri rct H1.3 Tn hiu c lng t honH1.4 Tn hiu s

1.2 Tn hiu ri rc1.2.1 nh ngha L tn hiu c th c biu din bng mt dy cc gi tr (thc hoc phc) vi phn t th n c k hiu l x(n). x = { x(n) } n = -...+ Thng thng tn hiu ri rc c c bng cch ly mu cc tn hiu lin tc trong thc t. Phng php ly mu thng gp l ly mu u tc l cc thi im ly mu cch nhau mt khong Ts gi l chu k ly mu.V d: Tn hiu v nhit l 1 tn hiu lin tc. Ti trm kh tng c 15 pht ngi ta ghi li nhit mt ln. Nh vy tc l thc hin thao tc ly mu tn hiu nhit vi chu k ly mu Ts = 15 pht, s liu thu c l tn hiu nhit ri rc.

1.2.2 Mt vi tn hiu ri rc quan trng Tn hiu xung n v:

H1.7 Xung n v

Tn hiu xung nhy bc n v:

0n11-12u(n)3H1.8 Xung nhy bc n v-2 Tn hiu hm s m:

0nx(n)-2-1 123H1.9 - Tn hiu hm s m vi 0 < a < 1 Tn hiu RectN

01234-1-2nu(n) H1.10 Tn hiu RectN Tn hiu tun honXt tn hiu x(n) ta ni rng tn hiu x(n) l tun hon vi chu k N nu: x(n) = x(n+N) = x(n+kN) vi mi n. Hnh v di y minh ho tn hiu tun hon vi chu k N = 4.

-6-5-4-3-2-186543210nGi tr N nh nht tho mn x(n) = x(n+N) c gi l chu k c bn ca tn hiu.Nhn xt: Mt tn hiu ri rc bt k c th biu din bi cng thc:

H thng (h/t)Thit b vt l, thit b sinh hc, hoc chng trnh thc hin cc php ton trn tn hiu nhm bin i tn hiu, rt trch thng tin, Thit thc hin php ton cn c gi l x l tn hiuV d Cc b lc t/h Cc b trch c trng thng tin trong t/h Cc b pht, thu, iu ch, gii iu ch t/h,

Mt h thng x l tn hiu s xc lp mi quan h gia tn hiu vo v tn hiu ra: y = T[x].

Tx(n)y(n)H1.5 M hnh mt h x l Phn loi h x l theo tn hiu vo v tn hiu ra: H ri rc: l h x l tn hiu ri rc. H tng t: l h x l tn hiu tng t.

LPFTn hiu voS&HADCDSPDACLPFTn hiu raH1.6 M hnh x l tn hiu s trong thc tTn hiu tng tTn hiu tng tTn hiu sTn hiu tng tTn hiu tng tTn hiu s LPF(Low Pass Filter): B lc thng thp loi b nhiu v m bo nh l Shannon. S&H(Sampling and Hold): Mch trch gi mu gi cho tn hiu n nh trong qu trnh chuyn i sang tn hiu s. ADC(Analog to Digital Converter): B chuyn i tng t thnh s. DAC(Digiatal to Analog Converter): B chuyn i s thnh tng t. DSP(Digital Signal Processing) X l tn hiu s.

u im C th lp trnh c D m phng, cu hnh - sn xut hng lot vi chnh xc cao Gi thnh h Tn hiu s d lu tr, vn chuyn v sao luNhc im Kh thc hin vi cc t/h c tn s cao1.1.2 nh l ly mu

1.1.3 H thng x l tn hiu s

Chng ta c th phn loi cc h thng theo chnh tn hiu cn x l. Theo , c cc loi h thng cn x l sau y:

Ch rng, v tn hiu s l mt trng hp ring ca tn hiu ri rc nn h thng ri rc cng c th x l tn hiu sH THNG X L S TN HIU TNG T:X l s tn hiu tng t l x l tn hiu tng t bng h thng s. thc hin vic ny, ta cn phi bin i tn hiu tng t thnh tn hiu s v sau khi x l dy kt qu c th c phc hi tr thnh tn hiu tng t. V d nh trng hp x l tn hiu thoi. Trong nhiu trng hp, mc tiu ca vic x l l trch ly cc tham s ca tn hiu hay cc thng ti cn thit t tn hiu . Khi , khng cn chuyn i tn hiu v dng tng t. V d: x l tn hiu rada hoc sonar. H thng x l s tn hiu tng t c trnh by trong hnh

Bin i A/D (Analog-to-Digital Conversion)Bin i A/D l bin i tn hiu tng t thnh tn hiu s. Bin i A/D c s khi nh sau:

Ly mu v gii mu

Ly mu l qu trnh bin i lin tc(tng t) sang tn hiu ri rc. Cnhiu cch ly mu mt tn hiu lin tc. Trong , thng dng nht l cch lymu tun hon (periodic sampling), cn gi l ly mu u (uniform sampling). l cch ly nhng mu bin tn hiu lin tc ti nhng thi im ri rccch u nhau mt khong thi gian TS, m ta gi l chu k ly mu. Nu xa(t) ltn hiu tng t ng vo b ly mu th tn hiu ri rc ng ra ca b ly mul xa(nTS) (Gi tt l tn hiu ly mu), n l s nguyn. M hnh vt l ca b lymu c minh ha trong hnh

Trong , b phn ly mu c m t nh l mt b kha c iu khinng m bi tn hiu xung ng h Ck c tn s l FS= 1/TS. x l bng kthut s hoc bng my tnh, thng thng tn hiu ri rc cn phi c lng tha c th biu din bin ca cc mu bng mt tp hu hn cc m nhphn. Tuy nhin, vic lng t ha v m ha khng th thc hin tc thi. Thngthng, tin trnh lng t ha v m ha mt mu c thc hin trong khongthi gian TS. V vy, gi tr ca ca mt mu phi c duy tr trong thi gian TS.y l chc nng ca b gia mu. B gia mu tiu biu l Zero-order-hold. Bly mu v gi mu kiu zero-order-hlod ny tng ng vi mt b iu chdy xung ch nht theo sau bi mt b lc tuyn tnh, m tn hiu ng ra ca n(Gi tt l tn hiu gi mu) c dng bc thang hnh 1.15.

Lng t ha v m ha (Quantizer and Coder) y l b bin i tn hiu ri rc sang tn hiu s c bin c biu dinbng cc m nh phn. Gi tr mi mu ca tn hiu ly mu c gn bi mt gitr c la chn t mt tp hu hn cc ga tr. Trong tin trnh m ha, mi gitr ri rc c gn bi mt m nh phn m bit, tng ng c 2m mc lng t.Nu bin ca tn hiu ly mu c chun ha trong khong -X0 x(n) X0thbc lng t ha (khong cch gia hai mc lng t k nhau) s l:D = 2X0/2m = X0/2m 1Bin i D/A (Digital to Analog Conversion) Trong nhiu ng dng thc t, tn hiu s sau khi c x l cn phi cphc hi li thnh tn hiu tng t. hi lm vic ny, ta cn c b bin i ssang tng t (D/A converter). Nguyn tc chung ca bin i D/A l ni cc im ri rc bng mt phngphp ni suy (Interpolation) no . Hnh 1.16 trnh by mt kiu bin i D/An gin, kiu xp x bc thang (staircase approximation), cn c gi lzero-order hold.

C nhiu kiu bin i D/A khc, nh: ni suy tuyn tnh (linearinterpolation), ni suy bc hai (quadratic interpolation),.... Vi mt tn hiu cbng tn hu hn, l thuyt ly mu s xc nh mt hnh thc ni suy ti u.1.1.2 nh l ly mu Cho mt tn hiu tng t bt k, vn l chn chu k ly mu TS hay tn sly mu FS nh th no cho hp l? Xu hng chung l chn tn s ly mu thp,bi v tn s ly mu cao s lm tng s mu, t lng php tnh trong qutrnh x l tn hiu s tng ln, ko di thi gian x l, ng thi lng b nh cnthit cng tng theo. Tuy nhin, nu tn s ly mu qu thp s xy ra hin tngbit d../Anh, khng th khi phc li tn hiu tng t mt cch chnh xc. Chngta s tr li vn ny trong chng 3, khi phn tch tn hiu trong min tn s, t chng minh nh l ly mu, m ta s pht biu sau y.

Tn hiu lin tc trong thc t c di hu hn (tn ti trong mt khongthi gian hu hn) l t hp tuyn tnh ca nhiu thnh phn hnh sin. Ta xt cctn hiu c bng tn hu hn, ngha l tn s cao nht trong bng tn c th xcnh. V d: tn hiu thoi c cc thnh phn tn s t vi trm Hz n 3KHz, tnhiu hnh c tn s cao nht l 6MHz. Nu ta bit thnh phn tn s cao nht Fmax, ta c th chn tn s ly muthch hp. nh ly ly mu c pht biu nh sau: nh l : Nu tn s cao nht cha trong mt tn hiu tng t xa(t) lFmax th tn hiu ch c th c khi phc mt cch chnh xc t cc mu ca nnu tn s ly mu FS 2Fmax,. cho gn, ta t Fmax = B. nh l trn cng ch ra rng xa(t) c th ckhi phc t cc mu xa(nTS) bng cch dng hm ni suy:

y xa(n/FS) = xa(nTS) = x(n) l cc mu ca xa(t). Nu tn s ly muFS=2Fmax=2B, th cng thc khi phc (1.91) tr thnh:

Tn s ly mu FS =2B = 2Fmax c gi l tn s Nyquist. Hnh 1.18 minhha mt cch bin i A/D l tng dng hm ni suy (1.90). Trong s hnh 1.12, mch lc trc c tc dng chng hin tng h danh.y l mt mch lc thng thp c chc nng lc b cc thnh phn tn s caohn FS/2, trong trng hp ph tn ca tn hiu vt qu kh nng ca b ly mu(khi ta phi chp nhn kt qu gn ng ca tn hiu ra). Ngay c khi thnhphn tn s cao nht ca tn hiu nh hn FS/2, nhiu tn s cao cng gy rahin tng h danh v cn phi lc b. Mch lc sau s trong hnh 1.12 cng l mt mch lc thng thp. N cchc nng lm trn (smoothing) sa dng tn hiu tng t thu c ng rachnh xc hn. BIU DIN TN HIUV H THNG RI RC TRONG MIN Z Chng 1 trnh by cch tnh p ng ca mt h thng trc tip t png xung ca n, bng cch tnh tng chp ca kch thch vi p ng xung. Cchtnh tng chp trc tip da vo cng thc nh ngha nh lm tn rt nhiuthi gian v cng sc. Hn na , trong thc t s mu khc khng ca kch thchv p ng xung l rt nhiu nn ta khng th tnh bng tay. Tuy nhin, phngphp tnh tng chp bng th nh trnh by cho ta mt thut ton ca chngtrnh tnh tng chp bng my tnh. Vic gii phng trnh sai phn tuyn tnh hs hng bng phng php qui cng ch c ngha khi s dng my tnh. K thut bin i l mt cng c hu hiu phn tch h thng LTI. Bini Z i vi tn hiu ri rc c vai tr tng t nh bin i Laplace i vi tnhiu lin tc, v chng c quan h ging nhau vi bin i Fourier. Tng chpca hai dy trong min thi gian s bin thnh tch ca hai bin i Z tng ngtrong min bin phc z. Tnh cht ny s lm n gin ha vic tnh p ng cah thng vi cc tn hiu vo khc nhau. Phng trnh sai phn tuyn tnh h shng cng c gii mt cch d dng hn khi dng cng c bin i Z. Nh ta s thy trong cc chng sau, bin i Fourier gia vai tr cha khatrong trong vic biu din v phn tch cc h thng ri rc. Tuy nhin, trong mts trng hp cn phi s dng dng tng qut ha ca bin i Fourier, lbin i Z.2.2 Cc khi nim v bin i Z2.2.1. BIN I Z ( THE Z - TRANSFORM): Bin i z ca mt dy x(n) c nh ngha nh l chui ly tha: