Xử lý tín hiệu số - Thầy Loan

*X L TN HIU S

*TI LIU THAM KHO Bi ging ny ! X l tn hiu s X l tn hiu s v lc s

*Chng 1TN HIU V H THNG RI RC

*Nhng ni dung cn nm vng:Chng 1Cc tn hiu ri rc c bit (xung n v, bc n v, hm m, tun hon)Cc php ton vi tn hiu ri rc (nhn vi h s, cng, php dch)Quan h vo-ra vi h TT-BB:Tn hiu vo (tc ng), tn hiu ra (p ng), p ng xungCch tnh tng chp y(n) = x(n) * h(n)Cc tnh cht ca h TT-BB nhn qu, n nhQuan h vo-ra thng qua PT-SP-TT-HSHH TT-BB xt trong min tn s:p ng tn s (p ng bin , p ng pha)Ph tn hiu (ph bin , ph pha)

*Nhng ni dung cn nm vng:Chng 2nh ngha bin i z (1 pha, 2 pha)Min hi t ca bin i zCc tnh cht ca bin i zPhng php tnh bin i z ngc (phn tch thnh cc phn thc hu t n gin)Cch tra cu bng cng thc bin i zng dng bin i z 1 pha gii PT-SPXt tnh nhn qu v n nh thng qua hm truyn t H(z).

*Nhng ni dung cn nm vng:Chng 3Phn loi b lc s (FIR, IIR)Phng php thc hin b lc s (phn cng, phn mm):- S khi- Lp trnh gii PT-SPCc thuc tnh ca b lc:Nhn qu, n nh, hm truyn t, p ng xung, p ng tn s (bin , pha), tnh cht lc (thng cao, thng thp, thng di, chn di)

*

Min thi gianMt phng zMin tn sT.h. vo x(n)T.h. ra y(n)p ng xung h(n)

y(n) = x(n) * h(n)Nhn qun nh(th hin qua p ng xung)X(z)= Z[x(n)]Y(z)= Z[y(n)]H(z)=Z[h(n)]=Y(z)/X(z)Y(z) = X(z). H(z)

Nhn qu:n nh:(V tr ca im cc ca H(z) so vi ng trn n v) Ph X(ejw)=F[x(n)]Ph Y(ejw)=F[y(n)]p ng tn sH(ejw)= Y(ejw)/ X(ejw)=F[h(n)]Y(ejw)= X(ejw). H(ejw)

*1.1 Khi nim v phn loiTn hiu l biu hin vt l ca thng tinV mt ton, tn hiu l hm ca mt hoc nhiu bin c lp. Cc bin c lp c th l: thi gian, p sut, cao, nhit Bin c lp thng gp l thi gian. Trong gio trnh s ch xt trng hp ny. Mt v d v tn hiu c bin c lp l thi gian: tn hiu in tim.

*Phn loi:Xt trng hp tn hiu l hm ca bin thi gian

Tn hiu tng t: bin (hm), thi gian (bin) u lin tc. V d: x(t)Tn hiu ri rc: bin lin tc, thi gian ri rc. V d: x(n)x(n)

*Phn loi tn hiuThi gian lin tcThi gian ri rcBin lintcBin ri rcTn hiu tng tTn hiu ri rcTn hiu lng t haTn hiu s

*X l s tn hiuTn hiutng tTn hiutng tTn hiusADCDAC

*Ti sao li tn hiu s ? c th x l t ng (bng my tnh) Gim c nhiu Cho php sao lu nhiu ln m cht lng khng thay i Cc b x l tn hiu s (DSP)khi c ch to hng lot c cht lng x l ng nht v cht lng x l khng thay i theo thi gian

*Bin i tng t-sLy mu sau lng t ha

Ly mu(ri rc ha thi gian)Lng t ha(ri rc ha bin )Fs >= 2Fmax (Fmax: tn s ln nht ca tn hiu)nh l Shannon (ly mu)Chu k ly mu TsTn s ly mu Fs = 1/Ts

*Mt s tn hiu ri rc c bitXung n v

*Tn hiu bc n v

*Tn hiu hm mx(n)=an

*Tn hiu tun honx(n)=x(n+N), N>0: chu kx(n)=sin[(2p/N)(n+n0)]

*1.3. Cc php ton vi tn hiu ri rc Php nhn 2 tn hiu ri rc Php nhn tn hiu ri rc vi h sx(n)aa x(n)

*1.3. Cc php ton vi tn hiu ri rc Php cng 2 tn hiu ri rc Php dchnu dch phi n0 mu, x(n) tr thnh y(n)y(n) = x(n-n0)

*1.3. Cc php ton vi tn hiu ri rcTr 1 muMt tn hiu ri rc bt k x(n) lun c thc biu dinDelay

*y(n) =x1(n-1)

*1.4. Phn loi cc h x l tn hiu ri rcx(n): tn hiu vo (tc ng)y(n): tn hiu ra (p ng)Phn loi da trn cc iu kin rng buc i viphp bin i T

y(n)=T[x(n)]H tuyn tnh nu tha mn nguyn l xp chng

*1.4. Phn loi cc h x l tn hiu ri rcT[ax1(n)+bx2(n)] =aT[x1(n)]+bT[x2(n)]=a y1(n) + b y2(n)Nu h tuyn tnh:y(n) = T[x(n)]

*

*1.4. Phn loi cc h x l tn hiu ri rcNu h bt bin theo thi gianTc ng d(n) cho p ng h(n)Tc ng d(n-k) cho p ng h(n-k)Vi h tuyn tnh bt bin (TTBB):h(n) l p ng xung ca h*: Php tng chp

*1.4. Phn loi cc h x l tn hiu ri rcV d H TTBB

*1.4. Phn loi cc h x l tn hiu ri rc di tn hiu: S lng mu khc 0 ca tn hiu Phn bit cc h TTBB da trn chiu di ca p ng xung FIR: H c p ng xung hu hn(Finite Impulse Response) IIR: H c p ng xung v hn(Infinite Impulse Response) Nng lng tn hiu

*1.4. Phn loi cc h x l tn hiu ri rcTnh tng chpV d 1Tn hiu vo v p ng xung ca h TTBBnh hnh v. Hy tnh tn hiu ra

*1.4. Phn loi cc h x l tn hiu ri rcTnh tng chpV d 1y(n)=x(0)h(n-0)+x(1)h(n-1)=0,5h(n)+2h(n-1)

*V d 2 n 0:Vi mi gi tr ca n:

*1.5.Tnh cht ca h TTBBGiao honKt hp

y(n)=x(n)*h(n)=h(n)*x(n)[y(n)*x(n)]*z(n)=y(n)*[x(n)*z(n)]

*1.5.Tnh cht ca h TTBB

Cc h tng ng

*1.5.Tnh cht ca h TTBBPhn phi

x(n)*(h1(n)+h2(n))=x(n)*h1(n)+ x(n)*h2(n)

*1.5.Tnh cht ca h TTBBH c nh v khng nhKhng nh: tn hiu ra ph thuc tn hiu vo cng thi im.V d y(n)=A.x(n)C nh: tn hiu ra ph thuc tn hiu vo nhiu thi im V d y(n) = x(n) x(n-1)

*1.5.Tnh cht ca h TTBBH ng nhtTn hiu ra bng tn hiu voy(n) = x(n)H A l o ca h B nu mc ni tip 2 h ny ta c 1 h ng nht

*1.5.Tnh cht ca h TTBBH o(A) v h kh o (B)

*1.5.Tnh cht ca h TTBB H nhn quTn hiu ra ch ph thuc tn hiu vo hin ti v qu khCha c tc ng th cha c p ngp ng khng xy ra trc tc ng Nu x(n) =0 vi n < n0 th y(n) =0 vi n < n0Nu h nhn qu th y(n) khngph thuc x(k) vi k >nh(n-k) = 0 vi k > n tc l h(n) = 0 vi n < 0

*V d p ng xung ca h n nh v khng n nh-5-4-3-2-1012345nh(n)-5-4-3-2-1012345nh(n)n nhKhng n nh

*1.6. p ng tn s ca h TTBBp ng tn s: cho bit tnh cht truyn t ca h i vi cc thnh phn tn s khc nhau ca tn hiu vo xt biu din tn s ca h TTBB, tc ng ca h c dng: H c p ng xung h(n)

*1.6. p ng tn s ca h TTBBp ng ca h:H(ejw) cho bit s truyn t ca h i vi mi tn s w nn H(ejw) l p ng tn s ca h.

*1.6. p ng tn s ca h TTBBH(ejw) l hm phc nn c th c biu dintheo phn thc, phn o:H(ejw)= HR(ejw) +jHI(ejw)hoc theo bin -pha:|H (ejw)|: p ng bin arg[H (ejw)]: p ng pha

*Nhn xt H(ejw) l hm lin tc theo w v tun hon theo wvi chu k 2p. Nu h(n) l thc, p ng bin i xngtrong khong 0 w 2p. Nu p ng xung l thc, ch cn xt khong tn s 0 w p.

*1.7. Php bin i Fourier ca tn hiu ri rc(1) c th c xem l biu din chui Fourier ca H(ejw)Cc h s ca chui l h(n)(1)(2)(1), (2) l cp bin i Fourier ca h(n)(1) l cng thc bin i Fourier thun (phn tch)(2) l cng thc bin i Fourier ngc (tng hp)

*PulseTone

*V d Xt mch lc thng thp l tngHy xc nh p ng xung h(n)

*Trng hp wC =p/2, fc = 1/4h(n)1

*Cc cng thc (1),(2) ng cho bt kdy no c th ly tng theo (1).Vy vi tn hiu x(n) bt k ta c:Theo tn s f:X(f) l hm phc ca bin thc f, tun hontheo f vi chu k = 1. X(f) = X(f+1)

*Ph bin v ph pha|X(f)|: Ph bin , arg[X(f)]: Ph phap ng xungp ng tn stn hiuph

*1.8. Mt s tnh cht c bn ca php bin i Fourier Tnh tuyn tnh Tnh tun honX(ejw) tun hon chu k 2pX(f) tun hon chu k l 1 Bin i Fourier ca tn hiu tr

*1.8. Mt s tnh cht c bn ca php bin i Fouriert n-n0 = mNhn xtTn hiu tr c ph bin khng thay icn ph pha dch i 1 lng wn0

*1.8. Mt s tnh cht c bn ca php bin i Fourier Nu x(n) thc:p ng bin l hm chn theo w|X(ejw)|=|X(e-jw)|p ng pha l hm l theo warg[X(ejw)]=-arg[X(e-jw)]c = a.b -> |c| = |a|.|b|arg[c] = arg[a] + arg[b]d = a/b -> |d| = |a|/|b|, arg[d] = arg[a] arg[b]

*1.9. Phng trnh sai phn tuyn tnhh s hng (PT-SP-TT-HSH) H tng t c quan h vo-ra theo phng trnh vi phn H ri rc c quan h vo-ra theo PT-SP-TT-HSH

*1.9. Phng trnh sai phn tuyn tnhh s hng (PT-SP-TT-HSH) Dng tng qutak, bk: cc h s ca PT-SP Trng hp N = 0So snh vi cng thc tng qut:H c p ng xung hu hn (FIR), hay h khng truy hi

*1.9. Phng trnh sai phn tuyn tnhh s hng (PT-SP-TT-HSH) Trng hp N > 0H c p ng xung v hn (IIR), hay h truy hi

*1.10. p ng tn s ca h biu din bng PT-SP-TT-HSHLy bin i Fourier c 2 v:p ng tn s xc nh bi cc h s ca PT-SP

*

*Bi tp chng 1 (1/3)Gi s x(n) = 0 vi n < 2 v n > 4. Vi mi tn hiu sau y, hy xc nh gi tr n cho tn hiu tng ng bng 0.a) x(n3)b) x(n+4)c) x(n)d) x(n+2)e) x(n2)Xt h S c tn hiu vo x(n) v tn hiu ra y(n). H ny c c bng cch mc h S1 ni tip vi h S2 theo sau. Quan h vora i vi 2 h S1 v S2 l:S1 : y1(n) = 2x1(n) + 4x1(n1)S2 :y2(n) = x2(n2) + (1/2)x2(n3)vi x1(n), x2(n) k hiu tn hiu vo.a) Hy xc nh quan h vora cho h Sb) Quan h vo ra ca h S c thay i khng nu thay i th t S1 v S2 (tc l S2 ni tip vi h S1 theo sau).

*Bi tp chng 1(2/3)Tn hiu ri rc x(n) cho nh hnh v sau. Hy v cc tn hiu:a) x(n4)b) x(3n)c) x(2n)d) x(2n+1)e) x(n)u(3n)f) x(n-1)u(3-n)g) x(n2) (n2)h) (1/2)x(n)+(1/2)(-1)nx(n)i) x((n-1)2)

*Bi tp chng 1(3/3)Cho x(n) = (n) + 2(n1) (n3) v h(n) = 2(n+1) + 2(n1)Hy tnh v v kt qu ca cc tng chp sau:a) y1(n) = x(n) * h(n)b) y2(n) = x(n+2) * h(n)H TT-BB c PT-SP: y(n)=(1/2)[x(n)-x(n-1)]Xc nh p ng xung ca hXc nh p ng tn s v v dng p ng bin

*Gii bi tp chng 1 (1/8)1. a) n-3 < -2 v n-3 >4. Vy n 7

*Gii bi tp chng 1 (2/8)a) x(n4) do x(n) tr (dch phi) 4 mub) x(3n): ly i xng x(n) qua n=0 c x(-n), sau dch x(-n) sang phi 3 mu c x(3-n)

*Gii bi tp chng 1 (3/8)d) x(2n+1) l x(n) ly ti cc thi im 2n+1 (ch khngphi do x(2n) dch tri 1 mu)e) x(n)u(3n): u(3-n) = 1 nu 3-n 0 tc l n 3u(3-n) = 0 nu 3-n 3Vy x(n)u(3n) = x(n) nu n 3x(n)u(3n) = 0 nu n > 3

*Gii bi tp chng 1 (4/8)3.f) x(n-1)u(3-n) l tch ca 2 tn hiu x(n-1) v u(3-n)g) x(n2) (n2) l tch ca 2 tn hiu x(n2) v (n2) h) (1/2)x(n)+(1/2)(-1)nx(n) = y(n)Nu n chn hoc n = 0:(-1)n = 1 nn y(n) = x(n)Nu n l :(-1)n = -1 nn y(n) = 0i) x((n-1)2) l x(n) ly ti cc thi im (n-1)2x(n-n0) do x(n) dch phi n0 mu (tr)x(n+n0) do x(n) dch tri n0 mu

*Gii bi tp chng 1 (5/8)4.x(n) = (n) + 2(n1) (n3) h(n) = 2(n+1) + 2(n1)a)y(n)=h(-1)x(n+1)+h(1)x(n-1)=2x(n+1)+2x(n-1)2x(n+1) = 2(n+1) + 4(n) 2(n2)2x(n-1) = 2(n-1) + 4(n2) 2(n4)y(n) = 2(n+1) + 4(n)+ 2(n-1) + 2(n2) 2(n4)

*Gii bi tp chng 1 (6/8)4.b)y(n)=h(-1)x(n+3)+h(1)x(n+1)=2x(n+3)+2x(n+1)y(n) = 2(n+3) + 4(n+2)+ 2(n+1) +2(n) 2(n2)

*Gii bi tp chng 1 (7/8)H TT-BB c PT-SP: y(n)=(1/2)[x(n)-x(n-1)]

Xc nh p ng xung ca hh(n)=y(n) khi x(n) = d(n) vy h(n)=(1/2)[d(n)-d(n-1)] Xc nh p ng tn s ca h

*Gii bi tp chng 1 (8/8)V dng p ng bin

*Chng 2PHP BIN I Z

*2.1. nh nghaBin i z ca tn hiu ri rc x(n) c nh ngha nh sau:

X(z) l hm phc ca bin phc z. nh ngha nh trnl bin i z 2 pha. Bin i z 1 pha nh sau: Xt quan h gia bin i z v bin i Fourier. Biu din bin phc z trong to ccz = rej

*2.1. nh nghaTrng hp c bit nu r = 1 hay |z|=1 biu thc trntr thnh bin i FourierBin i z tr thnh bin i Fourier khi bin ca bin z bng 1, tc l trn ng trn c bn knh bng 1 trongmt phng z. ng trn ny c gi l ng trn n v.

*2.1. nh ngha

*iu kin tn ti bin i z Min gi tr ca z chui ly tha trong nhngha bin i z hi t gi l min hi t. p dng tiu chun C-si xc nh min hi t Chui c dngs hi t nutha mn iu kin p dng tiu chun C-si cho X2(z)

*V d 1. Cho tn hiu x(n)=u(n). Hy xc nh bin i z v min hi t.vi |z|>1Rx-=1Rx+=V d 2. Cho tn hiu x(n)=anu(n). Hy xc nh bin i z v min hi t.vi |z|>|a|Rx-=|a|Rx+=im khng: z = 0im cc: z = aMin hi t khng cha im cc

*Bin i z thunBin i z ngc

*2.2. Php bin i z ngcp dng nh l C-siG: ng cong khp kn bao gc ta trn mt phng zNhn (1) vi v ly tch phn:(1)

*2.3. Mt s tnh cht ca bin i z Tnh tuyn tnhMin hi t ca X(z) t nht s l giao ca 2 min hi t ca X1(z) v X2(z)Rx- = max[Rx1-,Rx2-]Rx+ = min[Rx1+,Rx2+]

*2.3. Mt s tnh cht ca bin i z Bin i z ca tn hiu tri bin m=n-n0

*

*2.3. Mt s tnh cht ca bin i z Bin i z ca tn hiu tr

*2.3. Mt s tnh cht ca bin i z Vi phn ca bin i zNhn 2 v vi - z Bin i z ca tng chpy(n)=x(n)*h(n)Y(z)=X(z).H(z)

*2.4. Mt s phng php tnh bin i z ngc Khai trin thnh cc phn thc hu t n ginV dChovi |z|>2. Tm x(n) ?Mu s c 2 nghim theo z-1: z-1=1 v z-1=1/2

*2.4. Mt s phng php tnh bin i z ngc Khai trin thnh cc phn thc hu t n ginBit rngVy x(n)=2.2nu(n)-u(n)=u(n)[2n+1-1]

*2.4. Mt s phng php tnh bin i z ngc Khai trin theo php chiaX(z) c dng l t s ca 2 a thc theo z. Tinhnh php chia a thc c tng mu ca x(n)V d

*2.4. Mt s phng php tnh bin i z ngc Khai trin theo php chiaz-1 1-1,414z-1+z-2

z-1 -1,414z-2+z-3 z-1+ 1,414z-2+ z-3- z-5-1,414 z-6 1,414z-2-z-3 1,414z-2-2z-3+ 1,414z-4z-3 - 1,414z-4z-3 - 1,414z-4 + z-5- z-5 - z-5 + 1,414z-6 z-7 - 1,414z-6 + z-7x(0)=0. x(1)=1. x(2)=1,414. x(3)=1. x(4)=0. x(5)=-1n

*Mt s cp bin i z thng dng (1/2)

Tn hiuBin i zMin hi td(n)1Ton mf zu(n)|z|>1-u(-n-1)|z|0, tr nu m < 0anu(n)|z|>|a|-anu(-n-1)|z|

*Mt s cp bin i z thng dng (2/2)

Tn hiuBin i zMin hi tnanu(n)|z|>|a|-nanu(-n-1)|z|1sin(Wn)u(n)|z|>1

*2.5. ng dng bin i z gii PT-SP Gii PT-SP: Bit PT-SP, bit tn hiu vo, tnh tn hiu raV dCho PT-SP y(n) = x(n) + ay(n-1)Bit:iu kin u y(-1) = KTn hiu vo x(n) = ejwnu(n)Hy xc nh tn hiu raLy bin i z 1 pha PT-SP:p dng cng thc tnh bin i z 1 pha ca tn hiu tr

*2.5. ng dng bin i z gii PT-SPY(z)=X(z)+az-1Y(z)+ay(-1)x(n) = ejwnu(n)Bin i z ngcp ng vi iu kin up ng qu p ng i vi tn hiu vo

*2.6. Hm truyn t ca h TT-BBy(n)=h(n)*x(n) Y(z) =H(z).X(z)H(z): Hm truyn ta) H(z) ca h nhn quH nhn qu nn h(n) = 0 vi n < 0H(z) hi t viMin hi t khng cha im cc, vy:Mi im cc ca h TT-BB nhn qu u nm trongng trn c bn knh

*2.6. Hm truyn t ca h TT-BBb) H(z) ca h n nhH n nh th p ng xung tha mn(1)Hm truyn t c xc nh theo:Nu (1) tha mn th H(z) hi t ngay c khi |z|=1Min hi t ca H(z) cha ng trn n vth h s n nh

*2.6. Hm truyn t ca h TT-BBc) H(z) ca h nhn qu v n nhTon b im cc ca h nhn qu v n nh phi nm bn trong ng trn n v.d) H(z) ca h c trng bi PT-SP-TT-HSHLy bin i z c 2 v ca PT-SP

*2.6. Hm truyn t ca h TT-BBBiu din H(z) qua cc im khng zr v cc im cc pk:

*Bi tp chng 2 (1/2)Cho tn hiu Hy tnh bin i z ca tn hiu ny bng cch dng:nh ngha bin i zTn hiu u(n) v tr ca u(n)Tnh bin i z ngc ca vi |z|>1/2ng dng bin i z 1 pha gii PT-SP:y(n)-(1/2) y(n-1)=x(n)-(1/2) x(n-1)Bit x(n) = d(n), y(-1)=0.

*Bi tp chng 2 (2/2)H TT-BB c PT-SP:y(n)=y(n-1)+y(n-2)+x(n-1) a) Xc nh hm truyn t, im khng, im ccb) Nhn xt tnh nhn qu, n nhc) Xc nh p ng xung sao cho h nhn qu

*Gii bi tp chng 2 (1/5)1.Tn hiu x(n):

*Gii bi tp chng 2 (2/5)2.3.Bin i z 1 pha c 2 v ca PT-SP:y(-1) = 0, x(-1)=0, X(z) = 1Y(z) = 1y(n)=d(n)

*Gii bi tp chng 2 (3/5)4.y(n)=y(n-1)+y(n-2)+x(n-1)a) Bin i z c 2 v:Y(z)=z-1Y(z)+z-2Y(z)+z-1X(z)H c 1 im khng ti z=0 v 2 im cc ti z=1,62;z=-0,62Nghim mu s:

*Gii bi tp chng 2 (4/5)4.b)

*Gii bi tp chng 2 (5/5)4.c)

*S = a0 + a1 + a2 + a3 + + aN-1ai = ai-1.qS = a0.(1-qN)/(1-q)

S = a0 + a1 + a2 + a3 + + aN-1+ai = ai-1.qS = a0./(1-q)

*Chng 3B LC S

*3.1. Khi nim Trong nhiu ng dng khc nhau, ta thng phi thay i bin ca cc thnh phn tn s khc nhau ca tn hiu hoc loi b i mt s thnh phn tn s no . Qu trnh x l nh vy i vi tn hiu c gi l lc. C th dng b lc tng t lc tn hiu s c khng ? B lc s: l b lc dng lc tn hiu s

*3.1. Khi nimp ng bin ca b lc thng thp Xt h TT-BB c PT-SPp ng xung ca h:p ng tn s ca h:

*3.2. B lc FIRN=0 M=1y(n)=h(0)x(n)+h(1)x(n-1) B lc FIR v IIRN=0: FIRN>0: IIRS khi

*3.2. B lc FIRconst h0 = 0.5; h1 = 0.5;var xn, xnt1, yn: real;begin xnt1 := 0; repeat (* Nhp tn hiu vo t bn phm *) write(Nhp tn hiu vo xn = ); readln(xn); (* Tnh tn hiu ra *) yn:= h0 * xn + h1 * xnt1; (* Tr tn hiu *) xnt1 := xn; until Ketthuc;end.

*3.2. B lc FIR Trng hp tng quth(0)

*3.3. B lc IIR H bc nhta0y(n)+a1y(n-1)=b0x(n)Gi thit a0 = 1y(n)=-a1y(n-1)+b0x(n)

*3.3. B lc IIR H bc haia0y(n)+a1y(n-1)=b0x(n)+b1x(n-1)Gi thit a0 = 1y(n)=-a1y(n-1)+b0x(n)+ b1x(n-1) =-a1y(n-1) + w(n)w(n)=b0x(n)+b1x(n-1)

*3.3. B lc IIR Tng qut (a0 = 1)

*3.3. B lc IIRb0Dng trc tip 1

*3.3. B lc IIR

*3.3. B lc IIR

*3.3. B lc IIRDngtrctip 2(chuntc)M>N

*3.4. Mc ni tip v song song cc h H(z) ca h phc tp thng c phn tch thnh tnghoc tch H(z) ca cc h n gin, tng ng vi vic mc song song hoc ni tip cc h n gin Mc ni tipC: Hng s

*3.4. Mc ni tip v song song cc h Mc song songD: Hng s

*3.5.Kho st h bc 1a0 = b0 = 1, a1 = -ay(n) a y(n-1) = x(n) Hm truyn t H(z) c 1 im khng ti z = 0 v 1 im cc ti z = a n nh: H n nh nu |a| 1 Nhn qu: h(n) = anu(n) nu |z| > |a| Phn nhn qu:h(n) = -anu(-n-1) nu |z| < |a| H nhn qu v n nh nu |a| < 1 p ng tn s H(ejw) = H(z)|z = ejw

*V d: p ng bin v phaa=0,5a=-0,5

*3.6.Kho st h bc 2a0 = b0 = 1y(n) + a1 y(n-1)+a2y(n-2) = x(n) Hm truyn t 1 im khng bc 2 ti z = 0 2 im cc

* n nh v nhn qu: |p1| < 1, |p2| < 1Ranh gii im cc thc v phc: Xt im cc thc:(*)(**) cho kt qu tng t

* Xt im cc phc:

*H n nh v nhn qu nu a1 v a2thuc min tam gic.

*V d: p ng bin v pha1)2)1) a1 = 1, a2 = 0,52) a1 = -1, a2 = 0,5

*V d:X l nh. nh qua b lc thng thp (lm trung bnh)

*V d:nh qua b lc thng cao (o hm)

*Bi tp chng 3 (1/2)1. H TT-BB c quan h vo ra:Xc nh p ng tn sXc nh v v dng p ng bin . Nhnxt tnh cht lc ca h.2. Hm truyn t ca b lc s c dng: H(z) = 1 + 2z-1 + 4z-3Xc nh PT-SP biu din quan h vo-raV s khi thc hin b lc

*

*Bi tp chng 3 (2/2)3. H TT-BB c hm truyn t:H(z)=(1+az-1)/(1+bz-1+cz-2) vi a,b,c l hng s.Xc nh quan h vo-ra ca hV s dng chun tc thc hin h.

*Gii bi tp chng 3 (1)1. a) p ng xung:p ng tn s:b) p ng bin :|H(ejw)|=(1/3)|1+2cosw|

*Gii bi tp chng 3 (2)2. a) H(z) = 1 + 2z-1 + 4z-3 = Y(z)/X(z)Y(z) = X(z) + 2z-1X(z) + 4z-3 X(z)y(n) = x(n) + 2x(n-1) + 4x(n-3)b)

*Chng 4PHP BIN I FOURIER RI RC

*4.1. Chui Fourier ri rc ca tn hiu ri rc tun hon(DFS: Discrete Fourier Serie)Xt tn hiu xp(n) tun hon vi chu k N:

xp(n) = xp(n+kN), k nguyn

Tn hiu ny khng biu din c bng bin i z nhng c th biu din bng chui Fourier thng qua hm e m phc vi cc tn s l bi ca tn s c bn 2p/N.y l tn hiu tun hon theo k vi chu k N. k = 0,1,2,,N-1

*4.1. Chui Fourier ri rc ca tn hiu ri rc tun honChui Fourier biu din tn hiu ri rc tun hon:Xc nh cc h s Xp(k) theo xp(n) da vo tnh cht trc chun:m: s nguynNhn 2 v xp(n) vi v ly tng t n=0 n N-1(1)

*4.1. Chui Fourier ri rc ca tn hiu ri rc tun honThay i th t ly tngk r = mN [] = 1, k r mN [] = 0k=r+mN v k < N m=0 v k = rS dng tnh cht trc chun ta c:Hoc l:Nhn xt Xp(k) tun hon theo k vi chu k N Cc cng thc (1), (2) l biu din chui Fourier ca tn hiu ri rc tun hon. (1): Tng hp. (2): Phn tch(2)

*4.1. Chui Fourier ri rc ca tn hiu ri rc tun hon Quan h vi bin i zXt 1 chu k ca xp(n):Mt khcvy

*V d: Hy tnh cc h s chui Fourier ca dy tn hiu tun hon sau

*4.2. Bin i Fourier ri rc ca tn hiu c di hu hn(DFT: Discrete Fourier Transform)Ta xt cch biu din mt tn hiu ri rc tun hon bng chui Fourier. Bng cch din gii thch hp ta cng c th dng cch biu din nh vy cho cc tn hiu c di hu hn.C th coi tn hiu c di hu hn N l tn hiu tun hon c chu k N trong mt chu k chnh l tn hiu c di hu hn

*4.2. Bin i Fourier ri rc ca tn hiu c di hu hn Cp cng thc DFTBin i thun (phn tch)Bin i ngc (tng hp)

*4.3. Bin i nhanh Fourier(FFT: Fast Fourier Transform) Tnh trc tip DFT cn N2 php nhn s phc v N(N-1) php cng s phc Thut gii FFT: phn tch DFT ca dy N s ln lt thnh DFT ca cc dy nh hn iu kin p dng thut gii: N = 2m. S lng php ton gim xung cn Nlog2N

*4.4. Cc hm ca s Ly ra on tn hiu c di N phn tch Tng ng nhn tn hiu vi hm w(n)w(n) = 1 trong on tn hiu c lyw(n) = 0 trong on tn hiu khng c lyx(n) = x(n).w(n) Mc nhin dng ca s ch nht !

*4.4. Cc hm ca sX(f) = X(f)*W(f) Tn hiu c phn tch c di hu hn gy ra X(f) X(f) c sai s khi tnh bin i Fourier gim sai s c th tng N Phng php hay dng l chn W(f) hay chn w(n) Ca s ch nht gy sai s ln nn thng dng cc ca s khc nh Hamming, Hanning, Kaiser, Blackman

*4.4. Cc hm ca s Hm ca s Hamming, Hanning:N=256

*1. Gi thit tn hiu x(n) l tng ca 2 tn hiu x1(n) v x2(n). x1(n) l tn hiu cosin c tn s gc l 0,1rad/s, x2(n) cng l tn hiu cosin c tn s gc l 0,4rad/s. Ngi ta dng b lc thng cao FIR c di p ng xung bng 3 vi gi thit h(0) = h(2) = v h(1) = trit tiu tn hiu x1(n) v cho qua hon ton tn hiu x2(n). Hy xc nh cc h s , v v s khi thc hin b lc FIR ny. 2. Hm truyn t ca h TTBB nhn qu c dng nh sau:

vi a l s thc.a. Xc nh gi tr ca a sao cho H(z) ng vi mt h n nhb. Ly 1 gi tr c bit ca a trong s cc gi tr ny, biu din cc im cc, im khng v min hi t. c. nh gi |H(f)|

*Bi tp ln (1/2)1.B lc s FIR c PT-SPHy lp trnh bng Pascal xc nh p ngxung ca b lc ny.

Khi to tn hiu tr = 0 (xnt1, xnt2, xnt3, xnt4)Gn xn = 1 (xung n v)B vng lp:- Tnh tn hiu ra yn (=hn) theo PT-SP- Tr tn hiu vo xn:xnt4 := xnt3;xnt3 := xnt2;xnt2 := xnt1;xnt1 := xn;( sau buc lp u tin phi gn xn := 0)KT vng lp

y(n)=x(n) + 2x(n-1)-3x(n-3)+5x(n-4)

*Bi tp ln (2/2 )2.B lc s IIR c cc h s nh sau:Hy lp trnh bng Pascal xc nh 100 mu u tin ca p ng xung ca b lc ny.

a0 1.0000 b0 0.0252a1-9.7023b1 -0.0615 a28.8979b20.0684a3-12.7653b3-0.0800a413.1148b40.0976a5-4.0608b5-0.0800a65.1226b60.0684a7 -1.7620b7-0.0615a80.3314b80.0252

*Cho tn hiu vo = xung n v, tnh tn hiu ra theo PT-SPBEGIN- Khi to cc tn hiu tr = 0 (xnt1,,xnt8,ynt1,,ynt8)- Gn xung n v xn = 1B vng lp- Tinh wn theo cng thc (1)

- Tnh y[n] theo cng thc (2)

- Tr tn hiu xn v yn(* Sau bc lp u tin phi gn xn = 0) KT vng lpEND

*Kt qu c dng

*BI TPH TT-BB c tn hiu vo x(n) = u(n) u(n-2), h(n) = u(n) u(n-2). Hy xc nh v v tn hiu ra y(n).

Cho h TT-BB c quan h vo ra: y(n) = x(n) + 3x(n-1) 2x(n-3) + 5x(n-4)a) Xc nh p ng xung ca hb) H c n nh khng ? Ti sao ?c) V s khi thc hin h.

3) Cho h TT-BB c PT-SP: y(n) = x(n) x(n -1) 0,5 y(n -1)a) Xc nh hm truyn tb) V im cc im khng ca h, xt tnh n nh v nhn quc) Xc nh p ng xung h nhn qu.