Chng 3:BIU DIN TN HIU V H THNG TRONG MIN TN S LIN TC3.1 BIN I FOURIER 3.2 CC TNH CHT BIN I FOURIER3.3 QUAN H GIA BIN I Z & F3.4 BIU DIN H THNG TRONG MIN TN S3.5 LY MU & KHI PHC TN HIU

3.1 BIN I FOURIER3.1.1 NH NGHA BIN I FOURIER:K hiu:

x(n) X() hay X() = F{x(n)} X() x(n) hay x(n) = F-1{X()} Trong : - tn s ca tn hiu ri rc, = Ts - tn s ca tn hiu lin tc Ts - chu k ly mu Bin i Fourirer ca x(n):

X() biu din di dng modun & argument:Nhn thy X() tun hon vi chu k 2, tht vy:Trong :- ph bin ca x(n)- ph pha ca x(n)p dng kt qu:Biu thc bin i F ngc:

V d 3.1.1: Tm bin i F ca cc dy:Gii:

3.1.2 IU KIN TN TI BIN I FOURIERVy, X() hi t th iu kin cn l:Cc tn hiu tha iu kin hi t l tn hiu nng lng, thy vy:Nu:

V d 3.1.2: Xt s tn ti bin i F ca cc dy:Gii:X2() khng tn tiX3() khng tn ti

3.2 CC TNH CHT BIN I FOURIERa) Tuyn tnhNu:Th:b) Dch theo thi gianNu:Th:

c) Lin hip phcV d 3.2.1: Tm bin i F ca dy:Gii:Nu:Th:p dng tnh cht dch theo thi gian:

d) o bin sGii: Nu:Th:V d 3.2.2: Tm bin i F ca dy:Theo v d 6.1.1, c kt qu:suy ra:

e) Vi phn trong min tn sGii: Theo v d 6.1.1:Nu:V d 6.2.3: Tm bin i F ca:Suy ra:Th:

f) Dch theo tn sGii: Theo v d 6.1.1:Nu:V d 3.2.4: Tm bin i F ca:Th:

g) Tch 2 dyTh:Nu:

g) Tng chp 2 dyTh:Nu:V d 3.2.4: Tm y(n)=x(n)*h(n), bit: x(n)=h(n)=(n+2)+(n-2)Gii:Theo v d 6.2.1, c kt qu:

- gi l ph mt nng lngg) Quan h ParsevalTh:Nu:(*)Biu thc (*) cn gi l quan h ParsevalNhn xt:Nu:Theo quan h Parseval, ta c: Vi:

TNG KT CC TNH CHT BIN I F

3.3 QUAN H GIA BIN I FOURIER & ZHay bin i Fourier chnh l bin i Z c ly trn vng trn n v theo bin s Nu ROC[X(z)] c cha /z/=1X()=X(z) vi z=ej Nu ROC[X(z)] khng cha /z/=1X() khng hi t

V d 3.3.1: Tm bin i Z & F ca cc dy:Gii:Do ROC[X1(z)] c cha /z/=1, nn:Do ROC[X2(z)] khng cha /z/=1, nn X2() khng tn ti

3.4 BIU DIN H THNG TTBB RI RC TRONG MIN TN S3.4.1 nh ngha p ng tn sMin n:Min :Y()=X()H()Nu H() biu din dng mdun v pha:

V d: 3.4.1: Tm H(), v p ng bin & pha, bit:Gii:Bin i Fourier ca h(n):h(n)=rect3(n)Vi

3.4.2 p ng tn s ca cc h thng ghp nia. Ghp ni tip Min : Min n:

b. Ghp song song Min : Min n:

3.4.3 p ng ra h thng vi tn hiu vo hm m phcV d: 3.4.3: Tm y(n) bit:Xt tn hiu vo c dng m phc: x(n)=Aejn

3.4.4 p ng ra h thng vi tn hiu vo hm cos,sinXt tn hiu vo c dng hm cos:Biu din p ng tn s di dng mun & pha:

Tng t vi tn hiu vo c dng hm sin:Ta cng c kt qu:

3.5 LY MU & KHI PHC TN HiU3.5.1 Khi nim ly mu tn hiuQu trnh ly mu tn hiu

Tc ly mu cng ln -> khi phc tn hiu cng chnh xc

3.5.2 Quan h gia tn s tn hiu ri rc v tng tTrong : - tn s ca tn hiu ri rc - tn s ca tn hiu tng t Ts - chu k ly mu

3.5.3 Quan h gia ph tn hiu ri rc v ph tn hiu tng tV d: 3.5.1: Hy v ph bin tn hiu ri rc, bit ph bin tn hiu tng t cho nh hnh v, vi cc tc ly mu: a)Fs>2FM b) Fs=2FM c) Fs

3.5.4 nh l ly muTn hiu tng t xa(t) c di ph hu hn (-FM ,FM) ch c th khi phc 1 cch chnh xc t cc mu xa(nTs) nu tc ly mu tha Fs 2FMV d 3.5.2: Xc nh tc Nyquist ca tn hiu tng t: Fs =2FM=FN: Tc (tn s) NyquistGii:Tn hiu c cc tn s: F1=1 kHz, F2=3 kHz, F3=6 kHzFM=max{F1, F2, F3}=6 kHz FN =2FM = 12 kHz

3.5.5 Khi phc li tn hiu tng t khi phc li tn hiu tng t xa(t) th ph ca tn hiu c khi phc phi ging vi ph ban u ca xa(t). V ph ca tn hiu ly mu l s lp li v hn ca ph tn hiu tng t, nn cn phi gii hn li bng cch ngi ta cho cc mu xa(nTs) i qua mch lc thng thp l tng trong iu kin tha nh l ly mu c p ng tn s:

Cng thc ni suy, cho php khi phc xa(t) t xa(nTs)