ET 2060 Biu din tn hiu v h thng LTI trn min tn sTS. ng Quang Hiuhttp://ss.edabk.org Trng i hc Bch Khoa H Ni Vin in t - Vin thng

2011-2012

Vai tr ca bin i Fourier

Quan trng trong ton hc, vt l v cc ngnh k thut c bit l x l tn hiu. Khi nim chui Fourier do Joseph Fourier gii thiu vo nm 1807, v sau c pht trin bi nhiu nh khoa hc ni ting khc. Phn loi:

Chui Fourier (FS) Chui Fourier ri rc theo thi gian (DTFS) Bin i Fourier (FT) Bin i Fourier ri rc theo thi gian (DTFT)

Bin i Fourier ri rc c th c thc hin nhanh (cc thut ton FFT).

Hm ring ca h thng LTI (1)Xt h thng LTI vi u vo l dy ly tha x[n] = e jn y [n] = trong , H(e ) =j k= k=

h[k]e j(nk) = e jn H(e j )

h[k]e jk

e jn - hm ring (eigenfunction) ca h thng LTI H(e j ) - tr ring (eigenvalue)

Hm ring ca h thng (2)e jn H(e j )e jn h[n]

Nu biu din u vo bt k theo cc hm ring x[n] =k

ak e jk n

th y [n] =k

ak H(e jk )e jk n

Khng phi thc hin php chp!!! H(e j ) gi l p ng tn s ca h thng

Outline

Chui Fourier cho tn hiu lin tc

Chui Fourier cho tn hiu ri rc

Chui Fourier (FS)Tn hiu x(t) tun hon vi chu k c bn T = biu din bi chui Fourier (FS) nh sau: x(t) = trong 1 ck = T0 k= 2 0

c th c

ck e jk0 t

T

x(t)e jk0 t dt

l cc h s FS ca x(t) (ck , ck thnh phn hi bc |k|). V d: Hy tm khai trin chui Fourier cho cc tn hiu sau vi chu k c bn 0 = 2/T . (a) x(t) = cos(0 t) (b) Xt trong mt chu k, x(t) = 1, |t| T0 0, T0 < |t| < T /2

iu kin tn ti FS

Cc iu kin Dirichlet: 1. x(t) b chn 2. x(t) c hu hn cc cc i v cc tiu trong mt chu k 3. x(t) c hu hn cc im gin on trong mt chu k Tn hiu c nng lng hu hn trn mt chu k: |x(t)|2 dt <

T

Dng biu din khc ca FS

a0 x(t) = + 2

k=1

ak cos(k0 t) + bk sin(k0 t)

Quan h gia ak , bk v ck ? Nu coi ck l mt dy ri rc: x(t) = k=

X [k]e jk0 t

Tnh cht tuyn tnh

Nu FS x(t) ak FS y (t) bk th FS x(t) + y (t) ak + bk

Tnh cht dch

Dch theo thi gian: FS x(t t0 ) e j0 t0 ck Dch tn s: FS e jk0 0 t x(t) ckk0 1, 0 < t T0 0, T0 < t < T

V d: Tm khai trin cho tn hiu tun hon vi chu k T x(t) =

o trc thi gian

FS x(t) ck

Nu x(t) chn? Nu x(t) l?

Tnh cht i xng

FS x (t) ck

Nu x(t) thc? Nu x(t) o?

Quan h Parseval

1 T

T

|x(t)| dt =

2

k=

|ck |2

ngha: FS bo ton cng sut ca tn hiu.

Outline

Chui Fourier cho tn hiu lin tc

Chui Fourier cho tn hiu ri rc

Dy tun hon v chui FourierDy x (n) (hoc x (n)N ) tun hon vi chu k N: x (n) = x (n + rN), Khai trin chui Fourier cho dy x (n): x (n) = k

n, r Z

ck e j

2 kn N

c im ca cc thnh phn tn s e j ej2 kn N

2 kn N

,

k Z?

= ej

2 (k+rN)n N

,

r Z ck+rNr

N1

x (n) = k=0

X (k)e j

2 kn N

,

X (k) =

Tnh X (k)?(i) Nhn c hai v vi e jN12 mn N

, tnh tng vi n = 0, (N 1)N1 N1

x (n)e n=0

j 2 mn N

=n=0 k=0

X (k)e j

2 (km)n N

(ii) i th t ly tng v phiN1

x (n)e n=0

j 2 mn N

N1

N1

=k=0

X (k)n=0

ej

2 (km)n N

(iii) Tnh trc giao:N1

ejn=0

2 (km)n N

=

N, khi 0, khi2 mn N

k m = rN k m = rN

N1

=n=0

x (n)e j

= N X (m)

Khi nim chui Fourier ri rc1 X (k) = NN1

x (n)e j n=0

2 kn N

X (k) tun hon vi chu k N. K hiu: X (k) hoc X (k)N . Cp DTFS (chui Fourier ri rc theo thi gian) cho dy tun hon:N1

1 X (k) = DTFS{(n)} = x N x (n) = DTFS1 {X (k)} = 2

x (n)e j n=0 N1 k=0

2 kn N

2 X (k)e j N kn

WN = e j N cng thc? Khi nim bin v pha: |X (k)|, arg{X (k)}.

V d(1) Tm khai trin Fourier ca dy x (n) = r =

(n rN) =

1, n = rN, 0, n = rN

r Z

(2) Cho x (n) l dy tun hon vi chu k N x (n) = Hy tm X (k). (3) Dy x (n) tun hon vi chu k N cng c th coi l mt dy tun hon c chu k 2N. Nu X1 (k)N = DTFS{ (n)N } v x 2 (k)2N = DTFS{ (n)2N }. Hy tnh X2 (k)2N theo X1 (k)N . X x 1, N n N + M, 0, n cn li n Z, M < N

Cc tnh cht ca chui Fourier ri rc(1) Tuyn tnh: DTFS{a1 x1 (n)N + a2 x2 (n)N } = a1 X1 (k)N + a2 X2 (k)N (2) Dch thi gian DTFS{ (n n0 )} = e j2kn0 X (k) x = W kn0 X (k)N

(3) Dch tn sk DTFS{e j2k0 n x (n)} = DTFS{WN 0 n x (n)} = X (k k0 )

Tnh cht i ngu

Nu th

DTFS{(n)} = X (k) x DTFS{X (n)} = 1 x (k) N

V d: Cho X (k) = Hy tm x (n)?

r =

(k rN)

Cc tnh cht i xng

(a) DTFS{ (n)} = X (k) x (b) DTFS{(n)} = X (k) x1 (c) DTFS{Re[ (n)]} = 2 [X (k) + X (k)] x (d) DTFS{ 1 [ (n) + x (n)]} = Re[X (k)] x

(e) Khi x (n) R

2

X (k) = X (k) Re[X (k)] = Re[X (k)] Im[X (k)] = Im[X (k)] |X (k)| = |X (k)| arg{X (k)} = arg{X (k)}

Chp tun honCho DTFS{1 (n)} = X1 (k) x DTFS{2 (n)} = X2 (k) x Nu X3 (k) = X1 (k)X2 (k) th x3 (n) =? Khi nim chp tun hon:N1

x3 (n)N = x1 (n)()N x2 (n) = Chng minh?

m=0

x1 (m)2 (n m) x

Cch tnh chp tun hon

Tm x3 (n0 ) vi n0 [0, (N 1)] (1) Ly i xng qua trc tung x2 (m) x2 (m) (2) Dch theo trc honh n0 mu (3) Nhn hai dy vn0 (m) = x1 (m)2 (n0 m) trong on x [0, (N 1)]

(4) Tnh tng cc phn t ca dy vn0 (m) trong on [0, (N 1)] x3 (n0 ) (5) Dy tun hon chu k N: x3 (n0 ) = x3 (n0 + rN), r Z.

Cc tnh cht khc

Tch ca hai dy? Tng quan tun hon ca hai dy?

T c!

Bi tp1. Cho tn hiu lin tc (tun hon) xc (t) c khai trin Fourier nh sau:9

xc (t) =k=9

ak e j2kt/10

3

trong cc h s ak = 0, |k| > 9. Tn hiu ny c ly 1 mu vi chu k T = 6 103 [s] to thnh dy x(n) = xc (nT ).(a) Dy x(n) c tun hon khng, nu c th chu k bao nhiu? (b) Hy tnh X (k) theo cc h s ak .

2. Dng hm freqz v freqs trong Matlab v p ng tn s ca h thng c m t bi phng trnh sai phn / vi phn tuyn tnh h s hng. 3. Vit chng trnh Matlab tnh DTFS v v ph bin , ph pha ca tn hiu. Ti u code.