Lecture 7 Transformations in frequency domain. 2. Fourier transformation theory in 1D

Transcription

1 Lecture 7 rasformatios i frequecy domai. Basic steps i frequecy domai trasformatio 2. Fourier trasformatio theory i D

2 Basic steps for filterig i the frequecy domai he most commo image trasform takes spatial data ad trasforms it ito frequecy data akes spatial data ad trasforms it ito frequecy data he trasformatio is doe by Fourier trasformatio 2

3 Complex umbers ad expressio 2 2+ iθ ( iθ) i θ ( ) θ ( ) θ e = = = + i = cosθ + isiθ!! (2 )! (2+ )! = 0 = 0 = 0 = 0 i = a+ ib = R(cosθ + isi θ) = Re i 2 a R = a + b 2,taθ = b θ R θ a b

4 Fourier series he Fourier trasform is a method of expressig a periodic fuctio with period 2 i terms of the sum of its projectios oto a set of basis fuctios π x π x {si( ),cos( ), = 0, ±, ± 2,...} Fourier series: f(x) is periodic [-, ] π x π x f( x) = a0 + acos( ) + bsi( ), 2 = = π x a0 = f( x) dx, a = f( x) cos( ) dx π x b = f( x)si( ) dx, =,2,3

5 Example of Fourier decompositio

6 Example by Maple =

7 Maple commads > f := piecewise(x < -, x+2, x <, x, x < 3, x-2); > plot(f, x = -3..3, discot=true); > a := It(x*cos(*Pi*x), x = -..); > a := it(x*cos(*pi*x), x = -..); > b := it(x*si(*pi*x), x = -..); > with(plots): > F := plot(f, x = -3..3, discot=true, color=black): > S := sum(b*si(*pi*x), =..): > S2 := sum(b*si(*pi*x), =..2): > S5 := sum(b*si(*pi*x), si( =..5): > S20 := sum(b*si(*pi*x), =..20): > F2 := plot({s,s2,s5,s20}, x = -3..3): > display({f,f2});

8 Fourier series i geeral form π π i x i x f( x) = ce, c = f( x) e dx, = 0, ±,... 2 = π m π i x i π x i m x Proof : f ( x) e dx = c e e dx = 2 c, = π i x c = f ( x) e dx, 0,,... 2 = ± π x π x Proof 2: f ( x ) = a 0 + a cos( ) + b si( ) 2 = = π x π x x i i π π x i i i = = π x x i π a i ib a + ib = a0 + a ( e + e ) + b ( e e ) = = π x i = a0 + e + e = = c e, a ib a + ib c0 = a 0, c =, > 0; c =, < m

9 Cotiue Whe > 0, a ib π x π x c = = f ( x )cos( ) dx i f ( x )si( ) dx πx πx = x)(cos( isi( dx f ( ( ) )) 2 = f( x) e 2 Whe < 0, π x i dx a ib x x c + π π ( ( )cos( ) ( )si( ) 2 2 ) = = f x dx + i f x dx π x πx πx i = f ( x)(cos( ) isi( )) dx= f( x) e dx 2 2

10 Fourier series ad Fourier trasformatio π π i x i x f( x) = ce, c = f( x) e dx, 0,,... 2 = ± = Cosider whe +, u =, Δ u = u+ u = 2 2 2πi x 2 2πiu x ( ) = 2 = ( ) = ( ) cu c f xe dx f xe dx f x e dx= F u 2πiux ( ) ( ) π i x 2πikx 2πiux f( x) = c e = (2 c ) e Δ k = (2 c ) e Δu = = 2 π iu x 2 π iux = cu ( ) e Δ u = Fue ( ) du = = 2πiux Defie F{ f ( x)} = F( u) = f( x) e dx the Fourier trasfomrario of f( x) - 2πiux Defie { Fu ( )} = Fue ( ) duiverse Fourier tra F sfomrario of F( u)

11 Fourier trasformatio Defie 2π iux f ( x)} = F( u) = f ( x) e dx F{ ( )} ( ) ( ) the Fourier trasfomrario of f ( x) 2π iux F u = F u e du - Defie F { ( )} ( ) the iverse Fourier trasfomrario of F( u )

12 Fourier rasform D Fourier: Every periodic fuctio f(x) ca be decomposed ito a set of si() ad cos() fuctios of differet frequecies, give by 2 iux ( ) = π ( ), ( ) = F{ ( )} F u f x e dx F u f x F(u) is called the Fourier trasformatio of f(x). F(u) =R(u)+iI(u) repesets magitudes cos ad si at frequecy u. So we say F(u) is i the Frequecy domai. Coversely, give F(u), we ca get f(x) back usig the iverse Fourier trasformatio F - 2πiux Fu Fue du { ( )} = ( ) - - F { Fu ( )} = F { F{ f( x)}} = f( x)

13 Fourier Spectrum Fourier decompositio: Fu ( ) = Ru ( ) + iiu ( ) Fourier spectrum: F ( u ) = 2 2 R ( u ) + I ( u ) Fourier phase: Decompositio: θ ( u) = ta ( I( u)/ R( u)) Fu ( ) = Fu ( ) exp( iθ )

14 Properties of Fourier trasformatio Liear F{ Af( x) + Bgx ( )} = AF{ f( x)} + BF{ gx ( )} 2πiux Af x + Bgx = Af x + Bgx e dx 2π iux 2πiux F { ( ) ( )} [ ( ) ( )] = Af ( x) e dx+ Bg( x) e dx = 2 2 ( ) iux iux A π π f xe dx+ B gxe ( ) dx = AF{ f( x)} + BF{ g( x)}

15 Fourier trasformatio ad Covolutio Covolutio heorem: assume F th { f ()} t = F( u), F{ h()} t = H( u) the F{ f () t h()} t = F( u) H( u) f( t) h( t) H( u) F( u) Proof f() t h() t H( u) F( u) f() t h() t = f( x) h( t x) dx i2πut f t ht = f xht xdxe dt i 2πut F { ( ) ( )} [ ( ) ( ) ] = f( x)[ h( t x) e dt] dx ( )[ i2πux i2πux ( ) ] ( ) ( ) ( ) ( ) = = = f x H u e dx H u f xe dx HuFu

16 Example f () t A, W /2 t W /2 = Sic(x)=si(x)/x 0, otherwise W /2 2πiut 2πiut F( u) = f ( t) e dt = Ae dt W /2 A 2 iut W /2 A iuw iuw π uw π π π si( ) = e e e AW i2π u = W /2 i2πu = πuw

17 Example 2 Impulse fuctio ad its Fourier trasformatio, t = 0 δ() t =, δ() t dt =, 0, i F f () t δ () t dt = f (0), f () t δ ( t t ) dt = f ( t ) 2πiut 2πiut { δ()} = ( ) = δ() = δ() t F u t e dt e t dt 2π iu 0 = e = 2πiut 2πiut F{ δ( t t )} = δ( t t ) e dt = e δ( t t ) dt 2π iut = e 0 = cos(2π π ut 0 i π ut0 com ( x) = δ ( x ) = ) si(2 ) F{ com ( x)} = F{ δ( x )} = (cos(2 πu ) isi(2 πu )) = =

18 Examples

19 Example: Discrete impulse fuctio Uit discrete impulse fuctio, t = 0 δ() t =, δ() t = 0, i 0 x= x= f ( x ) δ ( x ) = f (0), f ( x) δ( x x ) = f ( x ) x= Impulse trai fuctio 0 0 S Δ () t = δ ( t Δ ) x=

20 Fourier trasformatio of impulse trai S () t = δ ( t Δ) Δ x= S t c e c S t e dt e Δ Δ Δ 2π 2π j Δ /2 j t Δ Δ 0 Δ() =, = () Δ /2 Δ = = = SΔ () t = e Δ = 2π j Δ 2π j t Δ F { e } = δ ( u ) Δ 2π j Δ Su ( ) = F { SΔ ( t)} = F { e } Δ = 2π j Δ = F { e } = δ ( u ) Δ Δ Δ = =

21 Samplig ad F of Sampled Fuctio Sampled fuctio ~ f () t = f ( x ) S = () t Δ f () t δ ( t Δ ) x=

22 Samplig ad F of Sampled Fuctio he value of each sample (stregth of the weighted impulse) f = f () t δ ( t k Δ ) dt = f ( k Δ ) k ~ ~ Fu ( ) = F { f( t)} = F { f( xs ) ( t)} = Fu ( )* Su ( ) = FxSu ( ) ( xdx ) Δ = Fx ( ) δ ( u ) dx Δ Δ = = Fx ( ) δ ( u t ) dx Δ Δ = = Fu ( ) Δ Δ =

23 he Samplig heorem Bad-limited fuctio f(t), its F F(u) = 0 whe u < -u max or u>u ~ max ~ Let f () t be the samplig fuctio of f(t), ad be its F ~ F ( u) Questio: if f(t) ca be recovered from f () t Samplig heorem: if the f(t) ca be recovered max Δ >2U Δ, umax u u Hu () = 0 otherwise F() u = H() u F() u ~ ~ ft () = F - { Fu ( )} = F - { HuFu ( ) ( )} ~ = ht ()* f() t max = f ( Δ)si c[( t Δ)/ Δ] =

24 he Samplig heorem Samplig: Over-samplig Critically-samplig Uder-samplig Aliasig: f(t) is corrupted 2U Δ > > max

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27 Discrete Fourier rasform(df) Derive DF from cotiuous F of sampled fuctio ~ ~ 2 π iu t F ( u ) = f ( t ) e d t = = = = f e 2 π iu t f ( t ) δ ( t Δ t ) e d t = Δ = = 2 π iu Δ 2 π iu t f ( t ) δ ( t t ) e d t m u =, m = 0,,..., M M Δ M 2 π im / M F m f e m M = 0 =, = 0,,..., =, = 0,,..., M M 2 π im / M f F m e M m = 0

28 DF M - -2 π xu / M ( ) ( ), 0,,..., - Fu = f e u= M x= 0 f x F u e x M M - 2 ixu / M ( ) = ( ) π, = 0,,..., 0 - M u = 0 F ( u ) = F ( u + km ) f ( x ) = f ( x + km ) M f ( x ) * h ( x ) = f ( m ) h ( x m ) m = 0

29 Matrix represetatio ω M = e 2πi M ( M ) F (0) ω... (0) M ωm ω M f 0 ( M ) F() ω... f() M ωm ωm = M M M M M M ( M ) 0 ( M ) ( M ) ( M ) FM ( ) ω... f( M ) M ωm ω M ( M ) f (0) ωm ωm... ω (0) M F f () 0 ( M )... F() = ωm ωn ωm M M M M M M ( M ) 0 ( M ) ( M ) ( M ) f( M ) ω... FM ( ) ω ω M M M ( M ) ω M ω M... ω M F(0) 0 ( M ) ω M ω M... ω F() M = M M M M M M ( M ) 0 ( M ) ( M ) ( M ) ω M ω M... ω FM ( ) M

30 Example 3 F (0) = f ( x ) = [ f (0) + f () + f (2) + f (3)] x = 0 = = 3 2 π i () x /4 0 iπ /2 iπ i 3 π /2 () = ( ) = = F f x e e e e e i x = π i ( 2 ) x / 4 0 iπ i 2 π i 3π (2) = ( ) = = F f x e e e e e x = π i ( 3 ) x / 4 0 i 3 π / 2 i 6 π / 2 i 9 π / 2 (3) = ( ) = e F f x e e e e x = π iu ( 0 ) 3 u = 0 u = = 3 2i f (0) = F ( u ) e = F ( u ) = [ 3 + 2i 3 2 i] = 4 4 4