Image Analysis. 3. Fourier Transform

Transcription

1 Image Analysis Image Analysis 3. Fourier Transform Lars Schmidt-Thieme Information Systems and Machine Learning Lab (ISMLL) Institute for Business Economics and Information Systems & Institute for Computer Science University of Hildesheim Course on Image Analysis, winter term /66 Image Analysis 1. Fourier Series Representation 2. The Fourier Transform 3. Discrete Signals 4. Discrete Fourier Transform 5. Two-dimensional Fourier Transforms 6. Applications Course on Image Analysis, winter term /66

2 Image Analysis / 1. Fourier Series Representation Periodic Functions f () Course on Image Analysis, winter term /66 Image Analysis / 1. Fourier Series Representation Periodic Functions f () Course on Image Analysis, winter term /66

3 Image Analysis / 1. Fourier Series Representation Periodic Functions A function f : R C is called T -periodic if f( + T ) = f() R Eample: the functions sin and cos are 2π-periodic. Eample: the function sin(ω) is 2π/ω-periodic. Course on Image Analysis, winter term /66 Image Analysis / 1. Fourier Series Representation Periodic Functions / Approimation f () ˆf() = 1 sin() Course on Image Analysis, winter term /66

4 Image Analysis / 1. Fourier Series Representation Periodic Functions / Approimation f () ˆf() = 1 sin() + 2 cos Course on Image Analysis, winter term /66 Image Analysis / 1. Fourier Series Representation Periodic Functions / Approimation f () ˆf() = 1 sin() + 2 cos + 2 sin 2 Course on Image Analysis, winter term /66

5 Image Analysis / 1. Fourier Series Representation Periodic Functions / Approimation f () ˆf() = 1 sin() + 2 cos + 2 sin cos 2 Course on Image Analysis, winter term /66 Image Analysis / 1. Fourier Series Representation Periodic Functions / Approimation f () ˆf() = 1 sin() + 2 cos + 2 sin cos sin 3 Course on Image Analysis, winter term /66

6 Image Analysis / 1. Fourier Series Representation Periodic Functions / Approimation f () ˆf() = 1 sin() + 2 cos + 2 sin cos sin cos 3 Course on Image Analysis, winter term /66 Image Analysis / 1. Fourier Series Representation Periodic Functions / Approimation f () ˆf() = 1 sin()+2 cos +2 sin 2+1 cos 2+3 sin 3+3 cos 3+2 sin 4 Course on Image Analysis, winter term /66

7 Image Analysis / 1. Fourier Series Representation Periodic Functions / Approimation f () ˆf() = 1 sin()+2 cos +2 sin 2+1 cos 2+3 sin 3+3 cos 3+2 sin 4+1 cos 4 Course on Image Analysis, winter term /66 Image Analysis / 1. Fourier Series Representation Fourier Series Representation Theorem (Fourier Series Representation). Any continuous, differentiable and T -periodic function f : R C can be written as f() = a a k cos kω + b k sin kω, ω := 2π T k=1 with coefficients a k, b k C, called a Fourier Series of f. How to compute Fourier coefficients a k, b k for a given function f? Jean Baptiste Joseph Fourier ( ), French Mathematician and Physicist Note. Actually, the class of functions that can be represented as Fourier series is much larger (see, e.g., [Kö90, p. 314]). Course on Image Analysis, winter term /66

8 Image Analysis / 1. Fourier Series Representation Trigonometric Addition Formulas Lemma (Trigonometric Addition Formulas). For all, y R: cos( + y) = cos cos y sin sin y sin( + y) = sin cos y + sin y cos Course on Image Analysis, winter term /66 Image Analysis / 1. Fourier Series Representation cos a cos b d = sin a sin b d = sin a cos b d = Some Trigonometric Integrals 1 2 ( 1 sin(a + b) + 2 a + b ) sin(a b), if a b a b sin(2a) + 2a, else 4a ( sin(a + b) a + b ) sin(a b), if a b a b sin(2a) 2a, else 4a 1 2 ( cos(a + b) + a + b ) cos(a b), if a b a b sin 2 (a), else 2a Course on Image Analysis, winter term /66

9 Image Analysis / 1. Fourier Series Representation Some Trigonometric Integrals The si formulas easily can be proven by differentiation, e.g., cos a cos b d =? 1 ( ) sin(a + b) sin(a b) + 2 a + b a b for a b. Derivation d d yields: cos a cos b =? 1 ( (a + b) cos(a + b) + 2 a + b = 1 (cos(a + b) + cos(a b)) 2 ) (a b) cos(a b) a b = 1 (cos a cos b sin a sin b + cos a cos( b) sin a sin( b)) 2 = 1 (cos a cos b sin a sin b + cos a cos b + sin a sin b) 2 = cos a cos b Course on Image Analysis, winter term /66 Image Analysis / 1. Fourier Series Representation Trigonometric Orthogonality Relations Let ω R +. The functions {sin kω k N, k > 0} {cos kω k N, k > 0} are pairwise othogonal with respect to f, g := +π/ω π/ω f()g()d i.e., for any two distinct such functions f, g f, g = +π/ω π/ω but f, f = π ω 0 f()g()d = 0 Course on Image Analysis, winter term /66

10 Image Analysis / 1. Fourier Series Representation Trigonometric Orthogonality Relations Proof: let f = cos kω and g = cos lω with k l, then: +π/ω [ ( 1 sin(k + l)ω sin(k l)ω f, g = cos kω cos lω d = + π/ω 2 (k + l)ω (k l)ω = 1 ( ) sin(k + l)π sin(k l)π sin (k + l)π sin (k l)π + 2 (k + l)ω (k l)ω (k + l)ω (k l)ω sin(k + l)π sin(k l)π = + (k + l)ω (k l)ω = 0 but [ sin(2kω) + 2kω f, f = 4kω ] +π/ω π/ω = )] +π/ω π/ω sin(2kπ) + 2kπ sin( 2kπ) + ( 2kπ) 4kω 4kω = π ω Course on Image Analysis, winter term /66 Image Analysis / 1. Fourier Series Representation Fourier Series Representation Theorem (Fourier Series Representation). Any continuous, differentiable and T -periodic function f : R C can be written as f() = a a k cos kω + b k sin kω, k=1 ω := 2π T with coefficients a k, b k C, called a Fourier Series of f. How to compute Fourier coefficients a k, b k for a given function f? a k = 1 +π/ω π/ω b k = 1 π/ω π/ω +π/ω π/ω f() cos kω d f() sin kω d Note. Actually, the class of functions that can be represented as Fourier series is much larger (see, e.g., [Kö90, p. 314]). Course on Image Analysis, winter term /66

11 Image Analysis / 1. Fourier Series Representation Fourier Series Representation Proof. a? k = 1 +π/ω π/ω = 1 π/ω = 1 π/ω π/ω +π/ω π/ω = 1 π π/ω ω a k =a k ( +π/ω π/ω f() cos kω d ( a 0 ) 2 + a l cos lω + b l sin lω cos kω d l=1 a 0 2 cos kω d + l=1 +π/ω + π/ω +π/ω π/ω a l cos lω cos kω d b l sin lω cos kω d ) Course on Image Analysis, winter term /66 Image Analysis / 1. Fourier Series Representation Fourier Representation / Rectangular function Let f be the 2π-periodic rectangular function 1, if ( π, 0) f() = 0, if { π, 0, π} +1, if (0, π) f () The Fourier representation of f is Proof: b k = 2 π a k = 2 π +π π +π π f() = 4 π f() sin k d = 2 +π π 2 sin k d = 4 π f() cos k d = 0 0 k N odd sin k k [ 1 ] π k cos k = { 4 4 (0 1) = πk (1 1) = 0, 4 πk πk, if k odd if k even in general, Fourier representations are infinite as in this eample! Course on Image Analysis, winter term /66

12 Image Analysis / 1. Fourier Series Representation Fourier Representation / Rectangular function f () Course on Image Analysis, winter term /66 Image Analysis / 1. Fourier Series Representation Fourier Representation / Rectangular function / k = 1 f () Course on Image Analysis, winter term /66

13 Image Analysis / 1. Fourier Series Representation Fourier Representation / Rectangular function / k = 5 f () Course on Image Analysis, winter term /66 Image Analysis / 1. Fourier Series Representation Fourier Representation / Rectangular function / k = 21 f () Course on Image Analysis, winter term /66

14 Image Analysis / 1. Fourier Series Representation Eulers Formula cos := n N( 1) n sin := n N( 1) n ep := n N n n! 2n (2n)! 2n+1 (2n + 1)! = 1 2 2! + 4 4! 6 6! + = 3 3! + 5 5! 7 7! + = ! + 3 3! + 4 4! + Lemma (Eulers formula). For C: e i = cos() + i sin(), with i := 1 the imaginary unit Proof: e i = (i) n n! n N = (i) 2n (2n)! + (i) 2n+1 (2n + 1)! n N n N = ( 1) n 2n (2n)! + i ( 1) n 2n+1 (2n + 1)! n N n N = cos() + i sin() Course on Image Analysis, winter term /66 Image Analysis / 1. Fourier Series Representation Comple Fourier Series Representation Theorem (Comple Fourier Series Representation). Any continuous, differentiable and T -periodic function f : R C can be written as f = c k e ikω, ω := 2π T k Z with coefficients c k C, called a Fourier Series of f. The coefficients of the comple Fourier series can be computed via c k = 1 +π/ω f() e ikω d 2π/ω π/ω Course on Image Analysis, winter term /66

15 Image Analysis / 1. Fourier Series Representation Comple Fourier Series Representation Proof. f = k Z c k e ikω with = k Z =c 0 + c k (cos kω + i sin kω) (c k + c k ) cos kω + (c k c k )i sin kω k=1 = a a k cos kω + b k sin kω k=1 and vice versa via: a 0 = 2c 0, a k = c k + c k, b k = i(c k c k ) c 0 = a 0 2, c k = 1 2 (a k ib k ), c k = 1 2 (a k + ib k ) Course on Image Analysis, winter term /66 Image Analysis 1. Fourier Series Representation 2. The Fourier Transform 3. Discrete Signals 4. Discrete Fourier Transform 5. Two-dimensional Fourier Transforms 6. Applications Course on Image Analysis, winter term /66

16 Image Analysis / 2. The Fourier Transform Fourier Transform Let f : R C be a function (that satisfies some regularity conditions). Then F : R C ω 1 2π + f()e iω d eists for each ω and is a continuous function called Fourier Transform of f (aka Fourier spectrum of f). One can show that (if F also satisfies some regularity conditions): f() = 1 2π + F (ω)e iω dω This is called Inverse Fourier Transform. We will write F(f) := F for the Fourier transform of a function f and F 1 (F ) for the inverse Fourier transform of a function F. Course on Image Analysis, winter term /66 Image Analysis / 2. The Fourier Transform Fourier Transforms / Eamples / Gaussian f () F () f() = 1 σ e 2 2σ 2 F () = e σ2 2 2 Course on Image Analysis, winter term /66

17 Image Analysis / 2. The Fourier Transform Fourier Transforms / Eamples / Uniform f () F () f() = δ( [ b, b]) F () = 2b sin(b) 2π Course on Image Analysis, winter term /66 Image Analysis / 2. The Fourier Transform Fourier Transforms / Eamples / Cosine cos () F(cos)() f() = cos(ω) F () = π (δ( ω) + δ( + ω)) 2 Course on Image Analysis, winter term /66

18 Image Analysis / 2. The Fourier Transform Fourier Transforms / Eamples / Sine sin () Im(F(sin)()) f() = sin(ω) F () = i π (δ( ω) δ( + ω)) 2 Course on Image Analysis, winter term /66 Image Analysis / 2. The Fourier Transform Properties function h Fourier transform H property name f F F f( ) inverse af + bg af + bg linearity f g F ()G() convolution 1 f()g() 2π F G multiplication f( a) e ia F () translation e ia f() F ( a) modulation f(/a) a F (a) scaling f F ( ) comple conjugate f() R F ( ) = F () hermitian symmetry Course on Image Analysis, winter term /66

19 Image Analysis 1. Fourier Series Representation 2. The Fourier Transform 3. Discrete Signals 4. Discrete Fourier Transform 5. Two-dimensional Fourier Transforms 6. Applications Course on Image Analysis, winter term /66 Image Analysis / 3. Discrete Signals The symbol Dirac Comb T () := n Z δ( T n) is called Dirac comb (aka impulse train, sampling function, Shah function) with sampling interval T. f() Lemma. The Fourier series of the Dirac comb T is T () = 1 e i2πk T T and its Fourier transform k Z F( T )() = 1 T 2π/T() = 1 T n Z δ( 2π T n) Course on Image Analysis, winter term /66

20 Image Analysis / 3. Discrete Signals Dirac Comb Proof: Obviously T is periodic with period T. Therefore f() = k Z c k e i2πk T with c k = 1 T = 1 T = 1 T +T +T/2 T/2 +T/2 T/2 = 1 T e i2πk T 0 T (y) e i2πk T y dy T (y) e i2πk T y dy δ(y) e i2πk T y dy = 1 T Course on Image Analysis, winter term /66 Image Analysis / 3. Discrete Signals Sampling f() f() Course on Image Analysis, winter term /66

21 Image Analysis / 3. Discrete Signals Sampling f() Sampling a function f at equidistant points T Z can be understood as f sampled () = f() T () Course on Image Analysis, winter term /66 Image Analysis / 3. Discrete Signals Fourier Transform of a Sampled Function Let f : R C be a function and f sampled be a sample of f with sampling period T. Then its Fourier transform F(f sampled ) is 2π/T -periodic and aggregates the Fourier transform F(f) over a 2π/T -periodic grid: F(f sampled )() = n Z F(f)( + n 2π T ) If F(f) vanishes for > π/t, then the Fourier transform F(f) is replicated in a period of the Fourier transform F(f sampled ). Otherwise replicas overlap and the Fourier transform becomes corrupted. This effect is called aliasing. Course on Image Analysis, winter term /66

22 Image Analysis / 3. Discrete Signals Fourier Transform of a Sampled Function The maimal occurring frequency ω ma := ma{ R, F(f)() 0} of the Fourier transform is called its bandwidth. The frequency ω s := 2π T of the sampling function is called its sampling frequency. Then the sampling frequency must be at least twice the bandwith: ω s > 2ω ma Course on Image Analysis, winter term /66 Image Analysis / 3. Discrete Signals Fourier Transform of a Sampled Function F(f) ω ma ω F(f_sampled1) ω ma ω ωs1 F(f_sampled2) ω ma ωs2 ω (cf. [BB08, p. 330]) Course on Image Analysis, winter term /66

23 Image Analysis / 3. Discrete Signals Fourier Transform of a Sampled Function Proof. F(f sampled )() =F(f T )() =F(f) F( T )() =F(f) 1 T 2π/T() =F(f) 1 δ( + n 2π T T ) n Z = 1 F(f) δ( + n 2π T T ) n Z = 1 F(f)( + n 2π T T ) n Z If < π/t and F(f)(y) vanishes for y > π/t, then F(f sampled )() = 1 T F(f)() Course on Image Analysis, winter term /66 Image Analysis / 3. Discrete Signals The Fourier Transform of a Discrete Function Let f be a discrete function: Then its Fourier transform is: F(f)(ω) = 1 2π f() = n Z y n δ( n), y n R = 1 2π = 1 2π n Z n Z f()e iω d y n δ( n) e iω d = 1 y n e iωn 2π n Z = 1 f(n) e iωn 2π n Z y n e iω δ( n)d i.e., a periodic function or equivalently: a function defined on an interval. Course on Image Analysis, winter term /66

24 Image Analysis 1. Fourier Series Representation 2. The Fourier Transform 3. Discrete Signals 4. Discrete Fourier Transform 5. Two-dimensional Fourier Transforms 6. Applications Course on Image Analysis, winter term /66 Image Analysis / 4. Discrete Fourier Transform Definition Let f : {0, 1,..., N 1} C be a finite discrete function, then F : {0, 1,..., N 1} C ω 1 N 1 N =0 = 1 N 1 N f()(cos(2π ω N ) i sin(2πω N )) =0 i2π ω f()e N is called discrete Fourier transform of f, denoted DFT(f). Then f() = 1 N 1 N ω=0 = 1 N 1 N ω=0 F (ω)(cos(2π ω N ) + i sin(2πω N )) i2π ω F (ω)e N This is called inverse discrete Fourier transform of f, denoted DFT 1 (F ).. Course on Image Analysis, winter term /66

25 Image Analysis / 4. Discrete Fourier Transform Eample DFT(f)(0) = =0 f() ω F (ω) i i i DFT i i i i i i i i i i i i DFT i i i i i f()(cos(2π 0 10 ) i sin(2π0 10 )) = 1 10 (cf. [BB08, p. 333]) 9 =0 f() = = Course on Image Analysis, winter term /66 Image Analysis / 4. Discrete Fourier Transform Eample DFT(f)(1) = 1 10 = =0 9 =0 f() ω F (ω) i i i DFT i i i i i i i i i i i i DFT i i i i i (cf. [BB08, p. 333]) f()(cos(2π 1 10 ) i sin(2π1 10 )) f() cos(π 5 ) i =0 f() sin(π ) = i 5 Course on Image Analysis, winter term /66

26 Image Analysis / 4. Discrete Fourier Transform Discrete Fourier Transform / Algorithm (naive) For C, denote R() its real part and I() its imaginary part, i.e., = R() + ii(), R(), I() R (for R one often also uses Re, for I also Im). To compute the discrete Fourier transform, one computes dft(f)(ω) for ω = 0,..., N 1 via: DFT(f)(ω) = 1 N 1 N =0 f()(cos(2π ω N ) i sin(2πω N )) = 1 N 1 (R(f()) + ii(f()))(cos(2π ω N N ) i sin(2πω N )) =0 = 1 N 1 N =0 N i N R(f()) cos(2π ω ) + I(f()) sin(2πω N N ) =0 R(f()) sin(2π ω ) + I(f()) cos(2πω N N ) Course on Image Analysis, winter term /66 Image Analysis / 4. Discrete Fourier Transform Discrete Fourier Transform / Algorithm (naive) 1 dft-naive(sequence f = (f() 0, f() 1 ) =0,...,N 1 ) : 2 N := length(f) 3 F := (F () 0, F () 1 ) =0,...,N 1 = (0, 0) =0,...,N 1 4 for ω := 0,..., N 1 do 5 c := (c 0, c 1 ) := (0, 0) 6 for := 0,..., N 1 do 7 c 0 := c 0 + f() 0 cos(2πω/n) + f() 1 sin(2πω/n) 8 c 1 := c 1 f() 0 sin(2πω/n) + f() 1 cos(2πω/n) 9 od 10 c 0 := c 0 / N 11 c 1 := c 1 / N 12 F (ω) := c 13 od 14 return F Course on Image Analysis, winter term /66

27 Image Analysis / 4. Discrete Fourier Transform Discrete Fourier Transform / Algorithm (naive) When computing the values of the discrete Fourier transform for different arguments ω, the cosine and sine functions repeatedly are callled with the same arguments: dft(f)(1): cos(2π 0/10) cos(2π 1/10) cos(2π 2/10) cos(2π 3/10) cos(2π 4/10) cos(2π 5/10) cos(2π 6/10) cos(2π 7/10) cos(2π 8/10) cos(2π 9/10) dft(f)(2): cos(2π 0/10) cos(2π 2/10) cos(2π 4/10) cos(2π 6/10) cos(2π 8/10) cos(2π 10/10) = cos(2π 0/10) cos(2π 12/10) = cos(2π 2/10) cos(2π 14/10) = cos(2π 4/10) cos(2π 16/10) = cos(2π 6/10) cos(2π 18/10) = cos(2π 8/10) Caching the epensive sine and cosine computations accelerates the algorithm! Course on Image Analysis, winter term /66 Image Analysis / 4. Discrete Fourier Transform Discrete Fourier Transform / Algorithm (naive, cached) 1 dft-naive-cached(sequence f = (f() 0, f() 1 ) =0,...,N 1 ) : 2 N := length(f) 3 for ω := 0,..., N 1 do 4 C(ω) := cos(2πω/n) 5 S(ω) := sin(2πω/n) 6 od 7 F := (F () 0, F () 1 ) =0,...,N 1 = (0, 0) =0,...,N 1 8 for ω := 0,..., N 1 do 9 c := (c 0, c 1 ) := (0, 0) 10 for := 0,..., N 1 do 11 c 0 := c 0 + f() 0 C(ω mod N) + f() 1 S(ω mod N) 12 c 1 := c 1 f() 0 S(ω mod N) + f() 1 C(ω mod N) 13 od 14 c 0 := c 0 / N 15 c 1 := c 1 / N 16 F (ω) := c 17 od 18 return F Course on Image Analysis, winter term /66

28 Image Analysis / 4. Discrete Fourier Transform Fast Fourier Transform (Gauss ca. 1805; Cooley/Tukey 1965) The naive algorithm for DFT still has compleity O(N 2 ). The Fast Fourier Transform algorithm is based on a decomposition of the DFT for sequences f of even length N: DFT(f)(ω) =DFT(f even )(ω mod N/2) + e i2πω/n DFT(f odd )(ω mod N/2) for ω = 0,..., N 1 and where f even () :=f(2), = 0,..., N/2 f odd () :=f(2 + 1) Course on Image Analysis, winter term /66 Image Analysis / 4. Discrete Fourier Transform Fast Fourier Transform / Proof Proof. N 1 i2π ω DFT(f)(ω) = f()e N = = = =0 N 1 =0 even N/2 1 =0 N/2 1 =0 N 1 i2π ω f()e N + i2π ω2 f(2)e N f even i2π ω ()e =0 odd N/2 1 + =0 i2π ω f()e N ω(2+1) i2π f(2 + 1)e N N/2 1 N/2 + e i2πω/n =0 =DFT(f even )(ω) + e i2πω/n DFT(f odd )(ω) f odd i2π ω ()e N/2 or more eaclty, as DFT(f even )(ω) is only defined for ω < N/2: =DFT(f even )(ω mod N/2) + e i2πω/n DFT(f odd )(ω mod N/2) Course on Image Analysis, winter term /66

29 Image Analysis / 4. Discrete Fourier Transform Fast Fourier Transform / Real version DFT(f)(ω) =DFT(f even )(ω mod N/2) + e i2πω/n DFT(f odd )(ω mod N/2) =DFT(f even )(ω mod N/2) and thus + (cos 2πω/N i sin 2πω/N)DFT(f odd )(ω mod N/2) R(DFT(f)(ω)) =R(DFT(f even )(ω mod N/2)) + cos 2πω/N R(DFT(f odd )(ω mod N/2)) + sin 2πω/N I(DFT(f odd )(ω mod N/2)) I(DFT(f)(ω)) =I(DFT(f even )(ω mod N/2)) + cos 2πω/N I(DFT(f odd )(ω mod N/2)) sin 2πω/N R(DFT(f odd )(ω mod N/2)) Course on Image Analysis, winter term /66 Image Analysis / 4. Discrete Fourier Transform Fast Fourier Transform / Algorithm 1 fft(sequence f = (f() 0, f() 1 ) =0,...,N 1 ) : 2 N := length(f) 3 if N is even 4 F := (F () 0, F () 1 ) =0,...,N 1 = (0, 0) =0,...,N 1 5 A := fft((f()) =0,2,4,...,N 2 ) 6 B := fft((f()) =1,3,5,,...,N 1 ) 7 for ω := 0,..., N 1 do 8 a := A(ω mod N/2) 9 b := B(ω mod N/2) 10 F (ω) 0 := a 0 + cos 2πω/N b 0 + sin 2πω/N b 1 11 F (ω) 1 := a 1 + cos 2πω/N b 1 sin 2πω/N b 0 12 od 13 return 14 else 15 F := dft-naive-cached(f) 16 fi 17 return F Course on Image Analysis, winter term /66

30 Image Analysis / 4. Discrete Fourier Transform Fast Fourier Transform / Outlook The computation of cos 2πω/N and sin 2πω/N also can be done recursively using the addition formulas. In this way, FFT best is applied to sequences of length 2 n (called radi-2 case). The FFT decomposition works with any factorization N = N 1 N 2 in a similar way, and thus also for sequences of length other than 2 n. FFT has compleity O(N log N) (if N is a power of 2). An early eperiment from 1969 reports a runtime of 13 1/2 hours for computing the DFT of a sequence of length 2048 by the naive method and 2.4 seconds using FFT. In practice, FFT is implemented in a linearized version avoiding eplicit recursions (see [CLRS03, p. 839]). Course on Image Analysis, winter term /66 Image Analysis / 4. Discrete Fourier Transform Types of Fourier Transforms type of function f sup f sup F(f) type of Fourier decomp. general (integrable) function R R (general) Fourier decomposition periodic function, function on interval interval I R Z Fourier series general (integrable) discrete function (sum of Diracs) Z interval I R discrete-time Fourier transform periodic discrete function, finite discrete function (finite sum of Diracs) finite I Z I discrete Fourier transform Course on Image Analysis, winter term /66

31 Image Analysis 1. Fourier Series Representation 2. The Fourier Transform 3. Discrete Signals 4. Discrete Fourier Transform 5. Two-dimensional Fourier Transforms 6. Applications Course on Image Analysis, winter term /66 Image Analysis / 5. Two-dimensional Fourier Transforms General Fourier Transform in 2D For two-dimensional functions f : R R C Fourier Transforms, Fourier Series and Discrete Fourier Transforms can be defined analogously. Let f : R R C be a function (that satisfies some regularity conditions). Then F : R R C (ω 1, ω 2 ) f(, y) e iω1 e iω2y dy d 2π eists for each (ω 1, ω 2 ) and is a continuous function called Fourier Transform of f (aka Fourier spectrum of f). One can show that (if F also satisfies some regularity conditions): f(, y) = F (ω 1, ω 2 )e iω1 e iω2y dω 1 dω 2 2π This is called Inverse Fourier Transform. We will write F(f) := F for the Fourier transform of a function f and F 1 (F ) for the inverse Fourier transform of a function F. Course on Image Analysis, winter term /66

32 Image Analysis / 5. Two-dimensional Fourier Transforms Bases in 2D Course on Image Analysis, winter term /66 Image Analysis / 5. Two-dimensional Fourier Transforms Fourier Series in 2D If f : R R C is (T1, T2)-periodic, i.e., f (, y) = f ( + T1, y + T2), y R then f can already be reconstructed from Z-many Fourier coefficients: 1 XX i 2π n i 2π my f (, y) = F (n, m) e T1 e T2 2π n Z m Z with 1 F (n, m) := 2π Z +T1 /2 Z +T2 /2 T1 /2 T2 /2 f (, y) e inω1 e imω2y dy d Course on Image Analysis, winter term /66

33 Image Analysis / 5. Two-dimensional Fourier Transforms If f is discrete, i.e., f(, y) = n Z Discrete Fourier Transform in 2D y n,m δ( T 1 n, y T 2 m), m Z then its Fourier transform is periodic: with f(, y) = 1 2π + + F (ω 1, ω 2 ) := 1 2π n Z m Z y n,m R F (ω 1, ω 2 ) e iω 1 e iω 2y dω 1 dω 2 f(n, m) e iω1n e iω 2m Course on Image Analysis, winter term /66 Image Analysis / 5. Two-dimensional Fourier Transforms Discrete Fourier Transform in 2D And finally, if f is discrete and finite, i.e., f(, y) = N 1 n=0 M 1 m=0 y n,m δ( n, y m), y n,m R then its Fourier transform is periodic and made from finitely many components: with f(, y) = F (ω 1, ω 2 ) := N 1 1 NM n=0 N 1 1 NM n=0 M 1 m=0 M 1 m=0 F (ω 1, ω 2 ) e iω 1 e iω 2y f(n, m) e iω 1n e iω 2m Course on Image Analysis, winter term /66

34 Image Analysis / 5. Two-dimensional Fourier Transforms Discrete Fourier Transform in 2D / Algorithm (naive, cached) 1 dft-2d-naive-cached(array f = (f() 0, f() 1 ) =0,...,N 1,y=0,...,M 1 ) : 2 for ω := 0,..., N 1 do 3 C 1 (ω) := cos(2πω/n) 4 S 1 (ω) := sin(2πω/n) 5 od 6 for ω := 0,..., M 1 do 7 C 2 (ω) := cos(2πω/m) 8 S 2 (ω) := sin(2πω/m) 9 od 10 F := (F () 0, F () 1 ) =0,...,N 1,y=0,...,M 1 = (0, 0) =0,...,N 1,y=0,...,M 1 11 for ω 1 := 0,..., N 1 do 12 for ω 2 := 0,..., M 1 do 13 c := (c 0, c 1 ) := (0, 0) 14 for := 0,..., N 1 do 15 for y := 0,..., M 1 do 16 C := C 1 (ω 1 mod N) C 2 (ω 2 y mod M) S 1 (ω 1 mod N) S 2 (ω 2 y mod M) 17 S := S 1 (ω 1 mod N) C 2 (ω 2 y mod M) + C 1 (ω 1 mod N) S 2 (ω 2 y mod M) 18 c 0 := c 0 + f() 0 C f() 1 S 19 c 1 := c 1 f() 0 S + f() 1 C 20 od 21 od 22 c 0 := c 0 / NM 23 c 1 := c 1 / NM 24 F (ω) := c 25 od 26 od 27 return F Course on Image Analysis, winter term /66 Image Analysis / 5. Two-dimensional Fourier Transforms Discrete Fourier Transform in 2D / Separability F (ω 1, ω 2 ) := 1 N 1 NM n=0 n=0 M 1 m=0 f(n, m) e iω 1n e iω 2m ( ) = 1 N 1 M 1 1 f(n, m) e iω 2m N M m=0 e iω 1n = 1 N 1 DFT((f(n, m)) m=1,...,m )(ω 2 ) e iω 1n N n=0 Course on Image Analysis, winter term /66

35 Image Analysis / 5. Two-dimensional Fourier Transforms Discrete Fourier Transform in 2D / FFT 1 fft-2d(array f = (f() 0, f() 1 ) =0,...,N 1,y=0,...,M 1 ) : 2 G := (G() 0, G() 1 ) =0,...,N 1,y=0,...,M 1 = (0, 0) =0,...,N 1,y=0,...,M 1 3 for ω 1 := 0,..., N 1 do 4 G(ω 1,.) := fft(f(ω 1, y) y=0,...,m 1 ) 5 od 6 F := (F () 0, F () 1 ) =0,...,N 1,y=0,...,M 1 = (0, 0) =0,...,N 1,y=0,...,M 1 7 for ω 2 := 0,..., M 1 do 8 F (., ω 2 ) := fft(g(, ω 2 ) =0,...,N 1 ) 9 od 10 return F Course on Image Analysis, winter term /66 Image Analysis / 5. Two-dimensional Fourier Transforms Power Spectrum The fourier spectrum of is a discrete N M gray-scale image f, i.e., with one channel, a discrete N M image F with comple intensity values, i.e., two channels. For visualization one usually shows the power spectrum defined as: F power (f)() := (RF(f)()) 2 + (IF(f)()) 2 The power spectrum measures the absolute value of the comple amplitude. The complementary information θ called phase is not shown. Course on Image Analysis, winter term /66

36 Image Analysis / 5. Two-dimensional Fourier Transforms What does Power Spectrum mean? Comple Coordinates Cartesian coordinates: Polar coordinates: Im r θ Re = (R, I) R R R =r cos θ I =r sin θ = (r, θ) R + 0 [0, 2π) r = (R) 2 + (I) 2 arctan(r/i), if > 0, y > 0 θ = arctan(r/i) + π, if y < 0 arctan(r/i) + 2π, if > 0, y < 0 = R + ii = r cos θ + ir sin θ = re iθ Course on Image Analysis, winter term /66 Image Analysis / 5. Two-dimensional Fourier Transforms Eample image f: power spectrum F power (f): Course on Image Analysis, winter term /66

37 Image Analysis / 5. Two-dimensional Fourier Transforms Displaying Fourier Power Spectra For displaying spectra, some further conventions are used: As the scale of many power spectra is dominated by a few large values, one usually plots log F power (f)() or (F power (f)()) 1 2 instead of the raw power spectrum values F power (f)(). Usually, the centered spectrum is shown, i.e., the intensities for and { N/2, N/2 + 1,..., 1, 0, 1,..., N/2 1, N/2} y { M/2, M/2 + 1,..., 1, 0, 1,..., M/2 1, M/2} instead of the intensities for 0, 1,..., N 1 and 0, 1,..., M 1. Course on Image Analysis, winter term /66 Image Analysis / 5. Two-dimensional Fourier Transforms Centered Spectrum A B C D Course on Image Analysis, winter term /66

38 Image Analysis / 5. Two-dimensional Fourier Transforms Centered Spectrum A B C A D A C B C B D A D B C D Course on Image Analysis, winter term /66 Image Analysis / 5. Two-dimensional Fourier Transforms Centered Spectrum original spectrum: A C centered spectrum: B D C D B A Course on Image Analysis, winter term /66

39 Image Analysis / 5. Two-dimensional Fourier Transforms Eample image f: power spectrum F power (f): Course on Image Analysis, winter term /66 Image Analysis / 5. Two-dimensional Fourier Transforms Symmetry of Fouier Power Spectra for Real Images Usually images are real, i.e., f() R (not C). For real functions, we know that F(f)( ) = F (f)() is hermitian (with := R i I). As C has the same radius as : r() = (R) 2 + (I) 2 =? r( ) = r(r ii) = (R) 2 + ( I) 2 for real functions F(f) and F (f) have the same radius and thus F power (f)( ) = F power (f)() i.e., the power spectrum is symmetric around the origin. Course on Image Analysis, winter term /66

40 Image Analysis / 5. Two-dimensional Fourier Transforms Symmetry of Fouier Power Spectra for Real Images image f: power spectrum F power (f): Spectra of real images are symmetric around the origin (red circle). So storing just half of the power spectrum is sufficient (e.g., above green line any line through the origin will do). Course on Image Analysis, winter term /66 Image Analysis / 5. Two-dimensional Fourier Transforms Another Eample image f: power spectrum F power (f): Course on Image Analysis, winter term /66

41 Image Analysis 1. Fourier Series Representation 2. The Fourier Transform 3. Discrete Signals 4. Discrete Fourier Transform 5. Two-dimensional Fourier Transforms 6. Applications Course on Image Analysis, winter term /66 Image Analysis / 6. Applications Low Pass Filters High frequencies are responsible for sharp edges, low frequencies for constant and slowly changing areas. Low pass filters retain only low frequencies ω ω ma, i.e., filter out high frequencies. and thus smooth / blur an image. Course on Image Analysis, winter term /66

42 Image Analysis / 6. Applications Low Pass Filters / Eample image f: power spectrum F power (f): Course on Image Analysis, winter term /66 Image Analysis / 6. Applications Low Pass Filters / Eample image f: power spectrum F power (f): Course on Image Analysis, winter term /66

43 Image Analysis / 6. Applications Low Pass Filters / Eample image f: power spectrum F power (f): Course on Image Analysis, winter term /66 Image Analysis / 6. Applications High Pass Filters High frequencies are responsible for sharp edges, low frequencies for constant and slowly changing areas. High pass filters retain only high frequencies ω ω min, i.e., filter out low frequencies. and thus sharpen an image and detect edges. Course on Image Analysis, winter term /66

44 Image Analysis / 6. Applications High Pass Filters / Eample image f: power spectrum F power (f): Course on Image Analysis, winter term /66 Image Analysis / 6. Applications High Pass Filters / Eample image f: power spectrum F power (f): Course on Image Analysis, winter term /66

45 Image Analysis / 6. Applications High Pass Filters / Eample image f: power spectrum F power (f): Course on Image Analysis, winter term /66 Image Analysis / 6. Applications Band Pass Filters High frequencies are responsible for sharp edges, low frequencies for constant and slowly changing areas. Band pass filters retain only frequencies ω [ω min, ω ma ] in a given interval (the frequency band), i.e., filter out low and high frequencies. and thus detect edges. Course on Image Analysis, winter term /66

46 Image Analysis / 6. Applications Band Pass Filters / Eample image f: power spectrum F power (f): Course on Image Analysis, winter term /66 Image Analysis / 6. Applications Band Pass Filters / Eample image f: power spectrum F power (f): Course on Image Analysis, winter term /66

47 Image Analysis / 6. Applications Reducing Periodic Noise image f: power spectrum F power (f): Course on Image Analysis, winter term /66 Image Analysis / 6. Applications Reducing Periodic Noise image f: power spectrum F power (f): Course on Image Analysis, winter term /66

48 Image Analysis / 6. Applications Reducing Periodic Noise image f: power spectrum F power (f): Periodic noise can be reduced by filtering out the frequencies belonging to the periodic noise pattern. This also can be understood as a simple method for inpainting. Course on Image Analysis, winter term /66 Image Analysis / 6. Applications Reducing Salt and Pepper Noise image f: power spectrum F power (f): Course on Image Analysis, winter term /66

49 Image Analysis / 6. Applications Reducing Salt and Pepper Noise image f: power spectrum F power (f): Reducing non-periodic noise patterns via frequency filters is difficult. Course on Image Analysis, winter term /66 Image Analysis / 6. Applications Deconvolution via Fourier Transform Assume, an image f has been corrupted by a convolution with a kernel k (e.g., blurred): g = k f If the kernel k is known, one can undo the convolution using the Fourier transform: F(g) = F(k f) = F(k) F(f) F(f) = F(g) F(k) ( ) F(g) f = F 1 F(f) = F 1 F(k) Course on Image Analysis, winter term /66

50 Image Analysis / 6. Applications Schedule Schedule until Christmas: net Tue., 9.12., no lecture. net Wed., 10.12, lecture. Tue., , no lecture. Wed., 17.12, lecture. Course on Image Analysis, winter term /66 Image Analysis / 6. Applications References [BB08] Wilhelm Burger and Mark J. Burge. Digital Image Processing, An Algorithmic Introduction Using Java. Springer, [CLRS03] Thomas H. Corman, Charles E. Leiserson, Ronald R. Rivest, and Clifford Stein. An Introduction to Algorithms. The MIT Press, [Kö90] Konrad Königsberger. Analysis 1. Springer, Course on Image Analysis, winter term /66