Image Processing

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recap assume a general periodic function f(x), w.l.o.g. period = 1. ⇔ assume a function such that f(x) = f(x + 1) then f(x) = a0. 2. + n. ∑ k=1 αk sin(2πk x + φk).
Image Processing Prof. Christian Bauckhage

outline lecture 08

recap

the Fourier transform – derivation part 2

the Fourier transform – derivation part 3

summary

recap

any L2 function f : R → R can be expressed as 1 f (x) = √ 2π

∞ Z

F(ω) eiωx dω −∞

where 1 F(ω) = √ 2π

∞ Z

f (x) e−iωx dx −∞

is called the Fourier transform of f (x)

recap

originally, the FT was developed as tool for solving (certain) differential equations it allows for reducing differentiation to multiplication F

f (x) −−−−→  d dx y F −1

F(ω)   y·iω

f 0 (x) ←−−−− iωF(ω)

recap

to begin to develop an understanding as to what the Fourier transform really is, we looked at finite Fourier series

recap

assume a general periodic function f (x), w.l.o.g. period = 1 ⇔ assume a function such that f (x) = f (x + 1)

recap

assume a general periodic function f (x), w.l.o.g. period = 1 ⇔ assume a function such that f (x) = f (x + 1) then a0 X + αk sin(2π k x + φk ) f (x) = 2 n

k=1

where a0 , αk , φk , and νk = 2π k ∈ R

recap

0

1

0

1

sin 2π k x, k = 1

0

sin 2π k x, k = 2

1

sin 2π k x, k = 3

0

1

P k

αk sin(νk x + φk )

recap

α=1

the expression f (x) =

n a0 X + αk sin(νk x+φk ) 2

α=

1 2

0

1

k=1

ν = 2π ν = 8π

involves amplitude αk

0

1

frequency νk phase φk

φ=0 φ = π4

0

1

recap

straightforward algebra yields a0 X f (x) = + αk sin(2π k x + φk ) 2 n

k=1

a0 X + = ak cos(2π k x) + bk sin(2π k x) 2 n

k=1

=

n X k=−n

ck ei 2π k x

recap

the coefficients ck are given by Z1 f (x) e−i 2π k x dx

ck = 0

recap

the coefficients ck are given by Z1 f (x) e−i 2π k x dx

ck = 0

instead of ck , we will also write ˆf (k)

recap

the coefficients ck are given by Z1 f (x) e−i 2π k x dx

ck = 0

instead of ck , we will also write ˆf (k) ˆf (k) is a complex valued function of k

recap

the coefficients ck are given by Z1 f (x) e−i 2π k x dx

ck = 0

instead of ck , we will also write ˆf (k) ˆf (k) is a complex valued function of k ˆf (k) is called the kth Fourier coefficient

recap

in summary, we have found

f (x) =

"Z n 1 X k=−n

0

# f (x) e−i 2π k x dx ei 2π k x

(1)

the Fourier transform – derivation part 2

question is our model “general” enough ?

question is our model “general” enough ? ⇔ given some finite n ∈ N, can we really write # "Z n 1 X f (x) e−i 2π k x dx ei 2π k x f (x) = k=−n

0

to express any periodic function of period one ?

question is our model “general” enough ? ⇔ given some finite n ∈ N, can we really write # "Z n 1 X f (x) e−i 2π k x dx ei 2π k x f (x) = k=−n

0

to express any periodic function of period one ?

answer no!

continuity

ei 2π k x (sin and cos) are continuous periodic functions

continuity

ei 2π k x (sin and cos) are continuous periodic functions the rectangle function Π(x) is periodic but not continuous

continuity

ei 2π k x (sin and cos) are continuous periodic functions the rectangle function Π(x) is periodic but not continuous

it is impossible to represent a discontinuous function as a finite sum of continuous functions

differentiability

ei 2π k x (sin and cos) are differentiable periodic functions

differentiability

ei 2π k x (sin and cos) are differentiable periodic functions the triangle function Λ(x) is periodic but not differentiable

differentiability

ei 2π k x (sin and cos) are differentiable periodic functions the triangle function Λ(x) is periodic but not differentiable

it is impossible to represent a non-differentiable function as a finite sum of continuous functions

bottom line

assume a periodic function f (x) if there is a discontinuity in any higher derivative of f (x), i.e. if there is lack of smoothness, the above arguments apply again

bottom line

assume a periodic function f (x) if there is a discontinuity in any higher derivative of f (x), i.e. if there is lack of smoothness, the above arguments apply again ⇔ any discontinuity in any higher derivative of f (x) prevents us from expressing f (x) in terms of (1) for some finite n

bottom line

assume a periodic function f (x) if there is a discontinuity in any higher derivative of f (x), i.e. if there is lack of smoothness, the above arguments apply again ⇔ any discontinuity in any higher derivative of f (x) prevents us from expressing f (x) in terms of (1) for some finite n

⇔ it takes very high frequencies to form edges

example (1): sharp edges require high frequencies

f (x)

f (x)

1

0

1

0

1

0

0

n=1

1

n=3 f (x)

f (x)

1

0

1

0

1

n=5

0

0

1

n=9

a function f (x) and Fourier series approximations using n = 1, 3, 5, 9

example (2): sharp edges require high frequencies

g(x)

g(x)

1

0

1

0

1

0

0

n=1

1

n=3 g(x)

g(x)

1

0

1

0

1

n=5

0

0

1

n=9

a function g(x) and Fourier series approximations using n = 1, 3, 5, 9

example (3): sharp edges require high frequencies

h(x)

h(x)

1

0

1

0

1

0

0

n=1

1

n=3 h(x)

h(x)

1

0

1

0

1

n=5

0

0

1

n=9

a function h(x) and Fourier series approximations using n = 1, 3, 5, 9

conclusion

putting aside mathematical rigor, we deduce that . . .

conclusion

putting aside mathematical rigor, we deduce that . . . to represent any periodic signal of period 1, we must consider infinite Fourier series

f (x) =

∞ X k=−∞

"Z

1

# f (x) e−i 2π k x dx ei 2π k x

0

(2)

question does (2) exists ? ⇔ can we compute (2) ? ⇔ does the series in (2) converge ?

question does (2) exists ? ⇔ can we compute (2) ? ⇔ does the series in (2) converge ?

answer let’s see . . .

claim (without proof)

suppose a periodic function f (x) of period 1 suppose the hypothesis of finite energy Z1

f (x) 2 dx < ∞

0

is valid for f (x), i.e. suppose f (x) ∈ L2

claim (without proof)

suppose a periodic function f (x) of period 1 suppose the hypothesis of finite energy Z1

f (x) 2 dx < ∞

0

is valid for f (x), i.e. suppose f (x) ∈ L2 then 2 Z1 X n i 2π k x ˆ f (k) e − f (x) dx −−−→ 0 n→∞ 0

k=−n

question what is the significance of the Fourier coefficient ˆf (k) =

Z1 f (x) e−i 2π k x dx 0

question what is the significance of the Fourier coefficient ˆf (k) =

Z1 f (x) e−i 2π k x dx 0

answer let’s see . . .

recall

the set C[a, b] of complex functions defined over the interval [a, b] forms a vector space

there is an inner product for f (x), g(x) ∈ C[a, b] Zb



f (x), g(x) = f (x) ¯ g(x) dx a

note

since f (x), e−i 2π k x ∈ C[0, 1]

note

since f (x), e−i 2π k x ∈ C[0, 1], the Fourier coefficient

ˆf (k) =

Z1 f (x) e−i 2π k x dx 0

Z1 f (x) ei 2π k x dx

= 0

= f (x), ei 2π k x

is an inner product

note

we have Z1 ei 2π k x e−i 2π m x dx

i 2π k x i 2π m x e ,e = 0

Z1 ei 2π (k−m) x dx

= 0

=

 1  1 i 2π (k−m) x  =0   i 2π (k−m) e 0 1

Z1 Z1    0  e dx = dx = x = 1 0

0

0

if k 6= m if k = m

note

in other words

i 2π k x i 2π m x e ,e =



0 if k 6= m 1 if k = m

note

in other words

i 2π k x i 2π m x e ,e =



0 if k 6= m 1 if k = m

⇒ the family of complex exponentials

ei 2π k x k∈Z

constitutes an orthonornmal basis of L2 [0, 1]

note

therefore, the Fourier series f (x) =

∞ X k=−∞

ˆf (x) ei 2π k x =

∞ X

f (x), ei 2π k x ei 2π k x

k=−∞

is a way of expressing f (x) in terms of the basis provided by the complex exponentials

note

this is a Hilbert space generalization of the familiar 2D case where, for a vector x = [x1 , x2 ]T , we discussed earlier that 2 X





x = x1 e1 + x2 e2 = x, e1 e1 + x, e2 e2 = x, ei ei i=1

note

this is a Hilbert space generalization of the familiar 2D case where, for a vector x = [x1 , x2 ]T , we discussed earlier that 2 X





x = x1 e1 + x2 e2 = x, e1 e1 + x, e2 e2 = x, ei ei i=1

more on this later!

the Fourier transform – derivation part 3

question what if f (x) is not periodic ?

question what if f (x) is not periodic ?

answer let’s see . . .

from Fourier series to Fourier transforms

in what follows, we will put aside mathematical rigor yet, if need be, we can make our arguments precise

(that is, we will now transit from Fourier series to Fourier transforms and our arguments largely appeal to intuition but we can rest assured that our results hold up to scrutiny)

key idea

view non-periodic (square integrable) functions over R as limiting cases of periodic ones that is, consider functions of period T and then let T → ∞

key idea

view non-periodic (square integrable) functions over R as limiting cases of periodic ones that is, consider functions of period T and then let T → ∞

this will lead to the Fourier transform and we will find that the Fourier transform generalizes the Fourier coefficient the inverse Fourier transform generalizes the Fourier series

transition to the Fourier transform (1)

if f (x) is of period T, the Fourier series becomes

f (x) =

∞ X k=−∞

ˆf (k) ei 2π k Tx =

∞ X k=−∞

ˆf (k) ei 2πT k x

transition to the Fourier transform (2)

if f (x) is of period T, the Fourier coefficient becomes

ˆf (k) = 1 T

ZT f (x) e−i

2π k T

x

dx

0

Z

T/2

1 = T

f (x) e−i −T/2

2π k T

x

dx

question why does the factor

1 T

appear ?

question why does the factor

1 T

appear ?

answer observe that ZT ZT

i 2π k x i 2π k x 2π k 2π k e T ,e T = ei T x e−i T x dx = 1 dx = T 0

⇒ the factor normalizes the inner product

0

transition to the Fourier transform (3)

introducing the angular frequency ω=

2π k T

we have k=

ωT 2π

transition to the Fourier transform (3)

introducing the angular frequency ω=

2π k T

we have k=

ωT 2π

we also have ∆k =

∆ωT 2π



∆ω =

2π ∆k T

transition to the Fourier transform (4)

as a function of ω, the Fourier coefficient becomes Z

T/2

˜f (ω) = 1 T

f (x) e−i ω x dx −T/2

transition to the Fourier transform (5.1)

in terms of ω, the Fourier series can be written as

f (x) =

∞ X k=−∞

˜f (ω) ei ω x

transition to the Fourier transform (5.2)

observe that k = −∞ . . . ∞ so that ω=−

2π ∞ 2π ∞ ... T T

transition to the Fourier transform (5.2)

observe that k = −∞ . . . ∞ so that ω=−

2π ∞ 2π ∞ ... = −∞ . . . ∞ T T

therefore f (x) =

∞ X k=−∞

˜f (ω) ei ω x =

∞ X ω=−∞

˜f (ω) ei ω x

transition to the Fourier transform (5.3)

observe that

f (x) =

∞ X

˜f (ω) ei ω x

ω=−∞ T/2 Z ∞ X 1 = f (x) e−i ω x dx ei ω x · 1 T ω=−∞ −T/2

T/2 Z ∞ X 1 f (x) e−i ω x dx ei ω x ∆k = T ω=−∞ −T/2

transition to the Fourier transform (5.4)

observe that T/2 Z ∞ X 1 f (x) = f (x) e−i ω x dx ei ω x ∆k T ω=−∞ −T/2

T/2 Z ∞ X 1 ∆ω T = f (x) e−i ω x dx ei ω x T 2π ω=−∞ −T/2

T/2 Z ∞ 1 X = f (x) e−i ω x dx ei ω x ∆ω 2π ω=−∞ −T/2

transition to the Fourier transform (5.5)

recall that ∆ω depends on T

∆ω =

2π ∆k 2π = T T

transition to the Fourier transform (6)

finally, letting T → ∞, we obtain

f (x) =

1 2π

∞ Z



∞ Z

f (x) e−i ω x dx ei ω x dω

 −∞

1 =√ 2π

∞ Z

−∞



−∞

  √1 2π

∞ Z

 f (x) e−i ω x dx ei ω x dω

−∞

definition Fourier transform of f (x)

1 F(ω) = √ 2π

∞ Z

f (x) e−i ω x dx −∞

inverse Fourier transform of F(ω)

1 f (x) = √ 2π

∞ Z

F(ω) ei ω x dω −∞

interpretation

the Fourier transform analyzes f (x) into its constituent parts — a continuous family of complex exponentials the inverse Fourier transform synthesizes f (x) from the constituent parts

summary

we now know about

the fact that Fourier coefficients are inner products the fact that Fourier series are basis expansions the Fourier transform and its inverse

exercises

read C. Bauckhage, Lecture Notes on Fourier Transforms (III), dx.doi.org/10.13140/2.1.4787.0245 C. Bauckhage, Lecture Notes on Fourier Transforms (IV), dx.doi.org/

exercises

assuming that f (x) is periodic of period T, show that ZT f (x) e 0

Z

T/2

−i

2π k T

x

f (x) e−i

dx = −T/2

2π k T

x

dx