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