[PPT] - , f m . . F , f = . PowerPoint Presentation

SLIDE 1

FAST FOURIER TRANSFORM

Lesson 4

SLIDE 2

We want to reinvestigate to the computation of

Fα,βf =

αθ, f
m

. . .

βθ, f
m
The DFT is an O
m2

algorithm, we instead want an O(m m) algorithm

The algorithm is based on carefully splitting the inner products for m = 2q in terms
f inner products for 2q−1 and iterating
We will also discuss convergence of the approximate Fourier series

SLIDE 3

SPLITTING INNER PRODUCTS

SLIDE 4

zj = θj = −ωj−1

m

m ωm = 2π/m m = 2n ωn = ω2

m

SLIDE 5

zj = θj = −ωj−1

m

m ωm = 2π/m m = 2n ωn = ω2

m

kθ, f
m = f(θ1)zk

1 + · · · + f(θm)zk m

SLIDE 6

zj = θj = −ωj−1

m

m ωm = 2π/m m = 2n ωn = ω2

m

kθ, f
m = f(θ1)zk

1 + · · · + f(θm)zk m

= (−1)k f(θ1) + f(θ2)ωk

m + f(θ3)ω2k m + · · · + f(θm)ω(m−1)k m

SLIDE 7

zj = θj = −ωj−1

m

m ωm = 2π/m m = 2n ωn = ω2

m

kθ, f
m = f(θ1)zk

1 + · · · + f(θm)zk m

= (−1)k f(θ1) + f(θ2)ωk

m + f(θ3)ω2k m + · · · + f(θm)ω(m−1)k m

= (−1)k

f(θ1) + f(θ3)ω2k

m + · · · + f(θm−1)ω(m−2)k m

+

(−1)kωk

m

f(θ2) + f(θ4)ω2k

m + · · · + f(θm)ω(m−2)k m

SLIDE 8

zj = θj = −ωj−1

m

m ωm = 2π/m m = 2n ωn = ω2

m

kθ, f
m = f(θ1)zk

1 + · · · + f(θm)zk m

= (−1)k f(θ1) + f(θ2)ωk

m + f(θ3)ω2k m + · · · + f(θm)ω(m−1)k m

= (−1)k

f(θ1) + f(θ3)ω2k

m + · · · + f(θm−1)ω(m−2)k m

+

(−1)kωk

m

f(θ2) + f(θ4)ω2k

m + · · · + f(θm)ω(m−2)k m

= (−1)k

f(θ1) + f(θ3)ωk

n + · · · + f(θm−1)ω(n−1)k n

+

(−1)kωk

m

f(θ2) + f(θ4)ωk

n + · · · + f(θm)ω(n−1)k n

SLIDE 9

zj = θj = −ωj−1

m

m ωm = 2π/m m = 2n ωn = ω2

m

kθ, f
m = f(θ1)zk

1 + · · · + f(θm)zk m

= (−1)k f(θ1) + f(θ2)ωk

m + f(θ3)ω2k m + · · · + f(θm)ω(m−1)k m

= (−1)k

f(θ1) + f(θ3)ω2k

m + · · · + f(θm−1)ω(m−2)k m

+

(−1)kωk

m

f(θ2) + f(θ4)ω2k

m + · · · + f(θm)ω(m−2)k m

= (−1)k

f(θ1) + f(θ3)ωk

n + · · · + f(θm−1)ω(n−1)k n

+

(−1)kωk

m

f(θ2) + f(θ4)ωk

n + · · · + f(θm)ω(n−1)k n

= n
kθ, f
n + ωk

m

kθ, f
θ + 2π

m

n

SLIDE 10

fh(θ) = f(θ + h 2π

m )

1, fm 1, f0n

1, f1n ,

SLIDE 11

fh(θ) = f(θ + h 2π

m )

1, fm 1, f0n

1, f1n ,
θ, f
m
θ, f0
n
θ, f1
n ,

SLIDE 12

fh(θ) = f(θ + h 2π

m )

1, fm 1, f0n

1, f1n ,
θ, f
m
θ, f0
n
θ, f1
n ,

. . .

(m−1)θ, f
m
(m−1)θ, f0
n
(m−1)θ, f1
n ,

m 2m

SLIDE 13

fh(θ) = f(θ + h 2π

m )

1, fm 1, f0n

1, f1n ,
θ, f
m
θ, f0
n
θ, f1
n ,

. . .

(m−1)θ, f
m
(m−1)θ, f0
n
(m−1)θ, f1
n ,

m 2m k = n

nθ, fh
n = (1)n 1, fhn

SLIDE 14

fh(θ) = f(θ + h 2π

m )

1, fm 1, f0n

1, f1n ,
θ, f
m
θ, f0
n
θ, f1
n ,

. . .

(m−1)θ, f
m
(m−1)θ, f0
n
(m−1)θ, f1
n ,

m 2m k = n

nθ, fh
n = (1)n 1, fhn

n 2n = m

kθ, f0
n
kθ, f1
n
k = 0, . . . , n 1

SLIDE 15

Now consider the case where m = 2q
We saw that the m constants
kθ, f
2q depend on the m constants
kθ, f0
2q−1

and

kθ, f1
2q−1

for k = 0, . . . , 2q−1 1 At the next level, we see that a similar expression holds true: 2q−1 kθ, fh

2q−1 = 2q−2

kθ, fh

2q−1 + ωk

2q

kθ, fh+2
2q−1
And so on, until we get the

constants and

SLIDE 16

Now consider the case where m = 2q
We saw that the m constants
kθ, f
2q depend on the m constants
kθ, f0
2q−1

and

kθ, f1
2q−1

for k = 0, . . . , 2q−1 1 At the next level, we see that a similar expression holds true: 2q−1 kθ, fh

2q−1 = 2q−2

kθ, fh

2q−1 + ωk

2q

kθ, fh+2
2q−1
And so on, until we get the m constants

1, f01 = f(θ1), · · · and 1, fm1 = f(θm)

SLIDE 17

m

kθ, fj
2q−r
k = 0, . . . , 2q−r − 1, j = 0, 2r − 1

m

kθ, fj
2q−r−1
k = 0, . . . , 2q−r−1 − 1, j = 0, 2r+1 − 1

c q (cm)q = cm 2 m

SLIDE 18

CONVERGENCE OF FOURIER SERIES

SLIDE 19

We state (without rigorous proof) some standard properties of Fourier series
If f 2, then the Fourier series converges in norm:
f

n

k=−n

ˆ fkk

2
If

, then , and the norms are equal:

In other words:

is equivalent to

SLIDE 20

We state (without rigorous proof) some standard properties of Fourier series
If f 2, then the Fourier series converges in norm:
f

n

k=−n

ˆ fkk

2
If f 2, then ˆ

f 2, and the norms are equal: f2 =

∞
k=−∞

ˆ fkk,

∞

j=−∞

ˆ fjj

∞
∞
In other words:

is equivalent to

SLIDE 21

We state (without rigorous proof) some standard properties of Fourier series
If f 2, then the Fourier series converges in norm:
f

n

k=−n

ˆ fkk

2
If f 2, then ˆ

f 2, and the norms are equal: f2 =

∞
k=−∞

ˆ fkk,

∞

j=−∞

ˆ fjj

=

∞

k,j=−∞

ˆ fk ˆ fj

k, j

=

∞

k=−∞
ˆ

fk

2

In other words: is equivalent to

SLIDE 22

We state (without rigorous proof) some standard properties of Fourier series
If f 2, then the Fourier series converges in norm:
f

n

k=−n

ˆ fkk

2
If f 2, then ˆ

f 2, and the norms are equal: f2 =

∞
k=−∞

ˆ fkk,

∞

j=−∞

ˆ fjj

=

∞

k,j=−∞

ˆ fk ˆ fj

k, j

=

∞

k=−∞
ˆ

fk

2

=

ˆ

f

2

In other words: f 2 is equivalent to ˆ f 2

SLIDE 23

If ˆ

f ∈ 1 then the Fourier series converges to a continuous function ˜ f This continuous function satisfies

f − ˜

f

2 = 0
Therefore: f continuous and ˆ

f ∈ 1 implies pointwise convergence of Fourier series to f

SLIDE 24

We now show the approximate Fourier series also converges pointwise provided

that ˆ f ∈ 1

We have

f() − fα,β() =

∞

k=−∞

ˆ fkkθ −

β

k=α

ˆ f m

k kθ ∞

β
Each term is bounded by 2, thus we have
Since

is bounded, this sum tends to zero as

SLIDE 25

We now show the approximate Fourier series also converges pointwise provided

that ˆ f ∈ 1

We have

f() − fα,β() =

∞

k=−∞

ˆ fkkθ −

β

k=α

ˆ f m

k kθ

=

∞

k=−∞

ˆ fkkθ −

β

k=α

(· · · + (−1)m ˆ fk−m + ˆ fk + (−1)m ˆ fk+m + · · · )kθ

Each term is bounded by 2, thus we have
Since

is bounded, this sum tends to zero as

SLIDE 26

We now show the approximate Fourier series also converges pointwise provided

that ˆ f ∈ 1

We have

f() − fα,β() =

∞

k=−∞

ˆ fkkθ −

β

k=α

ˆ f m

k kθ

=

∞

k=−∞

ˆ fkkθ −

β

k=α

(· · · + (−1)m ˆ fk−m + ˆ fk + (−1)m ˆ fk+m + · · · )kθ = · · · + ˆ fα−1((α−1)θ + (−1)mβθ) + ˆ fβ+1((β+1)θ + (−1)mαθ) + ˆ fβ+2((β+2)θ + (−1)m(α+1)θ) + · · ·

Each term is bounded by 2, thus we have
Since

is bounded, this sum tends to zero as

SLIDE 27

We now show the approximate Fourier series also converges pointwise provided

that ˆ f ∈ 1

We have

f() − fα,β() =

∞

k=−∞

ˆ fkkθ −

β

k=α

ˆ f m

k kθ

=

∞

k=−∞

ˆ fkkθ −

β

k=α

(· · · + (−1)m ˆ fk−m + ˆ fk + (−1)m ˆ fk+m + · · · )kθ = · · · + ˆ fα−1((α−1)θ + (−1)mβθ) + ˆ fβ+1((β+1)θ + (−1)mαθ) + ˆ fβ+2((β+2)θ + (−1)m(α+1)θ) + · · ·

Each term is bounded by 2, thus we have

|f() − fα,β()| ≤ 2

α−1
k=−∞
ˆ

fk

+

∞

k=β+1
ˆ

fk

Since ∞

k=−∞

ˆ

fk

is bounded, this sum tends to zero as m → ∞

SLIDE 28

SMOOTHNESS IMPLIES DECAY

SLIDE 29

|θ| (π2 − θ2)4 θ

3
2
1

1 2 3 2000 4000 6000 8000

3
2
1

1 2 3 0.5 1.0 1.5 2.0 2.5 3.0

3
2
1

1 2 3 0.6 0.7 0.8 0.9 1.0

θ

Let's assume that ˆ

fk decays algebraically fast like λ

In particular, assume there's a constant M such that, for all k,

|fk| (|k| + 1)λ < M

Differentiability implies decaying coefficients

SLIDE 30

400
200

200 400 10-16 10-12 10-8 10-4 1

20
10

10 20 10-15 10-11 10-7 0.001

400
200

200 400 10-8 10-5 0.01 10

|θ| (π2 − θ2)4 θ

k

Let's assume that ˆ

fk decays algebraically fast like λ

In particular, assume there's a constant M such that, for all k,

|fk| (|k| + 1)λ < M

Differentiability implies decaying coefficients

SLIDE 31

Denote the class of periodic continuous functions on T by C[T]
Denote the class of k-times differentiable function on T by Ck[T]

In other words, f ∈ Ck[T] if f ∈ C[T], f is differentiable and f ∈ Ck1[T] : If then , and hence its Fourier series converges:

SLIDE 32

Denote the class of periodic continuous functions on T by C[T]
Denote the class of k-times differentiable function on T by Ck[T]

In other words, f ∈ Ck[T] if f ∈ C[T], f is differentiable and f ∈ Ck1[T] : If f ∈ C2[T] then ˆ f ∈ 1, and hence its Fourier series converges: f() =

k=

ˆ fkkθ

=

n n

k=n

ˆ fkkθ

.

SLIDE 33

f is differentiable, so we can integrate by parts (twice):

π

f(θ)kθ θ = − 1 k π

π

f(θ) [kθ]

Since

is continuous

Therefore,

converges.

SLIDE 34

f is differentiable, so we can integrate by parts (twice):

π

f(θ)kθ θ = − 1 k π

π

f(θ) [kθ] = −f(π)kπ − f(−π)kπ k + 1 k π

π

f (θ)kθ θ

Since

is continuous

Therefore,

converges.

SLIDE 35

f is differentiable, so we can integrate by parts (twice):

π

f(θ)kθ θ = − 1 k π

π

f(θ) [kθ] = −f(π)kπ − f(−π)kπ k + 1 k π

π

f (θ)kθ θ = 1 k π

π

f (θ)kθ θ = 1 k2 π

π

f (θ) [kθ]

Since

is continuous

Therefore,

converges.

SLIDE 36

f is differentiable, so we can integrate by parts (twice):

π

f(θ)kθ θ = − 1 k π

π

f(θ) [kθ] = −f(π)kπ − f(−π)kπ k + 1 k π

π

f (θ)kθ θ = 1 k π

π

f (θ)kθ θ = 1 k2 π

π

f (θ) [kθ] = − 1 k2 π

π

f (θ)kθ θ

Since

is continuous

Therefore,

converges.

SLIDE 37

f is differentiable, so we can integrate by parts (twice):

π

f(θ)kθ θ = − 1 k π

π

f(θ) [kθ] = −f(π)kπ − f(−π)kπ k + 1 k π

π

f (θ)kθ θ = 1 k π

π

f (θ)kθ θ = 1 k2 π

π

f (θ) [kθ] = − 1 k2 π

π

f (θ)kθ θ

Since f is continuous
π

π

f (θ)kθ θ

≤

π

|f (θ)| θ < C

Therefore,

converges.

SLIDE 38

f is differentiable, so we can integrate by parts (twice):

π

f(θ)kθ θ = − 1 k π

π

f(θ) [kθ] = −f(π)kπ − f(−π)kπ k + 1 k π

π

f (θ)kθ θ = 1 k π

π

f (θ)kθ θ = 1 k2 π

π

f (θ) [kθ] = − 1 k2 π

π

f (θ)kθ θ

Since f is continuous
π

π

f (θ)kθ θ

≤

π

|f (θ)| θ < C

Therefore,
k=
ˆ

fk

≤
ˆ

f0

+ C

π

k=1

k2 converges.

SLIDE 39

Define the algebraic decaying coefficient space norms p

by

fp

λ =

...

3 2 1 2 3 ...

f
p

=

∞
k=−∞

(|k| + 1)p |fk|p 1/p

If f p

, it means the coefficients must decay

SLIDE 40

Suppose f ∈ C3[T]. Then we can integrate by parts again:

π

f()kθ = − 1 k2 π

π

f ()kθ = 1 k3 π

π

f () [kθ] = f ()kπ − f (−)kπ k3 − 1 k3 π

π

f ()kθ = 1 k3 π

π

f ()kθ = O

k3
In other words f ∈ C3[T] implies Ff ∈ 1

1

By induction (Why?), it follows that f ∈ Cd+2[T] implies Ff ∈ 1

d

Smoothness implies fast decay and fast convergence