f k k = k , f k f ( ) k = k - - PowerPoint PPT Presentation

DISCRETE FOURIER TRANSFORM

Lesson 4

• We saw last lecture that Fourier series can be thought of as a projection onto the

complex exponentials, using the continuous 2 inner product: f(θ) ∼

∞

• k=−∞

ˆ fkkθ =

∞

• k=−∞
• kθ, f
• kθ
• We will see now that the approximate Fourier series can also be thought of as a

projection onto the complex exponentials, using the discretized inner product: fα,β(θ) =

β

• k=α

ˆ f m

k kθ = β

• k=α
• kθ, f
• m kθ

Recall m = β − α + 1 so that the number of coefficients match the number

• f quadrature points
• We will further view the map from the values

to the approxi- mate coefficients as a linear operator (i.e., matrix), which is called the discrete Fourier transform

• This will be used to see that the approximate Fourier series interpolates

at the quadrature points

• We saw last lecture that Fourier series can be thought of as a projection onto the

complex exponentials, using the continuous 2 inner product: f(θ) ∼

∞

• k=−∞

ˆ fkkθ =

∞

• k=−∞
• kθ, f
• kθ
• We will see now that the approximate Fourier series can also be thought of as a

projection onto the complex exponentials, using the discretized inner product: fα,β(θ) =

β

• k=α

ˆ f m

k kθ = β

• k=α
• kθ, f
• m kθ

Recall m = β − α + 1 so that the number of coefficients match the number

• f quadrature points
• We will further view the map from the values f(θ1), . . . , f(θm) to the approxi-

mate coefficients f m

α , . . . , f m β as a linear operator (i.e., matrix), which is called the

discrete Fourier transform

• This will be used to see that the approximate Fourier series interpolates f at the

quadrature points

• Recall the discrete orthogonal properties of complex exponentials:
• kθ, jθ

m = (−1)j−k

for j − k = . . . , −2m, −m, 0, m, 2m, . . .

• kθ, jθ

m = 0

• therwise
• When we restrict our attention to αθ, . . . , βθ this implies that the complex

exponentials are orthogonal (assuming that m ≥ β − α + 1)

• Thus we get that the approximate Fourier series is a projection

Because we are in the finite dimensional setting, we immediately see that

β

• k=α
• kθ, fα,β
• kθ = fα,β(θ)

EXAMPLE

• 3
• 2
• 1

1 2 3 q

• 1.0
• 0.5

0.5 1.0

m = 5

5θ = 5θ + −5θ 2

• 3
• 2
• 1

1 2 3 q

• 1.0
• 0.5

0.5 1.0

m = 5

Re Im True function Approximate Fourier series

• For f(θ) = 5θ, we have

2

• k=−2
• kθ, f
• 5 kθ = −1
• Increasing

to we still haven't resolved

• Finally, for

we resolve the function

EXAMPLE

5θ = 5θ + −5θ 2

Re Im

• 3
• 2
• 1

1 2 3 q

• 1.0
• 0.5

0.5 1.0

m = 10

• 3
• 2
• 1

1 2 3 q

• 1.0
• 0.5

0.5 1.0

m = 10

• For f(θ) = 5θ, we have

2

• k=−2
• kθ, f
• 5 kθ = −1
• Increasing m to 10 we still haven't resolved

4

• k=−5
• kθ, f
• 10 kθ = −5θ
• Finally, for

we resolve the function

EXAMPLE

5θ = 5θ + −5θ 2

Re Im

• 3
• 2
• 1

1 2 3 q

• 1.0
• 0.5

0.5 1.0

m = 100

• 3
• 2
• 1

1 2 3 q

• 1.0
• 0.5

0.5 1.0

m = 100

• For f(θ) = 5θ, we have

2

• k=−2
• kθ, f
• 5 kθ = −1
• Increasing m to 10 we still haven't resolved

4

• k=−5
• kθ, f
• 10 kθ = −5θ
• Finally, for m > 10 we resolve the function

49

• k=−50
• kθ, f
• 100 kθ = f(θ)

DISCRETE FOURIER TRANSFORM

• We want to interpret the map from (exact) values

f m :=

• f(θ1)

. . . f(θm)

• to approximate Fourier coefficients

ˆ f α,β :=

• ˆ

f m

α

. . . ˆ f m

β

• as a linear operator
• We can rewrite the approximate coefficients in terms of

:

• We want to interpret the map from (exact) values

f m :=

• f(θ1)

. . . f(θm)

• to approximate Fourier coefficients

ˆ f α,β :=

• ˆ

f m

α

. . . ˆ f m

β

• as a linear operator
• We can rewrite the approximate coefficients in terms of f m:

ˆ f m

k

=

• kθ, f
• m = 1

m

m

• j=1

f(θj)−kθj = 1 m

• −kθ1, . . . , −kθm

f m

• It follows that

ˆ f α,β = Fα,βf m for the discrete Fourier transform matrix Fα,β := 1 m    −αθ1 · · · −αθm . . . ... . . . −βθ1 · · · −βθm   

• We want to interpret the map from (exact) values

f m :=

• f(θ1)

. . . f(θm)

• to approximate Fourier coefficients

ˆ f α,β :=

• ˆ

f m

α

. . . ˆ f m

β

• as a linear operator
• We can rewrite the approximate coefficients in terms of f m:

ˆ f m

k

=

• kθ, f
• m = 1

m

m

• j=1

f(θj)−kθj = 1 m

• −kθ1, . . . , −kθm

f m

INTERPOLATION

• We say that g interpolates f at the points θ = (θ1, . . . , θm) if

f(θ) = g(θ)

• In other words, f(θi) = g(θi)

for i = 1, . . . , m

• Problem: choose coefficients

so that

• interpolates

at

• We say that g interpolates f at the points θ = (θ1, . . . , θm) if

f(θ) = g(θ)

• In other words, f(θi) = g(θi)

for i = 1, . . . , m

• Problem: choose coefficients ck so that

g(θ) =

β

• k=α

ckkθ interpolates f at θ

• Represent the coefficients as a length m vector:

c =

• cα

. . . cβ

• The Vandermonde matrix is the map from coefficients to values at θ:

V =

• αθ | · · · | βθ
• =
• αθ1

· · · βθ1 . . . ... . . . αθm · · · βθm

• Therefore,
• What coefficients interpolate

at ? Precisely

• Represent the coefficients as a length m vector:

c =

• cα

. . . cβ

• The Vandermonde matrix is the map from coefficients to values at θ:

V =

• αθ | · · · | βθ
• =
• αθ1

· · · βθ1 . . . ... . . . αθm · · · βθm

• Therefore,

g(θ) =

β

• k=α

ckkθ = V c

• What coefficients interpolate f at θ? Precisely

c = V −1f(θ)

• We now show that F = V −1
• In other words, the discrete Fourier transform interpolates f at the points θ
• This follows immediately:

FV = 1 m

• −θ
• .

. . −θ

• θ | · · · | θ
• .

. . ... . . .

• .

. . ... . . .

• We now show that F = V −1
• In other words, the discrete Fourier transform interpolates f at the points θ
• This follows immediately:

FV = 1 m

• −θ
• .

. . −θ

• θ | · · · | θ
• = 1

m

• θ

θ · · ·

• θ

θ . . . ... . . .

• θ

θ · · ·

• θ

θ

• .

. . ... . . .

• We now show that F = V −1
• In other words, the discrete Fourier transform interpolates f at the points θ
• This follows immediately:

FV = 1 m

• −θ
• .

. . −θ

• θ | · · · | θ
• = 1

m

• θ

θ · · ·

• θ

θ . . . ... . . .

• θ

θ · · ·

• θ

θ

• =
• ,

m

· · ·

• ,

m

. . . ... . . .

• ,

m

· · ·

• ,

m

• = I

ALIASING FORMULA

• Define the following p norms:

f1 =

• k=

|fk| , f2 =

• k=

|fk|2, f∞ =

• <k<

|fk| for f =

• . . . , f1, f0, f1, . . .

C More generally, we can define

• The norm defines the

spaces:

• if

Absolutely converging Fourier series

• Define the following p norms:

f1 =

• k=

|fk| , f2 =

• k=

|fk|2, f∞ =

• <k<

|fk| for f =

• . . . , f1, f0, f1, . . .

C More generally, we can define fp =

• k=

|fk|p 1/p

• The norm defines the p spaces:

f p if fp <

Absolutely converging Fourier series

• Provided ˆ

f ∈ 1, we know that the Fourier series converges pointwise f() =

∞

• k=−∞

ˆ fkkθ We will come back to the technicalities of this statement

• We then have the following

ˆ f m

k

=

• kθ, f
• m =
• kθ,

∞

• j=−∞

ˆ fjjθ

• m

∞

• Provided ˆ

f ∈ 1, we know that the Fourier series converges pointwise f() =

∞

• k=−∞

ˆ fkkθ We will come back to the technicalities of this statement

• We then have the following

ˆ f m

k

=

• kθ, f
• m =
• kθ,

∞

• j=−∞

ˆ fjjθ

• m

=

∞

• j=−∞

ˆ fj

• kθ, jθ

m

• Provided ˆ

f ∈ 1, we know that the Fourier series converges pointwise f() =

∞

• k=−∞

ˆ fkkθ We will come back to the technicalities of this statement

• We then have the following

ˆ f m

k

=

• kθ, f
• m =
• kθ,

∞

• j=−∞

ˆ fjjθ

• m

=

∞

• j=−∞

ˆ fj

• kθ, jθ

m

= · · · + ˆ fk−2m + (−1)m ˆ fk−m + ˆ fk + (−1)m ˆ fk+m + fk+2m + · · ·

STANDARDIZED FORM

(α, β) Fα,β α = −m/2

• β = m/2 − 1

m α = −(m − 1)/2

• β = (m − 1)/2

m mF0,m−1 Fα,β F0,m−1

• Recall the expression

ˆ f m

k = · · · + ˆ

fk−2m + (−1)m ˆ fk−m + ˆ fk + (−1)m ˆ fk+m + ˆ fk+2m + · · ·

• It follows that

ˆ f m

k+m = (−1)m ˆ

f m

k

• In other words, for

, times times

• Recall the expression

ˆ f m

k = · · · + ˆ

fk−2m + (−1)m ˆ fk−m + ˆ fk + (−1)m ˆ fk+m + ˆ fk+2m + · · ·

• It follows that

ˆ f m

k+m = (−1)m ˆ

f m

k

• In other words, for α < 0 < β,

times times

• Recall the expression
• It follows that
• In other words, for

, Fα,βf =

• ˆ

f m

α

. . . ˆ f m

−1

ˆ f m . . . ˆ f m

β

• =
• (−1)m ˆ

f m

α+m

. . . (−1)m ˆ f m

m−1

ˆ f m . . . ˆ f m

β

• times

times

• Recall the expression

ˆ f m

k = · · · + ˆ

fk−2m + (−1)m ˆ fk−m + ˆ fk + (−1)m ˆ fk+m + ˆ fk+2m + · · ·

• It follows that

ˆ f m

k+m = (−1)m ˆ

f m

k

• In other words, for α < 0 < β,

Fα,βf =

• ˆ

f m

α

. . . ˆ f m

−1

ˆ f m . . . ˆ f m

β

• =
• (−1)m ˆ

f m

α+m

. . . (−1)m ˆ f m

m−1

ˆ f m . . . ˆ f m

β

• =
• (−1)m ˆ

f m

β+1

. . . (−1)m ˆ f m

m−1

ˆ f m . . . ˆ f m

β

• times

times

• Recall the expression

ˆ f m

k = · · · + ˆ

fk−2m + (−1)m ˆ fk−m + ˆ fk + (−1)m ˆ fk+m + ˆ fk+2m + · · ·

• It follows that

ˆ f m

k+m = (−1)m ˆ

f m

k

• In other words, for α < 0 < β,

Fα,βf =

• ˆ

f m

α

. . . ˆ f m

−1

ˆ f m . . . ˆ f m

β

• =
• (−1)m ˆ

f m

α+m

. . . (−1)m ˆ f m

m−1

ˆ f m . . . ˆ f m

β

• =
• (−1)m ˆ

f m

β+1

. . . (−1)m ˆ f m

m−1

ˆ f m . . . ˆ f m

β

• =

β+1 times

• −α times
• . . .

(−1)m ... (−1)m 1 ... 1

• F0,m−1