(joint work with Chris Sansing)

Gabor Frames and Directional Time-Frequency Analysis

(joint work with Chris Sansing)

Loukas Grafakos

[email protected]

University of Missouri

http://www.math.missouri.edu/∼loukas

Marriage

Gabor functions

ψ_m,τ(x) = e2πim·xψ(x − τ )

Ridgelets (Candès)

Marriage

Gabor functions

ψ_m,τ(x) = e2πim·xψ(x − τ )

Ridgelets (Candès)

ψa,t,u(x) = a−1/2ψ((x · u − t)/a)

Marriage

Gabor functions

ψ_m,τ(x) = e2πim·xψ(x − τ )

Ridgelets (Candès)

ψa,t,u(x) = a−1/2ψ((x · u − t)/a)

Marriage

Gabor functions

ψ_m,τ(x) = e2πim·xψ(x − τ )

Ridgelets (Candès)

ψa,t,u(x) = a−1/2ψ((x · u − t)/a)

General Idea

Radon transform

Rf (u, t)

General Idea

Radon transform

Rf (u, t)

Time-Frequency Analysis

The Radon Transform

An integral transform which is sensitive to direction.

Rf (u, s) = Z

u·x=s

The Radon Transform

An integral transform which is sensitive to direction.

Rf (u, s) = Z

u·x=s

The Radon Transform

An integral transform which is sensitive to direction.

Rf (u, s) = Z

u·x=s

f (x)dx

The Radon Transform

An integral transform which is sensitive to direction.

Rf (u, s) = Z

u·x=s

f (x)dx

• _{This transform was introduced in 1917 by} Johann Radon.

The Radon Transform

An integral transform which is sensitive to direction.

Rf (u, s) = Z

u·x=s

f (x)dx

• _{This transform was introduced in 1917 by} Johann Radon.

First attempt at a directional TF representation

Let g ∈ S(R) be some window function (i.e. Gaussian). Define

First attempt at a directional TF representation

Let g ∈ S(R) be some window function (i.e. Gaussian). Define

gm,t(s) = e2πim(s−t)g(s − t)

First attempt at a directional TF representation

Let g ∈ S(R) be some window function (i.e. Gaussian). Define

gm,t(s) = e2πim(s−t)g(s − t)

First attempt at a directional TF representation

Let g ∈ S(R) be some window function (i.e. Gaussian). Define

gm,t(s) = e2πim(s−t)g(s − t)

First attempt at a directional TF representation

Let g ∈ S(R) be some window function (i.e. Gaussian). Define

gm,t(s) = e2πim(s−t)g(s − t)

gm,t,u(x) = gm,t(u · x) = ridge functions

Z Sn−1 Z R Z R

First attempt at a directional TF representation

Let g ∈ S(R) be some window function (i.e. Gaussian). Define

gm,t(s) = e2πim(s−t)g(s − t)

The Radon transform presents itself naturally

When we pair our function f ∈ L1 ∩ L2(Rn) with g ∈ S(R) the Radon transform appears:

The Radon transform presents itself naturally

When we pair our function f ∈ L1 ∩ L2(Rn) with g ∈ S(R) the Radon transform appears:

hf, gm,t,ui = hR_uf, gm,ti

hf, gm,t,ui = Z

Rn

f (x)e−2πim(u·x−t)g(u · x − t)dx =

Z

R

Z

u·x=s

Attempt at Continuous Representation

Proposition 1 (Reproduction of the back-projection)

Let f ∈ L1 ∩ L2(Rn). Given functions g, ψ ∈ S(R), we have

Z Sn−1 Z R Z R

hf, gm,t,uiψm,t,udmdtdu = hg, ψi B(f )

where

Bf (x) = Z

Sn−1

Attempt at Continuous Representation

Proposition 1 (Reproduction of the back-projection)

Let f ∈ L1 ∩ L2(Rn). Given functions g, ψ ∈ S(R), we have

Z Sn−1 Z R Z R

hf, gm,t,uiψm,t,udmdtdu = hg, ψi B(f )

where

Bf (x) = Z

Sn−1

Ruf (u · x)du = R∗Rf (x)

Here R∗ is the adjoint of the Radon transform given by

R∗h(x) = Z

Sn−1

Attempt at Continuous Representation

Proposition 1 (Reproduction of the back-projection)

Let f ∈ L1 ∩ L2(Rn). Given functions g, ψ ∈ S(R), we have

Z Sn−1 Z R Z R

hf, gm,t,uiψm,t,udmdtdu = hg, ψi B(f )

where

Bf (x) = Z

Sn−1

Ruf (u · x)du = R∗Rf (x)

Here R∗ is the adjoint of the Radon transform given by

R∗h(x) = Z

Sn−1

h(x · u, u) du

which satisfies for a function h on the cylinder R × Sn−1

What is

B(f )

?

What is

B(f )

?

What is

B(f )

?

B(f ) = R∗R(f ) = c f ∗ | · |−1
(x)

• _{We are “off by a filter.”}

• _{The blurring effect is caused by the weight which amplifies low} (blurry) frequencies and attenuates the higher (sharp)

What is

B(f )

?

Weighted Gabor Ridge Functions

Weighted Gabor Ridge Functions

We need to introduce a filter (weight) to achieve perfect reconstruction. For this reason we define

Gm,t(s) = Dn−1 2 (g

m,t_)(s),

Weighted Gabor Ridge Functions

We need to introduce a filter (weight) to achieve perfect reconstruction. For this reason we define

Gm,t(s) = Dn−1 2 (g

m,t_)(s),

for some g ∈ S(R)

where m, t are real numbers and

Dn−1 2 (g

Weighted Gabor Ridge Functions

We need to introduce a filter (weight) to achieve perfect reconstruction. For this reason we define

Gm,t(s) = Dn−1 2 (g

m,t_)(s),

for some g ∈ S(R)

where m, t are real numbers and

Dn−1 2 (g

m,t_{) = ( dgm,t(σ)|σ|}n−1 2 )∨

Also define the weighted Gabor ridge functions

Weighted Gabor Ridge Functions

We need to introduce a filter (weight) to achieve perfect reconstruction. For this reason we define

Gm,t(s) = Dn−1 2 (g

m,t_)(s),

for some g ∈ S(R)

where m, t are real numbers and

Dn−1 2 (g

m,t_{) = ( dgm,t(σ)|σ|}n−1 2 )∨

Also define the weighted Gabor ridge functions

Exact Continuous Representation

Proposition 2 (Continuous Representation)

Given f ∈ L1 ∩ L2(Rn), and G_m,t,u, Ψ_m,t,u weighted Gabor ridge functions, we have 2f (x) = 1 hg, ψi Z Sn−1 Z R Z R

Exact Continuous Representation

Proposition 2 (Continuous Representation)

Given f ∈ L1 ∩ L2(Rn), and G_m,t,u, Ψ_m,t,u weighted Gabor ridge functions, we have 2f (x) = 1 hg, ψi Z Sn−1 Z R Z R

hf, G_m,t,uiΨ_m,t,udmdtdu

1 hg, ψi Z Sn−1 Z R Z R

hf, G_m,t,uiΨ_m,t,udmdtdu

= 1 hg, ψi Z Sn−1 Z R Z R

hR_uf, Gm,tiΨ_m,t,udmdtdu

= 1 hg, ψi Z Sn−1 Z R Z R hDn−1 2 (Ruf ), g m,t_iD n−1 2 (ψ m,t_{)(u · x)dmdtdu} = Dn−1 2 D n−1 2 Z Sn−1

Filtered Backprojection formula

Filtered Backprojection formula

f (σu)|σ|n−1e2πiσ(u·x)dσdu = 2 Z Rn b f (ξ)e2πiξxdξ = 2f (x)

This continuous representation is valid for any g, ψ ∈ S(R) with

Filtered Backprojection formula

f (σu)|σ|n−1e2πiσ(u·x)dσdu = 2 Z Rn b f (ξ)e2πiξxdξ = 2f (x)

This continuous representation is valid for any g, ψ ∈ S(R) with

Filtered Backprojection formula

f (σu)|σ|n−1e2πiσ(u·x)dσdu = 2 Z Rn b f (ξ)e2πiξxdξ = 2f (x)

This continuous representation is valid for any g, ψ ∈ S(R) with

Filtered Backprojection formula

f (σu)|σ|n−1e2πiσ(u·x)dσdu = 2 Z Rn b f (ξ)e2πiξxdξ = 2f (x)

This continuous representation is valid for any g, ψ ∈ S(R) with

Filtered Backprojection formula

f (σu)|σ|n−1e2πiσ(u·x)dσdu = 2 Z Rn b f (ξ)e2πiξxdξ = 2f (x)

This continuous representation is valid for any g, ψ ∈ S(R) with

Semi-discrete Reproduction

Theorem 3 (Semi-discrete Reproduction)

There exist g, ψ ∈ S(R) and 0 < α, β < 1 such that for f ∈ L1 ∩ L2(Rn)

f = 1 2 Z Sn−1 X m∈Z X t∈Z

Semi-discrete Reproduction

Theorem 3 (Semi-discrete Reproduction)

There exist g, ψ ∈ S(R) and 0 < α, β < 1 such that for f ∈ L1 ∩ L2(Rn)

f = 1 2 Z Sn−1 X m∈Z X t∈Z

hf, Gβm,αt,uiΨβm,αt,udu

Semi-discrete Reproduction

Theorem 3 (Semi-discrete Reproduction)

There exist g, ψ ∈ S(R) and 0 < α, β < 1 such that for f ∈ L1 ∩ L2(Rn)

f = 1 2 Z Sn−1 X m∈Z X t∈Z

hf, Gβm,αt,uiΨβm,αt,udu

Semi-discrete Representation

Semi-discrete Representation

Integrating over Sn−1 _{we deduce} Z Sn−1 D_n−1(R_uf )(u · x) du = Z Sn−1 X m∈Z X t∈Z hDn−1 2 (Ruf ), g αm,βt_iD n−1 2 (ψ αm,βt_{)(u · x) du} or 2f (x) = Z Sn−1 X m∈Z X t∈Z

Semi-discrete Frame Identity

We also have the identity

Semi-discrete Frame Identity

We also have the identity

Semi-discrete Frame Identity

We also have the identity

Examples of Reconstruction

Examples of Reconstruction

Examples of Reconstruction

Examples of Reconstruction

Examples of Reconstruction

Half-filtered Operators

Half-filtered Operators

We introduce the following half-filtered operators: • _{Half-filtered Radon transform}

Rf = Dn−1

2 (Rf (u, ·))(t) = D n−1

Half-filtered Operators

We introduce the following half-filtered operators: • _{Half-filtered Radon transform}

Half-filtered Operators

We introduce the following half-filtered operators: • _{Half-filtered Radon transform}

Rf = Dn−1 2 (Rf (u, ·))(t) = D n−1 2 (Rf )(u, t) • _{Half-filtered Backprojection} R∗g = Z Sn−1 Dn−1 2 (g)(u, u · x) du

Operator/Inverse Operator relationship

R∗

Representation of the associated frame operator

Theorem 4 (Representation formula analogous to Walnut’s)

For f ∈ L1 ∩ L2(Rn), g, ψ ∈ W (R) the frame operator

Sg,ψf = Z Sn−1 X m∈Z X t∈Z

hf, Gβm,αt,uiΨβm,αt,udu

Representation of the associated frame operator

Theorem 4 (Representation formula analogous to Walnut’s)

For f ∈ L1 ∩ L2(Rn), g, ψ ∈ W (R) the frame operator

Sg,ψf = Z Sn−1 X m∈Z X t∈Z

hf, Gβm,αt,uiΨβm,αt,udu

can be written as S_g,ψf = R∗ Qf where Qf = β−1 X r∈Z G_r(s)Rf (s − r β)

Functional Spaces measuring Gabor ridge coefficients

Let 1 ≤ p, q ≤ ∞ and g be a fixed window. Recall the modulation spaces with norm

Functional Spaces measuring Gabor ridge coefficients

Let 1 ≤ p, q ≤ ∞ and g be a fixed window. Recall the modulation spaces with norm

khk_Mp,q(R) = Z R Z R |Vgh(x, ξ)|p dx q/p dξ !1/q = kVghkLp,q(R2)

Functional Spaces measuring Gabor ridge coefficients

Let 1 ≤ p, q ≤ ∞ and g be a fixed window. Recall the modulation spaces with norm

khk_Mp,q(R) = Z R Z R |Vgh(x, ξ)|p dx q/p dξ !1/q = kVghkLp,q(R2)

One is tempted to define spaces the measure the TF coefficients of the Radon transform in the following way:

kf k_Ωp,q,r(Rn) =

Z

kR_uf kr_Mp,q(R) du

1/r

More appropriate definition: Weighted Spaces

Ω

A function f is in the space Ωp,q,r_s (Rn) if the following norm is finite:

More appropriate definition: Weighted Spaces

Ω

A function f is in the space Ωp,q,r_s (Rn) if the following norm is finite:

kf k_Ωp,q,r s (Rn) = Z Sn−1 k( dR_uf ω_cs)∨ kr_Mp,q(R)du 1/r , where ωs = (|σ| n−1 2 (1 + |σ|2) s 2)∨.

Also, we can create a discrete (unweighted) version.

k{a_βm,αt,u}k_ωp,q,r(Z×Z×Sn−1) =

Z

Sn−1

k{a_βm,αt,u}kr_ℓp,qdu

More appropriate definition: Weighted Spaces

Ω

A function f is in the space Ωp,q,r_s (Rn) if the following norm is finite:

kf k_Ωp,q,r s (Rn) = Z Sn−1 k( dR_uf ω_cs)∨ kr_Mp,q(R)du 1/r , where ωs = (|σ| n−1 2 (1 + |σ|2) s 2)∨.

Also, we can create a discrete (unweighted) version.

k{a_βm,αt,u}k_ωp,q,r(Z×Z×Sn−1) =

Z

Sn−1

k{a_βm,αt,u}kr_ℓp,qdu

More appropriate definition: Weighted Spaces

Ω

A function f is in the space Ωp,q,r_s (Rn) if the following norm is finite:

kf k_Ωp,q,r s (Rn) = Z Sn−1 k( dR_uf ω_cs)∨ kr_Mp,q(R)du 1/r , where ωs = (|σ| n−1 2 (1 + |σ|2) s 2)∨.

Also, we can create a discrete (unweighted) version.

Analysis and Synthesis operators

Introduce an analysis operator Cg by

Analysis and Synthesis operators

Introduce an analysis operator Cg by

C_g(f ) = hf, G_βm,αt,ui _m,t,u

and a synthesis operator D_ψ by

Analysis and Synthesis operators

Introduce an analysis operator Cg by

C_g(f ) = hf, G_βm,αt,ui _m,t,u

and a synthesis operator D_ψ by

D_ψ({c_m,t,u}) = Z Sn−1 X m∈Z X t∈Z c_βm,αt,uΨ_βm,αt,udu

Analysis and Synthesis operators

Introduce an analysis operator Cg by

C_g(f ) = hf, G_βm,αt,ui _m,t,u

and a synthesis operator D_ψ by

D_ψ({c_m,t,u}) = Z Sn−1 X m∈Z X t∈Z c_βm,αt,uΨ_βm,αt,udu

Analysis and Synthesis operators

Introduce an analysis operator Cg by

C_g(f ) = hf, G_βm,αt,ui _m,t,u

and a synthesis operator D_ψ by

D_ψ({c_m,t,u}) = Z Sn−1 X m∈Z X t∈Z c_βm,αt,uΨ_βm,αt,udu

• _{The associated frame operator is Sg,ψ = Dψ ◦ Cg.} • _{The operators Cg and Dψ are adjoint to each other.}

Analysis and Synthesis operators

Introduce an analysis operator Cg by

C_g(f ) = hf, G_βm,αt,ui _m,t,u

and a synthesis operator D_ψ by

D_ψ({c_m,t,u}) = Z Sn−1 X m∈Z X t∈Z c_βm,αt,uΨ_βm,αt,udu

Boundedness of the Analysis and Synthesis Operators

Theorem 5 (Boundedness of C_g)

If g ∈ M1,1(R), then C_g : Ωp,q,r → ωp,q,r for 1 ≤ p, q, r ≤ ∞ and

Boundedness of the Analysis and Synthesis Operators

Theorem 5 (Boundedness of C_g)

If g ∈ M1,1(R), then C_g : Ωp,q,r → ωp,q,r for 1 ≤ p, q, r ≤ ∞ and

kCgkΩp,q,r_→ωp,q,r ≤ C(α, β)kVggkW (L1) independently of p, q and r. Here

Boundedness of the Analysis and Synthesis Operators

Theorem 5 (Boundedness of C_g)

If g ∈ M1,1(R), then C_g : Ωp,q,r → ωp,q,r for 1 ≤ p, q, r ≤ ∞ and

kCgkΩp,q,r_→ωp,q,r ≤ C(α, β)kVggkW (L1) independently of p, q and r. Here

kgk_{W (L}1₎ = X k∈Z ess sup x∈[0,1] |g(x + k)| Theorem 6 (Boundedness of D_ψ)

Boundedness of the Frame Operator on

Ω

Corollary  Boundedness of S_g,ψ = D_ψ ◦ C_g 

If g, ψ ∈ M1,1(R), then the semi-discrete frame operator S_g,ψ is bounded on

Ωp,q,r for all 1 ≤ p, q, r ≤ ∞ and α, β > 0 with the following norm estimate:

Frame Extension to

Ω

Theorem 7  Extension of frame to Ωp,q,r  Assume g, ψ ∈ M1,1(R) and

Sg,ψ = I on L2(Rn). Then f = 1 2 Z Sn−1 X m∈Z X t∈Z

Frame Extension to

Ω

Theorem 7  Extension of frame to Ωp,q,r  Assume g, ψ ∈ M1,1(R) and

Sg,ψ = I on L2(Rn). Then f = 1 2 Z Sn−1 X m∈Z X t∈Z

hf, G_βm,αt,uiΨ_βm,αt,udu

Also, there are constants A, B > 0 such that for all f ∈ Ω_p,q,r

Applications

Applications

Applications

Applications

Image Enhancing