Weighted Gabor frames - univie.ac.at · Weighted Gabor frames Anna Grybo´s [email protected] December 6, 2006 Anna Grybo´s [email protected] Weighted Gabor frames

IntroductionWeighted Frames

ExamplesMore Examples

Conclusions

Numerical Harmonic Analysis Group

Weighted Gabor frames

Anna [email protected]

December 6, 2006

Anna Grybos [email protected] Weighted Gabor frames



Conclusions

DefinitionsGabor frameMatlab realisation

Definition

A family (gi )i∈I in a Hilbert space H is called a FRAME if thereexist constants A,B > 0 such that for all f ∈ H

A‖f ‖2 ≤∑i∈I

|〈f , gi 〉|2 ≤ B‖f ‖2 (1)

The frame bounds A and B are infimum and supremum,respectively, of the eigenvalues of the frame operator S , defined as:

Sf =∑i∈I

〈f , gi 〉gi (2)

Anna Grybos http://nuhag.eu



Conclusions


Dual Frame

For any frame dual frames (gi ) exist, allowing the expansion of fas follows:

f =∑i∈I

〈f , gi 〉gi =∑i∈I

〈f , gi 〉gi (3)

The canonical dual frame (gi ) is given by S−1(gi ).




Conclusions


Time-Frequency Shifts

Tx f (t) = f (t − x)

Mωf (t) = f (t) · e2πitω

Gabor Frame

Gabor Frame over the regular lattice Λ = aZd × bZd is the family(gm,n) generated by the shifted versions of one atom g .

gm,n := MmbTnag




Conclusions


From H to Cn

Discrete Gabor Frame is a set of the form:

(gm,k)m,k := {MmbTkag ,m = 1, . . . , n/b, k = 1, . . . , n/a}

Matlab toolbox (Peter Soendergaard)

G = gabbasp(atom, gapt, gapf )

Gabbasp generates a Gabor frame for a given atom and a givenlattice.




Conclusions


From H to Cn

Redundancy

Lattice constants a and b chosen so that

1 ≤ red =na ·

nb

n=

n

ab




Conclusions


From H to Cn

Lattice constants (a, b) for n = 144

(a, b) (12, 9) (9, 8) (6, 4) (4, 4) (4, 3)

red 1.33 2 6 9 12


From H to Cn



Conclusions

Weighted FrameTheorem

Weighted Frame

Weighted Frame

Let G = (gi )i∈I be a frame in Cn and w = {wi ∈ R, i ∈ I} be avector of scalars.

When WG = (wi · gi )i∈I is the frame?




Conclusions


Theorem

Let G = (gi )i∈I be a frame in Cn and let w be a bounded vector ofpositive scalars wi ∈ R, i ∈ I . Then the family WG = (wi · gi )i∈I isalso a frame.

Proof

For w ∈ [r1, r2] and wi > 0 ∀i , r1, r2 ∈ R, 0 < r1 ≤ r2:∑i∈I

|〈f ,wigi 〉|2 =∑i∈I

|wi |2|〈f , gi 〉|2

≤ r22∑i∈I

|〈f , gi 〉|2 ≤ r22B‖f ‖2




Conclusions


Proof continued

On the other hand:∑i∈I

|〈f ,wigi 〉|2 =∑i∈I

|wi |2|〈f , gi 〉|2

≥ r12∑i∈I

|〈f , gi 〉|2 ≥ r12A‖f ‖2

Therefore:

r12A‖f ‖2 ≤

∑i∈I

|〈f ,wigi 〉|2 ≤ r22B‖f ‖2


Weighted Frame

Weighted Frame

For w = const the reverse weight x = 1/w .



Conclusions

SettingsExample 1Example 2Example 3

General settings for the examples:

Atom g is chosen to be Gaussian of length n = 144. The weightw ∈ [0.5, 2] and x = 1/w ∈ [0.5, 2].

Lattice constants (a, b) for n = 144

(a, b) (12, 9) (9, 8) (6, 4) (4, 4) (4, 3)

red 1.33 2 6 9 12


Settings

The difference between DWG and x · DG is measured inFroebenius and Operator norms.

Example 1

The weight w is equal 2 on the 3x3 block and 1 on the rest,w ∈ [0.5, 2] and x = 1/w ∈ [0.5, 2].

Example 1

Example 1

Example 1



Conclusions


Example 1

The difference between DWG and x · DG

(a, b) (12, 9) (9, 8) (6, 4) (4, 4) (4, 3)

red 1.33 2 6 9 12Froebenius 0.8610 0.6788 0.3422 0.2540 0.2069Op.norm 0.5209 0.3591 0.1981 0.1646 0.1340


Example 2

The weight w has values equal 2 on the 3x3 block and randomfrom [0.5, 2] on the rest, x = 1/w ∈ [0.5, 2].



Conclusions


Example 2

The difference between DWG and x · DG in second example.

(a, b) (12, 9) (9, 8) (6, 4) (4, 4) (4, 3)



Example 3

The weights w and x

The weight w has random values from [0.5, 2], x = 1/w ∈ [0.5, 2].



Conclusions


Example 3


(a, b) (12, 9) (9, 8) (6, 4) (4, 4) (4, 3)



Example 4

The weights w and x

The weight w = 0.75 ∗ sin α + 1.25 applied along Time,α ∈ (0, 2π) x = 1/w ∈ [0.5, 2].

Example 4

Example 4



Conclusions

Example 4Example 5Example 6

Example 4


(a, b) (12, 9) (9, 8) (6, 4) (4, 4) (4, 3)



Example 5

The weights w and x

The weight w = 0.75 ∗ sin α + 1.25 applied along Time andFrequency, α ∈ (0, 2π) x = 1/w ∈ [0.5, 2].

Example 5



Conclusions


Example 5


(a, b) (12, 9) (9, 8) (6, 4) (4, 4) (4, 3)



Let’s take a look again at first example.

The weight w was equal 2 on the 3x3 block and 1 on the rest,w ∈ [0.5, 2] and x = 1/w ∈ [0.5, 2].



Conclusions


The weight w was equal 2 on the 3x3 block and rw on the rest,with rw ∈ {1; 1.3; 1.5; 1.75; 1.9} and x = 1/w .


rw 1 1.3 1.5 1.75 1.9

Froebenius 0.8610 0.4654 0.2888 0.1241 0.0458Op.norm 0.5209 0.2766 0.1687 0.0713 0.0265


Example 6



Conclusions

Conclusions

The procedure of finding the reverse weight x by x = 1/w workswell for the weights w constant and ”close” to constant.

The error depends on the range of weight values.

For varying weights another approach should be used — BestApproximation by Gabor Multipliers.




Conclusions

Thank you for your attention!


Documents

Weighted Gabor frames - univie.ac.at · Weighted Gabor frames Anna Grybo´s [email protected] December 6, 2006 Anna Grybo´s [email protected] Weighted Gabor frames