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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
IntroductionWeighted Frames
ExamplesMore Examples
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
IntroductionWeighted Frames
ExamplesMore Examples
Conclusions
DefinitionsGabor frameMatlab realisation
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 ).
Anna Grybos http://nuhag.eu
IntroductionWeighted Frames
ExamplesMore Examples
Conclusions
DefinitionsGabor frameMatlab realisation
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
Anna Grybos http://nuhag.eu
IntroductionWeighted Frames
ExamplesMore Examples
Conclusions
DefinitionsGabor frameMatlab realisation
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.
Anna Grybos http://nuhag.eu
IntroductionWeighted Frames
ExamplesMore Examples
Conclusions
DefinitionsGabor frameMatlab realisation
From H to Cn
Redundancy
Lattice constants a and b chosen so that
1 ≤ red =na ·
nb
n=
n
ab
Anna Grybos http://nuhag.eu
IntroductionWeighted Frames
ExamplesMore Examples
Conclusions
DefinitionsGabor frameMatlab realisation
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
Anna Grybos http://nuhag.eu
From H to Cn
IntroductionWeighted Frames
ExamplesMore Examples
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?
Anna Grybos http://nuhag.eu
IntroductionWeighted Frames
ExamplesMore Examples
Conclusions
Weighted FrameTheorem
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
Anna Grybos http://nuhag.eu
IntroductionWeighted Frames
ExamplesMore Examples
Conclusions
Weighted FrameTheorem
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
Anna Grybos http://nuhag.eu
Weighted Frame
Weighted Frame
For w = const the reverse weight x = 1/w .
IntroductionWeighted Frames
ExamplesMore Examples
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
Anna Grybos http://nuhag.eu
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
IntroductionWeighted Frames
ExamplesMore Examples
Conclusions
SettingsExample 1Example 2Example 3
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
Anna Grybos http://nuhag.eu
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].
IntroductionWeighted Frames
ExamplesMore Examples
Conclusions
SettingsExample 1Example 2Example 3
Example 2
The difference between DWG and x · DG in second example.
(a, b) (12, 9) (9, 8) (6, 4) (4, 4) (4, 3)
red 1.33 2 6 9 12Froebenius 4.4812 4.5021 3.1142 2.5221 2.1702Op.norm 1.1224 0.8231 0.4236 0.3119 0.2622
Anna Grybos http://nuhag.eu
Example 3
The weights w and x
The weight w has random values from [0.5, 2], x = 1/w ∈ [0.5, 2].
IntroductionWeighted Frames
ExamplesMore Examples
Conclusions
SettingsExample 1Example 2Example 3
Example 3
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 4.4804 4.4310 2.9691 2.5523 2.1905Op.norm 1.0715 0.8371 0.3949 0.3153 0.2713
Anna Grybos http://nuhag.eu
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
IntroductionWeighted Frames
ExamplesMore Examples
Conclusions
Example 4Example 5Example 6
Example 4
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 2.2042 1.5063 0.8476 0.6920 0.5993Op.norm 0.6066 0.3074 0.1278 0.1015 0.0879
Anna Grybos http://nuhag.eu
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
IntroductionWeighted Frames
ExamplesMore Examples
Conclusions
Example 4Example 5Example 6
Example 5
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 1.0747 0.7547 0.4042 0.2973 0.2762Op.norm 0.2946 0.1755 0.0618 0.0353 0.0390
Anna Grybos http://nuhag.eu
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].
IntroductionWeighted Frames
ExamplesMore Examples
Conclusions
Example 4Example 5Example 6
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 .
The difference between DWG and x · DG
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
Anna Grybos http://nuhag.eu
Example 6
IntroductionWeighted Frames
ExamplesMore Examples
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.
Anna Grybos http://nuhag.eu
IntroductionWeighted Frames
ExamplesMore Examples
Conclusions
Thank you for your attention!
Anna Grybos http://nuhag.eu