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Signal Reconstruction from its Spectrogram. Radu Balan IMAHA 2010, Northern Illinois University, April 24, 2010. Overview. Problem formulation Reconstruction from absolute value of frame coefficients Our approach Embedding into the Hilbert-Schmidt space Discrete Gabor multipliers - PowerPoint PPT Presentation
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Signal Reconstruction from its Spectrogram
Radu Balan
IMAHA 2010, Northern Illinois University, April 24, 2010
2/23
Overview
1. Problem formulation
2. Reconstruction from absolute value of frame coefficients
3. Our approach– Embedding into the Hilbert-Schmidt space– Discrete Gabor multipliers– Quadratic reconstruction
4. Numerical example
3/23
1. Problem formulation
• Typical signal processing “pipeline”:
Analysis Processing SynthesisIn Out
Features:Relative low complexity O(Nlog(N))On-line version if possible
4/23
<·,gi>x
Analysis Synthesis
Ii
iigc ˆc y
Hx
ighc
Ilc
,
)(
i
2
Iiiigcy
Hy
The Analysis/Synthesis Components:
Example: Short-Time Fourier Transform
fkfk gxc ,, , )()( /2, kbtgetg Fiftfk
)(
10,);,(2 ZlH
ZZFfZkfkI F
)()( 2, kbtgetg kbtiffk
5/23
*
=
fft fft
=
*
Data frame index (k)
f
ck,0
ck,F-1ck+1,F-1
ck+1,0
g(t)
x(t+kb:t+kb+M-1) x(t+kb+M:t+kb+2M-1)
x(t+kb)g(t) x(t+(k+1)b)g(t)
6/23
*
=
ifft ifft
=*
ĝ(t)
ck,0
ck,F-1ck+1,F-1
ck+1,0
+
7/23
Problem: Given the Short-Time Fourier Amplitudes (STFA):
we want an efficient reconstruction algorithm:Reduced computational complexityOn-line (“on-the-fly”) processing
1
/2,, )()(,
Mkb
kbt
Fiftfkfk ekbtgtxgxd
|.| Reconstructionck,f dk,f x
8/23
• Where is this problem important:– Speech enhancement– Speech separation– Old recording processing
9/23
• Setup: – H=En , where E=R or E=C
– F={f1,f2,...,fm} a spanning set of m>n vectors
• Consider the map:
• Problem 1: When is N injective?• Problem 2: Assume N is injective, Given c=N(x)
construct a vector y equivalent to x (that is, invert N up to a constant phase factor)
2. Reconstruction from absolute value of frame coefficients
mkk
mn fxxNREN
1
,)( , ~/:
1||scalar somefor ,~ zzyxyx
10/23
Theorem [R.B.,Casazza, Edidin, ACHA(2006)]
For E = R :• if m 2n-1, and a generic frame set F, then N is
injective;• if m2n-2 then for any set F, N cannot be injective;• N is injective iff for any subset JF either J or F\J
spans Rn.• if any n-element subset of F is linearly independent,
then N is injective; for m=2n-1 this is a necessary and sufficient condition.
mkk
mn fxxNRRN
1
,)( , ~/:
1scalar somefor ,~ zzyxyx
11/23
Theorem [R.B.,Casazza, Edidin, ACHA(2006)]
For E = C :• if m 4n-2, and a generic frame set F, then N is
injective.• if m2n and a generic frame set F, then the set of
points in Cn where N fails to be injective is thin (its complement has dense interior).
mkk
mn fxxNRCN
1
,)( , ~/:
1||scalar somefor ,~ zzyxyx
12/23
3. Our approach
• First observation:
fkfkgx
HSgxgxfkfk
ggyyKxxyyK
KKKKtrgxd
fk
fkfk
,,
*2
,2,
,)( , ,)(
,,
,
,,
Hilbert-Schmidt
Signal space: l2(Z)
x KxK
nonlinearembedding
Kgk,f
E=span{Kgk,f}
Hilbert-Schmidt: HS(l2(Z))
FZZFfZkfkIZlH 10,);,( , )(2
Ifkkbtgetg kbtiffk ),( , )()( 2
,Recall:
13/23
• Assume {Kgk,f} form a frame for its span, E. Then the projection PE can be written as:
where {Qk,f} is the canonical dual of {Kgk,f} .
fk
HSfkgE QKP
fk ,,
,,
)(
,)(
,
,,
1,
,
fk
fkfk
gfk
fkg
HSg
KSQ
KKXXSX
Frame operator
14/23
• Second observation:since:
it follows:
)()(: , )()(: where /2
,
bthtThhTthetMhhM
gTMgFit
kffk
** : , : where
,
TXTXXMXMXX
KK gkf
g fk
15/23
• However:
• Explicitely:
0,0,
S and
SSSfk
fk
kbtkbt
Fttif
ttfk QeQ
21
21
21 ,0,0/)(2
,,
16/23
Short digression: Gabor Multipliers
• Goes back to Weyl, Klauder, Daubechies• More recently: Feichtinger (2000), Benedetto-
Pfander (2006), Dörfler-Toressani (2008)
LatticeggmmG
dggm
,)( :Multiplierabor
,m : Multiplier STFT 2
Theorem [F’00] Assume {g , Lattice} is a frame for L2(R).Then the following are equivalent:1. {<.,g>g,Lattice} is a frame for its span, in HS(L2(R));2. {<.,g>g,Lattice} is a Riesz basis for its span, in HS(L2(R));3. The function H does not vanish,
)( , ,)()(2
LatticeDualGroupeggeeHLattice
17/23
• Return to our setting. Let
Theorem Assume {gk,f}(k,f)ZxZF is a frame for l2(Z).
Then
1. is a frame for its span in HS(l2(Z)) iff for each mZF, H(,m) either vanishes identically in , or it is never zero;
2. is a Riesz basis for its span in HS(l2(Z)) iff for each mZF and , H(,m) is never zero.
Zk
fkZf
F
mfki
ggemHF
2
,
2
,),(
Fg ZZfkKfk
),(;,
18/23
• Third observation. Under the following settings:– For translation step b=1;– For window support supp(g)={0,1,2,...,L-1}– For F2L
• The span of is the set
of 2L-1 diagonal band matrices.
Fg ZZfkKfk
),(;,
g
LgLgg
ggg
Lggggg
K g
000000
0)1()1()0(0
00
0)1()1()0(0
0)1()0()1()0()0(0
000000
2
2
2
19/23
• The reproducing condition (i.e. of the projection onto E) implies that Q must satisfy:
bandttXQgXgfk
ttttfkfkfk 21,
,,,,, , and X allfor , ,2121
By working out this condition we obtain:
1
02
2
,0,0)()(
1
d
epgpg
e
FQ
p
ip
it
tt
20/23
• The fourth observation:We are able now to reconstruct up to L-1 diagonals of Kx.
This means we can estimate
11
2,,, Lttttt xxxxx
Assuming we already estimated xs for s<t,we estimate xt by a minimization problem:
2
,
2
1,11,
2ˆˆmin JttxJtJttxtttxx KxxwKxxwKx
for some JL-1 and weights w1,...,wJ.
Remark: This algorithm is similar to Nawab, Quatieri, Lim [’83]IEEE paper.
21/23
Reconstruction Scheme
• Putting all blocks together we get:
IFFT
|ck,0|2
|ck,F-1|2
W0
WL-1
0ˆtz
1ˆ Ltz
LeastSquareSolver
ttx̂
Stage 1 Stage 2
t
t
ttzQzW
,0,01)(
22/23
3. Numerical Example
23/23
Conclusions
All is well but ...
• For nice analysis windows (Hamming, Hanning, gaussian) the set {Kgk,f} DOES NOT form a frame for its span! The lower frame bound is 0. This is the (main) reason for the observed numerical instability!
• Solution: Regularization.