signal reconstruction from its spectrogram
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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.
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