a
i
that's doesn't
a uh as i would like to apologise for in
but do you a low for not be here but the and the
problems coming here today
i your in the and work in the same subject and in a file
a to with much but the and
also so with the U
and uh in this talk okay we will uh
present the some results about
a classical problem problems speced of that addition
and look at the problem uh looking at that vacation and seen that there is a sparse a
the prior available
and we learn a lot of the problem of sparse like the position
we we present some sufficient condition
which allows us to know that the solution is unique
we prove use the would be we i would it would be give a sketch of the proof
and we present our reconstruction algorithm which works we really well
or or or that the position is we one known problem
where are we want to
we want to
uh uh recover a signal X of and no not from a control version of it the but using the
autocorrelation function
which is a convolution between the signal and its time for
we in the same problem in that the main where we have a a you to let this indicates a
for this form
where we have that the correlation function and therefore we have described also value of that well correlation
oh a of the signal
if to a cover the phase information from this uh
um um an information
we can sum or role
is is not really a only my mother got all in there are an application in which this problem is
a a real haven't
first of all of this bunch channel estimation if you have a channel H of fan
uh which is a no we would like to get
uh the closest response in order to improve the communication over channel
and we can only some me data for the input X and and measure the output
if we sum me white noise in the channel using the filter a a we know that
they so uh that what is gonna be that
absolute value of the uh with the of the channel
therefore
if if a cover the face C we got a of the impulse response of the channel
and the problem so
and of the five is a problem or rising up text
and then it back to got we are it as the next expect a lot of E
if where where we have a an cool and want them this and the fact that the shape because the
shape
yeah the term and the time means the function in biology most of the most cases
then for with that been i the monocle and putting it into a crystal a a we that and we
put the quickly a an X ray B
what we measure what we can measure that the faction that of the crystal which by the backs slow it's
that have with us from of the crystal
but the probably as that
if you we if we use for the rest a for the feel more C C D's
we can only measure think this of the light and therefore we lose the "'cause" information if we the domain
and if we can recover the fees we can also recover them
and
uh they original money
yeah a a result is really come on in uh which is a a really one no it's its a
uh this fact that that the addition that the same
we have a the correlation function that the main
uh we've tried to find the remotes and does that the main using for addition
but the problem is uh a how to assign each got or fruits so between the but inside and outside
the unit equal to the signal X of that
here since we assume the signal X of that to be a real their roots up using incredible
if we are aware that our signal is a a i mean phase is a we we can take just
the roots it's set of the unit to call
and we are sure that that of destruction is gonna be unique
is a good solution
otherwise if we don't have this prior about minimum phase a
we can't we have just to
try all the possible combination between inside and outside roots
you know to find our possible solution and therefore for is a common or number of possible solution
and uh there are known all algorithms to find a good so uh the resolution
what we can say that since the the tool because which we are interested rest have a sparse prior because
channel else usually can be modelled using multi about
fading channels which are sparse in the for the main
and also i them sink with still lower
results are sparse because the model just you want to so that is that is a complete
and the for and interesting to understand
which implies that sparsity in that
uh
reconstruction construction up to the an extension of the solution of the set of the vision problem
and moreover if we can say something about the you a be
because without
uh mean of a solution the number of solutions can or you want to avoid this situation
a a small hint about why spice D can be helpful can be here from these example when we have
a sparse signal
made of for example you have four components out of them
and we
a good the that the main and we embarked the route with the respect you
root outside only so that the sink was not a room
what that is that most likely if you try to you the month of a image that the uh
signal is not a more sparse that for that is a strong correlation to is a sparsity
and uh yeah yeah
and the remote in that the main
is also a negative to result that is a a a really one know where if we have to uh
a signal which are are a sparse signal
therefore for all the components are you from these that from each other
if if should down sampled sequence C K and D K
and the same old correlation function also that room
signal X N and Y and have the same at or some function and the for we know that the
solutions of taken a more
in this paper we present a a a a to look a a a a a a the which stays
that up to you condition sufficient condition the construction an unique
and when we are a far the term unique a clearly we talk about a make up the sign change
of a a a a a no shift
at time reversal because this information are completely hidden than to the phase information the for one with the face
is lost
we can recover anymore
and this to condition are
the following one the first or first the support of the autocorrelation correlation function
that's a so we have a signal X which a discrete and real often said before
and it may have that for a likely combination of the amplitudes of the signal X and
we use the coefficient in that course of function
which for as the it's
uh a some information
we want them like that in order to describe is mathematically we take that
and to get a function of the support of X and
and we check that or some function uh indicator function
it that of support
of the
to of course some function at the same
the condition one is satisfied
in are all the support function of a of a and is a a a small or or equal to
this
this condition is not really started all
because if you pick out to the weeks N which are normal is to boot the uniform is to board
all these to put in a nice way
it the was usually uh would probably probability one
the second condition or a first to uh from a but which is used in the
right extraction and that proof of the in city
and it's and in this way where we have a a a a and by M yeah a matrix
and we choose the columns uh in a particular way
we take the support of that core function is still that
and we on the columns point but this support
from this support to be build this frame a that is a or which is called at
a particle i think we have to this frame is that you want every possible back the which is in
the range of a must be unique to the mine from the that values
this work this condition to as been a third can investigated by but i "'cause" that's and that in in
this in a set of paper regarding a three maybe to the frame extraction from method
and uh they proposals also some algorithms which of this problem but they works only for some particular kind of
meat she's which are really to
this uh a big are we also impose a
not that algorithm little with these general and works for a possible
uh frame
our main result is the following suppose that can one a commission to are
uh us this fight i then for a possible signal
which is supported on the indicator function one and
we can uniquely need the a
signal X N from that course some function
in the following two slide so we give a small a uh sketch of the proof of the theorem
because it allows us to define a a a at the same time a algorithm of construction
and will do starting the finding a a new sequence G M which is in the range of a for
a to we define before
and
we are cool a G M from its up through by it is possible but condition to
and then we saw was more right one problem
which allows us to of being a a a a partial information of the free that's from up so uh
to some feasibility ambiguities
i we so this phase is by alignment
we take the inverse free of just from of the result and the proof is gonna be
in
so we have a
uh the input to lower our thing is that that uh are that's absolute values of the fourier efficient of
and
and we have only their i need to solve this
uh sequence
and initially we define a new sequence G M which is a multiplication of two free coefficient
and K E we don't know this sequence because we don't know that terms what we know the absolute values
of the signal
and we can also prove that this sequence are in this uh are in the same support of the
autocorrelation function and by conditional one also on the support of
and the support S
therefore this vector a is in the range of the frame of uh and that is a operator a
and that condition to we also the a cover this
but is is sick we just the find
a the known pretty C C using a conditions
now with that obtained that that which is the squared absolute value of the for those for which is given
at the beginning of the problem
and then new see C M with the uh phase factor C
we can be a this matrix one for every possible index a
if you factor right this may we can obtain easily to vectors
which are the same and the for these matrix of is a one
and that's a vector or uh the again but or are the our
solution one F and that's and plus one up to a a phase a
X C
since since we D E is uh uh this solution a C D is uh
given that to a phase and B did that them or
for a possible index and we have this phase ambiguity
therefore we have a a and we should is B with D C last
i i that is a big at that time for a possible index M
first a uh before obtaining the solution need to solve these make it is
you close so we take
and a solution for any coefficient that um help put the solution the row and we and then in this
way is is
the solution for the first problem is which of the second problem
and you can see that in this case for example data
uh
uh the double seat signal to that
and therefore we can define a new sequence in which is a line is exploited
and then new segment is F and Q that
where there is only one is um
i'm be D
taking the bus with a some of these uh uh sequence yeah allows us to recover ten
how to the no she does we define in the beginning
while the remaining problem is that yes we know that we can cover from the from operator at that
C C M from them the needed
the problem is that
we don't know how
yeah we propose a algorithm with
this of the probably in a high dimensional space
is i in just is a particle or best space where the problem is a trace of multiplication duplication to
make six
and or to C is we start first from the definition of G M which is a multiplication between the
m-th row of V
and that unknown vector C
if you take the square or so but you of of this a sequence
we obtain that this is a trace of to this is a a and that i am and C
a am is no because is the frame operator we have built and the beginning
why this is a no
and am are to right are matrices
and these are going to the previous definition of the bird space
it's a it's a a set of in recreation
therefore we can have to solve this a a set of your creation
but to be problematic because tended to mean
the for what we can do is that
and we can exploit the fact is a matrix C is a a rank one make
so uh that are them we propose a solve of these is a problem of recovering a a G M
from eight samples was also but use
goes uh as follows
we first project the solution until
that you know a fine space the fine by the set of you know which
and then and we take the estimation of their uh C matrix
uh uh which is a rank one using as D in this case but you can use any possible a
mix of the problem
we you need to make these to process to until convergence
but mean here that proving convergence for a possible signal is not uh we don't have a proof for yet
because this that design of comics
and therefore we are not sure about the correct uh we can prove convergence
but in practice in the signal a to sparse
is uh uh uh uh the calm the i agree that leads to the correct solution
a the last of like to present a small example but we take a signal
and
which is a uh six sparse signal
which is a a which satisfies going to one and two and therefore as unique solution
and we generate the input and correlation sequence and we try to recover
this C know from that for some sequence
this kind of algorithm works in this case that works in the most of the cases
when the sparsity is a a a a at most around ten percent of the
or the uh that and uh length of the signal
so this paper we investigate the problem of spec that put addition from
uh and another point of view
what the our prior is not a mean phase but just sparsity
which is a a a a a a uh
very good prior in our to where we are interested in sparse channels or increased still
a solution in that are is not an eight but including this sparsity prior
we can and if i i uh we can define and you the are which and two sufficient condition for
which we can prove the you of the solution
a and the proof for that you need the a we can also the i've
a a an algorithm which in practice works where well
and at the moment
uh we don't have any come just proves but we're working on
oh
you
have one to down
well was very interesting have spectral factorization rates a city
a a i guess the number of questions might be a one is what do you think is the problem
i you're indicating that is you
i have a less less tasks
signal that your your algorithm is gonna be able to handle at side what one because
uh basically
we had uh doubt question function is not a morse is sparse echo uh as a node of K square
therefore it gets less possible so that's sparse
is also trying to use all the algorithms
one that of course function is almost full use up the problem is because it's really hard to do we
need to five different come
and and it's really hard to the find a good a dish algorithm because these are collected in nature
side of the spec of some problem
the for it's we hard to the it's a five in a proper way
okay can another obvious you is no noise there uh
sound applications where you would have a clean what a correlation function that there are as where you were
a he looked at what happens then
no although the one that we just looking at this quite we had you have a low very case where
you have come signal
because an X the lower fee you have basically
uh can a signal and you measure
when a discrete the because of the been see this show that was crystal
and uh they are you have a also all the problems so as uh i'm using and the gets the
more complicated the or guy you work on this problem but it's not so so for more than
i i trying to rest in and that's and each problem and he is a lot of structural information that's
traditionally used in that hearing yeah
yeah i
and the you probably need to bring in
no yeah but that nobody really use the the by of the sparse
the use different priors
are we believe that to using this possible was problem can you yeah but you can
we we are starting are so different that will are present in was the fee
but for none of the algorithms that are
um um
let's say that are not many proves of convergence of or
uh a good the uh to bit the results
use so the does the
so they just the uh that's a sec the uh working in practise
and for it's really hard to define money course so not or
uh uh what time or what they will i
okay i i i i i'm guessing um
a a lot of the structural constraints that use my have sound
a interpretation
a because you now have certain kinds of my molecules stick together
yeah yeah but that's even for a we are looking more and more that it was more more to the
beginning
i i because a big more close you don't use a standard a lot of can use different techniques like
uh
ask uh a and a a number of the little scattering a you a more complicated techniques so where you
don't have a you know where more information so okay
why if you try to check for uh a model in the model
but keep the model uh i
you would like to use the methods
which i
at the moment is basically a projection that we that mean was of the mean right in the constraints which
are a it's a kind of like with a little in image extraction
uh
is it a from problem most of information
okay
or i thanks very that's
i think we have to maybe which
could you know
but
so
i and
uh could you never to a little bit and how you construct your on the basis matrix P can send
new
i do you did come you so you don't know that you got to the signal them and use and
is a very you need to a a a way of constructing matrix it
the support of the virtual function
as to for the for for for while i support as the support of a course of function
and i condition or on it's also support of on is uh in to get out the correlation of the
a function
i is one of no because we measured of or from function
that for we got the column of the for testing there
i would be this frame operator
oh i i also a in the previous question
why for that out a crucial part is not a anymore
we don't have any more than a and uh
frame and i a period and the for you
where the partial
for for is
i think you know much i think