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