0:00:13 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