0:00:13 | a |
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0:00:21 | i |

0:00:29 | that's doesn't |

0:00:30 | a uh as i would like to apologise for in |

0:00:34 | but do you a low for not be here but the and the |

0:00:36 | problems coming here today |

0:00:38 | i your in the and work in the same subject and in a file |

0:00:41 | a to with much but the and |

0:00:43 | also so with the U |

0:00:45 | and uh in this talk okay we will uh |

0:00:47 | present the some results about |

0:00:49 | a classical problem problems speced of that addition |

0:00:52 | and look at the problem uh looking at that vacation and seen that there is a sparse a |

0:00:57 | the prior available |

0:00:58 | and we learn a lot of the problem of sparse like the position |

0:01:01 | we we present some sufficient condition |

0:01:04 | which allows us to know that the solution is unique |

0:01:07 | we prove use the would be we i would it would be give a sketch of the proof |

0:01:10 | and we present our reconstruction algorithm which works we really well |

0:01:16 | or or or that the position is we one known problem |

0:01:19 | where are we want to |

0:01:21 | we want to |

0:01:23 | uh uh recover a signal X of and no not from a control version of it the but using the |

0:01:27 | autocorrelation function |

0:01:29 | which is a convolution between the signal and its time for |

0:01:32 | we in the same problem in that the main where we have a a you to let this indicates a |

0:01:37 | for this form |

0:01:38 | where we have that the correlation function and therefore we have described also value of that well correlation |

0:01:43 | oh a of the signal |

0:01:44 | if to a cover the phase information from this uh |

0:01:48 | um um an information |

0:01:49 | we can sum or role |

0:01:52 | is is not really a only my mother got all in there are an application in which this problem is |

0:01:55 | a a real haven't |

0:01:57 | first of all of this bunch channel estimation if you have a channel H of fan |

0:02:00 | uh which is a no we would like to get |

0:02:03 | uh the closest response in order to improve the communication over channel |

0:02:07 | and we can only some me data for the input X and and measure the output |

0:02:11 | if we sum me white noise in the channel using the filter a a we know that |

0:02:15 | they so uh that what is gonna be that |

0:02:17 | absolute value of the uh with the of the channel |

0:02:20 | therefore |

0:02:22 | if if a cover the face C we got a of the impulse response of the channel |

0:02:26 | and the problem so |

0:02:28 | and of the five is a problem or rising up text |

0:02:30 | and then it back to got we are it as the next expect a lot of E |

0:02:34 | if where where we have a an cool and want them this and the fact that the shape because the |

0:02:38 | shape |

0:02:39 | yeah the term and the time means the function in biology most of the most cases |

0:02:43 | then for with that been i the monocle and putting it into a crystal a a we that and we |

0:02:47 | put the quickly a an X ray B |

0:02:50 | what we measure what we can measure that the faction that of the crystal which by the backs slow it's |

0:02:54 | that have with us from of the crystal |

0:02:57 | but the probably as that |

0:02:58 | if you we if we use for the rest a for the feel more C C D's |

0:03:02 | we can only measure think this of the light and therefore we lose the "'cause" information if we the domain |

0:03:06 | and if we can recover the fees we can also recover them |

0:03:11 | and |

0:03:11 | uh they original money |

0:03:15 | yeah a a result is really come on in uh which is a a really one no it's its a |

0:03:20 | uh this fact that that the addition that the same |

0:03:22 | we have a the correlation function that the main |

0:03:24 | uh we've tried to find the remotes and does that the main using for addition |

0:03:28 | but the problem is uh a how to assign each got or fruits so between the but inside and outside |

0:03:33 | the unit equal to the signal X of that |

0:03:36 | here since we assume the signal X of that to be a real their roots up using incredible |

0:03:41 | if we are aware that our signal is a a i mean phase is a we we can take just |

0:03:45 | the roots it's set of the unit to call |

0:03:47 | and we are sure that that of destruction is gonna be unique |

0:03:50 | is a good solution |

0:03:53 | otherwise if we don't have this prior about minimum phase a |

0:03:56 | we can't we have just to |

0:03:57 | try all the possible combination between inside and outside roots |

0:04:00 | you know to find our possible solution and therefore for is a common or number of possible solution |

0:04:05 | and uh there are known all algorithms to find a good so uh the resolution |

0:04:10 | what we can say that since the the tool because which we are interested rest have a sparse prior because |

0:04:15 | channel else usually can be modelled using multi about |

0:04:19 | fading channels which are sparse in the for the main |

0:04:22 | and also i them sink with still lower |

0:04:23 | results are sparse because the model just you want to so that is that is a complete |

0:04:28 | and the for and interesting to understand |

0:04:30 | which implies that sparsity in that |

0:04:32 | uh |

0:04:34 | reconstruction construction up to the an extension of the solution of the set of the vision problem |

0:04:38 | and moreover if we can say something about the you a be |

0:04:41 | because without |

0:04:42 | uh mean of a solution the number of solutions can or you want to avoid this situation |

0:04:47 | a a small hint about why spice D can be helpful can be here from these example when we have |

0:04:51 | a sparse signal |

0:04:52 | made of for example you have four components out of them |

0:04:55 | and we |

0:04:57 | a good the that the main and we embarked the route with the respect you |

0:05:00 | root outside only so that the sink was not a room |

0:05:03 | what that is that most likely if you try to you the month of a image that the uh |

0:05:08 | signal is not a more sparse that for that is a strong correlation to is a sparsity |

0:05:12 | and uh yeah yeah |

0:05:14 | and the remote in that the main |

0:05:17 | is also a negative to result that is a a a really one know where if we have to uh |

0:05:21 | a signal which are are a sparse signal |

0:05:23 | therefore for all the components are you from these that from each other |

0:05:26 | if if should down sampled sequence C K and D K |

0:05:29 | and the same old correlation function also that room |

0:05:32 | signal X N and Y and have the same at or some function and the for we know that the |

0:05:36 | solutions of taken a more |

0:05:39 | in this paper we present a a a a to look a a a a a a the which stays |

0:05:43 | that up to you condition sufficient condition the construction an unique |

0:05:48 | and when we are a far the term unique a clearly we talk about a make up the sign change |

0:05:52 | of a a a a a no shift |

0:05:54 | at time reversal because this information are completely hidden than to the phase information the for one with the face |

0:05:59 | is lost |

0:06:00 | we can recover anymore |

0:06:03 | and this to condition are |

0:06:04 | the following one the first or first the support of the autocorrelation correlation function |

0:06:09 | that's a so we have a signal X which a discrete and real often said before |

0:06:13 | and it may have that for a likely combination of the amplitudes of the signal X and |

0:06:17 | we use the coefficient in that course of function |

0:06:20 | which for as the it's |

0:06:21 | uh a some information |

0:06:23 | we want them like that in order to describe is mathematically we take that |

0:06:27 | and to get a function of the support of X and |

0:06:30 | and we check that or some function uh indicator function |

0:06:33 | it that of support |

0:06:34 | of the |

0:06:35 | to of course some function at the same |

0:06:37 | the condition one is satisfied |

0:06:39 | in are all the support function of a of a and is a a a small or or equal to |

0:06:43 | this |

0:06:45 | this condition is not really started all |

0:06:46 | because if you pick out to the weeks N which are normal is to boot the uniform is to board |

0:06:50 | all these to put in a nice way |

0:06:52 | it the was usually uh would probably probability one |

0:06:55 | the second condition or a first to uh from a but which is used in the |

0:07:00 | right extraction and that proof of the in city |

0:07:02 | and it's and in this way where we have a a a a and by M yeah a matrix |

0:07:06 | and we choose the columns uh in a particular way |

0:07:10 | we take the support of that core function is still that |

0:07:12 | and we on the columns point but this support |

0:07:16 | from this support to be build this frame a that is a or which is called at |

0:07:21 | a particle i think we have to this frame is that you want every possible back the which is in |

0:07:24 | the range of a must be unique to the mine from the that values |

0:07:28 | this work this condition to as been a third can investigated by but i "'cause" that's and that in in |

0:07:33 | this in a set of paper regarding a three maybe to the frame extraction from method |

0:07:38 | and uh they proposals also some algorithms which of this problem but they works only for some particular kind of |

0:07:43 | meat she's which are really to |

0:07:45 | this uh a big are we also impose a |

0:07:48 | not that algorithm little with these general and works for a possible |

0:07:51 | uh frame |

0:07:54 | our main result is the following suppose that can one a commission to are |

0:07:57 | uh us this fight i then for a possible signal |

0:08:00 | which is supported on the indicator function one and |

0:08:03 | we can uniquely need the a |

0:08:05 | signal X N from that course some function |

0:08:08 | in the following two slide so we give a small a uh sketch of the proof of the theorem |

0:08:13 | because it allows us to define a a a at the same time a algorithm of construction |

0:08:18 | and will do starting the finding a a new sequence G M which is in the range of a for |

0:08:22 | a to we define before |

0:08:24 | and |

0:08:25 | we are cool a G M from its up through by it is possible but condition to |

0:08:30 | and then we saw was more right one problem |

0:08:32 | which allows us to of being a a a a partial information of the free that's from up so uh |

0:08:37 | to some feasibility ambiguities |

0:08:39 | i we so this phase is by alignment |

0:08:41 | we take the inverse free of just from of the result and the proof is gonna be |

0:08:45 | in |

0:08:46 | so we have a |

0:08:47 | uh the input to lower our thing is that that uh are that's absolute values of the fourier efficient of |

0:08:52 | and |

0:08:53 | and we have only their i need to solve this |

0:08:55 | uh sequence |

0:08:58 | and initially we define a new sequence G M which is a multiplication of two free coefficient |

0:09:02 | and K E we don't know this sequence because we don't know that terms what we know the absolute values |

0:09:06 | of the signal |

0:09:08 | and we can also prove that this sequence are in this uh are in the same support of the |

0:09:12 | autocorrelation function and by conditional one also on the support of |

0:09:16 | and the support S |

0:09:18 | therefore this vector a is in the range of the frame of uh and that is a operator a |

0:09:23 | and that condition to we also the a cover this |

0:09:25 | but is is sick we just the find |

0:09:28 | a the known pretty C C using a conditions |

0:09:33 | now with that obtained that that which is the squared absolute value of the for those for which is given |

0:09:37 | at the beginning of the problem |

0:09:38 | and then new see C M with the uh phase factor C |

0:09:42 | we can be a this matrix one for every possible index a |

0:09:45 | if you factor right this may we can obtain easily to vectors |

0:09:49 | which are the same and the for these matrix of is a one |

0:09:51 | and that's a vector or uh the again but or are the our |

0:09:55 | solution one F and that's and plus one up to a a phase a |

0:09:59 | X C |

0:10:00 | since since we D E is uh uh this solution a C D is uh |

0:10:04 | given that to a phase and B did that them or |

0:10:06 | for a possible index and we have this phase ambiguity |

0:10:10 | therefore we have a a and we should is B with D C last |

0:10:13 | i i that is a big at that time for a possible index M |

0:10:16 | first a uh before obtaining the solution need to solve these make it is |

0:10:20 | you close so we take |

0:10:22 | and a solution for any coefficient that um help put the solution the row and we and then in this |

0:10:27 | way is is |

0:10:28 | the solution for the first problem is which of the second problem |

0:10:31 | and you can see that in this case for example data |

0:10:33 | uh |

0:10:34 | uh the double seat signal to that |

0:10:37 | and therefore we can define a new sequence in which is a line is exploited |

0:10:41 | and then new segment is F and Q that |

0:10:43 | where there is only one is um |

0:10:46 | i'm be D |

0:10:47 | taking the bus with a some of these uh uh sequence yeah allows us to recover ten |

0:10:51 | how to the no she does we define in the beginning |

0:10:54 | while the remaining problem is that yes we know that we can cover from the from operator at that |

0:10:59 | C C M from them the needed |

0:11:01 | the problem is that |

0:11:03 | we don't know how |

0:11:04 | yeah we propose a algorithm with |

0:11:06 | this of the probably in a high dimensional space |

0:11:08 | is i in just is a particle or best space where the problem is a trace of multiplication duplication to |

0:11:13 | make six |

0:11:14 | and or to C is we start first from the definition of G M which is a multiplication between the |

0:11:19 | m-th row of V |

0:11:20 | and that unknown vector C |

0:11:23 | if you take the square or so but you of of this a sequence |

0:11:26 | we obtain that this is a trace of to this is a a and that i am and C |

0:11:30 | a am is no because is the frame operator we have built and the beginning |

0:11:33 | why this is a no |

0:11:36 | and am are to right are matrices |

0:11:38 | and these are going to the previous definition of the bird space |

0:11:41 | it's a it's a a set of in recreation |

0:11:44 | therefore we can have to solve this a a set of your creation |

0:11:47 | but to be problematic because tended to mean |

0:11:50 | the for what we can do is that |

0:11:52 | and we can exploit the fact is a matrix C is a a rank one make |

0:11:56 | so uh that are them we propose a solve of these is a problem of recovering a a G M |

0:12:00 | from eight samples was also but use |

0:12:02 | goes uh as follows |

0:12:04 | we first project the solution until |

0:12:06 | that you know a fine space the fine by the set of you know which |

0:12:10 | and then and we take the estimation of their uh C matrix |

0:12:13 | uh uh which is a rank one using as D in this case but you can use any possible a |

0:12:17 | mix of the problem |

0:12:20 | we you need to make these to process to until convergence |

0:12:23 | but mean here that proving convergence for a possible signal is not uh we don't have a proof for yet |

0:12:28 | because this that design of comics |

0:12:30 | and therefore we are not sure about the correct uh we can prove convergence |

0:12:34 | but in practice in the signal a to sparse |

0:12:37 | is uh uh uh uh the calm the i agree that leads to the correct solution |

0:12:42 | a the last of like to present a small example but we take a signal |

0:12:46 | and |

0:12:47 | which is a uh six sparse signal |

0:12:50 | which is a a which satisfies going to one and two and therefore as unique solution |

0:12:54 | and we generate the input and correlation sequence and we try to recover |

0:12:59 | this C know from that for some sequence |

0:13:00 | this kind of algorithm works in this case that works in the most of the cases |

0:13:04 | when the sparsity is a a a a at most around ten percent of the |

0:13:08 | or the uh that and uh length of the signal |

0:13:12 | so this paper we investigate the problem of spec that put addition from |

0:13:16 | uh and another point of view |

0:13:18 | what the our prior is not a mean phase but just sparsity |

0:13:21 | which is a a a a a a uh |

0:13:23 | very good prior in our to where we are interested in sparse channels or increased still |

0:13:28 | a solution in that are is not an eight but including this sparsity prior |

0:13:31 | we can and if i i uh we can define and you the are which and two sufficient condition for |

0:13:36 | which we can prove the you of the solution |

0:13:38 | a and the proof for that you need the a we can also the i've |

0:13:41 | a a an algorithm which in practice works where well |

0:13:44 | and at the moment |

0:13:46 | uh we don't have any come just proves but we're working on |

0:13:49 | oh |

0:14:07 | you |

0:14:16 | have one to down |

0:14:20 | well was very interesting have spectral factorization rates a city |

0:14:25 | a a i guess the number of questions might be a one is what do you think is the problem |

0:14:29 | i you're indicating that is you |

0:14:31 | i have a less less tasks |

0:14:34 | signal that your your algorithm is gonna be able to handle at side what one because |

0:14:39 | uh basically |

0:14:41 | we had uh doubt question function is not a morse is sparse echo uh as a node of K square |

0:14:47 | therefore it gets less possible so that's sparse |

0:14:50 | is also trying to use all the algorithms |

0:14:52 | one that of course function is almost full use up the problem is because it's really hard to do we |

0:14:57 | need to five different come |

0:14:59 | and and it's really hard to the find a good a dish algorithm because these are collected in nature |

0:15:04 | side of the spec of some problem |

0:15:06 | the for it's we hard to the it's a five in a proper way |

0:15:09 | okay can another obvious you is no noise there uh |

0:15:12 | sound applications where you would have a clean what a correlation function that there are as where you were |

0:15:17 | a he looked at what happens then |

0:15:19 | no although the one that we just looking at this quite we had you have a low very case where |

0:15:22 | you have come signal |

0:15:24 | because an X the lower fee you have basically |

0:15:27 | uh can a signal and you measure |

0:15:29 | when a discrete the because of the been see this show that was crystal |

0:15:33 | and uh they are you have a also all the problems so as uh i'm using and the gets the |

0:15:37 | more complicated the or guy you work on this problem but it's not so so for more than |

0:15:41 | i i trying to rest in and that's and each problem and he is a lot of structural information that's |

0:15:46 | traditionally used in that hearing yeah |

0:15:48 | yeah i |

0:15:49 | and the you probably need to bring in |

0:15:51 | no yeah but that nobody really use the the by of the sparse |

0:15:54 | the use different priors |

0:15:56 | are we believe that to using this possible was problem can you yeah but you can |

0:16:00 | we we are starting are so different that will are present in was the fee |

0:16:04 | but for none of the algorithms that are |

0:16:06 | um um |

0:16:07 | let's say that are not many proves of convergence of or |

0:16:10 | uh a good the uh to bit the results |

0:16:12 | use so the does the |

0:16:14 | so they just the uh that's a sec the uh working in practise |

0:16:18 | and for it's really hard to define money course so not or |

0:16:22 | uh uh what time or what they will i |

0:16:24 | okay i i i i i'm guessing um |

0:16:27 | a a lot of the structural constraints that use my have sound |

0:16:31 | a interpretation |

0:16:33 | a because you now have certain kinds of my molecules stick together |

0:16:36 | yeah yeah but that's even for a we are looking more and more that it was more more to the |

0:16:40 | beginning |

0:16:41 | i i because a big more close you don't use a standard a lot of can use different techniques like |

0:16:46 | uh |

0:16:47 | ask uh a and a a number of the little scattering a you a more complicated techniques so where you |

0:16:51 | don't have a you know where more information so okay |

0:16:54 | why if you try to check for uh a model in the model |

0:16:56 | but keep the model uh i |

0:16:59 | you would like to use the methods |

0:17:00 | which i |

0:17:01 | at the moment is basically a projection that we that mean was of the mean right in the constraints which |

0:17:06 | are a it's a kind of like with a little in image extraction |

0:17:09 | uh |

0:17:10 | is it a from problem most of information |

0:17:12 | okay |

0:17:13 | or i thanks very that's |

0:17:15 | i think we have to maybe which |

0:17:19 | could you know |

0:17:22 | but |

0:17:28 | so |

0:17:30 | i and |

0:17:32 | uh could you never to a little bit and how you construct your on the basis matrix P can send |

0:17:36 | new |

0:17:38 | i do you did come you so you don't know that you got to the signal them and use and |

0:17:42 | is a very you need to a a a way of constructing matrix it |

0:17:47 | the support of the virtual function |

0:17:49 | as to for the for for for while i support as the support of a course of function |

0:17:54 | and i condition or on it's also support of on is uh in to get out the correlation of the |

0:17:58 | a function |

0:17:59 | i is one of no because we measured of or from function |

0:18:02 | that for we got the column of the for testing there |

0:18:06 | i would be this frame operator |

0:18:08 | oh i i also a in the previous question |

0:18:10 | why for that out a crucial part is not a anymore |

0:18:13 | we don't have any more than a and uh |

0:18:16 | frame and i a period and the for you |

0:18:18 | where the partial |

0:18:19 | for for is |

0:18:25 | i think you know much i think |