0:00:14 | good afternoon everyone |
---|---|

0:00:15 | yeah a hand i got each |

0:00:17 | said |

0:00:18 | uh |

0:00:19 | oh |

0:00:20 | hmmm |

0:00:21 | location |

0:00:23 | what |

0:00:23 | i |

0:00:25 | this work |

0:00:26 | uh is a joint with them |

0:00:29 | a a to patrol boat and a cup or |

0:00:36 | and was sort the uh with the short back |

0:00:39 | uh on the system that you're talking about |

0:00:41 | uh we will make should use the optimization them match to that to are going to uh exploit here which |

0:00:47 | is the cramer-rao around uh bound um metric |

0:00:51 | uh uh we're going to use this one two |

0:00:54 | uh formulate the power allocation optimization problem |

0:00:57 | where the objective is to minimize the total transmitted power in a a |

0:01:01 | my more L multiple radar uh a vector |

0:01:05 | uh |

0:01:06 | to get an efficient to |

0:01:07 | solution for that we use the the make the composition for that |

0:01:11 | and it will use a some american analysis to show how uh |

0:01:16 | uh the power location |

0:01:17 | uh is generated |

0:01:19 | and finally some concluding remarks |

0:01:23 | a a target a localisation uh yeah estimation mean-square square error is known to be lower bounded by the cramer-rao |

0:01:29 | lower bound especially if we talk about that maximum likelihood estimator |

0:01:33 | and uh based on this metric it has been shown in the past that um |

0:01:37 | a system with widely the separated |

0:01:40 | uh uh and most people uh uh or |

0:01:42 | uh |

0:01:43 | and the and systems |

0:01:45 | using a coherent or non-coherent processing |

0:01:48 | uh offers advantages in terms of the estimation is square |

0:01:52 | and |

0:01:53 | that you're see again is proportional to the product of the number of transmit and receive antenna |

0:01:59 | i i in general though if we expand this dependency the mean square error depends on |

0:02:04 | as a set and the number of transmit and receive radars |

0:02:07 | but it also depends on the geometric metric layout |

0:02:10 | of of the transmit and receive radars with respect to the target |

0:02:13 | uh it depends on the uh signal effective bandwidth |

0:02:17 | and on the signal to noise ratio and that brings us to the |

0:02:20 | transmit power which were going to focus in this store |

0:02:25 | oh |

0:02:25 | that the the of that we're looking at is new |

0:02:28 | a widely separated |

0:02:30 | multiple radar system |

0:02:32 | that is uh mobile |

0:02:34 | and we see more and more application like that |

0:02:37 | uh one example is the |

0:02:39 | ground surveillance radar |

0:02:41 | where we have a |

0:02:42 | yeah a mounted and vehicles |

0:02:45 | that are spread the along borders we can see them mean um yeah pro controls and things like that |

0:02:52 | a in this type of |

0:02:53 | uh application |

0:02:55 | yeah it makes sense to be more conscious |

0:02:57 | about are the use of resources |

0:03:00 | and uh uh what you cover in uh |

0:03:03 | this work is |

0:03:05 | uh resource uh awareness in terms of the transmitted power |

0:03:11 | so our objective here |

0:03:12 | as we can see here used to minimize |

0:03:15 | the total transmitted power such that the predetermined estimation mean-square square error is a thing |

0:03:20 | uh uh white and the transmit power |

0:03:22 | at each a station within an acceptable range |

0:03:25 | so well we not that yeah |

0:03:27 | extending the |

0:03:29 | the number of of |

0:03:31 | uh rate hours |

0:03:32 | provides a higher accuracy in practical we need some level of your see which can can serve as a trash |

0:03:37 | threshold |

0:03:38 | and the question is how do we minimize the powers |

0:03:41 | the system before form a a at the threshold that we one |

0:03:45 | uh and this is what we going to do |

0:03:49 | so this is the system that we talking in in this figure |

0:03:51 | uh |

0:03:53 | an example of such a system and we keep it very general |

0:03:57 | in terms that |

0:03:58 | our readers can be |

0:04:00 | you or transmit or or cu |

0:04:02 | a or or can be vote i mean each one of this point can be a transmit receive a radar |

0:04:07 | and assumption something that the uh |

0:04:08 | are the information jointly so |

0:04:11 | from all of this element |

0:04:13 | and you have a target uh |

0:04:15 | here that we want to estimate its location |

0:04:17 | specifically a of before |

0:04:20 | with assume we have a a a as a C M transmit radars and receive radar |

0:04:24 | the target is modelled as an extended target |

0:04:27 | with the center mass located uh |

0:04:29 | position |

0:04:30 | S |

0:04:31 | we use of and all signal and assume we have |

0:04:34 | M and and the had propagation path |

0:04:37 | uh the transmitted power vector is given here as P of T uh |

0:04:41 | for each one of the transmitting ten |

0:04:45 | uh as we know the |

0:04:46 | time delay of propagation of each one of |

0:04:49 | pat |

0:04:50 | uh a time is a function of the range from to transmitting uh uh radar |

0:04:55 | uh to the target |

0:04:56 | and from the |

0:04:58 | target to the receive radar also |

0:05:00 | tao |

0:05:01 | and and basically a a measure those time delays for example if we use this one |

0:05:05 | as a transmit |

0:05:06 | to the target |

0:05:07 | and received here this would be |

0:05:10 | uh |

0:05:11 | this propagation that would be proportional to the |

0:05:13 | range sample |

0:05:16 | uh this |

0:05:17 | brings us to the received signal |

0:05:19 | on the specific path and pat to "'em" and |

0:05:22 | and we see that i went to that you take into account uh a in our model which we have |

0:05:27 | here off i of and basically is proportional to |

0:05:30 | uh the path so it was that the path loss |

0:05:33 | uh P of T in P of M T X is the transmitted power |

0:05:37 | a a or friends then it's is a complex coefficient |

0:05:39 | basically takes into account the |

0:05:41 | uh rate cross section on to M and |

0:05:44 | a plus and any phase offsets and this path |

0:05:48 | uh we have your uh uh uh delayed |

0:05:50 | time delayed version of the transmitted power |

0:05:53 | that's transmitted signals or and |

0:05:55 | oh a white gaussian the |

0:05:57 | voice |

0:05:59 | um |

0:06:00 | we actually a defined all of this |

0:06:02 | so we can find some metric you said that |

0:06:04 | the constrained our system are giving in terms of |

0:06:07 | square |

0:06:07 | so we need to find a a a a metric that labour enable us to |

0:06:11 | uh represent this man |

0:06:14 | for this were using the cramer-rao bound |

0:06:16 | i where we using the trace of the cramer-rao bound metrics |

0:06:20 | uh two |

0:06:21 | provide the bound on the mean square on the X action |

0:06:25 | and and the white direction one |

0:06:27 | uh a the previous work |

0:06:29 | trade between the two so |

0:06:31 | uh |

0:06:31 | optimising one of them |

0:06:33 | we just uh maximise the that |

0:06:37 | um |

0:06:38 | the the around on that because it was developed in previous studies it's not and you result |

0:06:42 | uh what we have to do your do is uh |

0:06:45 | re |

0:06:46 | state it |

0:06:47 | so it can be used to optimize power |

0:06:50 | so what we did here is we talk |

0:06:53 | the original expression and defined it as a sum |

0:06:56 | of some |

0:06:57 | elements here multiplied by the pope power transmitted by transmitter am and have end of the |

0:07:03 | i if we go one step forward and you can see your by the way |

0:07:06 | that the elements of this matrix |

0:07:08 | are dependent on off a age |

0:07:11 | which were uh what the code channel correct for state on path and man |

0:07:15 | and um |

0:07:16 | we have you the location of the transmit and receive radars with respect to the target |

0:07:21 | uh uh incorporated through this expression which are basically cosine and sign |

0:07:26 | of the angles between the transmitter and receiver to the target |

0:07:30 | uh the vector of a in this case |

0:07:33 | you use the target location X Y |

0:07:35 | and the channel |

0:07:36 | vector eight |

0:07:39 | uh using this type of uh |

0:07:41 | expression and us was us to |

0:07:43 | expressed the trace of the cramer-rao bound |

0:07:46 | in the form that you can see here |

0:07:48 | well basically we have some vector B |

0:07:50 | multiplying the vector of power |

0:07:52 | and in the denominator we have |

0:07:55 | a a uh metrics eight |

0:07:57 | that |

0:07:58 | second second order |

0:07:59 | expression for the the same power |

0:08:01 | and it you can see that basically be and any incorporate |

0:08:05 | all the existing system |

0:08:07 | a just the geometric spread the channel |

0:08:10 | that fading and so forth so these are |

0:08:12 | oh coming in to play through uh metrics as |

0:08:15 | metrics a and vector B |

0:08:18 | now that we have an expression for the cramer-rao bound we can formulate |

0:08:22 | position |

0:08:24 | uh a and as we said our objective is |

0:08:27 | uh uh that given a a predetermined threshold |

0:08:30 | but is if a mean square error or |

0:08:32 | uh we would like to optimize |

0:08:34 | uh |

0:08:35 | uh the the pa out basically um |

0:08:38 | minimize the total contrast |

0:08:40 | and this is the mathematical formulation for that |

0:08:43 | so we we minimize the total transmitted power |

0:08:46 | a a given a specific threshold are the cramer-rao bound where |

0:08:50 | uh we use of previous uh |

0:08:52 | estimate of the target location in age |

0:08:54 | to calculate |

0:08:55 | C |

0:08:57 | and also need for some limitation under transmitted power we we assume as you we transmit the minimal power |

0:09:03 | uh P T X minimum and the maximum power uh uh uh |

0:09:08 | P M T X max |

0:09:13 | ah |

0:09:13 | taking |

0:09:14 | just go back to second this this is obviously an an uh nonlinear optimization problem |

0:09:19 | uh |

0:09:20 | due to the structure of C |

0:09:23 | the trace of C |

0:09:24 | and what we're doing the is basically um relaxation of the region problem and we using the expression that we |

0:09:29 | just developed previously at these solo using a vector B in metrics say |

0:09:33 | and you get this type of um |

0:09:35 | expression for the optimization problem |

0:09:38 | ah |

0:09:39 | no for this problem since is a non-convex |

0:09:42 | problem |

0:09:44 | uh we decided to go um using uh the like you on and uh the K can take a kick |

0:09:49 | it T conditions to find a us uh |

0:09:52 | until a solution |

0:09:54 | so next uh in the bottom here you see the lagrangian and uh a function |

0:09:59 | for this optimization problem |

0:10:01 | where we incorporate the objective |

0:10:03 | the first equality constraint multiplied by |

0:10:05 | no no and we have the two |

0:10:08 | yeah uh sets of uh inequality constraint multi by by you and you |

0:10:13 | uh the cake the condition formulated here |

0:10:17 | uh uh uh where you see that basically this expression is by just by long down "'cause" our |

0:10:24 | um |

0:10:25 | a train here are equally equal to constraint was uh uh metric uh a one parameter |

0:10:31 | um |

0:10:32 | to solve that |

0:10:34 | we take one step |

0:10:36 | i had and we basically or |

0:10:39 | and the constraint on P max |

0:10:41 | a mean by choosing me you when you E close to the zero |

0:10:45 | we in Z want all those two um |

0:10:48 | equation |

0:10:49 | and we |

0:10:50 | uh |

0:10:51 | get from this set we get |

0:10:53 | the three questions that we have here |

0:10:55 | and this has a have an analytical solution a very simple analytical so |

0:11:01 | then it could can solution is given here |

0:11:04 | and |

0:11:05 | what you see by ignoring uh |

0:11:07 | for for temporal ignoring down |

0:11:11 | the restriction and the power |

0:11:13 | is is that the optimal power allocation |

0:11:16 | has uh uh basically a um a levelling mechanism here |

0:11:20 | one of "'em" though |

0:11:22 | and one are all of them and the uh what it does has to be E and by the the |

0:11:26 | the |

0:11:27 | um |

0:11:28 | by the way B E and eight E represent be in a good we had previously we just |

0:11:33 | use |

0:11:33 | a uh the uh last |

0:11:35 | estimate to make we have the location but the channel to actually calculate the of so it's an actually a |

0:11:39 | value based on estimate |

0:11:42 | and you can see that basically what it does |

0:11:45 | it moves |

0:11:46 | it we levels |

0:11:47 | the elements of B |

0:11:49 | B |

0:11:50 | uh we uh we do a value inversely proportional to one that's quite at on that |

0:11:55 | start here |

0:11:56 | and one of the star you can see that |

0:11:58 | this levelling mechanism |

0:12:01 | incorporates it's a |

0:12:03 | mixture |

0:12:04 | of what element that naturally uh |

0:12:07 | um |

0:12:08 | i think the system such as the location the channel |

0:12:11 | the uh propagation loss and so forth |

0:12:15 | uh and an important thing that you know this different is different for communication |

0:12:20 | uh a system or as passive sensor system in this case it we have a transmitter |

0:12:25 | that ready it's energy |

0:12:27 | the he's reflected back to the targets so there is a cross dependency between the |

0:12:31 | a selected |

0:12:33 | power level at the specific the transmitter |

0:12:36 | and uh a signal that we get at all |

0:12:39 | and receiver |

0:12:40 | so when we that |

0:12:42 | specific transmit it fact |

0:12:45 | and propagation path |

0:12:46 | and the are this is why we get you few more complex value for long |

0:12:51 | um |

0:12:53 | we can see here data |

0:12:55 | uh |

0:12:56 | uh a fact of the two track actual uh that to be |

0:12:59 | introduced |

0:13:00 | i do think about this solution is as i just a |

0:13:03 | we can or the constraint |

0:13:05 | part |

0:13:05 | right so we can get an analytical solution here but we can be |

0:13:09 | outside |

0:13:11 | the ability of our sets them in terms of transmit and receive a a transmit power mean and max |

0:13:18 | so |

0:13:18 | well this gives a something inside of how |

0:13:21 | the power is distributed between the the different transmitters |

0:13:24 | uh oh |

0:13:25 | we were looking for something for uh and then the could way to get |

0:13:29 | more solution feasible solution |

0:13:32 | so we |

0:13:33 | oh when |

0:13:34 | and |

0:13:35 | yeah used uh the composition that that's and basically what we did in this uh approach |

0:13:40 | since were looking for mean |

0:13:42 | transmitted power |

0:13:43 | we can use a boundary |

0:13:45 | a points |

0:13:46 | to fine |

0:13:49 | it's solutions |

0:13:50 | so for example if we looking into the minimal value that each transmitter can you |

0:13:56 | we can |

0:13:58 | take a scenario where we take for example one transmitter |

0:14:01 | make these transmitter transmit the minimum value |

0:14:04 | and then calculate all the other |

0:14:07 | uh in my one analytically |

0:14:10 | but for this we need to mathematically formulate it |

0:14:13 | such that we can separate between |

0:14:15 | a group of transmitter |

0:14:17 | where actually enforce either a minimal maximum value on them |

0:14:21 | and the set of transmit that we analytically calc |

0:14:26 | and basically the uh structures that you see here |

0:14:29 | be one one B to one |

0:14:30 | do you and B two |

0:14:32 | are we we organise are vector are transmitted vector so the first portion |

0:14:37 | of this vector or uh you see here as the P T X one |

0:14:41 | represents a uh the one to our car |

0:14:44 | and P P X two what are the one that we enforce and we enforce force K elements |

0:14:49 | to be on the bound |

0:14:52 | now for the boundaries we |

0:14:55 | are are a select think in the minimum |

0:14:57 | what we can select |

0:14:58 | maximum a minimal points |

0:15:00 | to not to know uh used to much |

0:15:03 | um |

0:15:04 | yeah i uh search unnecessary search |

0:15:06 | we evaluate what would be the power in case of uniform and any |

0:15:10 | uh some of uh the powers that is beyond this a uniform power allocation we are not even investigating that |

0:15:17 | uh doing so and you have the details in the paper or of how this is the derive but we |

0:15:21 | basically the a a a a an analytical form |

0:15:23 | to calculate the remaining vector that we did not enforce any boundary point on |

0:15:30 | and you see the same |

0:15:31 | uh a structure all |

0:15:34 | a levelling |

0:15:35 | um |

0:15:36 | mechanism |

0:15:37 | uh that works again on |

0:15:40 | the the lot of the uh |

0:15:42 | uh metrics is uh B and that uh vectors |

0:15:45 | a a a a a capital B and a vector B |

0:15:48 | that |

0:15:48 | that to represent and the system structure |

0:15:52 | and uh we have a more complex um |

0:15:56 | a calculation for on the squared but uh again this is simple a uh and a solution once we have |

0:16:02 | the form a let's take that takes a little bit longer |

0:16:04 | that the uh uh resulting in question a very simple to you |

0:16:10 | uh and basically what what it gives that sees the set all |

0:16:14 | optimization problem that can be used to be either or or you can use this to the processing to get |

0:16:19 | uh |

0:16:20 | the solutions here or you can um |

0:16:22 | send a have them |

0:16:24 | calculated that the different receivers where the information is uh available |

0:16:29 | uh the only information each one of these sound problems needs |

0:16:32 | to calculate is the uh a a a a a a a estimated location of the target X line |

0:16:37 | and that's to channel age and all the rest of the |

0:16:41 | data are is uh existing data are related to the structure of the system |

0:16:46 | uh so each one of this problem basically K means that we take |

0:16:50 | K elements and put them on the boundaries |

0:16:53 | uh K going from one to in minus one |

0:16:56 | and each one of these we get |

0:16:58 | an optimal set of solution |

0:17:00 | which the minimum one is transmitted to the fusion center |

0:17:04 | which are select the mean one |

0:17:05 | one out |

0:17:10 | a a to see how this um |

0:17:12 | i'll go with them |

0:17:13 | how this uh method work we we use a few scenarios here |

0:17:18 | oh okay so as want to four for and the left side are cases where we assume the distances |

0:17:24 | from the uh um elements to the target or equal i would basically a human a the effect of about |

0:17:29 | five |

0:17:31 | a a a a a on the left side and right sides or a case five to eight |

0:17:35 | um generate different |

0:17:37 | this and if i would use that the the right hand side the the right hand side it |

0:17:43 | uh |

0:17:43 | all channels like equal to one we can actually |

0:17:46 | uh |

0:17:47 | have an option of seeing what |

0:17:49 | how that the geometric effect |

0:17:51 | uh uh what do the do you could uh they'll pay yeah affect how it fact the power distribution between |

0:17:57 | the transmitter still |

0:17:58 | um the right hand side will help us um |

0:18:01 | understand that |

0:18:03 | which chose a a a few are possible for the channel or as i said uh one of them is |

0:18:08 | all the trend channels are perfect in terms of um |

0:18:12 | uh a target rcs |

0:18:14 | uh the second one has to |

0:18:17 | a good transmitters this other one has to be good transmitter |

0:18:21 | and |

0:18:22 | the question |

0:18:23 | before we you know |

0:18:25 | before we we we we we go forward is |

0:18:28 | you know of a valid question would be why not just take the expression we had previously |

0:18:33 | find a uniform power allocation and use it i mean we have an expression we can easily calculate what would |

0:18:39 | be the total power for uniform case then you have it here |

0:18:42 | and what we don't axis compare |

0:18:45 | optimally uh i don't think that the power to the to the scenario or just using uniform |

0:18:50 | and you see the results for case one case for using H two which H two means that |

0:18:55 | one and two are a one and for uh one and uh |

0:18:59 | transmitter one and five are the bad |

0:19:02 | you can see here that the total you the four would be one sixty two |

0:19:06 | people were compared to nineteen which has a fifty six percent saving power |

0:19:11 | so when compared to uniform power allocation be the same mean square error we save here about around fifty percent |

0:19:17 | by adapting the power |

0:19:19 | and not using uniform allocation |

0:19:22 | i this an i where these to we are the best we can see that |

0:19:25 | uh basically doesn't need to be transmitted together a performance |

0:19:28 | so you can see that |

0:19:30 | it which was different transmit based on geometry |

0:19:33 | well that they are uh a um the it it looks to uh white and uh that i aperture of |

0:19:38 | the um |

0:19:39 | a a a a a a the of the set of |

0:19:41 | transmitted that it uses |

0:19:43 | again you see the saving compared to a uniform a case |

0:19:46 | uh this is case as five to eight where we don't have the |

0:19:51 | lost or channel was but the only think that these a fact and you see that even when only in |

0:19:55 | terms of distance |

0:19:57 | there is a point in in using power location |

0:20:00 | it's still same some power |

0:20:03 | uh so two |

0:20:06 | and the summarise everything i well we look |

0:20:09 | into a resource away way operation of this to put multiple radar system |

0:20:14 | a by minimizing the total radiating eighteen power a a uh uh a a for a given in score trash |

0:20:18 | well |

0:20:19 | the optimization problem was solved to domain the composition at all do which basically generated |

0:20:25 | probably set of optimization problem that can be distributed |

0:20:29 | and in terms of processing |

0:20:30 | uh the power allocation expression we've level levelling |

0:20:33 | uh mechanism a a which gives the since like to how the system actually |

0:20:37 | um um i look at the power and we also showed that you for power allocation is not necessary or |

0:20:43 | optimal and that a |

0:20:45 | adapting the power |

0:20:46 | i in the way we suggested is uh offering saving in terms of power |

0:20:53 | i |

0:20:58 | i |

0:21:05 | hmmm |

0:21:08 | yeah |

0:21:13 | a |

0:21:21 | okay well |

0:21:22 | so i so was really you |

0:21:24 | sounds to switch are you just surface i'm to from the one into a is that correct |

0:21:29 | so |

0:21:30 | actually some of the points on the boundaries |

0:21:32 | yeah yeah but so to have to constrain so you are for my like to transform or and that you |

0:21:37 | want to a problems also four |

0:21:39 | you how missions you really have |

0:21:41 | do with |

0:21:42 | or |

0:21:42 | to |

0:21:43 | oh |

0:21:44 | something like that right right and the you from the which are on the bottom were |

0:21:47 | and also so of the gene that you wait until you |

0:21:51 | so to that to just low red |

0:21:52 | oh okay just one subject |

0:21:55 | thank you |

0:21:57 | yes |

0:21:58 | oh |

0:21:59 | oh |

0:22:00 | yeah |

0:22:02 | a a well we assume in this case we using the cramer-rao bound when we tracking a target and you |

0:22:08 | assume you have a uh uh uh a um you track the target to file the target then you have |

0:22:11 | some estimate on it and you keep and tracking it and you want to keep a tracking it in a |

0:22:15 | resource away way man so you use every time the previous estimate |

0:22:18 | to adapt the power |

0:22:21 | i |