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  • Is it possible that we override global variables?

    - by Ram Moj
    I have this function: function example(y) global TICTOC; tic TICTOC=5; toc end and I expect TICTOC=5 change the result of toc, since TICTOC is a global variable in tic and toc functions; but this is not the case; Does anyone know the reason? I like to know the answer, because I 'm worried to declare a global variable, which it's name has been declared global in some other functions, I'm not aware of. I saw this function in matlab 2008b help function tic % TIC Start a stopwatch timer. % TIC; any stuff; TOC % prints the time required. % See also: TOC, CLOCK. global TICTOC TICTOC = clock; function t = toc % TOC Read the stopwatch timer. % TOC prints the elapsed time since TIC was used. % t = TOC; saves elapsed time in t, does not print. % See also: TIC, ETIME. global TICTOC if nargout < 1 elapsed_time = etime(clock, TICTOC) else t = etime(clock, TICTOC); end thanks.

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  • How can I correctly calculate the direction for a moving object?

    - by Jakub Hampl
    I'm solving the following problem: I have an object and I know its position now and its position 300ms ago. I assume the object is moving. I have a point to which I want the object to get. What I need is to get the angle from my current object to the destination point in such a format that I know whether to turn left or right. The idea is to assume the current angle from the last known position and the current position. I'm trying to solve this in MATLAB. I've tried using several variations with atan2 but either I get the wrong angle in some situations (like when my object is going in circles) or I get the wrong angle in all situations. Examples of code that screws up: a = new - old; b = dest - new; alpha = atan2(a(2) - b(2), a(1) - b(1); where new is the current position (eg. x = 40; y = 60; new = [x y];), old is the 300ms old position and dest is the destination point. Edit Here's a picture to demonstrate the problem with a few examples: In the above image there are a few points plotted and annotated. The black line indicates our estimated current facing of the object. If the destination point is dest1 I would expect an angle of about 88°. If the destination point is dest2 I would expect an angle of about 110°. If the destination point is dest3 I would expect an angle of about -80°.

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  • Out-of-memory algorithms for addressing large arrays

    - by reve_etrange
    I am trying to deal with a very large dataset. I have k = ~4200 matrices (varying sizes) which must be compared combinatorially, skipping non-unique and self comparisons. Each of k(k-1)/2 comparisons produces a matrix, which must be indexed against its parents (i.e. can find out where it came from). The convenient way to do this is to (triangularly) fill a k-by-k cell array with the result of each comparison. These are ~100 X ~100 matrices, on average. Using single precision floats, it works out to 400 GB overall. I need to 1) generate the cell array or pieces of it without trying to place the whole thing in memory and 2) access its elements (and their elements) in like fashion. My attempts have been inefficient due to reliance on MATLAB's eval() as well as save and clear occurring in loops. for i=1:k [~,m] = size(data{i}); cur_var = ['H' int2str(i)]; %# if i == 1; save('FileName'); end; %# If using a single MAT file and need to create it. eval([cur_var ' = cell(1,k-i);']); for j=i+1:k [~,n] = size(data{j}); eval([cur_var '{i,j} = zeros(m,n,''single'');']); eval([cur_var '{i,j} = compare(data{i},data{j});']); end save(cur_var,cur_var); %# Add '-append' when using a single MAT file. clear(cur_var); end The other thing I have done is to perform the split when mod((i+j-1)/2,max(factor(k(k-1)/2))) == 0. This divides the result into the largest number of same-size pieces, which seems logical. The indexing is a little more complicated, but not too bad because a linear index could be used. Does anyone know/see a better way?

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  • Computing, storing, and retrieving values to and from an N-Dimensional matrix

    - by Adam S
    This question is probably quite different from what you are used to reading here - I hope it can provide a fun challenge. Essentially I have an algorithm that uses 5(or more) variables to compute a single value, called outcome. Now I have to implement this algorithm on an embedded device which has no memory limitations, but has very harsh processing constraints. Because of this, I would like to run a calculation engine which computes outcome for, say, 20 different values of each variable and stores this information in a file. You may think of this as a 5(or more)-dimensional matrix or 5(or more)-dimensional array, each dimension being 20 entries long. In any modern language, filling this array is as simple as having 5(or more) nested for loops. The tricky part is that I need to dump these values into a file that can then be placed onto the embedded device so that the device can use it as a lookup table. The questions now, are: What format(s) might be acceptable for storing the data? What programs (MATLAB, C#, etc) might be best suited to compute the data? C# must be used to import the data on the device - is this possible given your answer to #1?

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  • find consecutive nonzero values

    - by thymeandspace
    I am trying to write a simple MATLAB program that will find the first chain (more than 70) of consecutive nonzero values and return the starting value of that consecutive chain. I am working with movement data from a joystick and there are a few thousand rows of data with a mix of zeros and nonzero values before the actual trial begins (coming from subjects slightly moving the joystick before the trial actually started). I need to get rid of these rows before I can start analyzing the movement from the trials. I am sure this is a relatively simple thing to do so I was hoping someone could offer insight. Thank you in advance -Lilly EDIT: Here's what I tried: s = zeros(size(x1)); for i=2:length(x1) if(x1(i-1) ~= 0) s(i) = 1 + s(i-1); end end display(S); for a vector x1 which has a max chain of 72 but I dont know how to find the max chain and return its first value, so I know where to trim. I also really don't think this is the best strategy, since the max chain in my data will be tens of thousands of values. Thanks for helping me edit, Steve. :)

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  • How could I eliminate the meter names with an 'x' after them?

    - by Rose Comete
    Hi I imported some Excel data into MatLab - it is a list of about 200 meter names with about 28 rows each, but the problem is that for each there is a duplicate for the other direction - with the same meter name with an 'x' after it. Does anyone have any ideas as to how I can eliminate these ones with an 'x' after wards? Attached is the part of my code that imports the data, but unfortunately I have not got enough points on this site yet to be allowed to upload data/photos x clear all fid=fopen('sue1.csv'); % Open the file sue1.csv and read it all and put it into an array data = textscan(fid,'%s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s','Delimiter',',','CollectOutput',1); fclose(fid) j = 1; k = 1; % j - turbine number, k - date number for i = 1:length(data{1,1}) % Run through all the data if strcmp(data{1,1}(i),'') == 0 meterold{j}(k,:) = data{1,1}(i,:); % if strcmp(data{1,1}(i),'MeterName') == 0 % nummeter{j}(k,:) = str2num(data{1,1}(i,3:end)); % end k = k + 1; else % These commands are followed in the strings match (empty line) k = 1; % Reset the day counter as we're back to the beginning j = j + 1; % Add one to the meter counter as we're now looking at % a new turbine end end

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  • Parallelize or vectorize all-against-all operation on a large number of matrices?

    - by reve_etrange
    I have approximately 5,000 matrices with the same number of rows and varying numbers of columns (20 x ~200). Each of these matrices must be compared against every other in a dynamic programming algorithm. In this question, I asked how to perform the comparison quickly and was given an excellent answer involving a 2D convolution. Serially, iteratively applying that method, like so list = who('data_matrix_prefix*') H = cell(numel(list),numel(list)); for i=1:numel(list) for j=1:numel(list) if i ~= j eval([ 'H{i,j} = compare(' char(list(i)) ',' char(list(j)) ');']); end end end is fast for small subsets of the data (e.g. for 9 matrices, 9*9 - 9 = 72 calls are made in ~1 s). However, operating on all the data requires almost 25 million calls. I have also tried using deal() to make a cell array composed entirely of the next element in data, so I could use cellfun() in a single loop: # who(), load() and struct2cell() calls place k data matrices in a 1D cell array called data. nextData = cell(k,1); for i=1:k [nextData{:}] = deal(data{i}); H{:,i} = cellfun(@compare,data,nextData,'UniformOutput',false); end Unfortunately, this is not really any faster, because all the time is in compare(). Both of these code examples seem ill-suited for parallelization. I'm having trouble figuring out how to make my variables sliced. compare() is totally vectorized; it uses matrix multiplication and conv2() exclusively (I am under the impression that all of these operations, including the cellfun(), should be multithreaded in MATLAB?). Does anyone see a (explicitly) parallelized solution or better vectorization of the problem?

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  • Filter design for audio signal.

    - by beanyblue
    What I am trying to do is simple. I have a few .wav files. I want to remove noise and filter out specific frequencies. I don't have matlab and I intend to write my own code for all the filters. Right now, I have a way to read the .wav file and dump out the structure into a text file. My questions are the following: Can I directly apply the digital filters on this sampled data?{ ie, can I directly do a convolution between my input samples and h(n) for the filter function that i choose?). How do I choose the number of coefficients for the Window function? I have octave, so if someone can point me to anything that gives me some idea on how to process the .wav file using octave, that would be great too. I want to be able to filter out the frequency and then listen to the sound again. Is this possible with octave? I'm just a beginner with these kinds of things, so please bear with me if my questions are too naive. Any help will be great.

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  • Have a trouble with the function roots

    - by user3707462
    Hey guys I have multiple problems with using function 'roots'. I Have to find zeros of 's^1000 + 1'. I made Y = zeros(1,1000) then manually changed the 1000th matrice to '1'. but then 'root' function does not work with it ! Another problem is that I am having trouble with matrix multiplication. The question is finding zeros(roots) of (s^6 + 6*s^5 + 15*s^4 + 20*s^3 + 15*s^2 + 6*s +1)*(s^6 + 6s^5 + 15*s^4 +15*s^2 +6*s +1) so i did: a = [1 6 15 20 15 6 1] b = [1 6 15 0 15 6 1] y = a.*b; roots(y) but this gives me -27.9355 + 0.0000i -8.2158 + 0.0000i 0.1544 + 0.9880i 0.1544 - 0.9880i -0.1217 + 0.0000i -0.0358 + 0.0000i where I calculate the original equation with wolfram then I have made matrix as : p = [1 12 66 200 375 492 524 492 375 200 66 12 1] roots(p) and this gives me : -3.1629 + 2.5046i -3.1629 - 2.5046i 0.3572 + 0.9340i 0.3572 - 0.9340i -1.0051 + 0.0000i -1.0025 + 0.0044i -1.0025 - 0.0044i -0.9975 + 0.0044i -0.9975 - 0.0044i -0.9949 + 0.0000i -0.1943 + 0.1539i -0.1943 - 0.1539i and I think the second solution is right (that is what wolfram alpha gave me) How would you answer these two questions through matlab guys?

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  • Upgrading from TFS 2010 RC to TFS 2010 RTM done

    - by Martin Hinshelwood
    Today is the big day, with the Launch of Visual Studio 2010 already done in Asia, and rolling around the world towards us, we are getting ready for the RTM (Released). We have had TFS 2010 in Production for nearly 6 months and have had only minimal problems. Update 12th April 2010  – Added Scott Hanselman’s tweet about the MSDN download release time. SSW was the first company in the world outside of Microsoft to deploy Visual Studio 2010 Team Foundation Server to production, not once, but twice. I am hoping to make it 3 in a row, but with all the hype around the new version, and with it being a production release and not just a go-live, I think there will be a lot of competition. Developers: MSDN will be updated with #vs2010 downloads and details at 10am PST *today*! @shanselman - Scott Hanselman Same as before, we need to Uninstall 2010 RC and install 2010 RTM. The installer will take care of all the complexity of actually upgrading any schema changes. If you are upgrading from TFS 2008 to TFS2010 you can follow our Rules To Better TFS 2010 Migration and read my post on our successes.   We run TFS 2010 in a Hyper-V virtual environment, so we have the advantage of running a snapshot as well as taking a DB backup. Done - Snapshot the hyper-v server Microsoft does not support taking a snapshot of a running server, for very good reason, and Brian Harry wrote a post after my last upgrade with the reason why you should never snapshot a running server. Done - Uninstall Visual Studio Team Explorer 2010 RC You will need to uninstall all of the Visual Studio 2010 RC client bits that you have on the server. Done - Uninstall TFS 2010 RC Done - Install TFS 2010 RTM Done - Configure TFS 2010 RTM Pick the Upgrade option and point it at your existing “tfs_Configuration” database to load all of the existing settings Done - Upgrade the SharePoint Extensions Upgrade Build Servers (Pending) Test the server The back out plan, and you should always have one, is to restore the snapshot. Upgrading to Team Foundation Server 2010 – Done The first thing you need to do is off the TFS server and then log into the Hyper-v server and create a snapshot. Figure: Make sure you turn the server off and delete all old snapshots before you take a new one I noticed that the snapshot that was taken before the Beta 2 to RC upgrade was still there. You should really delete old snapshots before you create a new one, but in this case the SysAdmin (who is currently tucked up in bed) asked me not to. I guess he is worried about a developer messing up his server Turn your server on and wait for it to boot in anticipation of all the nice shiny RTM’ness that is coming next. The upgrade procedure for TFS2010 is to uninstal the old version and install the new one. Figure: Remove Visual Studio 2010 Team Foundation Server RC from the system.   Figure: Most of the heavy lifting is done by the Uninstaller, but make sure you have removed any of the client bits first. Specifically Visual Studio 2010 or Team Explorer 2010.  Once the uninstall is complete, this took around 5 minutes for me, you can begin the install of the RTM. Running the 64 bit OS will allow the application to use more than 2GB RAM, which while not common may be of use in heavy load situations. Figure: It is always recommended to install the 64bit version of a server application where possible. I do not think it is likely, with SharePoint 2010 and Exchange 2010  and even Windows Server 2008 R2 being 64 bit only, I do not think there will be another release of a server app that is 32bit. You then need to choose what it is you want to install. This depends on how you are running TFS and on how many servers. In our case we run TFS and the Team Foundation Build Service (controller only) on out TFS server along with Analysis services and Reporting Services. But our SharePoint server lives elsewhere. Figure: This always confuses people, but in reality it makes sense. Don’t install what you do not need. Every extra you install has an impact of performance. If you are integrating with SharePoint you will need to run this install on every Front end server in your farm and don’t forget to upgrade your Build servers and proxy servers later. Figure: Selecting only Team Foundation Server (TFS) and Team Foundation Build Services (TFBS)   It is worth noting that if you have a lot of builds kicking off, and hence a lot of get operations against your TFS server, you can use a proxy server to cache the source control on another server in between your TFS server and your build servers. Figure: Installing Microsoft .NET Framework 4 takes the most time. Figure: Now run Windows Update, and SSW Diagnostic to make sure all your bits and bobs are up to date. Note: SSW Diagnostic will check your Power Tools, Add-on’s, Check in Policies and other bits as well. Configure Team Foundation Server 2010 – Done Now you can configure the server. If you have no key you will need to pick “Install a Trial Licence”, but it is only £500, or free with a MSDN subscription. Anyway, if you pick Trial you get 90 days to get your key. Figure: You can pick trial and add your key later using the TFS Server Admin. Here is where the real choices happen. We are doing an Upgrade from a previous version, so I will pick Upgrade the same as all you folks that are using the RC or TFS 2008. Figure: The upgrade wizard takes your existing 2010 or 2008 databases and upgraded them to the release.   Once you have entered your database server name you can click “List available databases” and it will show what it can upgrade. Figure: Select your database from the list and at this point, make sure you have a valid backup. At this point you have not made ANY changes to the databases. At this point the configuration wizard will load configuration from your existing database if you have one. If you are upgrading TFS 2008 refer to Rules To Better TFS 2010 Migration. Mostly during the wizard the default values will suffice, but depending on the configuration you want you can pick different options. Figure: Set the application tier account and Authentication method to use. We use NTLM to keep things simple as we host our TFS server externally for our remote developers.  Figure: Setting your TFS server URL’s to be the remote URL’s allows the reports to be accessed without using VPN. Very handy for those remote developers. Figure: Detected the existing Warehouse no problem. Figure: Again we love green ticks. It gives us a warm fuzzy feeling. Figure: The username for connecting to Reporting services should be a domain account (if you are on a domain that is). Figure: Setup the SharePoint integration to connect to your external SharePoint server. You can take the option to connect later.   You then need to run all of your readiness checks. These check can save your life! it will check all of the settings that you have entered as well as checking all the external services are configures and running properly. There are two reasons that TFS 2010 is so easy and painless to install where previous version were not. Microsoft changes the install to two steps, Install and configuration. The second reason is that they have pulled out all of the stops in making the install run all the checks necessary to make sure that once you start the install that it will complete. if you find any errors I recommend that you report them on http://connect.microsoft.com so everyone can benefit from your misery.   Figure: Now we have everything setup the configuration wizard can do its work.  Figure: Took a while on the “Web site” stage for some point, but zipped though after that.  Figure: last wee bit. TFS Needs to do a little tinkering with the data to complete the upgrade. Figure: All upgraded. I am not worried about the yellow triangle as SharePoint was being a little silly Exception Message: TF254021: The account name or password that you specified is not valid. (type TfsAdminException) Exception Stack Trace:    at Microsoft.TeamFoundation.Management.Controls.WizardCommon.AccountSelectionControl.TestLogon(String connectionString)    at System.ComponentModel.BackgroundWorker.WorkerThreadStart(Object argument) [Info   @16:10:16.307] Benign exception caught as part of verify: Exception Message: TF255329: The following site could not be accessed: http://projects.ssw.com.au/. The server that you specified did not return the expected response. Either you have not installed the Team Foundation Server Extensions for SharePoint Products on this server, or a firewall is blocking access to the specified site or the SharePoint Central Administration site. For more information, see the Microsoft Web site (http://go.microsoft.com/fwlink/?LinkId=161206). (type TeamFoundationServerException) Exception Stack Trace:    at Microsoft.TeamFoundation.Client.SharePoint.WssUtilities.VerifyTeamFoundationSharePointExtensions(ICredentials credentials, Uri url)    at Microsoft.TeamFoundation.Admin.VerifySharePointSitesUrl.Verify() Inner Exception Details: Exception Message: TF249064: The following Web service returned an response that is not valid: http://projects.ssw.com.au/_vti_bin/TeamFoundationIntegrationService.asmx. This Web service is used for the Team Foundation Server Extensions for SharePoint Products. Either the extensions are not installed, the request resulted in HTML being returned, or there is a problem with the URL. Verify that the following URL points to a valid SharePoint Web application and that the application is available: http://projects.ssw.com.au. If the URL is correct and the Web application is operating normally, verify that a firewall is not blocking access to the Web application. (type TeamFoundationServerInvalidResponseException) Exception Data Dictionary: ResponseStatusCode = InternalServerError I’ll look at SharePoint after, probably the SharePoint box just needs a restart or a kick If there is a problem with SharePoint it will come out in testing, But I will definatly be passing this on to Microsoft.   Upgrading the SharePoint connector to TFS 2010 You will need to upgrade the Extensions for SharePoint Products and Technologies on all of your SharePoint farm front end servers. To do this uninstall  the TFS 2010 RC from it in the same way as the server, and then install just the RTM Extensions. Figure: Only install the SharePoint Extensions on your SharePoint front end servers. TFS 2010 supports both SharePoint 2007 and SharePoint 2010.   Figure: When you configure SharePoint it uploads all of the solutions and templates. Figure: Everything is uploaded Successfully. Figure: TFS even remembered the settings from the previous installation, fantastic.   Upgrading the Team Foundation Build Servers to TFS 2010 Just like on the SharePoint servers you will need to upgrade the Build Server to the RTM. Just uninstall TFS 2010 RC and then install only the Team Foundation Build Services component. Unlike on the SharePoint server you will probably have some version of Visual Studio installed. You will need to remove this as well. (Coming Soon) Connecting Visual Studio 2010 / 2008 / 2005 and Eclipse to TFS2010 If you have developers still on Visual Studio 2005 or 2008 you will need do download the respective compatibility pack: Visual Studio Team System 2005 Service Pack 1 Forward Compatibility Update for Team Foundation Server 2010 Visual Studio Team System 2008 Service Pack 1 Forward Compatibility Update for Team Foundation Server 2010 If you are using Eclipse you can download the new Team Explorer Everywhere install for connecting to TFS. Get your developers to check that you have the latest version of your applications with SSW Diagnostic which will check for Service Packs and hot fixes to Visual Studio as well.   Technorati Tags: TFS,TFS2010,TFS 2010,Upgrade

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  • How do I figure out which version of OpenCL comes with my nvidia-current-dev package?

    - by levesque
    I am running Ubuntu 10.04 with an nvidia geforce GT 240 card. I installed nvidia-current-dev with apt-get, thus I got a bunch of headers and lib files for OpenCL. However, I am unable to figure out if I have OpenCL 1.0 or 1.1 (could be 1.2, though I doubt it). Does anyone know a way to figure out which version of OpenCL comes with the nvidia-current-dev package (version: 195.36.24-0ubuntu1~10.04.2)? If not, is there another place where I could get an answer?

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  • Microsoft sponsor de la Fondation Linux ? L'éditeur figure parmi les Sponsors Gold de la conférence LinuxCon Europe

    Microsoft sponsor de la Fondation Linux ? L'éditeur figure parmi les Sponsors Gold de la conférence LinuxCon Europe Microsoft, sponsor de la fondation Linux ou en passe de l'être ? C'est en tout cas ce que laisse présager la liste des sponsors de l'événement LinuxCon Europe 2012 autour de l'écosystème Linux qui s'est déroulé la semaine dernière à Barcelone en Espagne et l'orientation vers l'ouverture de la société. La liste publiée sur le site de l'événement permet de constater que la firme de Redmond figure parmi les sponsors Gold de l'événement au même titre que HP, Red Hat ou encore Samsung. [IMG]http://rdonfack.develop...

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  • Rotation Matrix calculates by column not by row

    - by pinnacler
    I have a class called forest and a property called fixedPositions that stores 100 points (x,y) and they are stored 250x2 (rows x columns) in MatLab. When I select 'fixedPositions', I can click scatter and it will plot the points. Now, I want to rotate the plotted points and I have a rotation matrix that will allow me to do that. The below code should work: theta = obj.heading * pi/180; apparent = [cos(theta) -sin(theta) ; sin(theta) cos(theta)] * obj.fixedPositions; But it wont. I get this error. ??? Error using == mtimes Inner matrix dimensions must agree. Error in == landmarkslandmarks.get.apparentPositions at 22 apparent = [cos(theta) -sin(theta) ; sin(theta) cos(theta)] * obj.fixedPositions; When I alter forest.fixedPositions to store the variables 2x250 instead of 250x2, the above code will work, but it wont plot. I'm going to be plotting fixedPositions constantly in a simulation, so I'd prefer to leave it as it, and make the rotation work instead. Any ideas? Also, fixed positions, is the position of the xy points as if you were looking straight ahead. i.e. heading = 0. heading is set to 45, meaning I want to rotate points clockwise 45 degrees. Here is my code: classdef landmarks properties fixedPositions %# positions in a fixed coordinate system. [x, y] heading = 45; %# direction in which the robot is facing end properties (Dependent) apparentPositions end methods function obj = landmarks(numberOfTrees) %# randomly generates numberOfTrees amount of x,y coordinates and set %the array or matrix (not sure which) to fixedPositions obj.fixedPositions = 100 * rand([numberOfTrees,2]) .* sign(rand([numberOfTrees,2]) - 0.5); end function obj = set.apparentPositions(obj,~) theta = obj.heading * pi/180; [cos(theta) -sin(theta) ; sin(theta) cos(theta)] * obj.fixedPositions; end function apparent = get.apparentPositions(obj) %# rotate obj.positions using obj.facing to generate the output theta = obj.heading * pi/180; apparent = [cos(theta) -sin(theta) ; sin(theta) cos(theta)] * obj.fixedPositions; end end end P.S. If you change one line to this: obj.fixedPositions = 100 * rand([2,numberOfTrees]) .* sign(rand([2,numberOfTrees]) - 0.5); Everything will work fine... it just wont plot.

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  • second y-axis on pcolor plot

    - by user1155751
    Is it possible to prodce a pcolor plot with 2 yaxis? Consider the following example: clear all temp = 1 + (20-1).*rand(365,12); depth = 1:12; time =1:365; data2 = 1 + (60-1).*rand(12,1); time2 = [28,56,84,124,150,184,210,234,265,288,312,342]; figure; pcolor(time,depth,temp');axis ij; shading interp hold on plot(time2,data2,'w','linewidth',3); Instead of plotting the second dataset on the same y axis I would like it to placed on its own y-axis. Is this possible?

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  • Why is this class re-initialized every time?

    - by pinnacler
    I have 4 files and the code 'works' as expected. I try to clean everything up, place code into functions, etc... and everything looks fine... and it doesn't work. Can somebody please explain why MatLab is so quirky... or am I just stupid? Normally, I type terminator = simulation(100,20,0,0,0,1); terminator.animate(); and it should produce a map of trees with the terminator walking around in a forest. Everything rotates to his perspective. When I break it into functions... everything ceases to work. I really only changed a few lines of code, shown in comments. Code that works: classdef simulation properties landmarks robot end methods function obj = simulation(mapSize, trees, x,y,heading,velocity) obj.landmarks = landmarks(mapSize, trees); obj.robot = robot(x,y,heading,velocity); end function animate(obj) %Setup Plots fig=figure; xlabel('meters'), ylabel('meters') set(fig, 'name', 'Phil''s AWESOME 80''s Robot Simulator') xymax = obj.landmarks.mapSize*3; xymin = -(obj.landmarks.mapSize*3); l=scatter([0],[0],'bo'); axis([xymin xymax xymin xymax]); obj.landmarks.apparentPositions %Simulation Loop THIS WAS ORGANIZED for n = 1:720, %Calculate and Set Heading/Location obj.robot.headingChange = navigate(n); %Update Position obj.robot.heading = obj.robot.heading + obj.robot.headingChange; obj.landmarks.heading = obj.robot.heading; y = cosd(obj.robot.heading); x = sind(obj.robot.heading); obj.robot.x = obj.robot.x + (x*obj.robot.velocity); obj.robot.y = obj.robot.y + (y*obj.robot.velocity); obj.landmarks.x = obj.robot.x; obj.landmarks.y = obj.robot.y; %Animate set(l,'XData',obj.landmarks.apparentPositions(:,1),'YData',obj.landmarks.apparentPositions(:,2)); rectangle('Position',[-2,-2,4,4]); drawnow end end end end ----------- classdef landmarks properties fixedPositions %# positions in a fixed coordinate system. [ x, y ] mapSize = 10; %Map Size. Value is side of square x=0; y=0; heading=0; headingChange=0; end properties (Dependent) apparentPositions end methods function obj = landmarks(mapSize, numberOfTrees) obj.mapSize = mapSize; obj.fixedPositions = obj.mapSize * rand([numberOfTrees, 2]) .* sign(rand([numberOfTrees, 2]) - 0.5); end function apparent = get.apparentPositions(obj) %-STILL ROTATES AROUND ORIGINAL ORIGIN currentPosition = [obj.x ; obj.y]; apparent = bsxfun(@minus,(obj.fixedPositions)',currentPosition)'; apparent = ([cosd(obj.heading) -sind(obj.heading) ; sind(obj.heading) cosd(obj.heading)] * (apparent)')'; end end end ---------- classdef robot properties x y heading velocity headingChange end methods function obj = robot(x,y,heading,velocity) obj.x = x; obj.y = y; obj.heading = heading; obj.velocity = velocity; end end end ---------- function headingChange = navigate(n) %steeringChange = 5 * rand(1) * sign(rand(1) - 0.5); Most chaotic shit %Draw an S if n <270 headingChange=1; elseif n<540 headingChange=-1; elseif n<720 headingChange=1; else headingChange=1; end end Code that does not work... classdef simulation properties landmarks robot end methods function obj = simulation(mapSize, trees, x,y,heading,velocity) obj.landmarks = landmarks(mapSize, trees); obj.robot = robot(x,y,heading,velocity); end function animate(obj) %Setup Plots fig=figure; xlabel('meters'), ylabel('meters') set(fig, 'name', 'Phil''s AWESOME 80''s Robot Simulator') xymax = obj.landmarks.mapSize*3; xymin = -(obj.landmarks.mapSize*3); l=scatter([0],[0],'bo'); axis([xymin xymax xymin xymax]); obj.landmarks.apparentPositions %Simulation Loop for n = 1:720, %Calculate and Set Heading/Location %Update Position headingChange = navigate(n); obj.robot.updatePosition(headingChange); obj.landmarks.updatePerspective(obj.robot.heading, obj.robot.x, obj.robot.y); %Animate set(l,'XData',obj.landmarks.apparentPositions(:,1),'YData',obj.landmarks.apparentPositions(:,2)); rectangle('Position',[-2,-2,4,4]); drawnow end end end end ----------------- classdef landmarks properties fixedPositions; %# positions in a fixed coordinate system. [ x, y ] mapSize; %Map Size. Value is side of square x; y; heading; headingChange; end properties (Dependent) apparentPositions end methods function obj = createLandmarks(mapSize, numberOfTrees) obj.mapSize = mapSize; obj.fixedPositions = obj.mapSize * rand([numberOfTrees, 2]) .* sign(rand([numberOfTrees, 2]) - 0.5); end function apparent = get.apparentPositions(obj) %-STILL ROTATES AROUND ORIGINAL ORIGIN currentPosition = [obj.x ; obj.y]; apparent = bsxfun(@minus,(obj.fixedPositions)',currentPosition)'; apparent = ([cosd(obj.heading) -sind(obj.heading) ; sind(obj.heading) cosd(obj.heading)] * (apparent)')'; end function updatePerspective(obj,tempHeading,tempX,tempY) obj.heading = tempHeading; obj.x = tempX; obj.y = tempY; end end end ----------------- classdef robot properties x y heading velocity end methods function obj = robot(x,y,heading,velocity) obj.x = x; obj.y = y; obj.heading = heading; obj.velocity = velocity; end function updatePosition(obj,headingChange) obj.heading = obj.heading + headingChange; tempy = cosd(obj.heading); tempx = sind(obj.heading); obj.x = obj.x + (tempx*obj.velocity); obj.y = obj.y + (tempy*obj.velocity); end end end The navigate function is the same... I would appreciate any help as to why things aren't working. All I did was take the code from the first section from under comment: %Simulation Loop THIS WAS ORGANIZED and break it into 2 functions. One in robot and one in landmarks. Is a new instance created every time because it's constantly printing the same heading for this line int he robot class obj.heading = obj.heading + headingChange;

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  • how to apply iddata into calculation?

    - by Sam
    I am trying to figure out how to combine the input and output data into the ARX model and then apply it into the BIC (Bayesian Information Criterion) formula. Below is the code that I am currently working on: for i=1:30; %% Set Model Order data=iddata(output,input,1); model = arx(data,[8 9 i]); yp = predict(model,data); ye = regress(data,yp{1,1}(1:4018,1)); M(i) = var(yp); BIC(i)=(N+i*(log(N)-1))/(N-i)*log(M(i)); end But it does not work. It keep on giving me an error something like below: "The syntax "Data{...}" is not supported. Use the "getexp" command to extract individual experiments from an IDDATA object." I did not understand what does that mean. Can someone explain it to me and where do I do wrong on my piece of code? Thanks in advance. Sam.

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  • Error in Ordinary Differential Equation representation

    - by Priya M
    UPDATE I am trying to find the Lyapunov Exponents given in link LE. I am trying to figure it out and understand it by taking the following eqs for my case. These are a set of ordinary differential equations (these are just for testing how to work with cos and sin as ODE) f(1)=ALPHA*(y-x); f(2)=x*(R-z)-y; f(3) = 10*cos(x); and x=X(1); y=X(2); cos(y)=X(3); f1 means dx/dt;f2 dy/dt and f3 in this case would be -10sinx. However,when expressing as x=X(1);y=X(2);i am unsure how to express for cos.This is just a trial example i was doing so as to know how to work with equations where we have a cos,sin etc terms as a function of another variable. When using ode45 to solve these Eqs [T,Res]=sol(3,@test_eq,@ode45,0,0.01,20,[7 2 100 ],10); it throws the following error ??? Attempted to access (2); index must be a positive integer or logical. Error in ==> Eq at 19 x=X(1); y=X(2); cos(x)=X(3); Is my representation x=X(1); y=X(2); cos(y)=X(3); alright? How to resolve the error? Thank you

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  • Problems with real-valued input deep belief networks (of RBMs)

    - by Junier
    I am trying to recreate the results reported in Reducing the dimensionality of data with neural networks of autoencoding the olivetti face dataset with an adapted version of the MNIST digits matlab code, but am having some difficulty. It seems that no matter how much tweaking I do on the number of epochs, rates, or momentum the stacked RBMs are entering the fine-tuning stage with a large amount of error and consequently fail to improve much at the fine-tuning stage. I am also experiencing a similar problem on another real-valued dataset. For the first layer I am using a RBM with a smaller learning rate (as described in the paper) and with negdata = poshidstates*vishid' + repmat(visbiases,numcases,1); I'm fairly confident I am following the instructions found in the supporting material but I cannot achieve the correct errors. Is there something I am missing? See the code I'm using for real-valued visible unit RBMs below, and for the whole deep training. The rest of the code can be found here. rbmvislinear.m: epsilonw = 0.001; % Learning rate for weights epsilonvb = 0.001; % Learning rate for biases of visible units epsilonhb = 0.001; % Learning rate for biases of hidden units weightcost = 0.0002; initialmomentum = 0.5; finalmomentum = 0.9; [numcases numdims numbatches]=size(batchdata); if restart ==1, restart=0; epoch=1; % Initializing symmetric weights and biases. vishid = 0.1*randn(numdims, numhid); hidbiases = zeros(1,numhid); visbiases = zeros(1,numdims); poshidprobs = zeros(numcases,numhid); neghidprobs = zeros(numcases,numhid); posprods = zeros(numdims,numhid); negprods = zeros(numdims,numhid); vishidinc = zeros(numdims,numhid); hidbiasinc = zeros(1,numhid); visbiasinc = zeros(1,numdims); sigmainc = zeros(1,numhid); batchposhidprobs=zeros(numcases,numhid,numbatches); end for epoch = epoch:maxepoch, fprintf(1,'epoch %d\r',epoch); errsum=0; for batch = 1:numbatches, if (mod(batch,100)==0) fprintf(1,' %d ',batch); end %%%%%%%%% START POSITIVE PHASE %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% data = batchdata(:,:,batch); poshidprobs = 1./(1 + exp(-data*vishid - repmat(hidbiases,numcases,1))); batchposhidprobs(:,:,batch)=poshidprobs; posprods = data' * poshidprobs; poshidact = sum(poshidprobs); posvisact = sum(data); %%%%%%%%% END OF POSITIVE PHASE %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% poshidstates = poshidprobs > rand(numcases,numhid); %%%%%%%%% START NEGATIVE PHASE %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% negdata = poshidstates*vishid' + repmat(visbiases,numcases,1);% + randn(numcases,numdims) if not using mean neghidprobs = 1./(1 + exp(-negdata*vishid - repmat(hidbiases,numcases,1))); negprods = negdata'*neghidprobs; neghidact = sum(neghidprobs); negvisact = sum(negdata); %%%%%%%%% END OF NEGATIVE PHASE %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% err= sum(sum( (data-negdata).^2 )); errsum = err + errsum; if epoch>5, momentum=finalmomentum; else momentum=initialmomentum; end; %%%%%%%%% UPDATE WEIGHTS AND BIASES %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% vishidinc = momentum*vishidinc + ... epsilonw*( (posprods-negprods)/numcases - weightcost*vishid); visbiasinc = momentum*visbiasinc + (epsilonvb/numcases)*(posvisact-negvisact); hidbiasinc = momentum*hidbiasinc + (epsilonhb/numcases)*(poshidact-neghidact); vishid = vishid + vishidinc; visbiases = visbiases + visbiasinc; hidbiases = hidbiases + hidbiasinc; %%%%%%%%%%%%%%%% END OF UPDATES %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% end fprintf(1, '\nepoch %4i error %f \n', epoch, errsum); end dofacedeepauto.m: clear all close all maxepoch=200; %In the Science paper we use maxepoch=50, but it works just fine. numhid=2000; numpen=1000; numpen2=500; numopen=30; fprintf(1,'Pretraining a deep autoencoder. \n'); fprintf(1,'The Science paper used 50 epochs. This uses %3i \n', maxepoch); load fdata %makeFaceData; [numcases numdims numbatches]=size(batchdata); fprintf(1,'Pretraining Layer 1 with RBM: %d-%d \n',numdims,numhid); restart=1; rbmvislinear; hidrecbiases=hidbiases; save mnistvh vishid hidrecbiases visbiases; maxepoch=50; fprintf(1,'\nPretraining Layer 2 with RBM: %d-%d \n',numhid,numpen); batchdata=batchposhidprobs; numhid=numpen; restart=1; rbm; hidpen=vishid; penrecbiases=hidbiases; hidgenbiases=visbiases; save mnisthp hidpen penrecbiases hidgenbiases; fprintf(1,'\nPretraining Layer 3 with RBM: %d-%d \n',numpen,numpen2); batchdata=batchposhidprobs; numhid=numpen2; restart=1; rbm; hidpen2=vishid; penrecbiases2=hidbiases; hidgenbiases2=visbiases; save mnisthp2 hidpen2 penrecbiases2 hidgenbiases2; fprintf(1,'\nPretraining Layer 4 with RBM: %d-%d \n',numpen2,numopen); batchdata=batchposhidprobs; numhid=numopen; restart=1; rbmhidlinear; hidtop=vishid; toprecbiases=hidbiases; topgenbiases=visbiases; save mnistpo hidtop toprecbiases topgenbiases; backpropface; Thanks for your time

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  • Problems with real-valued deep belief networks (of RBMs)

    - by Junier
    I am trying to recreate the results reported in Reducing the dimensionality of data with neural networks of autoencoding the olivetti face dataset with an adapted version of the MNIST digits matlab code, but am having some difficulty. It seems that no matter how much tweaking I do on the number of epochs, rates, or momentum the stacked RBMs are entering the fine-tuning stage with a large amount of error and consequently fail to improve much at the fine-tuning stage. I am also experiencing a similar problem on another real-valued dataset. For the first layer I am using a RBM with a smaller learning rate (as described in the paper) and with negdata = poshidstates*vishid' + repmat(visbiases,numcases,1); I'm fairly confident I am following the instructions found in the supporting material but I cannot achieve the correct errors. Is there something I am missing? See the code I'm using for real-valued visible unit RBMs below, and for the whole deep training. The rest of the code can be found here. rbmvislinear.m: epsilonw = 0.001; % Learning rate for weights epsilonvb = 0.001; % Learning rate for biases of visible units epsilonhb = 0.001; % Learning rate for biases of hidden units weightcost = 0.0002; initialmomentum = 0.5; finalmomentum = 0.9; [numcases numdims numbatches]=size(batchdata); if restart ==1, restart=0; epoch=1; % Initializing symmetric weights and biases. vishid = 0.1*randn(numdims, numhid); hidbiases = zeros(1,numhid); visbiases = zeros(1,numdims); poshidprobs = zeros(numcases,numhid); neghidprobs = zeros(numcases,numhid); posprods = zeros(numdims,numhid); negprods = zeros(numdims,numhid); vishidinc = zeros(numdims,numhid); hidbiasinc = zeros(1,numhid); visbiasinc = zeros(1,numdims); sigmainc = zeros(1,numhid); batchposhidprobs=zeros(numcases,numhid,numbatches); end for epoch = epoch:maxepoch, fprintf(1,'epoch %d\r',epoch); errsum=0; for batch = 1:numbatches, if (mod(batch,100)==0) fprintf(1,' %d ',batch); end %%%%%%%%% START POSITIVE PHASE %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% data = batchdata(:,:,batch); poshidprobs = 1./(1 + exp(-data*vishid - repmat(hidbiases,numcases,1))); batchposhidprobs(:,:,batch)=poshidprobs; posprods = data' * poshidprobs; poshidact = sum(poshidprobs); posvisact = sum(data); %%%%%%%%% END OF POSITIVE PHASE %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% poshidstates = poshidprobs > rand(numcases,numhid); %%%%%%%%% START NEGATIVE PHASE %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% negdata = poshidstates*vishid' + repmat(visbiases,numcases,1);% + randn(numcases,numdims) if not using mean neghidprobs = 1./(1 + exp(-negdata*vishid - repmat(hidbiases,numcases,1))); negprods = negdata'*neghidprobs; neghidact = sum(neghidprobs); negvisact = sum(negdata); %%%%%%%%% END OF NEGATIVE PHASE %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% err= sum(sum( (data-negdata).^2 )); errsum = err + errsum; if epoch>5, momentum=finalmomentum; else momentum=initialmomentum; end; %%%%%%%%% UPDATE WEIGHTS AND BIASES %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% vishidinc = momentum*vishidinc + ... epsilonw*( (posprods-negprods)/numcases - weightcost*vishid); visbiasinc = momentum*visbiasinc + (epsilonvb/numcases)*(posvisact-negvisact); hidbiasinc = momentum*hidbiasinc + (epsilonhb/numcases)*(poshidact-neghidact); vishid = vishid + vishidinc; visbiases = visbiases + visbiasinc; hidbiases = hidbiases + hidbiasinc; %%%%%%%%%%%%%%%% END OF UPDATES %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% end fprintf(1, '\nepoch %4i error %f \n', epoch, errsum); end dofacedeepauto.m: clear all close all maxepoch=200; %In the Science paper we use maxepoch=50, but it works just fine. numhid=2000; numpen=1000; numpen2=500; numopen=30; fprintf(1,'Pretraining a deep autoencoder. \n'); fprintf(1,'The Science paper used 50 epochs. This uses %3i \n', maxepoch); load fdata %makeFaceData; [numcases numdims numbatches]=size(batchdata); fprintf(1,'Pretraining Layer 1 with RBM: %d-%d \n',numdims,numhid); restart=1; rbmvislinear; hidrecbiases=hidbiases; save mnistvh vishid hidrecbiases visbiases; maxepoch=50; fprintf(1,'\nPretraining Layer 2 with RBM: %d-%d \n',numhid,numpen); batchdata=batchposhidprobs; numhid=numpen; restart=1; rbm; hidpen=vishid; penrecbiases=hidbiases; hidgenbiases=visbiases; save mnisthp hidpen penrecbiases hidgenbiases; fprintf(1,'\nPretraining Layer 3 with RBM: %d-%d \n',numpen,numpen2); batchdata=batchposhidprobs; numhid=numpen2; restart=1; rbm; hidpen2=vishid; penrecbiases2=hidbiases; hidgenbiases2=visbiases; save mnisthp2 hidpen2 penrecbiases2 hidgenbiases2; fprintf(1,'\nPretraining Layer 4 with RBM: %d-%d \n',numpen2,numopen); batchdata=batchposhidprobs; numhid=numopen; restart=1; rbmhidlinear; hidtop=vishid; toprecbiases=hidbiases; topgenbiases=visbiases; save mnistpo hidtop toprecbiases topgenbiases; backpropface; Thanks for your time

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  • Find location using only distance and range?

    - by pinnacler
    Triangulation works by checking your angle to three KNOWN targets. "I know the that's the Lighthouse of Alexandria, it's located here (X,Y) on a map, and it's to my right at 90 degrees." Repeat 2 more times for different targets and angles. Trilateration works by checking your distance from three KNOWN targets. "I know the that's the Lighthouse of Alexandria, it's located here (X,Y) on a map, and I'm 100 meters away from that." Repeat 2 more times for different targets and ranges. But both of those methods rely on knowing WHAT you're looking at. Say you're in a forest and you can't differentiate between trees, but you know where key trees are. These trees have been hand picked as "landmarks." You have a robot moving through that forest slowly. Do you know of any ways to determine location based solely off of angle and range, exploiting geometry between landmarks? Note, you will see other trees as well, so you won't know which trees are key trees. Ignore the fact that a target may be occluded. Our pre-algorithm takes care of that. 1) If this exists, what's it called? I can't find anything. 2) What do you think the odds are of having two identical location 'hits?' I imagine it's fairly rare. 3) If there are two identical location 'hits,' how can I determine my exact location after I move the robot next. (I assume the chances of having 2 occurrences of EXACT angles in a row, after I reposition the robot, would be statistically impossible, barring a forest growing in rows like corn). Would I just calculate the position again and hope for the best? Or would I somehow incorporate my previous position estimate into my next guess? If this exists, I'd like to read about it, and if not, develop it as a side project. I just don't have time to reinvent the wheel right now, nor have the time to implement this from scratch. So if it doesn't exist, I'll have to figure out another way to localize the robot since that's not the aim of this research, if it does, lets hope it's semi-easy.

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  • Find location using only distance and bearing?

    - by pinnacler
    Triangulation works by checking your angle to three KNOWN targets. "I know the that's the Lighthouse of Alexandria, it's located here (X,Y) on a map, and it's to my right at 90 degrees." Repeat 2 more times for different targets and angles. Trilateration works by checking your distance from three KNOWN targets. "I know the that's the Lighthouse of Alexandria, it's located here (X,Y) on a map, and I'm 100 meters away from that." Repeat 2 more times for different targets and ranges. But both of those methods rely on knowing WHAT you're looking at. Say you're in a forest and you can't differentiate between trees, but you know where key trees are. These trees have been hand picked as "landmarks." You have a robot moving through that forest slowly. Do you know of any ways to determine location based solely off of angle and range, exploiting geometry between landmarks? Note, you will see other trees as well, so you won't know which trees are key trees. Ignore the fact that a target may be occluded. Our pre-algorithm takes care of that. 1) If this exists, what's it called? I can't find anything. 2) What do you think the odds are of having two identical location 'hits?' I imagine it's fairly rare. 3) If there are two identical location 'hits,' how can I determine my exact location after I move the robot next. (I assume the chances of having 2 occurrences of EXACT angles in a row, after I reposition the robot, would be statistically impossible, barring a forest growing in rows like corn). Would I just calculate the position again and hope for the best? Or would I somehow incorporate my previous position estimate into my next guess? If this exists, I'd like to read about it, and if not, develop it as a side project. I just don't have time to reinvent the wheel right now, nor have the time to implement this from scratch. So if it doesn't exist, I'll have to figure out another way to localize the robot since that's not the aim of this research, if it does, lets hope it's semi-easy.

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  • How would i down-sample a .wav file then reconstruct it using nyquist? - in matlab [closed]

    - by martin
    This is all done in MatLab 2010 My objective is to show the results of: undersampling, nyquist rate/ oversampling First i need to downsample the .wav file to get an incomplete/ or impartial data stream that i can then reconstuct. Heres the flow chart of what im going to be doing So the flow is analog signal - sampling analog filter - ADC - resample down - resample up - DAC - reconstruction analog filter what needs to be achieved: F= Frequency F(Hz=1/s) E.x. 100Hz = 1000 (Cyc/sec) F(s)= 1/(2f) Example problem: 1000 hz = Highest frequency 1/2(1000hz) = 1/2000 = 5x10(-3) sec/cyc or a sampling rate of 5ms This is my first signal processing project using matlab. what i have so far. % Fs = frequency sampled (44100hz or the sampling frequency of a cd) [test,fs]=wavread('test.wav'); % loads the .wav file left=test(:,1); % Plot of the .wav signal time vs. strength time=(1/44100)*length(left); t=linspace(0,time,length(left)); plot(t,left) xlabel('time (sec)'); ylabel('relative signal strength') **%this is were i would need to sample it at the different frequecys (both above and below and at) nyquist frequency.*I think.*** soundsc(left,fs) % shows the resaultant audio file , which is the same as original ( only at or above nyquist frequency however) Can anyone tell me how to make it better, and how to do the various sampling at different frequencies?

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  • MATLAB: What is an appropriate Data Structure for a Matrix with Random Variable Entries?

    - by user12707
    I'm working in an area that is related to simulation and trying to design a data structure that can include random variables within matrices. I am currently coding in MATLAB. To motivate this let me say I have the following matrix: [a b; c d] I want to find a data structure that will allow for a, b, c, d to be either real numbers or random variables. As an example, let's say that a = 1, b = -1, c = 2 but let d be a normally distributed random variable with mean 20 and SD 40. The data structure that I have in mind will give no value to d. However, I also want to be able to design a function that can take in the structure, simulate an uniform(0,1), obtain a value for d using an inverse CDF and then spit out an actual matrix. I have several ideas to do this (all related to the MATLAB icdf function) but would like to know how more experienced programmers would do it. In this application, it's important that the structure is as "lean" as possible since I will be working with very very large matrices and memory will be an issue.

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