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  • Config deployment on multiple servers.

    - by user66601
    I have multiple servers in WEB cluster (identical configuration for all of them, despite the IP) How do you deploy changes in configs on multile servers? I make the new config, then create config per every server (placing correct IP), and next: upload them on every server, replacing old ones (rsync over ssh) set on every server a job which reloads webserver at the same time (servers use ntp). - this done by issuing commands by script (to save time for logging in) before adding a job for server reload - there's checksum test of the config on the server) - an a notification in case of fail How do you see such method? What should be the "professional way :) ? (I don't say my way doesn't work... it works and saves my time not used for logging on every webserver.) Regards,

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  • Remote Debugging and deployment of a Project Built Locally

    - by Abhishek Gupta
    We work on our projects locally, and deploy them on remote servers for testing. This is currently done via git commits/push/pulls. But the problem here is that most of the commit contain errors and/or break the code in significant manner due to the lack of testing. Is there a way we can deploy the code on the remote server without using git commits, or some sort of temporary commits or patch or any other mechanism? And only commit, whenever it is important.

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  • MATLAB intersection of 2 surfaces

    - by caglarozdag
    Hi everyone, I consider myself a beginner in MATLAB so bear with me if the answer to my question is an obvious one. Phi=0:pi/100:2*pi; Theta=0:pi/100:2*pi; [PHI,THETA]=meshgrid(Phi,Theta); R=(1 + cos(PHI).*cos(PHI)).*(1 + cos(THETA).*cos(THETA)); [X,Y,Z]=sph2cart(THETA,PHI,R); surf(X,Y,Z); %display hold on; x1=-4:.1:4; [X1,Y1] = meshgrid(x1); a=1.8; b=0; c=3; d=0; Z1=(d- a * X1 - b * Y1)/c; shading flat; surf(X1,Y1,Z1); I have written this code which plots a 3d cartesian plot of a plane intersecting a peanut shaped object at an angle. I need to get the intersection of these on 2D (going to be the outline of a peanut, but a bit skewed since the intersection happens at an angle), but don't know how. Thanks

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  • Problem in matlab with too many outputs

    - by Ben Fossen
    I am writing a Matlab program for simpson's rule I keep getting an error about to many outputs when the program gets to left_simpson = Simpson(a,c,(e1)/2,level, level_max); What is wrong with settinf left_simpson to Simpson(a,c,(e1)/2,level, level_max);? function Simpson(a,b,e1,level, level_max) level = level + 1; h = b - a; c = (a+b)/2; one_simpson = h*(f(a) + 4*f(c) + f(b))/6; d = (a+c)/2; e = (c+b)/2; two_simpson = h*(f(a) + 4*f(d) + 2*f(c) + 4*f(e))/2; if level >= level_max disp('h') simpson_result = two_simpson; disp('maximum levels reached') disp(simpson_result); if abs(two_simpson - one_simpson) < 15*e1 simpson_result = two_simpson + (two_simpson - one_simpson)/15; else left_simpson = Simpson(a,c,(e1)/2,level, level_max); right_simpson = Simpson(c,b,(e1)/2,level, level_max); simpson_result = left_simpson + right_simpson; end end

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  • matlab precision determint problem

    - by ldigas
    I have the following program format compact; format short g; clear; clc; L = 140; J = 77; Jm = 10540; G = 0.8*10^8; d = L/3; for i=1:500000 omegan=1.+0.0001*i; a(1,1) = ((omegan^2)*(Jm/(G*J))*d^2)-2; a(1,2) = 2; a(1,3) = 0; a(1,4) = 0; a(2,1) = 1; a(2,2) = ((omegan^2)*(Jm/(G*J))*d^2)-2; a(2,3) = 1; a(2,4) = 0; a(3,1) = 0; a(3,2) = 1; a(3,3) = ((omegan^2)*(Jm/(G*J))*d^2)-2; a(3,4) = 1; a(4,1) = 0; a(4,2) = 0; a(4,3) = 2; a(4,4) = ((omegan^2)*(Jm/(G*J))*d^2)-2; if(abs(det(a))<1E-10) sprintf('omegan= %8.3f det= %8.3f',omegan,det(a)) end end Analytical solution of the above system, and the same program written in fortran gives out values of omegan equal to 16.3818 and 32.7636 (fortran values; analytical differ a little, but they're there somewhere). So, now I'm wondering ... where am I going wrong with this ? Why is matlab not giving the expected results ? (this is probably something terribly simple, but it's giving me headaches)

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  • Matlab-Bisection-Newton-Secant , finding roots?

    - by i z
    Hello and thanks in advance for your possible help ! Here's my problem: I have 2 functions f1(x)=14.*x*exp(x-2)-12.*exp(x-2)-7.*x.^3+20.*x.^2-26.*x+12 f2(x)=54.*x.^6+45.*x.^5-102.*x.^4-69.*x.^3+35.*x.^2+16.*x-4 Make the graph for those 2, the first one in [0,3] and the 2nd one in [-2,2]. Find the 3 roots with accuracy of 6 decimal digits using a) bisection ,b) newton,c)secant.For each root find the number of iterations that have been made. For Newton-Raphson, find which roots have quadratic congruence and which don't. What is the main common thing that roots with no quadratic congruence (Newton's method)? Why ? Excuse me if i ask silly things, but i'm asked to do this with no Matlab courses and I'm trying to learn it myself. There are many issues i have with this exercise . Questions : 1.I only see 2 roots in the graph for the f1 function and 4-5 (?) roots for the function f2 and not 3 roots as the exercise says. Here's the 2 graphs : http://postimage.org/image/cltihi9kh/ http://postimage.org/image/gsn4sg97f/ Am i wrong ? Do both have only 3 roots in [0,3] and [-2,2] ? Concerning the Newton's method , how am i supposed to check out which roots have quadratic congruence and which not??? Accuracy means tolerance e=10^(-6), right ?

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  • matlab precision determinant problem

    - by ldigas
    I have the following program format compact; format short g; clear; clc; L = 140; J = 77; Jm = 10540; G = 0.8*10^8; d = L/3; for i=1:500000 omegan=1.+0.0001*i; a(1,1) = ((omegan^2)*(Jm/(G*J))*d^2)-2; a(1,2) = 2; a(1,3) = 0; a(1,4) = 0; a(2,1) = 1; a(2,2) = ((omegan^2)*(Jm/(G*J))*d^2)-2; a(2,3) = 1; a(2,4) = 0; a(3,1) = 0; a(3,2) = 1; a(3,3) = ((omegan^2)*(Jm/(G*J))*d^2)-2; a(3,4) = 1; a(4,1) = 0; a(4,2) = 0; a(4,3) = 2; a(4,4) = ((omegan^2)*(Jm/(G*J))*d^2)-2; if(abs(det(a))<1E-10) sprintf('omegan= %8.3f det= %8.3f',omegan,det(a)) end end Analytical solution of the above system, and the same program written in fortran gives out values of omegan equal to 16.3818 and 32.7636 (fortran values; analytical differ a little, but they're there somewhere). So, now I'm wondering ... where am I going wrong with this ? Why is matlab not giving the expected results ? (this is probably something terribly simple, but it's giving me headaches)

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  • MATLAB Easter Egg Spy vs Spy

    - by Aqui1aZ3r0
    I heard that MATLAB 2009b or earlier had a fun function. When you typed spy in the console, you would get something like this: http://t2.gstatic.com/images?q=tbn:ANd9GcSHAgKz-y1HyPHcfKvBpYmZ02PWpe3ONMDat8psEr89K0VsP_ft However, now you have an image like this:http://undocumentedmatlab.com/images/spy2.png I'd like it if I could get the code of the original spy vs spy image FYI, there's a code, but has certain errors in it: c = [';@3EA4:aei7]ced.CFHE;4\T*Y,dL0,HOQQMJLJE9PX[[Q.ZF.\JTCA1dd' '-IorRPNMPIE-Y\R8[I8]SUDW2e+' '=4BGC;7imiag2IFOQLID8''XI.]K0"PD@l32UZhP//P988_WC,U+Z^Y\<2' '&lt;82BF>?8jnjbhLJGPRMJE9/YJ/L1#QMC$;;V[iv09QE99,XD.YB,[_]=3a' '9;CG?@9kokc2MKHQSOKF:0ZL0aM2$RNG%AAW\jw9E.FEE-_G8aG.d]_W5+' '?:CDH@A:lpld3NLIRTPLG=1[M1bN3%SOH4BBX]kx:J9LLL8H9bJ/+d_dX6,' '@;DEIAB;mqmePOMJSUQMJ2\N2cO4&TPP@HCY^lyDKEMMN9+I@+S8,+deY7^' '8@EFJBC<4rnfQPNPTVRNKB3]O3dP5''UQQCIDZ_mzEPFNNOE,RA,T9/,++\8_' '9A2G3CD=544gRQPQUWUOLE4^P4"Q6(VRRIJE[n{KQKOOPK-SE.W:F/,,]Z+' ':BDH4DE>655hSRQRVXVPMF5_Q5#R>)eSSJKF\ao0L.L-WUL.VF8XCH001_[,' ';3EI<eo ?766iTSRSWYWQNG6$R6''S?*fTTlLQ]bp1M/P.XVP8[H9]DIDA=]' '?4D3=FP@877jUTSTXZXROK7%S7(TF+gUUmMR^cq:N9Q8YZQ9_IcIJEBd_^' '@5E@GQA98b3VUTUY*YSPL8&T)UI,hVhnNS_dr;PE.9Z[RCaR?+JTFC?e+' '79FA?HRB:9c4WVUVZ+ZWQM=,WG*VJ-"gi4OTes-XH+bK.#hj@PUvftDRMEF,]UH,UB.TYVWX,e\' '9;ECAKTY< ;eWYXWX\:)YSOE.YI,cL/$ikCqV1guE/PFL-^XI-YG/WZWXY1+]' ':AFDBLUZ=jgY[ZYZ-<7[XQG0[K.eN1&"$K2u:iyO9.PN9-_K8aJ9_]]82[' '?CEFDNW\?khZ[Z[==8\YRH1\M/!O2''#%m31Bw0PE/QXE8+R9bS;da^]93\' '@2FGEOX]ali[][9(ZSL2]N0"P3($&n;2Cx1QN9--L9,SA+T< +d_:4,' 'A3GHFPY^bmj\^]\]??:)[TM3^O1%Q4)%''oA:D0:0OE.8ME-TE,XB,+da;5[' '643IGQZ_cnk]_^]^@@;5\UN4_P2&R6*&(3B;E1<1PN99NL8WF.^C/,a+bY6,' '7:F3HR[dol^_^AA<6]VO5Q3''S>+'');CBF:=:QOEEOO9_G8aH6/d,cZ[Y' '8;G4IS\aep4_a-BD=7''XP6aR4(T?,(5@DCHCC;RPFLPPDH9bJ70+0d\\Z' '9BH>JT^bf45ba.CE@8(YQ7#S5)UD-)?AEDIDDD/QKMVQJ+S?cSDF,1e]a,' ':C3?K4_cg5[acbaADFA92ZR8$T6*VE.*@JFEJEEE0.NNWTK,U@+TEG0?+_bX' ';2D@L9dh6\bdcbBEGD:3[S=)U7+cK/+CKGFLIKI9/OWZUL-VA,WIHB@,`cY']; i = double(c(:)-32); j = cumsum(diff([0; i])< =0) + 1; S = sparse(i,j,1)'; spy(S)

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  • How to hide zero values in bar3 plot in MATLAB

    - by Doresoom
    I've got a 2-D histogram (the plot is 3D - several histograms graphed side by side) that I've generated with the bar3 plot command. However, all the zero values show up as flat squares in the x-y plane. Is there a way I can prevent MATLAB from displaying the values? I already tried replacing all zeros with NaNs, but it didn't change anything about the plot. Here's the code I've been experimenting with: x1=normrnd(50,15,100,1); %generate random data to test code x2=normrnd(40,13,100,1); x3=normrnd(65,12,100,1); low=min([x1;x2;x3]); high=max([x1;x2;x3]); y=linspace(low,high,(high-low)/4); %establish consistent bins for histogram z1=hist(x1,y); z2=hist(x2,y); z3=hist(x3,y); z=[z1;z2;z3]'; bar3(z) As you can see, there are quite a few zero values on the plot. Closing the figure and re-plotting after replacing zeros with NaNs seems to change nothing: close z(z==0)=NaN; bar3(z)

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  • Assigning figure size to a figure with a given handle (MATLAB)

    - by James
    Hi, is there a way to assign the outerposition property of a figure to a figure with a given handle? For example, if I wanted to define a figure as say figure 1, I would use: figure(1) imagesc(Arrayname) % I.e. any array I can also change the properties of a figure using the code: figure('Name', 'Name of figure','NumberTitle','off','OuterPosition',[scrsz(1) scrsz(2) 700 700]); Is there a propertyname I can use to assign the outerposition property to the figure assigned as figure 1? The reason I am asking this is because I am using a command called save2word (from the MATLAB file exchange) to save some plots from a function I have made to a word file, and I want to limit the number of figures I have open as it does this. The rest of the code I have is: plottedloops = [1, 5:5:100]; % Specifies which loops I want to save GetGeometry = getappdata(0, 'GeometryAtEachLoop') % Obtains a 4D array containing geometry information at each loop NumSections = size(GetGeometry,4); %Defined by the fourth dimension of the 4D array for j = 1:NumSections for i = 1:plottedloops P = GetGeometry(:,:,i,j); TitleSize = 14; Fsize = 8; % Save Geometry scrsz = get(0,'ScreenSize'); %left, bottom, width height figure('Name', 'Geometry at each loop','NumberTitle','off','OuterPosition',[scrsz(1) scrsz(2) 700 700]); This specifies the figure name, dims etc., but also means multiple figures are opened as the command runs. % I have tried this, but it doesn't work: % figure(0, 'OuterPosition',[scrsz(1) scrsz(2) 700 700]); imagesc(P), title('Geometry','FontSize', TitleSize), axis([0 100 0 100]); text(20,110,['Loop:',num2str(i)], 'FontSize', TitleSize); % Show loop in figure text(70,110,['Section:',num2str(j)], 'FontSize', TitleSize);% Show Section number in figure save2word('Geometry at each loop'); % Saves figure to a word file end end Thanks

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  • Matlab Simulation: Point (symbol) Moving from start point to end point and back

    - by niko
    Hi, I would like to create an animation to demonstrate LDPC coding which is based on Sum-Product Algorithm So far I have created a graph which shows the connections between symbol nodes (left) and parity nodes (right) and would like to animate points travelling from symbol to parity nodes and back. The figure is drawn by executing the following method: function drawVertices(H) hold on; nodesCount = size(H); parityNodesCount = nodesCount(1); symbolNodesCount = nodesCount(2); symbolPoints = zeros(symbolNodesCount, 2); symbolPoints(:, 1) = 0; for i = 0 : symbolNodesCount - 1 ji = symbolNodesCount - i; scatter(0, ji) symbolPoints(i + 1, 2) = ji; end; parityPoints = zeros(parityNodesCount, 2); parityPoints(:, 1) = 10; for i = 0 : parityNodesCount - 1 ji = parityNodesCount - i; y0 = symbolNodesCount/2 - parityNodesCount/2; scatter(10, y0 + ji) parityPoints(i + 1, 2) = y0 + ji; end; axis([-1 11 -1 symbolNodesCount + 2]); axis off %connect vertices d = size(H); for i = 1 : d(1) for j = 1 : d(2) if(H(i, j) == 1) plot([parityPoints(i, 1) symbolPoints(j, 1)], [parityPoints(i, 2) symbolPoints(j, 2)]); end; end; end; So what I would like to do here is to add another method which takes start point (x and y) and end point as arguments and animates a travelling circle (dot) from start to end and back along the displayed lines. I would appreciate if anyone of you could show the solution or suggest any useful tutorial about matlab simulations. Thank you!

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  • Naive Bayes matlab, row classification

    - by Jungle Boogie
    How do you classify a row of seperate cells in matlab? Atm I can classify single coloums like so: training = [1;0;-1;-2;4;0;1]; % this is the sample data. target_class = ['posi';'zero';'negi';'negi';'posi';'zero';'posi']; % target_class are the different target classes for the training data; here 'positive' and 'negetive' are the two classes for the given training data % Training and Testing the classifier (between positive and negative) test = 10*randn(25, 1); % this is for testing. I am generating random numbers. class = classify(test,training, target_class, 'diaglinear') % This command classifies the test data depening on the given training data using a Naive Bayes classifier Unlike the above im looking at wanting to classify: A B C Row A | 1 | 1 | 1 = a house Row B | 1 | 2 | 1 = a garden Can anyone help? Here is a code example from matlabs site: nb = NaiveBayes.fit(training, class) nb = NaiveBayes.fit(..., 'param1',val1, 'param2',val2, ...) I dont understand what param1 is or what val1 etc should be?

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  • MATLAB: impoint getPosition strange behaviour

    - by tguclu
    I have a question about the values returned by getPosition. Below is my code. It lets the user set 10 points on a given image: figure ,imshow(im); colorArray=['y','m','c','r','g','b','w','k','y','m','c']; pointArray = cell(1,10); % Construct boundary constraint function fcn = makeConstrainToRectFcn('impoint',get(gca,'XLim'),get(gca,'YLim')); for i = 1:10 p = impoint(gca); % Enforce boundary constraint function using setPositionConstraintFcn setPositionConstraintFcn(p,fcn); setColor(p,colorArray(1,i)); pointArray{i}=p; getPosition(p) end When I start to set points on the image I get results like [675.000 538.000], which means that the x part of the coordinate is 675 and the y part is 538, right? This is what the MATLAB documentation says, but since the image is 576*120 (as displayed in the window) this is not logical. It seemed to me like, somehow, getPosition returns the y coordinate first. I need some clarification on this. Thanks for help

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  • How to interpolate in MatLab

    - by G Sam
    I have a 1x1 Matrix of points which specifies speed of a drive with respect to time. This speed changes throughout the operation; which means that the difference between two points is changing. To give you an example: M = [1; 2; 3; 5; 7; 9; 11; 15; 19]. (Only that this is a 892x1 matrix) I want to make this matrix twice as long (so changing the relative speed per timestep), while retaining the way the speeds change. Eg: M' = [1; 1.5; 2; 2.5; 3; 4; 5; 6; 7; 8; 9; 10; 11; 13; 15; 17; 19]. Is there an easy way to do this in MatLab? So far I have tried upsampling (which fills the time step with zeros); interp (which fills it with low-pass interpolation. Thanks!

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  • MATLAB setting matrix values in an array

    - by user324994
    I'm trying to write some code to calculate a cumulative distribution function in matlab. When I try to actually put my results into an array it yells at me. tempnum = ordered1(1); k=2; while(k<538) count = 1; while(ordered1(k)==tempnum) count = count + 1; k = k + 1; end if(ordered1(k)~=tempnum) output = [output;[(count/537),tempnum]]; k = k + 1; tempnum = ordered1(k); end end The errors I'm getting look like this ??? Error using ==> vertcat CAT arguments dimensions are not consistent. Error in ==> lab8 at 1164 output = [output;[(count/537),tempnum]]; The line to add to the output matrice was given to me by my TA. He didn't teach us much syntax throughout the year so I'm not really sure what I'm doing wrong. Any help is greatly appreciated.

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  • cell and array in Matlab

    - by Tim
    Hi, I am a little confused about the usage of cell and array in Matlab. I would like to hear about your understandings. Here are my observations: (1). array can dynamically adjust its own memory to allow dynamic number of elements, while cell seems not act in the same way. a=[]; a=[a 1]; b={}; b={b 1}; (2). several elements can be retrieved from cell, while they seem not from array. a={'1' '2'}; figure, plot(...); hold on; plot(...) ; legend(a{1:2}); b=['1' '2']; figure, plot(...); hold on; plot(...) ; legend(b(1:2)); % b(1:2) is an array, not its elements, so it is wrong with legend. Are these correct? What are some other different usages between the cell and array? Thanks and regards!

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  • Combining two data sets and plotting in matlab

    - by bautrey
    I am doing experiments with different operational amplifier circuits and I need to plot my measured results onto a graph. I have two data sets: freq1 = [.1 .2 .5 .7 1 3 4 6 10 20 35 45 60 75 90 100]; %kHz Vo1 = [1.2 1.6 1.2 2 2 2.4 14.8 20.4 26.4 30.4 53.6 68.8 90 114 140 152]; %mV V1 = 19.6; Acm = Vo1/(1000*V1); And: freq2 = [.1 .5 1 30 60 70 85 100]; %kHz Vo1 = [3.96 3.96 3.96 3.84 3.86 3.88 3.88 3.88]; %V V1 = .96; Ad = Vo1/(2*V1); (I would show my plots but apparently I need more reps for that) I need to plot the equation, CMRR vs freq: CMRR = 20*log10(abs(Ad/Acm)); The size of Ad and Acm are different and the frequency points do not match up, but the boundaries of both of these is the same, 100Hz to 100kHz (x-axis). On the line of CMRR, Matlab says that Ad and Acm matrix dimensions do not agree. How I think I would solve this is using freq1 as the x-axis for CMRR and then taking approximated points from Ad according to the value on freq1. Or I could do function approximations of Ad and Acm and then do the divide operator on those. I do not know how I would code up this two ideas. Any other ideas would helpful, especially simpler ones. Thanks

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  • Import & modify date data in MATLAB

    - by niko
    I have a .csv file with records written in the following form: 2010-04-20 15:15:00,"8.9915176259e+00","8.8562623697e+00" 2010-04-20 15:30:00,"8.5718021723e+00","8.6633827160e+00" 2010-04-20 15:45:00,"8.4484844117e+00","8.4336586330e+00" 2010-04-20 16:00:00,"1.1106980342e+01","8.4333062208e+00" 2010-04-20 16:15:00,"9.0643470589e+00","8.6885660103e+00" 2010-04-20 16:30:00,"8.2133517943e+00","8.2677822671e+00" 2010-04-20 16:45:00,"8.2499419380e+00","8.1523501983e+00" 2010-04-20 17:00:00,"8.2948492278e+00","8.2884797924e+00" From these data I would like to make clusters - I would like to add a column with number indicating the hour - so in case of the first row a value 15 has to be added in a new row. The first problem is that calling a function [numData, textData, rawData] = xlsread('testData.csv') creates an empty matrix numData and one-column textData and rawData structures. Is it possible to create any template which recognizes a yyyy, MM, dd, hh, mm, ss values from the data above? What I would basically like to do with these data is to categorize the values by hours so from the example row of input: 2010-04-20 15:15:00,"8.9915176259e+00","8.8562623697e+00" update 1: in Matlab the line above is recognized as a string: '2010-04-26 13:00:00,"1.0428104753e+00","2.3456394130e+00"' I would want this to be the output: 15, 8.9915176259e+00, 8.8562623697e+00 update 1: a string has to be parsed Does anyone know how to parse a string and retrieve a timestamp, value1 (1.0428104753e+00) and value2 (2.3456394130e+00) from it as separate values?

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  • Converting Single dimensional vector or array to two dimension in Matlab

    - by pac
    Well I do not know if I used the exact term. I tried to find an answer on the net. Here is what i need: I have a vector a = 1 4 7 2 5 8 3 6 9 If I do a(4) the value is 4. So it is reading first column top to buttom then continuing to next .... I don't know why. However, What I need is to call it using two indices. As row and column: a(3,2)= 4 or even better if i can call it in the following way: a{3}(2)=4 What is this process really called (want to learn) and how to perform in matlab. I thought of a loop. Is there a built in function Thanks a lot Check this: a = 18 18 16 18 18 18 16 0 0 0 16 16 18 0 18 16 0 18 18 16 18 0 18 18 0 16 0 0 0 18 18 0 18 18 16 0 16 0 18 18 >> a(4) ans = 18 >> a(5) ans = 18 >> a(10) ans = 18 I tried reshape. it is reshaping not converting into 2 indeces

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  • Matlab fft function

    - by CTZStef
    The code below is from the Matlab 2011a help about fft function. I think there is a problem here : why do they multiply t(1:50) by Fs, and then say it's time in millisecond ? Certainly, it happens to be true in this very particular case, but change the value of Fs to, say, 2000, and it won't work anymore, obviously because of this factor of 2. Right ? Quite misleading, isn't it ? What do I miss ? Fs = 1000; % Sampling frequency T = 1/Fs; % Sample time L = 1000; % Length of signal t = (0:L-1)*T; % Time vector % Sum of a 50 Hz sinusoid and a 120 Hz sinusoid x = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t); y = x + 2*randn(size(t)); % Sinusoids plus noise plot(Fs*t(1:50),y(1:50)) title('Signal Corrupted with Zero-Mean Random Noise') xlabel('time (milliseconds)') Clearer with this : fs = 2000; % Sampling frequency T = 1 / fs; % Sample time L = 1000; % Length of signal t2 = (0:L-1)*T; % Time vector f = 50; % signal frequency s2 = sin(2*pi*f*t2); figure, plot(fs*t2(1:50),s2(1:50)); % NOT good figure, plot(t2(1:50),s2(1:50)); % good

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  • MATLAB: svds() not converging

    - by Paul
    So using MATLAB's svds() function on some input data as such: [U, S, V, flag] = svds(data, nSVDs, 'L') I noticed that from run to run with the same data, I'd get drastically different output SVD sizes from run to run. When I checked whether 'flag' was set, I found that it was, indicating that the SVDs had not converged. My normal system here would be that if it really needs to converge, I'd do something like this: flag = 1 svdOpts = struct('tol', 1e-10, 'maxit', 600, 'disp', 0); while flag: if svdOpts.maxit > 1e6 error('There''s a real problem here.') end [U, S, V, flag] = svds(data, nSVDs, 'L', svdOpts) svdOpts.maxit = svdOpts.maxit*2 end But from what I can tell, when you use 'L' as the third argument, the fourth argument is ignored, meaning I just have to deal with the fact that it's not converging? I'm not even really sure how to use the 'sigma' argument in place of the 'L' argument. I've also tried reducing the number of SVDs calculated to no avail. Any help on this matter would be much appreciated. EDIT While following up on the comments below, I found that the problem had to do with the way I was building my data matrices. Turned out I had accidentally inverted a matrix and had an input of size (4000x1) rather than (20x200), which was what was refusing to converge. I also did some more timing tets and found that the fourth argument was not, in fact, being ignored, so that's on me. Thanks for the help guys.

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  • Error in bisection method code in Matlab

    - by Amanda Collins
    I need to write a proper implementation of the bisection method, which means I must address all possible user input errors. Here is my code: function [x_sol, f_at_x_sol, N_iterations] = bisection(f, xn, xp, eps_f, eps_x) % solving f(x)=0 with bisection method % f is the function handle to the desired function, % xn and xp are borders of search, % f(xn)<0 and f(xp)>0 required, % eps_f defines how close f(x) should be to zero, % eps_x defines uncertainty of solution x if(f(xp) < 0) error('xp must be positive') end; if(f(xn)>0) error('xn must be negative') end; if (xn >= xp) error ('xn must be less than xp') end; xg=(xp+xn)/2; %initial guess fg=f(xg); % initial function evaluation N_iterations=1; while ( (abs(fg) > eps_f) & (abs(xg-xp) > eps_x) ) if (fg>0) xp=xg; else xn=xg; end xg=(xp+xn)/2; %update guess fg=f(xg); %update function evaluation N_iterations=N_iterations+1; end x_sol=xg; %solution is ready f_at_x_sol=fg; if (f_at_x_sol > eps_f) error('No convergence') end and here is the error message I receive when I try to test this in Matlab: >> bisection(x.^2, 2, -1, 1e-8, 1e-10) Attempted to access f(-1); index must be a positive integer or logical. Error in bisection (line 9) if(f(xp)<0) I was attempting to see if my error codes worked, but it doesn't look like they do. I get the same error when I try to test it on a function that should work.

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  • Stop writing blank line at the end of CSV file (using MATLAB)

    - by Grant M.
    Hello all ... I'm using MATLAB to open a batch of CSV files containing column headers and data (using the 'importdata' function), then I manipulate the data a bit and write the headers and data to new CSV files using the 'dlmwrite' function. I'm using the '-append' and 'newline' attributes of 'dlmwrite' to add each line of text/data on a new line. Each of my new CSV files has a blank line at the end, whereas this blank line was not there before when I read in the data ... and I'm not using 'newline' on my final call of 'dlmwrite'. Does anyone know how I can keep from writing this blank line to the end of my CSV files? Thanks for your help, Grant EDITED 5/18/10 1:35PM CST - Added information about code and text file per request ... you'll notice after performing the procedure below that there appears to be a carriage return at the end of the last line in the new text file. Consider a text file named 'textfile.txt' that looks like this: Column1, Column2, Column3, Column4, Column 5 1, 2, 3, 4, 5 1, 2, 3, 4, 5 1, 2, 3, 4, 5 1, 2, 3, 4, 5 1, 2, 3, 4, 5 Here's a sample of the code I am using: % import data importedData = importdata('textfile.txt'); % manipulate data importedData.data(:,1) = 100; % store column headers into single comma-delimited % character array (for easy writing later) columnHeaders = importedData.textdata{1}; for counter = 2:size(importedData.textdata,2) columnHeaders = horzcat(columnHeaders,',',importedData.textdata{counter}); end % write column headers to new file dlmwrite('textfile_updated.txt',columnHeaders,'Delimiter','','newline','pc') % append all but last line of data to new file for dataCounter = 1:(size(importedData.data,2)-1) dlmwrite('textfile_updated.txt',importedData.data(dataCounter,:),'Delimiter',',','newline','pc','-append') end % append last line of data to new file, not % creating new line at end dlmwrite('textfile_updated.txt',importedData.data(end,:),'Delimiter',',','-append')

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  • Determining two lines in MATLAB

    - by Sanyi
    Hi! I am making the game planarity in MATLAB, and i am stuck where the program have to determine that is there any lines that are in inscision with each other. Any ideas how could the program find the inscision points. Heres my source for the game: > clear; clc; a=1; b=10; x=floor(rand(5,1)*(b-a+1)+a); y=floor(rand(5,1)*(b-a+1)+a); fig1=figure; set(fig1,'Color','w','Position',[100 100 500 500]) plot(x,y,'o','MarkerSize',12,'MarkerFaceColor','r'); axis([0 11 0 11]); axis off grid on; hold on; FirstPlayerLines=[]; SecondPlayerLines=[]; for a1=1:+1:2 for a2=5:-1:2 plot([x(a1) x(a2)],[y(a1) y(a2)],'Color','black','LineWidth',3); end end waitforbuttonpress

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  • Matlab GUI: How to Save the Results of Functions (states of application)

    - by niko
    Hi, I would like to create an animation which enables the user to go backward and forward through the steps of simulation. An animation has to simulate the iterative process of channel decoding (a receiver receives a block of bits, performs an operation and then checks if the block corresponds to parity rules. If the block doesn't correspond the operation is performed again and the process finally ends when the code corresponds to a given rules). I have written the functions which perform the decoding process and return a m x n x i matrix where m x n is the block of data and i is the iteration index. So if it takes 3 iterations to decode the data the function returns a m x n x 3 matrix with each step is stired. In the GUI (.fig file) I put a "decode" button which runs the method for decoding and there are buttons "back" and "forward" which have to enable the user to switch between the data of recorded steps. I have stored the "decodedData" matrix and currentStep value as a global variable so by clicking "forward" and "next" buttons the indices have to change and point to appropriate step states. When I tried to debug the application the method returned the decoded data but when I tried to click "back" and "next" the decoded data appeared not to be declared. Does anyone know how is it possible to access (or store) the results of the functions in order to enable the described logic which I want to implement in Matlab GUI?

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