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  • Metric 3d reconstruction

    - by srand
    I'm trying to reconstruct 3D points from 2D image correspondences. My camera is calibrated. The test images are of a checkered cube and correspondences are hand picked. Radial distortion is removed. After triangulation the construction seems to be wrong however. The X and Y values seem to be correct, but the Z values are about the same and do not differentiate along the cube. The 3D points look like as if the points were flattened along the Z-axis. What is going wrong in the Z values? Do the points need to be normalized or changed from image coordinates at any point, say before the fundamental matrix is computed? (If this is too vague I can explain my general process or elaborate on parts)

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  • openmp in mex : stackoverflow error

    - by Edwin
    i have got the following fraction of code that getting me the stack overflow error #pragma omp parallel shared(Mo1, Mo2, sum_normalized_p_gn, Data, Mean_Out,Covar_Out,Prior_Out, det) private(i) num_threads( number_threads ) { //every thread has a new copy double* normalized_p_gn = (double*)malloc(NMIX*sizeof(double)); #pragma omp critical { int id = omp_get_thread_num(); int threads = omp_get_num_threads(); mexEvalString("drawnow"); } #pragma omp for //some parallel process..... } the variables declared in the shared are created by malloc. and they consumes with large amount of memory there are 2 questions regarding to the above code. 1) why this would generate the stack overflow error( i.e. segmentation fault) before it goes into the parallel for loop? it works fine when it runs in the sequential mode.... 2) am i right to dynamic allocate memory for each thread like "normalized_p_gn" above? Regards Edwin

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  • half-sine pulse shaping

    - by kos
    hi, i wanted to know what is the pulse shape of the modem.oqpskmod? and if it is not half-sine pulse shape, how is it possible to make it half-sine pulse shape as it is stated in ieee 802.15.4(zigbee) standard where it shows it as follows p(t)=sin(pi*t/2*Tc) if 0<=t<=2Tc p(t)=0 if otherwise ? thanks a lot!

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  • best way to get a vector from sparse matrix

    - by niko
    Hi, I have a m x n matrix where each row consists of zeros and same values for each row. an example would be: M = -0.6 1.8 -2.3 0 0 0; 0 0 0 3.4 -3.8 -4.3; -0.6 0 0 3.4 0 0 In this example the first column consists of 0s and -0.6, second 0 and 1.8, third -2.3 and so on. In such case I would like to reduce m to 1 (get a vector from a given matrix) so in this example a vector would be (-0.6 1.8 -2.3 3.4 -3.8 -4.3) Does anyone know what is the best way to get a vector from such matrix? Thank you!

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  • Octave: Multiple submatrices from a matrix

    - by fbrereto
    I have a large matrix from which I would like to gather a collection of submatrices. If my matrix is NxN and the submatrix size is MxM, I want to collect I=(N - M + 1)^2 submatrices. In other words I want one MxM submatrix for each element in the original matrix that can be in the top-left corner of such a matrix. Here's the code I have: for y = 1:I for x = 1:I index = (y - 1) * I + x; block_set(index) = big_mat(x:x+M-1, y:y+M-1) endfor endfor The output if a) wrong, and b) implying there is something in the big_mat(x:x+M-1, y:y+M-1) expression that can get me what I want without needing the two for loops. Any help would be much appreciated

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  • How can I select the pixels from an image in opencv?

    - by ajith
    This is refined version of my previous question. Actually I want to do following operation... summation for all k|(i,j)?wk [(Ii-µk)*(Ij-µk)], where wk is a 3X3 window, µk is the mean of wk, Ii & Ij are the intensities of the image at i and j. I dont know how to select Ii & Ij separately from an image which is 2 dimensional[Iij]...or does the equation mean anything else?

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  • A dynamic array of class "landmark", inside another single class "landmarks"

    - by pinnacler
    I'm working on a robot localization simulator and I created a class called "landmark". The end result is going to be a robot that is always centered and always faces the top of the screen. As it turns, the birds eye view map will rotate around the robot. To accomplish this, I'm assuming I can rotate one class and have all elements inside rotate as well. So, the landmark class has properties x,y, label, and radius. This is suppose to simulate a tree location in a forest. To test everything, I need "forest data," and I wrote a script to generate 100 trees in a 100m x 100m area. The script automatically generates values within an acceptable range for x,y, radius. The generated data is stored in an object called tempForest and is 100x3. Ideally, I want to create a class called "landmarks" (plural) that has 100 landmark instances inside. How would I instantiate 100 instances of landmark in one instance of landmarks using that randomly generated data? Ideally, I'd just type treeBeacons = landmarks(); and it would randomly populate 100 (user definable, set in config file) instances with x, y, radius data. I'm not sure how to deal with a dynamic array of class "Landmark", inside another single class "landmarks." Any ideas?

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  • I need to write a program that reads angles in radians from an input disk and converts them in degre

    - by Amadou
    Write a program that reads angles in radians from an input disk le and converts them into degrees, minutes, and seconds. Output should be written into another le. A sample input le could be: # this is a comment # your program should be able to skip comment lines # and blank lines # input radian numbers could be seperated by blanks 0.0 1.0 # or by a newline 3.141593 6.0

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  • problems in trying ieee 802.15.4 working from msk

    - by asel
    Hi, i took a msk code from dsplog.com and tried to modify it to test the ieee 802.15.4. There are several links on that site for ieee 802.15.4. Currently I am getting simulated ber results all approximately same for all the cases of Eb_No values. Can you help me to find why? thanks in advance! clear PN = [ 1 1 0 1 1 0 0 1 1 1 0 0 0 0 1 1 0 1 0 1 0 0 1 0 0 0 1 0 1 1 1 0; 1 1 1 0 1 1 0 1 1 0 0 1 1 1 0 0 0 0 1 1 0 1 0 1 0 0 1 0 0 0 1 0; 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 1 1 1 0 0 0 0 1 1 0 1 0 1 0 0 1 0; 0 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 1 1 1 0 0 0 0 1 1 0 1 0 1; 0 1 0 1 0 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 1 1 1 0 0 0 0 1 1; 0 0 1 1 0 1 0 1 0 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 1 1 1 0 0; 1 1 0 0 0 0 1 1 0 1 0 1 0 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 1; 1 0 0 1 1 1 0 0 0 0 1 1 0 1 0 1 0 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1; 1 0 0 0 1 1 0 0 1 0 0 1 0 1 1 0 0 0 0 0 0 1 1 1 0 1 1 1 1 0 1 1; 1 0 1 1 1 0 0 0 1 1 0 0 1 0 0 1 0 1 1 0 0 0 0 0 0 1 1 1 0 1 1 1; 0 1 1 1 1 0 1 1 1 0 0 0 1 1 0 0 1 0 0 1 0 1 1 0 0 0 0 0 0 1 1 1; 0 1 1 1 0 1 1 1 1 0 1 1 1 0 0 0 1 1 0 0 1 0 0 1 0 1 1 0 0 0 0 0; 0 0 0 0 0 1 1 1 0 1 1 1 1 0 1 1 1 0 0 0 1 1 0 0 1 0 0 1 0 1 1 0; 0 1 1 0 0 0 0 0 0 1 1 1 0 1 1 1 1 0 1 1 1 0 0 0 1 1 0 0 1 0 0 1; 1 0 0 1 0 1 1 0 0 0 0 0 0 1 1 1 0 1 1 1 1 0 1 1 1 0 0 0 1 1 0 0; 1 1 0 0 1 0 0 1 0 1 1 0 0 0 0 0 0 1 1 1 0 1 1 1 1 0 1 1 1 0 0 0; ]; N = 5*10^5; % number of bits or symbols fsHz = 1; % sampling period T = 4; % symbol duration Eb_N0_dB = [0:10]; % multiple Eb/N0 values ct = cos(pi*[-T:N*T-1]/(2*T)); st = sin(pi*[-T:N*T-1]/(2*T)); for ii = 1:length(Eb_N0_dB) tx = []; % MSK Transmitter ipBit = round(rand(1,N/32)*15); for k=1:length(ipBit) sym = ipBit(k); tx = [tx PN((sym+1),1:end)]; end ipMod = 2*tx - 1; % BPSK modulation 0 -> -1, 1 -> 1 ai = kron(ipMod(1:2:end),ones(1,2*T)); % even bits aq = kron(ipMod(2:2:end),ones(1,2*T)); % odd bits ai = [ai zeros(1,T) ]; % padding with zero to make the matrix dimension match aq = [zeros(1,T) aq ]; % adding delay of T for Q-arm % MSK transmit waveform xt = 1/sqrt(T)*[ai.*ct + j*aq.*st]; % Additive White Gaussian Noise nt = 1/sqrt(2)*[randn(1,N*T+T) + j*randn(1,N*T+T)]; % white gaussian noise, 0dB variance % Noise addition yt = xt + 10^(-Eb_N0_dB(ii)/20)*nt; % additive white gaussian noise % MSK receiver % multiplying with cosine and sine waveforms xE = conv(real(yt).*ct,ones(1,2*T)); xO = conv(imag(yt).*st,ones(1,2*T)); bHat = zeros(1,N); bHat(1:2:end) = xE(2*T+1:2*T:end-2*T); % even bits bHat(2:2:end) = xO(3*T+1:2*T:end-T); % odd bits result=zeros(16,1); chiplen=32; seqstart=1; recovered = []; while(seqstart<length(bHat)) A = bHat(seqstart:seqstart+(chiplen-1)); for j=1:16 B = PN(j,1:end); result(j)=sum(A.*B); end [value,index] = max(result); recovered = [recovered (index-1)]; seqstart = seqstart+chiplen; end; %# create binary string - the 4 forces at least 4 bits bstr1 = dec2bin(ipBit,4); bstr2 = dec2bin(recovered,4); %# convert back to numbers (reshape so that zeros are preserved) out1 = str2num(reshape(bstr1',[],1))'; out2 = str2num(reshape(bstr2',[],1))'; % counting the errors nErr(ii) = size(find([out1 - out2]),2); end nErr/(length(ipBit)*4) % simulated ber theoryBer = 0.5*erfc(sqrt(10.^(Eb_N0_dB/10))) % theoretical ber

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  • Matlap Simulation

    - by iva123
    I have to make a simulation for Aircraft Cabin Pressure in Mathlap. But I could not find any characteristic equation on the web. Does anyone know where to start a characterisctic equation for matlap or can anyone suuggest me a book for it ?

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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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  • Simulink: Specifying trajectory

    - by stanigator
    I would like to use jtraj to specify a trajectory in a Simulink model. Below are what I attempted to retrieve in the command prompt: Q0 = [1 1 0]; Q1 = [1+0.5*cos(2*20) 1+0.5*sin(2*20) 0]; t = 0:0.1:20; [Q, Qd, Qdd] = jtraj(Q0, Q1, t); However, I don't know how to include such trajectory data in the Simulink model easily. Any comments? Thanks in advance.

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  • Simulink sim of rician channel ber process

    - by bob
    Hi, I'm learning simulink and I want to use the rician channle block from the communications blockset. I'm told I need to change the format format. Would anyone have some sample code where they used the rician channels in simulink to model a bit error rate process?

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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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  • Project Euler problem 214, How can i make it more efficient?

    - by Once
    I am becoming more and more addicted to the Project Euler problems. However since one week I am stuck with the #214. Here is a short version of the problem: PHI() is Euler's totient function, i.e. for any given integer n, PHI(n)=numbers of k<=n for which gcd(k,n)=1. We can iterate PHI() to create a chain. For example starting from 18: PHI(18)=6 = PHI(6)=2 = PHI(2)=1. So starting from 18 we get a chain of length 4 (18,6,2,1) The problem is to calculate the sum of all primes less than 40e6 which generate a chain of length 25. I built a function that calculates the chain length of any number and I tested it for small values: it works well and fast. sum of all primes<=20 which generate a chain of length 4: 12 sum of all primes<=1000 which generate a chain of length 10: 39383 Unfortunately my algorithm doesn't scale well. When I apply it to the problem, it takes several hours to calculate... so I stop it because the Project Euler problems must be solved in less than one minute. I thought that my prime detection function might be slow so I fed the program with a list of primes <40e6 to avoid the primality test... The code runs now a little bit faster, but there is still no way to get a solution in less than a few hours (and I don't want this). So is there any "magic trick" that I am missing here ? I really don't understand how to be more efficient on this one... I am not asking for the solution, because fighting with optimization is all the fun of Project Euler. However, any small hint that could put me on the right track would be welcome. Thanks !

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  • MATrix LABoratory interview questions?

    - by Shane
    I programmed in MATrix LABoratory for many years, but switched to using R exclusively in the past few years so I'm a little out of practice. I'm interviewing a candidate today who describes himself as an expert? What interview questions that I should be asking?

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  • Removing an array dimension where the elements sum to zero

    - by James
    Hi, I am assigning a 3D array, which contains some information for a number of different loadcases. Each row in the array defines a particular loadcase (of which there are 3) and I would like to remove the loadcase (i.e. the row) if ALL the elements of the row (in 3D) are equal to zero. The code I have at the moment is: Array = zeros(3,5) % Initialise array Numloadcases = 3; Array(:,:,1) = [10 10 10 10 10; 0 0 0 0 0; 0 0 0 0 0;]; % Expand to a 3D array Array(:,:,2) = [10 10 10 10 10; 0 0 0 0 0; 0 0 0 0 0;]; Array(:,:,3) = [10 10 10 10 10; 0 0 0 0 0; 0 0 20 0 0;]; Array(:,:,4) = [10 10 10 10 10; 0 0 0 0 0; 0 0 20 0 0;]; % I.e. the second row should be removed. for i = 1:Numloadcases if sum(Array(i,:,:)) == 0; Array(i,:,:) = [] end end At the moment, the for loop I have to remove the rows causes an indexing error, as the size of the array changes in the loop. Can anyone see a work around for this? Thanks

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  • Uiimport does not save variable to base workspace

    - by Tim
    I tried using uiimport to load a file to the base workspace.....It worked first time....but after trying again after a while...I wasnt seeing the variable in the base work space. I used the default variable name which is given by 'uiimport". This was the command I used: uiimport(filename) And two variables where created by default..."data" and "textdata"(which is the header)....but now when i run it is no longer saved in the base workspace I do not want to assign a variable to the uiimport like so... K = uiimport(filename) assignin(base,'green',K) I do not want to do that because My dataset has a text header and the data itself, and doing this would assign both "textdata" and "data" to "green" variable How would I be able to get the dimensions of ONLY the "data" in green and how would I pass only "data"(which is in the green variable in the workspace.."rmbr"...the green variable holds both "data" and "textdata") to another function. I was able to do all this when the uiimport automatically saved the variables in the base workspace....but somehow now it doesn't. I would appreciate any help or tips on this matter

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  • filter that uses elements from two arrays at the same time

    - by Gacek
    Let's assume we have two arrays of the same size - A and B. Now, we need a filter that, for a given mask size, selects elements from A, but removes the central element of the mask, and inserts there corresponding element from B. So the 3x3 "pseudo mask" will look similar to this: A A A A B A A A A Doing something like this for averaging filter is quite simple. We can compute the mean value for elements from A without the central element, and then combine it with a proper proportion with elements from B: h = ones(3,3); h(2,2) =0; h = h/sum(h(:)); A_ave = filter2(h, A); C = (8/9) * A_ave + (1/9) * B; But how to do something similar for median filter (medfilt2 or even better for ordfilt2)

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  • SOM for Detection

    - by Tim
    I would like to know how I can use SOM for disease detection. Given a lung cancer dataset, how can SOM be applied for detection, there are certain terminologies like, sensitivity, specificity and accuracy percentages....are there ways to calculate all these with the SOM algorithm? I would appreciate answers from anyone who can shed more light on this

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