I have an interference fringe pattern. I applied Fourier transform on it which gave an image with few very sharp peaks corresponding to each spatial frequency. My question is that is there any way(algorithm) to detect coordinates(pixel value) corresponding to these intense peaks?
Please note that there are more than one peaks(but not many) with different magnitudes.
You can use the built-in findpeaks, for example:
[pks,locs] = findpeaks(data)
returns local maxima or peaks, pks, in the input data, with the indices of the local peaks locs. The input data requires a row or column vector with real-valued elements with a minimum length of three. findpeaks compares each element of data to its neighboring values. If an element of data is larger than both of its neighbors or equals Inf, the element is a local peak. If there are no local maxima, pks will be an empty vector.
To set a threshold use any of the following handles: 'MINPEAKHEIGHT' , 'MINPEAKDISTANCE', or , 'THRESHOLD'
for example:
[pks,locs] = findpeaks(data,'MINPEAKHEIGHT',10)
Related
I want to do a comparison of 2 audio files (each audio file is speaking "ba a ta") with the existing function in matlab called Dynamic Time Warping (DTW). Before doing a dynamic time warping, I get an array/vector from the Fast Fourier Transform (FFT) functions available in matlab, my code so far (my matlab filename: test.m):
fftRecording1 = fft(audioread('C:\Users\handy\Documents\MATLAB\my_recording_1.wav'));
fftRecording2 = fft(audioread('C:\Users\handy\Documents\MATLAB\fajar.wav'));
dist = dtw(fftRecording1, fftRecording2);
When I try the DTW function there is an error because the length (row) of the array/vector 2 file is different. Error message:
Error using dtw (line 82)
The number of rows between X and Y must be equal when X and Y are matrices
Error in test (line 3)
dist = dtw(fftRecording1, fftRecording2);
contents of the fftRecording1 and fftRecording2 variables
My question is: before do the FFT and DTW, how do step by step normalize so that the length (row) 2 audio files is equal? or there are other ways to make the data length (row) 2 audio files is equal?
According to dtw's documentation:
To stretch the inputs, dtw repeats each element of x and y as many times as necessary. If x and y are matrices, then dist stretches them by repeating their columns. In that case, x and y must have the same number of rows.
In your case your columns represent the audio channels, with the rows representing the quantity to be aligned (i.e. the reverse of what dtw is expecting). To setup the inputs according to what dtw expect, simply transpose the inputs:
dist = dtw(transpose(fftRecording1), transpose(fftRecording2));
Dynamic Time Warping does not need the input sequences to be of same length. DTW is actually used to find similarity between two different time aligned sequences.
No, they don’t need to have the same length in a time-related-sense. They need to have the same number of dimensions (2D Signal, 3D Signal,...) which is equivalent to their number or rows. The whole idea of DTW is to match similar contents which might be stretched to different lengths - so there would absolutely be no point in requiring the inputs to have the same length.
Related to your question: just call the dtw with the transposed of your signals and you will get a proper result.
dtw(signal1’, signal2’);
You should apply the DTW on the original signals rather than the fourier transforms. The FFT transfers the signal from time to frequency domain. So instead of warping signal1 in order to match signal2, you are warping frequencies when using FFT before DTW. The amplitude of the fourier transform depends on the number of points in the considered FFT-Time-Window. From my point of view there is absolutely no point in applying DTW on a fourier transform.
I have three variables, e.g., latitude, longitude and temperature. For each latitude and longitude, I have corresponding temperature value. I want to plot latitude v/s longitude plot in 5 degree x 5 degree grid , with mean temperature value inserted in that particular grid instead of occurring frequency.
Data= [latGrid,lonGrid] = meshgrid(25:45,125:145);
T = table(latGrid(:),lonGrid(:),randi([0,35],size(latGrid(:))),...
'VariableNames',{'lat','lon','temp'});
At the end, I need it somewhat like the following image:
Sounds to me like you want to scale your grid. The easiest way to do this is to smooth and downsample.
While 2d histograms also bin values into a grid, using a histogram is not the way to find the mean of datapoints in a smooth grid. A histogram counts the occurrence of values in a set of ranges. In a 2d example, a histogram would take the input measurements [1, 3, 3, 5] and count the number of ones, the number of threes, etc. A 2d histogram will count occurrences of pairs of numbers. (You might want to use histogram to help organize a measurements taken at irregular intervals, but that would be a different question)
How to smooth and downsample without the Image Processing Toolbox
Keep your data in the 2d matrix format rather than reshaping it into a table. This makes it easier to find the neighbors of each grid location.
%% Sample Data
[latGrid,lonGrid] = meshgrid(25:45,125:145);
temp = rand(size(latGrid));
There are many tools in Matlab for smoothing matrices. If you want to have the mean of a 5x5 window. You can write a for-loop, use a convolution, or use filter2. My example uses convolution. For more on convolutional filters, I suggest the wikipedia page.
%% Mean filter with conv2
M = ones(5) ./ 25; % 5x5 mean or box blur filter
C_temp = conv2(temp, M, 'valid');
C_temp is a blurry version of the original temperature variable with a slightly smaller size because we can't accurately take the mean of the edges. The border is reduced by a frame of 2 measurements. Now, we just need to take every fifth measurement from C_temp to scale down the grid.
%% Subsample result
C_temp = C_temp(1:5:end, 1:5:end);
% Because we removed a border from C_temp, we also need to remove a border from latGrid and lonGrid
[h, w] = size(latGrid)
latGrid = latGrid(5:5:h-5, 5:5:w-5);
lonGrid = lonGrid(5:5:h-5, 5:5,w-5);
Here's what the steps look like
If you use a slightly more organized, temp variable. It's easier to see that the result is correct.
With Image Processing Toolbox
imresize has a box filter method option that is equivalent to a mean filter. However, you have to do a little calculation to find the scaling factor that is equivalent to using a 5x5 window.
C_temp = imresize(temp, scale, 'box');
I am trying to find the point that is at a minimum distance from the candidate set. Z is a matrix where the rows are the dimension and columns indicate points. Computing the inter-point distances, and then recording the point with minimum distance and its distance as well. Below is the code snippet. The code works fine for a small dimension and small set of points. But, it takes a long time for large data set (N = 1 million data points and dimension is also high). Is there an efficient way?
I suggest that you use pdist to do the heavy lifting for you. This function will compute the pairwise distance between every two points in your array. The resulting vector has to be put into matrix form using squareform in order to find the minimal value for each pair:
N = 100;
Z = rand(2,N); % each column is a 2-dimensional point
% pdist assumes that the second index corresponds to dimensions
% so we need to transpose inside pdist()
distmatrix = squareform(pdist(Z.','euclidean')); % output is [N, N] in size
% set diagonal values to infinity to avoid getting 0 self-distance as minimum
distmatrix = distmatrix + diag(inf(1,size(distmatrix,1)));
mindists = min(distmatrix,[],2); % find the minimum for each row
sum_dist = sum(mindists); % sum of minimal distance between each pair of points
This computes every pair twice, but I think this is true for your original implementation.
The idea is that pdist computes the pairwise distance between the columns of its input. So we put the transpose of Z into pdist. Since the full output is always a square matrix with zero diagonal, pdist is implemented such that it only returns the values above the diagonal, in a vector. So a call to squareform is needed to get the proper distance matrix. Then, the row-wise minimum of this matrix have to be found, but first we have to exclude the zero in the diagonals. I was lazy so I put inf into the diagonals, to make sure that the minimum is elsewhere. In the end we just have to sum up the minimal distances.
I am solving a funcion which uses the moving average filter to remove noise. How can I determine index and value of first and second negative peak after I apply the filter to input data?
Use findpeaks on the negative of your data, then extract the first two elements to extract the first and second indices of where the negative peaks are located. Supposing your signal was stored in f, you would simply do:
[peaks, locs] = findpeaks(-f);
p = peaks(1:2);
loc = locs(1:2);
findpeaks works by finding local maxima. If you want to find local minima (i.e. negative peaks), you would apply findpeaks to the negative of your signal so that the local minima become local maxima, then apply the same algorithm. loc would contain the first two locations of where the negative peaks are, while p will determine those negative peak amplitudes.
However, you'll probably need to play around with the input parameters to findpeaks, instead of using the default ones to suit your data, but this should be enough to get you started.
Sidenote
If you don't have access to findpeaks, take a look at this post that I wrote to find peaks for FFT data. The data is different, but the overall logic is the same. However, this finds all peaks - both local maxima and minima. If you just want to find the minima, simply look at the negative of the signal rather than the absolute value.
I have a 3d matrix of 100x100x100. Each point of that matrix has assigned a value that corresponds to a certain signal strength. If I plot all the points the result is incomprehensible and requires horsepower to compute, due to the large amount of points that are painted.
The next picture examplify the problem (in that case the matrix was 50x50x50 for reducing the computation time):
[x,y,z] = meshgrid(1:50,1:50,1:50);
scatter3(x(:),y(:),z(:),5,strength(:),'filled')
I would like to plot only the highest values (for example, the top 10). How can I do it?
One simple solution that came up in my mind is to asign "nan" to the values higher than the treshold.
Even the results are nice I think that it must be a most elegant solution to fix it.
Reshape it into an nx1 vector. Sort that vector and take the first ten values.
num_of_rows = size(M,1)
V = reshape(M,num_of_rows,1);
sorted_V = sort(V,'descend');
ind = sorted_V(1:10)
I am assuming that M is your 3D matrix. This will give you your top ten values in your matrix and the respective index. The you can use ind2sub() to get the x,y,z.