I've a big issue with continuous wavelet transform. I've created this signal
t = 0:1/2000:1-1/2000;
dt = 1/2000;
x1 = sin(50*pi*t).*exp(-50*pi*(t-0.2).^2);
x2 = sin(50*pi*t).*exp(-100*pi*(t-0.5).^2);
x3 = 2*cos(140*pi*t).*exp(-50*pi*(t-0.2).^2);
x4 = 2*sin(140*pi*t).*exp(-80*pi*(t-0.8).^2);
x = x1+x2+x3+x4;
And its rappresentation on time is
Than I computed its Fourier Transform in the classic way
Ts =1/Fs;
N = length(x);
t = 0:Ts:Ts*N-Ts;
FTx = fft(x,N);
S = (abs(FTx).^2)/N; %amplitude
f_FT = (0:Fs/N:Fs-Fs/N);
S = S(1:N/2); % i reject half signal
f_FT = f_FT(1:N/2);
And its rappresentation is
I computed the continuous wavelet transform using the function cwtft
s0 = 2;
a0 = 2^(1/32);
scales = (s0*a0.^(32:7*32)).*dt;
cwtx = cwtft({x,dt},'Scales',scales,'Wavelet',{'bump',[4 0.9]});
figure;
contour(t,cwtx.frequencies,abs(cwtx.cfs))
xlabel('Seconds'), ylabel('Hz');
grid on;
title('Analytic CWT using Bump Wavelet')
hcol = colorbar;
hcol.Label.String = 'Magnitude';
I don't know what magnitude rappresent, and specially why it differs so much with the values obtained with the classic FFT. Is there a way to convert it?
Thank you so much
There are two dimensions present here: Time and Frequency (Scale). The third dimension, which is magnitude or amplitude, shows the intensity of your frequencies (scales, which in wavelets scales are the inverse of frequencies) in those local times.
I guess it is much simpler than what I just explained. You are basically looking at a similar image to these three dimensions from the top in your last image in the question:
Source
Machine Learning with Signal Processing Techniques
Related
I am wondering whether I could broadcast the operation below without using a for loop.
X1 = torch.rand(B,d)
X2 = torch.rand(B,d)
X3 = torch.rand(B,d)
X = [X1,X2,X3]
w = torch.rand(3)
res = 0
for i in range(3):
res += X[i] * w[i]:
An important thing to note here is that w is a parameter that I need to optimize. So I need to track its gradient.
Any help would be appreciated.
I just don't know how to speed up this operation without ruining the gradient map.
I have decomposed some time series data using a custom FFT implementation. By design my FFT implementation gives me a set of cos and sine waves that I can then sum together to regenerate the original signal. This works well without issue, so I know that the extracted sine and cos waves are correct in terms of amplitude, period and phase.
The data I am using has 1024 samples which gives me the properties of 512 cos waves and 512 sine waves (eg the amplitude, phase and period data for each wave).
To save on data storage I am trying to find/understand the mathematical relationship between the amplitudes of the waves. Instead of having to save every amplitude for every sine and cos wave I would like to simply save some coefficients that I can later use to rebuild the amplitudes in code.
FFT Sine Waves with Amplitudes
From the above image you can see that there is a set of Power curve coefficients that roughly fit the amplitude data, however for my use case this is not accurate enough.
As I have all the source data along with the generated properties of each wave, is there a simple formula that I can use or a transform I can perform to generate the amplitudes in code after I have performed the FFT? I know that the amplitudes are related to the real and imaginary values however I cannot store all the real and imaginary values either due to space requirements.
As an example of how I am saving this issue for the period data, I have found that the period of each wave is simply Math.Power(waveIndex, -1). So for the wave periods I do not have to store the data, I can simply regenerate in code.
I cannot currently find a relationship between the amplitudes within the sine wave or even a relationship between cos and sine amplitudes, however the theory and math behind FFT is beyond me so I am hoping that there is a simply formula or concept I can implement.
Following the replies I have added the below code that I use to get the sine and cos wave values, this code snippet may help those replying.
internal void GetSineAndCosWavesBasic(double[] outReal, double[] outImag, int numWaves, out double[,] sineValues, out double[,] cosValues)
{
// the real and imag values from Cooley-Tukey decimation-in-time radix-2 FFT are passed in
// and we want to generate the cos and sine values for each sample for each wave
var length = outReal.Length;
var lengthDouble = (double)length;
var halfLength = lengthDouble / 2.0;
sineValues = new double[numWaves, length];
cosValues = new double[numWaves, length];
var Pi2 = 2 * Math.PI;
for (var waveIdx = 0; waveIdx < numWaves; waveIdx++)
{
for (var sampleIdx = 0; sampleIdx < length; sampleIdx++)
{
// first value case and middle value case
var reX = outReal[waveIdx] / halfLength;
if (sampleIdx == 0)
{
reX = outReal[waveIdx] / lengthDouble;
}
else if (sampleIdx == halfLength)
{
reX = outReal[waveIdx] / lengthDouble;
}
// precompute the value that gets sine/cos applied
var tmp = (Pi2 * waveIdx * sampleIdx) / lengthDouble;
// get the instant cos and sine values
var valueCos = Math.Cos(tmp) * reX;
var valueSin = Math.Sin(tmp) * (-outImag[waveIdx] / halfLength);
// update the sine and cos values for this wave for this sample
cosValues[waveIdx, sampleIdx] = valueCos;
sineValues[waveIdx, sampleIdx] = valueSin;
}
}
}
And the below is how I get the magnitude and phase values, although I do not currently use those anywhere.
internal void CalculateMagAndPhaseBasic(double[] outReal, double[] outImag, out double[] mag, out double[] phase)
{
// the real and imag values from Cooley-Tukey decimation-in-time radix-2 FFT are passed in
// and we want to generate the magnitude and phase values
var length = outReal.Length;
mag = new double[(length / 2) +1];
phase = new double[(length / 2) + 1];
for (var i = 0; i <= length / 2; i++)
{
mag[i] = Math.Pow((outReal[i] * outReal[i]) + (outImag[i] * outImag[i]), 0.5);
phase[i] = Math.Atan2(outImag[i], outReal[i]);
}
}
Actually the fft just returns you complex coefficients S(w)=a+jb
For an N point fft, abs(S(w)) * 2/N will be (close to) the amplitude of the sinusoidal component at frequency w.
This assumes that the sinusoidal component has a frequency close to the center of the fft bin, otherwise the power will be "split" between two adjacent bins.
And that the frequency you're interested in is present through all the fft window.
The output of an FFT has the same number of degrees of freedom as the input. There is no simple formula (other than the FFT itself) that relates the FFT results to just each other, as all of the FFT outputs can change if any of the FFT inputs changes.
The relationship between the sine and cosine of each FFT complex bin result is related to the phase of the sinusoidal input component at that frequency (of the bin center), circularly relative to the start and end. If the phase changes, so can both the sine and cosine component. See: atan2()
Suppose I have an image A, I applied Gaussian Blur on it with Sigam=3 So I got another Image B. Is there a way to know the applied sigma if A,B is given?
Further clarification:
Image A:
Image B:
I want to write a function that take A,B and return Sigma:
double get_sigma(cv::Mat const& A,cv::Mat const& B);
Any suggestions?
EDIT1: The suggested approach doesn't work in practice in its original form(i.e. using only 9 equations for a 3 x 3 kernel), and I realized this later. See EDIT1 below for an explanation and EDIT2 for a method that works.
EDIT2: As suggested by Humam, I used the Least Squares Estimate (LSE) to find the coefficients.
I think you can estimate the filter kernel by solving a linear system of equations in this case. A linear filter weighs the pixels in a window by its coefficients, then take their sum and assign this value to the center pixel of the window in the result image. So, for a 3 x 3 filter like
the resulting pixel value in the filtered image
result_pix_value = h11 * a(y, x) + h12 * a(y, x+1) + h13 * a(y, x+2) +
h21 * a(y+1, x) + h22 * a(y+1, x+1) + h23 * a(y+1, x+2) +
h31 * a(y+2, x) + h32 * a(y+2, x+1) + h33 * a(y+2, x+2)
where a's are the pixel values within the window in the original image. Here, for the 3 x 3 filter you have 9 unknowns, so you need 9 equations. You can obtain those 9 equations using 9 pixels in the resulting image. Then you can form an Ax = b system and solve for x to obtain the filter coefficients. With the coefficients available, I think you can find the sigma.
In the following example I'm using non-overlapping windows as shown to obtain the equations.
You don't have to know the size of the filter. If you use a larger size, the coefficients that are not relevant will be close to zero.
Your result image size is different than the input image, so i didn't use that image for following calculation. I use your input image and apply my own filter.
I tested this in Octave. You can quickly run it if you have Octave/Matlab. For Octave, you need to load the image package.
I'm using the following kernel to blur the image:
h =
0.10963 0.11184 0.10963
0.11184 0.11410 0.11184
0.10963 0.11184 0.10963
When I estimate it using a window size 5, I get the following. As I said, the coefficients that are not relevant are close to zero.
g =
9.5787e-015 -3.1508e-014 1.2974e-015 -3.4897e-015 1.2739e-014
-3.7248e-014 1.0963e-001 1.1184e-001 1.0963e-001 1.8418e-015
4.1825e-014 1.1184e-001 1.1410e-001 1.1184e-001 -7.3554e-014
-2.4861e-014 1.0963e-001 1.1184e-001 1.0963e-001 9.7664e-014
1.3692e-014 4.6182e-016 -2.9215e-014 3.1305e-014 -4.4875e-014
EDIT1:
First of all, my apologies.
This approach doesn't really work in the practice. I've used the filt = conv2(a, h, 'same'); in the code. The resulting image data type in this case is double, whereas in the actual image the data type is usually uint8, so there's loss of information, which we can think of as noise. I simulated this with the minor modification filt = floor(conv2(a, h, 'same'));, and then I don't get the expected results.
The sampling approach is not ideal, because it's possible that it results in a degenerated system. Better approach is to use random sampling, avoiding the borders and making sure the entries in the b vector are unique. In the ideal case, as in my code, we are making sure the system Ax = b has a unique solution this way.
One approach would be to reformulate this as Mv = 0 system and try to minimize the squared norm of Mv under the constraint squared-norm v = 1, which we can solve using SVD. I could be wrong here, and I haven't tried this.
Another approach is to use the symmetry of the Gaussian kernel. Then a 3x3 kernel will have only 3 unknowns instead of 9. I think, this way we impose additional constraints on v of the above paragraph.
I'll try these out and post the results, even if I don't get the expected results.
EDIT2:
Using the LSE, we can find the filter coefficients as pinv(A'A)A'b. For completion, I'm adding a simple (and slow) LSE code.
Initial Octave Code:
clear all
im = double(imread('I2vxD.png'));
k = 5;
r = floor(k/2);
a = im(:, :, 1); % take the red channel
h = fspecial('gaussian', [3 3], 5); % filter with a 3x3 gaussian
filt = conv2(a, h, 'same');
% use non-overlapping windows to for the Ax = b syatem
% NOTE: boundry error checking isn't performed in the code below
s = floor(size(a)/2);
y = s(1);
x = s(2);
w = k*k;
y1 = s(1)-floor(w/2) + r;
y2 = s(1)+floor(w/2);
x1 = s(2)-floor(w/2) + r;
x2 = s(2)+floor(w/2);
b = [];
A = [];
for y = y1:k:y2
for x = x1:k:x2
b = [b; filt(y, x)];
f = a(y-r:y+r, x-r:x+r);
A = [A; f(:)'];
end
end
% estimated filter kernel
g = reshape(A\b, k, k)
LSE method:
clear all
im = double(imread('I2vxD.png'));
k = 5;
r = floor(k/2);
a = im(:, :, 1); % take the red channel
h = fspecial('gaussian', [3 3], 5); % filter with a 3x3 gaussian
filt = floor(conv2(a, h, 'same'));
s = size(a);
y1 = r+2; y2 = s(1)-r-2;
x1 = r+2; x2 = s(2)-r-2;
b = [];
A = [];
for y = y1:2:y2
for x = x1:2:x2
b = [b; filt(y, x)];
f = a(y-r:y+r, x-r:x+r);
f = f(:)';
A = [A; f];
end
end
g = reshape(A\b, k, k) % A\b returns the least squares solution
%g = reshape(pinv(A'*A)*A'*b, k, k)
I have calculated 8 Gabor filters with Theta rotation m*PI/8.
Parameters of the Gabor kernel given as input to OpenCv cv2.getGaborKernel:
ksize = 11, theta = m*PI/8 lambd = 16/3 sigma = (5.09030 * 8.0) / (3.0 * PI) gamma = 0.5890 psi = 0
kernel = cv2.getGaborKernel(ksize = (ksize,ksize), sigma = sigma,
theta = theta, lambd = lambd,
gamma = gamma, psi = psi)
The parameters are designed according to "Features Extraction using a Gabor filter family", Zhen, Zhao, Wang.
The formula adopted is the one of the third family of Gabor filters.
The 8 filters obtained are:
The original image is:
The images obtained by filtering the images are:
They are calculated with cv2.filter2D
fimg = cv2.filter2D(img, cv2.CV_64F, kernel)
Why the gabor filters with theta = 0 and theta = PI / 2.0 have a really different continuous component compared to the others?
It does not really make sense to me.
The reason was the PSI param that I set to 0. The problem is immediatly fixed is psi is kept at the default value PI/2.
I'm trying to implement Otsu binarization technique on document images such as the one shown:
Could someone please tell me how to implement the code in MATLAB?
Taken from Otsu's method on Wikipedia
I = imread('cameraman.tif');
Step 1. Compute histogram and probabilities of each intensity level.
nbins = 256; % Number of bins
counts = imhist(I,nbins); % Each intensity increments the histogram from 0 to 255
p = counts / sum(counts); % Probabilities
Step 2. Set up initial omega_i(0) and mu_i(0)
omega1 = 0;
omega2 = 1;
mu1 = 0;
mu2 = mean(I(:));
Step 3. Step through all possible thresholds from 0 to maximum intensity (255)
Step 3.1 Update omega_i and mu_i
Step 3.2 Compute sigma_b_squared
for t = 1:nbins
omega1(t) = sum(p(1:t));
omega2(t) = sum(p(t+1:end));
mu1(t) = sum(p(1:t).*(1:t)');
mu2(t) = sum(p(t+1:end).*(t+1:nbins)');
end
sigma_b_squared_wiki = omega1 .* omega2 .* (mu2-mu1).^2; % Eq. (14)
sigma_b_squared_otsu = (mu1(end) .* omega1-mu1) .^2 ./(omega1 .* (1-omega1)); % Eq. (18)
Step 4 Desired threshold corresponds to the location of maximum of sigma_b_squared
[~,thres_level_wiki] = max(sigma_b_squared_wiki);
[~,thres_level_otsu] = max(sigma_b_squared_otsu);
There are some differences between the wiki-version eq. (14) in Otsu and the eq. (18), and I don't why. But the thres_level_otsu correspond to the MATLAB's implementation graythresh(I)
Since the function graythresh in Matlab implements the Otsu method, what you have to do is convert your image to grayscale and then use the im2bw function to binarize the image using the threhsold level returned by graythresh.
To convert your image I to grayscale you can use the following code:
I = im2uint8(I);
if size(I,3) ~= 1
I = rgb2gray(I);
end;
To get the binary image Ib using the Otsu's method, use the following code:
Ib = im2bw(I, graythresh(I));
You should get the following result:
Starting out with what your initial question was implementing the OTSU thresolding its true that MATLAB's graythresh function is based on that method
The OTSU's method considers the threshold value as the valley between two peaks that is one of the foreground pixels and the other of the background pixels
Pertaining to your image which seems like a historical manuscript found this paper that compares all the methods that could be used for thresholding document images
You can also download and read up sauvola thresholding from here
Good luck with its implementation =)
Corrected MATLAB Implementation (for 2d matrix)
function [T] = myotsu(I,N);
% create histogram
nbins = N;
[x,h] = hist(I(:),nbins);
% calculate probabilities
p = x./sum(x);
% initialisation
om1 = 0;
om2 = 1;
mu1 = 0;
mu2 = mode(I(:));
for t = 1:nbins,
om1(t) = sum(p(1:t));
om2(t) = sum(p(t+1:nbins));
mu1(t) = sum(p(1:t).*[1:t]);
mu2(t) = sum(p(t+1:nbins).*[t+1:nbins]);
end
sigma = (mu1(nbins).*om1-mu1).^2./(om1.*(1-om1));
idx = find(sigma == max(sigma));
T = h(idx(1));