I am trying to apply a high-pass filter to a signal (column or row vector) consisting of 1-pixel-wide lines taken from a black-and-white image. I know the resolution of the image (res in the code below, given in mm/pixel). How can I filter these line data in MATLAB to discard certain low frequencies (waviness) or large wavelengths, say >10 mm, using a Butterworth filter or any other?
Line data are not centered at zero.
Fs = 1; % I do not know if this assumption is correct for the image.
Fn = Fs/2; % Nyquist frequency.
lambda = 10; % Cut-off wavelength in mm, given.
samples_in_lambda = lambda/res; % divide by resolution to get samples.
fc = 1/samples_in_lambda; % Cut-off frequency from lambda.
I tried : [z, p, k] = butter(9, fc/fn, 'high'); % I see the filter is high pass on plotting.
Can I filter the line data using the above given and assumed values? If not, is there a way that I can filter the data using a cut-off wavelength?
The highest linear spatial frequency you can represent without aliasing is 1 wave cycle per 2 pixels. This means a spatial Nyquist frequency of 1 wave cycle per 2*(res*1e-3) meters, or 1000/(res*2) reciprocal meters. (Confront this with temporal frequencies, which are measured in reciprocal seconds a.k.a. hertz).
In terms of wavelengths: the shortest wave you can represent without aliasing is 2 pixels long per wave cycle. This means a spatial "Nyquist wavelength" of res*2e-3 meters. (Confront this with temporal "wavelengths" a.k.a. periods, which are measured in seconds.)
If you want to set a cutoff wavelength of 10 mm, that corresponds to a spatial frequency of 100 reciprocal meters. Since the butter() function takes as its second input argument (Wn, the cutoff frequency) an arbitrary fraction of the (spatial) Nyquist frequency (the MATLAB documentation calls it "half the sampling rate"), you merely need to set Wn=100/(1000/(res*2)), i.e. Wn=res/5.
Even though your definition of the spatial sampling frequency is not quite correct (unless you are intentionally measuring it in reciprocal pixels), your final result ended up being equivalent to Wn=res/5, so you should be fine using the call to butter() that you indicated.
I am a newbie to SOMs and am using the Matlab SOM package to examine sea level pressure over time. My 2D input array is (row x column): pressure (function of latitude and longitude) x time. When training is complete, and I plot the SOM weight positions, I get the following:
Is this correct? All of the weight position plots that I see are not so 1:1, so my plot appears strange.
Here is my code (note: code will not execute, just for conceptual purposes)
slp = somedata; % dim: 30200 x 1550 [pressure x time]
% Calculate mean for each location over time
mean_slp = nanmean(slp,2);
% Calculate anomalies for each location over time
slp_anom = nan((i2-i1+1)*(j2-j1+1),nfiles);
for i = 1:time
slp_anom(:,i) = slp3(:,i) - mean_slp(i,1);
end
% Normalize data
[slp_anom2,PS] = mapminmax(slp_anom);
net = selforgmap([4 4]);
net.trainParam.epochs = 1000;
net = train(net,slp_anom2);
I appreciate any and all feedback. Thanks!
What is a Weight positions plot in the context of the SOM algorithms's training?
How SOM are trained
The SOM algorithm essentially computes a set of prototype/codebook vectors of the same dimension as the input data. It does so by initializing # neurons according to some rule (random, PCA, etc.) inside the input space and then shifting their positions inside the input space so as to minimize a distance metric under the constraint of a neighborhood function that determines the influence of data point in the neurons' receptive field at each iteration.
Weight Position Plot
The Weight Positions Plot is a 3D plot (!) so you need to use the rotate 3D tool to be able to make sense of the map.
What you then see, depending on dimensionality, is a collection of pale-blue dots and red lines. The pale blue dots are the projections of the neuron positions according onto the two dimensions selected for the plot that have been shifted around the input space.
So the plot would look differently depending on which Weight Dimensions (aka. Input columns) you choose to compute the plot. Matlab typically chooses the first two input columns.
What about the water pressure?
I cannot help here as your dataset appear to be extremely wide for a simple pressure/time vector. Are the columns representing different measurement points on the globe? If so, you should ask yourself what would be gained by having a SOM model of that. What would you do when you got a new vector from a new timestamp? What would you like to do with it? What additinoal information would you gain?
I want to evaluate the grid quality where all coordinates differ in the real case.
Signal is of a ECG signal where average life-time is 75 years.
My task is to evaluate its age at the moment of measurement, which is an inverse problem.
I think 2D approximation of the 3D case is hard (done here by Abo-Zahhad) with with 3-leads (2 on chest and one at left leg - MIT-BIT arrhythmia database):
where f is a piecewise continuous function in R^2, \epsilon is the error matrix and A is a 2D matrix.
Now, I evaluate the average grid distance in x-axis (time) and average grid distance in y-axis (energy).
I think this can be done by Matlab's Image Analysis toolbox.
However, I am not sure how complete the toolbox's approaches are.
I think a transform approach must be used in the setting of uneven and noncontinuous grids. One approach is exact linear time euclidean distance transforms of grid line sampled shapes by Joakim Lindblad et all.
The method presents a distance transform (DT) which assigns to each image point its smallest distance to a selected subset of image points.
This kind of approach is often a basis of algorithms for many methods in image analysis.
I tested unsuccessfully the case with bwdist (Distance transform of binary image) with chessboard (returns empty square matrix), cityblock, euclidean and quasi-euclidean where the last three options return full matrix.
Another pseudocode
% https://stackoverflow.com/a/29956008/54964
%// retrieve picture
imgRGB = imread('dummy.png');
%// detect lines
imgHSV = rgb2hsv(imgRGB);
BW = (imgHSV(:,:,3) < 1);
BW = imclose(imclose(BW, strel('line',40,0)), strel('line',10,90));
%// clear those masked pixels by setting them to background white color
imgRGB2 = imgRGB;
imgRGB2(repmat(BW,[1 1 3])) = 255;
%// show extracted signal
imshow(imgRGB2)
where I think the approach will not work here because the grids are not necessarily continuous and not necessary ideal.
pdist based on the Lumbreras' answer
In the real examples, all coordinates differ such that pdist hamming and jaccard are always 1 with real data.
The options euclidean, cytoblock, minkowski, chebychev, mahalanobis, cosine, correlation, and spearman offer some descriptions of the data.
However, these options make me now little sense in such full matrices.
I want to estimate how long the signal can live.
Sources
J. Müller, and S. Siltanen. Linear and nonlinear inverse problems with practical applications.
EIT with the D-bar method: discontinuous heart-and-lungs phantom. http://wiki.helsinki.fi/display/mathstatHenkilokunta/EIT+with+the+D-bar+method%3A+discontinuous+heart-and-lungs+phantom Visited 29-Feb 2016.
There is a function in Matlab defined as pdist which computes the pairwisedistance between all row elements in a matrix and enables you to choose the type of distance you want to use (Euclidean, cityblock, correlation). Are you after something like this? Not sure I understood your question!
cheers!
Simply, do not do it in the post-processing. Those artifacts of the body can be about about raster images, about the viewer and/or ... Do quality assurance in the signal generation/processing step.
It is much easier to evaluate the original signal than its views.
I need a matlab code for a perceptual hashing algorithm descried here:
http://www.hackerfactor.com/blog/index.php?/archives/432-Looks-Like-It.html
Basically I want this to remove deatails in an image and only leave the major structure components information.
To do so, I think I need the following steps:
1. Reduce the DCT. Suppose the DCT is 32x32 (), just keep the top-left 8x8. Those represent the lowest frequencies in the picture.
Compute the average value. Like the Average Hash, compute the mean DCT value (using only the 8x8 DCT low-frequency values and excluding the first term since the DC coefficient can be significantly different from the other values and will throw off the average).
Further reduce the DCT. Set the 64 hash bits to 0 or 1 depending on whether each of the 64 DCT values is above or below the average value. The result doesn't tell us the actual low frequencies; it just tells us the very-rough relative scale of the frequencies to the mean. The result will not vary as long as the overall structure of the image remains the same; this can survive gamma and color histogram adjustments without a problem.
reconstruct image after the processing.
Anyone can help on any one of above steps?
I have tried some code that gives some results (in the below link), it is not yet perfect:
https://stackoverflow.com/questions/26748051/extract-low-frequency-from-dct-coeffecients-of-an-image-in-matlab
Try this:
% read image
I = imread('cameraman.tif');
% cosine transform and reduction
d = dct2(I);
d = d(1:8,1:8);
% compute average
a = mean(mean(d));
% set bits, here unclear whether > or >= shall be used
b = d > a;
% maybe convert to string:
string = num2str(b(:)');
I've been playing around a little with the Exocortex implementation of the FFT, but I'm having some problems.
Whenever I modify the amplitudes of the frequency bins before calling the iFFT the resulting signal contains some clicks and pops, especially when low frequencies are present in the signal (like drums or basses). However, this does not happen if I attenuate all the bins by the same factor.
Let me put an example of the output buffer of a 4-sample FFT:
// Bin 0 (DC)
FFTOut[0] = 0.0000610351563
FFTOut[1] = 0.0
// Bin 1
FFTOut[2] = 0.000331878662
FFTOut[3] = 0.000629425049
// Bin 2
FFTOut[4] = -0.0000381469727
FFTOut[5] = 0.0
// Bin 3, this is the first and only negative frequency bin.
FFTOut[6] = 0.000331878662
FFTOut[7] = -0.000629425049
The output is composed of pairs of floats, each representing the real and imaginay parts of a single bin. So, bin 0 (array indexes 0, 1) would represent the real and imaginary parts of the DC frequency. As you can see, bins 1 and 3 both have the same values, (except for the sign of the Im part), so I guess bin 3 is the first negative frequency, and finally indexes (4, 5) would be the last positive frequency bin.
Then to attenuate the frequency bin 1 this is what I do:
// Attenuate the 'positive' bin
FFTOut[2] *= 0.5;
FFTOut[3] *= 0.5;
// Attenuate its corresponding negative bin.
FFTOut[6] *= 0.5;
FFTOut[7] *= 0.5;
For the actual tests I'm using a 1024-length FFT and I always provide all the samples so no 0-padding is needed.
// Attenuate
var halfSize = fftWindowLength / 2;
float leftFreq = 0f;
float rightFreq = 22050f;
for( var c = 1; c < halfSize; c++ )
{
var freq = c * (44100d / halfSize);
// Calc. positive and negative frequency indexes.
var k = c * 2;
var nk = (fftWindowLength - c) * 2;
// This kind of attenuation corresponds to a high-pass filter.
// The attenuation at the transition band is linearly applied, could
// this be the cause of the distortion of low frequencies?
var attn = (freq < leftFreq) ?
0 :
(freq < rightFreq) ?
((freq - leftFreq) / (rightFreq - leftFreq)) :
1;
// Attenuate positive and negative bins.
mFFTOut[ k ] *= (float)attn;
mFFTOut[ k + 1 ] *= (float)attn;
mFFTOut[ nk ] *= (float)attn;
mFFTOut[ nk + 1 ] *= (float)attn;
}
Obviously I'm doing something wrong but can't figure out what.
I don't want to use the FFT output as a means to generate a set of FIR coefficients since I'm trying to implement a very basic dynamic equalizer.
What's the correct way to filter in the frequency domain? what I'm missing?
Also, is it really needed to attenuate negative frequencies as well? I've seen an FFT implementation where neg. frequency values are zeroed before synthesis.
Thanks in advance.
There are two issues: the way you use the FFT, and the particular filter.
Filtering is traditionally implemented as convolution in the time domain. You're right that multiplying the spectra of the input and filter signals is equivalent. However, when you use the Discrete Fourier Transform (DFT) (implemented with a Fast Fourier Transform algorithm for speed), you actually calculate a sampled version of the true spectrum. This has lots of implications, but the one most relevant to filtering is the implication that the time domain signal is periodic.
Here's an example. Consider a sinusoidal input signal x with 1.5 cycles in the period, and a simple low pass filter h. In Matlab/Octave syntax:
N = 1024;
n = (1:N)'-1; %'# define the time index
x = sin(2*pi*1.5*n/N); %# input with 1.5 cycles per 1024 points
h = hanning(129) .* sinc(0.25*(-64:1:64)'); %'# windowed sinc LPF, Fc = pi/4
h = [h./sum(h)]; %# normalize DC gain
y = ifft(fft(x) .* fft(h,N)); %# inverse FT of product of sampled spectra
y = real(y); %# due to numerical error, y has a tiny imaginary part
%# Depending on your FT/IFT implementation, might have to scale by N or 1/N here
plot(y);
And here's the graph:
The glitch at the beginning of the block is not what we expect at all. But if you consider fft(x), it makes sense. The Discrete Fourier Transform assumes the signal is periodic within the transform block. As far as the DFT knows, we asked for the transform of one period of this:
This leads to the first important consideration when filtering with DFTs: you are actually implementing circular convolution, not linear convolution. So the "glitch" in the first graph is not really a glitch when you consider the math. So then the question becomes: is there a way to work around the periodicity? The answer is yes: use overlap-save processing. Essentially, you calculate N-long products as above, but only keep N/2 points.
Nproc = 512;
xproc = zeros(2*Nproc,1); %# initialize temp buffer
idx = 1:Nproc; %# initialize half-buffer index
ycorrect = zeros(2*Nproc,1); %# initialize destination
for ctr = 1:(length(x)/Nproc) %# iterate over x 512 points at a time
xproc(1:Nproc) = xproc((Nproc+1):end); %# shift 2nd half of last iteration to 1st half of this iteration
xproc((Nproc+1):end) = x(idx); %# fill 2nd half of this iteration with new data
yproc = ifft(fft(xproc) .* fft(h,2*Nproc)); %# calculate new buffer
ycorrect(idx) = real(yproc((Nproc+1):end)); %# keep 2nd half of new buffer
idx = idx + Nproc; %# step half-buffer index
end
And here's the graph of ycorrect:
This picture makes sense - we expect a startup transient from the filter, then the result settles into the steady state sinusoidal response. Note that now x can be arbitrarily long. The limitation is Nproc > 2*min(length(x),length(h)).
Now onto the second issue: the particular filter. In your loop, you create a filter who's spectrum is essentially H = [0 (1:511)/512 1 (511:-1:1)/512]'; If you do hraw = real(ifft(H)); plot(hraw), you get:
It's hard to see, but there are a bunch of non-zero points at the far left edge of the graph, and then a bunch more at the far right edge. Using Octave's built-in freqz function to look at the frequency response we see (by doing freqz(hraw)):
The magnitude response has a lot of ripples from the high-pass envelope down to zero. Again, the periodicity inherent in the DFT is at work. As far as the DFT is concerned, hraw repeats over and over again. But if you take one period of hraw, as freqz does, its spectrum is quite different from the periodic version's.
So let's define a new signal: hrot = [hraw(513:end) ; hraw(1:512)]; We simply rotate the raw DFT output to make it continuous within the block. Now let's look at the frequency response using freqz(hrot):
Much better. The desired envelope is there, without all the ripples. Of course, the implementation isn't so simple now, you have to do a full complex multiply by fft(hrot) rather than just scaling each complex bin, but at least you'll get the right answer.
Note that for speed, you'd usually pre-calculate the DFT of the padded h, I left it alone in the loop to more easily compare with the original.
Your primary issue is that frequencies aren't well defined over short time intervals. This is particularly true for low frequencies, which is why you notice the problem most there.
Therefore, when you take really short segments out of the sound train, and then you filter these, the filtered segments wont filter in a way that produces a continuous waveform, and you hear the jumps between segments and this is what generates the clicks you here.
For example, taking some reasonable numbers: I start with a waveform at 27.5 Hz (A0 on a piano), digitized at 44100 Hz, it will look like this (where the red part is 1024 samples long):
So first we'll start with a low pass of 40Hz. So since the original frequency is less than 40Hz, a low-pass filter with a 40Hz cut-off shouldn't really have any effect, and we will get an output that almost exactly matches the input. Right? Wrong, wrong, wrong – and this is basically the core of your problem. The problem is that for the short sections the idea of 27.5 Hz isn't clearly defined, and can't be represented well in the DFT.
That 27.5 Hz isn't particularly meaningful in the short segment can be seen by looking at the DFT in the figure below. Note that although the longer segment's DFT (black dots) shows a peak at 27.5 Hz, the short one (red dots) doesn't.
Clearly, then filtering below 40Hz, will just capture the DC offset, and the result of the 40Hz low-pass filter is shown in green below.
The blue curve (taken with a 200 Hz cut-off) is starting to match up much better. But note that it's not the low frequencies that are making it match up well, but the inclusion of high frequencies. It's not until we include every frequency possible in the short segment, up to 22KHz that we finally get a good representation of the original sine wave.
The reason for all of this is that a small segment of a 27.5 Hz sine wave is not a 27.5 Hz sine wave, and it's DFT doesn't have much to do with 27.5 Hz.
Are you attenuating the value of the DC frequency sample to zero? It appears that you are not attenuating it at all in your example. Since you are implementing a high pass filter, you need to set the DC value to zero as well.
This would explain low frequency distortion. You would have a lot of ripple in the frequency response at low frequencies if that DC value is non-zero because of the large transition.
Here is an example in MATLAB/Octave to demonstrate what might be happening:
N = 32;
os = 4;
Fs = 1000;
X = [ones(1,4) linspace(1,0,8) zeros(1,3) 1 zeros(1,4) linspace(0,1,8) ones(1,4)];
x = ifftshift(ifft(X));
Xos = fft(x, N*os);
f1 = linspace(-Fs/2, Fs/2-Fs/N, N);
f2 = linspace(-Fs/2, Fs/2-Fs/(N*os), N*os);
hold off;
plot(f2, abs(Xos), '-o');
hold on;
grid on;
plot(f1, abs(X), '-ro');
hold off;
xlabel('Frequency (Hz)');
ylabel('Magnitude');
Notice that in my code, I am creating an example of the DC value being non-zero, followed by an abrupt change to zero, and then a ramp up. I then take the IFFT to transform into the time domain. Then I perform a zero-padded fft (which is done automatically by MATLAB when you pass in an fft size bigger than the input signal) on that time-domain signal. The zero-padding in the time-domain results in interpolation in the frequency domain. Using this, we can see how the filter will respond between filter samples.
One of the most important things to remember is that even though you are setting filter response values at given frequencies by attenuating the outputs of the DFT, this guarantees nothing for frequencies occurring between sample points. This means the more abrupt your changes, the more overshoot and oscillation between samples will occur.
Now to answer your question on how this filtering should be done. There are a number of ways, but one of the easiest to implement and understand is the window design method. The problem with your current design is that the transition width is huge. Most of the time, you will want as quick of transitions as possible, with as little ripple as possible.
In the next code, I will create an ideal filter and display the response:
N = 32;
os = 4;
Fs = 1000;
X = [ones(1,8) zeros(1,16) ones(1,8)];
x = ifftshift(ifft(X));
Xos = fft(x, N*os);
f1 = linspace(-Fs/2, Fs/2-Fs/N, N);
f2 = linspace(-Fs/2, Fs/2-Fs/(N*os), N*os);
hold off;
plot(f2, abs(Xos), '-o');
hold on;
grid on;
plot(f1, abs(X), '-ro');
hold off;
xlabel('Frequency (Hz)');
ylabel('Magnitude');
Notice that there is a lot of oscillation caused by the abrupt changes.
The FFT or Discrete Fourier Transform is a sampled version of the Fourier Transform. The Fourier Transform is applied to a signal over the continuous range -infinity to infinity while the DFT is applied over a finite number of samples. This in effect results in a square windowing (truncation) in the time domain when using the DFT since we are only dealing with a finite number of samples. Unfortunately, the DFT of a square wave is a sinc type function (sin(x)/x).
The problem with having sharp transitions in your filter (quick jump from 0 to 1 in one sample) is that this has a very long response in the time domain, which is being truncated by a square window. So to help minimize this problem, we can multiply the time-domain signal by a more gradual window. If we multiply a hanning window by adding the line:
x = x .* hanning(1,N).';
after taking the IFFT, we get this response:
So I would recommend trying to implement the window design method since it is fairly simple (there are better ways, but they get more complicated). Since you are implementing an equalizer, I assume you want to be able to change the attenuations on the fly, so I would suggest calculating and storing the filter in the frequency domain whenever there is a change in parameters, and then you can just apply it to each input audio buffer by taking the fft of the input buffer, multiplying by your frequency domain filter samples, and then performing the ifft to get back to the time domain. This will be a lot more efficient than all of the branching you are doing for each sample.
First, about the normalization: that is a known (non) issue. The DFT/IDFT would require a factor 1/sqrt(N) (apart from the standard cosine/sine factors) in each one (direct an inverse) to make them simmetric and truly invertible. Another possibility is to divide one of them (the direct or the inverse) by N, this is a matter of convenience and taste. Often the FFT routines do not perform this normalization, the user is expected to be aware of it and normalize as he prefers. See
Second: in a (say) 16 point DFT, what you call the bin 0 would correspond to the zero frequency (DC), bin 1 low freq... bin 4 medium freq, bin 8 to the highest frequency and bins 9...15 to the "negative frequencies". In you example, then, bin 1 is actually both the low frequency and medium frequency. Apart from this consideration, there is nothing conceptually wrong in your "equalization". I don't understand what you mean by "the signal gets distorted at low frequencies". How do you observe that ?