I'm trying to compare Matlab fft of a cosinus with two different zero padding. I thought that it wouldn't change the frequency response but when I superimpose the two curves, the frequencies are not the same. I suppose that there is something wrong with the way I do my two fft?
Fe = 8000;
F = 1680;
w = 2*pi*F;
N = 50;
P = 50;
T = 1/Fe;
t = (0:T:P*T);
x = real(exp(i*w*t))
x_reduced = x(1:P)
X = fft(x_reduced,N)
N = 1000;
Y = fft(x_reduced,N)
plot(abs(Y))
hold on
plot(abs(X),'*')
Thanks in advance
plot((0:999)/1000*Fe,abs(Y))
hold on
plot((0:49)/50*Fe,abs(X),'*')
You may need to align the frequencies of both cases.
When you pad the FFT you change the resolution of each bin (you are effectively interpolating between bins), so while the corresponding frequencies are still the same of course the actual mapping to bin indices will change. If you were to scale the two FFT plots so that the horizontal axes for both line up (i.e. bin 0 aligned on both, and bin 50 aligned with bin 1000) then the plots would match.
Related
I'm working on an application to determine from an image the degree of alignment of a fiber network. I've read several papers on this issue and they basically do this:
Find the 2D discrete Fourier transform (DFT = F(u,v)) of the image (gray, range 0-255)
Find the Fourier Spectrum (FS = abs(F(u,v))) and the Power Spectrum (PS = FS^2)
Convert spectrum to polar coordinates and divide it into 1º intervals.
Calculate number-averaged line intensities (FI) for each interval (theta), that is, the average of all the intensities (pixels) forming "theta" degrees with respect to the horizontal axis.
Transform FI(theta) to cartesian coordinates
Cxy(theta) = [FI*cos(theta), FI*sin(theta)]
Find eigenvalues (lambda1 and lambda2) of the matrix Cxy'*Cxy
Find alignment index as alpha = 1 - lamda2/lambda1
I've implemented this in MATLAB (code below), but I'm not sure whether it is ok since point 3 and 4 are not really clear for me (I'm getting similar results to those of the papers, but not in all cases). For instance, in point 3, "spectrum" is referring to FS or to PS?. And in point 4, how should this average be done? are all the pixels considered? (even though there are more pixels in the diagonal).
rgb = imread('network.tif');%513x513 pixels
im = rgb2gray(rgb);
im = imrotate(im,-90);%since FFT space is rotated 90º
FT = fft2(im) ;
FS = abs(FT); %Fourier spectrum
PS = FS.^2; % Power spectrum
FS = fftshift(FS);
PS = fftshift(PS);
xoffset = (513-1)/2;
yoffset = (513-1)/2;
% Avoid low frequency points
x1 = 5;
y1 = 0;
% Maximum high frequency pixels
x2 = 255;
y2 = 0;
for theta = 0:pi/180:pi
% Transposed rotation matrix
Rt = [cos(theta) sin(theta);
-sin(theta) cos(theta)];
% Find radial lines necessary for improfile
xy1_rot = Rt * [x1; y1] + [xoffset; yoffset];
xy2_rot = Rt * [x2; y2] + [xoffset; yoffset];
plot([xy1_rot(1) xy2_rot(1)], ...
[xy1_rot(2) xy2_rot(2)], ...
'linestyle','none', ...
'marker','o', ...
'color','k');
prof = improfile(F,[xy1_rot(1) xy2_rot(1)],[xy1_rot(2) xy2_rot(2)]);
i = i + 1;
FI(i) = sum(prof(:))/length(prof);
Cxy(i,:) = [FI(i)*cos(theta), FI(i)*sin(theta)];
end
C = Cxy'*Cxy;
[V,D] = eig(C)
lambda2 = D(1,1);
lambda1 = D(2,2);
alpha = 1 - lambda2/lambda1
Figure: A) original image, B) plot of log(P+1), C) polar plot of FI.
My main concern is that when I choose an artificial image perfectly aligned (attached figure), I get alpha = 0.91, and it should be exactly 1.
Any help will be greatly appreciated.
PD: those black dots in the middle plot are just the points used by improfile.
I believe that there are a couple sources of potential error here that are leading to you not getting a perfect alpha value.
Discrete Fourier Transform
You have discrete imaging data which forces you to take a discrete Fourier transform which inevitably (depending on the resolution of the input data) have some accuracy issues.
Binning vs. Sampling Along a Line
The way that you have done the binning is that you literally drew a line (rotated by a particular angle) and sampled the image along that line using improfile. Using improfile performs interpolation of your data along that line introducing yet another potential source of error. The default is nearest neighbor interpolation which in the example shown below can cause multiple "profiles" to all pick up the same points.
This was with a rotation of 1-degree off-vertical when technically you'd want those peaks to only appear for a perfectly vertical line. It is clear to see how this sort of interpolation of the Fourier spectrum can lead to a spread around the "correct" answer.
Data Undersampling
Similar to Nyquist sampling in the Fourier domain, sampling in the spatial domain has some requirements as well.
Imagine for a second that you wanted to use 45-degree bin widths instead of the 1-degree. Your approach would still sample along a thin line and use that sample to represent 45-degrees worth or data. Clearly, this is a gross under-sampling of the data and you can imagine that the result wouldn't be very accurate.
It becomes more and more of an issue the further you get from the center of the image since the data in this "bin" is really pie wedge shaped and you're approximating it with a line.
A Potential Solution
A different approach to binning would be to determine the polar coordinates (r, theta) for all pixel centers in the image. Then to bin the theta components into 1-degree bins. Then sum all of the values that fall into that bin.
This has several advantages:
It removes the undersampling that we talked about and draws samples from the entire "pie wedge" regardless of the sampling angle.
It ensures that each pixel belongs to one and only one angular bin
I have implemented this alternate approach in the code below with some false horizontal line data and am able to achieve an alpha value of 0.988 which I'd say is pretty good given the discrete nature of the data.
% Draw a bunch of horizontal lines
data = zeros(101);
data([5:5:end],:) = 1;
fourier = fftshift(fft2(data));
FS = abs(fourier);
PS = FS.^2;
center = fliplr(size(FS)) / 2;
[xx,yy] = meshgrid(1:size(FS,2), 1:size(FS, 1));
coords = [xx(:), yy(:)];
% De-mean coordinates to center at the middle of the image
coords = bsxfun(#minus, coords, center);
[theta, R] = cart2pol(coords(:,1), coords(:,2));
% Convert to degrees and round them to the nearest degree
degrees = mod(round(rad2deg(theta)), 360);
degreeRange = 0:359;
% Band pass to ignore high and low frequency components;
lowfreq = 5;
highfreq = size(FS,1)/2;
% Now average everything with the same degrees (sum over PS and average by the number of pixels)
for k = degreeRange
ps_integral(k+1) = mean(PS(degrees == k & R > lowfreq & R < highfreq));
fs_integral(k+1) = mean(FS(degrees == k & R > lowfreq & R < highfreq));
end
thetas = deg2rad(degreeRange);
Cxy = [ps_integral.*cos(thetas);
ps_integral.*sin(thetas)]';
C = Cxy' * Cxy;
[V,D] = eig(C);
lambda2 = D(1,1);
lambda1 = D(2,2);
alpha = 1 - lambda2/lambda1;
I use both Matlab and OpenCV to produce Grayscale histogram, divided into 10 bins.
In OpenCV, each bin has equal range (i.e. [0,25], [26,51], [52,77], ...).
However, in Matlab, the bin sizes are not equal (I guess it's related to some theory about different sensitivity to intensity changes between lower and higher values).
These different results make big trouble for me.
Is there an option to use calcHist with equal bin sizes? (Of course except for the option of implementing it myself...)
Answering my own question with a self-implemented function:
function h = fixedSizeBinnedHist(grayImg, numBins)
binSize = 256 / numBins;
binnedImg = floor(double(grayImg) / binSize);
maxVal = max(binnedImg(:));
numLeadingZeros = min(binnedImg(:));
numTrailingZeros = numBins - maxVal - 1;
% First, computing histogram for the existing range
h = hist(double(binnedImg(:)), maxVal - numLeadingZeros + 1);
leading = zeros(1, numLeadingZeros);
trailing = zeros(1, numTrailingZeros);
% Finally attaching needed zeros in both sides, so the histogram is in the requested size
h = [leading h trailing];
end
I am trying to compare the FFT of exp(-t^2) to the function's analytical fourier transform, exp(-(w^2)/4)/sqrt(2), over the frequency range -3 to 3.
I have written the following matlab code and have iterated on it MANY times now with no success.
fs = 100; %sampling frequency
dt = 1/fs;
t = 0:dt:10-dt; %time vector
L = length(t); %number of sample points
%N = 2^nextpow2(L); %necessary?
y = exp(-(t.^2));
Y=dt*ifftshift(abs(fft(y)));
freq = (-L/2:L/2-1)*fs/L; %freq vector
F = (exp(-(freq.^2)/4))/sqrt(2); %analytical solution
%Y_valid_pts = Y(W>=-3 & W<=3); %compare for freq = -3 to 3
%npts = length(Y_valid_pts);
% w = linspace(-3,3,npts);
% Fe = (exp(-(w.^2)/4))/sqrt(2);
error = norm(Y - F) %L2 Norm for error
hold on;
plot(freq,Y,'r');
plot(freq,F,'b');
xlabel('Frequency, w');
legend('numerical','analytic');
hold off;
You can see that right now, I am simply trying to get the two plots to look similar. Eventually, I would like to find a way to do two things:
1) find the minimum sampling rate,
2) find the minimum number of samples,
to reach an error (defined as the L2 norm of the difference between the two solutions) of 10^-4.
I feel that this is pretty simple, but I can't seem to even get the two graphs visually agree.
If someone could let me know where I'm going wrong and how I can tackle the two points above (minimum sampling frequency and minimum number of samples) I would be very appreciative.
Thanks
A first thing to note is that the Fourier transform pair for the function exp(-t^2) over the +/- infinity range, as can be derived from tables of Fourier transforms is actually:
Finally, as you are generating the function exp(-t^2), you are limiting the range of t to positive values (instead of taking the whole +/- infinity range).
For the relationship to hold, you would thus have to generate exp(-t^2) with something such as:
t = 0:dt:10-dt; %time vector
t = t - 0.5*max(t); %center around t=0
y = exp(-(t.^2));
Then, the variable w represents angular frequency in radians which is related to the normalized frequency freq through:
w = 2*pi*freq;
Thus,
F = (exp(-((2*pi*freq).^2)/4))*sqrt(pi); %analytical solution
Hi i'm having a problem where I have a dataset which ranges between -10^3 to 10^3
I need to be able to plot this as with a log scale but semilogy cannot plot negative values
Say for example my data is:
x = [-3,-2,-1,0,1,2,3];
y = [-1000,-100,-10,1,10,100,1000];
(or in general y=sign(x).*10.^abs(x);)
How can I plot this in MATLAB with a log scale? If possible It would be great if the log scale ticks could be on the Y-axis too
Use your actual data as labels, but scale the plotted data with log10.
% data
x = -3:0.1:3;
y = sign(x).*10.^abs(x);
% scaling function
scale = #(x) sign(x).*log10(abs(x));
N = 7; % number of ticks desired
% picking of adequate values for the labels
TickMask = linspace(1,numel(y),N);
YTickLabels = y(TickMask);
% scale labels and plotdata, remove NaN ->inconsistency, do you really want that?
YTick = scale( YTickLabels );
Y = scale(y);
YTick(isnan(YTick)) = 0;
Y(isnan(Y)) = 0;
% plot
plot(x,Y)
set(gca,'YTick',YTick,'YTickLabels',YTickLabels)
grid on
For N = 7:
For N = 11
How to find a valid value for N?
The following function (thanks to gnovice) will return all possible values you could choose for N:
n = numel(x);
N = find(rem(n./(1:n), 1) == 0) + 1;
about the semilogy-style labels: by adding the following line before the plot:
YTickLabels = cellfun(#(x) ['10^' num2str(x)], num2cell(YTick),'UniformOutput',false)
you could at least achieve something like this:
not beautiful and not generic, but a good point to start for you.
The reason you can't make a logarithmic axis that crosses zero, is that it doesn't make sense!
Since a logarithmic scale is generally displayed as eg. 100 - 10 - 1 - 1/10 - 1/100 - ..., you would need an infinite amount of space to make the axis cross zero.
How about this:
x=logspace(-3,3);
y=sign(x).*10.^abs(x);
loglog(x,y)
#thewaywewalk has already given a beautiful solution to it. The one I'm suggesting is an epsilon improvement on it. If you make two changes
(a) Define a new MATLAB function signia that basically extracts the sign before a number.
function value = signia(x)
if(x>=0)
value = '';
else
value = '-';
end
and (b) make this little change that instead of
YTickLabels = cellfun(#(x) ['10^' num2str(x)], num2cell(YTick),'UniformOutput',false)
you use
YTickLabels = cellfun(#(x) [signia(x) '10^{' num2str(x) '}'], num2cell(YTick),'UniformOutput',false);
(notice the presence of curly braces), you'll get an improvement in the Y ticks display. I got the following.
enter image description here
I have a vector of data, which contains integers in the range -20 20.
Bellow is a plot with the values:
This is a sample of 96 elements from the vector data. The majority of the elements are situated in the interval -2, 2, as can be seen from the above plot.
I want to eliminate the noise from the data. I want to eliminate the low amplitude peaks, and keep the high amplitude peak, namely, peaks like the one at index 74.
Basically, I just want to increase the contrast between the high amplitude peaks and low amplitude peaks, and if it would be possible to eliminate the low amplitude peaks.
Could you please suggest me a way of doing this?
I have tried mapstd function, but the problem is that it also normalizes that high amplitude peak.
I was thinking at using the wavelet transform toolbox, but I don't know exact how to reconstruct the data from the wavelet decomposition coefficients.
Can you recommend me a way of doing this?
One approach to detect outliers is to use the three standard deviation rule. An example:
%# some random data resembling yours
x = randn(100,1);
x(75) = -14;
subplot(211), plot(x)
%# tone down the noisy points
mu = mean(x); sd = std(x); Z = 3;
idx = ( abs(x-mu) > Z*sd ); %# outliers
x(idx) = Z*sd .* sign(x(idx)); %# cap values at 3*STD(X)
subplot(212), plot(x)
EDIT:
It seems I misunderstood the goal here. If you want to do the opposite, maybe something like this instead:
%# some random data resembling yours
x = randn(100,1);
x(75) = -14; x(25) = 20;
subplot(211), plot(x)
%# zero out everything but the high peaks
mu = mean(x); sd = std(x); Z = 3;
x( abs(x-mu) < Z*sd ) = 0;
subplot(212), plot(x)
If it's for demonstrative purposes only, and you're not actually going to be using these scaled values for anything, I sometimes like to increase contrast in the following way:
% your data is in variable 'a'
plot(a.*abs(a)/max(abs(a)))
edit: since we're posting images, here's mine (before/after):
You might try a split window filter. If x is your current sample, the filter would look something like:
k = [L L L L L L 0 0 0 x 0 0 0 R R R R R R]
For each sample x, you average a band of surrounding samples on the left (L) and a band of surrounding samples on the right. If your samples are positive and negative (as yours are) you should take the abs. value first. You then divide the sample x by the average value of these surrounding samples.
y[n] = x[n] / mean(abs(x([L R])))
Each time you do this the peaks are accentuated and the noise is flattened. You can do more than one pass to increase the effect. It is somewhat sensitive to the selection of the widths of these bands, but can work. For example:
Two passes:
What you actually need is some kind of compression to scale your data, that is: values between -2 and 2 are scale by a certain factor and everything else is scaled by another factor. A crude way to accomplish such a thing, is by putting all small values to zero, i.e.
x = randn(1,100)/2; x(50) = 20; x(25) = -15; % just generating some data
threshold = 2;
smallValues = (abs(x) <= threshold);
y = x;
y(smallValues) = 0;
figure;
plot(x,'DisplayName','x'); hold on;
plot(y,'r','DisplayName','y');
legend show;
Please do not that this is a very nonlinear operation (e.g. when you have wanted peaks valued at 2.1 and 1.9, they will produce very different behavior: one will be removed, the other will be kept). So for displaying, this might be all you need, for further processing it might depend on what you are trying to do.
To eliminate the low amplitude peaks, you're going to equate all the low amplitude signal to noise and ignore.
If you have any apriori knowledge, just use it.
if your signal is a, then
a(abs(a)<X) = 0
where X is the max expected size of your noise.
If you want to get fancy, and find this "on the fly" then, use kmeans of 3. It's in the statistics toolbox, here:
http://www.mathworks.com/help/toolbox/stats/kmeans.html
Alternatively, you can use Otsu's method on the absolute values of the data, and use the sign back.
Note, these and every other technique I've seen on this thread is assuming you are doing post processing. If you are doing this processing in real time, things will have to change.