I have a problem, which is how it is applied Fourier Transform (ftt) on the image (or how enhance image used Fourier Transform)
when i run my program
the input was fingerprint image
the output was white image
the problem is the output should be fingerprint image after enhance used Fourier Transform not white image
F=fft2( I );
factor=abs(F).^F;
block =ifft2(factor);
R= fftshift(block);
I hope finding some help
The exponentiation of F.^F seems to be a big number, so it is above the upper value and matlab slice it to be the upper value.
% Calculating fft2
fft2im = fft2(double(im));
% Taking the spectrum with log scaling
fft2im = log(1+(abs(fft2im)));
% Putting DC in the middle:
spectrum = fftshift(fft2im);
% finding maximum in spectrum:
maximum = max(max(spectrum));
% scaling maximum to 255 and minimum to 0:
spectrum = 255*spectrum/maximum;
% Casting to uint8 to be able to display:
spectrum = uint8(spectrum);
imshow(spectrum);
Related
I want to generate white noise in matlab.
clc;
clear;
mu=0;
sigma=1;
noise= sigma *randn(1,10)+mu
I can generate with this code but I guess I am not using noise power value. Can someone help me?
There are dedicated functions from MATLAB to add white Gaussian noise: wgn() and awgn().
sig = ones(100,1)
% add white Gaussian noise
snr = 50; % signal-to-noise ratio
sig_wgn = awgn(sig,snr,'measured')
if you want to do it yourself, note that rand() only returns numbers in [0,1], so you will need to scale + shift it to 2*(rand(,10)-0.5) to get a symmetric output that you than can scale to your range
I'm working on a filtered back projection algorithm using the central slice theorem for a homework assignment and while I understand the theory on paper, I've run into an issue implementing it in Matlab. I was provided with a skeleton to follow to do it but there is a step that I think I'm maybe misunderstanding. Here is what I have:
function img = sampleFBP(sino,angs)
% This step is necessary so that frequency information is preserved: we pad
% the sinogram with zeros so that this is ensured.
sino = padarray(sino, floor(size(sino,1)/2), 'both');
% diagDim should be the length of an individual row of the sinogram - don't
% hardcode this!
diagDim = size(sino, 2);
% The 2DFT (2D Fourier transform) of our image will start as a matrix of
% all zeros.
fimg = zeros(diagDim);
% Design your 1-d ramp filter.
rampFilter_1d = abs(linspace(-1, 1, diagDim))';
rowIndex = 1;
for nn = angs
% Each contribution to the image's 2DFT will also begin as all zero.
imContrib = zeros(diagDim);
% Get the current row of the sinogram - use rowIndex.
curRow = sino(rowIndex,:);
% Take the 1D Fourier transform the current row - be careful, as it's
% necessary to perform ifftshift and fftshift as Matlab tends to
% place zero-frequency components of a spectrum at the edges.
fourierCurRow = fftshift(fft(ifftshift(curRow)));
% Place the Fourier-transformed sino row and place it at the center of
% the next image contribution. Add the ramp filter in Fourier domain.
imContrib(floor(diagDim/2), :) = fourierCurRow;
imContrib = imContrib * fft(rampFilter_1d);
% Rotate the current image contribution to be at the correct angle on
% the 2D Fourier-space image.
imContrib = imrotate(imContrib, nn, 'crop');
% Add the current image contribution to the running representation of
% the image in Fourier space!
fimg = fimg + imContrib;
rowIndex = rowIndex + 1;
end
% Finally, just take the inverse 2D Fourier transform of the image! Don't
% forget - you may need an fftshift or ifftshift here.
rcon = fftshift(ifft2(ifftshift(fimg)));
The sinogram I'm inputting is just the output of the radon function on a Shepp-Logan phantom from 0 to 179 degrees. Running the code as it is now gives me a black image. I think I'm missing something in the loop where I add the FTs of rows to the image. From my understanding of the central slice theorem, what I think should be happening is this:
Initialize an array the same size as the what the 2DFT will be (i.e., diagDim x diagDim). This is the Fourier space.
Take a row of the sinogram which corresponds to the line integral information from a single angle and apply a 1D FT to it
According to the Central Slice Theorem, the FT of this line integral is a line through the Fourier domain that passes through the origin at an angle that corresponds to the angle at which the projection was taken. So to emulate that, I take the FT of that line integral and place it in the center row of the diagDim x diagDim matrix I created
Next I take the FT of the 1D ramp filter I created and multiply it with the FT of the line integral. Multiplication in the Fourier domain is equivalent to a convolution in the spatial domain so this convolves the line integral with the filter.
Now I rotate the entire matrix by the angle the projection was taken at. This should give me a diagDim x diagDim matrix with a single line of information passing through the center at an angle. Matlab increases the size of the matrix when it is rotated but since the sinogram was padded at the beginning, no information is lost and the matrices can still be added
If all of these empty matrices with a single line through the center are added up together, it should give me the complete 2D FT of the image. All that needs to be done is take the inverse 2D FT and the original image should be the result.
If the problem I'm running into is something conceptual, I'd be grateful if someone could point out where I messed up. If instead this is a Matlab thing (I'm still kind of new to Matlab), I'd appreciate learning what it is I missed.
The code that you have posted is a pretty good example of filtered backprojection (FBP) and I believe could be useful to people who wanted to learn the basis of FBP. One can use the function iradon(...) in MATLAB (see here) to perform FBP using a variety of filters. In your case of course, the point is to learn the basis of the central slice theorem and so finding a short cut is not the point. I have also learned a lot and refreshed my knowledge through answering to your question!
Now your code has been perfectly commented and describes the steps that need to be taken. There are a couple of subtle [programming] issues that need to be fixed so that the code works just fine.
First, your image representation in Fourier domain may end up having a missing array due to floor(diagDim/2) depending on the size of the sinogram. I would change this to round(diagDim/2) to have complete dataset in fimg. Be aware that this may lead to an error for certain sinogram sizes if not handled correctly. I would encourage you to visualize fimg to understand what that missing array is and why it matters.
Second issue is that your sinogram needs to be transposed to be consistent with your algorithm. Hence an addition of sino = sino'. Again, I do encourage you to try the code without this to see what happens! Note that zero padding must be happened along the views to avoid artifacts due to aliasing. I will demonstrate an example for this in this answer.
Third and most importantly, imContrib is a temporary holder for an array along fimg. Therefore, it must maintain the same size as fimg, so
imContrib = imContrib * fft(rampFilter_1d);
should be replaced with
imContrib(floor(diagDim/2), :) = imContrib(floor(diagDim/2), :)' .* rampFilter_1d;
Note that the Ramp filter is linear in frequency domain (thanks to #Cris Luengo for correcting this error). Therefore, you should drop the fft in fft(rampFilter_1d) as this filter is applied in the frequency domain (remember fft(x) decomposes the domain of x, such as time, space, etc to its frequency content).
Now a complete example to show how it works using the modified Shepp-Logan phantom:
angs = 0:359; % angles of rotation 0, 1, 2... 359
init_img = phantom('Modified Shepp-Logan', 100); % Initial image 2D [100 x 100]
sino = radon(init_img, angs); % Create a sinogram using radon transform
% Here is your function ....
% This step is necessary so that frequency information is preserved: we pad
% the sinogram with zeros so that this is ensured.
sino = padarray(sino, floor(size(sino,1)/2), 'both');
% Rotate the sinogram 90-degree to be compatible with your codes definition of view and radial positions
% dim 1 -> view
% dim 2 -> Radial position
sino = sino';
% diagDim should be the length of an individual row of the sinogram - don't
% hardcode this!
diagDim = size(sino, 2);
% The 2DFT (2D Fourier transform) of our image will start as a matrix of
% all zeros.
fimg = zeros(diagDim);
% Design your 1-d ramp filter.
rampFilter_1d = abs(linspace(-1, 1, diagDim))';
rowIndex = 1;
for nn = angs
% fprintf('rowIndex = %g => nn = %g\n', rowIndex, nn);
% Each contribution to the image's 2DFT will also begin as all zero.
imContrib = zeros(diagDim);
% Get the current row of the sinogram - use rowIndex.
curRow = sino(rowIndex,:);
% Take the 1D Fourier transform the current row - be careful, as it's
% necessary to perform ifftshift and fftshift as Matlab tends to
% place zero-frequency components of a spectrum at the edges.
fourierCurRow = fftshift(fft(ifftshift(curRow)));
% Place the Fourier-transformed sino row and place it at the center of
% the next image contribution. Add the ramp filter in Fourier domain.
imContrib(round(diagDim/2), :) = fourierCurRow;
imContrib(round(diagDim/2), :) = imContrib(round(diagDim/2), :)' .* rampFilter_1d; % <-- NOT fft(rampFilter_1d)
% Rotate the current image contribution to be at the correct angle on
% the 2D Fourier-space image.
imContrib = imrotate(imContrib, nn, 'crop');
% Add the current image contribution to the running representation of
% the image in Fourier space!
fimg = fimg + imContrib;
rowIndex = rowIndex + 1;
end
% Finally, just take the inverse 2D Fourier transform of the image! Don't
% forget - you may need an fftshift or ifftshift here.
rcon = fftshift(ifft2(ifftshift(fimg)));
Note that your image has complex value. So, I use imshow(abs(rcon),[]) to show the image. A couple of helpful images (food for thought) with the final reconstructed image rcon:
And here is the same image if you comment out the zero padding step (i.e. comment out sino = padarray(sino, floor(size(sino,1)/2), 'both');):
Note the different object size in the reconstructed images with and without zero padding. The object shrinks when the sinogram is zero padded since the radial contents are compressed.
I would like to create bitmap for any sine grating. How could I accomplish this.
I am doing that:
n = 64;
[X,Y] = meshgrid(linspace(-2*pi,2*pi,n));
sinewave2D=sin(5*X);
plot(sinewave2D(1,:))
imagesc(sinewave2D)
imwrite(sinewave2D,'sine.bmp')
imshow('sine.bmp')
But "sinewave2D" has negative values and bmp accepts values between 0-255 so how can I accomplish that and have my gratings that I need.
It sounds as though the problem is how to deal with negative numbers.
As the programmer, this is your choice!
One method would be to normalize your sine wave to fit in the output range, as follows:
% shift digital counts so the minimum is at zero
sinewave2D = sinewave2D - min(sinewave2D(:);
% scale digital counts so the maximum is at 255
sinewave2D = sinewave2D / max(sinewave2D(:) * 255;
% then, write the image
imwrite(uint8(sindwave2D), 'sine.bmp');
The problem with this method, of course, is if the dynamic range of the sine wave changes, these changes will not be reflected in the output image! It also centers zero at a nonzero digital count, so zero will appear gray in the bitmap.
I am using fft2 to compute the Fourier Transform of a grayscale image in MATLAB.
What is the common way to plot the magnitude of the result?
Assuming that I is your input image and F is its Fourier Transform (i.e. F = fft2(I))
You can use this code:
F = fftshift(F); % Center FFT
F = abs(F); % Get the magnitude
F = log(F+1); % Use log, for perceptual scaling, and +1 since log(0) is undefined
F = mat2gray(F); % Use mat2gray to scale the image between 0 and 1
imshow(F,[]); % Display the result
Here is an example from my HOW TO Matlab page:
close all; clear all;
img = imread('lena.tif','tif');
imagesc(img)
img = fftshift(img(:,:,2));
F = fft2(img);
figure;
imagesc(100*log(1+abs(fftshift(F)))); colormap(gray);
title('magnitude spectrum');
figure;
imagesc(angle(F)); colormap(gray);
title('phase spectrum');
This gives the magnitude spectrum and phase spectrum of the image. I used a color image, but you can easily adjust it to use gray image as well.
ps. I just noticed that on Matlab 2012a the above image is no longer included. So, just replace the first line above with say
img = imread('ngc6543a.jpg');
and it will work. I used an older version of Matlab to make the above example and just copied it here.
On the scaling factor
When we plot the 2D Fourier transform magnitude, we need to scale the pixel values using log transform to expand the range of the dark pixels into the bright region so we can better see the transform. We use a c value in the equation
s = c log(1+r)
There is no known way to pre detrmine this scale that I know. Just need to
try different values to get on you like. I used 100 in the above example.
So I have this image 'I'. I take F = fft2(I) to get the 2D fourier transform. To reconstruct it, I could go ifft2(F).
The problem is, I need to reconstruct this image from only the a) magnitude, and b) phase components of F. How can I separate these two components of the fourier transform, and then reconstruct the image from each?
I tried the abs() and angle() functions to get magnitude and phase, but the phase one won't reconstruct properly.
Help?
You need one matrix with the same magnitude as F and 0 phase, and another with the same phase as F and uniform magnitude. As you noted abs gives you the magnitude. To get the uniform magnitude same phase matrix, you need to use angle to get the phase, and then separate the phase back into real and imaginary parts.
> F_Mag = abs(F); %# has same magnitude as F, 0 phase
> F_Phase = cos(angle(F)) + j*(sin(angle(F)); %# has magnitude 1, same phase as F
> I_Mag = ifft2(F_Mag);
> I_Phase = ifft2(F_Phase);
it's too late to put another answer to this post, but...anyway
# zhilevan, you can use the codes I have written using mtrw's answer:
image = rgb2gray(imread('pillsetc.png'));
subplot(131),imshow(image),title('original image');
set(gcf, 'Position', get(0, 'ScreenSize')); % maximize the figure window
%:::::::::::::::::::::
F = fft2(double(image));
F_Mag = abs(F); % has the same magnitude as image, 0 phase
F_Phase = exp(1i*angle(F)); % has magnitude 1, same phase as image
% OR: F_Phase = cos(angle(F)) + 1i*(sin(angle(F)));
%:::::::::::::::::::::
% reconstruction
I_Mag = log(abs(ifft2(F_Mag*exp(i*0)))+1);
I_Phase = ifft2(F_Phase);
%:::::::::::::::::::::
% Calculate limits for plotting
% To display the images properly using imshow, the color range
% of the plot must the minimum and maximum values in the data.
I_Mag_min = min(min(abs(I_Mag)));
I_Mag_max = max(max(abs(I_Mag)));
I_Phase_min = min(min(abs(I_Phase)));
I_Phase_max = max(max(abs(I_Phase)));
%:::::::::::::::::::::
% Display reconstructed images
% because the magnitude and phase were switched, the image will be complex.
% This means that the magnitude of the image must be taken in order to
% produce a viewable 2-D image.
subplot(132),imshow(abs(I_Mag),[I_Mag_min I_Mag_max]), colormap gray
title('reconstructed image only by Magnitude');
subplot(133),imshow(abs(I_Phase),[I_Phase_min I_Phase_max]), colormap gray
title('reconstructed image only by Phase');