Using interp1 to resample an image, results in NaN values.
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Hello all,
I've run into an issue with a assignment I am doing. For this assignment it states: "Now, use an interpolation method (in Matlab®, ‘interp1’) to interpolate, i.e. resample, the spectrum from its original sampling to the linearized k-space sampling." During my tries I have come up with the following:
A = 807.6660554599425;
B = 0.06950484773532092;
C = -7.056743126149048e-6;
D = -3.62624279344136e-11;
N = linspace(1, 1024, 1024);
lambda = A*N.^3 + B*N.^2 + C*N + D;
k_n = (2*pi)./lambda;
max = max(k_n);
min = min(k_n);
maxmin = linspace(max,min,1024);
F_resampled = interp1(raw_plak, maxmin);
F_k = fft(F_resampled);
F4 = log(abs(F_k));
Where A through D are device specific settings and 'raw_plak' is the original image given by:
raw_plak = im2double(imread('plakband.tif'));
When using the interp1 fuction returns an array full of NaN values. What am I doing wrong?
Sorry if this has been answered already, I have looked but could not find an answer to my problem.
With regards.
2 件のコメント
Star Strider
2022 年 1 月 15 日
The assignment says to resample a spectrum. (To me, that implies a vector.)
Where does the image (a 2D or 3D array) enter into this?
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Image Analyst
2022 年 1 月 15 日
@Star Strider spectra can be 2-D. For example a diffraction pattern is a 2-D spectrum. You can get a spatial frequency spectrum from an image using fft2().
Because they say use interp1, that implies the input is 1-D. But since you say raw_plak is a 2-D image gotten from imread(), you'll need to use interp2() not interp1(). If that doesn't get rid of the nan's, then attach 'plakband.tif' in a .zip file along with your m-file with the paperclip icon.
Don't destroy the max and min values by overwriting the functions with values like you did here:
max = max(k_n);
min = min(k_n);
3 件のコメント
Star Strider
2022 年 1 月 15 日
@Image Analyst — I am well aware of that!
However the initial problem was to use interp1 (that interpolates vectors) on an image that was not provided.
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