Timeline for Gaussian Function Fourier Transform (Matlab)
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 23, 2016 at 11:11 | vote | accept | Merin | ||
| Apr 22, 2016 at 21:45 | answer | added | xoron | timeline score: 1 | |
| Apr 22, 2016 at 13:45 | comment | added | Chu | With k=0, x=1, and with k=1, x=0.007; so the x values are very small wrt the one at k=0. Hence the transformed spectrum looks like that of an impulse ie flat. | |
| Apr 22, 2016 at 13:17 | comment | added | Neil_UK | My MATLAB's a bit rusty. I'm happy that as k is vector, k.^2 produces a k squared vector, but I'm not sure about the a., as a is a scalar. One way or another however, you've got to get the simple x=a_guassian right before you do the difficult bit of interpreting its FFT. <quick edit>Ah! how about dividing k by n, so that the argument is sigma=5 at the end of the vector, rather than at the first bin away from the middle? <\quick edit> | |
| Apr 22, 2016 at 13:09 | comment | added | Merin |
@VicenteCunha But how could that be? I explicitly defined a Gaussian function x=exp(-a.*k.^2);. I did try lower \$ a\$ values but I am still getting the same features on the graph.
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| Apr 22, 2016 at 12:14 | comment | added | Vicente Cunha | Didn't inspect the code, but from your plot you are effectively sampling an impulse function (with a sine FT pair, hence the imaginary oscillations) instead of a gaussian. You should start plots with a lower a coefficient or a higher time resolution ("sampling rate"). | |
| Apr 22, 2016 at 12:06 | history | asked | Merin | CC BY-SA 3.0 |