Recent posts
https://software.intel.com/en-us/recent/799702
enFFT with modified fourier coeffitients
https://software.intel.com/en-us/forums/topic/367666
<p>Dear MKL forum,<br />I solve such a problem. Can you help me please?<br />Lets have a function Y=∑ <sub>k=−∞</sub><sup>∞</sup> iY<sub>n</sub>e<sup>ikπy</sup> and then I have a function which is defined as X=∑<sub>k=−∞</sub><sup>∞</sup> ik<sup>2</sup>Y<sub>n</sub>e<sup>ikπy</sup>.</p>
<p>I know the <em>Y</em>. The <em>i</em> is imaginary unit.</p>
<p>How can I compute the <em>X</em>? I think I do the FFT on <em>Y</em> and obtain thus the Y<sub>n</sub>, right? And then I think I will do the backward FFT of function defined as f=ik<sup>2</sup>Y<sub>n</sub>. But what have I do with the summation index <em>k</em> here in the function <em>f</em>?</p>
<p>It is right that FFT(ik<sup>2</sup>Y<sub>n</sub>)=X?</p>
<p>I'm not sure absolutely what to do with <em>k</em> when the FFT sum is summated per <em>k</em>. Or can I change something in MKL FFT directly?</p>
Mon, 11 Feb 13 02:07:10 -0800mu k.367666FFT Trigonometric transform - zeroes f(0)=0 and f(n)=0
https://software.intel.com/en-us/forums/topic/367651
<p>Hi,<br /> I have a small question about FFT trigonometric in MKL. I have a array of data I wish to transform. But I can't use exponential FFT, I need to use the sine trigonometric transformation. The algorithm wants from the that the array of data begins with zero and ends with zero.<br /> f(0)=0<br /> f(n)=0</p>
<p> But my natural data are not of such kind. Can I do this? : My array A length is m. I will create an array F which has length n=m+2 and I will copy the array A into F. I will define F(0)=0, F(1:m)=A and F(n)=0<br /> Will it work? Will be the transformation right? <br /> And then after the transformation, the transformed F, what should I do? Now I would like to have the values in A but the lenght of A is smaller than F. Can I copy A=F(1:m) and the points F(0) and F(n) not use? Will the array A than the sine FFT of A at the beginning?<br /> I hope so but I'm not sure if I'm not mistaken.<br /> Many thanks</p>
Mon, 11 Feb 13 01:00:16 -0800mu k.367651