title: CUDA cufftPlan2d plan size question
body: <p>I'm studying the code behind the convolutionFFT2D example of the Nvidia CUDA sdk, but I don't get the point of this line:</p>
  
  <pre><code>cufftPlan2d(&amp;fftPlan, fftH, fftW/2, CUFFT_C2C);
  </code></pre>
  
  <p>Apparently this initializes a complex plane for the FFT to be running in, but I don't see the point of dividing the plan width by 2.</p>
  
  <p>Just to be precise: the fftH and fftW are rounded values for imageX+kernelX+1 and imageY+kernelY+1 dimensions (just for speed reasons). I know that in the frequency domain you usually have a positive component and a symmetric negative component of the same frequency.. but this sounds like cutting half of my image data away.. </p>
  
  <p>Can someone explain this to me a little better? I've never used a FFT (I just know the theory behind a fourier transformation)</p>