The Discrete Fourier transform Pair

• Change from $H$ to $G$ to denote our switch to a discrete Fourier transform and assume a dimension $L$ for the problem
  

  $$G\left(\frac{p}{L}\right) \equiv {\cal D}(g) = \sum_{q=-n}^{n-1} g\left(\frac{qL}{2n}\right) e^{-i2\pi pq/2n}$$

  $$<tex2html_comment_mark>383 p = -n, -n+1, ..., 0, ..., n-2,n-1$$ (10.12)

  
  
  where can assume the spatial inverse relationship that we are interested in where we sample points in k-space $\Delta k = 1/L$ and by assuming $L= 2n \Delta x$ get the previous, because $\Delta k \cdot \Delta x = 1/2n$ or write in a form that looks like finite sum relating to the continuous integral equations
  

  $$G(p\Delta k) = \sum_{q=-n}^{n-1} g(q \Delta x) e^{-i2\pi p q \Delta x \Delta k}$$ (10.13)

  
  

• Can also do an inverse Discrete Fourier transform
  

  $$g\left(\frac{qL}{2n}\right) \equiv {\cal D}^{-1}(G) = \frac{1}{2n} \sum_{p=-n}^{n-1}G\left(\frac{p}{L}\right) e^{i2\pi pq/2n}$$

  $$q = -n, -n+1, ..., 0, ..., n-2,n-1$$ (10.14)

  
  

• Do have to show that all of this makes sense in HW.

• Discrete Fourier transform pair are an exact pair, so if your image wasn't mucked with, without your knowledge, and you have a real image, can Fourier transform back to re-obtain the k-space data that made the image.