Resolution and the Discrete Fourier Transform

• What about $\Delta x$?

• $x$-space image is also going to be sampled and truncated. For periodicity reasons, we are going to truncate at $L$, as assume that we collect $2n'$ points, such that
  

  $$L = 2n' \Delta x$$ (11.14)

  
  

• If we are going to use a Discrete Fourier transform, then $n = n'$. This implies that the smallest discernible voxel that we will be able to resolve is given by
  

  $$\Delta x = L/2n$$ (11.15)

  
  

• If we took our definitions for $\Delta x$ and $\Delta k$ back into the equations from the last chapter, we can come up with the measured signal and measured spin density being a discrete Fourier transform pair
  

  $$s(p/L) = \Delta x \sum_{q=-n}^{n-1} \hat{\rho}(qL/2n) e^{-i \pi pq/n}$$

  $$\hat{\rho}(qL/2n) = \Delta k \sum_{p=-n}^{n-1} s(p/L) e^{i\pi pq/n}$$ (11.16)

  
  
  where can define
  

  $$\hat{\rho}_{MRI}(qL/2n) = \hat{\rho}(qL/2n)\Delta x$$ (11.17)

  
  
  because our spin density is just a proportionality to the real spin density, so what does one more multiplication matter, to get
  

  $$s(p/L) = \sum_{q=-n}^{n-1} \hat{\rho}_{MRI}(qL/2n) e^{-i \pi pq/n}$$

  $$\hat{\rho}_{MRI}(qL/2n) = \frac{1}{2n} \sum_{p=-n}^{n-1} s(p/L) e^{i\pi pq/n}$$ (11.18)

  
  

  (IN-CLASS QUESTION)