Resolution

• Still the question remains. Know that a point object gets spread. Because a sinc is involved, actually gets spread a long way.
• What is the resolution that I can expect of the MRI experiment?
• going to have to pick a measure for resolution, and stick with it. In fact, resolution isn't as important in medicine as detectability. It is often more important that something can be detected than that it can be resolved. We will talk about visibility in chapter with contrast.
• The effective measure of resolution that used in the book is the point spread function evaluated at zero divided by the area under the point spread function. Think of a boxcar function, divide the area by the height and you have the width.
  
  
  
  
  
  
  
• This definition pre-supposes that all filters are peaked at the origin, and fall off away from $x=0$
• Mathematically, and using another Fourier transform property we have the definition that
  

  $$\mbox{spatial resolution of filter} \equiv \Delta x_{filter} \equiv \mbox{\lq blur' of filter} = \frac{1}{h_{filter}(0)}\int_{-\infty}^\infty dx\; h_{filter}(x)$$

  $$= \frac{H_{filter}(0)}{h_{filter}(0)}$$ (13.7)

  
  

• This implies that in the best case in MRI we need to understand what the resolution associated with $h_{w,s}(x)$ is where we are only going to do the evaluation over one period, and not over all space.
  

  $$\Delta x_{MRI} \equiv \mbox{window and sampling resolution with no additional filtering}$$

  $$= \frac{\displaystyle\sum_{p=-n}^{n-1} \int_{-L/2}^{L/2}dx e^{i2\pi p\Delta kx}} {\displaystyle\sum_{p'=-n}^{n-1}1}$$

  $$= \frac{1}{2n}\sum_{p=-n}^{n-1}L\delta_{p0}$$

  $$= \frac{1}{2n\Delta k} = \frac{1}{W} = \Delta x = \mbox{the Fourier pixel size}$$ (13.8)