Point Spread Function

• Consider a filter to be anything that multiplies the MRI signal.
• Point spread function is the Fourier transform of a filter. Why: Assume $\rho(x) = \delta(0)$. Using the convolution theorem in this case implies that $\hat{\rho}(x)$ will be the Fourier transform of the filter.
• Point spread function answers the question, how much blurring would occur if you are trying to image a point.
  
  
  
  
  
  
  
• Data always contains the effect of truncation and sampling. What does this imply? Windowing and sampling function can be described as
  

  $$H_{ws}(k) \equiv \Delta k\,{\rm rect}\left(\frac{k+\frac{1}... ...Delta k) = \Delta k \sum_{p=-n}^{n-1} \delta (k-p\Delta k)$$ (13.1)

  
  

• Remember multiplication in one domain is convolution in the other,
  

  $$\hat{\rho}(x) = \rho (x) * h_{ws}(x)$$ (13.2)

  
  

• so to find effect in $x$-space on a point function convolve this function with $\delta$ function, which just gives back Fourier transform of $H_{ws}(k)$
  

  $$h_{ws}(x) = \Delta k \sum_{p=-n}^{n-1}e^{i2\pi p\Delta kx}$$ (13.3)

  
  

• Doing this sum out gives
  

  $$h_{ws}(x) ={W} \frac{{{\rm sinc}}(\pi Wx)}{{{\rm sinc}}(\pi \Delta kx)}e^{- i\pi \Delta k \, x }$$ (13.4)

  
  

• This is a sinc like function, as we might expect, knowing that the Fourier transform of a rect is a sinc.
• Also note that the wider the sampling window $W$ becomes, the narrower the sinc becomes, positive thing that as we decrease $\Delta x$ the point-spread gets narrower.