Non-Uniform Sampling

• Lots of reasons for non-uniform sampling
  • sample during gradient ramps
  • use non-trapezoidal gradient waveforms
  • any benefits to sampling different parts of $k$-space differently
• Lots of tricky bits to watch out for when doing non-uniform sampling
  • What if gradient isn't acting exactly as you think it should be?
  • Longer gradient waveform plays on, the larger that errors may become
  • other system error may be introduced
• Consider a case where every other line of data is sampled at different $k$-space distance, i.e., $... \Delta k_{1} - \Delta k_{2} - \Delta k_{1} ...$, or has some other error that effects every other line.
• Know that the Fourier (discrete or continuous) transform is linear
  

  $${\cal D}(g_{1}(q) + g_{2}(q)) = {\cal D}(g_{1}(q)) + {\cal D}(g_{2}(q))$$ (12.2)

  
  

• Therefore, can treat this case by considering two functions sampled with sampling interval $2\Delta k$
  

  $$u_e(k) = \Delta k \sum_{m=-\infty}^{\infty}\delta (k-2m\Delta k)$$ (12.3)

  $$u_{o,i}(k) = \Delta k\sum_{m=-\infty}^{\infty}\delta (k-(2m+1-\alpha ) \Delta k)$$ (12.4)

  
  

• Assume again that we are imaging a delta function, only need to consider the Fourier transform of these two functions to figure out what image will look like
• Know the Fourier transform of sampling function, and can find Fourier transform of any shifted sampling function using the Fourier transform shift theorem
  

  $$U_e(x) = {\cal F}^{-1}\{u_e(k)\} = \Delta k \sum_{q=-\infty}^{\infty}\delta\left(2\Delta k\,x-q\right)$$

  $$= \frac{1}{2}\sum_{q=-\infty}^{\infty}\delta\left(x-q\frac{L}{2}\right)$$ (12.5)

  
  
  apply Fourier transform shift to figure out shifted result
  

  $$U_{o,i}(x) = \frac{1}{2}\sum_{q=-\infty}^{\infty} \delta \left(x - q \frac{L}{2} \right)e^{i \pi q (1-\alpha)}$$ (12.6)

  
  

• If $\alpha$ is zero, then when you sum these functions, all of the odd q terms will cancel, and you get the sampling function in $x$-space that you expect which only repeats with interval L
• However, if $\alpha$ isn't zero then you will get things repeating at $L/2$ and not $L$ which is what you expect for a function sampled at $2\Delta k$.
• putting in $\rho(x)$ instead of just looking at the sampling gives
  

  $$\hat{\rho}_{o,i}(x) = \frac{1}{2}\left[\rho\left(x+\frac{L}{2}\right) e^{-i\pi(1-\alpha)}+ \rho (x)+\rho\left(x-\frac{L}{2}\right) e^{i\pi(1-\alpha)}\right]$$

  $$= \frac{1}{2}\rho(x) + \frac{1}{2}\left[\rho_{r}\left(x+\frac{L}{2}... ...{-i\pi(1-\alpha)}+ \rho_{l}\left(x-\frac{L}{2}\right) e^{i\pi(1-\alpha)}\right]$$

  $$\hspace{1in} -L/2<x<L/2$$ (12.7)

  
  

• If you see things folded into the image, could be mistake in FOV choice, or could be a (N/2) sampling error. Need not be a simple mistake in FOV choice.