What Does Sampling Imply for the Reconstructed Image?

• If you are able to sample for ever, then the measured signal is the product of the ideal signal and sampling function.
  

  $$s_{\infty}(k) \equiv s(k) \cdot u(k)$$

  $$= \Delta k \sum_{p = -\infty}^{\infty} s(p \Delta k) \delta (k - p \Delta k)$$ (11.1)

  
  

• Therefore, the reconstructed image $\hat{\rho}_{\infty}(x)$ is going to be the convolution of this product
  

  $$\hat{\rho}_\infty(x) = \rho(x) * U(x)$$

  $$= \int_{-\infty}^{\infty} dk\, s_{\infty}(k) e^{i2\pi k x}$$

  $$= \int_{-\infty}^{\infty} dk \left[ \Delta k \sum_{p = -\infty}^{\infty} s(p\Delta k) \delta(k - p \Delta k) \right] e^{i2\pi k x}$$

  $$= \Delta k \sum_{p = -\infty}^{\infty} s(p\Delta k) e^{i2\pi p \Delta k\, x}$$ (11.2)

  
  

• So, now have Fourier series. $x \Delta k$ can have any value, which just means going around the same circle in a complex exponent, so this is going to be a periodic function.
  

  $$\hat{\rho}_\infty(x) = \hat{\rho}_\infty(x + 1/\Delta k)$$ (11.3)

  
  

• The periodicity of the reconstructed spin density is related to the $k$-space sampling interval by
  

  $$1/\Delta k \equiv L \equiv {\rm FOV}$$ (11.4)

  
  

• If I have an object with length $L_{object}$, such that
  

  $$L_{object} > 1/\Delta k \equiv L$$ (11.5)

  
  
  then the repetitions of the object will overlap, call this aliasing.