Physics of Image Formation

Question 1

Why is quadrature detection needed in MRI?

Answer 1

A single measurement cannot distinguish between magnetisation precessing above or below the reference frequency. Cos(a)=Cos(-a).

Question 2

How is quadrature detection achieved?

Answer 2

Measure the signal relative to two axed at 90deg to each other. The real (x) channel shows the magnitude and the imaginary (y) channel shows the direction. The signal is no described in complex notation.

Question 3

What is the complex notation for an MR signal?

Answer 3

S(r)=M_0.cos{gamma.G(r).r.t}+i.M_0.sin{gamma.G(r).r.t}

Question 4

What is the exponential notation for an MR signal?

Answer 4

S(r)=M_0.exp(-i.{gamma.G(r).r.t})

Question 5

What assumption is made about the complex and exponential forms of the signal?

Answer 5

G(r) is time invariant.

Question 6

How is the exponential notation changed if G(r) is time dependent?

Answer 6

Signal phase, phi, is given by:
phi(t)=int[0-t].(gamma.G(r,t).r.dt)
So:
s(t)=int{M_0(r).exp(-i.gamma.int[0-t]G(r,t).r.dt)}.dr

Question 7

How does k-space simplify the time-dependent exponential notation?

Answer 7

k=int[0-t’].(gamma.G(r,t).r.dt)
So:
s(k)=int{M_0(r).exp(-ikr)}.dr

Question 8

How does the sampled signal relate to the proton density?

Answer 8

The sampled signal as a function of k is a direct Fourier Transform of the proton density.

Question 9

How does the theoretical signal derivation differ from real life signal?

Answer 9

The theoretical equation describes a continuous signal, in reality, we sample discretely and so use the discrete Fourier Transform.

Question 10

What is the discrete version of k in a constant gradient?

Answer 10

K_n=Sigma[j=0,j=n]{gamm.Gr.j.dt}

Question 11

Why is it necessary to have a negative gradietn before sampling in k-space?

Answer 11

k=0 is the origin on k-space. if we want to sample negative regions of k-space we need to have a negative k. As k depend on G and t, and t can’t be negative we need a negative gradient.

Question 12

What is the Nyquist criterion?

Answer 12

To fully realise the data, we must sample at least twice within one cycle of the data.

Question 13

How is the Field of view related to k-space separation?

Answer 13

Phase accumulation: phi(dt)=gamm.Gx.dt
Maximum phase allowed in one dwell time: phi_max=gam.Gx.dt=pi (Nyquist criterion)
Rearranging: x= pi/(gamm.Gdt)=pi/dk
FOV is2x: 
FOV = 2*pi/dk

Question 14

How is the resolution related to the width of k-space?

Answer 14

res = FOV/Np (Np= No. of data points)
res = 2*pi/dk.Np
Np.dk is the width of k-space

Question 15

What problems arise from the defined frequency range of MRI in the presence of noise and how is it removed?

Answer 15

Noise has a much wider frequency range and so will alias in the image and increase in amplitude.
By filtering any signal before it enters the signal processing to just the MRI frequency range the noise will not alias.

Question 16

What is the sampling bandwidth?

Answer 16

The frequency separation of magnetisation at the edge of the FOV.

Question 17

What is the equation for the sampling bandwidth or sweep width?

Answer 17

SW=gamma.G.FOV/2*pi=1/dt

Question 18

Why can we filter to avoid aliasing in the FE gradient but not the PE gradient?

Answer 18

Not sampling a continuous frequency in PE, we are detecting a pseudo-frequency and so filtering can’t work.

Question 19

What is the ideal RF pulse for slice selection and what is its associated Fourier Transform?

Answer 19

• Top hat function (square slice) fully excites within the frequency range and no excitation outside.
• Fourier transform is a sinc function (sin(x)/x)

Question 20

How is the slice selection function created? What two waves are convolved together and what do they each effect?

Answer 20

A carrier sine wave at the Larmor frequency corresponding to the centre of the required slice is amplitude modulated with a sine(x) function that adjusts the slice thickness.

Question 21

Why is the rewinding gradient needed after the slice selection gradient?

Answer 21

Precession in combined B1 and G field causes dephasing. Need to rewind the phase to compensate for this.

Question 22

What is the equation for slice thickness?

Answer 22

Ts=(2*pi.BW)/(gamma.G_ss)

Question 23

How does the RF pulse affect the slice thickness?

Answer 23

Long pulse - Thin slice

Short pulse - Thick slice

Question 24

Why is impossible to create a perfect top-hat RF pulse?

Answer 24

The sinc function that drives it is an infinite function and must be truncated in the real world.
Truncation alters the frequency spectrum and so the slice is no longer rectangular.

Question 25

What filter can be used to suppress the ringing from a truncated sinc function?

Answer 25

Hamming.

Question 26

How can you speed up the acquisition from a multi-slice image?

Answer 26

Slice interleaving. As magnetisation outside the selected slice remains unexcited, you can excite this while waiting for the first slice to fully relax.

Question 27

What problems arise if you try to interleave slices 1,2,3,4 and so on?

Answer 27

As Rf pulse is not fully square there is cross talk (excited protons in adjacent slices).
In practice, it is safer to interleave 1,3,5,7,2,4,6,8.

Question 28

What is the main limitation for full 3D imaging?

Answer 28

As we have two phase encode gradients our scan time is N_ssXN_encxTR (N_ss matrix in slice select direction, N_enc matrix in-plane). And so it is only feasible for short TR scans.

Question 29

What is the effect in k-space of the 180 refocusing echo in a spin echo sequence?

Answer 29

An 180 pulse translates the point through the k-space origin.

Question 30

Why would we want to deliberately destroy signal?

Answer 30

Poor 180 pulse profiles mean that not every proton will experience an 180 pulse across the slice. There the pulse angle is 90 this could be excited on further RF pulse and create new, erroneous signal.

Question 31

How do crusher gradients remove the signal?

Answer 31

The signal is dephased until the net signal is zero.

Question 32

Why are crusher gradient placed symmetrically at either side of an 180 refocusing pulse?

Answer 32

The required M_xy generated by the 90 deg pulse is dephased by the first pulse and then rephrased after the 180 flip. as it experiences both gradients.
Any unwanted M_xy generated by the 180 pulse will only be dephased as it only experiences the second gradient.