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1 EE225E/BIOE265 Spring 2013 Principles of MRI Miki Lustig Assignment 8 Solutions 1. Nishimura 7.1 P)jc%z 5 ivcr3. 1. I Due Wednesday April 10th, 2013 (a Scrhon5 R2iwd b 0 ZOms Xn,s r cx > qs 4-4 8ni6 4 1E IDrAS boms 1 3E ton5 C- () c 0. S 1

2 2. In a spin-echo pulse sequence the 180 pulse refocuses the magnetization produced by the 90 excitation. Since the 180 is never perfect, it will will also excite magnetization. This produces additional signals, as is shown on the left below: s(t) Parasitic FID Spin Echo?? yy Ideal Image These signals, called parasitic FID s, will produce image artifacts that depend on the acquisition method. In each of the examples below, draw (i) where the parasitic FID ends up in k-space, and (ii) the image artifact it produces. Assume that the ideal image is as shown above on the right, and that signal only comes from the excited slice. a) In this case the phase encode and the readout dephaser are before the 180. Gz Gy Gx A/D

3 Solution: The signal from the 90 is properly encoded by the phase encode and readout gradients, and produces an image as usual. The k-space data, and the reconstructed image look like: k y k x The parasitic signal from the 180 does not see the phase encode gradient, or the dephaser. It only sees the readout gradient, and this is the same for every acquisition. The k-space data is then just the positive k x part of the k y = 0 k-space line. The k y = 0 line is the Fourier transform of the projection of the object along the x axis. The fact that we only have the positive spatial frequencies means this projection will be blurred. The k-space data and the image of the parasitic will then look like k y Parasitic k x The total image is the sum of the ideal image and the image from the parasitic, Parasitic 3

4 b) The same pulse sequence as in (1), with the addition that the 180 is inverted every other phase encode step. Gz Gy Gx A/D 90 /-180 Solution: Again, the image of the signal from the 90 will be properly encoded, since a 180 or a -180 will both produce the same spin echo. The difference is in the effect of phase cycling the 180 (alternating the sign). The premise of the problem is that the 180 is not perfect. If the 180 is actually a 179, then alternating the sign of the 180 will alternate the sign of the parasitic signal. The parasitic still only sees the readout gradient, so it is still the Fourier transform of the projection of the object along the x-axis. However, the sign of the k-space data alternates every other line. This modulates the parasitic signal to the edge of the FOV when the image is reconstructed: k y k x Parasitic When combined with the ideal image from the 90 degree excitation, the result is Parasitic The signal for the parasitic can then be suppressed simply by zeroing out one line in the image. This is a very common acquisition scheme in practice. c) The same pulse sequence as in (2), with the phase encode and readout dephaser after the

5 90 /-180 Gz Gy Gx A/D Solution: Again, the image of the signal from the 90 will be properly encoded. In this case the parasitic sees the phase encode gradient, and the dephaser and readout. This means that the parasitic echo is fully encoded, and will produce an image when reconstructed. However, the phase cycling of the 180 will cause this image to be shifted by half of the FOV. The k-space data and the reconstructed image from the parasitic are: k y k x Parasitic When combined with the ideal image from the 90 degree excitation, the result is Parasitic Note that since the parasitic is encoded in y, it can overlap the ideal image. It is much harder to suppress than in (b) above. This is unfortunate, since in general you want to keep the imaging gradients as close together in time as possible, in order to minimize flow and motion artifacts. 5

6 3. MRI Artifacts and Debugging In this question you will be given several pairs of images. The images may exhibit artifacts which will be indicated by arrows/circles. Answer the questions as best as you can. First describe the artifact and then provide as much information as possible on the source/s of the artifact and why it appears this way. There might be several possible right answers. A final answer without explanation will not be credited. a) Example: What is the source of the artifact in the left image and what is the difference in the acquisition/processing of the two? Solution: The artifact in the image appears as the back of the brain aliasing over and wrapping to the front. This is typical when the prescribed FOV in the phase encode is smaller than the FOV of the object. In that case the sampling density does not meet the Nyquist rate. The phase encode is obviously Anterior-Posterior. The image on the right does not exhibit aliasing. One way to avoid aliasing is to increase the number of phase encodes while keeping the resolution the same. The other way is to swap the readout and phase-encode directions. It seems that the latter approach of swapping the readout direction was used since the back of the skull as well as the nose are cropped. Choosing a larger FOV in the phase encode would have shown the entire nose and skull. Source of Artifact: aliasing in the phase-encode direction (A/P) Difference: readout and phase encode directions swapped 6

7 b) What is the source of the artifact in the left image and what is the difference in the acquisition/processing of the two? Solutions: Something has gone wrong in the geometry of the left guy s head! It has been compressed in space with respect to the rest of the FOV. This is a sagital scan with a very large FOV. For such large of a FOV, the gradients non-linearity start to show up. The gradients will be much weaker at the top and bottom and will result in a geometric distortion. This is fixed in postprocessing by applying interpolation to correct for the distortion in the right image. Source of artifact: Gradient non-linearity Difference:The right image was grad-warped and interpolated to correct for the non-linearities. c) What is the source of the artifact in the right image and what is the difference in the acquisition/processing of the two? Source of artifact: The fat in the right image is shifted right with respect to water. This is a result of a low-bandwidth/pixel readout. The readout direction is RL Difference: The readout bandwidth in the left image is higher (shorter readout) and the fat does not have time to accrue much phase and appears to not shift by much. (There s still a slight shift though! d) What is the source of the artifact in the two images and what is the difference in the acquisition/processing of the two? 7

8 Solutions: There s appear to be ghosting artifacts that seem to be originating from the eyes. This is an artifact of eye motion during a scan. Motion during an acquisition will create inconsistent data in the slow-direction (the phase encode). Therefore the phase encode direction in the left image is A/P and R/L in the right image. Source of artifact: Eye motion during the scan Difference:Readout-phase encode directions are swapped. e) What is the source of the artifact in the left image and what is the difference in the acquisition/processing of the two? (Ignore the contrast difference between the two) Solutions: The right image exhibits loss of signal next to air-tissue interfaces near the sinuses and the ear canal. This is a result of T2 decay typical for a gradient echo sequence. The right image does not exhibits the loss of signal. This can be due to much shorter echo time, but there s significant T2 weighting in the image that makes this infeasible. The other explanation is a Spin-echo sequence! Source of artifact: Loss of signal due to T2 intravoxel dephasing in a gradient echo sequence Difference:The right image does not exhibits loss of signal and is most likely a spin-echo sequence 8

9 4. Contrast Preparation: Consider the following sequence of two θ degrees tip-angle pulses separated by T seconds. Θ T Θ a) Given that the equilibrium magnetization is M 0 ẑ, derive an expression for the M z component of the magnetization immediately following the second as a function of T 2, T, and θ. You can neglect T 1 recovery (since T 1 T ) and off-resonance. Solutions: For easy notation we will define E 2 = e T T 2, C θ = cos(θ) and S θ = sin(θ) [ Cθ S θ ] [ E2 0 ] [ Cθ S θ ] [ 0 ] M(T ) = S θ C θ [ Cθ S θ 0 0 ] [ E2 S θ M 0 S θ ] C θ M 0 = S θ C θ C θ M 0 So, the M z (T ) component is, M z (T ) = (C 2 θ E 2 S 2 θ )M 0 M z = ( ) cos 2 (θ) e T T 2 sin 2 (θ) M 0 b) Given T = T 0, find the flip angle θ for which the M z component is zero for spins with a desired T 2 transverse relaxation value. Solutions: 0 = C 2 θ E 2 S 2 θ Cθ 2 Sθ 2 = E 2 tan(θ) = E 1/2 2 θ = atan(e 1/2 2 ) ( ) θ = atan e T 2T 2 9

10 MR Angiography is an important tool in assessing vascular diseases in patients. Often, T 1 shortening Gadolinium contrast agents are used in combination with short TR sequences to increase the blood-muscle contrast. However, using Gadolinium based contrast agents can result in a life threatening syndrome, called NSF, in patients that have renal disease. Consider the contrast preparation imaging paradigm below. For each preparation sequence a single phase encode is collected. Also, assume T R T 1. Gz Gy Gx Prep Sequence Imaging Sequence 90 Prep Sequence 90 TR The T 1 /T 2 of blood are 1000/220ms and the T 1 /T 2 of muscle are 870/50ms. We would like to design a non-contrast enhanced preparation pulse that will ideally have good blood signal and no muscle signal at all. In addition, we would like the preparation pulse to not be much longer than 50ms. c) Based on your previous derivations, design a preparation sequence that nulls the muscle signal while producing signal from blood. Draw the sequence pointing out the relevant parameters. (Extra points will be given for those coming with solutions that are insensitive to off-resonance). What is the blood M z magnetization after the prep-pulse? Solutions: The previous questions have taught us that we can null the signal of a particular T 2 specie. In this part we need to design a preparation sequence that will null the muscle signal while still producing blood signal. Unfortunately, we can not use inversion recovery based on the T 1 since that would take much longer than than 50ms. Using an excitation recovery would probably not leave much blood signal. Therefore we are going to use the T 2 prep-pulse from before. For this preparation sequence we need to choose the flip angle θ and T such that the muscle signal is nulled and in addition the blood signal is as high as possible. T is a free parameter, but it can not be longer than 50ms. Calculating the θ for T = 10ms, 25ms, 35ms, 50ms we get, θ = 47.86, 52.09, 54.83, with corresponding M z = 0.075M 0, 0.18M 0, 0.24M 0, 0.31M 0. So, T = 50ms gives the highest blood signal with θ = Of course this pulse is VERY sensitive to B 0 inhomogeneity because spins would be precessing during the waiting time and the second pulse will have different effect on different resonances. This can be mitigated by applying a spin-echo 180 pulse such that all the spins will refocus exactly at the time the second pulse is applied. However, if we use a single refocusing, we need to change the sign of the second so it will have the same effect. Alternatively we can use two refocusing pulses. Each 180 refocusing pulses should have crushers gradients on both sides to prevent a parasitic FID. A very important part of the question is to realize that at the end of this preparation sequence there remains is a significant transverse magnetization. To remove it, we need to apply a large gradient crusher after the second. Here is an acceptable solution: 10

11 Θx Prep Sequence T Θx Imaging Sequence 90 Gz Gy Gx Here are two acceptable B 0 insensitive solutions: Gz Gy Gx Prep Sequence Θx 180y -Θx T/2 T/2 Imaging Sequence 90 Gz Gy Gx Prep Sequence 180y T/4 T/2 T/4 Θx 180y Θx Imaging Sequence 90 Blood M zprep = 0.31M 0 11

12 5. Consider the EPI and spiral trajectories in fig in Nishimura. a) Assume T ms readout gradients and that the gradients start 1ms from the peak of the excitation. What is the echo-time for EPI and what is the echo time for the spiral trajectory? Solution: The echo time is the time that the trajectory crosses DC, or the center of k-space. This is the time in which the majority of the contrast of the image is being determined as most of the energy of images lies in the origin of k-space. EPI is a symmetric trajectory and the crossing of the center of k-space is in the middle of the readout. Therefor the echo time for EPI is 1 T/2ms. On the other hand, the spiral trajectory always starts at the origin, therefore the echo time is 1ms. b) What is the k-space weighting due to T2 decay in EPI and spiral for a gradient echo sequence. (Assume decay happens only in the slow direction.) Solution: Both readouts start 1ms after the middle of the and are T ms long. Let t = 0 the start of the readout. The T2 weighting as a function of time is: W (t) = e 1t T 2 Assuming the decay happens only in the slow direction k y (ignoring readout k x ) then, k y (t) = k max t T/2 T/2 and So, t = k yt 2k max T/2. ( W (k y ) = exp 1 t ) T2 ( = exp 1 T2 ) exp ( T 2T 2 ( )) ky 1 k max The decay during the spiral sequence can be approximated as a radial decay with, k r = k max t/t and So, t = T k r k max. ( W (k r ) = exp 1 ) ( T2 exp T k r T 2 k max ). 12

13 The k-space weighting is shown below: c) How would your answer change if the sequence is a spin-echo. What s the problem for spiral spin-echo imaging? Solution: In a spin-echo, T2 is refocused. Therefore, we can define t = 0 as the spin-echo time. For EPI, we get symmetric weighting in k y, ( W (k y ) = exp T 2T2 ( )) ky k max For spiral, we still have to start at the spin echo so we get rid of some of the constant T2 decay from the mid pulse to the start of the spiral, but the k-space weighting is the same as the gradient echo where the edges of k-space will be heavily T2 weighted, much more then the EPI! ( W (k r ) = exp T k ) r T2 k. max 6. Matlab Simulation: 2DFT Spin-Echo Sequence. In this exercise we are going to modify the sequence you designed in HW 5 Q6c to a spin-echo sequence. For that, we need to insert a 180 refocusing pulse. We are going to insert the refocusing pulse between the phase encodes (and the readout prewinder) and the readout. 13

14 a) Modify the GRE sequence to be a spin echo sequence. Add an appropriate hard refocusing pulse and modify the gradient timings such that the echo-time is TE=6ms. In your design assume maximum B1 max = 0.16G. Plot the pulse sequence diagram. b) Simulate the pulse sequence using the data from HW5. In the simulation make the refocusing pulse 170 to simulate B1 imperfection. In addition, to simulate off-resonance add a constant gradient field of 0.01G to Gy. (simply calculate the appropriate phase encode waveform and add 0.01 to it.) The constant gradient should be on through the entire pulse sequence! Reconstruct and plot magnitude and phase images. Can you see the effects of the parasitic FID from the refocusing pulse? do you see any phase due to off-resonance? c) Modify the sequence to overcome the artifact from the parasitic FID by chopping the refocusing pulse (±170 ). Reconstruct magnitude and phase images. Comment on the artifact and also on the phase of fat and water. 14