Principles of MRI Miki Lustig, Spring 2011

EE225E/BIOE265

Midterm Exam #2

NAME:

You have 1 hour & 60 minutes for 3 problems.

• Please do preliminary calculations on your own scratch paper. (Do not hand in.)

• Show enough (neat) work in the clear spaces on this exam to convince us that you derived,not guessed, your answers.

• Put your final answers in the boxes at the bottom of the page.

You can use your course notes, homework sets and solutions, calculator, and the Nishimura book.

Problem Score

1

2

3

Total

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Problem 1. MRI Artifacts and Debugging (80 points)

In this question you will be given several pairs of images. The images may exhibit artifacts whichwill be indicated by arrows/circles. Answer the questions as best as you can. First describe theartifact and then provide as much information as possible on the source/s of the artifact and why itappears this way. There might be several possible right answers. A final answer without explanationwill not be credited.

a) Example: What is the source of the artifact in the left image and what is the difference inthe acquisition/processing of the two?

Solution:

The artifact in the image appears as the back of the brain aliasing over and wrapping to thefront. This is typical when the prescribed FOV in the phase encode is smaller than the FOVof the object. In that case the sampling density does not meet the Nyquist rate. The phaseencode is obviously Anterior-Posterior.

The image on the right does not exhibit aliasing. One way to avoid aliasing is to increasethe number of phase encodes while keeping the resolution the same. The other way is toswap the readout and phase-encode directions. It seems that the latter approach of swappingthe 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

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b) What is the source of the artifact in the left image and what is the difference in the acquisi-tion/processing of the two?

Solutions:

Something has gone wrong in the geometry of the left guy’s head! It has been compressedin 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 bemuch weaker at the top and bottom and will result in a geometric distortion. This is fixed inpostprocessing by applying interpolation to correct for the distortion in the right image.

Source of artifact: Gradient non-linearityDifference: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 acqui-sition/processing of the two?

Source of artifact: The fat in the right image is shifted right with respect to water. This is aresult of a low-bandwidth/pixel readout. The readout direction is RLDifference: The readout bandwidth in the left image is higher (shorter readout) and the fatdoes not have time to accrue much phase and appears to not shift by much. (There’s still aslight shift though!

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d) What is the source of the artifact in the two images and what is the difference in the acqui-sition/processing of the two?

Solutions:

There’s appear to be ghosting artifacts that seem to be originating from the eyes. This is anartifact of eye motion during a scan. Motion during an acquisition will create inconsistentdata in the slow-direction (the phase encode). Therefore the phase encode direction in theleft image is A/P and R/L in the right image.

Source of artifact: Eye motion during the scanDifference: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 acquisi-tion/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 theear canal. This is a result of T ∗

2 decay typical for a gradient echo sequence. The right imagedoes not exhibits the loss of signal. This can be due to much shorter echo time, but there’ssignificant T2 weighting in the image that makes this infeasible. The other explanation is aSpin-echo sequence!

Source of artifact: Loss of signal due to T ∗2 intravoxel dephasing in a gradient echo sequence

Difference:The right image does not exhibits loss of signal and is most likely a spin-echosequence

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Problem 2. MRI Sequences and Contrast (100 points)

In my free time, I went ahead and scanned 7 bottles of water doped with different concentrationsof Copper-Sulphate and Manganese-Chloride. The different doping results in significant variationsin T1 and T2 relaxations. However, the proton-density is the same.

You will be presented with results of experiments and will be asked questions on the relaxationparameters and the type of sequences and sequence parameters that were used. All the sequenceshad the same slice thickness, same FOV and same resolution. All the acquisitions were singleaxial slice.

Below are two results that are obtained with two different receiver coils using a gradient echosequence with TE = 1.1ms, TR = 8000ms and a 90 degree flip angle. The bottles are numberedfor future reference.

1

2 3

4

5

6 7

1

2 3

4

5

6 7

TE=1.1ms TR=8000ms, flip=90TE=1.1ms TR=8000ms, flip=90

8 Channel Head Coil Body Coil

a) What is the dominant source of contrast weighting in the body coil image? Briefly explain.

Solutions: The bottles were scanned using a very long TR in which the longitudal magneti-zation is fully recovered and there should not be any T1 weighting. The echo time, TE=1ms,is very short therefore there shouldn’t be any (or much) T ∗

2 weighting as well. The contrastin the image is therefore generated only by the proton density.

Contrast: Short TE and long TR would result in proton-density weighting.

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b) Describe the main differences between the two images and briefly explain the sources of them.

Solutions:

There are two main differences between the images. The first is that the body-coil image seemsto be more noisy than the 8-channel array image. At the same time, the body-coil image isvery uniform whereas the 8-channel array image has intensity variations in the image.

The main source of noise in MR is body noise. The body coil is sensitive to the entireobject and therefore receives much more noise than the restricted FOV of each of the 8-channel array elements of the head coil. At the same time, the 8-channel array elementsdo not exhibit uniform sensitivity in space which is manifested by intensity variation whensquare-root-of-sum-of-squares combination is performed.

Differences: The 8-channel coil has better signal to noise because each of the small elementspick up less body noise. The 8 channel coil does not have uniform sensitivity. Positions closeto the elements have much higher signal intensity.

The images below are the result of scanning with Saturation-Recovery and Inversion-Recoveryspin echo sequences with different parameters.

Saturation-recovery Spin-echoTE=10 TR=1500 TE=30 TR=1500 TE=60 TR=1500 TE=120 TR=1500

TE=10 TR=34 TE=10 TR=100 TE=10 TR=300 TE=10 TR=600

Inversion Recovery Spin-EchoTE=10 TR=300 IR=50TE=120 TR=8000 IR=1200

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c) Roughly rate the T1 and T2 values of the water in the bottles. For T2 choose from the list: 6, 12, 30, 50, 100, 250, and 600ms. For T1 choose from: 50,135, 250,700 ,1000, 1400, and1700ms. You should be using very few calculations and lots of common sense.

Solutions: I have to admit that I had a mistake in listing a 70ms T1 value. I blame spotlightwhich uses base-10 for the log function....

In this question it was pretty obvious that the top row is T2-weighted and the bottom rowis T1-weighted. The left IR sequence is both T1 and T2 weighted, where as the right is onlyT1-weighted.

It is pretty straightforward to estimate the T2 components just by looking at the order thebottle disappear. The T1 was a bit trickier. This is where the IR sequence could have cometo the rescue. For TR� T1, TI ≈ 0.68T1. In the left IR image bottles 2,3,4 and 6 are dark.At TE = 200ms we should expect 2,3,4 to be nulled, but not 6. Therefore we can concludethat bottle 6 is nulled by the IR pulse and, T1 ≈ 1200/0.68 ≈ 1700.

In the right IR image, bottle 2 is nulled, but bottle 3 still has signal. Therefore we concludethat bottle 2 is nulled by the IR and T1 ≈ 70 the closest in the list is T1 = 50.

It is much harder to figure out the TR for bottles 1, 3 and 4. Therefore we accepted allreasonable answers.

T1 =

T1 =

T1 = 1000

T1 = 700

T2 = 600

T2 = 250

T1 =

T2 =

T1 =

T2 = T1 =

T2 =

T2 =

T2 =

1

2 3

4

5

67

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Consider the following acquired image,

TR=3000

Sequence = ?

TE=40

d) Which pulse sequence/s and parameters could have produced the above image?(choose one)Backup your proposal with evidence from the image!

Solutions:

Compared to the T2-weighted spin-echo at TE = 30 and TE = 60 we see here either anadditional spatially variant decay or a geometric distortion. The image is therefore likely tobe a gradient echo sequence with strong T ∗

2 weighting. Another option that I accepted is EPIsince the decay does appear like geometric distortion, though in EPI pileup of signal shouldhave resulted in bright spots that are not present here. The true answer is gradient echo.

Pulse Sequence and parameters: Gradient EchoEvidence: T ∗

2 decay close to the edges of the bottles due magnetic susceptibility variationcaused by the air-bottle interface.

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Now, consider the following acquired image,

TE=10 TR=3000

Sequence = ?

e) Which pulse sequence/s and parameters could have produced the above image? (choose one)Backup your proposal with evidence from the image!

Solutions:

The image exhibits blurring in the top-down direction that seems to be T2 dependent. Thelong T2 bottles look sharp, but the short T2 bottles are significantly blurred. This couldbe a result of a multi-echo sequence such as EPI of fast-spin-echo (FSE). There isn’t anyapparent T ∗

2 effects so the sequence is probably spin-echo EPI or FSE. There isn’t any apparentgeometric distortion, so it is less likely to be spin-echo EPI and more likely to be an FSEsequence. The readout is in the Left-right direction, which does not see the decay and theslow-direction is top-bottom. The true answer is FSE, but I accepted those who said it wasEPI, reducing a few points if spin-echo was not mentioned.

Pulse Sequence and parameters: Fast Spin echo with readout Left-right

Evidence: T2 dependent blurr in the top-down direction. No apparent T ∗2 decay and

no apparent geometric distortion.

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Problem 3. Contrast Preparation (80 points)

Consider the following sequence of two θ degrees tip-angle RF pulses separated by T seconds.

Θ Θ

T

a) Given that the equilibrium magnetization is M0z, derive an expression for the Mz componentof the magnetization immediately following the second RF as a function of T2, T , and θ. Youcan neglect T1 recovery (since T1 � T ) and off-resonance.

Solutions:

For easy notation we will define E2 = e− T

T2 , Cθ = cos(θ) and Sθ = sin(θ)

M(T+) =

[Cθ Sθ−Sθ Cθ

] [E2 00 0

] [Cθ Sθ−Sθ Cθ

] [0M0

]

=

[Cθ Sθ−Sθ Cθ

] [E2SθM0

CθM0

]

So, the Mz(T+) component is,

Mz(T+) = (C2θ − E2S

2θ )M0

Mz =

(cos2(θ)− e−

TT2 sin2(θ)

)M0

b) Given T = T0, find the flip angle θ for which the Mz component is zero for spins with adesired T2 transverse relaxation value.

Solutions:

0 = C2θ − E2S

2θ

C2θ

S2θ

= E2

tan(θ) = E−1/22

θ = atan(E−1/22 )

θ = atan

(e

T2T2

)

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MR Angiography is an important tool in assessing vascular diseases in patients. Often, T1shortening Gadolinium contrast agents are used in combination with short TR sequences toincrease the blood-muscle contrast. However, using Gadolinium based contrast agents canresult in a life threatening syndrome, called NSF, in patients that have renal disease.

Consider the contrast preparation imaging paradigm below. For each preparation sequence asingle phase encode is collected. Also, assume TR� T1.

PrepSequence

Imaging Sequence

RF

Gz

Gy

Gx

90

PrepSequence

90

TR

The T1/T2 of blood are 1000/220ms and the T1/T2 of muscle are 870/50ms. We would liketo design a non-contrast enhanced preparation pulse that will ideally have good blood signaland no muscle signal at all. In addition, we would like the preparation pulse to not be muchlonger than 50ms.

c) Based on your previous derivations, design a preparation sequence that nulls the musclesignal while producing signal from blood. Draw the sequence pointing out the relevant pa-rameters. (Extra points will be given for those coming with solutions that are insensitive tooff-resonance). What is the blood Mz magnetization after the prep-pulse?

Solutions:

The previous questions have taught us that we can null the signal of a particular T2 specie.In this part we need to design a preparation sequence that will null the muscle signal whilestill producing blood signal. Unfortunately, we can not use inversion recovery based on theT1 since that would take much longer than than 50ms. Using an excitation recovery wouldprobably not leave much blood signal. Therefore we are going to use the T2 prep-pulse frombefore.

For this preparation sequence we need to choose the flip angle θ and T such that the musclesignal 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 weget, θ = 47.86◦, 52.09◦, 54.83◦, 58.76◦ with corresponding Mz = −0.075M0, − 0.18M0, −0.24M0, − 0.31M0. So, T = 50ms gives the highest blood signal with θ = 58.76◦.

Of course this pulse is VERY sensitive to B0 inhomogeneity because spins would be precess-ing during the waiting time and the second RF pulse will have different effect on differentresonances. This can be mitigated by applying a spin-echo 180 pulse such that all the spinswill refocus exactly at the time the second RF pulse is applied. However, if we use a sin-gle refocusing, we need to change the sign of the second RF so it will have the same effect.Alternatively we can use two refocusing RF pulses. Each 180 refocusing pulses should havecrushers gradients on both sides to prevent a parasitic FID.

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A very important part of the question is to realize that at the end of this preparation sequencethere remains is a significant transverse magnetization. To remove it, we need to apply a largegradient crusher after the second RF.

Here is an acceptable solution:

RF

Gz

Gy

Gx

Prep Sequence Imaging Sequence90Θx ΘxT

Here are two acceptable B0 insensitive solutions:

RF

Gz

Gy

Gx

Prep Sequence Imaging Sequence90Θx 180y -ΘxT/2 T/2

RF

Gz

Gy

Gx

Prep Sequence Imaging Sequence90Θx 180y Θx180yT/2T/4 T/4

Blood |Mzprep| = 0.31M0

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