The Vectors Process - Case Study Example

TUT 3 Look at the animation on Screen 7 of Module 4. Comment on the ment that some vectors are moving in the reverse direction. How does this relate to precession and to omega-0, which is a function of 
B0? Although precession is always in the same direction, omega0, its exact frequency is adjusted using T2 and its associated effects. Due to this adjustment, some vectors precess at a little less than omega0 and actually move backwards. Those vectors that precess at a higher frequency than omega0, move forwards in the rotating frame view. 2)   Look at the image on Screen 11 of Module 4. Comment on the signal intensity of at least 5 different tissue types in this image. Suggest how the intensity may relate to the physico-chemical environment of each tissue type. The difference in the T2 images was because of the movement of molecules which control the correlation of time. Also, since air has no protons and proton density, it therefore has no signals in T1 and T2 weighted. The brightest signal seen comes from the CSF and the aqueous humor of the eye where molecules move around unrestricted. Bone tissue on the other hand has the least mobility and is labeled MR invisible. When compared to the white matter, the gray matter is brighter because it has less structure and so increased molecular mobility. Muscle tissue has a shorter T2 and is darker in a T2 weighted image. Lastly, the T2 image of fat, which appeared a bit bright, is usually around 50msec, and since it is a large and slow molecule, it should produce a dark image.   3)   What is an fid, and why do you think it has been given this name? FID measures the decaying magnetization vector and is free because it occurs when there is no motion disturbance. When the magnetization vector rotates in a plane that is perpendicular to the B1 direction of the receiver coil, voltage is induced. Since there are no photons absorbed or emitted, it doesn’t measure emitted or transmitted radiation. The energy that is produced through excitation is lost because of the transverse and longitudinal relaxation. TUT 4 1.  Describe in your own words the meaning of the term "phase coherence". Discuss how phase coherence is lost.  Can lost phase coherence be recovered - if so, under what circumstances and how?  Phase coherence takes place in the transverse plane and is described as "the degree to which the individual magnetization vectors align in the same direction".  The net magnetization or the sum of these vectors produces MR signals while precessing within the RF receiver coil. During dephasing, the degree of vectors alignment is reduced, the signal is lost and the net magnetization decays. If 2 vectors are aligned in the same direction within the transverse plane, then they are in-phase. After a very short 90 degree pulse, all the vectors will be in-phase due to perfect alignment. Due to varying angular precession rates caused by spin-spin relaxation and field homogeneity effects, these vectors will loose coherence and will dephase. If the spin vectors are produced from nuclei with different chemical shift, they too will dephase because of the different precession rates. This dephasing can be recovered in the same way as T2 and T2 dephasing.   2. The spin-echo method is usually initiated with a pi/2(+x) pulse. Consider what would happen if a pi/4(+x) was used instead - would a spin-echo form? (look at animation 5.3 as a guide for visualizing the trajectory of the magnetization)   A pi/4 excitation pulse creates some transverse magnetization when generated at a tip angel of 45 degrees. The dephased magnetization is rephased by a pi pulse resulting in a spin echo of a slightly reduced magnetization than pi/2. Also, the transverse component of pi.4 pulse is 0.71 Mo because its magnetization from a pulse of theta degrees is Mo.sin (theta). So the echo that is produced by a subsequent pi pulse would reduce in intensity by approximately 30%.   3.  Having correctly completed Question 5.1 of Module 5, construct a four-average (NEX=4) spin-echo experiment phase cycle which may be used to correct for deviations of the pi pulse from an exact pi spin flip.  Start each scan with a pi/2(+x) pulse and vary the pi-pulse phase from one pass of the sequence to the next. What should the phase of the "receiver" be in each repetition to allow the signal to add or accumulate with each acquisition?  How does your phase cycle compensate for imperfect pi pulses? Hint:  A possible first scan is shown in the table below - simply fill in the gaps! Phase of        Phase of 
pi/2 pulse      pi pulse   magnetization    receiver   (also valid) +x                  +y                  +y                    +y          +x Phase of        Phase of pi/2 pulse      pi pulse   magnetization    receiver   (also valid) +x                +y          +y                +y          +x +x                -y          +y                +y          +x +x                +x          -y                -y          -x +x                -x          -y                -y          -x The phases change in a controlled manner from one pass to the other instead of keeping each pi/2 and pi pulse the same. This helps to compensate for the imperfections in pi pulses and the pulse angel. Also, the sign of the magnetization and the corresponding receiver only change when the pi pulse phase changes by 90 degrees as seen in x to y. In the first cycle the receiver phase is random and will add if the relationship in successive scans is constant. Using the convention that pi/2 pulse of +x phase produces +y magnetization from initial z-magnetization and also, a +y phase pulse produces -x phase magnetization: RF phase        magnetisation +x                  +z    -->     +y -x                  +z    -->     -y +y                  +z    -->     -x -y                  +z    -->     +x    It is seen that the direction of rotation, as seen from the direction of the pulse, is always clockwise around the axis of the RF. TUT 5 1.  Show mathematically why the residual magnetization for a TR = T1/2 is reduced to 40%. 2. If S/N is a function of the amount of transverse magnetization and there is a limit to the length of time that a subject can stay in the magnet, discuss the choice of TR for a proton density image using a spin echo sequence.  The choice of TR will be long to diminish T1 effect for a proton density image because PD images require a longer TR. However, the optimum selection of MR parameters will have to be compromised. A true PD will require a TR of 8000msec. Excluding CSF; we can get a PD image of grey and white matter with a much shorter TR. Since, the decay is exponential, we can also compromise with 2000msec. For a TR of 2*T1, the steady state signal is only reduced by 13%.   3. Comment on the relative T1 of the CSF, grey and white matter, based on the images in Fig 6.2 on screen 10.  Fig 6.2 is T1 weighted and since the white matter has a shorter T, it appears bright. CSF has a longer T1 and so appears dark. Also, the T1 for gray matter is between CSF and white matter so its signal intensity will be between these two as well. 4) In T1-weighted spin echo, the short T1 species are most dependent on TR with the long T1 species being the least affected. Although, contrast is lost, by increasing TR the S/N of short T1 species improves dramatically. The S/N for the CSF does not change and stays at 0. The TR needs to be increased greatly in order to prevent the saturation of the CSF. 5. Use the answer from Question 4 to calculate the TI, as a function of T1, for the null point in an Inversion  Recovery experiment. The null point Mz = 0 is when Null-point when 2e**(-TI/T1) = 1 So e**(-TI/T1) = 0.5 -TI/T1 = ln(0.5) TI/T1 = -ln(0.5) = ln2 so TI=T1*ln2 = 0.69*T1 6. Why is the grey/white contrast reversed between the two images in Fig 6.3 on screen 13? Since, the TI time is different, the signal from different tissue will be null at different TI. Also, with TI of about 600 msec the CSF should be nulled in the IR images. The IR sequences are either in a negative or positive Mz, the contrast can reverse for small changes in one parameter, TI. In the first image, although both grey and white matters are in negative Mz, the grey is more negative. With the second image we observe, that although both tissues are in positive Mz, the white is more positive.   TUT 6 1) The water molecule contains two protons. Why does the proton spectrum of water only have one peak? If you studied the proton MR spectrum of fat, what features do you think you might observe? Water has two identical protons resonating at the same frequency because of which there isn’t any spin-spin coupling. Often times this is because the protons are attached to different nuclei. It can also be that protons attached to the same carbon have differing environments because of nearby atoms influencing them differently. For protons, coupling happens when the nuclei are separated by 4 or less chemical bonds where the effect gets transmitted through the bonding electrons. With water, both of the protons are attached to the same oxygen atom. This allows the water molecule to rapidly rotate and move so that the two attached protons experience the same environment. Animal fat is composed of triglycerides and carbon chains 18 carbons long made up mostly of CH2 groups. The double bonds in these molecules result in ether linkages and unsaturated carbons. The largest peak seen in the fat spectrum is due to the main CH2 peak and 1.5ppm from -(CH2)n – chain. The last CH3 group also gives off a peak, although at a lower frequency. Also, at a slightly higher frequency, to the right of the CH2, the -CH2 will be observed to attach to the olefinic carbons. Spin-spin coupling splits the peaks further and the CH3 protons, although identical, will attach to the non-identical protons on the nearby CH2. 2. On screen 11 of module 8, a water suppression pulse for MRI is described. If you wished to perform water suppressed MR spectroscopy at 1.5T, and wanted to observe the myo-Inositol peak at 3.4ppm (given that water occurs at 4.8ppm), what RF pulse parameters would you recommend, and at what frequency would you apply it? The suppression pulse should be applied at the water frequency to avoid exciting the inositol. Also, the bandwidth of the pulse must be small enough to make sure that the myo-Inositol isn’t suppressed. Difference in chemical shift is 4.8-3.4 = 1.4ppm = 89Hz at 63.9MHz for 1.5T. Therefore the pulse will need to have a bandwidth less than 178Hz. An appropriate pulse would be a 40msec 3-cycle sinc pulse with 33msec being the approximate calculated answer.   As seen from the solutions of exercises on screen 8 and11 of mod 8: Bandwidth of sinc pulse BW = 2n/tx10**-3  where t = pulse time  t = 2n/BW For BW = 178 Hz, and using a 3 cycle sinc pulse (n=3), this gives  t = 6/178 = 0.0337  t = 33.7 msec  What about the effect of using a different number of cycles for the sinc Pulse? To go from 3 to 2 cycles for the same pulse length, less power would be required to get a 90 degree pulse. Also, the excitation bandwidth would be reduced.   3-Describe 2 ways of halving the slice thickness and discuss the hardware limitations of each method, and the potential effect on image contrast. The first method would be to increase the gradient strength which would require stronger gradients, better gradient amplifiers, more eddy currents and more noise. The second would be to decrease the excitation profile (by increasing pulse duration) which would require longer pulses -> longer minimum TE and TR -> reduced S/N, lower T1 contrast. With the same FOV, TE and TR, the two methods for halving slice thickness should give the same result and have the same S/N and contrast. Both experiments excite the same slab of spins with the only difference being the potential for chemical shift error, like water versus fat, where the higher gradient strength will produce a reduced shift. 4. If you use gradients to halve the slice thickness, what happens to the offset frequency for each slice. On older imaging systems, the acquisition was always performed at the slice selection frequency, especially for early attempts at FSE. What effect might this have on the acquired image? (this is not an easy question, but try to consider the implications of acquiring the signal with the reference frequency set to the slice offset frequency) Since the gradient is stronger, the offset frequency for each slice increases. To achieve the same slice position when doubling the gradient strength, the offset frequencies for each slice must also be doubled. Although, the S/N will reduce because of halving the thickness, there will be no effect on T2 or the image contrast. This is because the reference frequency for demodulation will be shifted, while the resonance frequency of the water in centre of the slice will go back to the frequency with no gradients. Therefore, the echo will be acquired off-resonance. The image will be shifted in the read direction equal to the amount matching the slice offset frequency. Also, the inability of the Fourier transform to differentiate between a shift in resonance and reference frequency was solved in early systems by getting double the field of view in the read direction. The appropriate part of the data was then selected depending on the acquisition frequency for the particular slice. This method was important for early FSE because early electronics frequency shifting was not phase coherent. An additional random phase component had to be added to each echo if the frequency was shifted between the acquisition and the 180 degree pulses. This led to phase encoding noise being too high. To solve this problem, all current systems use digital phase shifting. Read More