Exam 8N080 - Introduction MRI

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1 Exam 8N080 - Introduction MRI Friday January 23 rd 2015, h For this exam you may use an ordinary calculator (not a graphical one). In total there are 6 assignments and a total of 65 points can be earned. You are allowed to take this exam with you afterwards. If you don t quite understand a certain assignment, don t hesitate to ask for a translation. Good luck! 1) An MRI operator wants to image an organ with a T 2 of 30 ms using a T 2-weighted spin-echo sequence. The tissue surrounding the organ has a T 2 of 50 ms. The proton density of the organ ρρ oooooooooo = 0.8, the proton density of the surrounding tissue ρρ tttttttttttt = 0.5. (total of 11 points) a. What is the approximate signal formula for a spin-echo sequence that is T 2-weighted (i.e. TR>>T 1)? (1 point) b. Sketch the signal amplitude of both the organ and the tissue as a function of the echo time TE and in the same graph, i.e. sketch SS tttttttttttt (TE) and SS oooooooooo (TE). (2 points) c. The contrast between the organ and the surrounding tissue is defined as the difference in signal intensity. Derive a formula for the contrast CC(TE). (2 points) d. Show that in this case we should choose TE as short as possible for best contrast. (3 points) e. Also explain why in MRI we can t take TE=0. (1 point) f. Now assume ρρ oooooooooo = ρρ tttttttttttt. What TE should we choose in this case for best contrast? (2 points) a) SS = ρρ ee TTTT TT2 1 ee TTTT TT1, and if we use TR>>T 1 we get SS ρρ ee TTTT TT2 b) See the figure below; intersecting curves since the organ has a higher ρρ but also a faster decay. 1

2 c) CC = ρρ oooooooooo ee TTTT/TT 2,oooooooooo ρρ tttttttttttt ee TTTT/TT 2,tttttttttttt. Notice that the signal difference can be both positive and negative; see the plot in b). d) If we calculate dddd dddddd = 0, we will find a good TE value (local maximum) to the right of the intersection of the curves. We get dddd dddddd = ρρ oooooooooo ee TTTT gggggggg TT2,oooooooooo + ρρ tttttttttttt TT 2,oooooooooo ee TTTT gggggggg TT 2,tttttttttttt 1 1 TT 2,oooooooooo TT 2,tttttttttttt = 0 which yields TTTT gggggggg = ln TT 2,ttttttttttttρρ oooooooonn TT2,oooooooooo ρρ / = tttttttttttt TT 2,tttttttttttt 73.6 mmmm. The contrast (absolute signal difference) at this value is SS tttttttttttt (TTTT) SS oooooooooo (TTTT) = Notice that this is smaller than the signal difference at TE=0, which is =0.3! So indeed it s a good TE value, but not the optimal one! We d better choose the shortest possible TE. (Note that if for technical reasons or so TE cannot be made shorter than around 25 ms, we d better aim for the calculated value of 73.6 ms, but no points are subtracted if you didn t calculate this critical value of 25 ms). e) We cannot take TE=0 because the echo can only occur is we apply gradients or refocussing RF-pulses after the excitation pulse. A certain amount of time is needed to apply these gradients/pulses. f) The equation that was derived in d) can now be used, ρρ oooooooooo and ρρ tttttttttttt now cross out since they are now equal. So we now get TTTT oooooooooooooo = ln TT 2,tttttttttttt / TT 2,oooooooooo 1 1 TT 2,oooooooooo TT 2,tttttttttttt =38.3 ms 2) For a certain scan we define a field of view (FOV) of 30x15 cm and we want to create an image of 256x128 pixels, see the figure below. The centre of our coordinate system (xx, yy) coincides with the middle of the FOV. The xx-direction is the read-out direction, the corresponding gradient strength is GG xx =7 mt/m. The gyromagnetic ratio γγ= MHz/T. (total of 11 points) a. When recording the echo s, the signal that comes from the receive coil is typically demixed before it is sampled. This demixing means that the so called carrier frequency ωω 0 is subtracted. Explain that this means that the signal actually becomes independent of the main magnetic field strength BB 0. (2 points) 2

3 b. In the remainder of this assignment we assume that the signal has already been demixed. What is the maximum signal frequency ff mmmmmm during read-out? (2 points) c. What sample frequency ff ss do we then minimally need? (1 point) d. What is the acquisition time of a single echo, TT eeeehoo? Hint: how many samples do we acquire per echo? (1 point) e. In this case we sample kk-space (kk xx, kk yy ) by taking samples in time (related to step size ddkk xx ) and by taking different values for GG yy (related to kk yy ). It is given that the step size dddd yy (related to dddd yy ) must be chosen such that the maximum phase step is π radians. Derive the value of dddd yy. The duration of the phase encoding gradient TT pppp =0.8 ms. (2 points) f. It is given that the repetition time TR of this sequence is 500 ms, and that we acquire 1 echo per excitation. What is the total acquisition time TT aaaaaa? (1 point) g. Assume we rotate the FOV 90 o around the origin. Should we now increase or decrease ff ss? And what about dddd yy? Explain your answers. (2 points) a) The precession frequency in the presence of gradients is generally given by ωω = γγ BB 0 + GG xx xx + GG yy yy + GG zz zz = ωω 0 + γγ GG xx xx + GG yy yy + GG zz zz. If we subtract ωω 0, the term that is dependent on BB 0 drops from the equation. (We can interpret this at looking at the spins from within the rotating frame of reference rather than the static laboratory frame). b) We can use ff = γγ BB. During read-out, GG 2ππ xx is on, and the local field then is highest at xx = FFFFFF xx /2. So we get ff mmmmmm = γγ GG 2ππ xx c) ff ss = 2ff mmmmmm =89.4 khz (Nyquist) FFFFFF xx 2 = 44.7 khz. d) TT eeeehoo = 256 tt ss = 256 = 2.9 ms, with tt ff ss the sample time. ss e) The largest phase steps are made at the borders of the FOV in phase encode direction, so in this case yy-direction. As an equation we then get ππ = γγ dddd FFFFFF yy yy which can be 2 1 rewritten to dddd yy = = 0.2 mt/m. γγffffff yy TT pppp f) TT aaaaaa = = 64 s g) ff ss can be decreased since the FOV gets narrower in read-out direction, so the maximum frequency during read out goes down as well. And from the equation in e) we can see that a larger FFFFFF yy entails a smaller phase encoding step size dddd yy. 3) Given is the schematic of a multi-echo sequence, see the figure below. The top row represents the RF pulses, the following rows indicate the switching of the 3 gradient systems, and the bottom row indicates the data acquisition (read-out; sampling indicated by vertical bars). (total of 10 points) 3

4 a. Indicate for each of the 3 gradients whether it is used as slice selection gradient, phase encoding gradient, or frequency encoding gradient. Explain your answers. (3 points) b. Is this sequence T 2 or T 2* weighted? Explain your answer. (1 point) c. Gradient lobes A and B are of the same magnitude, but opposing sign. Explain why gradient lobe B is twice as long. (1 point) d. Copy the following schematic to your answering sheet and draw the kk-space trajectory of this sequence. Indicate time instances 1-9 clearly in your sketch. Make sure to use the fact that the positive GG yy gradient lobes have the same magnitude as the negative lobe, but only 1/3 the width (time duration). (3 points) e. In which direction do you expect most relaxation-induced artefacts in the image, phase encoding direction or frequency encoding direction? Explain your answer. (2 points) a) GG xx is the slice selection gradient since it is on during the application of the RF pulse; GG yy is the phase encoding gradient since its on before each read out; GG zz is the frequency 4

5 encoding gradient since it is on during data-acquisition (read-out), as can be seen from the timing of the sampling on the lower time line. b) This is a gradient echo sequence without refocussing RF pulses, so it s a T 2-weighted sequence c) So that the dephasing caused by gradient lobe A is compensated exactly halfway gradient lobe B, since phase is proportional to gradient amplitude and duration (φφ = γγγγγγ). This means that the echo that occurs during the application of B has maximal amplitude in the middle of the read-out d) See the sketch below. Note that the last echo is at height kk yy = 0 because the cumulative effect of the yy-gradient is zero for this third echo. e) In phase encoding direction, since most relaxation occurs between echo s (time interval TR) rather than within echo s (time interval shorter than TE). 4) Given is an MRI sequence with a repetition time TR of 100 ms. We are imaging a tissue with a T 1 of 150 ms and we use a flip angle α=30 o. The equilibrium magnetization value MM 0 =1. See the schematic below. Each time point (i) is directly prior or after an α-pulse, and the duration of these pulses may be neglected (they are very short). (total of 8 points) a. Calculate MM zz directly after the first α-pulse; time point (1) in the schematic. (1 point) b. This MM zz now has until the next α-pulse to recover towards MM 0. Draw this recovery process MM zz (tt) and derive the corresponding formula. (3 points) c. Use this formula to find the value of MM zz directly prior to the second α-pulse, i.e. time point (2). (1 point) d. The second α-pulse is now applied. Again, MM zz has a period TR to recover towards MM 0. Find the value of MM zz directly prior to the third α-pulse, i.e. time point (3). (2 points) 5

6 e. Now find the amplitude of the transversal magnetization MM xx,yy at time point (4). (1 point) a) MM zz (tt 1 ) = cos(αα) MM b) See the figure below. This is T 1 relaxation starting at cos(αα) MM 0 and recovering towards MM 0. The formula is MM zz (tt) = MM 0 + (cos(αα) MM 0 MM 0 )ee tt/tt 1, with tt the time after the α-pulse. c) Just plug in tt = TTTT in the equation: MM zz (TTTT)=0.93 d) We have a similar recovery process as sketched in b), only now we start at 0.93 MM 0. Deriving a similar equation and again plugging in tt = TTTT gives MM zz (tt 3 ) = e) MM TT (tt 4 ) = sin(αα) ) The following questions are about the kk-space. (total of 13 points) a. Briefly explain what the kk-space is. Use the terms raw data, image, spatial frequency and Fourier transform in your answer. (3 points) b. Link the following spatial frequency images A-E to their kk-space positions 1-5. (2 points) c. Often MR images are based of full kk-space coverage. For a certain application, fast imaging is needed, and only part of the kk-space (in this case 50%) is acquired, see the figure below. Which partial kk-space coverage A-E would you prefer? Motivate your answer, and for each other option give a reason why you didn t choose it. It is given 6

7 that the horizontal direction is the frequency encoding direction, and the vertical direction is the phase encoding direction. (4 points) d. In the above figure, one can find 2 pairs of partial kk-spaces which have the same acquisition time. Which 2 pairs are these? Explain your answer. (2 points) e. Given below is an actual kk-space. Explain why most signal intensity is found in the middle. (2 points) kk-space: bright (lots of signal) near the origin a) The kk-space represents the raw data of an MRI scanner. It is a way to store the data (samples) from the individual echo s in a 2D matrix. The kk-space is linked to the MR image via a Fourier transform. It represents the spatial frequencies which are present in the image; the brighter a certain position in kk-space, the stronger the corresponding spatial frequency. b) The further away from the center of kk-space, the higher the spatial frequency A: large number of periods in both xx and yy 5 B: Moderate number of periods in xx 3 C: One period in xx, two periods in yy 2 7

8 D: One period in both xx and yy 1 E: One period in yy 4 c) For speed up, we need to reduce the number of phase encodes. This means we want partial coverage in yy-direction, and this applies to kk-spaces A, C and E. However, in E the phase encoding steps are too large, which would cause aliasing (fold-in). The difference between A and C is that in A we only loose high frequency content (details) in yydirection, if this is undesirable than C is the better option, although its acquisition time is slightly higher. B and D don t offer a reduction in number of phase encodes and therefore don t give a shorter acquisition time. Furthermore, D again gives aliasing (this time because the sample frequency is too low). d) As said, the number of lines in yy-direction corresponds to the number of echo s, which is a measure for the acquisition time. The number of lines in yy-direction is the same for A and E (half the number of echo s as in the full kk-space), and for B and D (partial readout of all echo s and undersampling of all echo s, respectively). e) Let s say yy again is the phase encoding direction and xx again is the read-out direction. Phase encoding introduces a phase difference in yy-direction, which corresponds to signal loss (dephasing). This dephasing is lowest for zero phase encoding (centre of kk-space in yy-direction), so that s were most signal is found along the kk yy axis. Furthermore, the echo intensity is highest at the middle of read-out (refocussing in xx-direction), which is the middle of kk-space in xx-direction. So indeed the center of k-space (low kk xx and low kk yy ) is brightest. 6) Traditional MR sequences (such as spin-echo) fill kk-space along rectangular pathways, using rectangular phase encoding and frequency encoding gradients. An alternative is provided below; frequency encoding and phase encoding here are a cosine and a sine of linearly increasing amplitude, respectively. Samples are acquired continuously, from the moment of the excitation pulse onwards. (total of 12 points) a. Sketch the kk-space trajectory of this sequence, with GG ffffffff and GG pphaaaaaa corresponding to kk xx and kk yy, respectively. (4 points) b. The 4 corners of kk-space are not filled by this sequence. Discuss what effect this has on the resulting image. (2 points) 8

9 c. Explain that for traditional image reconstruction, the data needs to be interpolated (this is called regridding ). (3 points) d. Is this sequence T 2 or T 2* weighted? (1 point) e. Explain what consequences these smooth gradient shapes have on hardware demands and scanner noise. (2 points) a. (4 pt) See the figure below b. (2 pt) The further away from the center of kk-space, the higher the corresponding spatial frequency. So if we don t fill the corners, we miss the highest spatial frequencies. However, notice that this also means that the resolution (level of detail) is highest in diagonal directions for rectangular kk-spaces. So effectively, the resolution is more homogenous for this trajectory (i.e. the same in all radial directions) c. (3 pt) The samples are not located on a grid, so we cannot directly apply a fast Fourier transform (FFT) for image reconstruction; this requires a 2D matrix as input (i.e. a rectangular grid of data). Hence the term regridding. d. (1 pt) T 2* weighted, since there are no refocussing pulses (only a single 90 o excitation pulse). e. (2 pt) It s quite difficult to create perfectly rectangular gradient shapes; due to the inductance of the gradient coils, some time is needed for the coil currents to start up/switch off, so the upslope/downslope is not perfectly vertical. This problem is much less in smoother gradient shapes in which coil current varies much slower. Furthermore, rapid switching of the gradients causes rapid flux change, which in turn induce eddy currents in the scanner plating. This causes mechanical vibrations and thus noise. This is less so for smooth gradient shapes (slower flux changes, less eddy currents, less noise). 9

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