Exam 8N080 - Introduction to MRI

Transcription

1 Exam 8N080 - Introduction to MRI Friday April , h For this exam you may use an ordinary calculator (not a graphical one). In total there are 5 assignments and a total of 50 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) We are using an RF pulse for slice selection at 3T. (total of 10 points) ( 1 H) = rad T -1 s -1 a. What is the energy difference between the spin up and the spin down state of 1 H nuclei at this field strength? Planck s constant h = J s, and the energy of electromagnetic radiation (radio waves) of frequency f (in Hz) is E = h f. (2 points) b. Calculate the RF pulse band width (BW) in Hz if we want to acquire slices in the zdirection of thickness 2 mm. The following gradient strength is given: G z = 7 mt/m. (2 points) c. For practical reasons, the RF pulses are typically truncated to finite length (in time). Explain why in case of strong truncation, it may be wise to leave a gap between consecutive slices, for example slices of 2 mm and slice spacing of 5 mm. (2 points) d. It is given that the duration of the 90 o excitation pulses we use is 40 µs, what is the RF pulse amplitude B 1 we then need? (2 points) e. Give a drawback and a benefit of using RF excitation pulses of less than 90 o. (2 points) a. ω = γb so f = γb/(2π). Combined with E = hf we get E = hγb 2π = /(2π) = J b. Δf = γg zδz 2π = 596 Hz c. Truncation results in spectral leakage. In other words, spins outside the theoretical rectangular slices are also excited. This means neighboring slices actually contain overlapping anatomy. To avoid this, slice gaps can be used. d. φ = γb 1 τ so π 2 = γb From this follows B 1 = mt e. Drawback: not all longitudinal magnetization is put into the transversal plane. Benefit: in fast sequences higher steady-state values can be reached for M T. 1

2 2) Given is the following MRI sequence. (total of 12 points) 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. Explain why there is no refocussing gradient lobe of opposing sign for the 180 o pulse. (2 points) c. Is this sequence T2 or T2* weighted? Explain your answer. (1 point) d. At what time point 1, 2 or 3 do we have least phase coherence in z-direction? Explain your answer. Ignore the effect of T2 or T2* relaxation. (2 points) e. At what time point 4, 5 or 6 do we have most phase coherence in y-direction? Explain your answer. Again ignore the effect of T2 or T2* relaxation. (2 points) f. Sketch the k-space trajectory of this sequence, assume the G x gradient is stepped from its most negative value to its most positive value. Make sure to number the successive k-lines in your sketch. (2 points) a. G x: phase encoding, because of the stepped gradients G y: frequency encoding. We see a second gradient lobe of opposing sign, this causes refocussing which creates the echo needed for read-out G z: slice selection: on during application of the RF pulses b. The second half of this gradient lobe already provides refocussing; because it is a 180 o pulse, the phase of the spins is inversed halfway this gradient, so the second half of this gradient induces refocussing instead of (further) dephasing (as is the case for e.g. a 90 o pulse) c. T2* weighted; this is a gradient echo (GE) sequence, the 180 o pulse is not a refocussing pulse as it is applied before (not after) the 90 o pulse. d. Since we may ignore spin-spin relaxation, we only need to take into account the effect of the gradients. At time point 1 the dephasing due to slice selection has not yet been compensated for, so at this time point we have least phase coherence (phase coherence is halfway restored at time point 2, and completely at time point 3) e. Time point 5; this is the read-out direction along which we have our echo. The echo is maximal halfway the read-out. 2

3 f. See the fig below 3) We are performing a scan in which we are interested in the contrast between 2 tissue types A and B. Of these tissue types we know ρ A =1, T 1,A =700 ms, T 2,A =40 ms, ρ B =0.8, T 1,B =1500 ms and T 2,B =30 ms. We use a sequence in which a 180 o pulse is applied a time interval TI before the 90 o pulse, see the figure of the previous assignment. We use TR >> T 1,B. (total of 8 points) a. What is the name of such a sequence? (1 point) b. Sketch the longitudinal magnetizations M z,a (TI) and M z,b (TI) of both tissue types in the same figure. (2 points) c. Calculate the TI for which the signal of tissue type A is nulled. (2 points) d. For the sake of the desired contrast, would it be better to null tissue type B instead? Explain your answer. (1 point) e. Let s say we have nulled tissue type A. What is the signal amplitude of tissue type B if we have TE=15 ms? (2 points) a. Inversion recovery b. See the fig below c. M z,a (TI A ) = 1 2e TI A T 1,A = 0 TI A = ln(2) T 1,A = 485 ms d. Yes, the signal difference is bigger then, this can already be seen from the figure. It can also be shown numerically: M z,b (TI A ) = M z,b (485 ms) =

4 The inversion time for B is: TI B = ln(2) T 1,B = 1040 ms and M z,a (TI B ) = So the signal amplitude is bigger is we null tissue type B. e. This is M z,b (TI A ) diminished by the T2 decay it experiences during TE after it is tipped into the transversal plane: e 15/30 = The minus sign is not of relevance here. 4) We are performing an MRI-scan with a field-of-view (FOV) that is tilted 20 o with respect to the G x direction, see the following figure. It is a square 20 cm FOV, and its center coincides with the scanner isocenter (0,0,0). The x-direction is the read-out direction and the strength of G x is 5 mt/m. The k-space we acquire has dimension 100 x 100. (total of 10 points) a. Calculate the maximum precession frequency f max (in Hz) in the rotating frame of reference (RFR) during read-out. (2 points) ( 1 H) = rad T -1 s -1 If you don t have an answer to this question, assume f max = 25 khz. b. What is the sampling frequency f s we thus need? (1 point) c. How long does it take to acquire an echo? (2 points) d. We choose the TR such that M z can recover towards at least 99% of its equilibrium value between excitations (90 o pulses). What can you say about the extent of T 1- weighting of the sequence? Explain your answer. (2 points) e. Calculate the TR value that is meant in question d). It is given that the T 1 of the tissue types that we are scanning varies between 800 and 1400 ms. Assume M z already reaches steady state after the first excitation. (3 points) a. The maximum precession frequency corresponds to the maximum x-position. This can be derived from the following figure: 4

5 cos(25 ) = x max /14.14 x max = 12,8 cm. We thus get f max = γg xx max = 27,3 2π khz b. f s = 2f max = 54,6 khz c. T acq = 100 t s = 100 = 1.8 ms f s Note that this is not the echo time (TE), because that is the time between the excitation pulse and the middle of the echo. d. Hardly any T1 weighting since Mz can almost completely recover between excitations e. The equation that needs to be solved is 1 e TR T1 = This has to hold for the slowest spins, so T1=1400 ms. We then get TR = T1 ln(0.01) = s 5) Given is the following MRI imaging sequence: (total of 10 points) a. Sketch the corresponding k-space trajectory. Clearly number each k-line. (3 points) b. What can you say about the k-space coverage of this sequence? Do we mainly acquire information on image contrast or on image details? Explain your answer. (1 point) c. We are imaging the simple object given below, the field of view (FOV) is also indicated. The middle of the FOV coincides with the scanner isocentre (0,0,0) and the x and y direction are the horizontal and the vertical direction, respectively. Sketch 5

6 the image we would get if we reconstruct the image from the data of the first k-line only. (2 points) d. Same question for the third k-line. (2 points) e. The echo time of this sequence is 30 ms, and for the tissue we are imaging we know that holds T 2=55 ms, T 2*=40 ms. Calculate the maximum echo amplitude for a flip angle of 10 o. The equilibrium magnetization M 0=1 and we assume that this flip angle is small enough for longitudinal magnetization M z to fully recover between excitations. (2 points) a. See the figure below b. These are radial k-lines, so the centre of k-space is sampled much more densely. This is where the low spatial frequencies and therefore the image contrast resides. c. The first k-line only samples spatial frequencies in x-direction, so it only resolves the image in x-direction. This would the following image: 6

7 d. Now the image is only resolved in y-direction: e. The fact that M z can fully recover between excitations means that directly pior to each excitation we have M z=m 0, so all echo s have the same amplitude (we do not reach a steady state value < M 0 after a few iterations). Directly after each 10 o pulse we therefor get a transversal magnetization M T (0) = M 0 sin(10 ) = sin (10 ). During the echo time we get T 2* decay (there are no refocussing pulses in the sequence), so finally we get M T (TE) = sin(10 ) e TE/T 2 = sin(10 ) e = THE END 7