Exam 8NC20-8NC29 - Introduction to NMR and MRI

Transcription

1 Exam 8NC-8NC9 - Inroducion o NMR and MRI Friday April 5, 8.-. h For his exam you may use an ordinary calculaor (no a graphical one). In oal here are 6 assignmens and a oal of 64 poins can be earned. You are allowed o ake his exam wih you aferwards. If you don quie undersand a cerain assignmen, don hesiae o ask for a ranslaion. Good luck! ) Afer pulse exciaion of nuclear spins, he ne magneizaion along he z-axis will recover as a resul of a process which is ermed longiudinal relaxaion or T relaxaion. When doing an NMR experimen, i is imporan o know he T relaxaion imes of he meabolies under invesigaion, for example o choose he opimal repeiion ime TR. In addiion, T relaxaion imes migh change in diseased saes and can herefore be informaive abou a paien s condiion. (oal of poins) a. Nuclear spin relaxaion is no a sponaneous process. For relaxaion o occur a flucuaing magneic field is required. Where is his magneic field coming from and wha causes he flucuaions of his magneic field? ( poins) The magneic field comes from he neighboring dipoles. The roaional umbling of he molecules causes he flucuaions of he magneic field ha is experienced by he dipole. b. In he figure below he specral densiy funcion J() is shown for nuclear spins in hree molecules: one wih slow umbling, one wih fas umbling and one wih an inermediae umbling rae. Which of he hree molecules has he longes T relaxaion ime for he given Larmor frequency (see figure)? Explain your answer. (3 poins) (Larmor frequency) The molecule wih slow umbling has he lowes probabiliy of finding a componen of he moion a he Larmor frequency, which is necessary for relaxaion o occur. Therefore, relaxaion will be leas efficien and T will be longes for he molecule wih slow umbling.

2 c. Derive expressions for he magneizaion M M, M, M x y z a ime poins,, 3 and 4 in he inversion-recovery (IR) experimen indicaed in he figure below. A ime M,,. (3 poins) poin you sar wih equilibrium magneizaion M inversion pulse deecion pulse 8 x 9 y 3 4 : M M M,, : M M M,, R : M M,, e R 3: M3 M e,, = equilibrium magneizaion along -z-axis T relaxaion R 4. R M M e e (cos,sin,) 4 shor : magneizaion along -x-axis long : magneizaion along +x-axis Signal is weighed by a funcion of R and ; T relaxaion ignored during acquisiion d. We would like o know if he T relaxaion imes of brain meabolies are affeced by epilepsy. Explain how you could deermine he T relaxaion imes of brain meabolies using he inversion recovery experimen of quesion c. ( poins) Measure differen experimens wih differen values for Deermine T by fiing a mono-exponenial funcion o he signal ampliudes versus or, in a quick-anddiry approach, deermine T from he zero-crossing (T = /ln). e. Unforunaely, he inversion recovery specra appear heavily disored by broad resonances from he skull. How could you modify he experimen in order o suppress hese broad signals? Make a drawing of your new pulse sequence and explain how he broad signals are suppressed. You do no have o derive expressions for he magneizaion a he differen ime poins in your pulse sequence. ( poins) 8 x 9 y 8 x The broad signals originae from compounds wih a shor T. Insead of reading ou he signal wih a 9-degree pulse, we could also read ou he signal wih a spin-echo sequence. If we choose he delays and hus he echo ime sufficienly long, hen he signals from he compounds wih shor T will be suppressed, because hey will have decayed due o T relaxaion.

3 ) We are scanning a paien wih suspeced MELAS, which sands for miochondrial encephalopahy wih lacic acidosis and sroke-like episodes. In his miochondrial disease, oxidaive energy producion is insufficien, leading o increased anaerobic meabolism and he accumulaion of lacae in he brain. We perform H NMR specroscopy in he paien s brain o see if we can deec lacae. The MR scanner has a field srengh of.35 Tesla (= MHz). For he deecion of he signal we use a single receiver coil. (oal of poins) a. In order o discriminae posiive and negaive frequencies one needs o acquire a complex signal. How could we acquire a complex signal wih he single receiver coil? Describe he differen seps and give expressions for he signal(s) generaed a each sep. (3 poins) We can sill measure a complex signal wih a single receiver coil by using elecronic manipulaion. The differen seps are: () spliing of he signal, () mixing wih he carrier frequency, (3) low-bandpass filering (and (4) combining he wo signals) (see scheme). cos real mixer low bandpass filer Re{S ()} S cos splier complex S () imaginary mixer filer Im{S ()} = carrier frequency S cos splier cos i cos cos sin mixer cos cos i cos sin i i cos sin sin lowbandpassfiler e i " e cos " i i sin b. Predic wheher he resonance frequency of he mehyl proons in lacae (he 3 proons aached o he uppermos C aom in he figure) is higher or lower han he resonance frequency of he mehine proon in lacae (he encircled proon). Explain your answer. ( poins) 3

4 H X 3 H C H A H C OH C O OH lacae The elecrons surrounding a nucleus can be regarded as small currens, giving rise o a magneic momen a he nucleus. Since his magneic momen opposes he exernal magneic field, he effecive magneic field a he nucleus is reduced, which is referred o as elecronic shielding. Differences in shielding lead o differen Larmor frequencies or differen chemical shifs. The elecronegaive oxygen aom shifs he elecron densiy away from he mehine proon, leading o reduced elecronic shielding and hus a higher resonance frequency han for he mehyl proons. Thus he mehyl proons have more elecronic shielding and a lower resonance frequency han he mehine proon. c. There is a J coupling ineracion beween he mehine and mehyl proons in lacae wih a J coupling consan of Hz. Make a skech of he NMR specrum ha you expec for lacae. Assume equal T and T relaxaion imes for all resonances. (3 poins) Drawing on scale: J AX = Hz = Hz / MHz * 6 =. ppm A X 3 AX AX 3 AX AX PPM d. Suppose ha you measure he specrum wih a specral widh of 4 ppm and 56 complex poins. Wha is he digial resoluion () of he specrum in Hz? ( poins) ( H) = rad T - s - The digial resoluion = SW/N. N=56 and SW=4 ppm. A a field srengh of.35t 4 ppm corresponds o 4-6 * (/) * *.35 = 4-6 * 6 = 4 Hz. So = 4/56 = 5.6 Hz 4

5 e. Unforunaely, he digial resoluion appears oo low o be able o observe he J coupling paern for lacae. Name wo mehods o improve he digial resoluion of our specrum. ( poins) () Decrease SW, so increase he sampling ime. () Increase N, so increase he number of sampling poins. (3) Perform zero-filling, i.e. exend he ime domain signal wih zeroes. 3) We are using an RF pulse for slice selecion a 3T. (oal of poins) ( H) = rad T - s - a. Wha is he energy difference beween he spin up and he spin down sae of H nuclei a his field srengh? Planck s consan h = J s, and he energy of elecromagneic radiaion (radio waves) of frequency f (in Hz) is E = h f. ( poins) b. Calculae he RF pulse band widh (BW) in Hz if we wan o acquire slices in he zdirecion of hickness mm. The following gradien srengh is given: G z = 7 mt/m. ( poins) c. For pracical reasons, he RF pulses are ypically runcaed o finie lengh (in ime). Explain why in case of srong runcaion, i may be wise o leave a gap beween consecuive slices, for example slices of mm and slice spacing of 5 mm. ( poins) d. I is given ha he duraion of he 9 o exciaion pulses we use is 4 µs, wha is he RF pulse ampliude B we hen need? ( poins) e. Give a drawback and a benefi of using RF exciaion pulses of less han 9 o. ( poins) a. ω = γb so f = γb/(π). Combined wih E = hf we ge E = hγb π = /(π) = J b. Δf = γg zδz π = 596 Hz c. Truncaion resuls in specral leakage. In oher words, spins ouside he heoreical recangular slices are also excied. This means neighboring slices acually conain overlapping anaomy. To avoid his, slice gaps can be used. d. φ = γb τ so π = γb 4 6. From his follows B =.47 mt e. Drawback: no all longiudinal magneizaion is pu ino he ransversal plane. Benefi: in fas sequences higher seady-sae values can be reached for M T. 4) Given is he following MRI sequence. (oal of poins) 5

6 a. Indicae for each of he 3 gradiens wheher i is used as slice selecion gradien, phase encoding gradien, or frequency encoding gradien. Explain your answers. (3 poins) b. Explain why here is no refocussing gradien lobe of opposing sign for he 8 o pulse. ( poins) c. Is his sequence T or T* weighed? Explain your answer. ( poin) d. A wha ime poin, or 3 do we have leas phase coherence in z-direcion? Explain your answer. Ignore he effec of T or T* relaxaion. ( poins) e. A wha ime poin 4, 5 or 6 do we have mos phase coherence in y-direcion? Explain your answer. Again ignore he effec of T or T* relaxaion. ( poins) f. Skech he k-space rajecory of his sequence, assume he G x gradien is sepped from is mos negaive value o is mos posiive value. Make sure o number he successive k-lines in your skech. ( poins) a. G x: phase encoding, because of he sepped gradiens G y: frequency encoding. We see a second gradien lobe of opposing sign, his causes refocussing which creaes he echo needed for read-ou G z: slice selecion: on during applicaion of he RF pulses b. The second half of his gradien lobe already provides refocussing; because i is a 8 o pulse, he phase of he spins is inversed halfway his gradien, so he second half of his gradien induces refocussing insead of (furher) dephasing (as is he case for e.g. a 9 o pulse) c. T* weighed; his is a gradien echo (GE) sequence, he 8 o pulse is no a refocussing pulse as i is applied before (no afer) he 9 o pulse. d. Since we may ignore spin-spin relaxaion, we only need o ake ino accoun he effec of he gradiens. A ime poin he dephasing due o slice selecion has no ye been compensaed for, so a his ime poin we have leas phase coherence (phase coherence is halfway resored a ime poin, and compleely a ime poin 3) e. Time poin 5; his is he read-ou direcion along which we have our echo. The echo is maximal halfway he read-ou. 6

7 f. See he fig below 5) We are performing a scan in which we are ineresed in he conras beween issue ypes A and B. Of hese issue ypes we know ρ A =, T,A =7 ms, T,A =4 ms, ρ B =.8, T,B =5 ms and T,B =3 ms. We use a sequence in which a 8 o pulse is applied a ime inerval TI before he 9 o pulse, see he figure of he previous assignmen. We use TR >> T,B. (oal of 8 poins) a. Wha is he name of such a sequence? ( poin) b. Skech he longiudinal magneizaions M z,a (TI) and M z,b (TI) of boh issue ypes in he same figure. ( poins) c. Calculae he TI for which he signal of issue ype A is nulled. ( poins) d. For he sake of he desired conras, would i be beer o null issue ype B insead? Explain your answer. ( poin) e. Le s say we have nulled issue ype A. Wha is he signal ampliude of issue ype B if we have TE=5 ms? ( poins) a. Inversion recovery b. See he fig below c. M z,a (TI A ) = e TI A T,A = TI A = ln() T,A = 485 ms 7

8 d. Yes, he signal difference is bigger hen, his can already be seen from he figure. I can also be shown numerically: M z,b (TI A ) = M z,b (485 ms) =.4475 The inversion ime for B is: TI B = ln() T,B = 4 ms and M z,a (TI B ) = So he signal ampliude is bigger is we null issue ype B. e. This is M z,b (TI A ) diminished by he T decay i experiences during TE afer i is ipped ino he ransversal plane: e 5/3 =-.. The minus sign is no of relevance here. 6) We are performing an MRI-scan wih a field-of-view (FOV) ha is iled o wih respec o he G x direcion, see he following figure. I is a square cm FOV, and is cener coincides wih he scanner isocener (,,). The x-direcion is he read-ou direcion and he srengh of G x is 5 mt/m. The k-space we acquire has dimension x. (oal of poins) a. Calculae he maximum precession frequency f max (in Hz) in he roaing frame of reference (RFR) during read-ou. ( poins) ( H) = rad T - s - If you don have an answer o his quesion, assume f max = 5 khz. b. Wha is he sampling frequency f s we hus need? ( poin) c. How long does i ake o acquire an echo? ( poins) d. We choose he TR such ha M z can recover owards a leas 99% of is equilibrium value beween exciaions (9 o pulses). Wha can you say abou he exen of T - weighing of he sequence? Explain your answer. ( poins) e. Calculae he TR value ha is mean in quesion 6d. I is given ha he T of he issue ypes ha we are scanning varies beween 8 and 4 ms. Assume M z already reaches seady sae afer he firs exciaion. (3 poins) a. The maximum precession frequency corresponds o he maximum x-posiion. This can be derived from he following figure: 8

9 cos(5 ) = x max /4.4 x max =,8 cm. We hus ge f max = γg xx max = 7,3 π khz b. f s = f max = 54,6 khz c. T acq = s = =.8 ms f s Noe ha his is no he echo ime (TE), because ha is he ime beween he exciaion pulse and he middle of he echo. d. Hardly any T weighing since Mz can almos compleely recover beween exciaions e. The equaion ha needs o be solved is e TR T =.99. This has o hold for he slowes spins, so T=4 ms. We hen ge TR = T ln(.) = s THE END 9