k B 2 Radiofrequency pulses and hardware

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1 1 Exra MR Problems DC Medical Imaging course April, 214 he problems below are harder, more ime-consuming, and inended for hose wih a more mahemaical background. hey are enirely opional, bu hopefully will make he mahemaicians happy. 1 Spin physics d) * By considering wo spin saes of energy E 1 and E 2, derive he Bolzmann equaion for he raio of spins in each sae. Answer: here are many ways of doing his, and his answer is by no means he bes. he Maxwell-Bolzmann disribuion f(e) gives you he probabiliy disribuion of energy in he sysem: f(e) Ω(E)e E/kB where Ω(E) is he degeneracy of sae E. We expec he degeneracy o be he same for each of he wo saes. herefore, he raio of populaions is going o be proporional o he raio of hese wo disribuions, i.e. f(e 1 ) f(e 2 ) = n = e E1/kB n e = E E 2/k B e k B e) * Deduce how he spin sae populaion raios change wih B in he regime where E/k B is small, i.e. find. How do you hink his compares o he B change in cos of an MR sysem wih magneic field? Answer: In he given regime, we are able o aylor expand he exponenial funcion, so ha he raio is approximaely 1 γb k B. We can rivially differeniae wih respec o B, and come o he conclusion ha signal is improved i.e. he raio decreased from uniy approximaely linearly wih B. (In fac, i s slighly worse han his he sign of he nex larges B 2 erm is going o be posiive, which reduces he signal we see). I is highly likely, however, ha he cos of an MR sysem does no scale linearly wih B. 2 Radiofrequency pulses and hardware e) ** [Nuers only] he way o undersand, in general, he effec of an RF pulse is o solve he Bloch equaions. hese describe he evoluion of magneisaion in maer, and are a phenomenological exension o a resul ha arises ou of Schrödinger equaions. Neglecing 1 and 2, he Bloch equaions can be wrien in he roaing reference frame as

2 2 dm x() dm y() dm z() G x B 1,y = γ G x B 1,x B 1,y B 1,x M x M y M z (1) where G = (G x, G y, G z ) is an applied linear magneic field gradien, x = (x(), y(), z()), and B 1 he applied RF field. Boh G and B 1 are funcions of ime. We now consider a small flip angle pulse, such ha M z M consan. By defining he ransverse magneisaion M xy = M x + im y, and he applied field B 1 = B 1,x + ib 1,y, show ha he firs wo componens of (1) can be wrien as dm xy = iγg xm xy + iγb 1 M. Answer: Muliply he second equaion by i and add he wo: (M x + im y ) = γg xm y γb 1,y M iγg xm x + iγb 1,x M = γg x(m y im x ) + γm ( B 1,y + ib 1,x ) = γg x( i 2 M y im x ) + γm (i 2 B 1,y + ib 1,x ) = γg x( i)(m x + im y ) + γm i(b 1,x + ib 1,y ) M xy = iγg xm xy + im γb 1 Now, if he sysem is iniially in he sae M = (,, M ), show ha his differenial equaion can be solved for he final magneisaion a a ime o yield M xy (x) = iγm B 1 ()e iγ G(s) x(s) ds. (2) Answer: his is a simple firs-order ODE, and jus requires an inegraing facor. he appropriae inegraing facor is e i γg(s) x ds see, e.g, physics or engineering firs year exbooks (Riley, Hobson & Bence or ryzig) for a derivaion. So, muliply boh sides of he DE by his inegraing facor, and he lef side is an exac derivaive. We herefore have: [M xy e i ] γ(s) ds = iγm B 1 ()e i γg(s) x ds. Inegraing: M xy e i γ(s) ds = im M xy = im e i γ(s) ds = im = im γb 1 ()e i γg(s) x ds γb 1 ()e i γg(s) x ds γb 1 ()e i γg(s) x ds e i γg(s) x ds. Combining limis: γb 1 ()e i γg(s) x ds If we now define a spaial frequency variable k() = γ G(s) ds, we can re-wrie (2) as somehing more pleasan. By wriing he exponenial facor as

3 3 an inegral over a hree-dimensional dela funcion, inerchanging he order of inegraion, and defining a new funcion p(k) = B 1 ()δ 3 (k() k) show ha he magneisaion response o an RF pulse is herefore M xy (x) = iγm p(k)e ix k dk (3) and sae wha his ells you abou he relaionship beween an RF pulse and is effec on maer. Answer: Le us define he spaial frequency variable k() by k() = γ hen we can re-wrie hings in erms of k: G(s) ds M xy = iγm B 1 ()e ik() x Now he funcion k paramerically describes a pah hrough spaial frequency space. We can express he exponenial as an inegral over k space by using he Dirac dela funcion, which, as you will recall, les us make subsiuions like f(a) = f(x)δ(x a) dx: M xy = iγm B 1 () δ 3 (k() k)e ik x dk We can now inerchange he order of inegraion freely (believe i or no, he dela funcion is differeniable and coninuous): { } M xy = iγm B 1 ()δ 3 (k() k) e ik x dk and define he funcion p(k) = such ha he overall response is M xy (x) = iγm B 1 ()δ 3 (k() k) p(k)e ix k dk. his makes he Fourier relaionship really bloody obvious, and we have derived he exciaion k-space formalism. Woo. he funcion p shows he explici weighing of exciaion k-space by he RF exciaion B 1 () and he rajecory aken.

4 4 5 Frequency Encoding Gradiens e) ** Consider a series of spins arranged on a one-dimensional laice a locaions x i. Le he densiy of spins a each sie be ρ i. We are now going o explore wha happens during a prephasing gradien and under a frequency encoding gradien, shown schemaically below. G FE i) When a prephasing gradien G F E,p () is played, wha will he precession frequency a each sie be? Answer: γ(b + G F E,P ()x i ) ii) In he roaing reference frame, show ha he phase accrued by spin i due o his gradien is given by φ p (x i, ) = γx i G F E,p () 2πx i k x,p where is he duraion of he prephasing gradien. Define k x,p. Answer: In he roaing reference frame B vanishes from he above equaion. We know ha ω = dφ, so i hen naurally follows ha φ p (x i, ) = γx i G F E,p () 2πx i k x,p wih k x,p = (γ/2π) G F E,p() being he k-space offse caused by he prephasing gradien. iii) he NMR signal S x,p observed is he sum of spin densiies, wih weighs given by he phase. Wrie down his sum. Answer: S x,p = n ρ xj e iφp(x i, ) j=1 iv) If we le he number of spins become coninuous, his sum generalises o S x,p = ρ(x)e iφp(x, ) dx where ρ(x) and φ p (x, ) are he coninuous analogies of hose funcions defined as above. Le us now see wha happens when we play a frequency encoding gradien lobe G F E () a his poin in ime. Show ha he NMR signal as a funcion of ime becomes

5 5 S() = ρ(x)e i(φ(x,) φp(x, )) dx (4) wih φ(x, ) defined analogously for he frequency encoding gradien as o he prephasing gradien. Answer: he only difference is ha now we sar acquiring phase (a an increasing rae) in he oher direcion, caused by he readou gradien. Everyhing is exacly he same, excep ha how much phase we ve go is a funcion of ime. his leads insanly o he equaion given. v) Wha happens when (φ(x, ) φ p (x, )) =? Answer: Everyhing is in phase, and he signal is jus he inegral of spin densiy in oher words, we have an echo. vi) As he prephasing and frequency encoding gradiens have opposie signs, show ha his happens a a ime echo defined by echo G F E,p () = G F E () Answer: A echo, he requiremen is for (φ(x, ) φ p (x)) =. Going back o wha acually generaes he phases, we see ha his is jus he same hing as echo G F E,p () = G F E () his defines he relaionship beween echo ime and he area under gradien lobes. Addiionally, (4) illusraes ha he observed signal is relaed o he Fourier ransform of spin densiy.