EE225E/BIOE265 Spring 213 Principles of MRI Miki Lustig Assignment 3 1 Finish reading Ch 4 2 Nishimura, Q 33 Solutions: 2D circularly symmetric objects can be expressed as m(r) and, G r = db z dr, r = x 2 + y 2 a) We are told that the gradient dbz dr is radial The field at a particular r distance away will be G r r (on a concentric ring of radius r) We can therefore write the signal equation in radial coordinates to integrate over all space (don t forget the extra r due to the Jacobian when transforming to radial coordinates): b) We are given, s(t) = define: = = 2π = 2π r 2π r iγr t m(r)e r k r = γ 2π r m(r)e iγr t Gr(τ)dτ rdrdθ 2π Gr(τ)dτ rdr m(r)e iγr t Gr(τ)dτ rdr t G r (τ)dτ m(r)e i2πr kr(t) rdr = 2πF {m(r) r} kr(t) δ(r R), G r (t) = G r, t > δ(r R)is an impulse ring that only has value when r = R From part (a) our baseband signal is: r 2πF {m(r) r} kr(t) = 2π δ(r R)e iγgr t r rdr = 2πRe iγgrtr where we used the sifting property of the dirac delta function c) We can represent a circular pillbox using a circ function circ(r) is a circular function with unit diameter r < 1/2 and unit amplitude In the general case of diameter D, m(r) = circ(r/d) dθ 1
From part (a) the baseband signal is s(t) = 2πF {circ(r/d)r} so, S(f) = F {s(t)} = F {2πF {circ(r/d)r}} = 2πcirc(r/D)r The frequency of the signal should map to radial position The amount of magnetization at each r position depends on 2πr So, to reconstruct, m(r) = S(f = γ/2πg rr) 2πr 2
3 Precession in Tops (brain teasers) (a) Explain why only spinning tops precess That is, explain why static tops will not ic not corcr c araf\ 3 wi m1ct L* L A1 5 oek f o Sfqt\c, op Di (t dvc&?p;ft aer t¾55 (fl 4 yv-t + (or9 r 4 K ink e ft rf j_ L 3, i lop s QnLr (fl)fflfl1aay11 s LS 9-4 - I L ( fctft ri) dl -7% L = => I Mfti 4te 3c4Qhbcij éd - ( ) QjAf h- t v,mycqf / 7 L -Al 3 -) -1-9 1 aror imm t 4! 4 -# ;it y x (b) Suppose you wanted to create an approximate macro version of the precessing quantum nuclear 3
magnetic moment Dr Bore thinks you could do this by embedding a permanent magnetic (PM) moment within a toy top Would you embed the PM perpendicular or parallel to the rotation axis? If you altered the applied external magnetic field, do you believe you would alter the mechanical precession frequency of the top that is, the precession would change from that determined merely by gravity? Could you create both positive and negative gyromagnetic ratios? Solution: To approximate a spin, the magnetic moment must be aligned with the angular momentum Therefore, the magnet should be aligned with the top Altering an external field can either contribute to, or subtract from the gravitational force The magnetic field can be altered to contradict the effect of gravity, thus stopping the precession If the magnetic field becomes a dominant factor, precession will change direction, effectively altering the gyromagnetic ratio 4
4 Magnetic field steps Suppose you placed two test tubes (each 5 ml) at 15 T and one test tube (5 ml) at 16 T Suppose both coils were visible with equal sensitivity to the RF coil Sketch the intensity of the received signal as a function of frequency 51 x1 G7 6 to Iiqur& out aè ih& h& Ht pthns t5n%, kjt3 U boj (ie tbt naifl Zb7 MH = 7 PAH 439 MH Li MF{?\othm 1sh vcus [ki *4NQte, üot & * g *tq 7, a& I5f will tr 5
5 Gradient field bandwidth Sketch the spectrum of received signal on a 15 T MRI scanner versus frequency for an object of diameter = 25 cm with a gradient field of peak ampitude 4 G/cm 51 x1 G7 6 to Iiqur& out aè ih& h& Ht pthns t5n%, kjt3 U boj (ie tbt naifl Zb7 MH = 7 PAH 439 MH Li MF{?\othm 1sh vcus [ki *4NQte, üot & * g *tq 7, a& I5f will tr 6
6 RF Excitation If we apply an RF waveform in addition to a gradient G, we can excite a slice through an object Assume that the RF envelope B 1 (t) is the 6 ms segment of a sinc(), as shown 1 1 2 3 4 5 6 t, ms Write an expression for B 1 (t), and its Fourier transform Assuming a gradient G of 94 G/cm, how wide is the excited slice? Solution: The RF waveform may be written where times are in ms The Fourier transform is B 1 (t) = sinc(2(t 3))rect((t 3)/6) F{B 1 (t)} = e j2π3f [ 1 2 rect(f/2) 6 sinc(6f) ] The phase term doesn t effect the slice width, so we ll ignore it The Fourier transform is a 2 khz wide rect convolved with a 1/6 khz wide sinc The result will be about 2 khz wide This is illustrated below, where the normalized functions are plotted, 15 rect(f/2) 15 sinc(6f) 15 rect(f/2)*6sinc(6f) 1 1 1 5 5 5 5 4 2 2 4 f, khz 5 4 2 2 4 f, khz 5 4 2 2 4 f, khz To find the width of the slice this pulse excites, first note that a gradient of 94 G/cm produces a linear frequency variation of γ G = (4257 khz/g)(94 G) = 4 khz/cm 2π The RF pulse has a spectrum that is 2 khz wide, so the excited slice will be in width This is a typical slice width 2 khz/(4 khz/cm) = 5 cm 7
7 Design of Time-Optimal Gradient Waveforms A key element in pulse sequence design in MRI is the design of the gradient waveforms A very common problem is to design a gradient waveform with a certain desired area To minimize the duration of the sequence, we often would like the waveform to be as short as possible However, the gradient pulse has to be realizable by the system and therefore must satisfy the system constraints of maximum gradient amplitude and slew-rate Solution: A minimum time solution is always either limited by the slew-rate or by the maximum gradient amplitude There are two cases we should consider depending on the desired gradient area: 2 area=gmax/smax 2 area>gmax/smax 2 area<gmax/smax t= 2Gmax / Smax When the desired area is smaller than G2 max then the gradient magnitude is never maxed out, so the solution is a triangle For an area larger than G2 max the solution is a trapezoid t1 t2 t3 G=[area Smax] 1/2 t=2[area/smax] 1/2 Δt= Gmax / Smax Δt= area/gmax - Gmax/Smax Δt= Gmax / Smax For a triangle we define t 1 = area and get: G(t) = For a trapezoid we define t 1 = Gmax, t 2 = area G max { Smax t, t t 1 2 area t, t 1 t 2t 1 and t 3 = area G max + Gmax and get, t, t t 1 G(t) = G ( max ), t 1 t t 2 area G max + Gmax t, t 2 t t 3 Given that the system is limited to maximum gradient amplitude G max = 4G/cm and a slew-rate of = dg(t) dt = 15G/cm/s (a) Find the shortest gradient waveform that has an area of t G(τ)dτ =8e-4 G*s/cm Draw the waveform Point out the maximum gradient, and its duration What is the shape of the waveform? 8
Solution: For the parameters we get that the desired area is smaller than G2 max =167e-4, so the solution is a triangle The maximum gradient is G = area = 3464G/cm and the duration is T = 2 area = 462ms (b) Find the shortest gradient waveform that has an area of t G(τ)dτ =16e-4 G*s/cm Draw the waveform Point out the maximum gradient, and its duration What is the shape of the waveform? Solution: Now, the desired area is bigger than G2 max =167e-4, so the waveform is a trapezoid with maximum gradient of 4G/cm and a duration of T = area G max + Gmax = 6667ms (c) Matlab assignment: In this part, we will write a matlab function to design minimum-time gradient waveforms This function will be used later in class, so make sure you get it right Write a function that accepts the desired gradient area (in G*s/cm), the maximum gradient amplitude (in G/cm), the maximum slew-rate (in G/cm/s) and sampling interval (in s) The function will return (a discrete) shortest gradient waveforms that satisfy the constraints: g = mintimegradientarea(area, Gmax, Smax, dt); Solution: There are many ways to implement this Here s an example of sampling the continuous function of the analytic gradient waveform solution function g = mintimegradientarea(area, Gmax, Smax, dt) % g = mintimegradientarea(area, Gmax, Smax, dt) A_triang = Gmax^2/Smax; if area <= A_triang disp( Triangle ) t1 = sqrt(area/smax); T = 2*t1; N = floor(t/dt); t = [1:N] *dt; idx1 = find(t < t1); idx2 = find(t >= t1); g = zeros(n,1); g(idx1) = Smax*t(idx1); g(idx2) = 2*sqrt(area*Smax)-Smax*t(idx2); else 9
end disp( Trapezoid ) t1 = Gmax/Smax; t2 = area/gmax; t3 = area/gmax + Gmax/Smax; T = t3; N = floor(t/dt); t = [1:N] *dt; idx1 = find(t < t1); idx2 = find((t>=t1) & (t < t2)); idx3 = find(t>=t2); g = zeros(n,1); g(idx1) = Smax*t(idx1); g(idx2) = Gmax; g(idx3) = (area/gmax + Gmax/Smax)*Smax - Smax*t(idx3); disp(sprintf( Maximum gradient:%fg/cm\nduration:%fms\n,max(g(:)),t(end)*1)); Plot the result of the function for area=6e-4, Gmax=4, Smax = 15, dt=4e-6; Plot the result of the function for area=6e-4, Gmax=1, Smax = 5, dt=4e-6; Solution: As seen below, gradients with higher slew-rate and maximum amplitude can produce the same gradient area faster 1