EE225E/BIOE265 Spring 2011 Principles of MRI. Assignment 5. Solutions

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1 EE225E/BIOE265 Spring 211 Principles of MRI Miki Lustig Handout Assignment 5 Solutions 1. Matlab Exercise: 2DFT Pulse sequence design. In this assignment we will write functions to design a 2DFT pulse sequence, and then simulate the design on a Bloch simulator. The first step is to design a readout gradient. The readout gradient is composed of a prewinder, and a readout part. In the readout, we are interested in having a portion of the gradient that will scan the desired k-space length (gradient area) in which the gradient waveform is constant. This will give us a steady linear scan in k-space. For the prewinder, we are only interested in generating a gradient area that is half the area of the readout part. This should be as fast as possible to minimize the scan time. In addition, the ramps for the readout part should also be as fast as possible. a. Write a function genreadoutgradient.m that designs a readout gradient given the sequence parameters and the system constraints. >> [gro,rowin] = genreadoutgradient(nf, FOVr, bwpp, Gmax, Smax, dt); The inputs to the function are : Nf is the number of frequency encodes. FOVr (in cm) is the desired field-of-view. bwpp (in Hz/pixel) is the desired bandwidth per pixel Gmax (in Gauss/cm) is the maximum gradient. Smax (in Gauss/cm/s) is the maximum slew-rate. dt (in s) is the duration for each sample. The outputs of the function are: gro - an array containing the gradient waveform. rowin - an array containing the indexes in gro that correspond to the readout portion of the gradient. This will be used to crop the interesting part of k-space for reconstruction. bwpp is something we have not discussed before. It basically defines the gradient amplitude we are going to use during the flat portion of the readout gradient. In essence, bwpp = γ 2π G F OV Nf. Now, the A/D has a sampling bandwidth of 1/dt. So, the effective number of digital readout samples may be higher than our desired Nf frequency encodes. This is OK, since after we get all the samples, we will filter them to a bandwidth of Nf*bwpp and subsample to get Nf samples. (Hint: You should first design a trapezoid that meets the criteria and then use the minimumtime-gradient function you wrote in previous homework to deign the prewinder. Remember to compensate for the ramp of the readout in the prewinder!!!!) 1

2 Solutions: There are many ways of implementing this. Here s one: function [gro,rowin] = genreadoutgradient(nf, FOVr, bwpp, Gmax, Smax, dt); %[gro,rowin] = genreadoutgradient(nf, FOVr, bwpp, Gmax, Smax, dt); gamma = 4257; res = FOVr/Nf; Wkx = 1/res; area = Wkx/gamma; G = bwpp/res/gamma; Tro = Wkx/gamma/G; Tramp = G/Smax; t1 = Tramp; t2 = t1+tro; T = Tramp*2+Tro; N = floor(t/dt); t = [1:N] *dt; idx1 = find(t < t1); idx2 = find((t>=t1) & (t < t2)); idx3 = find(t>=t2); gro = zeros(n,1); gro(idx1) = Smax*t(idx1); gro(idx2) = G; gro(idx3) = T*Smax - Smax*t(idx3); areatrapz = (T+Tro)*G/2; % area of readout trapezoid gpre = mintimegradientarea(areatrapz/2, Gmax, Smax, dt); rowin = length(gpre) idx2; gro = [-gpre(:);;gro(:)]; 2

3 b. To design a phase encode gradient, we only need to design the gradient for the largest phaseencode and then scale it accordingly for the others. Write a function genpegradient.m that designs the gradient phase encode gradient for the largest phase encode and a phase encode table to scale it. >> [grpe, petable] = genpegradient(np, FOVp, Gmax, Smax, dt); The inputs to the function are : Np is the number of phase encodes. FOVp (in cm) is the desired phase-encode field-of-view. Gmax (in Gauss/cm) is the maximum gradient. Smax (in Gauss/cm/s) is the maximum slew-rate. dt (in s) is the duration for each sample. The outputs of the function are: grpe - an array containing the gradient waveform. petable - Npe x 1 array containing the phase encode table to scale the phase-encode gradient for each phase-encode. The array entries should be bounded between [-1 : 1] Solutions: There are several choices how to distribute the phase encodes. In this case, I chose to distribute them such that there isn t a phase encode with amplitude zero. This way, k-space is sampled around the DC line and kx=ky= is not sampled. This is often done in practice to improve the dynamic range, since the DC point has a very high amplitude. The code uses the mintimegradientarea function from previous homework. Here s the code: function [grpe, petable] = genpegradient(np, FOVp, Gmax, Smax, dt) gamma=4257; kmax = 1/(FOVp/Np)/2; area = kmax/gamma; grpe = mintimegradientarea(area, Gmax, Smax, dt); petable = [Np/2-.5:-1:-Np/2+.5]. /(Np/2); 3

4 Now that we have a way to generate the waveforms of a 2DFT sequence, we will simulate such a sequence for a distribution of spins. Download the file hw5 img.mat from the class website. This file contains the arrays dp [7715x2], mx [7715x1], my[7715x1], and mz[7715x1],. These array represents the positions of 7715 spins in space and their magnetization. We will now image them! c. Design a 2DFT sequence with readout/phase-encode FOV of 14/7 cm, Nf/Np of 64/32 (giving a resolution of 2.2mm. Use a bandwidth per pixel of about Khz/cm. Use dt = 4µs, Gmax=4G/cm and Smax=15G/cm/s. Design a hard-pulse RF with 9 degree excitation to use with the gradient sequence. Plot k-space by integrating the gradient waveforms. Make sure it makes sense! Solutions: Here s the code: Nf = 64; Np = 32; Nrf = 92; FOVr = 14; FOVp = 7; Gmax = 4; Smax = 15; dt = 4e-6; bwpp = ; gamma = 4257; flip = 9; [gx,rowin] = genreadoutgradient(nf, FOVr, bwpp, Gmax, Smax, dt); [gpe,petable] = genpegradient(np, FOVp, Gmax, Smax, dt); RF9 = ones(nrf,1)*(flip/36)/(nrf*dt*gamma); gy = zeros(length(gx),1); gy(1:length(gpe)) = gpe; load hw5_img.mat G = []; res = zeros(np,length(rowin)); for n=1:np g = [gx(:),gy(:)*(-petable(n))]; G(:,n) = g*[1;i]; % simulate Excitation first [mx1,my1,mz1] = bloch(rf9,rf9*,dt,1,1,,dp,,mx,my,mz); end % simulate Readout [mx1,my1,mz1] = bloch(gx*,g,dt,1,1,,dp,2,mx1,my1,mz1); mxy = sum(mx1,2) + i*sum(my1,2); res(n,:) = mxy(rowin); figure, plot(cumsum(g*dt*gamma)); 4

5 figure, plot(real(g), b ); hold on, plot(imag(g), r ); im = crop(ifft2c(res),[np,nf]); figure, imshow(abs(im),[]); Here s the resulting gradient waveform: And the k-space trajectory: DC line is not sampled Samling the DC line is also fine. 5

6 d. Simulation: Simulate the sequence acquisition using the Bloch simulator one phase encode at a time (The simulation takes about 1-2sec per phase encode). Use T1=T2=1. The output of the simulator needs to be integrated across all the spins to get the signal. The code for the simulation part should look like: >>... >> g = [gro, gpe*phtable(n)]; >> [mx,my,mz] = bloch(rf,g,4e-6,1,1,,dp,2,mx,my,mz); >> mxy = sum(mx,2) + sqrt(-1)*sum(my,2); >>... You will find that the number of readout samples is bigger than Nr because we sampled at 25Khz (dt = 4µs). Scanners also sample at that rate and then apply a digital filter to get the desired number of readout points. There are two options, to do the filtering and subsampling of k-space or cropping the image. For simplicity, we will take the 2nd approach. take a 2D centered IFFT of the resulting k-space data. Crop the image to the desired FOV. You should be able to read something. If you can t,... then something is wrong. Submit a plot of the gradient waveforms and the image. Enjoy! Solutions: The resulting image is: 6

7 e. Reduce the FOV by a factor of 1.5 in the phase encode and repeat the scan. What do you get? submit the image. Solutions: In this case, the gradient area for the phase encode is twice as big. We need to make sure that the it does not overlap with the readout portion. Fortunately it doesn t. The resulting image is aliased. The polarity or the phase of the alias depends on the exact phase encode scheme ( if the DC line is sampled or not). If the DC line is sampled, the alias is positive. If it is not, then it is negative. Here it is negative, so some of the signal cancels. And here s when DC line is sampled: 7

8 2. Nishimura 6.5 Solution: This question demonstrates a very cool way of storing encoding information in M z. The advantage in storing information in the longitudal direction is the long relaxation time of T 1, which is usually much longer than T 2. This way, encoded information can be stored for long time periods before decoded. Since we ignore relaxation, the solution of the Bloch equation is a series of rotations. The first RF pulse produces a 9 degree rotation around the x axis pulse. The z gradient produces a z dependent rotation around the z axis, α = γg z zt. Finally, the 2nd RF pulse produces a 9 degree rotation around the x axis. The series of rotations is: M = R x ( 9)R z (γg z zt )R x (9)M So, if the magnetization starts at M = [,, M z ] T then after the first RF we get, After the gradient we get, After the final RF we get, cos(γg z zt ) sin(γg z zt ) sin(γg z zt ) cos(γg z zt ) M z M z sin(γg z zt ) M z cos(γg z zt ) = M z = M z =. M z sin(γg z zt ) M z cos(γg z zt ) M z sin(γg z zt ) M z cos(γg z zt ) So, M z is cosine modulated in z. This is in fact encoding the real part of a phase-encode in M z! Here is a SpinBench simulation plot, where the gradient produced a phase of 1 cycles/cm (γg z T = 1).. Gx Gy Gz RF 1 cycles/cm Mz Time ZPosition (mm) 8

9 3. If we use a 9 degree excitation for imaging, all of the initial M z magnetization is used up by each excitation, and we have to wait for it to rebuild before we can acquire each phase encode. An alternative is to use a sequence of small-tip-angle excitations θ i for i = 1 : N. The first pulse θ 1 produces a transverse magnetization M xy,1 = M sin θ 1 and leaves a longitudinal magnetization M z,1 = M cos θ 1 After a short interval T R, the repetition time, the transverse magnetization has decayed away, but M z,1 remains (T 1 is long compared to T R, and T 2 is short compared to T R ). We can then apply a second pulse θ 2 to create a second transverse magnetization leaving a second longitudinal magnetization M xy,2 = M z,1 sin θ 2 M z,2 = M z,1 cos θ 2. We continue this way for N pulses until M z,n =, and all of the magnetization has been used up. a) Show the remarkable result that we can choose θ i so that each transverse magnetization is M xy,i = M N. For example, with nine pulses, we could produce nine transverse magnetizations each of amplitude of M /3. This seems like three times more magnetization that we are entitled to! Solution: The easiest way to demonstrate this is to assume it is true, and work backwards from the final magnetization. For convenience we ll assume M = 1 since it is just a scaling constant. At the end the transverse mangentization M xy,n = 1/N, and the longitudinal magentization M z,n = (all of the magnetization has been used up). The magnetization just before the n th pulse was then M z,n 1 = 1/N. The transverse magnetization is M xy,n = 1/N, the same as all of the others. The total magnetization is then M n 1 = ( M xy,n Mz,n 1 2 = 2/N. This was along +z before the n 1 pulse, so M z,n 2 = / 2/N. We can then work backwords to find that M z, = / N/N = 1, which is consistent with the initial conditions. In general the magnetization before the n th pulse is M n = ( ) T,, (N (n 1))/N and after the n th pulse is M + n = ( ) T, 1/N, (N n)/n A plot of the trajectory of the magnetization during this sequence of pulses is shown below. 9

10 N/N z 2/N 1/N 1/N y where the curved arcs are the excitation pulses, and the straight lines back to the z axis are due to T 2 relaxation. b) Find the θ i sequence that produces this sequence of transverse magnetizations. Solution: The flip angle to apply for the n t h pulse is the angle of the magnetization after the n th pulse, or ( θ n = tan 1 1/ ) ( ) N = tan 1 1. N n/ N N n 1