k y 2k y,max k x 2k x,max

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1 EE225E/BIOE265 Spring 2013 Principles of MRI Miki Lustig Assignment 5 Solutions Due March 6th Finish reading Nishimura Ch For the 16 turn spiral trajectory, plotted below, what is the a) Spatial resolution, and b) given that k x,max = k y,max = 2.5 cycles/cm. Assume that the sampling rate along the spiral trajectory is not limiting. k y 2k y,max k x 2k x,max Solution: The resolution and are the same in each dimension, so let W k = W kx = W ky, and δ = δ x = δ y, etc. a) The k-space extent is This gives a spatial resolution of W k 2k max = (2)(2.5 cycles/cm) = 5 cycles/cm δ = 1 W k = 1 = 0.2 cm = 2 mm. 5 cycles/cm b) The is determined by the sampling interval in the radial direction. This is k = 2k max /(2 16) = (5 cycles/cm)/32 = 0.16 cycles/cm where we have used the fact that an N turn spiral crosses (samples) any diameter 2N times. The F OV is then F OV = 1 δ = 1 = 6.4 cm cycles/cm 1

2 3. Consider the 2DFT pulse sequence, shown below on the left, with the following timing RF G z G y 2 ms G x 4 ms A/D The amplitude of the readout gradient is 0.94 G/cm, as is the maximum of the phase encode gradient. Also, 256 samples are acquired during the readout, and 256 phase encode steps are used. The initial magnetization is fully relaxed for each acquisition. The RF pulse is a +90 rotation about +x. The object being imaged is shown above on the right. It is a triangle that is larger than half the in each dimension. The reconstructed image is the magnitude of the inverse FT of the sampled data. (a) What are the resolution and of the pulse sequence? Solution:The k-space extent in the readout direction is W kx = γ 2π G xτ x = (4.257 khz/g)(0.94 G/cm)(4 ms) = 16 cycles/cm This corresponds to a resolution The sampling in k x is δ x = 1 W kx = 1 = cm 16 cycles cm k x = W kx N r = 16 cycles/cm 256 = cycles/cm and the F OV in x is then F OV x = 1 k x = Along the y axis, the k-space extent is 1 = 16 cm cycles/cm W ky = 2k y,max = γ 2π G yτ y = 2(4.257 khz/g)(0.94 G/cm)(2 ms) = 16 cycles/cm which is the same as in the x dimension, so the resolution is again δ y = cm Since the number of phase encodes is the same as the number of readout samples, the is also the same in x and y, so F OV y = 16 cm. 2

3 (b) Sketch the image that would be produced if we just used the even numbered phase encodes. Solution: This only effects the y dimension. If we only use the even numbered phase encodes, the k-space extent remains the same, so the resolution δ y stays the same. The spacing of the phase encodes is now twice as large so that the new F OV y,n is half of what it was. k y,n = 2 k y F OV y,n = 1 k y,n = 1 2 k y = F OV y /2 Since F OV y is now half as big, the image replicas are twice as close, and overlap the original. If we reconstruct using the acquired even phase encodes, and zero out the odd phase encodes, we get the image of the left. If we reconstruct using N p /2 phase encodes, the result is the image on the right. /2 /2 (c) What does the image look like if we doubled the x gradient, and used the original phase encode gradient? Solution: This only effects the x dimension. If we double G x, then W kx = γ 2π G xτ x is also doubled. This means that δ x = 1 W kx is halved. The pixels are twice as small in x. Since we ve kept the same number of samples, the in x has also been halved. Since the readout direction is the axis where we have the anti-aliasing filter, there is no aliasing. If we reconstruct the image using a 256x256 2DFT the image looks like the one on the left below. If we compensate for the pixel size, we get the image on the right. /2 /2 3

4 (d) What does the image look like if we doubled the maximum y gradient, and used the original readout gradient? Solution: This only effects the y dimension. Doubling G y doubles W ky, and halves δ y. In addition the k-space sample spacing k y,n is doubled, so the F OV y is halved. If we reconstruct by doing a 256x256 2DFT, and don t compensate for pixel size, the result is the image on the left. If we do compensated for the smaller pixel size in y, we get the image on the right. /2 /2 (e) Assume that the sign of the RF is alternated every other phase encode, so that it produces a +90 rotation about the +x axis on the even phase encodes, and a 90 rotation on the odd phase encodes. We reconstruct as usual with an inverse FT. What does the image look like? Solution: There are many ways to look at this. The alternating RF pulse results in the signal for every other phase encode being multiplied byt ±1, so that the new acquired data is M xy,n (u k x, v k y ) = M xy (u k x, v k y )( 1) v where M xy (u, v) is the original sampled data, and M xy,n (u, v) is the new data. The new data is the original data multiplied by ( 1) v, where v is the integer index of the phase encode. One approach is to use the modulation theorem. Multiplication in the spatial frequency domain goes to convolution in the image domain, so the result will be the original image convolved with the 2D inverse Fourier transform of ( 1) v. We can also just work through the problem directly, which is what we will do here. As we showed in class, M xy,n (aδ x, bδ y ) = M xy (u k x, v k y )( 1) v e j2π(au/nr+bv/np) u v where u is summed from N r /2 + 1 to N r /2, and v is summed from N p /2 + 1 to N p /2. If we write ( 1) v as e ivπ, and collect terms, M xy,n (aδ x, bδ y ) = ( ) M xy (u k x, v k y )e jvπ e j2π au Nr + bv Np u v = ( ) M xy (u k x, v k y )e j2π( v 2 ) e j2π au Nr + bv Np u v = ( ( ) ) M xy (u k x, v k y )e j2π au Nr + b Np + 1 v 2 u v 4

5 = ( ) M xy (u k x, v k y )e j2π au Nr + (b+np/2)v Np u v = M xy,n (aδ x, (b + N p /2)δ y ) The new reconstructed image is the original image shifted by N p /2 pixels in y, or one-half of the. Note that since we are sampled in spatial frequency, the image domain is perioidic. As we shift the object out of the in one dimension, the next replicated image comes in from the opposite direction. The image that results is shown below: (f) Now consider the case where we are using the original acquisition gradients, but are imaging sodium, which has a γ/2π of khz/g. What is the resolution and? What does the image now look like? Solution: Since γ is now small by a factor of 4.257/1.126 = 3.78, almost 4, the same gradient waveform produces one quarter the spatial frequency encoding. This means that W kx and W ky are both reduced to one quarter of the original, and that the resolution element δ x and δ y are now both four times as large. The k-space sampling is also reduce to one quarter the original, so the is now four times as large. The result is an image that looks like this:

6 (16%) 2. Consider a conventional 2DFT sequence that produces the image 4. Artifacts in 2DFT (From Midterm I 2012) left (within the square). Consider a conventional 2DFT sequence that produces the image shown below (within the square). y y displayed! image! matrix x conventional part b) a) If the scan is repeated but with all gradient amplitudes scaled by a factor of 1.25 (this is the only change), sketch the resultant image. Note that the dumb (inflexible) reconstruction computer blindly takes the inverse FFT of the raw data and displays the image matrix. (a) If the scan is repeated but with all gradient amplitudes sc 1.25 (this is the only change), sketch the resultant image. N Since the gradients are scaled by 1.25, the effective is scaled down by 1.25 and the resolution is increased by The object lies at the edge of the, so aliasing will occur in the phase encode direction. In the readout direction, the anti-aliasing filter will crop the image. The result is displayed below: (inflexible) reconstruction computer blindly takes the inver data and displays the image matrix. (b) Another 2DFT sequence is applied and the image shown abo Draw the resulting image here: displayed. Explain what y might have changed in the pulse seq to the conventional sequence. displayed image matrix x 6

7 b) Now, the scan is repeated with the phase encode gradients turned off (this is the only change), sketch the resultant image. Again assume a dumb reconstruction computer. Since the phase encodes are off, all of the readouts will be the same. This means that there is no variation in the y direction and so the reconstruction is going to show signal only on the x axis. The intensity of the signal on the x axis is going to be the Fourier transform of a single readout through the center of k-space. This turns out of course to be a projection through the object (integrating across y). The result is shown below: Draw the resulting image here: y displayed image matrix x c) Now, the scan is repeated, but the imaging plane is rotated counterclockwise by 45, sketch the resultant image. Again assume a dumb reconstruction computer. Note, that only the imaging plane is rotated, not the sample!. When the imaging plane is rotated, it is like rotating the square box 45 degrees counterclockwise, while keeping the image the same. In that case, part of the object will be outside of the in the readout direction, and will be cropped. The result is below: Draw the resulting image here: The imaging plane y The displayed reconstructed image y x x readout readout 7

8 d) Another 2DFT sequence is applied and the image shown below gets displayed. Explain what might have changed in the pulse sequence as compared to the conventional sequence. y displayed image matrix x The image seems to be squeezed in the phase encode direction, or in fact there s more in y. The image is also squares, so the amount of phase encodes and feequency encodes has not changed. This means that the gradients in y were scaled by a factor of 1/2. This will result in double the in y, and half the resolution which will result in the figure above. Phase encode gradients are scaled by 1/2. 8

9 5. The Rings Trajectory (from midterm I 2012) 1 Consider the following pulse sequence : Each repetition is supposed to trace a single angular ring trajectory in k-space. The trajectory is designed by determining the parameters for the outmost ring and then scaling the gradients to trace the inter rings. We would like to use it to scan a circular object with a of 25.6cm at a spatial resolution of 1mm. For this question assume that the sampling interval is T = 4µs and the maximum gradient amplitude is limited to G x, G y < 4 G/cm. Recall sin(at)dt = 1 a cos(at), and cos(at)dt = 1 a sin(at). a) What are W k, the extent of the trajectory in k-space, and k, the minimum required spacing between samples in k-space? The extent of k-space is determined by the resolution, which is W k = 1/δ = 1/0.1 = 10 cm 1. the sampling spacing is determined by the and is k = 1/ = 10/256 cm 1. W k = 10cm 1 k = 10/256 cm 1 1 This is a tribute to Holden Wu, who s PhD thesis was on the ring trajectory 9

10 b) For the outmost ring, what are T P E, G P E, G RO, and T RO that result in the fastest scan that does not violate the and gradient amplitude constraints? The outmost ring has a diameter of 10cm 1, therefore its radius is 5cm 1. This is also the area of the prewinder. Since we do not collect data in the pre winder we can go as fast as possible, so we can set G P E = 4G/cm. From the area requirement γ 2π G P ET P E = 5 we get, T P E = 0.294ms. From part (a), the maximum sampling distance is k = 10/256cm 1. This puts a restriction on G RO such that γ 2π G RO T < k, which results in G RO = 2.294G/cm. Recall that the k-space radius of the outmost ring is 5cm 1, and that G x (t) = G RO sin( 2π T RO t ). Therefore, k x (t ) = t 0 γ 2π G RO sin( 2π T RO τ)dτ = γ 2π γ T RO 2π 2π G RO = 5cm 1 and T RO = 3.217ms T RO 2π G RO cos( 2π T RO t ). From this, we get that T P E = 0.294ms G P E = 4G/cm G RO = 2.294G/cm T RO = 3.217ms c) How many rings, N, are required to cover k-space? Short answer: The outermost ring radius is 5cm 1 and the distance between rings should be less than k = This results in N = 5/ = 128. Long answer: One problem with this is that the inner most ring will have radius of k and therefore the inter ring would violate Nyquist. In that case we need to add 1 more sample at DC. This is however quite wasteful, so instead the inter most ring diameter can be set to k. N = 128 would result in the our most ring being 5 k/2. Both N = 128 and N = 129 are good answers. N = 128 or, N =

11 d) You scan a point object located at x = 10cm. Unfortunately, due to systematic errors, the start and end of the A/D window is delayed by 10 samples with respect to the gradient waveforms. Assuming a dumb reconstruction computer that is not aware of the delay, draw is the image that is going to be reconstructed? Comment on the sensitivity of rings to delays. The important thing to notice is that a delay between the A/D window and the gradient would result in a pure rotation in k-space. Since the A/D window is delayed, the result is a counterclockwise rotation. The angle of rotation is θ = 2π T RO 10 T This results in θ = rotation and the object being displaced to x = 10 cos(θ) = 9.97cm and y = 10 sin(θ) = 0.78cm. reconstructed Actual [9.97,0.78] o 11

12 6. 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, r, bwpp, Gmax, Smax, dt); The inputs to the function are : Nf is the number of frequency encodes. r (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!!!!) 12

13 There are many ways of implementing this. Here s one: function [gro,rowin] = genreadoutgradient(nf, r, bwpp, Gmax, Smax, dt); %[gro,rowin] = genreadoutgradient(nf, r, bwpp, Gmax, Smax, dt); gamma = 4257; res = r/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(:);0;gro(:)]; 13

14 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, p, Gmax, Smax, dt); The inputs to the function are : Np is the number of phase encodes. p (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] 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=0 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, p, Gmax, Smax, dt) gamma=4257; kmax = 1/(p/Np)/2; area = kmax/gamma; grpe = mintimegradientarea(area, Gmax, Smax, dt); petable = [Np/2-0.5:-1:-Np/2+0.5]. /(Np/2); 14

15 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 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=15000G/cm/s. Design a hard-pulse RF with 90 degree excitation to use with the gradient sequence. Plot k-space by integrating the gradient waveforms. Make sure it makes sense! Here s the code: Nf = 64; Np = 32; Nrf = 92; r = 14; p = 7; Gmax = 4; Smax = 15000; dt = 4e-6; bwpp = ; gamma = 4257; flip = 90; [gx,rowin] = genreadoutgradient(nf, r, bwpp, Gmax, Smax, dt); [gpe,petable] = genpegradient(np, p, Gmax, Smax, dt); RF90 = ones(nrf,1)*(flip/360)/(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(rf90,rf90*0,dt,100,100,0,dp,0,mx,my,mz); end % simulate Readout [mx1,my1,mz1] = bloch(gx*0,g,dt,100,100,0,dp,2,mx1,my1,mz1); mxy = sum(mx1,2) + i*sum(my1,2); res(n,:) = mxy(rowin); figure, plot(cumsum(g*dt*gamma)); 15

16 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. 16

17 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=100. 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-6,100,100,0,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 250Khz (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. 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! The resulting image is: 17

18 e. Reduce the by a factor of 1.5 in the phase encode and repeat the scan. What do you get? submit the image. 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: 18