Assignment 2 Solutions

Transcription

1 EE225E/BIOE265 Spring 2012 Principles of MRI Miki Lustig Assignment 2 Solutions 1. Read Nishimura Ch Non-Uniform Sampling. A student has an assignment to monitor the level of Hetch-Hetchi reservoir in Yosemite National Park. In order to meet the Nyquist rate, he has to drive and measure the water level at least once a day. Driving every day for 3 hours one way is no fun, so he came up with a new plan. In the new plan he visits the reservoir every 2 days. He arrives at yosemite early in the day and records the level of the reservoir. Then, he goes on a 6 hours hike and before heading back he records the level again. The samples he collects are g 1 [n] = g(nt ) and g 2 [n] = g(nt + 1/4), where T = 2days and g(t) is the continuous water level of the lake. Therefore, ĝ 1 (t) = g(t) 1 T X( t T ) and, ĝ 2 (t) = g(t) 1 T X( t T T ) where T = 1/4 We will show in this question that it is possible to reconstruct the continuous signal, even though the Nyquist rate appears to be violated. However, there are some costs associated with this. We will solve the problem in the frequency domain. Solution: The basic setup in this question is that instead of uniform sampling at the Nyquist rate we have non-uniform sampling, which on-average samples at the required rate (one sample too close, one too far). T ΔT T (a) g(t) is band limited. If T = 1 [days] is the Nyquist rate, what is the highest frequency fnyq? Solution: The highest frequency is (1/T)/2 which is 1/2[day 1 ]. Since we sample at a rate T = 2 [days], both Ĝ1(f) and Ĝ2(f) are periodic with a period 1/2 [day 1 ]. We would like to express recover G(f) from Ĝ1(f) and Ĝ2(f). Since G(f) is band limited between 1/2 < f < 1/2, we need only focus on that interval. 1

2 (b) For the interval 0 f 1/2[days 1 ] express Ĝ1(f) as a function of G(f). Solution: Computing the Fourier transform of ĝ 1 (t) yields: Ĝ 1 (f) = G(f) X(T f) So, G(f) is replicated every 0.5 interval. From the above figure we see that between 0 and 0.5 there are only two replicas. Therefore, Ĝ 1 (f) = G(f) + G(f 1/2) 0 f 1/2 (c) For the interval 0 f 1/2 [days 1 ] express Ĝ2(f) as a function of G(f). Solution: Computing the Fourier transform of ĝ 2 (t) yields: ( ) Ĝ 1 (f) = G(f) e i2πt T f X(T f) In a similar way we get a replication every 0.5 interval, but now each replica will have a different phase. For T = 2, T = 1/4, f = 1/2 we get that the phase of the replica centered at f = 1/2 is e i2πt T f = e iπ/2 = i. From the above figure we see that between 0 and 0.5 there are only two replicas. Therefore, Ĝ 1 (f) = G(f) ig(f 1/2) 0 f 1/2 (d) Use the result from (b)+(c) to find G(f) as a function of Ĝ 1 (f) and Ĝ2(f) for 0 f 1/2 [days 1 ]. 2

3 Solution: We can express the two equations in matrix form: [ ] [ 1 1 G(f) 1 i G(f 1/2) We then get that, [ and G(f) G(f 1/2) ] G(f) = iĝ1(f) + Ĝ2(f) i + 1 ] = = 1 [ i 1 i (e) Repeat (b)-(d) for the interval 1/2 f 0 [days 1 ]. Solution: In a similar way, G(f) = iĝ1(f) + Ĝ2(f) i + 1 [ Ĝ1 (f) Ĝ 2 (f) ] ] [ Ĝ1 (f) Ĝ 2 (f) ] 0 f 1/2 1/2 f 0 (f) In order to recover G(f) you had to invert a set of linear equations. Express the set of equations in (b)-(c) in matrix form. Plot the condition number (use the cond function in Matlab) for different shifts, e.g., 0 < T < 1/2. The condition number is the ratio between the highest and lowest eigen-values and indicates how much noise amplification occurs when inverting the matrix. From the plot, what can you make a conclusion on the cost of non-uniform sampling? Solutions: For T = 1/2 we have uniform sampling at the Nyquist rate and therefore a condition number equal to 1. However, as T is smaller and smaller our condition number worsens and the penalty is sensitivity to noise Condition number vs T Condition number T 3

4 3. Interpolation in k-space: In this problem we will see the effect of linear interpolation in k-space. This is a more involved version of the question that was given in the midterm in The following image was resampled in the vertical direction in k-space by linear interpolation at 3 4 the original rate. Interpolation in k-space corresponds to finer k-space sampling and therefor a larger FOV. So, the FOV of the interpolated image (right) is larger in the vertical dimension. There are two distinct artifacts that appear in the interpolated image and not in the original. The interpolated image is apodized (no longer has flat magnitude). In addition, there are some aliasing artifacts that appear as replication of the image in the interpolated dimension. These artifacts are the result of using a linear interpolation as opposed to a better interpolation scheme. In the following questions we will learn the source of these artifacts through a simple 1D example. You are given the following 1D signal that you want to sample in k-space. In this question, f(x) is the space signal and F (k) is its k-space. f(x) x, [cm] a) You decide to collect 256 k-space samples of F (k) with k = cycles/cm. Express ˆf(x), the result of sampling F (k) in k-space, as a function of f(x). Draw the signal ˆf(x). Emphasize any artifacts in the drawing and point out the differences with the original f(x). Solution: There are going to be two effects. Blurr due to the finite extent in k-space and replication due to the discrete sampling. Since k = and the number of samples is 256 the total extent in k-space is W kz = = 8 cycles/cm. The finite extent can therefore be written 4

5 as a multiplication with the rect function (k/8). The discrete sampling at rate of k = cycles/cm can be written as a multiplication with the Shah function 8 X( 8 k). So, we can write ˆF (k) as, ˆF (k) = F (k) X(256 8 k) (k 8 ). Taking the inverse Fourier transform yields, ˆf(x) = f(x) 8sinc(8x) X( 8 }{{} 256 x) blurr }{{} replication = 8 (f(x) 8sinc(8x)δ(x 32n) 256 n= ^ f(x) blurr ringing 1/ x, [cm] b) You prepare to show the results from part (a) to your adviser, however, you suddenly remember that he asked you to sample k-space at a finer interval of k = cycles/cm. Not wanting to repeat the experiment, you decide to interpolate ˆF (k) and resample it accordingly. Write an expression for F (k), the resampled signal, assuming linear interpolation. Express F (k) as a function of ˆF (k), HINT: { linear interpolation is a convolution with an appropriately scaled triangle function Λ(ak) = 1 ak ak < 1 0 otherwise Solution: 8 Linear interpolation is a convolution with a triangle function. Our samples are spaced at 256 cycles/cm interval, so the convolution is with Λ( k). The result Λ( 8 k) ˆF (k) is a continuous function. In order to resample, we need to multiply the continuous function with an impulse 6 train at 256 cycles/cm interval. Therefore we get that, ( F (k) = Λ( 256 ) 8 k) ˆF 256 (k) 6 X(256 6 k) c) Write an expression for f(x), the image domain representation of F (k) from part (b). First express f(x) as a function of ˆf(x) and then as a function of f(x). Draw f(x), pointing out any artifacts due to the interpolation and resampling. (HINT: First try to draw the effect of the interpolation, then draw the effect of the resampling.) Solution: For this question we need to know that the Fourier transform of a triangle function is, F 1 {Λ(k)} = sinc 2 (x). So, F 1 {Λ(ak)} = 1 a sinc2 (x/a). This can also be derived from the relation Λ(k) = (k) (k). 5

6 So, the convolution and resampling yields, ( 8 f(x) = ˆf(x) 256 sinc2 ( 8 ) 256 x) X( x). and, f(x) = Drawinf ˆf(x) sinc2 ( 8 256x) we get, ~ f(x) (( f(x) 8sinc(8x) X( 8 ) 256 x) sinc2 ( 8 ) 256 x) X( x) x, [cm] in which the replicas have not been fully suppressed and the signal heavily apodized. Convolving the result with X( 6 256x) we get, x, [cm] ~ f(x) aliasing x, [cm] d) Linear interpolation can be considered as interpolating with a poor low-pass filter. How would the result change if we use a filter with 0.1% ripple in the pass-band and stop band for the interpolation? Draw the result pointing out the differences. Solutions: A TBW=8 windowed sinc is a better filter and suppresses the replicas much better as illustrated below: 6

7 ~ f(x) slight apodization x, [cm] The replication of the signal will have negligible aliasing within the field of view! ~ f(x) almost no aliasing x, [cm] 4. Matlab assignment: 2D Fourier transforms: Use the function publish to make your report. In this assignment we will play with 2D Fourier transforms. As mentioned in class, in MRI the (0,0) coordinates both in k-space and of the image are located in the center. The functions fft2 and ifft2 in Matlab place the origin of the axis at the corner of the image. We will therefore need to write new functions that compute the centered DFT. (a) Program the functions F=fft2c(f) and f=ifft2c(f). Recall that >> F = ifftshift(fft2(fftshift(f))); >> f = fftshift(ifft2(ifftshift(f))); (b) Rectangles in k-space. Download the functions mysinc.m and myjinc.m. The functions implement the sinc and jinc function as described in class. They accepts a vector (or matrix) of inputs and return the results in the same format as the input. The Fourier transform of a rectangle with length 18[cm] and amplitude 1 is F (k) = 18 2 sinc(18k x )sinc(18k y ) We would like to sample the rectangle in k-space while achieving a resolution of 1mm and a 25.6cm Field of view. Recall from class that the resolution x = 1 W kx where W kx is the extent in k-space. Also recall that FOV x = 1 k x where k x is the spacing between samples in k-space. For 1mm, we get that W kx = 10 and k x = 10/256. 7

8 To sample the function we will create a grid using the function meshgrid. First, let s create the intervals: >> Wkx = 10; >> Wky = 10; >> Nx = 256; >> Ny = 256; >> kx_itvl = [ -Nx/2:Nx/2-1]/Nx * Wkx; >> ky_itvl = [ -Ny/2:Ny/2-1]/Ny * Wky; Now, we will create the grid and evaluate the function >> [kx, ky] = meshgrid(kx_itvl, ky_itvl); >> F = 18*18*mysinc(18*kx).*mysinc(18*ky); Since the sample lie on a grid, we can reconstruct the data using the inverse fft. In the inverse fft, the data is scaled by 1/(NxNy) = 1/(256 2 ). On the other hand, each point in k-space represents an area of k x k y = 10 2 / So, for the ifft to approximate the inverse Fourier transform we need to multiply the result by a 100. This is however not very important in MRI. The actual scaling of an image in MRI depends on many things and can vary significantly between scans. The contrast in the image is the important information. We will do it here just so the rectangle will have an amplitude of 1. But again, this is not so important. >> f = 100*ifft2c(F); >> figure(1), imshow(abs(f),[]), title( f ) >> figure(2), imshow(angle(f),[]), title( The phase of f ); Validate that the size of the rectangle is 180 pixels. In a similar way we can sample a circle in k-space. Here s an example of a circle with a diameter of 20cm: >> r = sqrt(kx.^2+ky.^2); >> F_circ = 20^2*myjinc(20*r); >> f_circ = 100*ifft2c(F_circ); >> figure(1), imshow(abs(f_circ),[]), title( f ) >> figure(2), imshow(angle(f_circ),[]), title( The phase of f ); To demonstrate the modulation property of the Fourier transform we will shift the k-space of the rectangle by 2 k x. We should expect 2 cycles of phase across the FOV in the x-axis. >> F_srect = 22*22*mysinc(22*(kx-10/256*2)).*mysinc(22*ky); >> f_srect = 100*ifft2c(F_srect); >> figure(1), imshow(abs(f_srect),[]), title( f ) >> figure(2), imshow(angle(f_srect),[]), title( The phase of f ); To shift a rectangle in the image domain, we need to modulate k-space accordingly. Here s an example of shifting a rectangle of length 5cm by 5cm in both axis. 8

9 >> F_shift = 10*10*mysinc(10*kx).*mysinc(10*ky).*exp(-i*2*pi*(5*kx+5*ky)); >> f_shift = 100*ifft2c(F_shift); >> figure(1), imshow(abs(f_shift),[]), title( f ) >> figure(2), imshow(angle(f_shift),[]), title( The phase of f ); (c) Write the function F = rectf(kx,ky,lx,ly,x0,y0). The function evaluates the Fourier transform of a rectangle. kx and ky are k-space coordinates matrices (can be generated by meshgrid for example). Lx and Ly are scalars corresponding to the length of the rectangle in the x and y axis. x0 and y0 are scalars corresponding to the coordinates of the center of the rectangle. Write the function F = circf(kx,ky,diam,x0,y0). The function evaluates the Fourier transform of a circle. kx and ky are k-space coordinates matrices (can be generated by meshgrid for example). diam is a scaler representing the diameter of the circle. x0 and y0 are scalars corresponding to the coordinates of the center of the circle. Using these functions create the function F = face(kx,ky). The function evaluates the Fourier transform of the following image: 1.5cm 3.5cm 1cm 1cm 20cm 4cm 1.5cm 2.5cm 10cm 3.5cm Intensity=0 Intensity=0.4 Intensity=0.8 (d) Sample and display the resulting image for the following parameters: i. FOV = cm, resolution = 4mm ii. FOV = 24 cm, resolution = 5mm iii. FOV = 25.6 cm, resolution = 1mm iv. FOV = 19.2 cm, resolution = 1mm What happens when the imaging FOV is smaller than the object? Solutions: The solutions are based on Frank Ong s hw solutions. Thanks Frank! 9

10 Table of Contents EE225e hw b... 1 Rectangle... 1 Circle... 3 Shifted k-space... 5 Modulated k-space d Sample and Display face d i) FOV = 102.4cm, res = 4mm d ii) FOV = 24cm, res = 5mm d iii) FOV = 25.6cm, res = 1mm d iv) FOV = 19.2cm, res = 1mm Aliasing happens when FOV is smaller than object EE225e hw2-3 1b close all clc clear Wkx = 10; Wky = 10; Nx = 256; Ny = 256; kx_itvl = [ -Nx/2:Nx/2-1]/Nx * Wkx; ky_itvl = [ -Ny/2:Ny/2-1]/Ny * Wky; [kx, ky] = meshgrid(kx_itvl, ky_itvl); F = 18*18*mysinc(18*kx).*mysinc(18*ky); Rectangle f = 100*ifft2c(F); figure(1), imshow(abs(f),[]), title(' f ') figure(2), imshow(angle(f),[]), title('the phase of f'); 1

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12 Circle r = sqrt(kx.^2+ky.^2); F_circ = 20^2*myjinc(20*r); f_circ = 100*ifft2c(F_circ); figure(1), imshow(abs(f_circ),[]), title(' f ') figure(2), imshow(angle(f_circ),[]), title('the phase of f'); 3

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14 Shifted k-space F_srect = 22*22*mysinc(22*(kx-10/256*2)).*mysinc(22*ky); f_srect = 100*ifft2c(F_srect); figure(1), imshow(abs(f_srect),[]), title(' f ') figure(2), imshow(angle(f_srect),[]), title('the phase of f'); 5

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16 Modulated k-space F_shift = 10*10*mysinc(10*kx).*mysinc(10*ky).*exp(-i*2*pi*(5*kx+5*ky)); f_shift = 100*ifft2c(F_shift); figure(1), imshow(abs(f_shift),[]), title(' f ') figure(2), imshow(angle(f_shift),[]), title('the phase of f'); 7

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18 1d Sample and Display face 1d i) FOV = 102.4cm, res = 4mm Wkx = 10/4; Wky = 10/4; Nx = 1024/4; Ny = 1024/4; kx_itvl = [ -Nx/2:Nx/2-1]/Nx * Wkx; ky_itvl = [ -Ny/2:Ny/2-1]/Ny * Wky; [kx, ky] = meshgrid(kx_itvl, ky_itvl); F = face(kx, ky); f = Wkx*Wky*ifft2c(F); figure(1), imshow(abs(f),[0 1]), title('face 1') 9

19 1d ii) FOV = 24cm, res = 5mm Wkx = 10/5; Wky = 10/5; Nx = 240/5; Ny = 240/5; kx_itvl = [ -Nx/2:Nx/2-1]/Nx * Wkx; ky_itvl = [ -Ny/2:Ny/2-1]/Ny * Wky; [kx, ky] = meshgrid(kx_itvl, ky_itvl); F = face(kx, ky); f = Wkx*Wky*ifft2c(F); figure(2), imshow(abs(f),[0 1]), title('face 2') 10

20 1d iii) FOV = 25.6cm, res = 1mm Wkx = 10; Wky = 10; Nx = 256; Ny = 256; kx_itvl = [ -Nx/2:Nx/2-1]/Nx * Wkx; ky_itvl = [ -Ny/2:Ny/2-1]/Ny * Wky; [kx, ky] = meshgrid(kx_itvl, ky_itvl); F = face(kx, ky); f = Wkx*Wky*ifft2c(F); figure(3), imshow(abs(f),[0 1]), title('face 3') 1d iv) FOV = 19.2cm, res = 1mm Wkx = 10; Wky = 10; Nx = 192; Ny = 192; kx_itvl = [ -Nx/2:Nx/2-1]/Nx * Wkx; ky_itvl = [ -Ny/2:Ny/2-1]/Ny * Wky; [kx, ky] = meshgrid(kx_itvl, ky_itvl); F = face(kx, ky); f = Wkx*Wky*ifft2c(F); figure(4), imshow(abs(f),[0 1]), title('face 4') 11

21 Aliasing happens when FOV is smaller than object. Published with MATLAB