EE225E/BIOE265 Spring 2013 Principles of MRI. Assignment 9 Solutions. Due April 29th, 2013

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1 EE5E/BIOE65 Spring 013 Principles of MRI Miki Lustig This is the last homework in class. Enjoy it. Assignment 9 Solutions Due April 9th, 013 1) In class when we presented the spin-echo saturation recovery pulse sequence, we assumed that the echo time T E was short relative to the repetition time T R, so we could ignore it. This is not always a good approximation, particularly if we want to measure T 1. In fact, after the 90 there will be some M z recovery. The 180 will invert this, and M z will continue to recover until the next 90. Let M z be the magnetization immediately before the RF T E/ T R M 0 M z M z Assume that T E is small compared to T 1, but not negligible, so that we can approximate 1 e T E/T 1 ) T E /T 1. This means that the M z recovery curves are approximately linear within a time T E after the 90. a) Find an approximate expression for the time when the magnetization cross the M z = 0 axis. If T E is short relative to T 1, we can linearize the recovery curves near M z = 0. As long as M z does get too big i.e. T E is short) the recovery curves are lines with a slope 1/T 1. Following the 90 at t = 0, the z magnetization evolves as M z t T 1 T E / T 1 T E T E t T E/ T 1 t T E T 1 1

2 where the constant scale factor M 0 has been suppressed in the plot. The z magnetization increases with a slope M 0 /T 1 until T E /, when it is M TE 0 T 1 ). This is inverted by the 180 to M TE 0 T 1 ). which again recovers linearly with a slope M 0 /T 1. The equation for this part of the recovery curve is ) t TE M z = M 0 T 1 This goes through zero at t = T E. b) This might be called a T 1 echo. Why? A spin echo refocuses spins no matter what their resonant frequencies are. In this case M z goes through zero at T E no matter what the T 1 is. This is illustrated below for the case of three different T 1 s M z T E t The 180 refocuses the different decay rates of the different T 1 s, just as the 180 refocuses different precession rates for a spin echo. b) Find an expression for M z using what you found in a). The z magnetization is zero at time T E, and then recovers for the remainder of the T R interval. The z magnetization just before the next RF pulse is then M z = 1 e T R T E T 1 )M 0 c) Compare M z calculated using your expression and the expression ignoring T E for the case where T R = 00 ms, T E = 0 ms, and T 1 = 800 ms. What is the percent error produced by neglecting T E in this case? If we ignore T E, If we include the effect of T E, The percent error is then M z = M 0 1 e T R/T 1 ) = M 0 1 e 00/800 ) = M 0 0.1). M z = M 0 1 e T R T E )/T 1 ) = M 0 1 e 180/800 ) = M ) = 9.8% ) Choosing Scan Parameters We will cover material for this on Tuesday.) You are designing a pulse sequence to image the upper abdomen. The tissues of interest are

3 Liver, T 1 = 600 ms, T = 50 ms. Spleen, T 1 = 1000 ms, T = 80 ms. Fat, T 1 = 350 ms, T = 60 ms. Gall Bladder, T 1 = 000 ms, T = 300 ms. This will be a spin echo acquisition, where you have to choose the repetition time T R, the echo time T E. In addition, you can also add an inversion recovery pulse if you need it, and specify the inversion time T I. Otherwise leave the T I blank. The minimum T E is 15 ms, and the minimum T R is 0 ms. In each case you only need to specify the scan parameters approximately. You only need reasonable values, not necessarily the optimum values. However, you need to describe the reasoning behind your choices. a) Choose the scan parameters so that you have good contrast between liver and spleen, and low signal from the gall bladder. The liver and spleen differ in both T 1 and T so either of these contrasts might work. If we think about T contrast long T R, long T E ), the spleen would be brighter than the liver. However we also see that the gall bladder would also be brighter still, since it has a T = 300 ms. This doesn t work unless we null it out with an inversion pulse. We only want to do this if we have to. Next, if we think about T 1 contrast T R <= T 1, short T E ) we see that the liver would be brighter than the spleen. In addition, the gall bladder is has a low signal since it has a very long T 1. This is the answer. A reasonable choice of paramters might be T R = 800 the average of the T 1 s for liver and spleen), and a minimum T E to maximize signal, and minimize T contrast. Any shorter T R would also work, provided it is longer than the T E. Note that a sequence with short T R and long T E doesn t work, because this produces both T 1 and T weigting, and these go in opposite directions. The result is no contrast. T R = 800 ms T E = 15 ms T I = none b) Choose the scan parameters so that you see only the gall bladder, with little signal from the other tissues. The gall bladder has a long T 1 and a long T. T 1 contrast by itself won t work, long T 1 s produce low signal, and that isn t what we want. We could use an inversion pulse, but we don t want to if we don t have to. Heavy T contrast will make the gall bladder the brighter than everything else, since it has the longest T by far. This is the answer. For T contrast we want a long T R and a long T E. If we choose T R = T 1 = 000 ms we will be reasonably SNR efficient at the cost of some T 1 weighting). Then we choose T E = T = 300 ms, which is much longer that the T s of all of the other tissues, resulting in almost no signal from these tissues. Other values of T R would also work, provided it is bigger than T E. 3

4 T R = 000 ms T E = 300 ms T I = none c) Choose the scan parameters so that the liver produces no signal at all, but the spleen is bright. In this case we want to completely eliminate the signal from liver, while maintaining some signal from the spleen. The T 1 s and T s of these tissues don t differ enough to do this with either simple T 1 contrast or T contrast alone. In this case we need to use an inversion-recovery pulse to null out the signal from liver. Many different T R s and T E s can be used, provided they don t lose the spleen signal. To maximize the spleen signal, a good choice is the minimum T E = 15 ms. The T R should be significantly longer than this to allow time for the recovery, and the inversion pulse. A reasonable choice might be T R = 1000 ms. Once these are decided, we need to figure out what the inversion time should be. This is in the book, and was one of the cases we worked through in class. The result is ) 1 + e T R /T 1 T I = T 1 log Substituting T R = 1000 ms, and T 1 = 600 ms, we get T I = 31 ms. There were many other acceptable answers, depending on what you chose for T R. T R = 1000 ms T E = 15 ms T I = 31 ms 3) Nishimura 7.3 Solutions: a. [ ] Ix, y) = K ρ x, y) 1 e T R/T 1x,y) e T E/T x,y) Let Cx, y) = I A x, y) I B x, y) then, d Cx, y) = 0 T R d [ = K ρ T R = K ρ d T R 1 e T R/T 1A ) ) ] e T E/T A 1 e T R/T 1B e T E/T B [ )] e T E/T A e T E/T A e T R/T 1A e T E/T A e T R/T 1B e T E/T B ] = K ρ [1/T 1Ae T E/T A e T R/T 1A 1/T 1B e T R/T 1B e T E/T B 1/T 1Ae T E/T A e T R/T 1A = 1/T 1B e T R/T 1B e T E/T B T 1B e T E/T A e T E/T B T 1A = e T R/T 1A e T R/T 1B 4

5 T 1B e T E1/T B 1/T A ) T 1A = e T R1/T 1A 1/T 1B ) lnt 1B /T 1A ) + T E1/T B 1/T A ) 1/T 1A 1/T 1B = T R b. T R = ln90/780) + 01/100 1/9) 1/780 1/90 = 757ms c. Using the same Cx, y) we take d T E and get after some algebra: T E = lnt B/T A ) + ln 1/T A 1/T B ) 1 e T R/T 1A 1 e T R/T 1B d. ) T E = ln100/9) + ln 1 e 3000/780 1 e 3000/90 = 116ms 1/9 1/100 4) In order to produce a very high resolution image of the brain, you specify a scan that provides 0.5 mm resolution over a 5.6 cm FOV. Unfortunately, the SNR is very low. a) You lowpass filter the image to increase the SNR, using an ideal lowpass filter in both x and y. If the resulting resolution is 0.5 mm in both x and y, what is the final SNR compared to the original image? The voxel volume has been double in both dimensions. However, the ideal lowpass filter throws away 3/4 of the data. The effective A/D time is now 1/4 of what it originally was. The new SNR is then SNR = ft 1, T, ρ) T A/D, δ x, δ y, δ z, = ft 1, T, ρ) T A/D,1 /4 δ x,1 )δ y,1 )δ z,1 = ft 1, T, ρ) T A/D,1 δ x,1 δ y,1 δ z,1 = SNR 1 b) How does this SNR compare to an image with 0.5 mm resolution, and the same total imaging time as the 0.5 mm resolution image? Assume the same A/D duration for each readout interval, and that the 0.5 mm scan averages to provide the same number of readouts. In this case the voxel volume doubles, but we get the benefit of the full scan time SNR 3 = ft 1, T, ρ) T A/D,3 δ x,3 δ y,3 δ z,3 = ft 1, T, ρ) T A/D,1 δ x,1 )δ y,1 )δ z,1 = 4fT 1, T, ρ) T A/D,1 δ x,1 δ y,1 δ z,1 = 4 SNR 1 5

6 The conclusion from a) and b) is that while we can recover somewhat in SNR by lowpass filtering, we can never do as well as acquiring the data at the final resolution directly. There is an opportunity cost is having collected high-spatial frequency data that you won t ultimately use. 5) You have a working DFT gradient-recalled echo pulse sequence that produces an image with an SNR 1 of 100. What is the new SNR after you make one of the following changes. a) Double the TBW of the slice selective RF pulse, while the RF pulse duration and the slice select gradient remain the same. Solution Doubling the TBW while keeping the pulse the same length the gradient is the same length) doubles the bandwidth of the RF. The gradient is the same strength, so the slice thickness is doubled, δ z, = δ z,1 The voxel volume has doubled, and every thing else has remained the same. SNR is proportional to voxel volume, so SNR = SNR 1 = 100) = 00 SNR = 00 b) Double the readout gradient strength, while keeping the A/D duration and sampling rate the same. Solution Doubling the readout gradient strength while keeping the duration the same doubles the extent in k-space. The halves the resolution in x, and halves the voxel volume. Since SNR is propotional to voxel volume SNR = 1 SNR 1 = 1 100) = 50. SNR = 50 c) Double both the readout gradient strength and the sampling rate, while halving the duration of the A/D window. Assume the anti-aliasing filter bandwidth matches the sampling rate. Solution In this case we cover the same extent in k-space in half the time. The resolution remains the same, but the A/D time is halved. Since SNR is proportional to the square root of the total A/D time, SNR = 1 SNR 1 = = 70.7 SNR =

7 d) Double the number of phase encodes, while keeping the maximum phase-encode gradient amplitude the same. Solution The maximum phase encode gradient remains the same, so the resolution remains the same. The total A/D time is doubled, since there are twice as many phase encodes. Since SNR is proportional to the square root of the total A/D time, SNR = SNR 1 = 100) = 141 SNR = 141 7

8 6) Nishimura 7.10 Solutions: Reference DFT has: 56 Phase encodes 10ms readout 56 points) Total imaging time T 0 FOV: F OV x0, F OV y0 Resolution: δ x0, δ y0 SNR=1 a) In this case, we have 18 phase encodes Imaging time cut by half. BW is the same. FOV the same δ y is doubled. SNR δ x δ y T = δ x0 δ y0 ) T 0 / = SNR 0 = b) 51 phase encodes same y-resolution, double F OV y. Imaging time is doubled, resolution the same, BW is the same. SNR δ x δ y T = δ x0 δ y0 T0 = SNR 0 = c) Here the resolution in both dimension is improved, so δ x and δ y are doubled. Same number of phase encodes and frequency encodes, so F OV x and F OV y are halved. We have the same number of frequency encoding points. The gradient amplitude is unchanged, but the k-space coverage is doubled. This means that the BW is halved twice as much time spent on each sample). SNR δ x δ y T = δ x0 /)δ y0 /) T 0 = 1/8SNR 0 = 1/8 7) Consider the Hadamard encoding of two slices as shown in Fig. 8. in Nishimura. What is the relative SNR of the two slices compared to imaging two slices sequentially? 8

9 8) Consider the EPI and spiral trajectories in fig. 8.1 in Nishimura. a) Assume T ms readout gradients and that the gradients start 1ms from the peak of the RF excitation. What is the echo-time for EPI and what is the echo time for the spiral trajectory? 9

10 The echo time is the time that the trajectory crosses DC, or the center of k-space. This is the time in which the majority of the contrast of the image is being determined as most of the energy of images lies in the origin of k-space. EPI is a symmetric trajectory and the crossing of the center of k-space is in the middle of the readout. Therefor the echo time for EPI is 1 + T/ms. On the other hand, the spiral trajectory always starts at the origin, therefore the echo time is 1ms. b) What is the k-space weighting due to T decay in EPI and spiral for a gradient echo sequence. Assume decay happens only in the slow direction.) Both readouts start 1ms after the middle of the RF and are T ms long. Let t = 0 the start of the readout. The T weighting as a function of time is: W t) = e 1+t T Assuming the decay happens only in the slow direction k y ignoring readout k x ) then, k y t) = k max t T/ T/ and So, t = k yt k max + T/. W k y ) = exp 1 + t ) T = exp 1 T ) exp T T )) ky + 1 k max The decay during the spiral sequence can be approximated as a radial decay with, k r = k max t/t and So, t = T k r k max. W k r ) = exp 1 ) T exp T k r T k max ). 10

11 The k-space weighting is shown below: c) How would your answer change if the sequence is a spin-echo. What s the problem for spiral spin-echo imaging? In a spin-echo, T is refocused. Therefore, we can define t = 0 as the spin-echo time. For EPI, we get symmetric weighting in k y, W k y ) = exp T T )) ky k max For spiral, we still have to start at the spin echo so we get rid of some of the constant T decay from the mid RF pulse to the start of the spiral, but the k-space weighting is the same as the gradient echo where the edges of k-space will be heavily T weighted, much more then the EPI! W k r ) = exp T k ) r T k. max 11