EE225E/BIOE265 Spring 211 Principles of MRI Miki Lustig Handout Assignment 6 Solutions 1. Nishimura 6.7 (Thanks Galen!) a) After the 9 y pulse, the spin is in the ˆx direction (using left-handed rotations). The first negative gradient lobe rotates about z by the angle θ a = and the second positive lobe rotates by θ b = T/2 = γg = γg T T/2 = +γg = γg ω a dt T/2 (y + vt) dt ( 1 2 y T + 1 ) 8 vt 2, ω b dt T (y + vt) dt T/2 ( 1 2 y T + 3 ) 8 vt 2. Since both rotations are about z, the angles add, and the net z rotation produced is θ = θ a + θ b = γg 4 vt 2. The magnetization vector just before the second 9 y pulse is then M = R z (θ)( M ˆx) = = cos θ sin θ sin θ cos θ 1 M cos θ M sin θ, M where θ = γg 4 vt 2. This vector has length M and direction ( cos θ, sin θ, ) b) After the second π/2 pulse, the magnetization vector is ( M = R y π ) M 2 1
= = = cos( π/2) sin( π/2) 1 sin( π/2) cos( π/2) 1 1 1 M sin θ M cos θ. M cos θ M sin θ M cos θ M sin θ c) From the above expression, the signal is proportional to sin θ. Therefore, zero signal occurs for θ = nπ. The smallest non-zero velocity occurs when θ = γg 4 vt 2 = π, or v = 4π γgt 2 d) For these values, the rotation produced by the bipolar gradient lobes is θ = γg 4 vt 2 = 36.8, so the magnetization just before the second 9 pulse is M cos θ M = M sin θ = M.8.6. This vector is entirely in the xy plane, so a 9 applied about an axis parallel to this vector will leave it in the xy plane and preserve the signal of the moving spin. 2
2. Design the following slice-selective excitation pulses. In each case, choose a time-bandwidth product that is a multiple of four. The maximum gradient strength is 4 G/cm. Assume the excitation pulse is a Hamming windowed sinc, as we described in class, and that we want the sharpest profile within the constraints. a) Design a pulse to excite a 3 mm slice. The pulse should be 1 ms in duration. Find the timebandwidth of the pulse, and the amplitude of the slice-select gradient. Solution: First, the maximum gradient we have is 4 G/cm, or A 3 mm slice then has a bandwidth of (4 G/cm)(4.257 khz/g) = 17 khz/cm. (17 khz/cm)(.3cm) = 5.1 khz. This is the maximum bandwidth that we will ever need with our gradient system. Next, we need to choose the time-bandwidth of our RF pulse. The maximum time bandwidth is (1 ms)(5.1 khz) = 5.1 We want this to be a multiple of 4, so we choose a time-bandwidth of 4. Then the bandwidth is BW = T BW/T = 4/(1 ms) = 4 khz. The gradient strength is then found by solving for G z. The result is G z = 4 khz ( γ 2π )( z) = γ 2π G z z = 4 khz 4 khz = 3.13 G/cm. (4.257 khz/g)(.3 cm) b) Design a pulse to excite a 8 cm slab. Assume we want to use a time-bandwidth 16 pulse, and that the shortest time we can play this waveform is 2 ms, due to limits on peak RF amplitude. What is the gradient amplitude? Solution: We know the time, and the time-bandwidth product, so the bandwidth of the pulse is BW = T BW/(T ) = 16/(2 ms) = 8 khz. We want this to correspond to an 8 cm slab, so Solving for G z we get 8 khz. = γ 2π G z z = (4.257 khz/g)g z (8 cm) G z =.235 G/cm. c) A 2 ms, time-bandwidth 8 pulse is to be used to excite a 1 cm slice centered at +12 cm from gradient isocenter (the zero point of the gradients). Find the gradient amplitude for this pulse, and the frequency for the RF pulse compared to the slice at isocenter. 3
Solution: The bandwidth of this pulse is If this is a 1 cm slice, then Solving for G z This corresponds to BW = 8/(2 ms) = 4 khz 4 khz = γ 2π G z z = (4.257 khz/g)g z (1 cm) G z =.94 G/cm. γ 2π G z = (4.257 khz/g)(.94 G/cm) = 4 khz/cm. Hence, in order to offset the slice by 12 cm, we need to increase the frequency of the RF transmitter by (4 khz/cm)(12 cm) = 48 khz. 3. a) We would like to use the waveform plotted below as an RF excitation pulse. Assume that the slice select gradient is applied during the pulse, and then inverted to refocus the slice. What is the duration of the refocusing gradient that produces the maximum signal? Solution: The refocusing gradient basically takes us backwards in time through the excitation pulse. If the refucusing gradient and the slice select gradient have the same amplitude, then refocusing by t ms takes us to a point t ms before the end of the pulse. The signal we receive is proportional to the amplitude of the pulse at that time. In this case, the largest signal corresponds to a time of 3.1 ms during the pulse, or.9 ms from the end of the pulse. Hence a refocusing interval of.9 ms produces the maximum signal. b) If we refocus for 2 ms, which is 1/2 the slice select gradient length, how does the signal compare to that of part (a)? Solution: If we refocus for 2 ms, this will bring us to the t = 2 ms point in the RF pulse, where the k-space weighting is zero. This refocusing interval produces no signal, which is much smaller than the answer for (a)..3.2 B1(t), G.1.1.5 1 1.5 2 2.5 3 3.5 4 Time, ms 4
4. Introduction: This assignment concerns typical Fourier transform designs of excitation pulses. This includes designing windowed sinc pulses, calculating the RF amplitude required, simulating slice profiles,designing a pulse for a specific application, and computing the relative SAR of a pulse sequence. For simulating the slice profile, you will need the bloch simulator. a. Design of Windowed Sinc RF Pulses Write an m-file that computes a Hamming windowed sinc pulse, given a time-bandwidth product, and number of samples. >> rf = wsinc(timebandwidth, samples) Write the mfile so that it scales the waveform to sum to one, sum(rf) = 1. Plot windowed sincs with TBW of 4, 8, and 12. The TBW=4 pulse is common for 18 degree pulses, the TBW of 8 is typical for excitation pulses, and a TBW of 12 or 16 is typical for slab select pulses. Solution:.15.1.5 TBW=4 TBW=8 TBW=12.5 1 2 3 4 5 6 7 8 9 1 b. Plot RF Amplitude For convenience, we will assume that the RF waveforms are normalized so that the sum of the RF waveform is the flip angle in radians. The sampled RF waveform can then be thought of as a sequence of small flips. This eliminates the need to explicitly consider the pulse duration in the design and simulation. However, it is sometimes important to compute the RF pulse amplitude in Gauss. In this problem you will write an m-file that takes a normalized RF pulse, and then, given a overall pulse length, scales the waveform to Gauss. First, generate a 3.2 ms, T BW = 8 windowed sinc RF pulse, and scale it to a π/2 flip angle >> rf = (pi/2)* wsinc(8,256); Then, write an m-file called rfscaleg that takes a normalized RF pulse and a pulse duration, and returns a waveform that is scaled to Gauss, >> rfs = rfscaleg(rf, pulseduration); Plot the pulse you generated, scaled to Gauss. Label the axes. What is the peak amplitude? Solution: One possible m-file for rfscaleg.m is 5
function rfs = rfscaleg(rf,t); % rfs = rfscaleg(rf,t) % % rf -- rf waveform, scaled so sum(rf) = flip angle % t -- duration of RF pulse in ms % rfs -- rf waveform scaled to Gauss % gamma = 2*pi*4.257; % khz/g dt = t/length(rf); rfs = rf/(gamma*dt); This takes the rotation produced by each sample, and determines what RF amplitude would be needed to produce that rotation in one sample dwell time. Note that the input RF is in radians, so we need the 2π factor in gamma. The scaled RF plot looks like.15 Amplitude, Gauss.1.5.5.5 1 1.5 2 2.5 3 Time, ms The peak amplitude is a about.147 G. c. Simulated Slice Profiles We will use the bloch simulator for simulating the RF pulse. Calculate the slice thickness of the T BW = 8 pulse from problem (b), based on the relations presented in class. Assume the slice select gradient is.6 G/cm. Simulate the RF pulse using > dp = linspace(-2,2,512). ; % simulate from -2cm to 2cm > mx = zeros(512,1); my=zeros(512,1);, mz = ones(512,1); > dt = 3.2e-3/256; > [mx,my,mz] = bloch(rfs,g,dt,1,1,,dp,,mx,my,mz); > mxy = mx+i*my; > figure, plot(dp,abs(mxy)), xlabel( cm ), ylabel( amplitude ); 6
Is the slice the expected width? Solutions: The slice width can be found from so T (BW ) = 8 BW = 8/(3.2 ms) = 2.5 khz The gradient strength is.6 G/cm or γ 2π.6 G/cm = 2.55 khz/cm. The slice thickness is then about 2.5 khz =.98 cm 2.55 khz/cm 1.2 1 Amplitude.8.6.4.2.2 2 1.5 1.5.5 1 1.5 2 position, cm This shows the half-amplitude width to be about 1 cm, very close to what we calculated. d. Design a Slab Select Pulse You are designing a 3D pulse sequence, and you need a slab select pulse in the z dimension. You have 6 ms for the pulse, and want it to be as sharp as possible. You also have a peak RF amplitude constraint of.17 G. 1. What is the highest time-bandwidth you can allow, given that the maximum flip angle will be 9 degrees? Solution: Trying a few values we get >> max(rfscaleg((pi/2)*wsinc(16,256),6)) ans =.1572 >> max(rfscaleg((pi/2)*wsinc(17,256),6)) ans =.1667 >> max(rfscaleg((pi/2)*wsinc(18,256),6)) ans =.1762 7
So T BW = 17 is OK, and T BW = 18 is not. We could go for more resolution if you want, and come put with a T BW = 17.35 that is pretty close to.17 G, but 16 or 17 is fine. 2. Assume we want the minimum slab thickness to be 8 cm. What is the gradient amplitude that this requires? Solution: Assuming T BW = 17, and a 6 ms pulse, the frequency bandwidth is BW = 17 = 2.83 khz 6 ms We want this to be 8 cm, so γ 2π G(8 cm) = 2.83 khz, and G = 2.83 khz =.83 G/cm (4.257 khz/g)(8 cm) Your answer will differ, depending on the TBW product you used. The important point to note is that this is very low, only about 35 Hz/cm. This means that patient susceptibility shifts are going to distort the slab. 8
3. Simulate the slice profile. How wide is the transition band compared to the passband? Assume that the passband edge is 95% of the middle of the passband, and the stopband edge is 5% of the passband. Solution: 1.8 Amplitude.6.4.2 8 6 4 2 2 4 6 8 position, cm 1.8 Amplitude.6.4.2 3 3.2 3.4 3.6 3.8 4 4.2 4.4 4.6 4.8 5 position, cm The slab thickness is about 8 cm which is what we designed for, and the transition width (neglecting the sidelobe) is about.7 cm. We d expect the transition width to be about 8 cm/ (17/2) =.94 cm. This is off somewhat, but is due to how exactly the transition width is defined. It is on the right order, though. e. Compute the Relative SAR of a Pulse Sequence SAR stands for specific absorption rate and is the amount of RF energy deposition in the body. Usually, this is calculated using software model that gives the SAR limit measured in terms of 1 ms rectangular 18 degree pulses ( hard 18 s). This depends on the patient weight, body part, and RF coil. The limit might be 1 hard 9
18s per second, for example. The power of a particular RF pulse is P = B1(t) 2 dt. The relative SAR of a particular RF pulse is it s power divided by the power in a 1 ms hard 18. The relative SAR of a pulse sequence is computed as the sum of the relative SAR s of each of the pulses, measured in equivalent hard 18 s. Write an mfile rsar.m which takes a normalized RF pulse and pulse duration, and returns the relative SAR, measured in equivalent hard 18 s >> eq18 = rsar(rf,t) Assume that you are developing a very fast sequence. The excitation pulse is a 1 ms TBW=2 windowed sinc. Assume you need a TR of 2.25 ms. What is the maximum flip angle you can allow, given that the SAR limit is 1 equivalent 18 s per second? Solution: The samples of the normalized RF pulse are γb 1 (t i ) t which is the flip angle in radians that each sample produces. What we want to compute is (γb 1 (t i )) 2 t and compare this to the value of a hard 18 An m-file computes this ratio is given below: i (γb 1 ) 2 t = π 2 (1 ms) = π 2 function npi = rsar(rf,t) % npi = rsar(rf,t) % % rf -- rf waveform, scaled so that sum(rf) = flip angle % t -- duration in ms % % npi -- number of equivalent hard, 1 ms, pi pulses % dt = t/length(rf); %convert radians to radians/ms rf = rf/dt; % compute power (radians^2)(ms) p = sum(rf.*rf)*dt; % normalize by value for a hard pi pulse npi = p/(pi*pi); 1
To answer the flip angle question, first define a TBW=2 RF pulse >> rf = wsinc(2,256); This gives an RF pulse that sums to 1 radian. The relative sar of a single 1ms pulse is >> rsar(rf,1) ans =.1754 With a TR of 2.25 ms, we are applying 444 pulses/second. We want the total relative sar to be 1. So we want to find a constant theta such that >> 444*rsar(theta*rf,1) gives an answer of 1. The value theta is then the maximum flip angle in radians. Experimenting, we can rapidly close in on an answer of θ = 1.135 radian = 65. 11