Bioengineering 280A: Announcements - Fall 2005

1. (11/07/05) Midterm: As a couple of students have pointed out, there is a typo in the wording of Problem 3. The sentence towards the end of the first paragraph should read: Similarly, the coverage in the readout direction is asymmetric, so that either the 32nd or 33rd ADC sample of each line (depending on odd or even line) coincides with k_x = 0.
   
   
2. (11/07/05) Midterm; For problem 1 of the midterm, 90x means 90 degree rotation about the x-axis, while 180y means 180 degree rotation about the y axis.
   
   
3. (11/07/05) Midterm: Some midterm clarifications/hints
   
   1) For problem 1, you may assume that the longitudinal magnetization is M0 prior to the first pulse.
   
   2) For problem 2, you may find it useful to make use of the hints given in class last Thursday (e.g. delta(at)=delta(t)/|a|), etc.)
   
   3) For problem 3, find the minimum time to play out each of the gradients, e.g. you can optimize gyd and gxd separately. I'll talk about this some more tomorrow at the beginning of class.
   
   
4. (11/08/05) Midterm:
   1) For problem 3, part b, the hint I meant to give in class is: (-1)^n = (-1)^(-n)
   
   2) For problem 2, there are a few clarifications and hints:
   a) As mentioned in class, assume that the peak of the slice profile corresponds to pi/2. e.g. a rect profile would be represented as (pi/2)*rect(z)
   
   b) In dealing with the triangle function, it's useful to first write it down as a function of z and then as a function of f. For example (1-|z|)---> (1-beta|f|) where beta is the scaling factor that takes you from z to f. After you have the expression in terms of f, write this as a convolution of two rect functions. In doing this, you will find that you need a scaling term that multiplies the convolution so that the convolution has the right height (e.g. M(f) =(pi/2)*b*rect(f/L)*rect(f/L) where b is the scaling term chosen so that M(f=0)=pi/2). You will find that this scaling term will help to cancel out terms that pop up when you take the inverse Fourier transform.
   
   c) In dealing with the convolution with shifted Dirac delta functions, it's easier to think about what's happning in the temporal frequency domain. For example, rect(z)*delta(z-z0) --> rect(beta*f)*delta(f-f0) where f0 is related to z0 by some scaling factor. Note that there is no scaling of the dirac delta function in f because we still want it to just shift the rect profile. Previously, I mentioned the use of delta(at) =delta(t)/|a| -- while this is still applicable, it turns out that the application is rather subtle in this case and it's easier to take the other approach.
   
   d) Making use of hints (b) and (c) should allow you to obtain an answer that has the correct units.
   
   3) For the gradients in problem 3, part a, you may optimize all the gradients individually. However, some students have stretched the shorter gradient to take advantage of "dead-time" -- e.g. made the width of gyd the same as the width of gxd and made the width of gyp to be the same width as the ramps in gxr. If you have already done a "stretched" solution, this is okay as long as you have fully stretched the gradient. If you submit a "stretched" solution, then please indicate so in the e-mail you send to me. If this doesn't make any sense, then you probably have not "stretched" your gradients and can ignore this message.