MRI Notes

1. Apply B0 field (1-3T)
2. Spins line up in the direction of B0 (lab frame Z-axis)
3. Apply (90 degree) RF pulse at resonance frequency
4. RF energy is absorbed by the proton and it precessess at 90 degrees from the B0 field. The precessing frequency (time to go around (360 degrees)) is the same in any orientation.
This precession frequency depends on the particular atomic nuclie and the applied field B0.
This is also known as resonance frequency.
The Mag field component of the RF field is perpendicular to the B0 field.
How do you know when to stop the RF signal?
Apply the RF signal for the right amount of time (until the precession is 90 deg).
5. When the RF is turned OFF, the proton goes back to the Z axis and re-aligns with the B0 field.
6. How long does it take to go back to Z axis?
depends on 2 factors - T1 and T2. These are called relaxation time constants.
T1: is the time taken to re-align with B0 field.
T2: is the time until which the spins of neighboring atoms are in sync. After this spin they go out of phase (dephase time).

Fourier Transform notes

If a function is not differentiable(not smooth), then it cannot be represented by a fourier transform(sum of sines and cosines) bcoz the sum of sines and cosines are differentiable(smooth).
Functions/signals like delta function, triangle wave, cannot be represented by sum of sines and cosines. Only way to represent is to use sum of sines/cosines from +infinity to -infinity, bcoz it take a large number of frequencies to make sharp corners(or any kind of corner).
Functions can be represented as sum of sines/cosines(trignometric). Sum from 1 to k.
or
sum of complex exponentials. These are special complex exponentials that satisfy symmetry (complex num + its conjugate = real number). Since they are symmetric, the sum will be from -n to +n.
Ref: stanford lec. on fourier transforms on youtube.
To represent general periodic signals we must consider infinite sums. Mathematically infinite, but practically consider terms only till the series converges.
Any non-smooth signal will generate infinitely many fourier coefficients.
Convergence:
When the signal is continuous - yes, converges.
When the singal is smooth - yes, converges.
When the signal is not smooth and has jump discontinuity(square wave) - converges* (if t0 is a point of discontinuity, then series converges to the average value at t0)


Fourier transform of a signal f(t) is F(T) = Integral -infinity to +infinity ( f(t).coswt-isinwt).
Integral -infinity to +infinity is nothing but sum : Sigma -infinity to +infinity.
FT does not recognize the direction of rotation of a vector. So the result will be in both +x and -x.
To find the FT of a signal f(t), just multiply f(t) by (cos wt-isinwt) for w = 1 thru infinity.
isinwt is the imaginary part. Just multiply f(t) by cos wt.
When you multiply f(t) by coswt (at w=1,2,3,4........), the result will be 0 for all frequencies except at freq of f(t).
For example if f(t)= cos 4t+cos9t, then the product of ( f(t)coswt) will be zero for all w except when w=4 and w=9. Those are the frequency components in f(t).
Ref: http://www.cis.rit.edu/htbooks/nmr/inside.htm