By Frank Luo
Part 2: The Spin Echo
- The Boltzmann Magetization M0=4kTNγ2ℏ2B0, then after elimination the units is J/T.
- The Polarization is P=2kTγℏB0, then after elimination we can get that P is a special number depends on the material, no SI units.
- The polarization is P=10051−49=0.02 .
- The magnet field strength should be B0=0.00000340.02≈5882T .
- The temperature should be T=0.02300×0.0000034=0.051K.
- Since the Boltzmann Magetization Equation is M=M0(1−e−T1t)e−T2t , so we can calculate the signal.
The signal of Tissue A : MA=M0(1−e−300150)e−2012.5=0.21 .
The signal of Tissue B : MB=M0(1−e−200150)e−4012.5=0.38 .
Surely Tissue B will deliver more signal.
- We have calculated that Tissue B will deliver more signal if both Tissue A and B has the same Boltzmann Magetization.
If Tissue A is 85% of Tissue B, then the Tissue A signal will become lesser, so Tissue B deliver more signal.
- Let function f(t)=M0A(1−e−T1ATR)e−T2At−M0B(1−e−T1BTR)e−T2Bt reprent the signal of time t.
Consider the function: f(t)=(1−e−200150)e−40t−(1−e−300150)e−20t reaches its PEAK at about t=16, so the TE should be TE=32ms.
If both tissues deliver the SAME signal, which means M0Ae−T2At=M0Be−T2Bt.
Put the data in, then we can get 4.1e−30t−3.7e−50t=0, solve it and get t≈7.7ms.
So the echo time is: TE=2×t≈15.4ms.
e−T2tS0=e−5030≈0.55, so the signal is 0.55mV.
If the magetic field of inhomogeneity of ΔB=1 ppm, the signal can be calculated by such equation:
Put the data in, we can e−5030e−1×267×0.000003≈0.43, so the signal amplitude is 0.43.
- We should delivered the 180° pulses at times 20ms and times 40ms if we wish to detect echoes at times 40ms and 80ms.
The signal will be e−5040≈0.45 at times 40ms and e−5080≈0.20 at times 80ms.
From the equation M=M0e−T2t, then we can solve e−60t≤0.1, the answer is t≈138.55ms.
Therefore, we can get the echoes at times 10ms, 30ms, 50ms, 70ms, 90ms, 110ms, 130ms, so we can get 7 echoes.
Part 3: Fourier Transform and K-Space
The amplitude of the FFT result is 15. We need to times N2 to get the correct answer.
The reason are as follows:
Both the positive and negative frequency contribute the answer, but we only use the positive, so have to multiply 2.
Each operation we have to sum once, so we need the result to multiply N1 to get the final answer.