The Answers of MRI Tutorial Videos

By Frank Luo

This is my answers of the MRI Tutorial Videos How MRI Works - Part 2: The Spin Echo and How MRI Works - Part 3:Fourier Transform and K-Space .

Part 2: The Spin Echo


Part 2 Questions 1

Part 2 Question 2


Question 1:

  1. The Boltzmann Magetization M0=Nγ22B04kTM_0 = \frac{N {\gamma}^2 \hbar^2 B_0}{4 k T}, then after elimination the units is J/TJ/T.
  2. The Polarization is P=γB02kTP = \frac{\gamma \hbar B_0}{2kT}, then after elimination we can get that PP is a special number depends on the material, no SI units.

Question 2:

  1. The polarization is P=5149100=0.02P = \frac{51-49}{100} = 0.02 .
  2. The magnet field strength should be B0=0.020.00000345882TB_0 = \frac{0.02}{0.0000034} \approx 5882T .
  3. The temperature should be T=300×0.00000340.02=0.051KT = \frac{300 \times 0.0000034}{0.02} = 0.051K.

Question 3:

  1. Since the Boltzmann Magetization Equation is M=M0(1etT1)etT2M = M_0(1- e^{-\frac{t}{T_1}}) e^{-\frac{t}{T_2}} , so we can calculate the signal.

The signal of Tissue AA : MA=M0(1e150300)e12.520=0.21M_A = M_0(1- e^{-\frac{150}{300}}) e^{-\frac{12.5}{20}} = 0.21 .
The signal of Tissue BB : MB=M0(1e150200)e12.540=0.38M_B = M_0(1- e^{-\frac{150}{200}}) e^{-\frac{12.5}{40}} = 0.38 .

Surely Tissue BB will deliver more signal.

  1. We have calculated that Tissue BB will deliver more signal if both Tissue AA and BB has the same Boltzmann Magetization.

If Tissue AA is 85%85\% of Tissue BB, then the Tissue AA signal will become lesser, so Tissue BB deliver more signal.

  1. Let function f(t)=M0A(1eTRT1A)etT2AM0B(1eTRT1B)etT2Bf(t) = M_{0A}(1- e^{-\frac{TR}{T_{1A}}})e^{-\frac{t}{T_{2A}}} - M_{0B}(1- e^{-\frac{TR}{T_{1B}}})e^{-\frac{t}{T_{2B}}} reprent the signal of time tt.

Consider the function: f(t)=(1e150200)et40(1e150300)et20f(t) = (1 - e^{-\frac{150}{200}}) e^{-\frac{t}{40}} - (1- e^{-\frac{150}{300}}) e^{-\frac{t}{20}} reaches its PEAK at about t=16t = 16, so the TETE should be TE=32msTE = 32ms.

Question 4:

If both tissues deliver the SAME signal, which means M0AetT2A=M0BetT2BM_{0A} e^{-\frac{t}{T_{2A}}} = M_{0B}e^{-\frac{t}{T_{2B}}}.

Put the data in, then we can get 4.1et303.7et50=04.1e^{-\frac{t}{30}} - 3.7e^{-\frac{t}{50}} = 0, solve it and get t7.7mst \approx 7.7ms.

So the echo time is: TE=2×t15.4msTE = 2 \times t \approx 15.4ms.

Question 5:

  1. etT2S0=e30500.55e^{-\frac{t}{T_2}}S_0 = e^{-\frac{30}{50}} \approx 0.55, so the signal is 0.55mV0.55mV.

  2. If the magetic field of inhomogeneity of ΔB=1\Delta B = 1 ppm, the signal can be calculated by such equation:

S(t)=S0etT2eγΔBtS(t) = S_0 e^{-\frac{t}{T_2}} e^{- \gamma \Delta B t}

Put the data in, we can e3050e1×267×0.0000030.43e^{-\frac{30}{50}} e^{-1 \times 267 \times 0.000003} \approx 0.43, so the signal amplitude is 0.430.43.

  1. 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 e40500.45e^{-\frac{40}{50}} \approx 0.45 at times 40ms and e80500.20e^{-\frac{80}{50}} \approx 0.20 at times 80ms.

Question 6:

From the equation M=M0etT2M = M_0 e^{-\frac{t}{T_2}}, then we can solve et600.1e^{-\frac{t}{60}} \le 0.1, the answer is t138.55mst \approx 138.55ms.

Therefore, we can get the echoes at times 10ms, 30ms, 50ms, 70ms, 90ms, 110ms, 130ms, so we can get 77 echoes.

Part 3: Fourier Transform and K-Space


Part 3 Question


The amplitude of the FFT\textit{FFT} result is 1515. We need to times 2N\frac{2}{N} to get the correct answer.

The reason are as follows:

  1. Both the positive and negative frequency contribute the answer, but we only use the positive, so have to multiply 22.

  2. Each operation we have to sum once, so we need the result to multiply 1N\frac{1}{N} to get the final answer.