MRI EXAM QUESTIONS 1 & 2:
QUESTION 1)
The Bloch equations describing the time evolution of the magnetization vector
M = Mx + My + Mz in a static magnetic field B0 are:
where M0 is the initial magnetization in the z direction prior to the RF excitation pulse, ω0 is the Larmor frequency, T1 is the spin-lattice relaxation time and T2 is the spin-spin relaxation time.
The general solutions of the Bloch equations are:
[1]
[2]
[3]
where c1, c2 and c3 are constants.
For an RF excitation pulse that tips the magnetization vector by an angle of θ from the z axis, write equations for the three components of the magnetization vector as functions of time t from the center of the RF pulse where we define t=0.
ANSWER:
The RF pulse tips the magnetization by an angle θ, therefore the initial conditions at time t=0 are:
[a]
[b]
To determine c1 and c2, solve [1] given:
1) the condition Mz=M0 at t=∞
→
2) the initial condition [a] at t=0:
→
To determine c3, solve [b] given [2] and [3] at t=0:
Therefore, the magnetization components at time t are:
QUESTION 2) Consider two protons (hydrogen nucleus) located at two different positions x1 and x2 within the sample, which is a human body. The magnetic field as a function of position x is: where B0 = 1.0 T (Tesla) and B is the static magnetic field caused by sample inhomogeneity. Proton 1 is located within static tissue and does not move, and Proton 2 is located within a blood vessel and is moving at a constant velocity v in the +y direction. The spin-echo pulse sequence diagram below shows the timing of gradients relative to the two RF pulses. Assume the RF pulses are impulses with zero width. During the readout data acquisition period (ADC) a magnetic field gradient is applied in the y direction with negative polarity during the prephasing lobe that precedes signal acquisition ( < t < 3), and with positive polarity during the readout lobe
(3 < t < 5).
The phase is computed from the equation
where γ = 42.58 MHz/Tesla.
Calculate the phase difference between Proton 1 and Proton 2 as a function of time t during the signal acquisition (ADC) period (3 < t < 5), where t=0 at the center of the π/2 pulse. Remember, the π pulse will invert any phase that has accumulated prior to its application at t=
ANSWER:
The magnetic field experienced by each proton is the superposition of field components arising from: the main field, the sample inhomogeneity and the applied gradient:
The total phase is therefore a linear combination of the phase shifts produced by each field component, and each can be considered separately.
Main field B0:
The phase produced by the main field can be ignored, since the signal is acquired in the rotating frame (demodulated). Also, we are computing the phase shift between protons which experience the same main field.
Sample inhomogeneity:
is constant with respect to time, since neither proton changes its x position (for Proton2, motion is along y only). The phase accumulation over time 0<t<is:
.
This phase is inverted by the 180 pulse: .
So, the phase during the post-180 time period t<5 which includes the ADC period is:
.
The phase difference between Proton1 and Proton2 due to inhomogeneity is then:
Applied Gradient
For Proton1, which is stationary at y=y1:
Proton2, which starts at y2 and moves with velocity v in the y direction:
The phase difference between Proton1 and Proton2 due to the applied gradient is:
The total phase shift between Protons 1 and 2 is therefore: