Appendix A - Quantitative MRI mOOC

1Derivation¶

From the MTR protocol in Brown et al., 2013 of the MTR section, $\alpha_{1}$ =15 deg and TR = 0.03 s, so assuming a T₁ at 1.5T (field strength that Brown used) of 0.55 s in healthy WM, so R₁ = 1.8, we can calculate the signal from Eq. 6.6 of an experiment with no MT pulse ( $\alpha_{2}$ = 0).

\\ S_{0}=0.087\frac{\left(1.8\cdot 0.03\right)}{\frac{0.087^{2}}{2}+0+1.8\cdot 0.03}A

(1)

S_{0}=0.0815A

(6A.1)

For an MT-weighted image, we get an equation as we don’t know $\alpha_{2}$ ,

S_{MT}=0.087\frac{\left(1.8\cdot 0.03\right)}{\frac{0.087^{2}}{2}+\frac{\alpha_{2}^{2}}{2}+1.8\cdot 0.03}A

(2)

S_{MT}=\frac{0.0047}{0.0578+\frac{\alpha_{2}^{2}}{2}}A

(6A.2)

To simplify (and for reasons seen later), let’s define $\delta=\alpha_{2}^{2}/2$ ,

S_{MT}=\frac{0.0047}{0.0578+\delta}A

(6A.3)

We’d like to calculate the contribution from the MT pulse, δ. We can do this by using the measured MTR value for this protocol, which we simulated for in the previous blog post and found to be ~0.46. We can now use the MTR equation and substitute the S₀ and S_MT, and the solve for δ.

\text{MTR}=\frac{\left(S_{0}-S_{MT}\right)}{S_{0}}\cdot 100

(3)

46=\frac{\left(0.0815A-S_{MT}\right)}{0.0815A}\cdot 100 \text{, (from 6A1)}

(4)

S_{MT}=0.044A\text{, (refactor)} \\

(5)

\left(\frac{0.0047}{0.0578+\delta}\right)A=0.044A\text{, (from 6A3)}

(6)

\frac{\left(0.0047\right)}{\left(0.0578+\delta\right)}=0.044\text{, (A cancels out)}

(7)

\delta=\left(\frac{0.0047}{0.044}\right)-0.0578\text{, (refactor)}

(8)

\delta=0.049

(6A.4)

References¶

Brown, R. A., Narayanan, S., & Arnold, D. L. (2013). Segmentation of magnetization transfer ratio lesions for longitudinal analysis of demyelination and remyelination in multiple sclerosis. Neuroimage, 66, 103–109.

Magnetization Transfer Saturation

Simulations

Magnetization Transfer Saturation

Appendix B