IMG-11

GRE signal (Ernst)

S = M0 sinα (1−E1)/(1−cosα E1) e^{−TE/T2*}, E1=e^{−TR/T1}.

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MRIErnst GRE

Governing equation

S=M_0\sin\alpha\dfrac{1-E_1}{1-\cos\alpha\,E_1}e^{-TE/T_2^*}

$\alpha \in \left[1, 90\right]$

where

M_0: M0 (a.u.)
\alpha: Flip angle (°)
TR: TR (ms)
T_1: T1 (ms)
TE: TE (ms)
T_2^*: T2* (ms)
S: GRE signal (a.u.)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

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①

Governing law — red

S=M_0\sin\alpha\dfrac{1-E_1}{1-\cos\alpha\,E_1}e^{-TE/T_2^*}

$\alpha \in \left[1, 90\right]$

$\alpha = 1$ $S = 0.015$

Along the curve

②

Domain of the variable — blue

\alpha_{\min} = 1,\quad \alpha_{\max} = 90

$\alpha \in \left[1, 90\right]$

$\alpha = 1$ $\alpha = 1$

Along the curve

③

Response curve — green

S=M_0\sin\alpha\dfrac{1-E_1}{1-\cos\alpha\,E_1}e^{-TE/T_2^*}

$\alpha \in \left[1, 90\right]$

$\alpha = 1$ $S = 0.015$

Along the curve

④

Rate of change — amber

\frac{\mathrm{d}S}{\mathrm{d}\alpha}

$\alpha \in \left[1, 90\right]$

$\alpha = 1.56$ $S^{\prime} = 0.014$

Along the curve

Lecture brief

Historical brief

Larmor precession, CT Beer projections, SNR and Nyquist sampling are why MRI and CT images exist as numbers. The lab computes frequency, dose and resolution limits. This sheet (IMG-11 — GRE signal (Ernst)) is the form associated with Ernst GRE. Working symbols: $M_0$ , $\alpha$ , $TR$ , $T_1$ , $TE$ , $T_2^*$ $\rightarrow$ $S$ . Ernst angle cos αE = E1 maximises steady-state signal. Spoiled GRE kills leftover Mxy.

Purpose

Purpose: compute

S

from

M_0

\alpha

TR

T_1

TE

T_2^*

in Medical imaging via

S=M_0\sin\alpha\dfrac{1-E_1}{1-\cos\alpha\,E_1}e^{-TE/T_2^*}

S = M0 sinα (1−E1)/(1−cosα E1) e^{−TE/T2*}, E1=e^{−TR/T1}. Use it when a real medical imaging question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

M_0 = 1.000\,\mathrm{a.u.}

\alpha = 15.000\,\mathrm{^{\circ}}

TR = 8.000\,\mathrm{ms}

T_1 = 1000.000\,\mathrm{ms}

TE = 3.000\,\mathrm{ms}

T_2^* = 30.000\,\mathrm{ms}

, the governing relation

S=M_0\sin\alpha\dfrac{1-E_1}{1-\cos\alpha\,E_1}e^{-TE/T_2^*}

yields

S = 0.0447\,\mathrm{a.u.}

. A flip angle, a TR, a steady-state arrow. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

M0 M_01 a.u.Flip angle \alpha15 °TR TR8 msT1 T_11000 msTE TE3 msT2* T_2^*30 ms

Outputs

GRE signal S0.0447 a.u.

Reading speed

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Narration of this film

A flip angle, a TR, a steady-state arrow.

Ernst angle cos αE = E1 maximises steady-state signal. Spoiled GRE kills leftover Mxy.

Reading speed

Watch on YouTube