Date: Sat, 23 Jan. 2021 18:15:41 +00:00
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In the FID pulse sequence the situation is quite simple. We apply an RF pulse to rotate the magnetization with 90 degrees so the transverse component will be maximal and the longitudinal component becomes zero. Then as the spins precess, the acquisition is turned on to measure the induced signal until it vanishes due to {EQUATION()}$T^*_2${EQUATION} decay. The sequence diagram of the FID is shown in Figure 1.
{img fileId="3171" thumb="y" rel="box[g]" width="720" imalign="center" align="center" desc=Figure 1. Sequence diagram of a free induction decay experiment. Left: RF pulse and signal in the laboraroty frame of reference. Right: RF pulse and signal in the rotating frame, i.e. the demodulated signal with zero offset frequency. Note that the laboratory signal on the left is only suggestive since the real Larmor oscillation frequency is far too large to display.}
As was mentioned in Relaxation section, the {EQUATION()}$T^*_2${EQUATION} decay in the FID experiment comes from two components, the {EQUATION()}$T_2${EQUATION} of spin-spin relaxation and the {EQUATION()}$T'_2${EQUATION} due to external field inhomogeneities. To get a clue of the practical values of this {EQUATION()}$T^*_2${EQUATION} we here present an estimation of the dephasing time {EQUATION()}$T'_2${EQUATION}.
Assume we have an external field of 1.5 T with a homogeneity of 1 point per million (ppm), and we are performing NMR experiment on protons. One can say that the FID signal disappears roughly when the spins have gained a phase difference of {EQUATION()}$\pi${EQUATION}. We can calculate the time $\tau$ needed for this from the following:
::{EQUATION()}
\label{T2*_estimate1}
\gamma ( \Delta B_0 ) \tau = \pi
{EQUATION}(1)::
::{EQUATION()}
\label{T2*_estimate2}
42.58 \mathrm{[MHz/T]}\times 2 \pi \times 1.5 \mathrm{[T]} \times 1 \mathrm{[ppm]} \times \tau= \pi
{EQUATION}(2)::
::{EQUATION()}
\label{T2*_estimate3}
\tau \approx 7.8 \hspace{2pt} \mathrm{ms}
{EQUATION}(3)::
As can be seen, an inhomogeneity as small as 1 ppm eliminates the signal in less than 10 ms in a device with the commonly used 1.5 T.