Two questions must now be answered; how
frequently does one sample the FID, and for how long? The answer lies in the
Nyquist Theorem, which states that to define a full (2p) cycle of a sine
wave, its intensity must be sampled at least twice during one cycle. We need
to sample at intervals which will therefore allow the highest frequency sine
wave to be sampled, which in the example set out above is actually 600 Hz.
Sampling must therefore occur every 1/2*600 = 0.0008333 seconds. For how
long? Well, to avoid losing information, our digital spectrum must eventually
be able to distinguish between two peaks in the NMR spectrum say 0.293 Hz
apart, similar in magnitude to the smallest couplings normally seen in
standard samples, and referred to as the required **digital resolution**
of the spectrum. This actually needs 2048 digital points in a spectrum that
will be eventually 600 Hz wide (600/2048 = 0.293). Why did we pick exactly
**0.293 Hz** and hence require exactly **2048** points? Because all
this digital information is going to be processed using a mathematical
technique known as Fourier Transformation (FT), and FTs are particularly fast
(*ie* they are FFTs) when the number of points processed is exactly 2^{n}
(n is an even integer, in this case 10);

F(w) = º f(t) e^{-iwt}dt (5)

Here, the FID = f(t), and F(w) is the same data expressed as
frequencies (=1/T) rather than as times and is called a **frequency domain
spectrum.** This is the form we know for "conventional" NMR spectra, ie
frequencies relative to an internal standard such as TMS. Equation (5) cannot
be integrated analytically, and numerical methods such as the Cooley-Tukey
algorithm have to be applied. The right hand side of equation (5) can also be
expressed as;

F(w) = ºf(t) cos -wt + º f(t)i sin -wt (6)

which means that F(w) has two
components, referred to as the real and imaginary parts. Each component
contains the same frequency data but with a different combination of phase
and amplitude. Only the real component is normally displayed. For 2*2^{n} points in f(t),
only 2^{n} unique
frequencies in F(w) are obtained. The total number of points to be
measured therefore to achieve an eventual resolution of 0.293Hz over a width
of 600 Hz actually corresponds to 2 * 600/0.293 = 4096 points, and our total
sampling time is 4096 * 0.0008333 = 3.41 seconds. At the end of this time,
the time constants defining the exponential decay of q means it is
normally close to zero, f(t) is effectively zero and the entire cycle can
start again with a new pulse.
Finally, we note that the time
constants involved in the exponential decay of q (T1 and T2) are
manifested in the Fourier Transformed function F(w) *via* 1/T,
which has the same unit (Hz) used to measure the width of an NMR peak. Put
another way, if **M** did not decay at all, T1 = T2 = ƒ and
F(w) would have infinitely narrow and hence lead to unobservable
lines, ie 1/ƒ = 0 Hz. This also explains why it is not a good idea
to clean NMR tubes with chromic acid. Traces of this (paramagnetic) substance
provide an excellent mechanism for **M** to relax, T1 ~ 0, and hence the
peaks are infinitely wide,* ie* also unobservable.

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Copyright (c) H. S. Rzepa and ICSTM Chemistry Department, 1994, 1995.