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 2n (n is an even integer, in this case 10);
F(w) = º f(t) e-iwtdt (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*2n points in f(t), only 2n 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 = and F(w) would have infinitely narrow and hence lead to unobservable lines, ie 1/ = 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.
Copyright (c) H. S. Rzepa and ICSTM Chemistry Department, 1994, 1995.