1
$\begingroup$

My question has two parts: One is general and has to do with the Fourier-Transformation, one has to do with Time-Domain NMR. Both parts are interlinked, of course.

I tried to find out, why people do Time-Domain (TD) NMR instead of Fourier-Transform (FT) NMR, which is up to my knowledge by far the standard.

Part 1 on TD-NMR

On my search on TD-NMR, I found a vendor that states on his website

...low magnetic field results in a low resolution, which is insufficient for obtaining Fourier-transformation frequency spectra.

Certainly, for Fourier-approaches, there has to be a certain number of sample points available to get good spectra. On the other hand, fitting the signal in time-domain with few data points is equally problematic. So, why is it done and is TD-NMR done with some kind of special fitting approach?

Part 2, general

The vendor stated that TD-NMR uses low magnetic fields, which also translates to a low signal-to-noise ratio. Is there a reason why signal processing should be done in TD only because of an SNR issue? I imagine, since the FT is a linear transformation, there should be no gain or loss in terms of SNR, right? Hence, it should not impede the choice of the domain that you would like to work in.

Improve this question
$\endgroup$
3
  • $\begingroup$ Data processing in the time or the frequency domain are equivalent, both have exactly the same SNR if done correctly. TD techniques are important in magnetic resonance because we are 1) dealing with a quantum system that evolves in the time domain and 2) practical implementations of NMR experiments are much more flexible when they are being done with TD methods. Any good textbook on NMR will contain dozens of TD methods that allow precision measurements on certain aspects of interactions. As for the vendor... it sounds like someone is trying to pull your wallet. Ignore that and get a textbook. $\endgroup$
    – CuriousOne
    Feb 10, 2016 at 11:27
  • $\begingroup$ 1) It would be helpful if you defined NMR for the reader. 2) While it is true that a standard FFT should have an output that one can then invert to go back to the time-domain, when processing/converting data it is rarely the case the the software/instrument keeps both the FFT'd result and the original signal. Generally, if one is only given the FFT'd result it is not possible to determine the original time series input. $\endgroup$
    – honeste_vivere
    Apr 13, 2016 at 13:14
  • $\begingroup$ You might be also interested to read about Fourier-transform spectroscopy (see, e.g., thsi question physics.stackexchange.com/a/619546/247642) $\endgroup$
    – Roger Vadim
    Apr 21, 2021 at 9:43

1 Answer 1

Reset to default
1
$\begingroup$

If you tip $M_0$ to XOY plane and recieve the signal, you'll get some complex data points.

Supposing 1024 points were sampled, there should be an array of complex data with 1024 elements.

The Fourier transform of these complex data is a frequency NMR spectra. The resolution of the spectra is decided by the SNR & # of points and the dwell time.

The Inverse Laplace Transform of the envelope of the complex data (the magnitude of each complex element is usually used) is a Time-Domain distribution map(For FID, ILT would produce a T2* distribution map).

Now we could come to the questions:

  1. Compared with TD NMR, FD-NMR need a much higher SNR and a more uniform B0 to produce high resolution results. If the spectral width are too wide, peaks would be more likely to overlap with each other. The spectra would be useless. Furthermore, FT is much more robust to noise than ILT. But they both need enough data points.
  2. Both can suffer with poor SNR.LF-NMR or TD-NMR are compact NMR system, the size are much smaller and they usually use perminate magnet as $B0$. There are no He or N needed. Both price and size are suitable for industrial integration. But using TD-NMR cannot indentify the chemical structure. They can only measue an inexact relaxation distribution map or diffusion map. Fourier-NMR equipements are expansive and the maintenance cost would be much higher than TD-NMR(which has no maintenance cost). But one can indentify and quantify an certain chemical structure using the frequency domain NMR spectra.
Improve this answer
$\endgroup$
1
  • $\begingroup$ Small size and low field do not distinguish FT-NMR instruments from TD-NMR instruments. This was true long ago, but now there are small low-field FT-NMR spectrometers with high resolution, even better than 1 Hz. TD-instruments do not require such high field homogeneity, but are adapted for larger samples, and a greater variety of samples, including solids. TD samples show faster relaxation, so that narrow spectral lines cannot be observed. The focus is instead on the distribution of relaxation rates. It is the different sample relaxation rates that leads to the different analysis approach. $\endgroup$
    – 10ppb
    Apr 21, 2021 at 11:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.