Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers and students. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Why do we typically take $n^2$ scans, even in the basic NMR experiment? I've heard various explanations, including that it helps the fast Fourier transform, and would like to know the real reason.

share|improve this question
up vote 6 down vote accepted

If you mean why do we process raw NMR data using $n^2$ data points, then @Klaus's answer is correct. Although, it is actually usually possible to record an arbitrary number of time domain data points; it is just the processing algorithm is limited to how many data points it can handle. You will either lose the extra time domain data (if using less frequency domain data points) or add zero filling (if using more frequency domain points).

If you mean the number of scans as the number of transients added together to give the final raw FID, then this comes about from the requirement for phase cycling a pulse sequence in order to remove unwanted NMR signals that may arise from

  • imperfections in hardware phase (giving rise to quad images)
  • coherent noise
  • generation of undesired magnetization arising in multi-pulse experiments

Here, though, the number of scans is not limited to $n^2$. Historically, the number of scans for a 1D spectrum would typically have been confined to a multiple of 8, as a phase cycle requirement. As you start to need to run greater numbers of scans due to sensitivity issues, then you would need to consider the accumulation of noise as well as signal, as @ron has alluded to (doubling the number of scans only give a $\sqrt{2}$ improvement in S/N; you need to run 4 times as many scans to double S/N). It is through this combined relationship that the number of scans, then, adheres to a $n^2$ increase. If 8 scans (required by phase cycling) didn't give sufficient S/N, then we need to double or quadruple the number of scans to see a real improvement in S/N. And so of for ns=16 or 32 or 64. For more complicated experiments, phase cycling may have required multiples of 16 or more. I remember running a dqf-COSY experiment which had a phase cycle of 24, so the number of scans had to be a multiple of 24. Yes, these experiments took many, many hours to run.

However, (there's always a however) in modern spectrometers (anything newer than about 20 years old), the phase cycling restrictions are largely removed through improvements in hardware design and implementation of gradient pulses. So now we see for most basic 1D experiments, such as a 1H, there is a minimum number of scans of 1. We run many high quality, high resolution, quantitative analyses of samples using long acquisition, single scan experiments. For multi-pulse experiments, there is still a requirement to phase cycle, but even for many 2D and 3D experiments, this is limited to multiples of 1, 2 or 4. Very few experiments these days require large numbers of scans to account for phase cycling.

share|improve this answer
So if I do say, 10 scans (n = 10 in brukeres) in a simple 1D, 1H experiment, then process it, would I be throwing away data points to get it to be 2^n? – Canageek Mar 19 '14 at 16:33
@Canageek - No - you are confusing two separate and independent parameters: Number of data points in the FID (on Bruker systems called TD), and the number of scans (on Bruker systems called ns). Each scan is collected over TD data points, and this is repeated ns times. Each scan (consisting of TD points) is added together to give the final FID. S/N increases as you increase the number of scans. Processing the final FID (of TD data points) is done by the FFT algorithm, and has limited allowed solutions, consisting of SI data points, and which can only handle $2^n$ data points. – long Mar 19 '14 at 22:23
The number of data points in the FID (TD) defines the resolution of the final processed (frequency domain) spectrum. Over a given sweep width, the more data points will result in higher resolution. – long Mar 19 '14 at 22:25
The number of scans defines the number of times you will repeat the acquisition and add all the individual FIDs together to give a final raw (time domain) FID. The final FID still contains TD data points. S/N improves at a rate of $/sqrt 2$ for every doubling of the number of scans, because noise also accumulates over time. For most 1D 1H experiments, the FID consists typically of, 16k, 32k or 64k points. In your example an experiment of 10 scans results in 10 individual FIDs (each of, say, 32k points) added together to give a final FID of 32k points. – long Mar 19 '14 at 22:32
Exactly. Simple 1D experiments such as 1H or 13C{1H} will have a phase cycle of 1, meaning you can do exactly as you say. Many other experiments still have phase cycling requirements which require the number of scans to be multiples of $x$ where $x$ is the phase cycle; this may be as little as 2, but for complicated experiments, and those not using gradient pulses, will still be multiples of 8 or 16, for example. For Bruker pulse programs, the number of required scans is listed in the pulse program (use edpul or edcpul to view), usually right near the bottom; it will be listed as $ns = x * n$. – long Mar 23 '14 at 19:46

The signal to noise ratio (S/N) is proportional to the square root of the number of scans. Hence the more scans, the better the S/N ration

share|improve this answer

The processing of the FID uses the Cooley-Tukey FFT algorithm, which can only process $2^n$ data points.

share|improve this answer
This is the reason why we typically process raw NMR data with n^2 data points, not why we record n^2 scans – long Mar 18 '14 at 21:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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