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I aquire typically 2D time traces which I process by a 2D Fourier transform. The aquisition is done line by line. First, I let the signal evolve for t1, then an action is performed that possibly changes its frequency and during t2 I sample the signal again in increments of dt:

    |  <<<----->>> | ------->
start       t1     A   t2

Using that scheme, I correlate the frequencies during t1 with those during t2.

However, my signal typically lives near the diagonal in the 2D FFT like this:

Example signal

I wonder whether the sampling could be adjusted to reduce the large areas in the 2D FFT that do not contain any information. Hopefully this would allow to reduce the aquisition time. Is there any sampling scheme, that allows to sample around the diagonal in a 2D FFT? It would kind of mean that the delta frequency between time period 1 and time period 2 are assumed to be small.

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  • $\begingroup$ To your question: Does the signal always occupy the diagonal or could it "evolve" at a different rate? And more generally, would it be possible to provide a little bit more context? I am not sure I understand what a "2D trace" looks like. It sounds like a 1D trace but repeatedly acquired at regular intervals over which the main feature shifts in time (?) $\endgroup$
    – A_A
    Commented Jul 16, 2020 at 9:22
  • $\begingroup$ Sorry that I left this a little unclear. The context is 2D spectroscopy in NMR. My signal is not always located on the diagonal, it is located around the diagonal. Otherwise, you are right a 1D scheme would be suitable. So to say, during time period t1 the signal evolves with a different frequency than during period 2. However, I know that the difference between the frequencies is small. Sampling t1 is expensive, while sampling t2 is cheap. Before the start trigger I have some overhead which is there for every single t1 sampling point but only once there for all t2 sampling points. $\endgroup$
    – P. Egli
    Commented Jul 16, 2020 at 14:57
  • $\begingroup$ OK, but are you interested in a special case? Because this is what you might be seeing here. That could possibly be optimised. Because you don't know what the response is going to be until you have probed for it. So, unless this is a special case, you really must wait for both "times" because the data you record is a function of both. By the way, you only record t2 not t1. t1 is the delay between the "probe" pulses. (?) $\endgroup$
    – A_A
    Commented Jul 16, 2020 at 16:31

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