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I have the following two discrete signals

Spktrain1 = {{12.853, 1}, {124.719, 1}, {236.349, 1}, {347.979, 1}};
Spktrain2 = {{1.502, 1}, {113.556, 1}, {225.186, 1}, {336.816, 1}};

I would like first to convolve them with a kernel function (that in my case is a Gaussian function) in order to make the signals continuous. Once I have them, I would like to cross-correlate them obtaining a function of the lag time, which I would like to plot.

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    $\begingroup$ Convolution is a linear operation, so you'd be better off writing out your formulas explicitly and then cross-correlating the kernel functions directly, instead of convolving first and then cross-correlating the result. In essence, interchange the order of sums and integrals to make things simpler. $\endgroup$
    – Roman
    Mar 1, 2021 at 18:19
  • $\begingroup$ Thanks for your answer. Unfortunately I need both of them, but I do not know how to do it in Mathematica. Could you please show me an example? $\endgroup$
    – VDF
    Mar 1, 2021 at 18:24

1 Answer 1

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Define a Gaussian kernel. The width is specified by s

kernel = #2 Exp[-(t - #1)^2/(2 s^2)] &;

Define a function to compute the correlation in terms of the convolution

correlate[x_, y_, t, T_] := Convolve[x, y /. t -> -t, t, T]

Test this by correlating two kernels at different times

correlate[kernel[t1, 1], kernel[t2, 1], t, T]
(* (E^(-((T - t1 + t2)^2/(4 s^2))) Sqrt[π])/Sqrt[1/s^2] *)

As expected the location of the peak is given by t1-t2.

Use a value of s=10 to spread the specified points and correlate.

correlation = 
 Block[{s = 10}, 
  correlate[Total[kernel @@@ Spktrain1], Total[kernel @@@ Spktrain2], 
   t, T]]
(* 1.98775*10^-113 E^((-1.61981 - 0.0025 T) T) + 
 1.98531*10^-48 E^((-1.06167 - 0.0025 T) T) + 
 2.55025*10^-48 E^((-1.06049 - 0.0025 T) T) + 
 1.7299*10^-10 E^((-0.503515 - 0.0025 T) T) + 
 1.94791*10^-10 E^((-0.502335 - 0.0025 T) T) + 
 1.94791*10^-10 E^((-0.502335 - 0.0025 T) T) + 
 12.9801 E^((0.055815 - 0.0025 T) T) + 
 25.9602 E^((0.055815 - 0.0025 T) T) + 
 12.8435 E^((0.056755 - 0.0025 T) T) + 
 7.54594*10^-16 E^((0.613965 - 0.0025 T) T) + 
 7.54594*10^-16 E^((0.613965 - 0.0025 T) T) + 
 5.81383*10^-16 E^((0.616085 - 0.0025 T) T) + 
 3.82712*10^-59 E^((1.17211 - 0.0025 T) T) + 
 2.32725*10^-59 E^((1.17424 - 0.0025 T) T) + 
 8.12734*10^-130 E^((1.73239 - 0.0025 T) T) *)

Plot the results

Plot[correlation, {T, -500, 500}, PlotRange -> All]

The peak of the correlation gives the delay between the two sets of samples

FindMaximum[correlation, {T, 0}]
(* {70.897, {T -> 11.21}} *)
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  • $\begingroup$ mikado thank you very much for your answer. So if I have understood correctly you convolve the two signals with the Kernel function in order to obtain a continuous function. but at the end do you calculate also they cross-correlation? $\endgroup$
    – VDF
    Mar 1, 2021 at 21:57
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    $\begingroup$ You are correct - I've convolved the smoothed functions rather than correlating them. Of course, correlation is convolution with one of the signals reversed. Perhaps you can work it out from here. $\endgroup$
    – mikado
    Mar 1, 2021 at 21:58
  • $\begingroup$ Mikado thank you again for further explanations. Yes I agree I have to reverse the second term in order to obtain the cross-correlation. I have simply applied Reverse[Total[kernel @@@ Spktrain2]], which I think it works $\endgroup$
    – VDF
    Mar 1, 2021 at 22:45
  • $\begingroup$ I have done same trials, and unfortunately I did not get the right results. Can you help me to solve it? I am confused about convolution and cross-correlation in terms of code. $\endgroup$
    – VDF
    Mar 2, 2021 at 9:37

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