# Convolution and cross-correlation of two discrete signals

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.

• 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. Mar 1, 2021 at 18:19
• 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?
– VDF
Mar 1, 2021 at 18:24

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}} *)

• 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?
– VDF
Mar 1, 2021 at 21:57
• 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. Mar 1, 2021 at 21:58
• 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
– VDF
Mar 1, 2021 at 22:45
• 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.
– VDF
Mar 2, 2021 at 9:37