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Generating my first function

num = 1000;
Amp = 0.05;
time = 1.5;
width = 0.1;
T = 1.5;
(*declaration of continous function*)
With[{w = width, T = time}, 
 pulse[x_] := 
  Cos[2*Pi*x/(w*T)]*(UnitStep[x + w*T/4] - UnitStep[x - w*T/4])]
(*funciton sampling*)
funX = Table[i, {i, -T/2, T/2, T/(num - 1)}];
fun1 = pulse /@ funX + Amp*RandomReal[{-0.5, 0.5}, num];
ListPlot[Transpose[{funX, fun1}], PlotRange -> All, Filling -> Axis, 
 Frame -> True, FrameLabel -> {"Time [s]", "Amplitude [V]", "Pulse"}, 
 PlotLegends -> {"Pulse"}, ImageSize -> Large]

enter image description here

and the second one

With[{\[Delta] = 0.1}, 
 ImpulseResponse[t_] := (1/(2.0*Pi*10.0*Sqrt[1 - \[Delta]*\[Delta]]))*
   Exp[-\[Delta]*2.0*Pi*10.0*t]*
   Sin[2.0*Pi*10.0*Sqrt[1 - \[Delta]*\[Delta]]*t]*HeavisideTheta[t]]
funTF = ImpulseResponse /@ funX;
ListPlot[Transpose[{funX, funTF}], Frame -> True, PlotRange -> All, 
 ImageSize -> Large]

enter image description here

Now the plan was to follow this discussion using

    konv = ListConvolve[Transpose[{funX, funTF}], Transpose[{funX, fun1}]]
    (*{{0.152085}}*)

but for some reason this doesn't work. The ListConvolve[] only returns one value instead of list. Any ideas why?

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You need to tell ListConvolve how to handle "end conditions". Since your sequences are about the same length, everything is an end condition. Hence:

ListPlot[ListConvolve[funTF, funX, {1, -1}]]

will give the circular convolution of the sequences funX and funTF. There are several other choices for the pair of numbers (like {1,-1}) or you can directly specify the desired length of the output. You can read more about these optional arguments in the Details section of the help file.

From your comment, I guess you are expecting linear (rather than circular) convolution. You can do this by zeropadding. For example:

zeropad = Flatten[{ConstantArray[0, Length[funTF]], funX, 
                   ConstantArray[0, Length[funTF]]}];
ListPlot[ListConvolve[funTF, zeropad, {1, -1}], PlotRange -> All]
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  • $\begingroup$ I think you meant ListConvolve[funTF, fun1, {1,-1}]; or num instead of {1,-1}.... but anway... I am not very satisfied with the result, because if you plot it ListPlot[ListConvolve[funTF, fun1, {1,-1}], PlotRange -> All, ImageSize -> Large, Filling -> Axis] you see an unexpected rising at the end which (in my opinion) should not be there and also isn't if you convolve continuous functions using Convolve[ImpluseResponse[t],pulse[t],t,x]. $\endgroup$ – skrat Dec 16 '16 at 14:47

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