How to plot the convolution of dirac delta series with a sine function [closed]

Question

I am new to mathematica. I do the convolution of dirac delta :(DiracDelta(x-10)" with "sine(t),t=0-pi". How to plot the output ? Theoretically, the full wave should appear at the location of the singularity.

Answer 1

Using convolution theorem http://en.wikipedia.org/wiki/Convolution_theorem that Fourier of the convolution of two functions is the same as multiplications of their Fourier transforms then

Use UnitStep to generate the time limited sin function to convolve with, like this

Plot[UnitStep[t] UnitStep[Pi - t] Sin[t], {t, -3 Pi, 3 Pi}]

and now apply the convolution theorem as above (earlier I forgot to InverseForurierTranform at the end, thanks to OleksandrR for noticing)

Clear[t, w];
f1 = DiracDelta[t - 10];
f2 = UnitStep[t] UnitStep[Pi - t] Sin[t];
y = FourierTransform[f1, t, w]  FourierTransform[f2 , t, w] ;
conv = InverseFourierTransform[y, w, t]

which gives

((1/Sign[10 - t] - 1/Sign[10 + Pi - t])*Sin[10 - t])/(2*Sqrt[2*Pi])

Plotting it

Plot[conv, {t, 0, 50}]

Using Convolve[] directly as suggested by OleksandrR below seems to be faster on V8.04.

Here is using Convolve[] directly. Much faster also. (I do not know why I did not try this first).

Clear[t, z];
f1 = DiracDelta[t - 10];
f2 = UnitStep[t] UnitStep[Pi - t] Sin[t];
conv2 = Convolve[f1, f2, t, z]
Plot[conv2, {z, 0, 50}, Exclusions -> None]

where conv2 above is

Piecewise[{{-Sin[10 - z], 10 <= z <= 10 + Pi}}, 0]

Hopefully not more errors in this now. It is nice that Mathematica offers many ways and functions to analyze a problem.