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Physics Wolfram Language

What's with Fourier Transforms in Mathematica?

Yaj Bhattacharya

Posted 10 years ago

What's the issue with Fourier Transforms in Mathematica? Why does it always have this DiracDelta? In[1]:= << "FourierSeries`" In[2]:= f[x_] := (2 b/\[Pi])^(3/4) E^-br^2 In[3]:= ft[k_] = FourierTransform[f[x], x, k] Out[3]= 2 b^(3/4) E^-br^2 (2/\[Pi])^(1/4) DiracDelta[k] In[4]:= ft[k_] = FourierTransform[f[x], x, k, FourierParameters -> {-1, 1}] Out[4]= b^(3/4) E^-br^2 (2/\[Pi])^(3/4) DiracDelta[k]

What's the issue with Fourier Transforms in Mathematica? Why does it always have this DiracDelta?

In[1]:= << "FourierSeries`"

In[2]:= f[x_] := (2 b/\[Pi])^(3/4) E^-br^2

In[3]:= ft[k_] = FourierTransform[f[x], x, k]

Out[3]= 2 b^(3/4) E^-br^2 (2/\[Pi])^(1/4) DiracDelta[k]

In[4]:= ft[k_] = 
 FourierTransform[f[x], x, k, FourierParameters -> {-1, 1}]

Out[4]= b^(3/4) E^-br^2 (2/\[Pi])^(3/4) DiracDelta[k]

POSTED BY: Yaj Bhattacharya

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David Keith

Posted 10 years ago

Hi Yaj, In this example, the argument is also constant because the bx is a single symbol. for b*x you need either the asterisk or, more commonly, a space between b and x. Mathematica outputs the delta when it is in fact correct. Consider: In[1]:= FourierTransform[Exp[- x^2/b^2], x, k] Out[1]= E^(-(1/4) b^2 k^2)/(Sqrt[2] Sqrt[1/b^2]) In[2]:= FourierTransform[UnitStep[x - 1] - UnitStep[x - 2], x, k] Out[2]= -((I E^(I k) (-1 + E^(I k)))/(k Sqrt[2 \[Pi]])) Best, David