Help with fourier transform

Hello.

How does this function do Fourier transform?

f(t)=(1/√2)e^(-t^2/2a^2)

A is a constant.

Answer 1

[Star Strider]

Try this:

syms a t w 
f(t)=(1/sqrt(2))*exp(-t^2/2*a^2)
F(w) = int(f*exp(1j*w*t), t);
F(w) = simplify(F, 'Steps', 050)

producing:

F(w) =
((-pi)^(1/2)*erf((2^(1/2)*(t*a^2*1i + w))/(2*(-a^2)^(1/2)))*exp(-w^2/(2*a^2)))/(2*(-a^2)^(1/2))

or:

[YiJing Pan]

Thanks.

How to draw this Fourier transform?

[Bruno Luong]

This results looks odd.The FT of a Gaussian is a Gaussian

http://www.cse.yorku.ca/~kosta/CompVis_Notes/fourier_transform_Gaussian.pdf

Not sure why MATLAB returns ERF term (actually I do know).

[Star Strider]

My pleasure.

This calculates the indefinite integral. You need to substitute the appropriate time values for ‘t’, and then evaluate it as you would for any integral. (If the ‘t’ values are symmetrical, for example [-T +T] some terms may cancel, resulting in a simpler expression for the definite integral.) Then plot it as a function of ‘w’ (actually ω or ). Add a constant of integration as well, if you want to.

It is also necessary to provide numerical values for ‘a’ and ‘T’:

syms a t T w 
f(t)=(1/sqrt(2))*exp(-t^2/2*a^2)
F(w) = int(f*exp(1j*w*t), t, -T, T);
F = subs(F,{a,T},{5, 10});
F(w) = simplify(F, 'Steps', 500)

a = 1;
figure
fplot(real(F(w)), [-20 20]*pi)
hold on
fplot(imag(F(w)), [-20 20]*pi)
hold off

The evaluated and substituted function is then:

Experiment to get the result you want.

[Yi-Jing Pan]

Thank you very much!!!!

I will try it !

[Star Strider]

My pleasure!

[Bruno Luong]

This formula is wrong. The ERF terms must be removed, otherwise it is not Fourier transform.