Hello! I am sttrugled with a question concerning Fourier transforms. The problem is as follows: I have a function initially defined as:
where \(\displaystyle \,\tau\,\) is a very small quantity. This has real and imaginary parts as:
If you take the Fourier transform of this, the first part is identically zero, whereas the second part is a Cauchy principal value integral, and the result in \(\displaystyle \,\omega\,\) space is:
\(\displaystyle \begin{eqnarray}
G(\omega)\,=\,-i\,\pi\,\nu\, sign(\omega)
\end{eqnarray}\)
Everything is fine till now. Now, suppose I want to calculate the following function related to \(\displaystyle G(\omega)\,\) as:
It is clear that for \(\displaystyle \,g=0,\,\) I recover the previous result, as expected. The problem arises now when I take the (inverse)Fourier trasform of \(\displaystyle \, H(\omega)\,\) to come back to \(\displaystyle \,t\,\) space. In this case, the result is as follows:
You can now see the problem, in the case \(\displaystyle \,g=0,\,\) the first term completely cancels! But it remains the other term, that will provide the \(\displaystyle \,G(\omega)\,\) I defined above. However, I'm concerned with the disappearance of the first term, since it is the relevant one and the Dirac delta in time completely disappears.
Can anyone see what is going on here? Does it have to be with the (arbitrary) definition of the Fourier transform of \(\displaystyle \,G(\omega)?\,\) What I mean here by arbitrary is that, inserting the \(\displaystyle \,\delta(t)\,\) term doesn't change the function in \(\displaystyle \,\omega\,\) when you Fourier transform. Any help is really appreciated here, thanks!!
\(\displaystyle \begin{eqnarray}
F_{0}(t)\,=\,\dfrac{-i\,\nu}{i\,t\, + \,sign(t)\,\tau}
\end{eqnarray}\)
F_{0}(t)\,=\,\dfrac{-i\,\nu}{i\,t\, + \,sign(t)\,\tau}
\end{eqnarray}\)
where \(\displaystyle \,\tau\,\) is a very small quantity. This has real and imaginary parts as:
\(\displaystyle \begin{eqnarray}
-i\, \pi\,\nu\,\delta(t)\,sign(t)\, -\,\dfrac{\nu}{t}
\end{eqnarray}\)
-i\, \pi\,\nu\,\delta(t)\,sign(t)\, -\,\dfrac{\nu}{t}
\end{eqnarray}\)
If you take the Fourier transform of this, the first part is identically zero, whereas the second part is a Cauchy principal value integral, and the result in \(\displaystyle \,\omega\,\) space is:
\(\displaystyle \begin{eqnarray}
G(\omega)\,=\,-i\,\pi\,\nu\, sign(\omega)
\end{eqnarray}\)
Everything is fine till now. Now, suppose I want to calculate the following function related to \(\displaystyle G(\omega)\,\) as:
\(\displaystyle \begin{eqnarray}
H(\omega)\,=\,\dfrac{-i\,\pi\,\nu \,sign(\omega)}{1\,-\,i\,g\,sign(\omega)}
\end{eqnarray}\)
H(\omega)\,=\,\dfrac{-i\,\pi\,\nu \,sign(\omega)}{1\,-\,i\,g\,sign(\omega)}
\end{eqnarray}\)
It is clear that for \(\displaystyle \,g=0,\,\) I recover the previous result, as expected. The problem arises now when I take the (inverse)Fourier trasform of \(\displaystyle \, H(\omega)\,\) to come back to \(\displaystyle \,t\,\) space. In this case, the result is as follows:
\(\displaystyle \begin{eqnarray}
H(t)\,=\,-i\,\pi\,\nu\, \dfrac{g}{1\,+\,g^{2}}\,\delta (t)\, -\,\dfrac{\nu}{(1\,+\,g^{2})\,t}
\end{eqnarray}\)
H(t)\,=\,-i\,\pi\,\nu\, \dfrac{g}{1\,+\,g^{2}}\,\delta (t)\, -\,\dfrac{\nu}{(1\,+\,g^{2})\,t}
\end{eqnarray}\)
You can now see the problem, in the case \(\displaystyle \,g=0,\,\) the first term completely cancels! But it remains the other term, that will provide the \(\displaystyle \,G(\omega)\,\) I defined above. However, I'm concerned with the disappearance of the first term, since it is the relevant one and the Dirac delta in time completely disappears.
Can anyone see what is going on here? Does it have to be with the (arbitrary) definition of the Fourier transform of \(\displaystyle \,G(\omega)?\,\) What I mean here by arbitrary is that, inserting the \(\displaystyle \,\delta(t)\,\) term doesn't change the function in \(\displaystyle \,\omega\,\) when you Fourier transform. Any help is really appreciated here, thanks!!
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