# Need an Identity for CTFT Polynomial Raised to an Exponent

I need to the find the inverse continuous time Fourier transform for unitary angular frequency of the following signal:

$e^{a\omega^2 - b\omega + c}$ where $a$ and $b$ and $c$ are real numbers and I want to find the corresponding time-domain signal.

I was able to get the $e^{a\omega^2 + c}$ term transformed, but I was unable to find a transform for $e^{- b\omega}$ that didn't result in a distribution.

Without doing math and looking at the symmetry, I want to say that the $e^{- b\omega}$ term does a complex time shift, but now my mind hurts. Can somebody look up the identity?

--edit-- This spectrum doesn't need to be symmetric, or causal, so expecting some solutions to have imaginary time could be correct. Can somebody comment on this?

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A general method that helps is to use the technique of completing the square so that $\exp(a\omega^2-b\omega+c)$ can be expressed in the form $$\exp\left((\alpha\omega-\beta)^2 + \gamma\right)$$ which is usually easier to deal with. In fact you seem to know how to handle terms of this form since you say that you were able to get an answer for $\exp(a\omega^2+c)$. –  Dilip Sarwate Nov 5 '12 at 13:31
@DilipSarwate Well, its possible to do exp(aw^2) by looking at the tables on wikipedia for the Gaussian :-) –  Mikhail Nov 6 '12 at 4:59
Read my comment on Hilmar's answer regarding restrictions on when the "tables on wikipedia for the Gaussian" can be used. –  Dilip Sarwate Nov 6 '12 at 13:03

Dilip's idea is an excellent suggestion. Let me slightly modify it. You can rewrite $$a\omega ^{2}-b\omega +c = a(\omega -\omega_{0} )^{2}+d$$ So the whole thing can be written as
$$e^{d}\cdot e^{a(\omega -\omega_{0} )^{2}}$$
$$x(t) = k\cdot e^{\tilde{a}t^{2}}\cdot e^{j\omega _{0}t}$$ where $k$, $\tilde{a}$, and $\omega _{0}$ are determined by the original a, b, and c. a better be a negative number, though, otherwise I don't think that the transform would be convergent.
As you say, $a$ better be a negative number. Indeed, if $a$ is positive, then "...essentially a Gaussian shifted in frequency..." no longer holds and the proposed form of the inverse transform is not valid either. –  Dilip Sarwate Nov 6 '12 at 2:20