Deconvolution of shifted gaussian function in the frequency range I have a signal defined as
$$A(t)\cdot\exp\left(-i\omega_0t\right)$$
with $A$ the envelope function and $\omega_0$ the carrier frequency. I would like to transfer this signal into the fourier space and back, but need a large amount of points in time to accurately resolve the carrier frequency (the larger the time window, the more points are necessary). To circumvent that issue I intend to use a different strategy by using
$$F(A(t)\cdot\exp\left(-i\omega_0t\right))=\widehat{A}(\omega)\star\delta\left(\omega - \omega_0\right)$$
After the amount of points necessary to resolve $A$ accurately is significantly smaller than to resolve $\omega_0$ that significantly speeds up my calculation. Nevertheless, to apply the inverse Fourier transformation with the original amount of points used for the transform from $A(t)\rightarrow\widehat{A}(\omega)$ I have to &quot;deconvolve&quot; this function again by convolving it with
$$\delta(\omega + \omega_0)$$
but would require having negative frequencies. Is that even possible? If not, are there other ways to &quot;shift&quot; the function back, such that I can use fewer points for my FFT?
It is naive, but you didn't tell why you don't use $s(t) = A(t) \cdot \exp(-i \omega_0 t)$, and calculate $A(t) = s(t) \cdot \exp(i \omega_9 t)$...
I named your signal $s(t)$ here.
Other not so naive things.
You mention number of points, so I assume you are discretizing the signal. And your Fourier transform is a finite length DFT. In that case the Fourier transform will have alias, if you sample rate is enough to capture all the information in $A(t)$ then you can simply calculate $S(\omega) = DFT(s(k T_s))$, and the $A(\omega) = S(\omega + \omega_0~ \textrm{mod}~ F_s)$. Where $T_s$ is the sampling period, and $F_s$ is the sampling frequency, thus $s(k T_s)$ is the k-th sample of your signal.