# 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 "deconvolve" 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 "shift" 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.

• I am not entirely sure if I understand everything correctly, thus I have to ask for clarification. Concerning your point 1: I have an equation for $A(t)$ and the equation for $\exp(-i\omega_0t)$, and therefore I can separate those two parts. I am not sure why I should use $s(t)$ in that case? Concerning 2: What is $s$, $k$ and $T_s$? May 14, 2021 at 20:28

Most importantly, applying a shift in frequency, which the OP is proposing to do, will have no effect on the number of points needed to achieve a desired frequency resolution. The frequency resolution is given by the total number of samples alone and is finest when no further windowing is applied and results in a bandwidth (as the equivalent noise bandwidth) that is the reciprocal of the time duration of the signal:

$$f_b = 1/T$$

Where $$f_b$$ is the equivalent noise bandwidth in Hz and $$T$$ is the time duration of the signal in seconds. When the signal is sampled at $$f_s$$ Hz, this results in a bandwidth of $$f_s/N$$ where $$N$$ is the total number of samples.