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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?

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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.

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  • $\begingroup$ 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 $T_s$? $\endgroup$
    – arc_lupus
    May 14 at 20:28
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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.

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