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If we have a exponential modified Gaussian function G(t) and we add the first derivative from G(t) to get a new function X(t).

X (t) = G + K*d/dt (G) ----(1) where K is a constant real number multiplier. What would be equivalent process to obtain X(t) in the Fourier domain?

Let me add some context: I am a chemist, sometimes we would like to reduce the width of a peak shaped signal which appears like G(t). The data is discrete e.g. time vs. light absorption. One approach to sharpen this peak is to add a small fraction of a first derivative as shown in (1). The resulting peak is slightly narrower than the original which helps in increasing the resolution if there are multiple peaks. I am curious if the same process (1), can also be done in the Fourier domain. What would be the functional form of (1) in Fourier domain? Later one can do an inverse and obtain X(t). Thanks.

Exponentially modified Gaussian

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If $X(t)=G(t)+KG'(t)$ then $\hat X(\xi)=\hat G(\xi)+2\pi i \xi K \hat G(\xi)=(1+2\pi i \xi K)\hat G(\xi)$, so the required transformation on the frequency domain is scaling by an affine function.

Note that some people define the Fourier with different normalization constants, so $\hat{f'}$ can also be given by $i\xi\hat f$. It depends on whether the Fourier transform is defined as $\int_{\Bbb R} f(x)e^{-ix\xi}\text dx$ or as $\int_{\Bbb R} f(x)e^{-2\pi i x\xi }\text dx$.

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  • $\begingroup$ Thanks, could you share a reference, where I can read more about it? Is zeta the same frequency omega and the hat (caret) represents the Fourier transform? $\endgroup$
    – ACR
    Commented May 22, 2019 at 4:20
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    $\begingroup$ Yes, $\xi$ is the frequency and $\hat f$ denotes the Fourier transform of $f$. I like fourier.eng.hmc.edu/e101/lectures/handout3/node2.html for references on the Fourier transform because they have proofs of some nice properties. $\endgroup$
    – GuPe
    Commented May 22, 2019 at 4:28
  • $\begingroup$ @ Guacho, This link was useful. The instrument output is digital data, we don't know the functional form, if I were to Fourier transform G(t), all we have to do is to multiply "each" discrete value by a constant (1+2*piiK) in the Fourier domain? $\endgroup$
    – ACR
    Commented May 22, 2019 at 16:29
  • $\begingroup$ Not a constant. You would have to multiply by $(1+2\pi i \xi)$ (or $1+i\xi$ if you use a different constant in the complex exponential) $\endgroup$
    – GuPe
    Commented May 22, 2019 at 18:01

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