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I have asked this at the Math SE also.

Predicating this on the definition of the continuous Fourier Transform preferred by most electrical engineers:

$$ X(f) \triangleq \mathscr{F} \Big\{ x(t) \Big\} \triangleq \int\limits_{-\infty}^{+\infty} x(t) \, e^{-i 2 \pi f t} \ \mathrm{d}t $$

and inverse:

$$ x(t) \triangleq \mathscr{F}^{-1} \Big\{ X(f) \Big\} = \int\limits_{-\infty}^{+\infty} X(f) \, e^{+i 2 \pi f t} \ \mathrm{d}f $$

Even with different signs on $i$, the elegant symmetry between the forward transform and inverse should be clear. And it makes remembering the Duality property, Parseval's theorem easy:

If $X(f) = \mathscr{F} \Big\{ x(t) \Big\}$, then $x(-f) = \mathscr{F} \Big\{ X(t) \Big\}$.

$$ X(0) = \int\limits_{-\infty}^{+\infty} x(t) \ \mathrm{d}t $$

$$ x(0) = \int\limits_{-\infty}^{+\infty} X(f) \ \mathrm{d}f $$

$$ \int\limits_{-\infty}^{+\infty} \Big| x(t) \Big|^2 \ \mathrm{d}t = \int\limits_{-\infty}^{+\infty} \Big| X(f) \Big|^2 \ \mathrm{d}f$$

NO nasty asymmetrical scaling factors to worry about!! (Just remember the $2\pi$ in the exponent.) This is why EE's like this definition of the Fourier Transform.

Given this definition, then the Fourier transform of the gaussian function (exponent scaled as shown) is itself:

$$ \mathscr{F} \Big\{ e^{- \pi t^2} \Big\} = e^{- \pi f^2} $$


So, harmonizing the parameters and symbols, the normalized Gabor "mother wavelet" (with this parameter $\mathcal{F}$) is simply a complex sinusoid with a gaussian "window" or "envelope":

$$\begin{align} w(t) &\triangleq e^{-\pi t^2} \, e^{i 2 \pi \mathcal{F} t } \\ &= e^{\pi (i\mathcal{F})^2} \left(e^{-\pi t^2} \, e^{i 2 \pi \mathcal{F} t } e^{-\pi (i\mathcal{F})^2} \right) \\ &= e^{-\pi \mathcal{F}^2} e^{-\pi (t-i\mathcal{F})^2} \\ \end{align}$$

The Fourier Transform isn't too hard to get:

$$ W(f) = e^{- \pi (f-\mathcal{F})^2} $$

My question is, in the literature, has this been generalized a little more and does this have a name? (Like where can I read about it?)

$$\begin{align} w(t) &\triangleq e^{-\pi t^2} \, e^{i \pi \beta t^2} \, e^{i 2 \pi \mathcal{F} t } \\ &= e^{-\pi (1 - i\beta) t^2} e^{i 2 \pi \mathcal{F} t } \\ &= e^{-\pi (\sqrt{1 - i\beta} \, t)^2} e^{i 2 \pi \mathcal{F} t } \\ &= e^{-\pi \mathcal{F}^2} e^{-\pi (\sqrt{1 - i\beta} \, t - i\mathcal{F})^2} \\ \end{align}$$

This makes this windowed sinusoid, a windowed "chirp" function. We need both the $\beta$ parameter and the $\mathcal{F}$ parameter because the scaling on the width of the window is still normalized to 1.

I believe the Fourier Transform is

$$ W(f) = \frac{1}{\sqrt{1 - i\beta}} \, e^{- \pi (f/\sqrt{1 - i\beta} -\mathcal{F})^2} $$

This can be generalized one step further by putting in an exponential "ramp" parameter $\lambda$

$$\begin{align} w(t) &\triangleq e^{-\pi t^2} \, e^{i \pi \beta t^2} \, e^{i 2 \pi \mathcal{F} t } e^{2 \pi \lambda t } \\ &= e^{-\pi (1 - i\beta) t^2} e^{i 2 \pi (\mathcal{F}-i\lambda) t } \\ &= e^{-\pi (\sqrt{1 - i\beta} \, t)^2} e^{i 2 \pi (\mathcal{F}-i\lambda) t } \\ &= e^{-\pi \mathcal{F}^2} e^{-\pi (\sqrt{1 - i\beta} \, t - i(\mathcal{F}-i\lambda))^2} \\ &= e^{-\pi \mathcal{F}^2} e^{-\pi (\sqrt{1 - i\beta} \, t - i\mathcal{F}-\lambda))^2} \\ \end{align}$$

And it looks like the Fourier Transform is

$$ W(f) = \frac{1}{\sqrt{1 - i\beta}} \, e^{- \pi (f/\sqrt{1 - i\beta} -\mathcal{F} + i \lambda)^2} $$

Does this generalization exist in the lit somewhere and, if so, can I read about it without a pay-wall?

Also, it appears to me that, in general, we can say that

$$ \mathscr{F} \Big\{ e^{a t^2 + b t + c} \Big\} = e^{A f^2 + B f + C} $$

where the constants $A$, $B$, and $C$ can be explicitly mapped from $a$, $b$, and $c$. It appears to me that the mapping is:

$$\begin{align} A &= \frac{\pi^2}{a} \\ B &= i \frac{\pi b}{a} \\ C &= c - \frac{b^2}{4a} - \tfrac{1}{2}\log\left(-\frac{a}{\pi}\right) \\ \end{align}$$

and the inverse mapping (which should be self-similar) is:

$$\begin{align} a &= \frac{\pi^2}{A} \\ b &= -i \frac{\pi B}{A} \\ c &= C - \frac{B^2}{4A} - \tfrac{1}{2}\log\left(-\frac{A}{\pi}\right) \\ \end{align}$$

Looks like $\Re\{a\}<0$ and $\Re\{A\}<0$ for the integrals to converge and for the $\log(\cdot)$ to be real and finite in the mapping.

This appears to be true for quadratics in the exponent. Is it also true for higher-order polynomials in the exponent? Does this also exist in the lit somewhere?

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  • $\begingroup$ Chirplet transform maybe? $\endgroup$ – geometrikal May 1 '18 at 10:46
  • $\begingroup$ yup @geometrikal. that's what the math guys told me. it funny but i hadn't heard the term "chirplet" before this. $\endgroup$ – robert bristow-johnson May 1 '18 at 18:03

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