STFT calculation with Gaussian Window

$$ f(t)=\exp(jat^2) \,\,\, and \,\,\, g(t)\,\,is\,\, a\,\, Gaussian\,\, Window:$$

$$ g(t)= \left (πσ^2\right)^{\frac{-1}{4}}\exp\left (\frac{-t^2}{2σ^2} \right ) , \,\,\,\,\,\,\left \|g(t) \right \|=1 $$ $$ $$

I want to find the STFT (Short-Time-Fourier-Transform) of f(t) and prove that:

$$ $$

$$Psf(u,\xi)=|Sf(u,\xi)|^2=\left (\frac{4πσ^2}{1+4α^2σ^4}\right )^{\frac{1}{2}}\exp\left (\frac{-σ^2(\xi-2au)^2}{1+4a^2σ^4} \right )$$

$$ $$

I started by calculating the Fourier Tranformation of f(t) and found that $$f(t)=\exp(jat^2)\leftrightharpoons K \cdot \exp\left ( \frac{-ω^2}{4α}\right)=F(ω) $$

where K is a constant

$$ $$

I am confused with the next steps i have to follow in order to make use of F(ω) in order to calculate STFT. Do i have to use the definition of STFT?:

$$Sf(u,\xi)= \langle\,f,g_{u\xi},\rangle= \int_{-\infty}^{\infty} f(t) \cdot g(t-u) \cdot e^\left (-j \,ξ \,t \right )dt $$

I tried to do so but i didn't manage to calculate the integral above. Is there an easier way to calculate STFT by using any properties?Any help is much appreciated!Thanks in advance!

Answer

so MJ, my approach to using the Gaussian window for the STFT is first to use the "ordinary frequency" definition of the continuous-time Fourier Transform:

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

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

and start with this really cool isomorph of the Fourier Transform:

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

and use the well known time-scaling and translation or frequency-scaling and translation theorems of the Fourier Transform to get you

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

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

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

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

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

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.