STFT calculation with Gaussian Window

Question

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

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

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 1

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.

Answer 2

okay, i am gonna change the variables a little and the names of variables, to make this more consistent with bone-head electrical engineers (which is me).

my signal is

$$ x(t) = e^{j \alpha t^2} $$

and my window is

$$ w(t-u) = \left( \pi\sigma^2 \right)^{-1/4} \ e^{\frac{-(t-u)^2}{2\sigma^2}} $$

the product is

$$\begin{align} x(t)w(t-u) &= e^{j \alpha t^2} \left( \pi\sigma^2 \right)^{-1/4} \ e^{\frac{-(t-u)^2}{2\sigma^2}} \\ \\ &= \left( \pi\sigma^2 \right)^{-1/4} \ e^{j\alpha t^2} e^{\frac{-(t-u)^2}{2\sigma^2}} \\ \\ &= \left( \pi\sigma^2 \right)^{-1/4} \ e^{j\alpha t^2} e^{\frac{-t^2+2tu-u^2}{2\sigma^2}} \\ \\ &= \left( \pi\sigma^2 \right)^{-1/4} \ e^{(-1/(2\sigma^2) + j\alpha)t^2 + tu/\sigma^2 - u^2/(2\sigma^2)} \\ \\ &= e^{(-1/(2\sigma^2) + j\alpha)t^2 + tu/\sigma^2 - u^2/(2\sigma^2) - \log(\pi\sigma^2)/4} \\ \\ &= e^{a t^2 + b t + c} \\ \end{align} $$

so

$$\begin{align} a &= \frac{-1}{2\sigma^2} + j\alpha\\ \\ b &= \frac{u}{\sigma^2} \\ \\ c &= -\tfrac{1}{4}\log(\pi\sigma^2) - \frac{u^2}{2\sigma^2}\\ \end{align} $$

then

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

and

$$\begin{align} A &= \frac{\pi^2}{a} \\ &= \frac{\pi^2}{\frac{-1}{2\sigma^2} + j\alpha} \\ &= \frac{-\pi^2}{\frac{1}{4\sigma^4} + \alpha^2} \left(\frac{1}{2\sigma^2} + j\alpha \right) \\ \\ B &= j \frac{\pi b}{a} \\ &= j \frac{\pi \frac{u}{\sigma^2}}{\frac{-1}{2\sigma^2} + j\alpha} \\ &= \frac{\pi u}{\alpha\sigma^2 + \tfrac{j}{2} } \\ \\ C &= c - \frac{b^2}{4a} - \tfrac{1}{2}\log\left(-\frac{a}{\pi}\right) \\ &= -\tfrac{1}{4}\log(\pi\sigma^2) - \frac{u^2}{2\sigma^2} - \frac{\left(\frac{u}{\sigma^2}\right)^2}{4(\frac{-1}{2\sigma^2} + j\alpha)} - \tfrac{1}{2}\log\left(-\frac{\frac{-1}{2\sigma^2} + j\alpha}{\pi}\right) \\ &= - \frac{u^2}{2\sigma^2} - \frac{\left(\frac{u}{\sigma^2}\right)^2}{4(\frac{-1}{2\sigma^2} + j\alpha)} - \tfrac{1}{2}\log\left(\frac{\frac{1}{2\sigma} - j\alpha\sigma}{\sqrt{\pi}}\right) \\ &= - \frac{u^2}{2\sigma^2}\left(1 - \frac{1}{1 - j2\alpha\sigma^2} \right) - \tfrac{1}{2}\log\left(\frac{\frac{1}{2\sigma} - j\alpha\sigma}{\sqrt{\pi}}\right) \\ \end{align}$$

simplify this to get an answer in terms of ordinary frequency $f$ and then substitute $ f = \frac{\xi}{2 \pi}$ and you will have your answer in terms of angular frequency $\xi$.