The Exponential Sine Sweep (ESS), according to Farina [1], can be described by the following formula:
$$x(t)=\sin\left(\frac{2\pi f_1 T}{R}\left(e^{\frac{t R}{T}} -1\right) \right)$$
where,
$t$ - time variable (seconds)
$T$ - the duration of the sine sweep. This can be viewed as the time it takes for the ESS to sweep from frequency $f_1$ to frequency $f_2$. (seconds)
$f_1$, $f_2$ - the start and final frequency, respectively (Hz)
$R = ln(\frac{f_2}{f_1})$ - exponential sweep rate
Now my question: what is the mathematical expression for the Continuous Fourier Transform of the ESS, i.e. what is:
$$X(j\omega) = \int_{-\infty}^{\infty}{x(t) e^{-j\omega t}} dt$$
I think it is possible to express the Continuous Fourier transformation as a LaPlace transformation by first assuming causality and absolutely integrability of the ESS. The LaPlace variable $s$ can then be replaced by $j\omega$ to obtain the Continuous Fourier Transform.
The LaPlace transformation can be expressed as:
$$X(s) = \int_{0}^{\infty}{x(t) e^{-st}} dt$$
However, this is as far as I'm able to solve the expression. I can't seem to figure out how to solve this integral further. Help is appreciated.
Thanks in advance!