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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!

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  • $\begingroup$ When you say that $T$ is the duration of the sine sweep, does that mean that $x(t)$ is defined as in your first equation just for $t\in [0,T]$ and is zero otherwise? $\endgroup$
    – Matt L.
    Aug 15, 2020 at 16:23
  • $\begingroup$ @MattL. I’ve given this some further thought. The variable $T$ can also be viewed as the time it takes for the ESS to sweep from frequency $f_1$ to $f_2$. When defined like this, the signal $x(t)$ is still defined for zero to infinity. Which might be easier to calculate? I’m not sure. $\endgroup$ Aug 16, 2020 at 7:36
  • $\begingroup$ OK, but honestly, I don't think that there's a closed form solution for the Fourier transform of that signal. You might be able to compute a good numerical approximation though, using the DFT. $\endgroup$
    – Matt L.
    Aug 17, 2020 at 8:14

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