# Fourier Transform of an Exponential Sine Sweep

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.

• 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? Aug 15, 2020 at 16:23
• @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. Aug 16, 2020 at 7:36