Prove that exponential sweep sine decay 6 dB per octave

Question

Exponential sweep sine (ESS) is a good excitation signal to measure the impulse response of an acoustic system, such as loudspeaker - room - microphone system. An ESS has the form of

$$ x(t) = \sin\Big[\frac{2\pi f_1 T}{R} (e^{tR/T} - 1) \Big] $$ where $t$ is the time variable in second, $T$ is the duration of the sine sweep, $f_1$ and $f_2$ are the start and final frequency in Hz, respectively, and $R=\ln(f_2/f_1)$ is the exponential sweep rate.

The signal has a constant magnitude over time, and the magnitude spectrum shows a decay, of 6 dB per octave to be precise. This magnitude decay must be compensated when the impulse response is calculated with the deconvolution procedure.

In this question Matt states that ESS doesn't have a close form Fourier transform. So my question is that is it possible to prove this 6 dB per octave decay. In this case we only need the magnitude of the Fourier transform of ESS.

Related question: Calculating the inverse filter for the (exponential) sine sweep Method

Answer 1

This JAES paper gives a close form of the Fourier transform of the synchronized swept-sine (SSS) signal which has the same form as the ESS

$$ x(t) = \sin \big\{2\pi f_1L \big[\exp(t/L) -1 \big]\big\} $$ where $$ L = \frac{1}{f_1} \mathrm{round}\left[\frac{\hat{T}f_1}{\ln(f_2/f_1)}\right] $$ and $\hat{T}$ is the approximate time length of $x(t)$.

The authors derive the Fourier transform of the analytic signal of the SSS: $$ z(t) = \exp\Big\{j2\pi f_1L \big[\exp(t/L) -1\big]\Big\} $$ and $$ Z(f) = \int_{-\infty}^\infty z(t) e^{-j2\pi ft} dt $$

After a long derivation, we have the final result

$$ Z(f) = \exp\left[ j2\pi fL\left( 1-\ln\frac{f}{f_1} \right) \right] \times \sqrt{\frac{L}{f}} \exp(j\frac{\pi}{4}) $$

The magnitude of $Z(f)$ is inversely proportional to $\sqrt{f}$, i. e. the Fourier spectrum falls down by 3 dB/octave, and the power spectrum decreases by 6 dB/octave.