I would like to understand how division of round-off errors can lead to large errors???

I am using a method to measure the frequency dependent dissipation in simulations performed by Discontinuous Galerkin(Finite element based) Time Domain method on wave propagation problems. To this end, I excite my domain with a Gaussian pulse and measure the propagated signal $y_1$ and $y_2$ at 2 monitoring points $z_1$ and $z_2$ of the domain. The discrete Fourier transform(DFT) $F_1(f)$ and $F_2(f)$ of both signals are computed and the dissipation $\alpha_{diss}$ is evaluated through :

$\alpha_{diss}(f) = \frac{log(|F_2(f)|/|F_1(f)|)}{z_2-z_1}$

If there is no dissipation, the amplitude spectrums $|F_1|$ and $|F_2|$ should theoretically be the exact same Gaussian $F_{th}$. However, due to finite length signal, the DFT makes round-off errors(different for $y_1$ and $y_2$). As a result, there is an increasing overall error in the measure of $\alpha_{diss}(f)$ which prevents the method to be used far away from the mid-band frequency $f_0$ of the Gaussian spectrum.

The figure below shows this error increase of $|\alpha_{diss}(f)-0|$(in log scale) in red. The blue and cyan curves show the round-off errors $F_2-F_{th}$ and $F_1-F_{th}$ respectively. The mid-band frequency is $f_0=2$ and Gaussian line-width $\Delta f_0=0.5$. We can clearly see that beyond $3\Delta f_0$ the evaluation of $\alpha_{diss}(f)$ is not reliable anymore.

I really wonder why is this happening?

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  • $\begingroup$ How is the amplitude of the "noise" in your output plots when you refine your mesh? $\endgroup$ – nicoguaro Oct 13 '17 at 15:23
  • $\begingroup$ I just realized that the error in $\alpha$ is actually related to the relative error of the DFT. Since far away from the mid-band frequency, the amplitude of the Gaussian signal is on the order of the absolute error $F-F_{th}$ the error in $\alpha$ keeps increasing. $\endgroup$ – Ronan Tarik Drevon Oct 15 '17 at 0:00

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