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I have the equation of the power spectral density $S(f)$. I seek $(H,f)$ of each sinusoid which $H$ is the magnitude and $f$ is the frequency.

How to calculate $H$ from $S(f)$?

$S(f)$ is the spectral density of a random signal; the Fourier Transform of the autocorrelation function of the time series.

To reproduce a time series, I need (H, T) for each regular sinusoidal component.

I calculated the time series eta(x,t) = sum(H/2 * cos(kx-2pift + phi))

The random phase (phi) with radians unit is randomly distributed between [-pi, pi]. Phi = 2pi(rand(1,length(F))-0.5), where F here is the wave frequency.


S(f) is the spectral density of random signal. So it was the fast Fourier transformation of the autocorrelation function of the time series.

to reproduce a time series, I need $(H, T)$ for each regular wave.

Than I calculate the time series eta (x,t) = sum(H/2 * cos(k*x-2*pi*f*t + phi))

The random phase (phi) with radians unit is randomly distributed between [-pi, pi]. Phi = 2*pi*(rand(1,length(F))-0.5), where F here is the wave frequency.

I wish it was more clearly.

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    $\begingroup$ you have $S(f)$. That is very close is what you describe you're looking for: it gives you the power for every frequency component at each frequency $f$. Not quite sure what the question then is? $\endgroup$ Commented Apr 12, 2022 at 11:21
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    $\begingroup$ Removing the discrete-signals tag, since this is not a discrete signal, and the python tag, since this is not involving Python. $\endgroup$ Commented Apr 12, 2022 at 11:22

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The power spectral density (PSD) is the power over a given resolution bandwidth (for example, within 1 Hz of bandwidth when the units for the PSD are given in "per Hz", such as a phase power spectral density given in $rad^2/Hz$. So what is reported for each value of the PSD is the total power over that bandwidth.

We could achieve that same PSD if there was exactly one sinusoid in each span of the resolution bandwidth, in which case the value reported is the rms magnitude squared for that sinusoid. Any phase information is lost in the power computation. So assuming the OP seeks the magnitude, it would be given by $\sqrt{2S(f)}$ for the peak fluctuation in the same units as the PSD (whether it be radians, frequency or any other unit), or simply $\sqrt{S(f)}$ for the rms fluctuation. The notation $S(f)$ typically refers to the one-sided PSD, which extends from DC and over positive frequencies, which is consistent with interpreting each value as being associated with a sinusoid such as $\cos(\omega(t)$ as described here, while two-sided PSD's are also used (typically given as $\mathscr{L}(f)$) in which case each value would be associated with $\frac{1}{2}e^{j\omega t}$.

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  • $\begingroup$ @Tarek Did I answer your question here or is there still some confusion? Please let me know $\endgroup$ Commented Apr 19, 2022 at 12:58

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