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Given a signal with zero mean and a standard deviation of 0.1 sampled at 5000 Hz.
What would be the Standard Deviation of its 1st, 2nd and 'n' derivative?

For instance, let's say we measure the distance to a constant range target. The STD of the sensor measurements of the range is 0.1 [Meter] (Assume the error is white) and its sampling rate is 5000 Hz.

What would be the standard deviation of the acceleration process?

Thanks.

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As pointed out in pichenettes' answer, you need to know more about the original signal $x(t)$. If the signal's power spectrum $S_x(\omega)$ is known then the power spectrum of its $n^{th}$ derivative is

$$S_y(\omega)=\omega^{2n}S_x(\omega)$$

The power of the $n^{th}$ derivative can then be computed as

$$E\{|y(t)|^2\}=\frac{1}{2\pi}\int_{-\infty}^{\infty}\omega^{2n}S_x(\omega)d\omega\tag{1}$$

(assuming convergence of the integral, e.g. by assuming that $x(t)$ is band-limited). Since the mean of the derivative is zero (even if the mean of the original signal isn't), the variance of $y(t)$ equals its power, and so the standard deviation is the square root of the expression in (1).

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  • $\begingroup$ Hi Matt, thank you for the answer. I thought about that, yet, that would be the answer for a continuous signal. Don't you think the discrete property of the measurements will have any effect on it? What would be the the STD of the discrete acceleration? $\endgroup$ – Royi Jul 15 '14 at 10:59
  • $\begingroup$ @Drazick: What do you mean by the derivative of a discrete signal? $\endgroup$ – Matt L. Jul 15 '14 at 11:17
  • $\begingroup$ I mean you only have discrete observations of the signal. I thought about the reconstruction of the PSD of the continuous signal, yet thought there is another way. $\endgroup$ – Royi Jul 15 '14 at 17:19
  • $\begingroup$ @Drazick: If the signals are appropriately band-limited (i.e. cut-off $\pi/T$ angular frequency) - and this is what you make sure when you sample anyway - then the power spectra of the discrete signals are equal to the power spectra of the original continuous signals, so everything in my answer remains valid. Just the integration interval changes to $(-\pi/T,\pi/T)$. $\endgroup$ – Matt L. Jul 16 '14 at 14:38
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It depends on the frequency content of the signal.

Consider sinusoidal signals with an amplitude of 1 and frequencies of 1Hz and 1kHz. Both have the same standard deviation, but the standard deviation of their derivatives are different - 1000 times as large for the latter.

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  • $\begingroup$ You can assume the signal is White Noise. $\endgroup$ – Royi Jul 15 '14 at 11:01
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Since the signal is discrete and the operation is linear it be formed using a Filter.

Assuming the signal is given by $ x \left[ n \right] $.
Then its derivative is given by:

$$ y \left[ n \right] = \frac{ x \left[ n \right] - x \left[ n - 1 \right] }{ \frac{1}{5000} } $$

Since $ x \left[ n \right] $ samples are independent the STD is given by:

$$ \operatorname{Var} \left( y \left[ n \right] \right) = \sqrt{ {5000}^{2} \cdot \left( {0.1}^{2} + {0.1}^{2} \right) } = 707.1068 $$

This will also match the calculation using a filter with corresponding coefficients.

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