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I'm reading 'The Scientist and Engineer's Guide to Digital Signal Processing'. It says for a noisy signal, the important parameter is not the deviation from the mean, but the power represented by the deviation from the mean.

So, if I were to consider Amplitude it would just be voltage deviations from the mean, but for power it becomes Voltage^2 deviations. Then an average is taken and a square root to give the standard deviation [It would just be an average deviation in the case of the Amplitude].

I'm trying to understand why considering just the Amplitude is not good enough? Why is it that when I'm dealing with noisy signals, I need to consider power calculations to try and reduce the noise?

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It's generally not true that the average deviation from the mean is not important or much less important than the power of the deviation. For sampled data, the mean deviation would be defined as

$$\epsilon=\frac{1}{N}\sum_{n=0}^{N-1}|x_n-\mu_x|\tag{1}$$

where $\mu_x$ is the mean of the data. This quantity is definitely relevant. However, the absolute value in $(1)$ makes it much less convenient to use than the equally relevant standard deviation

$$\sigma=\sqrt{\frac{1}{N}\sum_{n=0}^{N-1}|x_n-\mu_x|^2}\tag{2}$$

which is differentiable, and, consequently, easier to handle analytically (such as in least squares fitting, etc.). (Note that in Eq. $(2)$ you might as well use a factor $1/(N-1)$ instead of $1/N$, but that is not relevant here).

Apart from the above argument based on mathematical convenience, it is true that in the case of uncorrelated noise sources, it is the noise variances (i.e., the squared standard deviations) that add up, and not the mean deviations.

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  • $\begingroup$ You may like to add $|(.)|$ around the square term to make it applicable for $x_n\in \mathbb{C}$ as well. $\endgroup$
    – Neeks
    Nov 24, 2015 at 12:59
  • $\begingroup$ I'm still unable to understand why we use power with regard to noisy signals, unfortunately. If energy was getting converted into different forms then I understand using a common standard [i.e. power] to make comparisons is logical. But, within a noisy, electrical signal - since, it's all electrical energy [flow of electrons], why not just use current for example? Square the current instead of the voltage..? $\endgroup$
    – JJT
    Nov 25, 2015 at 10:52
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    $\begingroup$ @JJT: The point is that we're working with a squared quantity for the above mentioned reasons (easier to handle mathematically, and variances of independent noise sources simply add up). It doesn't matter if it's voltage, current, or square-feet. In signal processing any squared quantity is referred to as power, because it's proportional to power. $\endgroup$
    – Matt L.
    Nov 25, 2015 at 12:35
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Independent noise sources add in power domain; whereas in the voltage domain they are incoherent. "Signals" add in the voltage domain. You might think on Parsevel's theorem as well. BTW: a good analogy is in linear algebra where straight lines add but vectors in different directions add in some form of Pythagoras theorem

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