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If a sampled voltage signal \$y[n]\$ is composed of its offset \$\bar y\$ plus noise \$x[n]\$; i.e. \$y[n] = \bar y + x[n]\$, are both following equations for the RMS value of the noise component \$\text{RMS}(x)\$ same?:

\begin{align} \text{RMS}(x) &= \text{RMS}(y - \bar y)\\ \text{RMS}(x) &= \sqrt{\text{RMS}(y)^2 - \bar y^2} \end{align}

  1. Are both equations above correct and same?
  2. For the equations to be correct does \$x[n]\$ have to be random in nature? What if \$x\$ has periodic component?
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    \$\begingroup\$ This reads like a homework question. What work have you already done? Are you sure this wouldn't be better suited on math.SE, as it seems like a numerical analysis question? \$\endgroup\$ – Hearth Feb 16 at 17:29
  • \$\begingroup\$ Have you read Bernard Widrow's book? \$\endgroup\$ – analogsystemsrf Feb 16 at 17:42
  • \$\begingroup\$ no, they are not same, which you could have simply tested by trying with two random numbers that you made up in your mind, for example \$\bar y = 2\$, RMS(x) =0. \$\endgroup\$ – Marcus Müller Feb 16 at 17:42
  • \$\begingroup\$ It depends on noise BW , spectral shape and crest factor Pk/avg and periodic content. Normally sampled noise is measured in Vpp as RMS could be 0V if noise spectrum exceeds sample interval spectrum, but periodic by root of sum-squared components. I would normally use Vpp signal/Vpp noise to determine SNR \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 16 at 17:57
  • \$\begingroup\$ @Hearth What make you to think it is homework? I'm not a student, so no its not homework. Just trying to find which equation is valid for quantifying rms noise of a sampled volatge when using MATLAB. \$\endgroup\$ – panic attack Feb 16 at 18:03
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Yes, those equations are correct and the same.

In the first equation, you just replace \$x\$ with its expression following from \$y=\bar{y}+x\$.

For the second equation, you can do the following: The square of the rms value of y is equal to $$ \begin{aligned} rms(y)^2\ &= \frac{1}{N}\sum_{n=1}^N (y[n])^2\\ &= \frac{1}{N}\sum_{n=1}^N (\bar{y} + x[n])^2 \\ &= \frac{1}{N}\sum_{n=1}^N (\bar{y}^2 + 2 \bar{y} x[n] + (x[n])^2) \end{aligned} $$

The first term (\$\frac{1}{N}\sum\limits_{n=1}^N \bar{y}^2\$) is just equal to \$\bar{y}^2\$.

The second term (\$\frac{1}{N}\sum\limits_{n=1}^N 2 \bar{y} x[n]\$) is zero since you can take the constant factor \$2 \bar{y}\$ out of the sum and \$\bar{x}=\frac{1}{N}\sum\limits_{n=1}^N x[n]\$ is zero.

The third term (\$\frac{1}{N}\sum\limits_{n=1}^N (x[n])^2\$) is the square of the rms value of x.

Thus $$ rms(y)^2 = \bar{y}^2 + rms(x)^2 $$ and thus $$ rms(x) = \sqrt{rms(y)^2 - \bar{y}^2} $$


The example of Marcus Müller in the comments is still valid: \$\bar{y}=2\$, \$rms(x)=0\$, and thus \$rms(y)=2\$. Note that an rms value is not the same as a standard deviation if the signal (\$y\$ here) has a non-zero mean value.

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  • \$\begingroup\$ Thanks a lot. Very neat solution. And any hint on the second question?:) \$\endgroup\$ – panic attack Feb 16 at 18:24
  • \$\begingroup\$ The only assumption is that x has zero mean. If x has a periodic component, then you need to make sure to measure an entire period (or multiple periods). \$\endgroup\$ – Koen Tiels Feb 16 at 19:41
  • \$\begingroup\$ Great but in practice given the samples it is I guess very hard to figure out to clip the entire period or even figure out what is composed of in time series. \$\endgroup\$ – panic attack Feb 16 at 21:21
  • \$\begingroup\$ If you take an fft (Fast Fourier transform) of your data, you can immediately see if the signal is periodic or not (if not, then you can observe spectral leakage (dsp.stackexchange.com/questions/10120/…)). \$\endgroup\$ – Koen Tiels Feb 16 at 21:27

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