# Noise quantification from sampled data

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?
• 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? – Hearth Feb 16 '19 at 17:29
• Have you read Bernard Widrow's book? – analogsystemsrf Feb 16 '19 at 17:42
• 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. – Marcus Müller Feb 16 '19 at 17:42
• 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 – Tony Stewart Sunnyskyguy EE75 Feb 16 '19 at 17:57
• @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. – pnatk Feb 16 '19 at 18:03

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.

• Thanks a lot. Very neat solution. And any hint on the second question?:) – pnatk Feb 16 '19 at 18:24
• 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). – Koen Tiels Feb 16 '19 at 19:41
• 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. – pnatk Feb 16 '19 at 21:21
• 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/…)). – Koen Tiels Feb 16 '19 at 21:27