Can anyone in a simple way explain why this formula works. It is the Root Sum of Squares where you square your values then add together then take the square root. It seems to be used quite a lot in electronics engineering e.g. for combining noise voltages, combining tolerances of components, adding risetimes of waveforms in the time domain together etc.

Here are a few examples:

  1. Two noise voltages of 1 nV/Hz2 added together give will be 1.414 nV/Hz2.

  2. Two resistors of 2% tolerance in a voltage divider will give an output voltage at the junction of both resistors of 2.828%.

  3. Two LPF filters in series where one produces a risetime of 10 ns risetime the following one 50ns risetime, produce a total risetime of 51 ns.

  • 2
    \$\begingroup\$ Not an expert: But I know that \$x^2\$ is what mathematicians would call a "well behaved" (infinitely differentiable) function, whereas \$\text{abs}(x)\$ is not even differentiable one time. That enables the use of mathematical analyses that would not be possible if we averaged absolute values instead of averaging squares. \$\endgroup\$ Nov 15, 2021 at 15:13
  • \$\begingroup\$ FYI, If you take \$N\$, the number of particles in a system, as an independent variable then entropy is a direct function of \$N\$ and its partial derivative (with respect to \$N\$) is \$-\frac{\mu}{k T}\$. Quantum mechanics, assuming only the existence of countable quantized stationary states can prove the thermal behavior of matter, an emergent phenomena resulting from large-\$N\$ statistics (doesn't exist at microscopic levels.) This fact is strong evidence for the existence of atoms and Poisson independence. The question has deep answers (not so shallow as seen in an electronics class.) \$\endgroup\$
    – jonk
    Nov 15, 2021 at 20:27
  • \$\begingroup\$ Note your 1. premise is wrong (unduly optimistic) if the noise voltages are correlated, as they often are. 2. Not sure where you get that from. If the resistors are equal the tolerance is 2%. If they are very much unequal it can approach 4% (or 0%). 3. approximately square root of sum of squares. If you investigate "gaussian" distributions more will become clear. \$\endgroup\$ Nov 16, 2021 at 0:01
  • 1
    \$\begingroup\$ The key here is that in each case, the root sum of squares is used for statistical analysis of uncorrelated random variables. For example, No. 2 talks about resistor tolerance. Tolerance is typically specified as “some number of standard deviations of these resistors (often 3 or 4) fall within this tolerance (2%).” Doing the root sum of squares on two resistors with that tolerance will then mean that the same number of standard deviations of the voltage divider will fall within the resulting tolerance (2.828%), but you could have a worst case outcome of 4%. It’s just a lower probability. \$\endgroup\$
    – Ryan
    Nov 16, 2021 at 2:10

4 Answers 4


It's Pythagoras' theorem applied to orthogonal entities.

Signal theory teaches us that almost all electronic signals can be thought of as a summation of orthogonal sine waves (entities).

Signals can therfore be seen as a geometrical segment/vector/hypotenuse of a right-angled triangle with infinitely many legs.


The square root of the summation of the square of all Fourier coefficients of a certain signal is the length of the hypotenuse.

That's why in calculus functions are often referred to as vectors.

  • \$\begingroup\$ And we can do this because we take the variables in question to be independent of each other. \$\endgroup\$ Nov 16, 2021 at 2:45

It is a common reoccurrence in engineering because engineering applications deal with real world processes that require viewing variables as random fluctuations. This is especially true of random noise, and also engineering tolerances, (as you mentioned) that have a “statistical spread”. This statistical spread can be usually generalised to be quantified as “variance”. Variance is a measure of dispersion, meaning it is a measure of how far a set of numbers is spread out from their average value. So you have electrical noise power which is actually power generated by a random electromagnetic process, then you have variance of component or dimensional tolerances that is also the cause of random component or dimensional imperfections due to random manufacturing imperfections, all varying with a particular “spread” from the average expected value.

So yes, it may be true that you can usually add noise powers as the sum V ² or as I ² (with Watts as a dimension) but they are actually just “variances” of a stochastic process with a particular spread (Gaussian or otherwise), and generally variances of whatever (noise, component tolerances, or other types of random phenomena).

Electrical noise is usually measured with an instrument that directly measures noise power (a spectrum analyser) in a given bandwidth, and from this you can take the square root to get the noise voltage or current, just as you would sum “variances” together and then take the square root to get the “deviation” of that random variable etc - all this providing the particular combined random fluctuations are not correlated! This is a first order assumption/approximation that is often done in Engineering to simply things but still retain some accuracy. If the random noises you are squaring, summing and then square-rooting happen to be corelated, then your above sum of square will not strictly hold true unless the correlation is small.

This would be the same for component tolerances etc. For the later, it is more common for component tolerances to be considered in worst-case tolerance analysis, in a traditional type of tolerance stack-up calculation. The individual variables are placed at their tolerance limits in order to make the measurement as large or as small as possible. Summing of variances and square rooting the sum makes sense if you know the tolerance “spread” (the variance or the standard deviation). Generally, components are specified with some known min and max values as tolerances.

Then on another note, there’s also the Pythagoras theorem c ² = a ² + b ² that kind of looks the same but is not a result of the root of the sum of the variance of random variables a and b. It’s got nothing to do with statistics and random stuff. Pythagoras was lucky that a and b happen to be uncorrelated ;-)

  • \$\begingroup\$ But how does this work with risetimes? \$\endgroup\$
    – Edba
    Nov 15, 2021 at 17:33
  • \$\begingroup\$ Actually you mentioned that for rise times. I think you're using one of those "rules of thumb" that are often quoted as an approximation, and based on very loose engineering, kind of like saying "use this rule of thumb" but take note its ±30% or something along those lines. Adding rise-times, and then more precisely the rise times after a chain of LPFs is starting to diverge from the first two truly random processes - adding noises and tolerances etc.. I've personally never added rise-times that way, and have never hear of this "rule of thumb". \$\endgroup\$
    – citizen
    Nov 16, 2021 at 8:34
  • \$\begingroup\$ The reason being is that LPFs have "transfer functions". An input signal, an output signal, and a system that "shapes" the input into an output in a deterministic way. This transfer function is what will give rise to a change in rise/fall time of the output, with respect to the input. You can have many different types of LPFs, like Butterworth LPFs, Chebyshev, Bessel, and variations of etc. They have different TFs even though they are LPFs per se, the rise/fall time of a Butterworth will differ from that of a Cheby even with the same cut-off frequency - in a deterministic way ... \$\endgroup\$
    – citizen
    Nov 16, 2021 at 8:40
  • \$\begingroup\$ The key word here is "deterministic" which makes the third point different (and also makes it stand out like a sore thumb) from the first two, which are results of randomness in a system ... \$\endgroup\$
    – citizen
    Nov 16, 2021 at 8:41

Because, in a noise calculation, you want the POWER difference of the noise elements or difference from ideal for tolerance. To change a resistance to power, it is inversely proportional to the square of resistance (assuming a constant voltage). Or change in voltage noise to power, it is proportional to square of voltage.

Using simple averages will give too much weight to smaller changes.


As already mentioned, one aspect is purely geometric: Computing the (2-) norm of a sum of orthogonal vectors is done by squaring the individual norms, adding and square-rooting.

A slightly different perspective on things like noise or tolerances is a statistical one: If we assume the noises (or the resistor tolerances) are normally distributed (which many times is not a bad assumption 1), then we can see the following: Let's say the two (independent) noises (or resistor tolerances) have the normal distributions N(m1, s1) and N(m2, s2) respectively with the means m1, m2 (for noise in signals we usually assume m1 = m2 = 0) and the standard deviations s1, s2. The standard deviations can be thought of as the amplitudes.

Then adding these noises results in a new distribution, it is again a normal distribution N(m, s) but with m=m1+m2 and s = sqrt(s1^2 + s2^2). 2

I'm sure there is also a geometric interpretation of this but I wanted to stress that this is based on some assumptions about the properties of those measurements and has a statistical meaning.

  • \$\begingroup\$ Examples 1. & 2. above are a result of stochastic processes (that which derives from randomness), whereas example 3. is deterministic. They cannot be simply lumped together, and one has to use the root sum square appropriately and with caution to avoid getting things wrong or confused ... \$\endgroup\$
    – citizen
    Nov 17, 2021 at 8:38

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