What's the meaning of "the resultant noise"?

[BeuysVonTelekraft]

I am reading The Scientist and Engineer's Guide to Digital Signal Processing

Here, he says:


As a starting point, the expression,|xi-μ|, describes how far the ith sample deviates (differs) from the mean. The average deviation of a signal is found by summing the deviations of all the individual samples, and then dividing by the number of samples, N. Notice that we take the absolute value of each deviation before the summation; otherwise the positive and negative terms would average to zero. The average deviation provides a single number representing the typical distance that the samples are from the mean. While convenient and straightforward, the average deviation is almost never used in statistics. This is because it doesn't fit well with the physics of how signals operate. In most cases, the important parameter is not the deviation from the mean, but the power represented by the deviation from the mean. For example, when random noise signals combine in an electronic circuit, the resultant noise is equal to the combined power of the individual signals, not their combined amplitude.

What's the meaning of "the resultant noise" and "combined power"?

 

[FvM]

Imagine "resultant noise" as result of a summing the voltage of two unrelated noise sources.

 

[BeuysVonTelekraft]

I imagined something like this, but i wasn't sure. If i sum the noisy signal, it's going to be equal to the power of this sum and not equal to the sum of the signal? 8-O

 

[FvM]

Yes, signals of same frequency and phase sum by magnitude, orthogonal signals, e.g. unrelated noise sources by power (total signal = squareroot of summed squares).

 

[BeuysVonTelekraft]

I didn't get what's the procedure here.

Is it \[{{S}_{1}}^{2} + {{S }_{2 }}^{2} + ... + {{ S}_{N }}^{2}\] or \[{({S }_{1 } + {S }_{2 } + ... + { S}_{ N})}^{2}\]?

I guess it's not the first, i did some experiments on mathematica and the \[{\sigma }^{2 }\] is different from total.

 

[rahul.6sept]

I wonder if we can consider it to be two or more Noise Figure in cascade.

 

[enjunear]

I didn't get what's the procedure here.

Is it \[{{S}_{1}}^{2} + {{S }_{2 }}^{2} + ... + {{ S}_{N }}^{2}\] or \[{({S }_{1 } + {S }_{2 } + ... + { S}_{ N})}^{2}\]?

I guess it's not the first, i did some experiments on mathematica and the \[{\sigma }^{2 }\] is different from total.

Agree with FvM, it should be:

\[\sqrt{{{S}_{1}}^{2} + {{S }_{2 }}^{2} + ... + {{ S}_{N }}^{2}}\]
(squareroot of the summed squares)

Squaring a signal gets rid of the phase dependency, effectively giving you "signal powers". Sum the powers, then take the sqrt to return you to a scale of "signal magnitude". (in laymans terms)

 

[BeuysVonTelekraft]

I did this on mathematica.

And i could see the first example do not follow.

 

[albbg]

Your code is wrong. The sum of squares is always positive or zero, then if all the numbers are reals the the sqrt is real.

 

[BeuysVonTelekraft]

I noticed, that's why i made two variables. p and g.

g is the right one, see that it's result is 5.549...

 

[albbg]

I don't know mathematica, but it seems that you have calculated the mean of a vector with mean=0. Since the mean is close to 0 but not exactly 0 (due to a finite element vector), in you case negative, then the sqrt is an imaginary number.
You should, instead, generate N vectors :

S1 = rand(), S2=rand() ... SN=rand()

sum them toghether

S = S1+S2+..+SN

Verify that the standard deviation of S is the same (or, better, is close to):

sqrt(stdev(S1^2) + stdev(S2^2) +..+ s(SN^2))

Scilab code:

// generates 6 vectors of different amplitudes //
r1=17*(0.5-rand(1000,1));
r2=31*(0.5-rand(1000,1));
r3=14*(0.5-rand(1000,1));
r4=46*(0.5-rand(1000,1));
r5=9*(0.5-rand(1000,1));
r6=44*(0.5-rand(1000,1));

// Resultant signal //
r = r1 + r2 + r3 +r4 + r5 + r6;

// RMS amplitude of the resultant signal
a=stdev(r)

// RMS amplitude estimate from single amplitudes
b=sqrt(stdev(r1.^2) + stdev(r2.^2) + stdev(r3.^2) + stdev(r4.^2) + stdev(r5.^2) + stdev(r6.^2))

I'll get, for example, a=21.3 and b=20.5

 

[BeuysVonTelekraft]

It's not possible to apply Standard Deviation on a sample alone.