The power of a signal is something different from the level of the signal. I'm not sure how to give a simple explanation of power, so here are a few key points:

• Power is not a linear function of the signal; when you double \, you don't double the power of \ - you quadruple it.
• Power does not depend on the polarity of the signal. A negative signal has the same amount of power as a positive signal.

There are two ways to talk about power:

• The instantaneous power of a signal is \. This is the power at time \ (ie: right now) and it doesn't depend on what happens to the signal before or after this moment. Notice that this formula fits both of the points from above.
• The average power of a signal is the average of the instantaneous power - if your signal has a power of \ half of the time and \ the other half, then the average power is \.

If you remember that the average of \ points is

then you can see that your formula is a calculation of the average power of \.

If the signal is of finite length (and if integral of its square during its definition interval is also finite) then its average power given by will be zero...

Btw, you may compute an instantaneous power of the signal, simply as the square of it ...

How can I calculate the total power? It is "the variance plus the mean squared", so I should just calculate $(\sum_{i}^N x[i]) / N$ for the signal of length $N$?

$(\sum_{i}^N x[i]\color{red}{^2}) / N$ is the average power, not the total energy. Just the sum $\sum_{i}^N x[i]\color{red}{^2}$ is what you're looking for.

Or just sum the PSD (like in this link)?

and

Should I perform this calculation in the time domain (just time series like above), or should I calculate FFT and PSD (Power Spectral Density) first and calculate on PSD (which is also just an array of numbers, if I understand correctly)?

The sum of the power in time domain (as above) and in frequency domain are the same, so that's just another way to compute the same (probably just more labor-intense).

I have seen quite a few formulas for the total power. For finite number of samples and discrete measurements I have seen (∑Nix[i]) divided by N, N+1, 2N or 2N+1. Which one should I use? Does it have something to do with bias/unbiased sampling, i.e. in statistics we add +1 to the divisor to get unbiased calculations?

Since you're really not looking for the power but the energy, neither. You don't have to divide at all.

"Concept: The instantaneous power of a signal is calculated as: (pleft( t right) = smallint {s^2}left( t right)dt) The average power will be: ({P_{avg}} = frac{1}{T}mathop smallint limits_0^T {s^2}left( t right)dt) Application: Given signal is, x(t) = cos(2πf0t) = cos(ω0t) Fundamental frequency = f0 Time period = T = 1f0 Average power, ({S_{avg}} = frac{1}{T}mathop smallint limits_0^T {cos ^2}left( {omega t} right)dt) ( = frac{1}{T}mathop smallint limits_0^T frac{1}{2}left( {1 + cos 2omega t} right)domega t) ( = frac{1}{T}left[ {frac{1}{2}t - frac{1}{{4omega }}sin 2omega t} right]_0^T = frac{1}{2} = 0.5;W)"

Use the buffer function to segment the signal into 1-second increments. Since you want 1-second samples, the vector length is simply ‘Fs’, so — y = randn(73113,1); % Create Signal To Test Code (One Channel) fs = 8192; % Sampling Frequency (Also Frame Length) N = size(y,1); % Signal Length time = linspace(0, N, N-1)/fs; % Time Vector y_b = bandpass(y,[300 2000],fs); % Filter Signal y_seg = buffer(y_b,fs); % Matrix Of Signal Segments y_pwr = sum(y_seg.^2); % Calculate Power figure for k = 1:numel(y_pwr)-1 tv = time((1:fs)+fs*(k-1)); subplot(5,2,k) plot(tv,y_seg(:,k)) grid title(sprintf('Segment %d Power %.1f',k,y_pwr(k))) end subplot(5,2,k+1) tv = time(fs*k:end); plot(tv,y_seg(1:numel(tv),k+1)) grid title(sprintf('Segment %d Power %.1f',k+1,y_pwr(k+1))) EDIT — (28 Apr 2021 at 12:40) Corrected typoographical error.