[SOLVED] Find RMS Voltage from FFT

[viniciushoff]

Hi
I want to calculate the RMS Voltage from a FFT of a signal sampling.
I found several sources which show the following equation:
\[Vrms = \sqrt{\sum_{k = 0}^{N/2} {(V^2re(k)+V^2im(k))}}\]
Where:
N = number of samples;
Vre(k) = are real parts of k^th harmonics of voltage;
Vim(k) = are imaginary parts of k^th harmonics of voltage;

But with this equation I not able to find the correct result.
So, I found in only one site this equation:
\[Vrms = \sqrt2*\frac{\sum_{k = 0}^{N/2} {\sqrt{(V^2re(k)+V^2im(k))}}}{N}\]
With the second equation I found correct results.

Why only the second equation works for me?
How do I deduce this equations?

 

[FvM]

The first equation corresponds to the nature of orthogonal signals, what's the reasoning for the second one?

 

[viniciushoff]

Hi FvM, I don´t know what´s the reasoning for the second one, this is the problem.
But when I use a vector of a signal which I created in Excel, so with known values, I calculate the right value of Vrms with the second equation.

 

[albbg]

The FFT decomposes the signal in frequency, generating a complex spectrum with both positive and negative frequencies, depending on both amplitude and phase relationships. We are now not interested in the phase.
Fixing a frequency, the absolute amplitude for positive frequency is the same of that in the negative part. So, in order to recover the amplitude, we can consider just the positive spectrum. In this case we will have the amplitude divided by 2. Furthermore, if the FFT is taken over N points the amplitude at the output of the FFT will be N times the actual amplitude for the considered frequency.
Then considering the positive spectrum of a N points FFT we will have to multiply by 2/N in order to recover the peak amplitude.
Then the peak amplitude of a given frequency, considering the complex spectrum, will be:

2/N*sqrt[re(V)^2+im(V)^2]

Now applying this to all, let's suppose K, frequencies we have to sum them in power, that is:

\[2/N*\sqrt{\sum_{n = 1}^K [{Re(Vn)}^{2}+{Im(Vn)}^{2}]}\]

since we want the RMS we have to divide by \[\sqrt{2}\], then:

\[\sqrt{2}/N*\sqrt{\sum_{n = 1}^K [{Re(Vn)}^{2}+{Im(Vn)}^{2}]}\]

With respect to your second equation, in my derivation, the square root and the sum are exchanged: are you sure to have written them in the correct order ? Furthermore the maximum index of the sum "K" and the number of points of the FFT "N" have not to be confused.

 

[viniciushoff]

Thank you albbg

I understood your derivation of second equation. But I'm sure that square root is within the sum. I tried to put out the square root and recalculate, but the results don´t match.
If the peak amplitude of a given frequency is 2/N*sqrt[re(V)^2+im(V)^2], the sum of terms should not contain the square root?

 

[albbg]

Could you post the samples of your signal in the time domain (just a txt file with one column) ? I'm also interested in the site where you saw the second equation.

 

[viniciushoff]

I attached a txt with the samples. There are 128 samples of one cycle with 220Vrms.
I'll look for the site and post here.

View attachment signal.txt

 

[FvM]

I get the right rms value (about 160) with the formula in post #6.

 

[viniciushoff]

Right, I was seeing the wrong values of signal e because of this I was confused with the equations.

Thank you!