relation between fft and rms

I'd like to clarify a fundamental issue: what is the relation between fft of a function and its rms value?

Following Matlab example of computing fft of a function:

%>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

...

NFFT = 2^nextpow2(L); % Next power of 2 from length of y

Y = fft(y,NFFT)/L;

f = Fs/2*linspace(0,1,NFFT/2+1);

% Plot single-sided amplitude spectrum.

plot(f,2*abs(Y(1:NFFT/2+1)))

...

%>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

what is the relation between 2*abs(Y(1:NFFT/2+1)) and rms(Y)?

Many thanks.

Answer 1

[Star Strider]

The fft is the (fast) Fourier transform of a signal. It transforms it from a time-comain signal (signal amplitude as a function of time) to a frequency-domain signal, expressing the amplitudes of various components in the signal with respect to their frequencies.

the RMS (root mean squared) value of a signal is a way of expressing its average (mean) power. It is the square root of the mean of the squared value of the signal. For simusiodal signals, the RMS value is 0.707 times the peak-to-peak amplitude of the signal.

For a signal vector s:

RMS = sqrt(mean(x.^2));

The total energy of a signal is preserved under the Fourier transform ( Parseval's theorem ), so the sum (or integral) of the square of a function is equal to the sum (or integral) of the square of its transform. The RMS would be the square root of that value.

[Lian]

Thank you Star.

[Star Strider]

My pleasure!

Answer 2

[Matt J]

For brevity, I define the half-spectrum as

Z=2*abs(Y(1:NFFT/2+1));

I think it helps to start with the identity

rms(Y)= sqrt(sum(abs(Y.^2)))/sqrt(NFFT)

For real-valued initial signal, y, the spectrum Y would be conjugate symmetric and so it would be approximately, but not exactly, true that

sum(abs(Y.^2)) =  2*sum( abs(Y(1:NFFT/2+1)).^2)
               =  sum(Z.^2)/2

It would be exact if the DC component Y(1) is zero.

Combining the above equations therefore leads to the approximation

rms(Y)= sqrt(sum(Z.^2))/(sqrt(2*NFFT));

EDIT:

Sorry, I just noticed that the above relation is true when NFFT is odd. When it is even, the relation is more complicated. When NFFT is odd, it is easy to verify the above relationship, though:

>> NFFT=31; Y=abs(fft(rand(1,NFFT))); Y(1)=0; Z=2*Y(1:ceil(NFFT/2));
>> rms(Y), sqrt(sum(Z.^2))/(sqrt(2*NFFT))

ans =

      1.4291

ans =

      1.4291

[Lian]

Thank you Matt. What about the relation to rms(y)?

According to Parseval's theorem, rms(y) should equal sqrt(sum(Y(1:NFFT/2+1).^2)), (Y is after /L), is it right? However I tested a signal and there always about 20% difference between the two. What is the problem?

The fft is not windowed and the DC component is 0.

[Matt J]

If length(y)=NFFT, then the relationship between rms(y) and rms(Y) is

rms(Y)=rms(y)/sqrt(NFFT)

This is true whether NFFT is odd or even and is readily verifiable by examples,

>> y=rand(1,10);  Y=fft(y)/10; rms(y)/sqrt(10), rms(abs(Y))

ans =

      0.1852

ans =

      0.1852

The relationship to the half spectrum then follows from my original answer above. If length(y)<NFFT, then things are more complicated.

[Big dream]

Thanks for your explanation

[Geo]

Hello Matt,

hope you are doing well. Based on your example/explanation that means that rms(Y)=/rms(y), right? Meaning that the RMS of the time domain gives a different value as the RMS of the frequency domain. Only with a difference of 1/sqrt(NFFT) though but still.

Then why are people trying to show that, rms(Y)=rms(y) because of the Parseval's theorem?

Thank you very much in advance.

Answer 3

[Adrian Sanz]

Hello, I got the case that length(y)<NFFT, how would be the relation? Actually I'm getting NFFT as

NFFT = 2^nextpow2(L);    
Y = fft(y,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2+1);