# Does dividing the magnitude spectrum of white noise by sqrt(2) give an RMS magnitude spectrum?

I understand that the RMS Amplitude of a sinusoidal signal is around 0.707 ($$\frac1{\sqrt2}$$) times the Peak Value, but this is not true for noise.

However, an FFT of a noise signal indicates magnitude in bins at equally spaced frequencies, which means the magnitude represented in each bin, is the magnitude of a periodic signal (sinusoid).

So, I assume that multiplying the magnitude of an FFT by 0.707, should give me RMS magnitude spectrum on the Y-axis, instead of the peak value. Am I correct in assuming this, or am I missing something?

## 1 Answer

However, an FFT of a noise signal indicates magnitude in bins at equally spaced frequencies, which means the magnitude represented in each bin, is the magnitude of a periodic signal (sinusoid).

No; it just means that when projecting the noise signal onto a sinusoidal signal, you don't get nothing.

Assuming your noise realization in each is independent identically distributed (usually the way we model noise!), then the (rotated) sum of such noise samples approximates a normal distribution, no matter how the (finite-variance) original distribution looked like. The magnitude in each bin is an approximation (because by definition, your DFT has finite length!) to the noise power density; in an uncorrelated noise process, that's the variance==power==rms².

• So does that mean, the square root of the power spectrum is actually an RMS spectrum, for an uncorrelated noise process? Or would it mean, the only way to obtain the RMS of the noise is to first obtain a moving RMS of the noise signal and then obtain the FFT? Jan 28, 2019 at 17:16
• I think my last sentence was pretty clear on that in terms of RMS². Jan 28, 2019 at 18:50
• Perhaps crystal clear to someone who knows what it all means right away, but terminology has a tendency to scare me sometimes, so I just wanted a clarification. Thank you for your answer, as well as your clarification. Jan 28, 2019 at 19:09