DC term in a periodogram estimate

[arve9066]

I was trying to understand the noise in switiching regulators and took 8192 samples of a 5V DC supply at 5Khz. If I calculate the 8k point FFT of this sequence, divide the absolute value by 8192 and square the expression, I will get the power right? It comes out as 14dB approximate. If I connect the dc supply then to a 50 ohm spectrum analyzer, would I see 14dB at zero frequency or 10*log10(5*5/50) = -3dB?

 

[KlausST]

Hi,

If I calculate the 8k point FFT of this sequence, divide the absolute value by 8192 and square the expression,

What "absolute value" do you mean?
An 8192 FFT gives 4096 "real" results and 4096 "imaginary" results.

***

It comes out as 14dB approximate.

"dB" is a relative value. It is the same as "501%"

***
Can you explain more clearely?

Klaus

 

[arve9066]

Say x is the L= 8192 long data vector of the 5V DC supply sampled at some frequency and I calculate Sx as Sx=abs(fft(x)/L).^2
The plot of fftshift(10*log10(Sx)) shows approx 14dB at center which is the power at DC or zero frequency as per the periodogram estimate definition

I am trying to understand if this will be the value that will show up in a 50 ohm spectrum analyzer at zero frequency if I hook up the same DC supply to it.

 

[arve9066]

Or in other words, what would I see at 0Hz if I connect a 5V DC supply to a spectrum analyzer..

 

[KlausST]

Hi,

As said before: dB is a relative value. One can not calculate an absolute value without reference form plain dB.

Maybe it is dBV, then it is dB related to 1V.

In your case this might be. Then 14dBV means 5V.

Klaus

 

[arve9066]

I get that dB is a relative value. But what I am looking for is what is the power a spectrum analyzer at 0Hz would show when I connect a 5V DC supply to it in dB (relative to 1W).

 

[KlausST]

Hi,

Then - to be correct - it should be "dBW"

https://en.wikipedia.org/wiki/Decibel

5V at 50 Ohms = 0.5W --> about -3dBW (This is what I expect)

Klaus