Measuring signals with spectrum analyzer

Hallo,

I tried to measure the noise of a DSP-output when no signal is applied
to the output. I measured the "rest"-noise with an oszilloscope. The
amplitude of the signal seems to be about 2mV. Then i wanted to
analyse the power spectrum of the noise. I connected the DSP-output to
a spectrum analyzer.
Now the problem is that i get different results when changing the
frequancy span of the spectrum analyzer. When i measure from 10Hz to
3.2KHz, the sprectrum of the measured noise lies at about -100dBVrms.
Increasing the span (10Hz to 25,6KHz) leads to a spectrum that lies at
about -90dBVrms. Compared to the previous measurement it is
approximately 10dB higher. Has anybody an explanation for this effect?
How can i ensure a precise measurement of the noise?

Thanks in advance

Jens

 

a écrit :

What you probably see (probably because we don't know which SA you have
and which measurement you did exactly) is this:
when you increase the frequency span, you also increase the SA IF
bandwidth (most SAs have frequency span, IF and video BW coupled in
order to make the measurement in a 'reasonable' time).

What you want is a noise power density (nV/rtHz or uV/rtHz) and what the
SA gives you is nV or uV. For each analyzed frequency it gives the total
noise power that is within the IF BW around the analyzed frequency (hope
that's clear enough). As you increase span, IF BW increases and so does
the noise power.

To convert to PSD you'll have to check in your manual how the noise BW
of your SA is related to IF BW (it may not be identical, but often is),
and then divide the noise power by the sqare root of the noise BW.

Then the obtained figures should be identical.

Some analyzers also have the possibility to directly do noise PSD
measurements at some spot frequencies during the sweep. Check your
manual for this.

 

Hello Fred,

i hope, i got you right.
But even if the IF (IF=input filter? antialiasing filter?) reduces the
bandwith when a lower span is chosen, nevertheless the lower
frequencies must be in the lower span with the same power for these
lower frequencies are not filtered and so the power should not be less
than with a wider span.
If that would be the case, then the same phenomena would occure with
much larger signals but it doesn't.
I'm i right? Or do i get you completely wrong?

Thanks,

Jens

 

I have some supplements:

Hi again,

Now i know what you mean with IF. I don't have a heterodyne Analyser.
I have an Analyser which is based on FFT (signal processing).
Hope that information helps.

Regards

Jens

 

a écrit :

In fact I was expecting that (from 25.6kHz top frequency) but this
changes nothing.
The bandwidth here is related to the frequency bin width, i.e.
span/number of bins. If you have it, check your manual.

 

a écrit :

IF = intermediate frequency amplifier. An heterodyne SA is nothing but a
precision receiver.

reduces the

I'm not sure I understand what you mean. Well, in fact I'm sure I don't
understand at all.

For a given spectral noise density, the wider the filter bandwidth, the
higher the power you'll observe at the filter output...

....and SA just display this power.

To get noise spectral *density* you have to relate the observed power to
the bandwidth.

For a discrete signal spectrum the effect is different: widening the IF
BW just (artificially) widens the displayed rays width.

Have a look at agilent app note (AN150 I think) "spectrum analysis basics".

 

All the figures look self consistent. Would seem you have something like a
20MHz bandwidth oscilloscope.
The 10dB change you're seeing, emphatically implies that the spectrum
analyser automatically increased it's measurement bandwidth by 10 times,
when you switched from the small 3200HZ span to the much wider 25600Hz.
You will have an option to override this 'auto bandwidth/sweep time'
function and switch to a 'manual' filter setting.
It a sensible world it should be described on page #1 of the operator manual

 

Hello Fred, hello John,

thank you for your help.
Fred: This Application Note helps a lot.
John: Currently i have no time to care about the problem but i will
pick your advice next week.

Again, thank you for your help.

Jens