I am trying to measure the maximum output power of an SDR transmitting a 20 MHz LTE signal with a spectrum analyzer.

Adjusting the resolution bandwidth (RBW) changes the power spectral density (PSD) of my signal, which seems to suggest there is a sweep spot for the RBW.

I have been doing some reverse engineering of the correct RBW trying to measure a PSD that would agree with the maximum transmit power stated on the SDR datasheet. Doing so, I found that the most similar estimate of the total power (obtained by integrating the PSD over the useful bandwidth of 18MHz, so excluding the guard band) is with a 10Hz RBW.

On the other hand, the spectrum analyzer would set 180kHz as automatic RBW, which leads to a dramatically different PSD (by around 50dB.)

What am I missing here? My device under test (DUT) is a NI USRP B210 while I am using a Keysight P5001A as a spectrum analyzer.

Edit: Please find below two screenshots obtained with RBW 10Hz and 10kHz, respectively.

enter image description here

enter image description here


1 Answer 1


The total power will be constant as you adjust the resolution bandwidth (RBW), but the power that is indicated with the marker on the spectrum analyzer at any given frequency in band will go down as RBW is reduced, when measuring signals that have occupied bandwidth greater than the resolution bandwidth. This is simply because the marker is indicating how much power is within that resolution bandwidth of what is effectively a bandpass filter to represent that frequency measurement, so if your reduce the bandwidth, you are reducing the amount of power that can get through that filter.

To estimate the total power in the modulated signal, given it has a flat response over a total frequency range $BW$ Hz with a marker indicating the power level is $P$ dBm, use:

$$P_{tot} = P_{marker}-10\log_{10}(RBW)+ 10\log_{10}(BW) $$

$$= P_{marker}+10\log_{10}(BW/RBW)$$

Where all powers given are dB quantities (such as dBm). The result of the first subtraction is a normalized power spectral density in units of dBm/Hz. The final addition of $10\log_{10}(BW)$ is the log approach of multiplying the power spectral density P/Hz by the total bandwidth in Hz resulting in total power P in dB units as dBm.

So for the OP’s screenshots we see (visually) a power level around -50 dBm for the 10 Hz resolution bandwidth, which is a power spectral density of $-50 -10\log_{10}(10) = -60$ dBm/Hz. The total bandwidth appears to be 18 MHz and thus the total power is estimated to be $-60 + 10\log_{10}(18E6) = -17.5$ dBm. In the other screen shot the resolution bandwidth was increased to 10KHz, which is a $10\log_{10}(1000)= 30 $ dB increase in bandwidth, so the marker power should have risen to -20 dBm (something is astray there, typically such measurements are within a few dB of the actual channel power).

The resolution bandwidth is effectively (or equivalently) describing a narrow bandpass filter that sweeps across your signal, and at each instant the power that gets through that filter is presented on the vertical axis of the display, while the center frequency of that filter is presented on the horizontal axis (as further detailed in a block diagram at this post). Thus if we increase the resolution bandwidth, more power gets through and the power displayed in dBm will go up. Using the formula above, we will see that the total power calculated will remain constant.

See page 6-8 of this app note (App Note 1303, old HP) for further details on additional corrections up to 2.5 dB when using the log power or lin power measurements from a spectrum analyzer in order to get "true-rms" power measurements assuming noise floors as limited by Additive White Gaussian Noise (AWGN). Most modern analyzers include a channel power measurement in a given bandwidth utility that already make these compensations, which I would recommend using.

  • $\begingroup$ Thanks for the material and explanation! There is one detail that I do not understand: why does not the formula depend on the RBW?. If I plugged in the marker value I read for different RBW's, I would actually get different values for $P_{tot}$. $\endgroup$ Apr 11, 2023 at 12:58
  • $\begingroup$ Can you add screen shots for your two cases and what you computed? Ultimately you are measuring a noise density in all cases which doesn’t change but there may be a further detail I missed when I see your screen shot and results. If you reduce the resolution bandwidth by a factor of 10, doesn’t your marker power go down by 10 dB? $\endgroup$ Apr 11, 2023 at 13:04
  • $\begingroup$ Makes much more sense now, thanks! I added two screenshots obtained with 10Hz and 10kHz RBW's. I tried applying the updated formula and I get slightly different results. Should I trust the lowest RBW? $\endgroup$ Apr 11, 2023 at 13:13
  • $\begingroup$ @AntonioAlbanese something doesn’t match up. You have the detector set for “peak”, is “rms” an option? Still I don’t understand the 10 dB error, see the computations I added $\endgroup$ Apr 11, 2023 at 13:29
  • $\begingroup$ Your screen shots say this is a 2 port VNA—- I’m not familiar with the P5001A, you can use it as a 1 port spectrum analyzer? $\endgroup$ Apr 11, 2023 at 13:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.