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I have a vector signal analyzer which is measuring an OFDM signal with a bandwidth of 20MHz.

The spectrum shows "Power Spectral Density [dBm/RBW per 100kHz]" on the Y-axis. The RBW is set for 100kHz. The average power (in time domain) is measured to be 10dBm.

I am trying to understand how to convert the PSD value to average power.

*I have added a few pictures to show my measurement. The first two are from the power spectral density of the 20Mhz signal being measured. The third image show information about the signal, including the average time domain power (sqrt(I^2 + Q^2))

Spectrum1Spectrum2AveragePower

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  • \$\begingroup\$ Normally, the density part of PSD means you're getting power per Hz, so you would multiply the PSD values by the bandwidth of interest, and see the total power in that frequency range. Your spectrum specifies RBW per 100kHz, so it may just be the number of bins, which is the frequency range divided by 100 kHz. \$\endgroup\$
    – pscheidler
    Feb 9, 2018 at 18:34
  • \$\begingroup\$ Maybe post a pic of your spec-a screen? I'm asking because [dBm/RBW per 100kHz] evaluates to dBm/Hz/Hz, not dBm/Hz, which is the correct unit for PSD. \$\endgroup\$
    – Selvek
    Feb 9, 2018 at 22:32
  • \$\begingroup\$ @Selvek I have added screenshots of my spectrum \$\endgroup\$
    – Prit1123
    Feb 14, 2018 at 18:33

3 Answers 3

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Convert PSD to dBm/Hz * Hz BW = dBm ( average sub-intervals if not flat)

Or convert dBm @ BW to 100kHz RBW

e.g.

20MHz BW @ 10dBm time domain (RMS power meter??) displayed as PSD @ 100kHz RBW normalized by = 10 log ( 100K/20M ) = -23dB

thus OFDM signal should average 10dBm - 23dB = -33 -13dBm over a 20MHz BW

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  • \$\begingroup\$ Careful with signs... 10dBm-23dB = -13dBm... \$\endgroup\$
    – Selvek
    Feb 14, 2018 at 20:02
  • \$\begingroup\$ Sr's moment's rate is rising at 65 \$\endgroup\$
    – Tony Stewart EE75
    Feb 14, 2018 at 20:43
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If you have a total power of 10 dBm that's 10 mW in real numbers. If that power was supplied equally from two halves of the spectrum the power in each half would be 5 mW.

If that power is supplied from equal 100 kHz sections contributing to a total bandwidth of 20 MHz, the power in each section is 50 uW or -43 dBm if I'm not mistaken (and understand your question).

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I still don't understand why your spec-a has such a weird, ambiguous axis label. It should be either dBm/RBW or dBm/100kHz...

But, assuming one of those two things, the total power in the signal is straightforward.

Your signal is roughly 18MHz of occupied bandwidth, meaning there are (18000kHz / 100kHz =) 180 frequency bins inside it. Each one is roughly -40dBm, or 0.0001mW, so the total power is (0.0001mW * 180 =) 0.018mW, or -17.4dBm. This agrees with the "power" measurement on the info screen, although the "power" measurement is more accurate because it will take the actual power per bin into account instead of just an eyeballed approximation.

Alternate math: 10*log10(18000kHz / 100kHz) = 22.6dB -40dBm + 22.6dB = -17.4dBm

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  • \$\begingroup\$ Thanks! That explains it very clearly! I am not sure why the labeling is so confusing, but it seems that they were trying to convey this measurement is in bins. To me it implies the following: ((dBm * 100KHz)/ RBW). And to get the average power, you would need to multiply that by B/100KHz as you showed in your calculation. \$\endgroup\$
    – Prit1123
    Feb 14, 2018 at 21:08

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