I have done some measurements with a Spectrum Analyzer of a square wave of 100kHz and amplitude equal to 1V. The resulting power spectral density is here shown:

enter image description here

The highest peaks represent the odd harmonics (even harmonics are very low since the input square wave is approximately odd). Peak values are expressed in dBm. Now my question is: how can I evaluate the voltage amplitude of the peaks?

I have done this reasoning. My signal generator has an output impedance of 50Ohm, and the Spectrum Analyzer has an input impedance of 50Ohm. This means that the equivalent circuit of the measurement is this one:

enter image description here

Let's consider the first peak: since 4.62 dBm corresponds to 2.52mW, I have done the following computation:

enter image description here

where the term 1/2 is due to the voltage divider between the two resistances of 50Ohm, and 1/sqrt(2) is due to the rms value. But the result of this computation does not agree with the ideal computation of the fourier transform of my square wave. In fact, the first peak of the fourier transform should be equal to:

2/pi = 0.64V

This last computation is due to the fact that the Fourier Series of a square wave is:

enter image description here

and so the first peak of the fourier transform is 4/pi * 1/2, since the Fourier transform of a sine is proportional to 1/2 * delta dirac.

Where is the mistake?

  • \$\begingroup\$ Isn't the measurement in the spectrum analyzer taken with a 50 Ohm load? Therefore, 4.62 dBm into 50 ohms, not 100. \$\endgroup\$ – BEE Oct 28 '19 at 19:57
  • \$\begingroup\$ Where is that I put 100Ohm instead of 50Ohm? \$\endgroup\$ – Kinka-Byo Oct 28 '19 at 21:57
  • \$\begingroup\$ Aren't you trying to find the voltage that the spectrum analyzer is measuring? If so, then why do you have the divide by 2? "where the term 1/2 is due to the voltage divider between the two resistances of 50Ohm" \$\endgroup\$ – BEE Nov 1 '19 at 15:23

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