In this link on page 218, just before the first paragraph ends, it says: " We can interpret the power axes as one-half the square [...] of the modulus of the complex envelope."

In the same paragraph, the average input power is stated to be $P_\text{in} = A^2/2$. Can somebody please explain to me, why there is no resistance ? The amplitude probably refers to a voltage signal. The RMS of a sinusoidal signal is $A/\sqrt(2)$. Since this is the square root of the mean, it makes sense to me, that the average input poewr is $A^2/2$. But still, I am confused about the lacking resistance. Especially, since in the second paragraph, it is stated, that the power axis will be graduated in units of $10\log(P_\text{in})$.

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    $\begingroup$ Often, there is some statement somewhere, according to "Amplitudes are normalized to a 1Ohm load" or something similar. Normalizations are very common in DSP, otherwise, we would also need to take care of the unit of impulse responses, time and so on. $\endgroup$ – Maximilian Matthé Feb 15 '17 at 6:34

You're dealing with digital signals. Power hence doesn't have a physical unit, just as the amplitude of the signal doesn't have one – where it originated, i.e. the ADC, it might have had one, but usually, that unit just doesn't matter in DSP and the fact that the digital numbers are proportional with some constant factor to a physical signal is totally sufficient.

If you later need to actually map a digital power to a physical power, you'd just calibrate (ie. measure with a known source of amplitude or power) and calculate that constant factor.

TL;DR: This is DSP. We don't care about physical units, so (mentioning the) resistance is futile.

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  • $\begingroup$ Is that why the frequency domain is shown to have units 'Amplitude' and 'Magnitude' instead of Volt-seconds? $\endgroup$ – Lewis Kelsey Sep 29 at 15:25

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