I am trying to understand the basics of IQ modulation. The amplitude of a demodulated IQ signal is given as $\sqrt{I^2 + Q^2}$ which is easy to understand from the phasor diagram.

Yet, I don't understand the IQ power: ${I^2 + Q^2}$ . Where is the load impedance in this expression ? I'd expect something like

$Z_{load} = 50 \Omega $

$P = \frac{I^2 + Q^2}{Z_{load}}$

What is missing in my understanding ?


1 Answer 1


Nothing is missing in your understanding. When we calculate power that way, it is assuming a $1 \Omega$ load and that amplitudes are given in volt, but almost nobody ever says that even in textbooks. Unless it is explicitly stated otherwise, I think it is safe to assume this.

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    $\begingroup$ Also note that for almost anything you ever do with IQ signals, the actual physical power doesn't matter – what matters are relative powers ("this signal is 10 times as strong as the noise") or amplitudes ("this signal has an amplitude 30 times that of the interferring tone"). $\endgroup$ Mar 18, 2020 at 15:04
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    $\begingroup$ Good point. Even if someone asks "What is the received power?", I think they never really mean that and the true question is what is the SNR, SINR or something other ratio. $\endgroup$
    – Engineer
    Mar 18, 2020 at 15:48
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    $\begingroup$ To the same point that I and Q are often expressed in many other units besides volts (in a digital system often in counts, or normalized to 1 being full scale digital, so continuing that point the units of power needn't necessarily be in Watts-- yet it is still a power quantity! $\endgroup$ Mar 18, 2020 at 15:53
  • $\begingroup$ From engineering perspective, I see your point. It's like a common reference that nobody cares. However, would you agree that in some cases this simplification can be problematic? Especially, in applied physics applications where incident power at DUT input and tranmission etc. must be known. So then, should one re-adjust the power measurements with Zload = 50 Ohm ? $\endgroup$
    – Krlngc
    Mar 18, 2020 at 16:49
  • $\begingroup$ @Krlngc Not necessarily; I would say that in all cases where an absolute power quantity is used (rather than a relative ratio as in dB) it is important that the units used be specified. Only if the units need be in Watts that the power measurement would be adjusted with Zload in ohms and magnitude in Volts. Or consider that the mangitude can be a current quantity in units of Amps in which case you would multiply by Zload in ohms. The main point that in all cases $I^2 + Q^2$ is a power quantity. $\endgroup$ Mar 19, 2020 at 2:29

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