# I/Q Signal Power Calculation

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 ?

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.
• @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. Mar 19, 2020 at 2:29