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I`m confused on how to calculate a signal power. For example, suppose I generate a complex: signal $$x[n] = (1 + j3)e^{j2 \pi k n/N}$$

The DSP theory says the average power of a periodic signal with period $N$ is: $$ \frac{1}{N} \sum \limits_{n=0}^{N-1} \left| x[n] \right|^2$$

Does that mean the signal power is independent of its sampling frequency? For example, suppose I have these $N$ samples within 2ms and another signal with these $N$ samples within $1ms$. The above formula for average power yields the same result, although from a continuous time perspective the second signal would present the same energy in less time, which would mean more power.

Is it correct?

Suppose a want to generate a signal with a given power in Watts and assume the sampling frequency is given, how should I control the signal power (through normalization), using the discrete signal average power formula or considering symbol period (in seconds)?

Thank you

igorauad
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    The formula that you've written is actually the average energy and not the average power. – Sudarsan Dec 09 '13 at 20:03
  • Assume a periodic sequence with period $N$. It is the average power. – igorauad Dec 09 '13 at 20:20
  • @igorauad check https://dsp.stackexchange.com/questions/9367/power-or-energy-signal/9369#9369 and https://www.dsprelated.com/freebooks/mdft/Rayleigh_Energy_Theorem_Parseval_s.html – VMMF Oct 20 '17 at 15:31

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