# Discrete signal power. Does it depend on sampling frequency?

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

• The formula that you've written is actually the average energy and not the average power. Dec 9, 2013 at 20:03
• Assume a periodic sequence with period $N$. It is the average power. Dec 9, 2013 at 20:20
• – VMMF
Oct 20, 2017 at 15:31