1
$\begingroup$

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

$\endgroup$
3

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.