I am trying to understand Energy and Power concepts as applied to discrete electrical signals.
Is the concept of power needed because periodic signals will have infinite energy?
Also, in the context of periodic signals, power is lim (N->Inf) (1/2N) (sum(x(n)*2) for n = 0 to N-1
So, the power number works out to be a finite number when a large number (summation term) is divided by another large number? (sorry, haven't been in touch with limits topic for quite some time)
It really comes down to the signal that you are analyzing. A signal by itself does not have either energy or power, at least not in a proper sense. I've read about measuring signal power and energy, but I have to question the wisdom in doing so. Just because you can calculate a number doesn't bestow any meaning on that number.
That being said there are circumstances, such as if you are measuring a Voltage applied to a resistive load, that the conventional equations, with which I am sure you are familiar, do apply. But we need to be careful in out terminology. Here, we would really be measuring power per Ohm, but I'm ok to just use the term power.
As such, the instantaneous power is equal to the instantaneous Voltage squared. Similarly, we can average the power of a given period or do any number of other operations to it if there were some reason, such as a periodic signal, to do so. I can't think of any reason you would want to calculate the average power over all time.
With regards to energy, what we would be talking about is actually a change in energy from one time to another, as opposed to an absolute number. Power is defined as the rate of change of energy with respect to time, so energy is the integral of power with respect to time plus some constant. Since that constant is impossible (and I would argue irrelevant) to know, all we can calculate from the power is change in energy relative to some point in time.
Based on this, to answer your first question, a periodic signal would result in an infinite amount of energy dissipated over all time. Whether or not power and/or energy are "needed" is entirely dependent on the application.
To answer your second question, the answer is yes. If I have a sinusoidal Voltage and I choose to measure it, the RMS doesn't change based on the sample rate, so long as there aren't any aliasing effects.
Thanks a lot.