Is this an energy or power signal?

Question

Is the given input signal x[n] an energy or power signal?

The image shows what I did so far. Is it correct?

Thank you!

EDIT:

I solved it again, please tell me if I did it correctly this time.

$$E= \lim\limits_{n \to \infty}\sum_{n=-N}^N = |4^nu[n]|$$ $$E= \lim\limits_{n \to \infty}\frac{1-4^{2(N+1)}}{1-4^2}$$ $$E=\infty$$

$$P= \lim\limits_{n \to \infty}\frac{1}{2N+1}\sum_{n=-N}^N = |4^nu[n]|$$ $$P= \lim\limits_{n \to \infty}\frac{1}{2N+1}(\frac{1-4^{2(N+1})}{1-4^2})$$ $$P=\infty$$

Since both P and E are infinite, then it is neither an energy or power signal.

Answer 1

As a complement to Matt's answer, on the intuition: $u[n]$ has value $1$ from $n=0$ on. So basically its energy will increase for ever, because it keeps adding ones for $n\ge0$. Then, you build another signal $x[n]$ that grows way faster because you multiply it with the exponential term $4^n$.

Therefore, only at the common sense level, one cannot expect it to be an energy signal. It is not feasible.

And then, goes the proof that the sum of the series actually diverges. I would suggest to use the simplest proofs, to avoid complexities, and formulae that are not valid (sum of geometric series for $a \ge1$).

$$\sum_{n=-N}^N |4^nu[n]|^2 \ge \sum_{n=0}^N 4^{2n}|u[n]|^2\ge \sum_{n=0}^N |u[n]|^2 \ge N+1$$ which diverges. Hence the signal is not energy. Something similar can be used for power.

Of course, as Matt said, if the exponential term becomes $4^{-n}$, this geometrical series then decreases fast to zero, and wins over the mere sum of ones in $u[n]$.

Answer 2

It's always good to do some plausibility checks on your results. The energy of $x[n]$ is computed by summing up the values $|x[n]|^2$. Each one of these values is non-negative, so if you end up with a negative value for the signal energy, you should ask yourself where you went wrong (unless your name is Ramanujan).

Check the conditions on the number $a$ for which your energy formula is valid. As a final note, if a signal is not an energy signal it is not necessarily a power signal.