# Is this an energy or power signal? 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.

• are you sure you don't mean: $$x[n] = 4^{-n} \ u[n] \qquad ?$$ > – robert bristow-johnson Oct 29 '20 at 5:07
• @robertbristow-johnson Yep, I'm sure! It's really x[n] = 4^n u[n]. Did I solve it correctly? – Rose Oct 29 '20 at 5:44
• @LaurentDuvalNoted! I retyped my edited solution using the math syntax :) – Rose Oct 29 '20 at 10:41
• You possibly forgot squares in your summation terms. I have updated the answer for a simple proof for the energy part. This could provide guidelines for the power part – Laurent Duval Oct 29 '20 at 11:08

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 $$u[n]$$.

• +1, away from abstract math, back to common sense :) – Matt L. Oct 29 '20 at 10:24
• Ohhh okay, thank you for this! Hopefully like you guys, I can call these things common sense too in the future :( – Rose Oct 29 '20 at 10:47

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.

• Oh yea, thank you! It turns out I can't use it cause a should be less than 1 otherwise the answer would be negative :( But ugh now I'm really stuck. How should I proceed? @@ – Rose Oct 29 '20 at 8:04
• @LaurentDuval: I believe that this is controversial, but anyway, those weird summations are usually called Ramanujan summations, whoever came up with them first. – Matt L. Oct 29 '20 at 10:20
• @LaurentDuval: Yes, sometimes it seems that Euler did it all, and what he didn't do, Gauss did ... – Matt L. Oct 29 '20 at 10:27
• Both of you are so awesome @@ – Rose Oct 29 '20 at 10:56
• @Rose: Your conclusion is correct, even though you forgot to square the magnitudes. – Matt L. Oct 29 '20 at 11:34