Hello. I would like to calculate the average power of the signal x[n] = n. I know the formula I need to use is P = lim(N→∞) (1/(2N+1)) ∑ [ n = -N, N] |x[n]|^2 . I have no idea how to apply it to just x[n] = n. I plug n in as x[n] to get n^2 and then what? Must I change the limits of summation to n = 0, N , or n = 0, ∞, or n = -∞, ∞, or something else? I solved two problems like this and both times n "disappeared" so it wasn't an issue.

 

You need to show your best attempt at solving YOUR homework. It sounds like you've got enough of a base to give it a decent shot. Don't worry about getting it right or wrong. What you present serves as a springboard for discussion so that we can spot where you are going right and were you are going wrong.

 

You need to show your best attempt at solving YOUR homework. It sounds like you've got enough of a base to give it a decent shot. Don't worry about getting it right or wrong. What you present serves as a springboard for discussion so that we can spot where you are going right and were you are going wrong.

I really don't know how to continue, I've written all I know. x[n] = n could be any signal I have no idea what it looks like, so I don't know what the limits would be.

P = lim(N→∞) (1/(2N+1)) ∑ [ n = -N, N] |x[n]|^2 = lim(N→∞) (1/(2N+1)) ∑ [ n = -N, N] |n|^2 = lim(N→∞) (1/(2N+1)) ∑ [ n = -N, N] n^2 = lim(N→∞) (1/(2N+1)) N +1 +n^2 -N = lim(N→∞) (1/(2N+1)) n^2 + 1 =(1/(2(∞)+1)) (∞)^2 + 1 = ∞

That's my best effort.

 

I really don't know how to continue, I've written all I know. x[n] = n could be any signal I have no idea what it looks like, so I don't know what the limits would be.

P = lim(N→∞) (1/(2N+1)) ∑ [ n = -N, N] |x[n]|^2 = lim(N→∞) (1/(2N+1)) ∑ [ n = -N, N] |n|^2 = lim(N→∞) (1/(2N+1)) ∑ [ n = -N, N] n^2 = lim(N→∞) (1/(2N+1)) N +1 +n^2 -N = lim(N→∞) (1/(2N+1)) n^2 + 1 =(1/(2(∞)+1)) (∞)^2 + 1 = ∞

That's my best effort.

This gives us something to start with, though to be honest I am quite tired and need to turn in and so probably won't offer much until sometime tomorrow, but others may well chime in before that.

The first thing I note is your statement that x[n] = n could be any signal. Why do you say that? If you have a signal y(t) = t, would you say that that could be any signal? No. It is a very clearly defined signal. Well, x[n] = n is the same signal except in a discrete world. What is x[10]? Why, it's 10. It can't be anything else. What is x[-213]? It's -213.

That might get you to a point where you can make a bit more progress.

 

A quick purusal of your final answer shows that it is correct (based on intuitive reasoning, so I could be wrong). As n grows, the amount of power at the start (n=-N) and step (n=+N) of the range grows without bound. In particular, as we extend the range the new power is higher than the average of the old range (since it is higher than the peak power in the old range) and hece the average power over the range increases as N increases. Therefore, as N goes to infinity, the average power goes to infinity.

Now, whether you got the right answer but by an invalid line of reasoning has yet to be determined.

 

This gives us something to start with, though to be honest I am quite tired and need to turn in and so probably won't offer much until sometime tomorrow, but others may well chime in before that.

The first thing I note is your statement that x[n] = n could be any signal. Why do you say that? If you have a signal y(t) = t, would you say that that could be any signal? No. It is a very clearly defined signal. Well, x[n] = n is the same signal except in a discrete world. What is x[10]? Why, it's 10. It can't be anything else. What is x[-213]? It's -213.

That might get you to a point where you can make a bit more progress.

Yes, you are right. This is the same as y=x on a normal graph.

On Wikipedia it says that to calculate ∑ [ n = 0, N] |n|^2 you just use the formula ∑ [ n = 0, N] |n|^2 = (1/6) (N(N+1)(2N+1)) however, I don't think I need to start from 0 but -N rather so I'm not sure if this formula still applies. I tried using the Wolfram computational engine to compute ∑ [ n = -N, N] |n|^2 and it gave me (1/3) (N(N+1)(2N+1)) why 1/3 and not 1/6 if wiki says it's suppose to be 1/6?

 

Split your range into two halves, one from -N to 0 and one from 0 to +N (and note that you are double counting n=0, which doesn't affect the sum but does affect the average if you aren't careful). Given that the square function is an even function, what does this tell you?

 

It may be worth visiting the concept of the summation of the squares of the integers.

http://www.trans4mind.com/personal_development/mathematics/series/sumNaturalSquares.htm

 

Yes, you are right. This is the same as y=x on a normal graph.

On Wikipedia it says that to calculate ∑ [ n = 0, N] |n|^2 you just use the formula ∑ [ n = 0, N] |n|^2 = (1/6) (N(N+1)(2N+1)) however, I don't think I need to start from 0 but -N rather so I'm not sure if this formula still applies. I tried using the Wolfram computational engine to compute ∑ [ n = -N, N] |n|^2 and it gave me (1/3) (N(N+1)(2N+1)) why 1/3 and not 1/6 if wiki says it's suppose to be 1/6?

Wolfram and Wikipedia and such are great resources and references, but you need to not just use them as a means of finding a quick equation to do a plug and chug with. The goal of problems like this are, in large part, for you to learn the underlying concepts. So I would recommend being sure that, at the end of the day, that you can sit down and start with the summation notation and solve for it in closed form.

 

I reviewed some summation material and found the correct way to compute the summation. Thank you for the help.

 

Congratulations! You will be surprised at some of the strange places you will find problems that being able to recognize that a summation process is at work and then being able to deal with it expidetiously will be quite a boon -- and impress the people you should be trying to impress, namely your boss and your customers.