Calculating Energies of x(n) Sequences: Even and Odd Relations Explained

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Hi, I'm sorry if I post this thread at the wrong section. I have a question regarding sequences.

Qn: if there is a sequence x(n) = (0.5)^n u(n), find the energies of the latter sequence of x(2n) and x(0.5n). Also, how the even and odd sequence of x(2n) and x(0.5n) relate to those of x(n)

first, I'm not sure what does it mean by "find the energies of the latter sequence of..." and second, i don't know how to relate those sequences to those of x(n). Is there anyone who can help me? Any help will be greatly appreciated.:smile:

Thanks!
 
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I'm not sure what is meant by "find the energies of the latter sequence" either.

How about starting with the sequence x(n) = 0.5^n u(n)? Can you write down an expression for the energy of this sequence? Can you evaluate the expression? Show us how far you are able to get.
 
hmm.. actually I'm totally clueless on how to derive the energy of the sequence. I mean what does it actually mean when they want the energy? :confused:

is there some kind of formula to it? this is because we just start a new quarter here, and the professor went on really fast. I tried to look at the textbook but there's no formula for that either. I'm so confused.

Thank you.
 
I assume this is for some sort of engineering course? If so, I believe the energy of a sequence x(n) is typically defined as

E = \sum_{n=-\infty}^\infty |x(n)|^2

Does that look familiar?
 
oh ya.. I think I've seen that somewhere, but I can't remember. Thanks for the recall.
So back to the question of the energy of x(n) = 0.5n u(n)
It means that it will be the sum from negative infinity to infinity of (1/4)^n

is that right?
 
oh I think I've figured the answer out already. Thanks! :smile:

so that's when x(n) = (1/2)nu(n). but what if the question is more complicated, something like x(n) = (1/2)(n-1) u(n-2) + (1/3)n u(n-1)? how do we compute the energy of the sequence then?

Thank you!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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