(New page: This page will show how to compute the Fourier transforms of CT and DT signals that have a power of absolute value (e.g. <math>(\frac{1}{2})^{|n|}</math>). First, I will show an example of...) |
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You can think of |n| in two different ways, but they both reach the same conclusion. | You can think of |n| in two different ways, but they both reach the same conclusion. | ||
| − | 1) Sum x[n]u[n], and x[n]u[-n-1]. | + | 1) Sum x[n]u[n], and x[n]u[-n-1].<br><br> |
| + | <math>X(w) = \sum_{n=-\infty}^{\infty} (\frac{1}{2j})^{-n}u[-n-1]e^{-jwn} + \sum_{n=-\infty}^{\infty} (\frac{1}{2j})^{n}u[n]e^{-jwn}</math><br><br> | ||
<math>X(w) = \sum_{n=-\infty}^{-1} (\frac{1}{2j})^{-n}e^{-jwn} + \sum_{n=0}^{\infty} (\frac{1}{2j})^{n}e^{-jwn}</math><br><br> | <math>X(w) = \sum_{n=-\infty}^{-1} (\frac{1}{2j})^{-n}e^{-jwn} + \sum_{n=0}^{\infty} (\frac{1}{2j})^{n}e^{-jwn}</math><br><br> | ||
Revision as of 14:39, 24 October 2008
This page will show how to compute the Fourier transforms of CT and DT signals that have a power of absolute value (e.g. $ (\frac{1}{2})^{|n|} $). First, I will show an example of Professor Mimi's, then I will solve a different problem.
$ x[n] = (\frac{1}{2j})^(|n|) $
$ X(w) = \sum_{n=-\infty}^{\infty} x[n]e^{-jwn} $
$ = \sum_{n=-\infty}^{\infty} (\frac{1}{2j})^{|n|}e^{-jwn} $
You can think of |n| in two different ways, but they both reach the same conclusion.
1) Sum x[n]u[n], and x[n]u[-n-1].
$ X(w) = \sum_{n=-\infty}^{\infty} (\frac{1}{2j})^{-n}u[-n-1]e^{-jwn} + \sum_{n=-\infty}^{\infty} (\frac{1}{2j})^{n}u[n]e^{-jwn} $
$ X(w) = \sum_{n=-\infty}^{-1} (\frac{1}{2j})^{-n}e^{-jwn} + \sum_{n=0}^{\infty} (\frac{1}{2j})^{n}e^{-jwn} $
Since there a u[n] functions in this method, it might be a little easier to set the bounds of the summation.
2) Or, just come to the conclusion that |n| = n if n>=0, and |n| = -n if n<0.
