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As an example, we can use the function <math>x(n) = e^{\omega_0 j n}</math>. | As an example, we can use the function <math>x(n) = e^{\omega_0 j n}</math>. | ||
To prove this, we do the following: | To prove this, we do the following: | ||
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<font size="4"> | <font size="4"> | ||
<math>x(n+N) = x(n)</math> | <math>x(n+N) = x(n)</math> | ||
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| − | <math>e^{\omega_0 j n} e^{ | + | <math>e^{\omega_0 j n} e^{\omega_0 j N} = e^{\omega_0 j n}</math> |
Revision as of 10:55, 28 September 2008
Periodic Functions
The definition of a periodic function given in class is as follows: The function x(n) is periodic if and only if there exists an integer N such that x(n+N) = x(n). The value of N is called the "period".
As an example, we can use the function $ x(n) = e^{\omega_0 j n} $. To prove this, we do the following:
$ x(n+N) = x(n) $
$ e^{\omega_0 j (n+N)} = e^{\omega_0 j n} $
$ e^{\omega_0 j n} e^{\omega_0 j N} = e^{\omega_0 j n} $
$ e^{\omega_0 j N} = 1 $
$ \cos(\omega_0 N) + j\sin(\omega_0 N) = 1 $
---Which is true if:
$ \omega_0 N = k2\pi $ (where k is an integer)
---at some point.
This leads to the conclusion that if $ {\omega_0 \over 2\pi} = {k \over N} $
or, put another way, $ {\omega_0} \over {2\pi} $ is a rational number, then the function is periodic.
Put yet another way: if the equation is of the form $ e^{\omega_0 j n} $ and $ \omega_0 $ is made up of $ \pi $ and a rational component (contains no irrationals besides $ \pi $) then the function is periodic.
