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Clearly, <math>x(t)</math> is periodic because there is a <math>T > 0</math> such that <math>x(t + T) = x(t)</math> for all <math>t</math>. An obvious choice for <math>T</math> would be <math>T = 2\pi</math>. Shifting <math>x(t)</math> by <math>2\pi</math> gives the original function since <math>2\pi</math> is the ''fundamental period'' of <math>x(t) = sin(t)</math> | Clearly, <math>x(t)</math> is periodic because there is a <math>T > 0</math> such that <math>x(t + T) = x(t)</math> for all <math>t</math>. An obvious choice for <math>T</math> would be <math>T = 2\pi</math>. Shifting <math>x(t)</math> by <math>2\pi</math> gives the original function since <math>2\pi</math> is the ''fundamental period'' of <math>x(t) = sin(t)</math> | ||
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| + | ==A Non-Periodic Function== | ||
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| + | <math>x[n] = sin(n)</math> | ||
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| + | At first glance, it would appear that <math>x[n]</math> is periodic following the same reasoning as above; however, because <math>x[n]</math> is a ''discrete'' function, this is not the case. The definition for a periodic discrete signal is that there exists an ''integer'' <math>N > 0</math> such that <math>x[n + N] = x[n]</math> for all <math>n</math>. The fundamental period of the sine function is <math>2\pi</math>, which is not an integer. Furthermore, any integer multiple of <math>2\pi</math> is not an integer. Therefore, <math>x[n] = sin(n)</math> is not periodic. | ||
Latest revision as of 20:21, 4 September 2008
A Periodic Function
$ x(t) = sin(t) $
Clearly, $ x(t) $ is periodic because there is a $ T > 0 $ such that $ x(t + T) = x(t) $ for all $ t $. An obvious choice for $ T $ would be $ T = 2\pi $. Shifting $ x(t) $ by $ 2\pi $ gives the original function since $ 2\pi $ is the fundamental period of $ x(t) = sin(t) $
A Non-Periodic Function
$ x[n] = sin(n) $
At first glance, it would appear that $ x[n] $ is periodic following the same reasoning as above; however, because $ x[n] $ is a discrete function, this is not the case. The definition for a periodic discrete signal is that there exists an integer $ N > 0 $ such that $ x[n + N] = x[n] $ for all $ n $. The fundamental period of the sine function is $ 2\pi $, which is not an integer. Furthermore, any integer multiple of $ 2\pi $ is not an integer. Therefore, $ x[n] = sin(n) $ is not periodic.
