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<math>x(t)=10cos(4\pi n + 2\pi)\!</math> | <math>x(t)=10cos(4\pi n + 2\pi)\!</math> | ||
| − | <math>K= | + | In order to find the period of the signal below, we need to find the smallest value of K that will make N an integer. |
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| + | <math>N = \frac{2\pi}{\omega_0} K \!</math> | ||
| + | |||
| + | <math>N = \frac{2\pi}{2\pi} K \!</math> | ||
| + | |||
| + | <math>N = \frac{2\pi}{2\pi} K \!</math> | ||
| + | |||
| + | <math>N = K\!</math> | ||
| + | |||
| + | So the smallest value that K can have is 1. | ||
| + | |||
| + | <math>K = 1\!</math> | ||
Revision as of 11:50, 26 September 2008
$ x(t)=10cos(4\pi n + 2\pi)\! $
In order to find the period of the signal below, we need to find the smallest value of K that will make N an integer.
$ N = \frac{2\pi}{\omega_0} K \! $
$ N = \frac{2\pi}{2\pi} K \! $
$ N = \frac{2\pi}{2\pi} K \! $
$ N = K\! $
So the smallest value that K can have is 1.
$ K = 1\! $
