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| − | <math>x(t)=10cos(4\pi n + 2\pi)\!</math> | + | <math>x(t)=10cos(4\pi n + 2\pi)+5sin(2\pi n + 4\pi)\!</math> |
| − | <math>K= | + | In order to find the period of the signal below, we need to find a value of K that will make N an integer. |
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| + | <math>N_1 = \frac{2\pi}{\omega_0} K \!</math> | ||
| + | |||
| + | <math>N_1 = \frac{2\pi}{4\pi} K \!</math> | ||
| + | |||
| + | <math>N_1 = 2K \!</math> | ||
| + | |||
| + | <math>N_2 = \frac{2\pi}{\omega_0} K \!</math> | ||
| + | |||
| + | <math>N_2 = \frac{2\pi}{2\pi} K \!</math> | ||
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| + | <math>N_2 = K \!</math> | ||
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| + | Since both numbers are integers before multiplying by K, we can just let K = 1. | ||
| + | |||
| + | <math>a_k = \frac{1}{N} \sum^{N-1}_{n = 0} X[n] e^{-jk\frac{2\pi}{N} n}</math> | ||
Latest revision as of 11:59, 26 September 2008
$ x(t)=10cos(4\pi n + 2\pi)+5sin(2\pi n + 4\pi)\! $
In order to find the period of the signal below, we need to find a value of K that will make N an integer.
$ N_1 = \frac{2\pi}{\omega_0} K \! $
$ N_1 = \frac{2\pi}{4\pi} K \! $
$ N_1 = 2K \! $
$ N_2 = \frac{2\pi}{\omega_0} K \! $
$ N_2 = \frac{2\pi}{2\pi} K \! $
$ N_2 = K \! $
Since both numbers are integers before multiplying by K, we can just let K = 1.
$ a_k = \frac{1}{N} \sum^{N-1}_{n = 0} X[n] e^{-jk\frac{2\pi}{N} n} $
