(New page: <math>x(t)=10cos(4\pi n + 2\pi)\!</math> <math>K=)
 
 
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<math>x(t)=10cos(4\pi n + 2\pi)\!</math>
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<math>x(t)=10cos(4\pi n + 2\pi)+5sin(2\pi n + 4\pi)\!</math>
  
<math>K=
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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>
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<math>N_1 = \frac{2\pi}{4\pi} K \!</math>
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<math>N_1 = 2K \!</math>
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<math>N_2 = \frac{2\pi}{\omega_0} K \!</math>
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<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.
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<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} $

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