(Guess the Periodic Signal)
 
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<math>x[n] = 1 + 3e^{j\pi n} \!</math>
 
<math>x[n] = 1 + 3e^{j\pi n} \!</math>
 
==Answer==
 

Latest revision as of 16:40, 26 September 2008

Guess the Periodic Signal

Find a DT signal that satisfies these properties:

1. period N=2

2. $ \sum_{n=0}^3 x[n] = 2 $

3. $ a_1=3 \! $

Solution

1. a period of 2 gives us

$ x[n] = \frac{1}{2}\sum_0^{N-1} a_ke^{jk\pi n} $

2. $ a_0 = 1 \! $

3. $ a_1 = 3 \! $

Giving us:

$ x[n] = 1 + 3e^{j\pi n} \! $

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