Revision as of 17:08, 25 September 2008 by Blaskows (Talk)

Guess the Signal

1. $ x[n] $ is periodic with N=8.

2. $ \sum_{n=2}^{9}(-1)^{\frac{3}{4}n}x[n]=4 $.

3. $ a_3=a_5 $.

4. $ \sum_{k=0}^{7}a_k=1 $.

So What Is x[n]?

Given the first property of the signal, we can rewrite the bounds of summation in the second property since the zeroth term is the same as the eighth term and the first term is the same as the ninth term. Furthermore, we can change the $ (-1)^{\frac{3}{4}n} $ in the summation to something a little easier to manipulate. This will leave us with

<math\sum_{n=2}^{9}(e)^{\frac{3}{4}j\pi n}x[n]=4</math>

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Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

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