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==So What Is x[n]?== | ==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 | + | 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 <math>(-1)^{\frac{3}{4}n}</math> 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> | ||
Revision as of 17:08, 25 September 2008
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>
