(New page: Suppose a DT signal x[n] satisfies 1. x[n] is periodic and period N=8. 2. <math>\sum_{n=0}^{7}x[n]=5</math> 3.<math>a_{k+3} = a_k</math>) |
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3.<math>a_{k+3} = a_k</math> | 3.<math>a_{k+3} = a_k</math> | ||
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| + | Find x[n]. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | Answer: | ||
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| + | from 1 we deduce that x[n]= <math>\sum_{n=0}^{8}a_k e^{-jk \frac {\pi}{4} n}=5</math> | ||
| + | |||
| + | from 2 we have <math>a_0 = avg = \frac {5}{8}</math> | ||
| + | |||
| + | from 3 we have <math>a_3 = a_6 = a_0 = \frac {5}{8}</math> | ||
Revision as of 11:49, 25 September 2008
Suppose a DT signal x[n] satisfies
1. x[n] is periodic and period N=8.
2. $ \sum_{n=0}^{7}x[n]=5 $
3.$ a_{k+3} = a_k $
Find x[n].
Answer:
from 1 we deduce that x[n]= $ \sum_{n=0}^{8}a_k e^{-jk \frac {\pi}{4} n}=5 $
from 2 we have $ a_0 = avg = \frac {5}{8} $
from 3 we have $ a_3 = a_6 = a_0 = \frac {5}{8} $
