| Line 1: | Line 1: | ||
Suppose a DT signal x[n] satisfies | Suppose a DT signal x[n] satisfies | ||
| − | 1. x[n] is periodic and period N= | + | 1. x[n] is periodic and period N=6. |
| − | 2. <math>\sum_{n=0}^{ | + | 2. <math>\sum_{n=0}^{5}x[n]=5</math> |
| − | 3.<math>a_{k+ | + | 3.<math>a_{k+2} = a_k</math> |
Find x[n]. | Find x[n]. | ||
| Line 13: | Line 13: | ||
Answer: | Answer: | ||
| − | from 1 we | + | from 1 we have that x[n]= <math>\sum_{n=0}^{5}a_k e^{-jk \frac {\pi}{3} n}</math> |
| − | from 2 we have <math>a_0 = avg = \frac {5}{ | + | from 2 we have <math>a_0 = avg = \frac {5}{6}</math> |
| − | from 3 we have <math> | + | from 3 we have <math>a_2 = a_4= a_0 = \frac {5}{6}</math> |
Revision as of 12:58, 25 September 2008
Suppose a DT signal x[n] satisfies
1. x[n] is periodic and period N=6.
2. $ \sum_{n=0}^{5}x[n]=5 $
3.$ a_{k+2} = a_k $
Find x[n].
Answer:
from 1 we have that x[n]= $ \sum_{n=0}^{5}a_k e^{-jk \frac {\pi}{3} n} $
from 2 we have $ a_0 = avg = \frac {5}{6} $
from 3 we have $ a_2 = a_4= a_0 = \frac {5}{6} $
