| Line 21: | Line 21: | ||
from 3 we have <math>a_2 = a_4= a_0 = \frac {5}{6}</math> | from 3 we have <math>a_2 = a_4= a_0 = \frac {5}{6}</math> | ||
| − | from 4, power of x[n] = <math>\frac {1}{6} \sum_{n=0}^{5} | + | from 4, power of x[n] = <math>\frac {1}{6} \sum_{n=0}^{5} \abs{x[n]}^2</math> |
| − | = <math>\sum_{n=0}^{5} | + | = <math>\sum_{n=0}^{5} \abs{a_k}^2</math> |
to get a minimum value, <math> a_1=a_3=a_5=0 </math> | to get a minimum value, <math> a_1=a_3=a_5=0 </math> | ||
Revision as of 12:02, 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 $
4. x[n] has minimum power among all signals that satisfy 1,2,3.
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} $
from 4, power of x[n] = $ \frac {1}{6} \sum_{n=0}^{5} \abs{x[n]}^2 $
= $ \sum_{n=0}^{5} \abs{a_k}^2 $
to get a minimum value, $ a_1=a_3=a_5=0 $
