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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} |x[n]|^2 | + | from 4, power of x[n] = <math>\frac {1}{6} \sum_{n=0}^{5} |x[n]|^2 = \sum_{n=0}^{5} |{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> | ||
| − | Thus <math>x[n]=\frac {5}{6} (1+e^{-jk \frac {2\pi}{3}+e^{-jk \frac {4\pi}{3})</math> | + | Thus <math>x[n]=\frac{5}{6} (1+e^{-jk \frac {2\pi}{3}}+e^{-jk \frac {4\pi}{3}})</math> |
Revision as of 12:05, 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} |x[n]|^2 = \sum_{n=0}^{5} |{a_k}|^2 $
to get a minimum value, $ a_1=a_3=a_5=0 $
Thus $ x[n]=\frac{5}{6} (1+e^{-jk \frac {2\pi}{3}}+e^{-jk \frac {4\pi}{3}}) $
