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From #2 gives the period of N=10, from that can deduce that the frequency <math>\ w = k\frac{2\pi}{10}</math> and if assume k=1 then the frequency <math>\ w=\frac{\pi}{5}</math> | From #2 gives the period of N=10, from that can deduce that the frequency <math>\ w = k\frac{2\pi}{10}</math> and if assume k=1 then the frequency <math>\ w=\frac{\pi}{5}</math> | ||
| − | #2 also gives the equation <math> | + | #2 also gives the equation <math> x[n] = \sum_{0}^{9} a_{k} e^{-jkn\pi /5 }</math> |
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| + | since a_{11} = 5 that means that a_1 also equals 5 | ||
| + | |||
| + | <math>\ a_5 = \frac{1}{10} \sum_{0}^{9} x[n]e^{-j\pi n}</math> | ||
| + | |||
| + | <math>\ a_5 = \frac{1}{10} \sum_{0}^{9} x[n](-1)^n</math> | ||
| + | |||
| + | <math>\ a_5 = \frac{1}{10}</math> | ||
| + | |||
| + | <math>\ x[n] = a_0 + 5e^{j\pi/ 5(1)n} +a_2e^{j\pi/ 5(2)n}+a_3e^{j\pi/ 5(3)n}+a_4e^{j\pi/ 5(4)n}+\frac{1}{10}e^{j\pi/ 5(5)n}</math> | ||
Latest revision as of 15:12, 26 September 2008
1. x[n] is a real and even signal
2. x[n] has period N = 10 and Fourier coefficients $ \ a_k $
3. $ \ a_{11} = 5 $
4. $ \ \frac{1}{10} \sum_{n=0}^{9}|x[n]|^2 = 50 $
From #2 gives the period of N=10, from that can deduce that the frequency $ \ w = k\frac{2\pi}{10} $ and if assume k=1 then the frequency $ \ w=\frac{\pi}{5} $
- 2 also gives the equation $ x[n] = \sum_{0}^{9} a_{k} e^{-jkn\pi /5 } $
since a_{11} = 5 that means that a_1 also equals 5
$ \ a_5 = \frac{1}{10} \sum_{0}^{9} x[n]e^{-j\pi n} $
$ \ a_5 = \frac{1}{10} \sum_{0}^{9} x[n](-1)^n $
$ \ a_5 = \frac{1}{10} $
$ \ x[n] = a_0 + 5e^{j\pi/ 5(1)n} +a_2e^{j\pi/ 5(2)n}+a_3e^{j\pi/ 5(3)n}+a_4e^{j\pi/ 5(4)n}+\frac{1}{10}e^{j\pi/ 5(5)n} $
