(New page: Periodic Signal 1. <math>N = 4\,</math> 2. <math>a_k = 0\,</math> for all |k|>1 3. <math>\sum_{n=0}^{6}x[n]=4</math> 4. <math>\sum_{n=0}^{3}(-1)^nx[n]=10</math> Inspection : Using In...) |
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<math>a_0=\frac{5}{8}</math> | <math>a_0=\frac{5}{8}</math> | ||
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| + | * [[HW4.5 Ronny Wijaya_ECE301Fall2008mboutin]] | ||
Latest revision as of 19:29, 26 September 2008
Periodic Signal
1. $ N = 4\, $
2. $ a_k = 0\, $ for all |k|>1
3. $ \sum_{n=0}^{6}x[n]=4 $
4. $ \sum_{n=0}^{3}(-1)^nx[n]=10 $
Inspection :
Using Info 1 :
$ x[n]=\sum_{n=0}^{6}a_ke^{jk(2\pi/4)n}\, $
$ x[n]=\sum_{n=0}^{6}a_ke^{jk(\pi/2)n}\, $
Using third info to find $ a_0\, $:
$ a_0=\frac{1}{16}\sum_{n=0}^{6}x[n]=\frac{1}{16} 4 = \frac{1}{4} $
Using fourth info, we can find $ a_1\, $:
$ a_0=\frac{1}{16}\sum_{n=0}^{6}(-1)^nx[n]\, $
$ a_0=\frac{1}{16}\sum_{n=0}^{6}x[n]e^{-jn\pi}\, $
$ a_0=\frac{5}{8} $
Source:
