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<math>a_0=\frac{5}{4}</math>
 
<math>a_0=\frac{5}{4}</math>
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Which says that the function is:
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 +
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<math>x[n]=\frac{1}{2}+\frac{5}{4}e^{jn\pi}\,</math>

Latest revision as of 11:06, 26 September 2008

Informations

1. $ N = 2\, $

2. $ a_k = 0\, $ for all |k|>1

3. $ \sum_{n=0}^{3}x[n]=2 $

4. $ \sum_{n=0}^{3}(-1)^nx[n]=5 $

Inspections

From the first information, we can directly subtitute N into:

$ x[n]=\sum_{n=0}^{3}a_ke^{jk(2\pi/2)n}\, $

$ x[n]=\sum_{n=0}^{3}a_ke^{jk\pi n}\, $

From the third information, we can find $ a_0\, $:

$ a_0=\frac{1}{4}\sum_{n=0}^{3}x[n]=\frac{1}{4} 2 = \frac{1}{2} $

From the fourth information, we can find $ a_1\, $:

$ a_0=\frac{1}{4}\sum_{n=0}^{3}(-1)^nx[n]\, $

$ a_0=\frac{1}{4}\sum_{n=0}^{3}x[n]e^{-jn\pi}\, $

$ a_0=\frac{5}{4} $

Which says that the function is:


$ x[n]=\frac{1}{2}+\frac{5}{4}e^{jn\pi}\, $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn