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<math>x[1] = \sum_{k = 0}^{1} a_k e^{jk\pi 1} = a_0 e^0 + a_1 e^{j\pi} = a_0 + a_1 e^{j\pi}</math> | <math>x[1] = \sum_{k = 0}^{1} a_k e^{jk\pi 1} = a_0 e^0 + a_1 e^{j\pi} = a_0 + a_1 e^{j\pi}</math> | ||
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| + | C) From C: | ||
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| + | x[0] + x[1] + x[2] = 1 | ||
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| + | We know that x[0] + x[1] = 0 since <font size = '4'><math>a_0 = 0</math></font> and the period is 2. | ||
Revision as of 15:49, 26 September 2008
Guessing a periodic signal based on a few properties given
Properties:
A) Fundamental Period = 2
B) $ a_0 = 0 $
C) $ \sum_{n = 0}^{2} x[n] = 1 $
Finding the signal based on the properties
A) From A:
N = 2.
$ x[n] = \sum_{k = 0}^{1} a_k e^{jk\pi n} $
$ a_k = \frac{1}{2} \sum_{n=0}^{1} x[n] e^{-jk\pi n} $
B) From B:
$ a_0 = \frac{1}{2} \sum_{n=0}^{1} x[n] = 0 $
Therefore 0.5x[0] + 0.5x[1] = 0
$ a_1 = \frac{1}{2} \sum_{n=0}^{1} x[n] e^{-j\pi n} = \frac{1}{2} [x[0] + x[1]e^{-j\pi}] $
$ x[0] = \sum_{k = 0}^{1} a_k e^{jk\pi 0} = \sum_{k = 0}^{1} a_k = a_0 + a_1 $
$ x[1] = \sum_{k = 0}^{1} a_k e^{jk\pi 1} = a_0 e^0 + a_1 e^{j\pi} = a_0 + a_1 e^{j\pi} $
C) From C:
x[0] + x[1] + x[2] = 1
We know that x[0] + x[1] = 0 since $ a_0 = 0 $ and the period is 2.
