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| − | Therefore, <font size = '4'><math>x[n] = a_0 e^0 + a_1 e^{j\pi n}</math></font> | + | Therefore, <font size = '4'><math>x[n]= \sum_{k = 0}^{1} a_k e^{jk\pi n} = a_0 e^0 + a_1 e^{j\pi n} = 1e^{j\pi n}</math></font> |
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| + | Note: Original signal I used to come up with these properties was <font size = '4'><math>x[n] = sin(3\pi n + \frac{\pi}{2})</math></font>, which can be simplified to <font size = '4'><math>x[n] = e^{j\pi n}</math></font> | ||
Latest revision as of 15:57, 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.
Therefore x[2] = 1. If x[2] is 1 then x[0] = 1.
From earlier $ x[0] = a_0 + a_1 $
Therefore $ a_1 = 1 $
Therefore, $ x[n]= \sum_{k = 0}^{1} a_k e^{jk\pi n} = a_0 e^0 + a_1 e^{j\pi n} = 1e^{j\pi n} $
Note: Original signal I used to come up with these properties was $ x[n] = sin(3\pi n + \frac{\pi}{2}) $, which can be simplified to $ x[n] = e^{j\pi n} $
