Difference between revisions of "HW4.5 Caleb Tan ECE301Fall2008mboutin" - Rhea

Revision as of 15:54, 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, font size = '4'>$ x[n] = a_0 e^0 + a_1 e^{j\pi n}< $</font>