Difference between revisions of "HW4.2 Jacob Pfister ECE301Fall2008mboutin" - Rhea

Revision as of 17:01, 23 September 2008

Periodic DT Signal

The following plot shows two periods of the periodic DT signal $ x[n] $, a sawtooth:

Fourier Series Coefficients

$ a_{k} = \frac{1}{N} \sum_{n=0}^{N-1} x[n] e^{-jk \frac{2 \pi}{N} n} $

From the plot above, N = 4:

$ a_{k} = \frac{1}{4} \sum_{n=0}^{3} x[n] e^{-jk \frac{\pi}{2} n} $

and:

$ x[0] = 0 $

$ x[1] = 1 $

$ x[2] = 2 $

$ x[3] = 1 $

$ x[4] = 0 $

Therefore:

$ a_{0} = \frac{1}{4} \sum_{n=0}^{3} x[n] = \frac{1}{4}(0+1+2+1) $

$ a_{0} = 1 $

$ a_{1} = \frac{1}{4} \sum_{n=0}^{3} x[n] e^{-j \frac{\pi}{2} n} = \frac{1}{4}[(0)(1)+(1)(0)+(2)(-1)+(1)(0)] $

$ a_{1} = -\frac{1}{2} $