Revision as of 17:30, 25 September 2008 by Tsafford (Talk)

DT Signal Fourier Coefficients

Let's make up a signal.

$ x[0] = 0 $

$ x[1] = 1 $

$ x[2] = 1 $

$ x[3] = 0 $

$ x[4] = x[0] $ etc, the function is periodic with period 4

Using the formula

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

Since the period is 4, N=4.

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

Now to find the fourier series coefficients:

$ a_0 = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j(0) \frac{\pi}{2} r} = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{0} = \frac{1}{4} (x[0] + x[1] + x[2] + x[3]) = \frac{1}{4} (0 + 1 + 1 + 0) = \frac{1}{2} $

$ a_1 = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j(1) \frac{\pi}{2} r} = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j \frac{\pi}{2} r} = \frac{1}{4} (x[0] e^{0} + x[1] e^{-j \frac{\pi}{2}} + x[2] e^{-j \pi}+ x[3] e^{-j \frac{3 \pi}{2} }) $

$ = \frac{1}{4} (0(0) + 1(-j) + 1*-1 + 0*i) = \frac{-1-j}{4} $

$ a_2 = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j(2) \frac{\pi}{2} r} = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j \pi r} = \frac{1}{4} (x[0] e^{0} + x[1] e^{-j \pi} + x[2] ee^{-j 2 \pi}+ x[3] e^{-j 3 \pi}) $

$ = \frac{1}{4} (0 + 1 * -1 + 1 * 1 + 0 ) = 0 $

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