(New page: ==CT Signal Fourier Coefficients== An easy signal to compute is a sine or cosine based function. I'm trying to look around and find something less trivial but all i can find is functions...) |
|||
| Line 1: | Line 1: | ||
| − | == | + | ==DT Signal Fourier Coefficients== |
| − | + | Let's make up a signal. | |
| + | |||
| + | <math> x[0] = 0 </math> | ||
| + | |||
| + | <math> x[1] = 1 </math> | ||
| + | |||
| + | <math> x[2] = 1 </math> | ||
| + | |||
| + | <math> x[3] = 0 </math> | ||
| + | |||
| + | <math> x[4] = x[0] </math> etc, the function is periodic with period 4 | ||
| + | |||
| + | Using the formula | ||
| + | |||
| + | <math>x[n] = \sum_{k=0}^{N-1} a_k e^{jk \frac{2 \pi}{N} n}</math>, where <math>a_k = \frac{1}{N} \sum_{r=0}^{N-1} x[r] e^{-jk \frac{2 \pi}{N} r}</math> | ||
| + | |||
| + | Since the period is 4, N=4. | ||
| + | |||
| + | <math>x[n] = \sum_{k=0}^{3} a_k e^{jk \frac{2 \pi}{4} n}</math>, where <math>a_k = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-jk \frac{2 \pi}{4} r}</math> | ||
Revision as of 16:57, 25 September 2008
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} $
