| (2 intermediate revisions by the same user not shown) | |||
| Line 6: | Line 6: | ||
<br> K equals 3 results in a integer. | <br> K equals 3 results in a integer. | ||
<br><math>N = \frac{2\pi}{3\pi} 3 \,</math> | <br><math>N = \frac{2\pi}{3\pi} 3 \,</math> | ||
| − | <br><math>N = | + | <br><br><math>N = 2 \,</math> |
| + | Prove that it is periodic: | ||
| + | x(0)= -3 ; x(1)=3 ; x(2)=-3 ; x(3) = 3 etc. | ||
| + | <br> | ||
| + | <br> | ||
| + | <math>a_k = \frac{1}{N} \sum^{N-1}_{n = 0} X[n] e^{-jk\frac{2\pi}{N} n}</math> | ||
| + | <br><br> | ||
| + | <math>a_k = \frac{1}{2} \sum^{1}_{n = 0} X[n] e^{-jk\pi n}</math> | ||
| + | <br><br> | ||
| + | <math>a_0 = \frac{1}{2} \sum^{1}_{n = 0} X[n] e^{0}</math> | ||
| + | <br><br> | ||
| + | <math>a_0 = \frac{1}{2} (-3 + 3) = 0</math> | ||
| + | <br><br> | ||
| + | <math>a_1 = \frac{1}{2} \sum^{1}_{n = 0} X[n] e^{-j\pi n}</math> | ||
| + | <br><br> | ||
| + | <math>a_1 = \frac{1}{2} (-3 * e^0 + 3 * e^{-j\pi})</math> | ||
| + | <br><br> | ||
| + | <math>a_1 = \frac{1}{2} (-3 * 1 + 3 * -1) = -3</math> | ||
Latest revision as of 08:36, 25 September 2008
DT signal
$ X[n] = 3\cos(3 \pi n + \pi)\, $
Find a value for k that makes N an integer:
$ N = \frac{2\pi}{3\pi} K \, $
K equals 3 results in a integer.
$ N = \frac{2\pi}{3\pi} 3 \, $
$ N = 2 \, $
Prove that it is periodic:
x(0)= -3 ; x(1)=3 ; x(2)=-3 ; x(3) = 3 etc.
$ a_k = \frac{1}{N} \sum^{N-1}_{n = 0} X[n] e^{-jk\frac{2\pi}{N} n} $
$ a_k = \frac{1}{2} \sum^{1}_{n = 0} X[n] e^{-jk\pi n} $
$ a_0 = \frac{1}{2} \sum^{1}_{n = 0} X[n] e^{0} $
$ a_0 = \frac{1}{2} (-3 + 3) = 0 $
$ a_1 = \frac{1}{2} \sum^{1}_{n = 0} X[n] e^{-j\pi n} $
$ a_1 = \frac{1}{2} (-3 * e^0 + 3 * e^{-j\pi}) $
$ a_1 = \frac{1}{2} (-3 * 1 + 3 * -1) = -3 $
