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<math>N_2sin = \frac{2\pi}{\pi} k = \frac{2}{1} k</math> | <math>N_2sin = \frac{2\pi}{\pi} k = \frac{2}{1} k</math> | ||
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<math>N_4sin = \frac{2\pi}{\frac{\pi}{2}} k = \frac{2}{\frac{1}{2}} k</math> | <math>N_4sin = \frac{2\pi}{\frac{\pi}{2}} k = \frac{2}{\frac{1}{2}} k</math> | ||
| + | |||
| + | |||
| + | Take <math>k = 1</math>, | ||
| + | |||
| + | <math>N_2sin = 2</math> | ||
| + | |||
| + | <math>N_4sin = 4</math>, so the overall fundamental period is | ||
| + | |||
| + | <math>N = 4</math> | ||
Revision as of 10:20, 25 September 2008
Define a periodic DT signal
I am going to choose a sine signal, since there have been many cosines done already.
DT signal: $ x[n] = 2\sin(\pi n + \pi) + 4\sin(\frac{\pi}{2} n + \pi)\, $
Now, each sine has its own period, and the fundamental period of the function is the greater of the separate periods.
$ N_2sin = \frac{2\pi}{\pi} k = \frac{2}{1} k $
$ N_4sin = \frac{2\pi}{\frac{\pi}{2}} k = \frac{2}{\frac{1}{2}} k $
Take $ k = 1 $,
$ N_2sin = 2 $
$ N_4sin = 4 $, so the overall fundamental period is
$ N = 4 $
