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| − | Take <math>k = 1\</math>, | + | Take <math>k = 1\,</math>, |
| − | <math>N_2sin = 2\</math> | + | <math>N_2sin = 2\,</math> |
| − | <math>N_4sin = 4\</math>, so the overall fundamental period is | + | <math>N_4sin = 4\,</math>, so the overall fundamental period is |
| − | <math>N = 4\</math> | + | <math>N = 4\,</math> |
Revision as of 11:22, 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\, $
