(→A Periodic DT Signal) |
(→A Periodic DT Signal) |
||
| Line 6: | Line 6: | ||
<br><br> | <br><br> | ||
| − | <math> x[n] = 1 + {1 \over 2j}[e^{j({2\pi \over N})n}-e^{-j({2\pi \over N})n}] + {5 \over 2}[e^{j({2\pi \over N})n}-e^{-j({2\pi \over N})n}] + {7 \over 2j}[e^{j({4\pi \over N}+{\pi \over 2}n)}-e^{-j({4\pi \over N} + {\pi \over 2})n}] | + | <math> x[n] = 1 + {1 \over 2j}[e^{j({2\pi \over N})n}-e^{-j({2\pi \over N})n}] + {5 \over 2}[e^{j({2\pi \over N})n}-e^{-j({2\pi \over N})n}] + {7 \over 2j}[e^{j({4\pi \over N}+{\pi \over 2}n)}-e^{-j({4\pi \over N} + {\pi \over 2})n}]</math> |
| + | <br><br> | ||
| + | The Fourth term on the right hand goes away since <math> e^{({\pi \over 2})n} \text{ and } e^{({\pi \over 2})n} </math> are zero. | ||
Revision as of 13:58, 26 September 2008
A Periodic DT Signal
$ x[n] = 1 + sin({2\pi \over N})n + 5cos({2\pi \over N})n + 7sin({4\pi \over N}n - {\pi \over 2}) $
The signal above x[n] is periodic with period N.
$ x[n] = 1 + {1 \over 2j}[e^{j({2\pi \over N})n}-e^{-j({2\pi \over N})n}] + {5 \over 2}[e^{j({2\pi \over N})n}-e^{-j({2\pi \over N})n}] + {7 \over 2j}[e^{j({4\pi \over N}+{\pi \over 2}n)}-e^{-j({4\pi \over N} + {\pi \over 2})n}] $
The Fourth term on the right hand goes away since $ e^{({\pi \over 2})n} \text{ and } e^{({\pi \over 2})n} $ are zero.
