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| − | <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} | + | <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> |
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| − | The Fourth term on the right hand goes away since <math> e^{ | + | The Fourth term on the right hand goes away since <math> e^{j{\pi \over 2}n} \text{ and } e^{-j{\pi \over 2}n} </math> are zero. |
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| + | <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}] </math> | ||
Revision as of 14:00, 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^{j{\pi \over 2}n} \text{ and } e^{-j{\pi \over 2}n} $ are zero.
$ 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}] $
