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== A Periodic DT Signal == | == A Periodic DT Signal == | ||
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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} | + | <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}n-{\pi \over 2})}-e^{-j({4\pi \over N}n - {\pi \over 2})}]</math> |
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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> | <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> | ||
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| − | <math>x[n] = 1 + ({1 \over 2j} + ) + | + | <math>x[n] = 1 + e^{j({2\pi \over N})n}[{1 \over 2j} + {5 \over 2}] - e^{-j({2\pi \over N})n}[{1 \over 2j} - {5 \over 2}] - ({7 \over 2j}e^{-j{\pi \over 2}})e^{-j({4\pi \over N}n)} + ({7 \over 2j}e^{j{\pi \over 2}})e^{j({4\pi \over N}n)}</math> |
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| + | <math>x[n] = 1 + e^{j({2\pi \over N})n}[{1 \over 2j} + {5 \over 2}] - e^{-j({2\pi \over N})n}[{1 \over 2j} - {5 \over 2}] - ({7 \over 2j}e^{-j{\pi \over 2}})e^{-j2({2\pi \over N}n)} + ({7 \over 2j}e^{j{\pi \over 2}})e^{j2({2\pi \over N}n)}</math><br><br><br><br><br> | ||
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| + | <math> a_0 = 1</math><br><math> a_1 = {1 \over 2j} + {5 \over 2} </math><br><math> a_{-1} = -{1 \over 2j} + {5 \over 2}</math><br><math> a_2 = {7 \over 2}</math><br><math> a_{-2} = {7 \over 2}</math><br><br> | ||
Latest revision as of 16:31, 26 September 2008
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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}n-{\pi \over 2})}-e^{-j({4\pi \over N}n - {\pi \over 2})}] $
$ 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}] $
$ x[n] = 1 + e^{j({2\pi \over N})n}[{1 \over 2j} + {5 \over 2}] - e^{-j({2\pi \over N})n}[{1 \over 2j} - {5 \over 2}] - ({7 \over 2j}e^{-j{\pi \over 2}})e^{-j({4\pi \over N}n)} + ({7 \over 2j}e^{j{\pi \over 2}})e^{j({4\pi \over N}n)} $
$ x[n] = 1 + e^{j({2\pi \over N})n}[{1 \over 2j} + {5 \over 2}] - e^{-j({2\pi \over N})n}[{1 \over 2j} - {5 \over 2}] - ({7 \over 2j}e^{-j{\pi \over 2}})e^{-j2({2\pi \over N}n)} + ({7 \over 2j}e^{j{\pi \over 2}})e^{j2({2\pi \over N}n)} $
$ a_0 = 1 $
$ a_1 = {1 \over 2j} + {5 \over 2} $
$ a_{-1} = -{1 \over 2j} + {5 \over 2} $
$ a_2 = {7 \over 2} $
$ a_{-2} = {7 \over 2} $
