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== DT LTI System Part a == | == DT LTI System Part a == | ||
| + | <br><br> | ||
| + | <math> h[n] = cos[{\pi \over 3} n]u[n]</math><br><br> | ||
| + | and the input signal, <br><br> | ||
| + | <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> | ||
| + | <math> H(e^{jw}) = \sum_{k=0}^{\infty} cos[{\pi \over 3} n]e^{-jwn} = \sum_{k=0}^{\infty} {1 \over 2}(e^{j{\pi \over 3}n} + e^{-j{\pi \over 3}n})e^{-jwn}</math><br><br><br> | ||
| + | :::<math> = \sum_{k=0}^{\infty} {1 \over 2} (e^{j({\pi \over 3} - w)n} + e^{-j({\pi \over 3} + w)n}) | ||
Revision as of 17:50, 26 September 2008
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DT LTI System Part a
$ h[n] = cos[{\pi \over 3} n]u[n] $
and the input signal,
$ 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)} $
$ H(e^{jw}) = \sum_{k=0}^{\infty} cos[{\pi \over 3} n]e^{-jwn} = \sum_{k=0}^{\infty} {1 \over 2}(e^{j{\pi \over 3}n} + e^{-j{\pi \over 3}n})e^{-jwn} $
- $ = \sum_{k=0}^{\infty} {1 \over 2} (e^{j({\pi \over 3} - w)n} + e^{-j({\pi \over 3} + w)n}) $
