(→DT LTI System Part a) |
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== DT LTI System Part a == | == DT LTI System Part a == | ||
<br><br> | <br><br> | ||
| − | <math> h[n] = | + | <math> h[n] = e^{-n}u[n]</math><br><br> |
and the input signal, <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>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} | + | <math> H(e^{jw}) = \sum_{k=0}^{\infty} e^{-n}e^{-jwn} = \sum_{k=0}^{\infty} e^{(-jw-1)n}</math><br><br><br> |
| − | :::<math> = \sum_{k=0}^{\infty} {1 \over | + | :::<math> = \sum_{k=0}^{\infty} [e^{(-jw-1)}]^n </math><br><br><br> |
| + | :::<math> = {1 \over 1 - e^{-jw-1}}</math><br><br><br> | ||
| + | Applying this to y[n],<br><br> | ||
| + | <math> y[n] = 1 + H(e^{{2\pi \over N}})e^{j({2\pi \over N})n}[{1 \over 2j}+{5 \over 2}] | ||
Revision as of 18:00, 26 September 2008
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DT LTI System Part a
$ h[n] = e^{-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} e^{-n}e^{-jwn} = \sum_{k=0}^{\infty} e^{(-jw-1)n} $
- $ = \sum_{k=0}^{\infty} [e^{(-jw-1)}]^n $
- $ = {1 \over 1 - e^{-jw-1}} $
- $ = \sum_{k=0}^{\infty} [e^{(-jw-1)}]^n $
Applying this to y[n],
$ y[n] = 1 + H(e^{{2\pi \over N}})e^{j({2\pi \over N})n}[{1 \over 2j}+{5 \over 2}] $
