(→DT LTI System Part a) |
(→DT LTI System Part a) |
||
| Line 5: | Line 5: | ||
<math> h[n] = e^{-n}u[n]</math><br><br> | <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}] - | + | <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 2}e^{-j2({2\pi \over N}n)} + {7 \over 2}e^{j2({2\pi \over N}n)}</math><br><br><br> |
<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> 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} [e^{(-jw-1)}]^n </math><br><br><br> | :::<math> = \sum_{k=0}^{\infty} [e^{(-jw-1)}]^n </math><br><br><br> | ||
:::<math> = {1 \over 1 - e^{-jw-1}}</math><br><br><br> | :::<math> = {1 \over 1 - e^{-jw-1}}</math><br><br><br> | ||
Applying this to y[n],<br><br> | Applying this to y[n],<br><br> | ||
| − | <math> y[n] = 1 + H(e^{j{2\pi \over N}})e^{j({2\pi \over N})n}[{1 \over 2j}+{5 \over 2}] + H(e^{-j{2\pi \over N}})e^{-j({2\pi \over N})n}[{1 \over 2j} - {5 \over 2}] - ( | + | <math> y[n] = 1 + H(e^{j{2\pi \over N}})e^{j({2\pi \over N})n}[{1 \over 2j}+{5 \over 2}] + H(e^{-j{2\pi \over N}})e^{-j({2\pi \over N})n}[{1 \over 2j} - {5 \over 2}] - H(e^{-j{4\pi \over N}}){7 \over 2}e^{-j({4\pi \over N}n)} + H(e^{j{4\pi \over N}}){7 \over 2}e^{j({4\pi \over N}n)}</math> |
Revision as of 18:08, 26 September 2008
Preview
This is only a preview; changes have not yet been saved! (????)
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 2}e^{-j2({2\pi \over N}n)} + {7 \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^{j{2\pi \over N}})e^{j({2\pi \over N})n}[{1 \over 2j}+{5 \over 2}] + H(e^{-j{2\pi \over N}})e^{-j({2\pi \over N})n}[{1 \over 2j} - {5 \over 2}] - H(e^{-j{4\pi \over N}}){7 \over 2}e^{-j({4\pi \over N}n)} + H(e^{j{4\pi \over N}}){7 \over 2}e^{j({4\pi \over N}n)} $
