| Line 4: | Line 4: | ||
For a linear system we have: | For a linear system we have: | ||
| − | <math>e^{j2t} \rightarrow [system] \rightarrow te^{-j2t}\!</math><br> | + | <math>e^{j2t}\rightarrow [system]\rightarrow te^{-j2t}\!</math><br> |
| − | <math>e^{-j2t} \rightarrow [system] \rightarrow te^{j2t}\!</math><br> | + | <math>e^{-j2t}\rightarrow [system]\rightarrow te^{j2t}\!</math><br> |
---- | ---- | ||
| Line 20: | Line 20: | ||
<math>cos(2t) = 0.5\times[cos(2t) + jsin(2t) + cos(2t) - jsin(2t)] = 0.5\times[e^{j2t} + e^{-j2t}]\!</math> | <math>cos(2t) = 0.5\times[cos(2t) + jsin(2t) + cos(2t) - jsin(2t)] = 0.5\times[e^{j2t} + e^{-j2t}]\!</math> | ||
| − | Since the system is linear we know that: | + | Since the system is linear, we know that: |
| + | |||
| + | <math>0.5\times[e^{j2t} + e^{-j2t}]\rightarrow[system]\rightarrow0.5[te^{j2t}+te^{-j2t}]\!</math> | ||
Revision as of 10:45, 19 September 2008
Given:
For a linear system we have:
$ e^{j2t}\rightarrow [system]\rightarrow te^{-j2t}\! $
$ e^{-j2t}\rightarrow [system]\rightarrow te^{j2t}\! $
To find the response of the system above we first note that
$ e^{j2t} = cos(2t) + jsin(2t)\! $
$ e^{-j2t} = cos(2t) - jsin(2t)\! $
We can use this to represent our input signal as follows:
$ cos(2t) = 0.5\times[cos(2t) + jsin(2t) + cos(2t) - jsin(2t)] = 0.5\times[e^{j2t} + e^{-j2t}]\! $
Since the system is linear, we know that:
$ 0.5\times[e^{j2t} + e^{-j2t}]\rightarrow[system]\rightarrow0.5[te^{j2t}+te^{-j2t}]\! $
