| Line 14: | Line 14: | ||
<math>e^{-j2t} = cos(2t) - jsin(2t)\!</math> | <math>e^{-j2t} = cos(2t) - jsin(2t)\!</math> | ||
| + | |||
We can use this to represent our input signal as follows: | We can use this to represent our input signal as follows: | ||
<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: | ||
Revision as of 10:55, 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}]\! $
So the response of the system to the input excitation $ cos(2t)\! $ is
$ 0.5[te^{j2t}+te^{-j2t}] = 0.5[tcos(2t) + jtsin(2t) + tcos(2t) - jsin(2t)] = tcos(2t)\! $
