(New page: ==The Basics of Linearity== The problem tells us that: <math> \exp(2jt) </math>) |
(→The Basics of Linearity) |
||
| Line 3: | Line 3: | ||
The problem tells us that: | The problem tells us that: | ||
| − | <math> \ | + | <math>\ e^{2jt} \longrightarrow sys \longrightarrow t e^{-2jt} </math> and that |
| + | <math>\ e^{-2jt} \longrightarrow sys \longrightarrow t e^{2jt} </math> | ||
| + | |||
| + | and asks us to find the output for an input of <math>\ \cos{2t} </math>. However due to Euler Identity, we can write <math>\ \cos{2t} </math> as a linear combination of the two exponential inputs above. Then as a result of the systems linearity, the output will be the just be the linear combination of the outputs of the two exponential above. | ||
| + | |||
| + | <math>\ \cos{2t}= 0.5 e^{2jt} +(- 0.5 e^{-2jt}) </math> | ||
| + | |||
| + | Therefore putting this into the system will yield: | ||
| + | |||
| + | <math>\ \cos{2t}= 0.5 e^{2jt} +(- 0.5 e^{-2jt}) \longrightarrow sys | ||
| + | |||
| + | \longrightarrow t(- 0.5 e^{-2jt})+ t 0.5 e^{2jt} = t(0.5 e^{2jt} +(- 0.5 e^{-2jt})) </math> | ||
| + | |||
| + | <math>\ = t \cos(2t) </math> | ||
Latest revision as of 15:46, 18 September 2008
The Basics of Linearity
The problem tells us that:
$ \ e^{2jt} \longrightarrow sys \longrightarrow t e^{-2jt} $ and that $ \ e^{-2jt} \longrightarrow sys \longrightarrow t e^{2jt} $
and asks us to find the output for an input of $ \ \cos{2t} $. However due to Euler Identity, we can write $ \ \cos{2t} $ as a linear combination of the two exponential inputs above. Then as a result of the systems linearity, the output will be the just be the linear combination of the outputs of the two exponential above.
$ \ \cos{2t}= 0.5 e^{2jt} +(- 0.5 e^{-2jt}) $
Therefore putting this into the system will yield:
$ \ \cos{2t}= 0.5 e^{2jt} +(- 0.5 e^{-2jt}) \longrightarrow sys \longrightarrow t(- 0.5 e^{-2jt})+ t 0.5 e^{2jt} = t(0.5 e^{2jt} +(- 0.5 e^{-2jt})) $
$ \ = t \cos(2t) $
