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Homework 3 Ben Horst: [[HW3.A Ben Horst _ECE301Fall2008mboutin| A]] :: [[HW3.B Ben Horst _ECE301Fall2008mboutin| B]] :: [[HW3.C Ben Horst _ECE301Fall2008mboutin| C]] | Homework 3 Ben Horst: [[HW3.A Ben Horst _ECE301Fall2008mboutin| A]] :: [[HW3.B Ben Horst _ECE301Fall2008mboutin| B]] :: [[HW3.C Ben Horst _ECE301Fall2008mboutin| C]] | ||
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| + | |||
| + | ==Answer== | ||
| + | If the system is linear (it is) then the following should be true: | ||
| + | |||
| + | Thus, if we say | ||
| + | |||
| + | x1 = exp(2jt) and x2 = exp(-2jt) | ||
| + | |||
| + | then | ||
| + | |||
| + | y1 = t exp(-2jt) and y2 = t exp(2jt) | ||
| + | |||
| + | |||
| + | That means that: | ||
| + | |||
| + | x1 + x2 ->|system|-> y1 + y2 | ||
| + | |||
| + | |||
| + | Since <math>\cos(2t) = {e^{j2t} + e^{-j2t} \over 2}</math> | ||
| + | |||
| + | Then we can suggest that <math>{1 \over 2t}x1 + {1 \over 2t}x2 -> |system| -> {1 \over 2t}y1 + {1 \over 2t}y2 </math> | ||
| + | |||
| + | This leads to the conclusion that the input of cos(2t) would be <math>{1 \over 2t}x1 + {1 \over 2t}x2</math> | ||
| + | |||
| + | Which, simplified, is: <math>\cos(2t) \over t</math>. | ||
Revision as of 09:57, 19 September 2008
Homework 3 Ben Horst: A :: B :: C
Answer
If the system is linear (it is) then the following should be true:
Thus, if we say
x1 = exp(2jt) and x2 = exp(-2jt)
then
y1 = t exp(-2jt) and y2 = t exp(2jt)
That means that:
x1 + x2 ->|system|-> y1 + y2
Since $ \cos(2t) = {e^{j2t} + e^{-j2t} \over 2} $
Then we can suggest that $ {1 \over 2t}x1 + {1 \over 2t}x2 -> |system| -> {1 \over 2t}y1 + {1 \over 2t}y2 $
This leads to the conclusion that the input of cos(2t) would be $ {1 \over 2t}x1 + {1 \over 2t}x2 $
Which, simplified, is: $ \cos(2t) \over t $.
