(New page: Homework 3 - '''A''' - '''B''' - '''C''')
 
 
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[[Homework 3_ECE301Fall2008mboutin]] - [[HW3.A Allen Humphreys_ECE301Fall2008mboutin|'''A''']] - [[HW3.B Allen Humphreys_ECE301Fall2008mboutin|'''B''']] - [[HW3.C Allen Humphreys_ECE301Fall2008mboutin|'''C''']]
 
[[Homework 3_ECE301Fall2008mboutin]] - [[HW3.A Allen Humphreys_ECE301Fall2008mboutin|'''A''']] - [[HW3.B Allen Humphreys_ECE301Fall2008mboutin|'''B''']] - [[HW3.C Allen Humphreys_ECE301Fall2008mboutin|'''C''']]
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We know that the system is linear, therefore, we can sum the inputs to equal the sum of outputs:
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 +
The input is
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: <math>x_{c}(t) = e^{2\times jt} + e^{-2\times jt} </math>
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 +
and the corresponding output is
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:<math>y_{c}(t) = t \times e^{-2jt} + t \times e^{2jt}</math>
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We know by Euler's formula:
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: <math>\cos 2t = {e^{2jt} + e^{-2jt} \over 2}</math>
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Finally, by the multiplication property of linear systems:
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 +
The input,
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:<math>  {x_{c}(t) \over 2} = \cos 2t = {e^{2\times jt} + e^{-2\times jt} \over 2} </math>
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 +
will yield the output:
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:<math>{y_{c}(t) \over 2} = {t e^{-2jt} + t e^{2jt} \over 2}</math>

Latest revision as of 13:29, 19 September 2008

Homework 3_ECE301Fall2008mboutin - A - B - C

We know that the system is linear, therefore, we can sum the inputs to equal the sum of outputs:

The input is

$ x_{c}(t) = e^{2\times jt} + e^{-2\times jt} $

and the corresponding output is

$ y_{c}(t) = t \times e^{-2jt} + t \times e^{2jt} $

We know by Euler's formula:

$ \cos 2t = {e^{2jt} + e^{-2jt} \over 2} $

Finally, by the multiplication property of linear systems:

The input,

$ {x_{c}(t) \over 2} = \cos 2t = {e^{2\times jt} + e^{-2\times jt} \over 2} $

will yield the output:

$ {y_{c}(t) \over 2} = {t e^{-2jt} + t e^{2jt} \over 2} $

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