(New page: == Part B: The Basics of Linearity == It is given that the input <math>e^{2jt} \!</math> gives the output <math>t*e^{-2jt} \!</math> and that the input <math>e^{-2jt} \!</math> gives the ...) |
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| − | We are asked to give the system's response to an input of <math>cos(2t) \!</math>. Recall that <math>cos(2t)=\frac{1}{2}(e^{2jt}+e^{-2jt}) \! </math>. Note we are given the systems response to the inputs <math>e^{2jt} \!</math> and <math>e^{-2jt} \!</math>. Thus, because the system is linear, we can easily see the system's output to this input will be <math>\frac{ | + | We are asked to give the system's response to an input of <math>cos(2t) \!</math>. Recall that <math>cos(2t)=\frac{1}{2}(e^{2jt}+e^{-2jt}) \! </math>. Note we are given the systems response to the inputs <math>e^{2jt} \!</math> and <math>e^{-2jt} \!</math>. Thus, because the system is linear, we can easily see the system's output to this input will be <math>\frac{1}{2}(te^{-2jt}+te^{2jt})=tcos(2t). |
Latest revision as of 12:08, 18 September 2008
Part B: The Basics of Linearity
It is given that the input $ e^{2jt} \! $ gives the output $ t*e^{-2jt} \! $ and that the input $ e^{-2jt} \! $ gives the output $ t*e^{2jt} \! $.
We are asked to give the system's response to an input of $ cos(2t) \! $. Recall that $ cos(2t)=\frac{1}{2}(e^{2jt}+e^{-2jt}) \! $. Note we are given the systems response to the inputs $ e^{2jt} \! $ and $ e^{-2jt} \! $. Thus, because the system is linear, we can easily see the system's output to this input will be $ \frac{1}{2}(te^{-2jt}+te^{2jt})=tcos(2t). $
