We are told that a system is linear and given inputs

$ \,x_1(t)=e^{2jt}\, $ yields $ \,y_1(t)=te^{-2jt}\, $

$ \,x_2(t)=e^{-2jt}\, $ yields $ \,y_2(t)=te^{2jt}\, $


The input

$ \,x(t)=\cos(2t)\, $

can be rewritten as

$ \,x(t)=\frac{e^{2jt}+e^{-2jt}}{2}\, $

$ \,x(t)=\frac{1}{2}e^{2jt}+\frac{1}{2}e^{-2jt}\, $

$ \,x(t)=\frac{1}{2}x_1(t)+\frac{1}{2}x_2(t)\, $


Since the system is linear, and by the definition of linearity, we can write the response as

$ \,y(t)=\frac{1}{2}y_1(t)+\frac{1}{2}y_2(t)\, $

$ \,y(t)=\frac{1}{2}te^{-2jt}+\frac{1}{2}te^{2jt}\, $

$ \,y(t)=t(\frac{e^{2jt}+e^{-2jt}}{2})\, $

$ \,y(t)=t\cos(2t)\, $

Comments on Other Answers

Talk:HW3.B Ben Laskowski_ECE301Fall2008mboutin

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