Given:

For a linear system we have:

$ e^{j2t}\rightarrow [system]\rightarrow te^{-j2t}\! $
$ e^{-j2t}\rightarrow [system]\rightarrow te^{j2t}\! $


To find the response of the system above we first note that

$ e^{j2t} = cos(2t) + jsin(2t)\! $

$ e^{-j2t} = cos(2t) - jsin(2t)\! $


We can use this to represent our input signal as follows:

$ cos(2t) = 0.5\times[cos(2t) + jsin(2t) + cos(2t) - jsin(2t)] = 0.5\times[e^{j2t} + e^{-j2t}]\! $


Since the system is linear, we know that:

$ 0.5\times[e^{j2t} + e^{-j2t}]\rightarrow[system]\rightarrow0.5[te^{j2t}+te^{-j2t}]\! $


So the response of the system to the input excitation $ cos(2t)\! $ is

$ 0.5[te^{j2t}+te^{-j2t}] = 0.5[tcos(2t) + jtsin(2t) + tcos(2t) - jsin(2t)] = tcos(2t)\! $

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett