Revision as of 20:50, 16 September 2008 by Jpfister (Talk)

Problem

A linear system’s response to $ e^{2jt} $ is $ te^{-2jt} $, and its response to $ e^{-2jt} $ is $ te^{2jt} $. What is the system’s response to $ \cos{(2t)} $?

Solution

If the system is linear, then the following is true:


For any $ x_{1}(t) \; \rightarrow \; y_{1}(t) $ and $ x_{2}(t) \; \rightarrow \; y_{2}(t) $

and any complex constants $ a $ and $ b $


then


$ ax_{1}(t) \; + \; bx_{2}(t) \; \rightarrow \; ay_{1}(t) \; + \; by_{2}(t) $


and "conveniently":

$ e^{2jt} \; + \; e^{-2jt} = \cos{(2t)} \; + \; j \sin{(2t)} \; + \; \cos{(-2t)} \; + \; j \sin{(-2t)} $           (by Euler's Formula)

$ =\cos{(2t)} \; + \; j \sin{(2t)} \; + \; \cos{(2t)} \; - \; j \sin{(2t)} $           ($ \cos{(-x)}=\cos{(x)} $ and $ \sin{(-x)}=-\sin{(x)} $)

$ =2\cos{(2t)} $


therefore:


$ \cos{(2t)} = \frac{1}{2}\cdot 2\cos{(2t)} = \frac{1}{2}(e^{2jt} \; + \; e^{-2jt}) $


and


$ \cos{(2t)} \rightarrow \frac{1}{2}(te^{2jt} \; + \; te^{-2jt}) $

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

$ \frac{1}{2}t[2\cos{(2t)}] $

$ t\cos{(2t)} $

Alumni Liaison

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

Dr. Paul Garrett