| Line 35: | Line 35: | ||
| − | + | and | |
| − | + | <math>\cos{(2t)} \rightarrow \frac{1}{2}(te^{2jt} \; + \; te^{-2jt})</math> | |
| + | |||
| + | <math>=\frac{1}{2}t(e^{2jt} \; + \; e^{-2jt})</math> | ||
| + | |||
| + | <math>\frac{1}{2}t[2\cos{(2t)}]</math> | ||
| + | |||
| + | <math>t\cos{(2t)}</math> | ||
Revision as of 21:50, 16 September 2008
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)} $
