| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| − | A linear system’s response to <math>e^{2jt}</math> is <math>te^{-2jt}</math>, and its response to <math>e^{-2jt}</math> is <math>te^{2jt}</math>. What is the system’s response to <math>cos(2t)</math>? | + | A linear system’s response to <math>e^{2jt}</math> is <math>te^{-2jt}</math>, and its response to <math>e^{-2jt}</math> is <math>te^{2jt}</math>. What is the system’s response to <math>\cos{(2t)}</math>? |
==Solution== | ==Solution== | ||
| Line 32: | Line 32: | ||
<math>\cos{(2t)} = \frac{1}{2}\cdot 2\cos{(2t)} = \frac{1}{2}(e^{2jt} \; + \; e^{-2jt})</math> | <math>\cos{(2t)} = \frac{1}{2}\cdot 2\cos{(2t)} = \frac{1}{2}(e^{2jt} \; + \; e^{-2jt})</math> | ||
| + | |||
| + | |||
| + | conclusion: | ||
| + | |||
| + | |||
| + | The respone to <math>\cos{(2t)}</math> is <math>\frac{1}{2}(e^{2jt} \; + \; e^{-2jt})</math> | ||
Revision as of 21:40, 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}) $
conclusion:
The respone to $ \cos{(2t)} $ is $ \frac{1}{2}(e^{2jt} \; + \; e^{-2jt}) $
