Line 3: | Line 3: | ||
We know that the system is linear, therefore, we can conclude the following about the given information. | We know that the system is linear, therefore, we can conclude the following about the given information. | ||
− | <math>x_{1}(t) = e^{2\times jt}</math> and <math>x_{2}(t) = e^{-2\times jt}</math> | + | : <math>x_{1}(t) = e^{2\times jt}</math> and <math>x_{2}(t) = e^{-2\times jt}</math> |
− | <math>y_{1}(t) = t \times e^{-2jt}</math> and <math>y_{2}(t) = t \times e^{2jt}</math> | + | : <math>y_{1}(t) = t \times e^{-2jt}</math> and <math>y_{2}(t) = t \times e^{2jt}</math> |
− | By linearity, | + | By linearity, |
+ | : <math> x_{1}+x_{2}=y_{1}+y_{2} </math>. Also, | ||
+ | : <math>x_{3} = x_{1} + x_{2}</math> and | ||
+ | : <math>y_{3} = t\times y_{1} + t\times y_{2}</math>. | ||
+ | |||
+ | By euler's formula: | ||
+ | |||
+ | : <math>\cos x = {e^{jx} + e^{-jx} \over 2}</math> |
Revision as of 18:51, 18 September 2008
Homework 3_ECE301Fall2008mboutin - A - B - C
We know that the system is linear, therefore, we can conclude the following about the given information.
- $ x_{1}(t) = e^{2\times jt} $ and $ x_{2}(t) = e^{-2\times jt} $
- $ y_{1}(t) = t \times e^{-2jt} $ and $ y_{2}(t) = t \times e^{2jt} $
By linearity,
- $ x_{1}+x_{2}=y_{1}+y_{2} $. Also,
- $ x_{3} = x_{1} + x_{2} $ and
- $ y_{3} = t\times y_{1} + t\times y_{2} $.
By euler's formula:
- $ \cos x = {e^{jx} + e^{-jx} \over 2} $