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So when the signal is run through the system the following is what is produced | So when the signal is run through the system the following is what is produced | ||
| − | <math>\frac{e^{i2t}+e^{-i2t}}{2}=\frac{te^{-i2t}+te^{i2t}}{2} = t\frac{e^{-i2t}+e^{i2t}}{2}</math> | + | <math>\frac{e^{i2t}+e^{-i2t}}{2}=\frac{te^{-i2t}+te^{i2t}}{2} = t\frac{e^{-i2t}+e^{i2t}}{2} = t\cos(2t)</math> |
Latest revision as of 07:09, 19 September 2008
The overall response for this system is
$ f(t) = tf(-t) $
To relate to the systems that were already use in example
$ \cos(2t) = \frac{e^{i2t}+e^{-i2t}}{2} $
So when the signal is run through the system the following is what is produced
$ \frac{e^{i2t}+e^{-i2t}}{2}=\frac{te^{-i2t}+te^{i2t}}{2} = t\frac{e^{-i2t}+e^{i2t}}{2} = t\cos(2t) $
