(New page: The overall response for this system is <math>f(t) = tf(-t)</math> To relate to the systems that were already use in example <math>\cos(2t) = \frac{e^{i2t}+e^{-i2t}}{2}</math> <math>\...) |
|||
| (One intermediate revision by the same user not shown) | |||
| Line 7: | Line 7: | ||
<math>\cos(2t) = \frac{e^{i2t}+e^{-i2t}}{2}</math> | <math>\cos(2t) = \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}</math> | + | 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} = 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) $
