(New page: <math>x(t) = e^{-3t} , t>3 \,</math> <math>x(t)= e^{-6t} , 0 \le t \le 3</math> <math>x(t)= 0 , t < 0 \,</math> <math>x(t)= e^{-3t} u(t-3) + e^{-6t}( u(t-3)-u(t))\,</math> <math>X(\omega)...) |
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| − | <math>x(t) = e^{-3t} , t>3 \,</math> | + | <math>x(t) = e^{-3t} , t>3 \,</math>, |
| − | <math>x(t)= e^{-6t} , 0 \le t \le 3</math> | + | <math>x(t)= e^{-6t} , 0 \le t \le 3</math>, |
<math>x(t)= 0 , t < 0 \,</math> | <math>x(t)= 0 , t < 0 \,</math> | ||
Revision as of 18:47, 8 October 2008
$ x(t) = e^{-3t} , t>3 \, $, $ x(t)= e^{-6t} , 0 \le t \le 3 $, $ x(t)= 0 , t < 0 \, $
$ x(t)= e^{-3t} u(t-3) + e^{-6t}( u(t-3)-u(t))\, $ $ X(\omega) = \int^\infty_\infty e^{-3t}e^{-j\omega t} dt + \int^2_0 e^{-6t}e^{-j\omega t} dt\, $ $ X(\omega) = \int^\infty_\infty e^{-(3+j\omega)t} dt + \int^3_0 e^{-(6+j\omega) t} dt\, $ $ X(\omega) = {\left. \frac{e^{-(j\omega + 3)t}}{-(j\omega +3)} \right]^{\infty}_0 } + {\left. \frac{e^{-(j\omega + 6)t}}{-(j\omega +6)} \right]^3_0 }\, $ $ X(\omega) = \frac{e^{-(3j\omega + 9)}}{j\omega +3} - \frac{e^{-(3j\omega + 18)t}}{-j\omega +6} + \frac{1}{6+j\omega} \, $ $ X(\omega) = \frac{e^{-(3j\omega + 9)}}{j\omega +3} + \frac{1 - e^{-(3j\omega + 18)t}}{-j\omega +6} \, $
