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<math>= -\frac{e^{-a}}{a+jw} [e^{-(a+jw)t}]^{\infty}_{1} </math> | <math>= -\frac{e^{-a}}{a+jw} [e^{-(a+jw)t}]^{\infty}_{1} </math> | ||
| − | <math>= -\frac{e^{-a}}{a+jw} [-e^{a+jw}]</math> | + | <math>= -\frac{e^{-a}}{a+jw} [-e^{-(a+jw)}]</math> |
<math>=\frac{e^{-(2a+jw)}}{a+jw}</math> | <math>=\frac{e^{-(2a+jw)}}{a+jw}</math> | ||
Revision as of 13:28, 8 October 2008
Let $ x(t) = e^{-a(t+1)} u(t + 1) $
$ \chi(w) = \mathcal{F} (x(t)) = \int^{\infty}_{-\infty} e^{-at}e^{-a} u(t + 1) e^{-jwt} dt $
$ = e^{-a} \int^{\infty}_{-1} e^{-at}.e^{-jwt} dt $
$ = e^{-a} \int^{\infty}_{-1}e^{-(a+jw)t} dt $
$ = -\frac{e^{-a}}{a+jw} [e^{-(a+jw)t}]^{\infty}_{1} $
$ = -\frac{e^{-a}}{a+jw} [-e^{-(a+jw)}] $
$ =\frac{e^{-(2a+jw)}}{a+jw} $
