(New page: Let <font size = '4'><math>x(t) = e^{-at} u(t)</math></font> <math>\chi(w) = \mathcal{F} (x(t)) = \int^{\infty}_{-\infty} e^{-at} u(t) e^{-jwt} dt</math> <math>= \int^{\infty}_{0} e^{-at...) |
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| − | Let <font size = '4'><math>x(t) = e^{- | + | Let <font size = '4'><math>x(t) = e^{-a(t+1)} u(t + 1)</math></font> |
| − | <math>\chi(w) = \mathcal{F} (x(t)) = \int^{\infty}_{-\infty} e^{-at} u(t) e^{-jwt} dt</math> | + | <math>\chi(w) = \mathcal{F} (x(t)) = \int^{\infty}_{-\infty} e^{-at}e^{-a} u(t + 1) e^{-jwt} dt</math> |
| − | <math>= \int^{\infty}_{ | + | <math>= e^{-a} \int^{\infty}_{-1} e^{-at}.e^{-jwt} dt</math> |
| − | <math>= \int^{\infty}_{ | + | <math>= e^{-a} \int^{\infty}_{-1}e^{-(a+jw)t} dt</math> |
| − | <math>= -\frac{ | + | <math>= -\frac{e^{-a}}{a+jw} [e^{-(a+jw)t}]^{\infty}_{0} </math> |
| − | <math>= -\frac{ | + | <math>= -\frac{e^{-a}}{a+jw} [-e^{a+jw}]</math> |
| − | <math>=\frac{ | + | <math>=\frac{e^{-(2a+jw)}}{a+jw}</math> |
Revision as of 13:27, 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}_{0} $
$ = -\frac{e^{-a}}{a+jw} [-e^{a+jw}] $
$ =\frac{e^{-(2a+jw)}}{a+jw} $
