Line 7: Line 7:
 
<math>= e^{-a} \int^{\infty}_{-1}e^{-(a+jw)t} dt</math>
 
<math>= e^{-a} \int^{\infty}_{-1}e^{-(a+jw)t} dt</math>
  
<math>= -\frac{e^{-a}}{a+jw} [e^{-(a+jw)t}]^{\infty}_{0} </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} $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn