Line 1: Line 1:
<math>x(t)=t^3 e^{-3t} </math>
+
<math>x(t)=e^{-3t} u(t-3) u(t+3) </math>
  
 
<math>X(w) = \int^{\infty}_{- \infty}x(t)e^{-jwt} dt</math>
 
<math>X(w) = \int^{\infty}_{- \infty}x(t)e^{-jwt} dt</math>
  
<math>= \int^{\infty}_{- \infty} t^3 e^{-3t} e^{-jwt} dt</math>
+
<math>= \int^{\infty}_{- \infty} e^{-3t} u(t-3) u(t+3) e^{-jwt} dt</math>
  
<math>= \int^{\infty}_{- \infty} t^3 e^{-(3 + jw)t} dt</math>
+
<math>= \int^{3}_{-3} e^{-(3 + jw)t} dt</math>
  
<math>[\frac{1}{3} t^4 \frac{e^{-(3 + jw)t}}{-(3 + jw)}]_{- \infty}^{\infty}</math>
+
<math>[\frac{e^{-(3 + jw)t}}{-(3 + jw)}]_{-3}^{3}</math>

Revision as of 09:39, 7 October 2008

$ x(t)=e^{-3t} u(t-3) u(t+3) $

$ X(w) = \int^{\infty}_{- \infty}x(t)e^{-jwt} dt $

$ = \int^{\infty}_{- \infty} e^{-3t} u(t-3) u(t+3) e^{-jwt} dt $

$ = \int^{3}_{-3} e^{-(3 + jw)t} dt $

$ [\frac{e^{-(3 + jw)t}}{-(3 + jw)}]_{-3}^{3} $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn