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<math>f(t)=j*5*sin(t)</math>
 
<math>f(t)=j*5*sin(t)</math>
  
<math>E_\infty = \int_{-\infty}^\infty |5*sin(t)|^2\,dt)</math>
+
<math>E_\infty = \int_{-\infty}^\infty |5sin(t)|^2\,dt)</math>
  
<math>E_\infty = \int_{-\infty}^\infty 25*sin(t)^2u(t)\,dt)</math>
+
<math>E_\infty = \int_{-\infty}^\infty 25sin(t)^2u(t)\,dt)</math>
  
 
<math>E_\infty = \int_{-\infty}^\infty 25(.5 + .5cos(2t)),dt)</math>
 
<math>E_\infty = \int_{-\infty}^\infty 25(.5 + .5cos(2t)),dt)</math>

Revision as of 03:55, 22 June 2009

$ f(t)=j*5*sin(t) $

$ E_\infty = \int_{-\infty}^\infty |5sin(t)|^2\,dt) $

$ E_\infty = \int_{-\infty}^\infty 25sin(t)^2u(t)\,dt) $

$ E_\infty = \int_{-\infty}^\infty 25(.5 + .5cos(2t)),dt) $

$ E_\infty =\frac{t}{2} + frac{sin(t)}{4}\bigg]_-infty^\infty) $

$ E_\infty =\infty-0 = \infty $

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010