Line 7: Line 7:
 
h(t) = 10delta(t-1)
 
h(t) = 10delta(t-1)
  
<math>H(s) = \int_{-\infty}^{\infty} h(t)e^{-st}</math>
+
<math>H(s) = \int_{-\infty}^{\infty}h(t)e^{-st}</math>
  
<math>H(s) = \int_{-\infty}^{\infty} 10delta(t-1)e^{-st}</math>
+
<math>H(s) = \int_{-\infty}^{\infty}10delta(t-1)e^{-st}</math>
  
<math>H(s) = 10\times\int_{-\infty}^{\infty} 10delta(t-1)e^{-st}</math>
+
<math>H(s) = 10\times\int_{-\infty}^{\infty}delta(t-1)e^{-st}</math>

Revision as of 12:00, 25 September 2008

CT LTI system:

y(t) = 10x(t-1)

plugging in delta(t) into the system we get:

h(t) = 10delta(t-1)

$ H(s) = \int_{-\infty}^{\infty}h(t)e^{-st} $

$ H(s) = \int_{-\infty}^{\infty}10delta(t-1)e^{-st} $

$ H(s) = 10\times\int_{-\infty}^{\infty}delta(t-1)e^{-st} $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva