Line 23: Line 23:
 
</math>
 
</math>
  
since  I got
+
since  I got  
 +
 
 +
<math>a_1 = \frac{3}{j}</math>
 +
 
 +
<math>a_2= \frac{-3}{j}</math>
 +
 
 +
<math>a_3 = 2 </math>
 +
 
 +
<math>a_4 = 2</math>

Revision as of 18:42, 25 September 2008

part A

$ e^{st}\rightarrow h(t)\rightarrow H(S)e^{st} $

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

assume

$ h(t) = 5u(t-3) $

$ H(s) = 5\int_3^\infty e^{-st}dt $

$ H(s) = \frac{-5}{s} $

$ 4 to \infty $

$ H(s)= \frac {5}{s} $


part B

$ h(t) = \sum_{-\infty} ^\infty a_k H(s) $

since I got

$ a_1 = \frac{3}{j} $

$ a_2= \frac{-3}{j} $

$ a_3 = 2 $

$ a_4 = 2 $

Alumni Liaison

BSEE 2004, current Ph.D. student researching signal and image processing.

Landis Huffman