Line 11: Line 11:
 
<math>\ b_{0} = 0 </math>
 
<math>\ b_{0} = 0 </math>
  
<math>\ b_{1} = (\frac{1 + 2j}{2}) (\frac{5}{1+jw}) </math>
+
<math>\ b_{1} = (\frac{1 + 2j}{2}) (\frac{5}{1+jw}) </math>
  
 
<math>\ b_{-1}= (\frac{1 + 2j}{2}) (\frac{5}{1 - j}) </math>
 
<math>\ b_{-1}= (\frac{1 + 2j}{2}) (\frac{5}{1 - j}) </math>
  
<math>\ b_{2} = \frac{5}{2j} \frac{5}{1 +5j} </math>
+
<math>\ b_{2} = \frac{5}{2j} \frac{5}{1+5j}         </math>
  
<math>\ b_{-2}= \frac{5}{2j} \frac{5}{1 - 5j}</math>
+
<math>\ b_{-2}= \frac{5}{2j} \frac{5}{1-5j}         </math>

Revision as of 17:56, 26 September 2008

$ \ h(t) = 5e^{-t} $

$ \ H(jw) = 5\int_0^{\infty} e^{-\tau}e^{-jw{\tau}}\,d{\tau} $

$ \ H(jw) = 5[-\frac{1}{1 + jw}e^{-\tau}e^{-jwr} ]^{\infty}_0 $

$ \ H(jw) = \frac{5}{1+ jw} $

So,

$ \ b_{0} = 0 $

$ \ b_{1} = (\frac{1 + 2j}{2}) (\frac{5}{1+jw}) $

$ \ b_{-1}= (\frac{1 + 2j}{2}) (\frac{5}{1 - j}) $

$ \ b_{2} = \frac{5}{2j} \frac{5}{1+5j} $

$ \ b_{-2}= \frac{5}{2j} \frac{5}{1-5j} $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett