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<math>\ h(t) = 5e^{-t} </math>
 
<math>\ h(t) = 5e^{-t} </math>
 +
  
 
<math>\ H(jw) = 5\int_0^{\infty} e^{-\tau}e^{-jw{\tau}}\,d{\tau}  </math>
 
<math>\ H(jw) = 5\int_0^{\infty} e^{-\tau}e^{-jw{\tau}}\,d{\tau}  </math>
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<math>\ H(jw) = \frac{5}{1+ jw} </math>
 
<math>\ H(jw) = \frac{5}{1+ jw} </math>
 +
  
 
So,  
 
So,  
  
<math>\ a_{0} = b_{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 - j}) </math>
 +
 
 +
<math>\ b_{2} = (\frac{5}{2j}) (\frac{5}{1+5j})          </math>
 +
 
 +
<math>\ b_{-2}= (\frac{5}{2j}) (\frac{5}{1-5j})          </math>
 +
 
 +
 
 +
So,
  
<math>\ b_{1} = a_{1} * (frac{1}{1+jw}
+
<math>\ y(t) = (\frac{1 + 2j}{2}) (\frac{5}{1+jw})e^{jt} + (\frac{1 + 2j}{2})e^{-jt} (\frac{5}{1 - j}) + (\frac{5}{2j}) (\frac{5}{1+5j})e^{j4t} + (\frac{5}{2j}) (\frac{5}{1-5j})e^{-j4t} </math>

Latest revision as of 18:04, 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}) $


So,

$ \ y(t) = (\frac{1 + 2j}{2}) (\frac{5}{1+jw})e^{jt} + (\frac{1 + 2j}{2})e^{-jt} (\frac{5}{1 - j}) + (\frac{5}{2j}) (\frac{5}{1+5j})e^{j4t} + (\frac{5}{2j}) (\frac{5}{1-5j})e^{-j4t} $

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To all math majors: "Mathematics is a wonderfully rich subject."

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