| Line 11: | Line 11: | ||
<math>\ a_{0} = 0 </math> | <math>\ a_{0} = 0 </math> | ||
| − | <math>\ b_{1} = | + | <math>\ b_{1} = \frac{1 + 2j}{2}* \frac{5}{1+jw} |
Revision as of 17:45, 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,
$ \ a_{0} = 0 $
$ \ b_{1} = \frac{1 + 2j}{2}* \frac{5}{1+jw} $
