| Line 15: | Line 15: | ||
<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 17:57, 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}) $
