| Line 9: | Line 9: | ||
So, | So, | ||
| − | <math>\ | + | <math>\ b_{0} = 0 </math> |
| − | <math>\ b_{1} = | + | <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> | ||
Revision as of 17:55, 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} $
