Difference between revisions of "HW4.3 Carlos Leon ECE301Fall2008mboutin" - Rhea

Revision as of 18:02, 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}) + (\frac{1 + 2j}{2}) (\frac{5}{1 - j}) + (\frac{5}{2j}) (\frac{5}{1+5j}) + (\frac{5}{2j}) (\frac{5}{1-5j})