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<math>\ h(t) = 5e^{-t} </math> | <math>\ h(t) = 5e^{-t} </math> | ||
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<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> | ||
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| + | So, | ||
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| + | <math>\ b_{0} = 0 </math> | ||
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| + | <math>\ b_{1} = (\frac{1 + 2j}{2}) (\frac{5}{1+jw}) </math> | ||
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| + | <math>\ b_{-1}= (\frac{1 + 2j}{2}) (\frac{5}{1 - j}) </math> | ||
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| + | <math>\ b_{2} = (\frac{5}{2j}) (\frac{5}{1+5j}) </math> | ||
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| + | <math>\ b_{-2}= (\frac{5}{2j}) (\frac{5}{1-5j}) </math> | ||
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| + | |||
| + | So, | ||
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| + | <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} $
