ECE108HW%239sol_06

ECE108HW%239sol_06 - ECE 108 HW#9 Solution 9.8 There are 3...

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ECE 108 HW#9— Solution 9.8 There are 3 possible ROC for . They are ( ) X s 3 31 1 σ <− −< <− −< Now we know that ( ) ( ) ( ) () ( ) 2 2 for 2 tt gt e xt e xt Gj Xs s j ωω == = + Therefore is in the ROC for . But this is only possible if the ROC of is ( ) X s 2 =− ( ) X s Because the ROC is between two poles, must be a two-sided signal ( ) x t 9.21 (d) 0 22 0 ts t t X s te e dt te dt −− −∞ =+ ∫∫ Now 2 1 at at at te dt e a + Then 0 00 0 0 0 21 st jt X s t e d e d t t e d e d t ee ss e e σσ ∞∞ −+ −∞ + + = +  ++   The first term converges at if . The second term converges at if 20 o r 2 <<
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2 . In this case the ROC is . The Laplace transform is then () 20 o r 2 σσ +> > 22 σ << 2 2 2 2 11 4 2 2 2 2 ss s Xs s s −+ +− =− + = = −+ + + Zeros: 0 s = Poles: . Because of the squared term, these are called second order poles. 12 2 and 2 pp == (f) ( ) ( ) ( ) tt xt te u t = Then 00 2 2 0 2 2 2 st ts t t jt X s te e dt te dt ee s ω −− −∞ −∞ −∞ +− = ∫∫ This term converges to zero at the lower limit if 2 −< ⇒ < Then [] 2 1 for Re 2 2 X s =< (g) 1 1 0 0 1 for all st st X se d t e s s = =
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ECE108HW%239sol_06 - ECE 108 HW#9 Solution 9.8 There are 3...

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