ECE2260F09_HW5p4soln

ECE2260F09_HW5p4soln - 3 3! = 1 s 1 s 3 = 1 s 4 If we...

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2260 F 09 Homework #5 prob 4 solution T RANSFORM P AIR : L { t n } = n ! s n + 1 P ROOF : We derive this transform pair by considering how we get 1 s n for values of n starting from n = 1. L { u ( t )} = 1 s We use the time-integral identity to obtain 1 s 2 , 1 s 3 , and so forth. L f ( τ ) d τ 0 t = F s ( ) s N OTE : τ is a variable of integration that plays the role of t to avoid confusion with the t in the upper limit of the integral. Thus, we have L u ( τ ) d τ 0 t = t = 1 s 1 s = 1 s 2 and L τ d τ 0 t = t 2 2 = 1 s 1 s 2 = 1 s 3 and L τ 2 2 dt 0 t = t
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Unformatted text preview: 3 3! = 1 s 1 s 3 = 1 s 4 If we continue, the pattern is the following: L t n n ! = 1 s n + 1 Multiplying by n ! yields the identity we seek. L t n { } = n ! s n + 1 N OTE : To complete a formal proof by induction, we may apply the time-integral identity to show the formula holds for t n + 1 ....
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