4e st dt. 0 e st dt. lim. f (t)e st dt. f (t) e st dt + 0. e (2 s)t dt + 0. e (2 s)4 1 = 1. = t s e st

Transcription

1 Worked Solution 8 Chapter : The Laplace Tranform 6 a F L] e t dt e t dt e t ] lim t e t e if > for > 6 c F L f t] f te t dt f t e t dt + e t dt + e t + f t e t dt e t dt ] e e ] 6 e F L f t] f te t dt f t e t e t dt + e t e t dt + e t dt + e t + ] e f t e t dt e t dt e ] 6 g Uing integration by part, F L f t] f te t dt te t dt + t e t e t dt e t dt +

2 8 The Laplace Tranform e + e ] e e ] e 7 a Uing formula 7, L t ]! for > 7 c Uing formula 8, L e 7t] 7 for > 7 7 e Uing formula 8, L e 7t] for > 7 8 a Lint] for > 8 c L7] L7 ] 7L] 7 7 for > 8 e e Linht] L t e t ] L e t e t] L e t] L e t] ] ] + ] For the above to hold, we need both > and > But ince >, it uffice to have > 8 g L 6e t + 8e t] 6L e t] + 8L e t] For the above to hold, we need both > and > But ince >, it uffice to have > 8 i For >, L cot int] Lcot] Lint]

3 Worked Solution 8 9 a L t ] / Ŵ + t / + Ŵ 5 / ] 9 c L t / Ŵ + / + a Ŵ + 5 / Ŵ Ŵ / 5 / Ŵ / π 5 / π 5 / { } { } if t < if t < tep t if t if t { } if t < f t if t b L f t] L tep t ] L] L tep t ] a L te t] L e t t Here, So, for any X >, f t e ] L e t f t ] F e ] F L f t] Lt] for > FX X, and we complete the computation we tarted above with L te t] F }{{ } FX X, X keeping in mind that we mut have X > ; that i, > c L e int] t L e t ] int L e f t ] F f t with F L f t] Lint] + 9 for > FX F X for X > for >

4 8 The Laplace Tranform So the firt line in our computation continue a L e int] t F + 9 for > e L e t t] L e t t / with f t F L f t] F Ŵ ] L e t f t ] F L t ] / Ŵ + / + / for > π / for > FX π X / for X > F π / for > So the firt line in our computation continue a L e t π t] F / for > a L t n e αt] L e αt t n Here, So, In particular, f t ] L e αt f t ] F α F L f t] L t n] n! n+ for > FX n! X n+ for X > F α n! α n+ for α >, and the computation tarted above are completed with L t n e αt] F α n! α n+ for > α c L e αt coωt ] L e αt ] coωt L e αt f t ] F α f t with F L f t] Lcoωt] + ω for > FX F α X X + ω for X > α α + ω for α >

5 Worked Solution 85 So the firt line in our computation continue a L e αt coωt ] F α α α + ω for > α a Uing the definition Ŵx u x e u du for x > along with integration by part, we have Ŵσ + u σ+ e u du u σ e u du u σ e u u lim u uσ e u + σ e + σ σ u σ e u du u σ e u du lim u uσ e u + σŵσ for σ > In addition, uing L Hôpital rule, it i eaily een that, for any real value σ, lim u uσ e u So, the firt term in the lat formula derived above for Ŵσ + vanihe, leaving u with Ŵσ + σŵσ for σ > a Clearly lim t + f t i finite, and the only dicontinuity on, i a jump dicontinuity at t So the function i piecewie continuou on, c Since int i continuou at every point on,, it i automatically i continuou at every point on, and ha a finite limit at t So the function i piecewie continuou on, e Since lim t π/ tant, the function i not piecewie continuou on, g Thi function blow up become infinite a t + So it i not piecewie continuou on, i Since lim t t, the function i not piecewie continuou on,

6 86 The Laplace Tranform k On any finite ubinterval of,, tairt i continuou at every point except at the finite number of integral value of t in the interval But thoe dicontinuitie are all finite jump Alo, lim t + tairt which i finite So tairt i piecewie continuou on, 5 Firt of all, ince gt i continuou at t t, lim gt gt lim gt t t + t t Thu, and f t gt lim f gt lim t t + t t + lim f t t t + lim f gt lim t t t t lim f t t t f t gt lim gt t t + lim gt t t lim f t gt t t + lim f t gt t t Hence, the jump in f g at t lim f gt lim f gt t t + t t lim f t gt t t + lim f t gt t t lim f t lim f t gt t t + t t the jump in f at t gt, 6 a Since e a e b whenever a b, we clearly have e t e t So e t i of exponential order for any whenever t and 6 c Applying the tet, it i clear that lim te t] e t t lim t te t if However, if < then e t a t, and a naive computation of the limit yield lim te t] e t lim t t te t,

7 Worked Solution 87 telling u that we mut compute thi limit via L Hôpital rule Doing o, te t] e t lim t lim t te t t lim t e t lim t e t lim t e t Hence, the tet applie and tell u that te t i of exponential order for any > 6 e By baic algebra, we clearly have int e t whenever t and So e t i of exponential order for any 7 a For convenience, let f t t α e σt Oberve that f i a continuou, differentiable function on, and that, ince σ and α are poitive, and f, lim t f t f t > for < t < By baic calculu, we know the maximum value of f on, mut then occur where the derivative i zero, f t αt α e σt σ t α e σt α σ t]t α e σt α σ t t α σ So t α σ i where f t i maximum on,, and thi maximum i M α,σ f α σ α σ α e σ α/ σ α α e α σ