Worked Solution Chapter : The Laplace Tranform 6 a F L4] 6 c F L f t] 4 4e t dt e t dt 4 e t 4 ] e t e 4 if > 6 e F L f t] 6 g Uing integration by part, f te t dt f t e t dt + e t dt + e t + 4 4 4 f te t dt f t e t e t dt + e t e t dt + e t dt + f t e t dt e t dt e ] 4 e t 4 + F L f t] 4 f t e t dt e t dt e 4 ] t e t f te t dt te t dt + e t dt for > e ] e 4 ] e t dt + e + e e ] e ] e
The Laplace Tranform 7 a Uing formula 7, L t 4 4! 4+ 4 5 4 5 7 c Uing formula 8, L e 7t 7 7 e Uing formula 9, L e i7t 8 a Lint] + + 9 i7 for > 8 c L7] L7 ] 7L] 7 7 for > 8 e e Linh4t] L 4t e 4t L e 4t e 4t L e 4t L e 4t ] 4 ] + 4 ] + 4 4 4 4 + 4 6 For the above to hold, we need both > 4 and > 4 But ince 4 > 4, it uffice to have > 4 8 g L 6e t + 8e t 6L e t + 8L e t 6 + 8 + 6 + 8 + For the above to hold, we need both > and > But ince >, it uffice to have > 8 i For >, 9 a L cot 4 int] Lcot] 4Lint] L t / Ŵ + t / + Ŵ 5 / + 4 8 + + 4 Ŵ + 5 / Ŵ 5 / π 5 / π 5 / 4 9 c L t / Ŵ + Ŵ Ŵ / / + / { } { } a if t < if t < tep t if t if t { } if t < f t if t
Worked Solution b L f t] L tep t L] L tep t e e ] a L te 4t L e 4t t Here, f t L e 4t f t F 4 F L f t] Lt] for > So, for any X >, FX X, and we complete the computation we tarted above L te 4t F }{{ 4 } FX X X 4, keeping in mind that we mut have 4 X > ; that i, > 4 c L e t int L e t int L e t f t F f t F L f t] Lint] for > + 9 FX F X + 9 + 9 So the firt line in our computation continue a L e t int F e L e t t L e t t / f t F L f t] F Ŵ for X > for > + 9 L e t f t F L t / Ŵ + / / + π / for > for > for > π FX X / for X > π F / for > So the firt line in our computation continue a L e t π t F / for >
4 The Laplace Tranform 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 L t n e αt n! F α for > α α n+ c L e αt coωt L e αt coωt L e αt f t F α f t F L f t] Lcoωt] for > + ω FX F α So the firt line in our computation continue a L e αt coωt F α a Uing the definition Ŵx X X + ω for X > α α + ω for α > along integration by part, we have Ŵσ + α α + ω for > α u x e u du for x > u σ+ e u du u σ e u du u σ e u u σ u σ e u du u uσ e u + σ e + σ u σ e u du u uσ e u + σŵσ for σ > In addition, uing L Hôpital rule, it i eaily een that, for any real value σ, u uσ e u
Worked Solution 5 So, the firt term in the lat formula derived above for Ŵσ + vanihe, leaving u Ŵσ + σŵσ for σ > 4 a Clearly t + f t i finite, and the only dicontinuity on, i a jump dicontinuity at t So the function i piecewie continuou on, 4 c Since int i continuou at every point on,, it i automatically i continuou at every point on, and ha a finite it at t So the function i piecewie continuou on, 4 e Since t π / tant, the function i not piecewie continuou on, 4 g Thi function blow up become infinite a t + So it i not piecewie continuou on, 4 i Since t, the function i not piecewie continuou on, t 4 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, t + tairt which i finite So tairt i piecewie continuou on, 5 Firt of all, ince gt i continuou at t t, gt gt gt t t + t t Thu, and Hence, t t + t t f gt t t + f t gt f t gt f t gt t t + t t + t t + f gt t t the jump in f g at t f t gt f t gt f t gt t t t t t t f gt f gt t t + t t f t gt t t + f t gt t t f t 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 whenever t and So e t i of exponential order for any
6 The Laplace Tranform 6 c Applying the tet, it i clear that ] te t e t te t if However, if < then e t a t, and a naive computation of the it yield ] te t e t te t, telling u that we mut compute thi it via L Hôpital rule Doing o, ] te t e t te t t e t e 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, 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 α e σ α/ σ σ α α e α σ