MA360Handout 3b

MA360Handout 3b - 1 MA360 Handout 3b(week of Jan 23 Sp2006...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 MA360 Handout 3b (week of Jan 23) Sp2006 Kallfelz Assignment: P 41 Sheng: 1,2,4,5( a), b), c)),6( b), c)), 7a) Second session of Class: Remaining Material in Ch I, Sheng Lemma 3b.1: lim s →∞ L [ f ( t )] = lim s →∞ F ( s ) = 0, for any continuous 1 f . Proof: () [ ] ( ) () dt e t f dt e t f t f L st st- ∞ ∞- ∫ ∫ ≤ = Moreover, for any continuous f ( t ), there exist M , α , t such that: () t Me t f α < for all t > t (i.e., any continuous function, no matter how fast-growing, is bounded above by an exponential function for sufficiently large values of t .) Hence, choose M , α such when t 0 = 0: () t Me t f α < for all t > 0 So: () [ ] ( ) () ( ) ( ) ( ) ( ) { } ( ) ( ) ( ) { } ( ) { } ( ) α α α α α α α α α α---- ∞ →- ∞ →-- ∞ → ∞--- ∞- ∞ ∞- =- =- = = = < ≤ =---- ∫ ∫ ∫ ∫ ∫ s M s s e s e d d s e d d t s d t s st t st st M M M dt e M dt e M dt e Me dt e t f dt e t f t f L d s t s 1 lim lim lim Hence: ( ) ( ) ( ) ( ) lim lim lim lim = ∴ = < ≤ ∞ →- ∞ → ∞ → ∞ → s F s F s F s s M s s s α The above Lemma gives us a valuable insight into the unique aspects of s – space (the domain space where the LTs live, i.e. the domain of the functions F ( s ), given that t – space is the domain space where f ( t ) live, i.e. the arguments of the LTs). Its ‘asymptotic’ behavior is such that it always vanishes, i.e. F is non-zero on a finite interval. We can use Lemma 3b.1 to prove, for example, Thm 10: () ( ) () [ ] ) : ( ), ( lim lim t f L s F where s sF t f s t = = ∞ → → + 1 This Lemma can be generalized also to piecewise continuous functions. 2 Note1: we’re taking a right-hand limit on the left hand side (in t-space) since the domain of f is nonnegative. Proof: According to Thm2, for any differentiable (therefore continous) f : () [ ] () [ ] ( ) f t f sL t f L dt d- = Now, obviously, (since f is continuous 2 ): ( ) ( ) t f f t + → = lim Hence: () [ ] () [ ] ( ) () [ ] () t f t f sL f t f sL t f L t dt d + →- =- = lim Taking the s →∞ limit of both sides: () [ ] ( ) [ ] () t f t f sL t f L t s dt d s + → ∞ → ∞ →- = lim lim lim Note 2: The last term on the right hand side is a constant with respect to s . Therefore its limiting value is the same throughout. 3 But according to Lemma3b.1, the left hand side vanishes, if we assume the derivative of f is continuous. Hence: () [ ] ( ) [ ] ( ) () () [ ] t f sL t f t f t f sL t f L s t t s dt d s ∞ → → → ∞ → ∞ → = ∴- = = + + lim lim lim lim lim Thm 11 is the analogous version, this time when: s →∞ , and t → + Both Theorems 10 & 11 should be viewed as tracking the limiting behavior of...
View Full Document

This note was uploaded on 11/18/2010 for the course MA 360 taught by Professor Williamm.kallfelz during the Spring '06 term at Capitol College.

Page1 / 7

MA360Handout 3b - 1 MA360 Handout 3b(week of Jan 23 Sp2006...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online