Reading assignment: In this chapter we will cover Sections Definition and the Laplace transform of simple functions

Transcription

1 Chapter 4 Laplace Tranform 4 Introduction Reading aignment: In thi chapter we will cover Section Definition and the Laplace tranform of imple function Given f, a function of time, with value f(t at time t, the Laplace tranform of f which i denoted by L(f (or F i defined by L(f( F ( e t f(t dt > ( Example 4 The Laplace tranform i linear: L(αf(t+βg(t αl(f(t+βl(g(t e t (αf(t + βg(t dt α e t f(t dt + β e t g(t dt f(t F ( e t dt e t 3 f(t e at for > a F ( e t e at dt e ( at dt ( a e ( at ( a 4 For a poitive integer n, f(t t n ( F ( L(t n e t t n dt e t t n dt ( ( n e t t n e t t n dt ( n ( n e t t n dt L(t n

2 ( n So we find that L(t n L(t n We can now ue thi formula over and over to ucceively reduce the power of t to obtain L(t n ( n L(t n ( n(n L(t n ( n! n ( n! L( n+ 5 f(t co(at To compute the Laplace tranform we will ue the Euler formula decribed in the note for Chapter 3 e iθ co(θ + i in(θ ( which implie that co(θ eiθ + e iθ Therefore, uing i and ( + ib( ib (ib + b, we can write ( e ibt + e ibt L(co(bt L ( ib + + ib ( ( + ib ( ib + ( + b ( + b ( ( + ib + ( ib ( + b ( + b So we arrive at L(co(bt ( + b

3 6 Given a function f(t find L(f (t we apply integration by part a follow L(f (t e t f (t dt f(te t ( e t f(t dt L(f(t f( or L(f (t L(f(t f( 7 Given a function f(t find L(f (t Thi can be eaily done by uing the previou formula L(f (t L(f (t f ( ( L(f(t f( f ( L(f(t f( f ( So we have L(f (t L(f(t f( f ( 8 f(t in(at To compute thi Laplace tranform we will ue two previou formula L(in(bt b L(co(bt [L(co(bt co(] b [ ( ] b ( + b [ ] b ( + b [ ] ( + b b ( + b b ( + b Therefore L(in(bt b ( + b Let u conider a few example of finding Laplace tranform Example 4 Find the Laplace tranform of f(t ( + e t To do thi we firt note that 3

4 f(t + e t + e 4t o we have L(f(t L( + e t + e 4t + ( + ( 4 Example 43 Find the Laplace tranform of f(t (co(t + in(t To do thi we firt note that f(t + in(t co(t + in(t o we have L(f(t L( + in(t + ( + 4 In order to do the next example we need one of the addition formula from trig in(α ± β in(α co(β ± in(β co(α co(α ± β co(α co(β in(α in(β Example 44 Find the Laplace tranform of f(t in(t + π/ f(t in(t co(π/ + in(π/ co(t co(t Therefore L(f(t L(co(t ( + 4 The Invere Laplace Tranform Given a function f(t the operation of taking the Laplace tranform i denoted by L(f(t F ( and the invere proce i denoted by L (F ( f(t The proce of computing the Laplace tranform of a function turn out to be more challenging than mot tudent like It involve lot of algebra and uing a table of Laplace tranform backward For example, if we were aked to find L (3/ 3 we would write L (3/ 3 3 L (/ 3 3 t 4

5 ince we know that L(t / 3 and we can adjut the contant to work out Mot generally thi proce will require the ue of the method of partial fraction Partial Fraction Thee note are concerned with decompoing rational function P ( Q( a M M + a M M + + a + a N + b N N + + b + b Note: We can (without lo of generality aume that the coefficient of N in the denominator i Alo in our intended application we will alway have M < N By the fundamental theorem of algebra we know that the denominator factor into a product of power of linear and quadratic term where the quadratic term correpond to complex root Namely, it can be written in the form ( r m ( r k m k ( α + α + β p ( α l + αl + βl p l, k l where m j + p j n j j The proce referred to a Partial Fraction i a method to reduce a complex rational function into a um of much impler term of the form c ( r j or c + d ( α + α + β j The mot important point i to learn how to deal with certain type of term that can appear But firt there i a pecial cae that arie and i worth a pecial attention Thi i the cae of non-repeated linear term I Nonrepeated Linear Factor If Q( ( r ( r ( r n and r i r j for i j P ( Q( A ( r + A ( r + + A n ( r n 5

6 II Repeated Linear Factor If Q( contain a factor of the form ( r m then you mut have the following term A ( r + A ( r + + A m ( r m III A Nonrepeated Quadratic Factor If Q( contain a factor of the form ( α + α + β ( α + β then you mut have the following term A + B ( α + α + β IV Repeated Quadratic Factor If Q( contain a factor of the form ( α + α + β m then you mut have the following term A + B ( α + α + β + A + B ( α + α + β + + A m + B m ( α + α + β m Let conider ome example of computing invere Laplace tranform: Example 45 Find ( L (! 3! L 3 t ( ( + Example 46 Find L 3 We need to firt expand the numerator to get So we have ( ( ( ( + L 3 L t t + 6 t3 6

7 ( Example 47 Find L We have (4 + ( L 4 ( (4 + L ( + /4 4 e t/4 ( 6 Example 48 Find L For thi example we have ( + 9 ( ( 6 L L ( + 9 ( + 9 L ( 3 ( + 9 co(3t in(3t ( Example 49 Find L 4 For thi example we firt note that ( + 3 ( + 3 ( + 3( o we have ( ( L 4 L ( ( + 3( In order to carry out thi invere Laplace tranform we mut ue partial fraction Since the denominator conit of non-repeated term we can ue the firt box (in the formula for partial fraction to write 4 ( + 3 A ( B ( To find A and B we can ue the cover-up method to find that A 3 and B o that ( L 4 ( + 3 ( ( 3 L + L 3e 3t + e t ( + 3 ( Now let olve an initial value problem uing the formula for the Laplace tranform of a derivitive Example 4 Find the olution of y y with y( Firt we take the Laplace tranform of both ide (uing Y L(y to get L(y L(y L( (Y Y Y ( 7

8 So to find y we need to take the invere Laplace tranform which require we do partial fraction ( y L (Y L ( ( ( L + L + e t ( Example 4 Find the olution of y + 5y + 4y with y( 3 and y ( Firt we take the Laplace tranform of both ide (uing Y L(y to get L(y + 5L(y + 4L(y ( Y 3 + 5(Y 3 + 4Y Solving for Y we get Y 3( + 5 ( Y 3( + 5 ( + 4( + So to find y we need to take the invere Laplace tranform which require we do partial fraction ( 3( + 5 y L (Y L ( + 4( ( ( 4 L + L ( + 4 ( + e 4t + 4e t 43 The Shift Theorem Two of the mot important and ueful reult we need to dicu are the firt and econd hift (or Tranlation theorem Theorem 4 (Firt Shift Theorem If L(f(t F ( then L(e at f(t F ( a for any real number a Proof L(e at f(t e t e at f(t dt e ( at f(t dt F ( a 8

9 Let conider ome example Example 4 Example 43 L(t e 3t L(t 3 L(e t in(t L(in(t + ( 3 3 ( Example 44 L(t(e t + e t L(t(e t + e 3t + e 4t L(te t + L(te 3t + L(te 4t ( + ( 3 + ( 4 In order to preent the econd hift theorem we fit need to dicu the unit tep function (or Heaviide function t < u(t t or the hifted unit tep function t < a u(t a t a With thi we can tate the econd hift theorem Theorem 4 (Second Shift Theorem If L(f(t F ( then L(u(t af(t a e a F ( for any a > 9

10 Proof L(u(t af(t a ( et τ t a e t u(t af(t a dt a e t f(t a dt e (τ+a f(τ dτ e a e τ f(τ dτ e a e t f(τ dt e a F ( A very ueful modification of the econd hift theorem how how to take the Laplace tranform of a unit tep function time a function that i not hifted Theorem 43 (Modified Second Shift Theorem If L(f(t F ( then L(u(t ag(t e a L(g(t + a for any a > Proof Let f(t g(t + a then we have f(t a g(t and we can ue the econd hift theorem to write L(u(t ag(t L(u(t af(t a e a L(f(t e a L(g(t + a Example 45 Find the Laplace tranform of t, t < f(t, t Firt we write f(t t u(t t and then compute L(f(t L(t u(t t L(t L(tu(t e L((t + e [ L(t + L( ] [ e + ]

11 Example 46 Find the Laplace tranform of, t < π f(t in(t, t π Firt we write f(t u(t π in(t and then compute where we have ued the addition formula L(f(t L(u(t π in(t e π L(in(t + π e π L( in(t e π ( + in(α + β in(α co(β + in(β co(α to get in(t + π in(t co(π + in(π co(t in(t Example 47 Find the Laplace tranform of f(t u(t e t t Ue the firt hift theorem to get rid of the e t and then apply the econd hift theorem L(f(t L(u(t e t t L(u(t t ( [ e L((t + ] [ ] ( e ( ( + ( Next we conider ome example of invere Laplace tranform uing the hift theorem ( Example 48 Find L We note that the denominator doe not factor Then we check and ee that the dicriminant i ( 4((5 < o we know the reult will involve ine and coine We proceed by completing the quare in the denominator ( L ( L ( + + ( e t L + e t in(t ( + 5 Example 49 Find L We note that the denominator doe not factor Then we check and ee that the dicriminant i (6 4((34 < o we know the reult

12 will involve ine and coine We proceed by completing the quare in the denominator ( ( + 5 L L ( So we ee that the denominator ha a hifted value but the numerator doe not To fix thi we write So we have + 5 [( + 3 3] + 5 ( + 3 ( ( + 5 ( + 3 L L ( ( ( + 3 L 5 ( ( L e 3t L ( + 5 e 3t co(5t 5 e 3t in(5t 5 ( e 3t L ( ( e Example 4 Find L Since thi function of contain an exponential e a + we know the invere Laplace tranform mut involve the econd hift theorem Namely we have ( ( e L L e + + So by the econd hift theorem backward we have ( ( e L u(t L + + (t In order to continue we need to ue partial fraction to implify matter Combining thee reult we have ( ( L L + e t ( + ( e L u(t [ e (t ] +

13 ( e Example 4 Find L 5 Since thi function of contain an exponential e a + 4 we know the invere Laplace tranform mut involve the econd hift theorem Namely we have ( ( e L 5 L e + 4 Now let olve an initial value problem u(t 5 co((t Example 4 Solve y y u(t and y(, y ( Firt we take the Laplace tranform of both ide (uing Y L(y to get L(y L(y e ( Y Y e Solving for Y we get Y e ( Y e ( ( + So to find y we need to take the invere Laplace tranform which require we do partial fraction Namely ( ( + A + B ( + C ( + We eaily find A, B and C o we have ( y L (Y L e ( ( + ( L e + L ( e ( u(t [ e (t + e (t ] ( + L e ( + Example 43 Solve y + 4y 4u(t π and y(, y ( Firt we take the Laplace tranform of both ide (uing Y L(y to get L(y + 4L(y 4 e π ( Y + 4Y 4 e π Solving for Y we get Y 4 e π ( + 4 Y e ( + 3

14 So to find y we need to take the invere Laplace tranform which require we do partial fraction Namely ( + A + B + C ( + We eaily find A, B and C o we have ( 4 e y L (Y L π ( + ( ( L π e + L e π ( + u(t π [ + co((t π] u(t π [co(t ] 44 Additional Operational Propertie In thi ection we cover everal very ueful theorem for Laplace tranform Theorem 44 If L(f(t F ( then L(t n f(t ( n dn d n F ( Proof L(t n f(t e t t n f(t dt te t [t (n f(t] dt d d L(t(n f(t Repeating thi calculation n time we arrive at L(t n f(t d d L(t(n f(t ( d d L(t(n f(t ( n dn d n L(t( f(t ( n dn d n F ( Example 44 Find the Laplace tranform of f(t t in(t Applying Theorem 44 we have L(t in(t d ( d ( + 4 ( 4 ( ( + 4 4

15 Example 45 Find the Laplace tranform of f(t te t co(t In thi cae we firt apply the firt hift theorem to write L(f(t L(te t co(t L(t co(t ( So we next need to compute L(t co(t and for thi we apply Theorem 44 to obtain L(t co(t d ( ( ( + ( d ( + ( + Finally then we arrive at L(f(t ( + (( + Example 46 Find the invere Laplace tranform f(t of F ( ln Theorem 44 a follow: o we have df d d (ln( + ln( 5 d ( + ( + We ue 5 L(tf(t df ( d + 5 tf(t e 5t e t or f(t e5t e t t Example 47 Find the invere Laplace tranform f(t of F ( ln Theorem 44 a follow: o we have df d d (ln( + ln( 5 d + 5 ( + We ue 5 L(tf(t df ( d + 5 tf(t e 5t e t 5

16 or f(t e5t e t t Example 48 Find the invere Laplace tranform f(t of F ( tan ( above theorem a follow: df d d ( ( tan d + We ue the o we have L(tf(t df ( d + tf(t in(t or f(t in(t t ( Theorem 45 If L(f(t F ( then L f(τ dτ F ( Proof Let g(t f(τ dτ o that g (t f(t L(f(t L(g (t L(g(t g( L(g(t Therefore, dividing by we have ( L f(τ dτ L(g(t L(f(t F ( Example 49 Find the invere Laplace tranform f(t of F ( previou reult and the fact that e 5t L (/( 5 we have Uing the ( 5 ( ( L L ( 5 ( 5 ( L L(e5t [ ] t e 5τ [ e 5t ] 5 5 e 5τ dτ 6

17 Convolution and their Application Definition 4 (Convolution Given two function f(t and g(t we define the convolution product by (f g(t f(τg(t τ dτ (3 Theorem 46 (Two Important Propertie (f g(t (g f(t, ie, Convolution multiplication i commutative If F ( L(f(t and G( L(g(t, then L(f g( F (G( Proof To how that (f g(t (g f(t we begin with the definition (f g(t f(τg(t τ dτ (Let w t τ dw dτ, τ t w t f(t wg(w ( dw f(t wg(w dw (g f(t If F ( L(f(t and G( L(g(t, then we want to how that L(f g( F (G( We have ( F (G( ( e τ f(τ dτ e w g(w dw ( e (τ+w f(τg(w dτ dw f(τ ( e (τ+w g(w dw dτ (Let t τ + w dt dw and w t τ ( f(τ e t g(t τ dt dτ τ ( Change the order of integration ( e t f(τg(t τ dτ dt L((f g(t 7

18 Remark 4 The econd part of the theorem give a very imple proof of an earlier reult ( L f(τ dτ L(( f(t L(L(f(t F ( Thi reult can be ueful in finding invere Laplace tranform ince ( F ( L Example 43 Find the invere Laplace tranform of For the firt problem we have f(τ dτ ( ( L L ( + L(in(t ( + and ( + in(τ dτ co(τ t co(t And, uing the above we can olve the econd problem a follow ( ( L L ( + L( co(τ ( co(τ dτ (τ in(τ t t in(t Example 43 Show that ( k L 3 in(kt kt co(kt (4 ( + k For thi problem will ue one of the addition formula from trig: co(a ± b co(a co(b in(a in(b which implie that in(a in(b [co(a b co(a + b] 8

19 By our main property of the convolution we have ( ( ( k L 3 k L k k ( + k ( + k ( + k k (in(kt (in(kt in(kτ in(k(t τ dτ k [co(k(τ t co(kt] dτ [ ] k in(kt t co(t in(kt kt co(kt k Example 43 Let u conider an example of finding a convolution by two different method The firt method will be to ue the definition of convolution Let f(t t and g(t e t and find h(t (f g(t Recall the f g g f o we have h(t (f g(t (t ve v dv ve t v dv e t ve v dv [ e t v ( e v ] dv e t [ t ve v e t [ te t + ( e v dv ] e v dv e t [ te t ( e t ] t + e t ] Next we compute the ame reult uing Laplace tranform We have h(t (f g(t o that H( F (G( L(tL(e t ( 9

20 We ue partial fraction to find ( h(t L ( ( A L + B + C ( We eaily find A, B, C ( L + + ( t + e t Example 433 Conider another example of finding a convolution Let f(t in(t and g(t co(t and find h(t (f g(t One way to compute the reulting integral i to ue a trig identity in(a ± b in(a co(b ± in(b co(a which, upon adding the two equation (one for plu and one for minu together, implie in(a co(b [ in(a + b + in(a b ] So for thi example we et a v and b t v and in(v co(t v [ in(t + in(v t ] So we have h(t (f g(t in(v co(t v dv [ ] in(t + in(v t dv t in(t + in(v t dv Let u v t du dv t in(t + 4 t in(t t in(u du

21 ince the econd integral i If we try to compute the ame reult uing Laplace tranform We have h(t (f g(t o that H( F (G( L(in(tL(co(t ( + Thi i already in partial fraction form o partial fraction will not help The only way out here i to ue a formula in the book (Example, page, Cullen and Zill, Advanced Engineering math, 4th Ed L(t in(t In the preent cae we then have (with k Volterra Integral Equation k ( + k ( h(t L ( ( + L ( + t in(t Laplace tranform method are particularly ueful in olving Volterra Integral equation An example of a Volterra equation i an equation in the form y(t f(t + k(t τ y(τ dτ Notice that the the integral term i a convolution, o if we apply the Laplace tranform to thi equation with Y ( L(y, F ( L(f and K( L(k we have Y ( F ( + K(Y ( which implie that Y ( F ( K( So to find y(t we only need to take the invere Laplace tranform of the right hand ide of the above equation Example 434 Conider the equation y(t t + y(t τ in(τ dτ

22 Applying the Laplace tranform we have Y ( ( + Y ( + which implie Y ( ( Applying the invere Laplace tranform to both ide we get ( ( y(t L L t 4 6 t3 Example 435 Conider a mixed integro-differential equation y (t + Applying the Laplace tranform we have y(t τe τ dτ, y( Y ( ( + Y ( + which implie Y ( ( + ( + + Applying the invere Laplace tranform to both ide we get y(t e t Periodic Function Theorem 47 If f(t i a periodic function with period T >, ie, f(t + T f(t for all t > then L(f(t T e t f(t dt e T Proof Aume that f(t i a periodic function with period T >, ie, f(t + T f(t for all t > Then we have

23 L(f(t T e t f(t dt e t f(t dt + ( Set t τ + T dt dτ T T T e t f(t dt + T e t f(t dt e (τ+t f(τ + T dτ e t f(t dt + e T e τ f(τ dτ e t f(t dt + e T L(f(t From thi we obtain L(f(t T e t f(t dt e T Example 436 Conider the function f(t t for < t < and f(t + f(t for all t Then f(t i periodic with period T and we have L(f(t e t t dt e In order to evaluate the integral in the numerator we need to ue integration by part ( e e t t t dt t dt ( e t t + e t dt e + ( e t e e + e e 3

24 So we have L(f(t e e ( e 45 The Dirac Delta Function In thi ection we cover application of Laplace tranform involving the Dirac Delta Function The firt thing you hould know i that the Dirac Delta Function i not a function Thi will become clear from it defining property What i it? It i a limit of function in a pecial ene which we now decribe Let a > be given and define, < t < a δ n (t a n, a t < a + /n, t a + /n Then we define the Dirac Delta Function upported at t a, δ(t a, applied to a continuou function f(t by the formula δ(t a f(t dt lim n δ n (t a f(t dt Theorem 48 For any continuou function f(t we have δ(t a f(t dt f(a Proof We need to how that lim n δ n (t a f(t dt f(a To thi end we firt define F (t f(τ dτ Then, from the fundamental theorem of calculu we have F (t f(t 4

25 We alo have lim n δ n (t a f(t dt lim n n a+/n a f(t dt lim n F (a + /n F (a /n F (a f(a Example 437 Conider a few integral involving the delta function: 3 (t δ(t dt in(3tδ(t π/ dt in(3π/ e t δ(t dt e Now we conider the Laplace tranform of the delta function Theorem 49 L (δ(t a e a Proof Firt we not that the function δ n (t a can be written in term of the unit tep function a δ n (t a n u(t a n u(t (a + /n So we have L(δ n (t a n L(u(t a n L(u(t (a + /n [ ] e a n e (a+/n [ ] e e a /n /n Now we note that lim n [ ] e /n /n 5

26 by L Hopital rule So we have [ ] e L(δ(t a lim L(δ n (t a e a /n lim e a n n /n Example 438 Conider the differential equation y + y δ(t π with y( and y ( Applying the Laplace tranform we have ( ( Y + Y e π Y e π + Applying the invere Laplace tranform and uing the econd hift theorem we have y(t u(t π in(t π u(t π in(t Example 439 Conider the differential equation y + 5y + 6y e t δ(t with y( and y ( 5 Applying the Laplace tranform we have ( Y (Y + 6Y e (+ ( Y ( e (+ Note that ( Y + 5 ( + ( + 3 o we have Y ( + 5 ( + ( e (+ ( + ( + 3 Next we apply partial fraction to each term to get ( + 5 ( + ( + 3 ( + + ( + 3 and ( + ( + 3 ( + ( + 3 Applying the invere Laplace tranform and uing the econd hift theorem we have y(t ( e t + e 3t + u(t ( e (t e 3(t 4/3 6

27 Table of Laplace Tranform f(t for t f L(f e t f(t dt e at a t n n! (n,, n+ t a Γ(a + a+ (a > in bt co bt b + b + b f (t L(f f( f (t L(f f( f ( f (n (t n L(f (n f( (n f ( f (n ( t n f(t ( n dn F d n ( e at f(t L(f( a t a e a u(t a t > a (f g(t u(t af(t a δ(t a f(t τg(τ dτ e a L(f( e a L(f g L(fL(g 7