Chapter 7: The Laplace Transform

Transcription

1 Chapter 7: The Laplace Tranform 王奕翔 Department of Electrical Engineering National Taiwan Univerity November 2, / 25 王奕翔 DE Lecture 1

2 Solving an initial value problem aociated with a linear differential equation: 1 General olution = complimentary olution + particular olution. 2 Plug in the initial condition to pecify the undetermined coefficient. Quetion: I there a fater way? 2 / 25 王奕翔 DE Lecture 1

3 In Chapter 4, 5, and 6, we majorly deal with linear differential equation with continuou, differentiable, or analytic coefficient. But in real application, ometime thi i not true. For example: E(t) E(t) L R t C Square voltage input: Periodic, Dicontinuou. Quetion: How to olve the current? How to deal with dicontinuity? 3 / 25 王奕翔 DE Lecture 1

4 In thi lecture we introduce a powerful tool: Laplace Tranform Invented by Pierre-Simon Laplace ( ). 4 / 25 王奕翔 DE Lecture 1

5 a i, i 1,..., n and the precribed initial condition y, y 1,..., y n 1, and G() i the Laplace tranform of g(t). * Typically, we put the two term in (11) over the leat common denominator and then decompoe the expreion into two or more partial fraction. Finally, the olution y(t) of the original initial-value problem i 1 y(t) {Y()}, where the invere tranform i done term by term. The procedure i ummarized in the diagram in Figure Overview of the Method Find unknown y(t) that atifie DE and initial condition Apply Laplace tranform Tranformed DE become an algebraic equation in Y() Solution y(t) of original IVP Apply invere Laplace 1 tranform Solve tranformed equation for Y() FIGURE Step in olving an IVP by the Laplace tranform The next example illutrate the foregoing method of olving DE, a well a partial fraction decompoition in the cae when the denominator of Y() contain a quadratic polynomial with no real factor. 5 / 25 王奕翔 DE Lecture 1

6 1 Laplace and Invere Laplace Tranform: Definition and Baic 6 / 25 王奕翔 DE Lecture 1

7 Definition of the Laplace Tranform Definition For a function f(t) defined for t, it Laplace Tranfrom i defined a F() := L {f(t)} := given that the improper integral converge. Note: Ue capital letter to denote tranform. e t f(t)dt, f(t) L F(), g(t) L G(), y(t) L Y(), etc. Note: The domain of the Laplace tranform F() (that i, where the improper integral converge) depend on the function f(t) 7 / 25 王奕翔 DE Lecture 1

8 Example of Laplace Tranform Example Evaluate L {1}. L {1} = = lim T e t (1)dt = lim T [ ] e t T When doe the above converge? >! T = lim T e t dt 1 e T. Hence, the domain of L {1} i >, and L {1} = 1. 8 / 25 王奕翔 DE Lecture 1

9 Example of Laplace Tranform Example Evaluate L {t}. L {t} = te t dt = lim T [ ] te t T = lim + T T T When doe the above converge? >! ( ) e t td 1 Te T e t dt = lim + 1 T L {1}. Hence, the domain of L {t} i >, and L {t} = / 25 王奕翔 DE Lecture 1

10 Laplace Tranform of t n L {t n } = n!, n =, 1, 2,..., > n+1 Proof: One way i to prove it by induction. We will how another proof after dicuing the Laplace tranform of the derivative of a function. 1 / 25 王奕翔 DE Lecture 1

11 Laplace Tranform of e at Proof: L { e at} = 1 a, > a L { e at} = = lim T e at e t dt = lim T [ ] e ( a)t T a When doe the above converge? a >! T e ( a)t dt 1 e ( a)t = lim T a Hence, the domain of L {e at } i > a, and L {e at } = 1 a. 11 / 25 王奕翔 DE Lecture 1

12 Laplace Tranform of in(kt) and co(kt) Proof: L {in(kt)} = L {in(kt)} = k 2 + k 2, L {co(kt)} = 2 + k 2, > in(kt)e t dt = [ in(kt)e t = [ in(kt)e t = ] ] + k When doe the above converge? >! = ( ) e t in(kt)d co(kt)e t dt + k L {co(kt)} [ ] in(kt)e t = 12 / 25 王奕翔 DE Lecture 1

13 Laplace Tranform of in(kt) and co(kt) Proof: L {in(kt)} = L {co(kt)} = k 2 + k 2, L {co(kt)} = 2 + k 2, > co(kt)e t dt = [ co(kt)e t = [ co(kt)e t = ] ] k When doe the above converge? >! = ( ) e t co(kt)d in(kt)e t dt k L {in(kt)} [ ] co(kt)e t = / 25 王奕翔 DE Lecture 1

14 Laplace Tranform of in(kt) and co(kt) L {in(kt)} = Proof: { Solve the above, we get the reult: k 2 + k 2, L {co(kt)} = 2 + k 2, > L {in(kt)} = k L {co(kt)} L {co(kt)} = 1 k L {in(kt)} L {in(kt)} = k L {co(kt)} = k 2 k2 2 L {in(kt)} = 2 + k 2 2 L {in(kt)} = k 2 = L {in(kt)} = k L {co(kt)} = k L {in(kt)} = 2 + k k 2 14 / 25 王奕翔 DE Lecture 1

15 Laplace Tranform i Linear Theorem For any α, β, f(t) L F(), g(t) L G(), L {αf(t) + βg(t)} = αf() + βg() Proof: It can be proved by the linearity of integral. Example Evaluate L {inh(kt)} and L {coh(kt)}. ( A: inh(kt) = 1 2 e kt e kt) (, coh(kt) = 1 2 e kt + e kt). Hence inh(kt) L 1 ( 1 2 k 1 ) k = + k 2 k 2, > k coh(kt) L 1 ( 1 2 k + 1 ) = + k 2 k 2, > k. 15 / 25 王奕翔 DE Lecture 1

16 Laplace Tranform of Some Baic Function f(t) F() Domain of F() t n n! n+1 > e at 1 a > a k in(kt) 2 + k 2 > co(kt) 2 + k 2 > inh(kt) coh(kt) k 2 k 2 2 k 2 > k > k 16 / 25 王奕翔 DE Lecture 1

17 Exitence of Laplace Tranform Theorem (Sufficient Condition for the Exitence of Laplace Tranform) If a function f(t) i piecewie continuou on [, ), and of exponential order, then L {f(t)} exit for > c for ome contant c. Definition A function f(t) i of exponential order if c R, M >, τ > uch that f(t) Me ct, t > τ. Note: If f(t) i of exponential order, then c R uch that for > c, lim t f(t)e t =. 17 / 25 王奕翔 DE Lecture 1

18 Exitence of Laplace Tranform Theorem (Sufficient Condition for the Exitence of Laplace Tranform) If a function f(t) i piecewie continuou on [, ), and of exponential order, then L {f(t)} exit for > c for ome contant c. Proof: For ufficiently large T > τ, we plit the following integral: T τ T f(t)dt = f(t)e t dt + f(t)e t dt. } {{ } τ } {{ } I 1 I 2 We only need to prove that I 2 converge a T : I 2 T τ f(t)e t dt = T τ f(t) e t dt T which converge a T for > c ince L { e ct} exit. τ Me ct e t dt, 18 / 25 王奕翔 DE Lecture 1

19 In thi lecture, we focu on function that are piecewie continuou on [, ), and of exponential order 19 / 25 王奕翔 DE Lecture 1

20 Laplace Tranform of Derivative Suppoe f(t) i continuou on [, ) and of exponential order, and f (t) i alo continuou on [, ), then the Laplace tranform of f (t) can be found a follow: L {f (t)} = e t f (t)dt = e t d (f(t)) = [ f(t)e t] + e t f(t)dt = L {f(t)} f(), > c Note: ince f(t) i of exponential order, f(t)e t a t for > c for ome contant c. Similarly, if f (t) i alo of exponential order, we can find L {f (t)} = L {f (t)} f () = 2 L {f(t)} f() f (). 2 / 25 王奕翔 DE Lecture 1

21 Laplace Tranform of Derivative Theorem If f, f,..., f (n 1) are continuou on [, ) and are of exponential order, and if f (n) (t) i piecewie continuou on [, ), then { } L f (n) (t) = n F() n 1 f() n 2 f () f (n 1) (), where F() := L {f(t)}. Example Evaluate L {t n }. A: Let f(t) = t n. Since f (n) (t) = n!, f (k) () = for any k n 1, uing the above theorem we get L {n!} = n F() = n! = F() = n! n / 25 王奕翔 DE Lecture 1

22 Invere Laplace Tranform L {f(t)} = F() L 1 {F()} = f(t) F(): Laplace tranform of f(t) f(t): invere Laplace tranform of F() Note: Invere Laplace tranform i alo linear: L 1 {αf() + βg()} = αf(t) + βg(t) 22 / 25 王奕翔 DE Lecture 1

23 Some Invere Laplace Tranform F() f(t) Domain of F() n! n+1 t n > 1 a e at > a k 2 + k 2 in(kt) > 2 + k 2 co(kt) > k 2 k 2 inh(kt) > k 2 k 2 coh(kt) > k 23 / 25 王奕翔 DE Lecture 1

24 Example Example Evaluate L 1 { Step 1: Decompoe } = Step 2: By the linearity of invere Laplace tranform, { } { } { } L = 2L 1 + 3L = 2 co 2t + 3 in 2t. 24 / 25 王奕翔 DE Lecture 1

25 Example Example Evaluate L 1 { ( + 3) 2 ( 1)( 2)( + 4) }. Step 1: Decompoe into partial fraction: F() := ( + 3) 2 ( 1)( 2)( + 4) = A 1 + B 2 + C + 4. We can find A, B, C by the following: [ ] ( + 3) 2 A = ( = 16 [ ] 1)( 2)( + 4) =1 5, B = ( + 3) 2 ( 1) ( = 25 2)( + 4) =2 6 [ ] ( + 3) 2 C = = ( 1)( 2) 1 ( + 4) = 4 3 Step 2: Linearity = f(t) = 16 5 et e2t e 4t. 25 / 25 王奕翔 DE Lecture 1