18.03 Lecture 26, April 9, 2010 Laplace Transform: definition and basic properties 1. Definition of LT; L[1] 2. Region of convergence 3. Powers 4. Linearity 5. s-shift rule 6. sines and cosines 7. t-domain and s-domain [1] Laplace Transform We continue to consider functions f(t) which are zero for t < 0 . (I may forget to multiply by u(t) now and then.) The Laplace transform takes a function f(t) (of "time") and uses it to manufacture another function F(s) (where s can be complex). It: [Slide] (1) makes explicit long term behavior of f(t) . (2) answers the question: if I know w(t), how can I compute p(s) ? (3) converts differential equations into algebraic equations. But we won't see these virtues right away. Definition: The Laplace transform of f(t) is the improper integral F(s) = integral_{0}^¥infty e^{-st} f(t) dt (formula subject to two refinements). We will often write f(t) ------> F(s) and L[f(t)] = F(s) (This notation isn't so good, because there's no room for "s" on the left.) For each value of s , F(s) is a weighted integral of f(t). Wnen s = 0 , for example, the Laplace integral is just the integral of f(t). When s > 0 , the values of f(t) for large t are given less weight. Each value of F(s) contains information about the whole of f(t). Example: f(t) = 1 : F(s) = integral_0^infty e^{-st} dt = lim_{T --> infty} e^{-st}/(-s) |^T_0 = (-1/s) (lim_{T --> infty} e^{-st} - 1). If s > 0 , e^{-st} ---> 0 as T ---> infty , and we get 1/s .If s < 0 , e^{-st} ---> infty as T ---> infty , and the improper integral diverges. Actually, I'll want s to be a complex number, not just a real number. With s = a+bi , !e^{st}| = |e^{at}e^{ibt}| = e^{at} , and the same argument shows that the integral converges for Re(s) > 0 . So this is our first calculation: 1 has Laplace transform 1/s [2] For a more general function f(t) , the integral will converge for a given value of s provided that |f(t)e^{-st}| can be integrated. This will happen if |f(t)| < |e^{st}| = e^{(Re s)t} at least for t large. So to make the integral converge *somewhere*, we make the assumption that f(t) is "of exponential order" if there is a constant M such that for large t , |f(t)| < e^{at} . In that case, the integral for F(s) converges for Re(s) >= a . Eg e^{t^2} has no Laplace transform. So in the definition I really meant: Refinement #1: "for Re(s) large." -- the "Region of Convergence." We'll see that the region of convergence contains important information and it's good practice to declare it when you state a Laplace transform: 1 has LT 1/s with region of convergence Re(s) > 0 . The expression obtained by means of the integration makes sense everywhere in C except for a few points -- like s = 0 here -- and this is how we define the Laplace transform for values of s whose real part is too small. This is called "analytic continuation." [3] Powers L[t^n] = int_0^infty t^n e^{-st} dt u = t^n dv = e^{-st} dt du = n t^{n-1} dt v = (-1/s) e^{-st} = t^n (-1/s) e^{-st} |_0^infty + (n/s) int_0^infty t^{n-1} e^{-st} dt = n/s L[t^{n-1}] . Start with n = 1 : L[t] = (1/s)(1/s) = 1/s^2 . n = 2 : L[t^2] = (2/s)(1/s^2) = 2/s^3n = 3 : L[t^3] = (3/s)(2/s^3) = 3!/s^4 ... L[t^n] = n!/s^{n+1} . The region of convergence is Re(s) > 0 . [4] These first computations can be exploited using general properties of the Laplace Transform. We'll develop quite a few of these rules, and in fact normally you will not be using the integral definition to compute Laplace transforms. Rule 1 (Linearity): af(t) + bg(t) ------> aF(s) + bG(s). This is clear from the linearity of the integral. Question 26.1. What is L[(1+t)^2] ? 1. (1/s + 1/s^2)^2 = 1/s^2 + 2/s^3 + 1/s^4 2. 1/s + 2/s^2 + 2/s^3 3. (1+t)(1/s + 1/s^2) 4. Don't know [5] Rule 2 (s-shift): If z is any complex number, L[e^{zt}f(t)] = F(s-z) with region of convergence Re(s) > Re(z) Here's the calculation: L[e^{zt}f(t)] = integral_0^infinity e^{zt} f(t) e^{-st} dt = integral_0^infinity f(t) e^{-(s-z)t} dt = F(s-z). Using f(t) = 1 and our calculation of its Laplace transform we find L[e^{zt}] = 1/(s-z). (*) [6] As usual we can get sinusoids out from the complex exponential. Using linearity and Euler's identity cos(omega t) = (1/2)(e^{i omega t} + e^{-i omega t}) we find L[cos(omega t)] = (1/2)((1/(s - i omega) + 1/(s + i omega)) = s/(s^2 + omega^2)Using sin(omega t) = (1/2i)(e^{i omega} - e^{-i omega}) we find L[sin(omega t)] = omega/(s^2 + omega^2). Both are convergent for Re(s) > 0 , since +- i omega t lie on the imaginary axis. [7] Two worlds ... [Slide] ------------------------------------------------------------------------ | | | The t domain | | | | t is real and positive | | | | functions f(t) are signals, i.e. functions, perhaps generalized | | | | ODEs relating them | | | | convolution | | | ------------------------------------------------------------------------ | ^ L | | L^{-1} v | ------------------------------------------------------------------------ | | | The s domain | | | | s is complex | | | | beautiful functions F(s) , often rational = poly/poly | | | | and algebraic equations relating them | | | | ordinary multiplication of functions | | | ------------------------------------------------------------------------MIT OpenCourseWare http://ocw.mit.edu 18.03 Differential Equations�� Spring 2010 For information about citing these materials or our Terms of Use, visit:
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