logging in or signing up laplace aSGuest140585 Download Post to : URL : Related Presentations : Let's Connect Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Copy embed code: Embed: Flash iPad Dynamic Copy Does not support media & animations Automatically changes to Flash or non-Flash embed WordPress Embed Customize Embed URL: Copy Thumbnail: Copy The presentation is successfully added In Your Favorites. Views: 94 Category: Education License: Some Rights Reserved Like it (0) Dislike it (0) Added: July 30, 2012 This Presentation is Public Favorites: 0 Presentation Description its very useful in degree courses. Comments Posting comment... Premium member Presentation Transcript Ch 6.1: Definition of Laplace Transform: Ch 6.1: Definition of Laplace Transform Many practical engineering problems involve mechanical or electrical systems acted upon by discontinuous or impulsive forcing terms. For such problems the methods described in Chapter 3 are difficult to apply. In this chapter we use the Laplace transform to convert a problem for an unknown function f into a simpler problem for F , solve for F , and then recover f from its transform F . Given a known function K ( s , t ), an integral transform of a function f is a relation of the formThe Laplace Transform: The Laplace Transform Let f be a function defined for t 0, and satisfies certain conditions to be named later. The Laplace Transform of f is defined as Thus the kernel function is K ( s , t ) = e - st . Since solutions of linear differential equations with constant coefficients are based on the exponential function, the Laplace transform is particularly useful for such equations. Note that the Laplace Transform is defined by an improper integral, and thus must be checked for convergence. On the next few slides, we review examples of improper integrals and piecewise continuous functions.Example 1: Example 1 Consider the following improper integral. We can evaluate this integral as follows: Note that if s = 0, then e st = 1. Thus the following two cases hold:Example 2: Example 2 Consider the following improper integral. We can evaluate this integral using integration by parts: Since this limit diverges, so does the original integral.Piecewise Continuous Functions: Piecewise Continuous Functions A function f is piecewise continuous on an interval [ a , b ] if this interval can be partitioned by a finite number of points a = t 0 < t 1 < … < t n = b such that (1) f is continuous on each ( t k , t k +1 ) In other words, f is piecewise continuous on [ a , b ] if it is continuous there except for a finite number of jump discontinuities.Example 3: Example 3 Consider the following piecewise-defined function f . From this definition of f , and from the graph of f below, we see that f is piecewise continuous on [0, 3].Example 4: Example 4 Consider the following piecewise-defined function f . From this definition of f , and from the graph of f below, we see that f is not piecewise continuous on [0, 3].Theorem 6.1.2: Theorem 6.1.2 Suppose that f is a function for which the following hold: (1) f is piecewise continuous on [0, b ] for all b > 0. (2) | f ( t ) | Ke at when t M , for constants a , K , M , with K , M > 0. Then t he Laplace Transform of f exists for s > a . Note: A function f that satisfies the conditions specified above is said to to have exponential order as t .Example 5: Example 5 Let f ( t ) = 1 for t 0. Then the Laplace transform F ( s ) of f is:Example 6: Example 6 Let f ( t ) = e at for t 0. Then the Laplace transform F ( s ) of f is:Example 7: Example 7 Let f ( t ) = sin( at ) for t 0. Using integration by parts twice, the Laplace transform F ( s ) of f is found as follows:Linearity of the Laplace Transform: Linearity of the Laplace Transform Suppose f and g are functions whose Laplace transforms exist for s > a 1 and s > a 2 , respectively. Then, for s greater than the maximum of a 1 and a 2 , the Laplace transform of c 1 f ( t ) + c 2 g ( t ) exists. That is, withExample 8: Example 8 Let f ( t ) = 5 e - 2 t - 3 sin(4 t ) for t 0. Then by linearity of the Laplace transform, and using results of previous examples, the Laplace transform F ( s ) of f is: You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.