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Unformatted text preview: Math 334 Lecture #27 6.5: Unit Impulse Functions and Dirac Delta Functions Example. How does the solution of the IVP y 00 + 4 y + 4 y = u 1 ( t ) u 1+ k ( t ) k , y (0) = 0 , y (0) = 0 , behave as k + ? [For k = 1, the IVP is that of the example in the Lecture #32.] [See Maple worksheet for animation of solution as k + .] An Idealized Impulse Force. The discontinuous function f k ( t ) = u 1 ( t ) u 1+ k ( t ) k describes the constant force of 1 /k applied on the interval [1 , 1 + k ]. [Sketch the graph of the discontinuous forcing function for several values of k , especial small values.] The impulse (or strength) of f k is its integral over [0 , ): I ( f k ) = Z u 1 ( t ) u 1+ k ( t ) k dt = 1 k Z 1+ k 1 dt = 1 for k > . What is the limit of the impulse of f k as k + ? lim k + I ( f k ) = 1 . For a fixed value of t 0, what is the limit of f k ( t ) as k + ? It is the function ( t 1) = lim k + f k ( t ) = if t 6 = 1 , if t = 1 ....
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 Spring '11
 Smith
 Math

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