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24 step and delta functions, step and impulse responses

# 24 step and delta functions, step and impulse responses -...

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18.03 Lecture #24 Nov. 2, 2009: notes Today we’ll look at something that sounds like a purely engineering idea; but by Wednesday, it will become the basis of the extremely theoretical idea of Green’s functions. So pay close attention: if you hate what is going on one minute, you may love it the next (and vice versa). The engineering idea is driving a system using step functions: an input that is turned on suddenly, like flipping a switch. Here’s the simplest example, taken from the supplementary notes IR.4. The unit step function is u 0 ( t ) = braceleftbigg 0 , t < 0 , 1 , t > 0 . (Unit step at 0) The value of u 0 at t = 0 is left undefined; for our purposes it will never matter. We want to use u as the input in a differential equation p ( D ) y = u 0 . (Linear ODE with unit step function forcing) I’ll assume that p is an n th degree polynomial with leading term 1: p ( D ) = D n + a n 1 D n 1 + · · · + a 1 D + a 0 . The idea is that the system is undriven for negative time; then beginning at t = 0, a unit input is applied. Since nothing is being done to the system before time 0, it is often natural to look at a solution that is zero for all negative time: y 0 ( t ) = 0 , ( t < 0) . (Starting conditions) Suppose p has degree n , so that we are looking at an n th order ODE. The starting conditions tell us what the derivatives of y 0 have to look like at time 0: y 0 (0) = 0 , y 0 (0) = 0 , . . . , y ( n 1) 0 (0) = 0 . (Step initial conditions) The jump discontinuity in u 0 at t = 0 forces the n th derivative of y 0 to have a jump discontinuity at t = 0: the starting conditions imply that lim t 0 - y ( n ) 0 ( t ) = 0 , but the differential equation and the initial conditions imply that lim t 0 + y ( n ) 0 ( t ) = 1 . (This uses the fact that p has leading coefficient 1; otherwise we’d get one over the leading coeffi- cient.) This discussion is meant to explain Theorem: response to unit step. Suppose p is a polynomial of degree n with leading coefficient 1 . Then the solution y 0 to (Linear ODE with unit step function forcing) and (Starting conditions) may be calculated as follows. For t 0 , y 0 ( t ) = 0 . For t 0 , y 0 ( t ) is the solution to p ( D ) y = 1 and (Initial conditions). The function y 0 has n - 1 continuous derivatives; the n th derivative jumps from 0 to 1 at t = 0 .

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