24 step and delta functions, step and impulse responses

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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 Fipping a switch. Here’s the simplest example, taken from the supplementary notes IR.4. The unit step function is u 0 ( t ) = b 0 , t < 0 , 1 , t > 0 . (Unit step at 0) The value of u 0 at t = 0 is left unde±ned; for our purposes it will never matter. We want to use u as the input in a di²erential 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 di²erential equation and the initial conditions imply that lim t 0 + y ( n ) 0 ( t ) = 1 . (This uses the fact that p has leading coe³cient 1; otherwise we’d get one over the leading coe³- cient.) This discussion is meant to explain Theorem: response to unit step. Suppose p is a polynomial of degree n with leading coeFcient 1 . Then the solution y 0 to (Linear ODE with unit step function forcing) and (Starting conditions) may be calculated as follows. ±or t 0 , y 0 ( t ) = 0 . ±or 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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/06/2010 for the course 18 18.03 taught by Professor Unknown during the Fall '09 term at MIT.

Page1 / 4

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online