c26mudd - 18.03 Muddy Card responses, April 14, 2006 1. A...

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18.03 Muddy Card responses, April 14, 2006 1. A number of people brought up the point made at the end of Lecture 25, on April w 12: how do we know what initial conditions yield the unit step or impulse responses? This is a tricky point and I did not explain it completely. One point to be made is this: in the case of the unit impulse response, when we have a n x ( n ) + · · · + a 0 x = δ ( t ), the solution x should not be too wild at t = 0 since I have to be able to differentiate it n times in order to apply the differential operator to it. If I wind up with a discontinuity in the ( n 2)nd derivative, for example, then when I differentiate once more I get a delta function, and we have not tried to understand what happens when you differentiate a delta function. So the derivatives up to the ( n 2)nd should be zero at t = 0 (since they are zero for t < 0). The ( n 1)st should be such that when I differentiate once more and multiply by a n I get the delta function: so it should increase
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This note was uploaded on 01/31/2011 for the course MAT 17A taught by Professor Staff during the Winter '08 term at UC Davis.

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