MIT18_03S10_c25

MIT18_03S10_c25 - 18.03 Class 25, April 7, 2010...

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, April 7, 2010 Convolution, or, the superposition of impulse response 1. General equivalent initial conditions for delta response 2. Two implications of LTI 3. Superposition of unit impulse responses 4. Convolution product 1. Equivalent for general delta-response ( t > 0 ) : p(D) x = a_n x^(n) + . ... + a_0 x ( a_n not zero) p(D) w = delta(t) with rest initial conditions specifies the *unit impulse response*. For t > 0 , this equation is equivalent (for t > 0 ) to a_n x^(n) + . .. + a_0 x = 0 with initial conditions x(0) = . .. = x^(n-2)(0) = 0 , x^(n-1)(0) = 1/a_n . Ex: n = 2 : mx" + bx' + kx = delta(t) , rest initial conditions, is equivalent to mx" + bx' + kx = 0 , x(0) = 0 , x'(0) = 1/m . [2] We work with a system modeled by an LTI operator p(D) = a_n D^n + . ... The time invariance has two important consequences: (a) If p(D) x = q(t) , then p(D) x' = q'(t) . This is because D p(D) = p(D) D (because the coefficients are constant) so p(D) x' = p(D) Dx = D p(D) x = D q = q' . In particular, since u'(t) = delta(t), v'(t) = w(t) : the derivative of the unit step response is the unit impulse response. e.g. D v = D ( u(t) (1/k) ( 1 - e^{-kt}) ) = e^{-kt} = w . (Notice that v is continuous: there are no delta terms in its generalized derivative.) (b) It doesn't really matter when you start the clock. So, for example, we found that the unit impulse response for D+kI is w(t) = u(t) e^{-kt} and the unit step response is v(t) = u(t) (1/k) ( 1 - e^{-kt} ) . By time invariance the solution to x' + kx = delta(t-a) with x(t) = 0 for t < a is x = u(t-a) e^{-k(t-a)}) . [3] We learn about a system by studying how it responds to various input
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MIT18_03S10_c25 - 18.03 Class 25, April 7, 2010...

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