MIT2_017JF09_p02

# MIT2_017JF09_p02 - 2 CONVOLUTION 2 3 Convolution The step...

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Unformatted text preview: 2 CONVOLUTION 2 3 Convolution The step function s(t) is deﬁned as zero when the argument is negative, and one when the argument is zero or positive: � s(t) = 0 if t < 0 1 if t ≥ 0 For the LTI systems whose impulse responses are given below, use convolution to determine the system responses to step input, i.e., u(t) = s(t). 1. h(t) = 1 The impulse response is the step function itself - it turns on to one as so on as the impulse is applied, and this makes it a pure integrator. We get for the response to step input �t y (t) = 0 �t = = �0t 0 s(τ )s(t − τ )dτ s(t − τ )dτ and the integrand is one because always t ≥ τ so dτ = t. You recognize this as the integral of the input step. 2. h(t) = sin(t) This impulse response is like that of an undamped second-order oscillator, having unity resonance frequency. �t y (t) = �0t = 0 s(t − τ ) sin(τ )dτ sin(τ )dτ = − cos(τ )|t = 1 − cos(t). 0 3. h(t) = 2 sin(t)e−t/4 This is a typical underdamped response for a second-order system - a sinusoid multi­ plied by a decaying exponential. We make the substitution and ﬁnd: �t y (t) = 0 s(t − τ )2 sin(τ )e−τ /4 dτ �t =2 = 0 sin(τ )e−τ /4 dτ � 32 � 1 − e−t/4 [sin(t)/4 + cos(t)] 17 MIT OpenCourseWare http://ocw.mit.edu 2.017J Design of Electromechanical Robotic Systems Fall 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . ...
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## This note was uploaded on 11/29/2011 for the course CIVIL 1.00 taught by Professor Georgekocur during the Spring '05 term at MIT.

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MIT2_017JF09_p02 - 2 CONVOLUTION 2 3 Convolution The step...

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