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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
�t y (t) = 0 �t =
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
�t y (t) = �0t = 0 s(t − τ ) sin(τ )dτ
sin(τ )dτ = − cos(τ )|t = 1 − cos(t).
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
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- Spring '05