lecture_23

lecture_23 - MIT OpenCourseWare http://ocw.mit.edu 2.004...

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 2.004 Dynamics and Control II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . m K B B 1 2 V v m s I j M ox x- 3- 2 Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Spring Term 2008 Lecture 23 1 Reading: Nise: 4.7, 4.8 1 Pole/Zero Cancelation: Consider the following mechanical system: The transfer function relating v m to V s is v m ( s ) B 1 s + K G ( s ) = = V s ( s ) ms 2 + ( B 1 + B 2 ) s + K which is clearly second-order. However, if the parameter values are m 1 = 1 / 3 kg, K = 2 N/m, B 1 = 1 N-s/m, B 2 = 2 / 3 N-s/m the transfer function becomes s + 2 3( s + 2) G ( s ) = = (1 / 3) s 2 + (5 / 3) s + 2 ( s + 2)( s + 3) with a coincident pole and zero at s = 2. 1 copyright c D.Rowell 2008 231 s + 2 3 s + 5 s + 6 x ( t ) V ( t ) = u ( t ) v ( t ) s s m Clearly, with these values, cancelation takes place in the transfer function and 3 G ( s ) = s + 3 This phenomenon is known as pole/zero cancelation . With these values the response will be identical to a first-order system with a pole at s = 3. Lets compute the step response of this system to understand what has happened. We start by breaking the system into two cascaded blocks as we have discussed previously: When the input V s ( t ) = u s ( t ), the unit-step (Heaviside) function, the intermediate variable x ( t ) is du s ( t ) x ( t ) = + 2 u s ( t ) = ( t ) + 2 u s ( t ) dt where ( t ) is the Dirac delta function. The system output v m ( t ) will therefore be v m ( t ) = y ( t ) + 2 y s ( t ) where y ( t ) is the impulse response of the all-pole system to the right, and y s ( t ) is its step response . You can show for yourself that y s ( t ) = . 5...
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lecture_23 - MIT OpenCourseWare http://ocw.mit.edu 2.004...

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