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# sol06 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering 2.004 Dynamics and Control II Fall 2007 Problem Set #6 Solution Posted: Friday, Oct. 26, ’07 1. Starting with the Supplement to Lecture 12 (“Supplement” for short,) modify the time–domain torque balance equation (3) on the motor shaft and force balance equation (5) on the mass to include the compliances. Answer: Since we have compliance terms in the system, the spring torque T c and force F c are T C ( t ) = C r θ ( t ) , F C ( t ) = C t x ( t ) . Note that the spring torque and force are proportional to displacement θ ( t ) and x ( t ) respectively, which are time integrals of ω ( t ) and v ( t ). Thus equations (3) and (5) will be t K m i ( t ) = ˙ ( t ) + ( t ) + C r ω ( τ ) + T p ( t ) , 0 t F p ( t ) = Mv ˙( t ) + f v v ( t ) + C t v ( τ ) dτ. 0 2. Laplace transform your new torque/force–balance equations and repeat the steps leading to equation (13) of the Supplement. The KVL equation should be identi- cal to (1–2) since we are neglecting the inductance, as we did in the Supplement. Obtain the transfer function V ( s ) /V s ( s ) of the compliant system without feed- back. After substituting C r = 2N · m / rad, C t = 10N / m and the remaining numerical values from the Supplement, your transfer function should be 0 . 3162 s . s 2 + 2 s + 3 You should use this transfer function in the rest of the problem set, even if you can’t get your algebra from question 1 to match it. Answer: Applying the Laplace transform to the previous results, we obtain Ω( s ) K m I ( s ) = Js Ω( s ) + D Ω( s ) + C r + T p ( s ) , s V ( s ) F p ( s ) = MsV ( s ) + f v V ( s ) + C t . s 1

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Using the above two relations, we obtain the Laplace transform of the current i ( t ) as J + r 2 M D + r 2 f v C r + r 2 C t 1 I ( s ) = s + + Ω( s ) . rK m rK m rK m s Combining this result with KVL, J + r 2 M D + r 2 f v C r + r 2 C t 1 V ( s ) V s ( s ) = s + + RV ( s ) + K v , rK m rK m rK m s r we end up with the transfer function given by rK m s V ( s ) R ( J + r 2 M ) = . V s ( s ) 2 D + r 2 f v + K m K v /R C r + r 2 C t s + s + J + r 2 M J + r 2 M Substituting the numerical values, we find 0 . 3162 s G ( s ) = . s 2 + 2 s + 3 3. Now consider the complete feedback system, which includes the differential ampli- fier with reference voltage input and velocity feedback. Write expressions for the open loop transfer function and the closed–loop transfer function of this system with the feedback gain K as a variable. Answer: Open loop transfer function G ( s ) is 0 . 3162 s . s 2 + 2 s + 3 Closed loop transfer function is 0 . 3162 Ks . s 2 + (2 + 0 . 3162 K ) s + 3 4. What is the order, type, and nature ( i.e. , overdamped, undamped, etc.) of this system? Back up your answer with a sketch of the open–loop zeros and poles. Answer: Obviously this system has two poles, so it is 2nd order system. The number of pure integrators is zero, thus it is Type 0. (The steady–state error is non–zero for step input and zero for ramp input.) The open loop system is underdamped, but the closed loop system transitions from underdamped to overdamped as the gain K increases because the poles are p 1 , 2 = (2 + 0 . 3162 K ) ± (2 + 0 . 3162 K ) 2 12) / 2 .
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