MEM355W04-05_HW4_2_sol

MEM355W04-05_HW4_2_sol - MEM 355: Solutions to HW#4:...

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MEM 355: Solutions to HW#4: (Winter 04-05) Problem 4.4: a) Transfer Function: Set R = 0, then E = -Y ) ( 1500 )] ( ) ( [ 600 ) ( ) 60 ( s W s Y s k s Y k s Y s I p - + - = + I p k s k s sw s W s Y 600 ) 10 1 ( 60 1500 ) ( ) ( 2 + + + - = b) For roots at 60 60 j ± - , which corresponds to 7200 120 0 60 ) 60 ( 2 2 2 + + = = + + s s s Comparing this to the denominator of Y/W above, we get 1 . 0 and , 12 = = p I k k Problem 4.8: a) With , 10 and , 1 = = = J H H y r the characteristic equation is . 0 10 . 0 10 1 10 1 ) ( 1 2 2 2 = + = + = + p p k s s k s s D No k p can yield a stable solution. b) : ) ( ) ( s k k s D D p + = Find type and error constants w.r.t. reference input. This is a unity feedback system, so we could look at )) ( ) ( 1 ( s D s G + to determine the type. Note that the char. Eqn. 0 10 2 = + + p D k s k s can be stabilized with appropriate choices of the gains. 2 10 ) ( 1 )) ( ) ( 1 ( s s k k s D s G D p + + = + Due to the double pole at the origin, this is a type 2 system. Consequently, from Table
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MEM355W04-05_HW4_2_sol - MEM 355: Solutions to HW#4:...

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