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Unformatted text preview: ECE 147A FEEDBACK CONTROL SYSTEMS  THEORY AND DESIGN F08 Midterm Exam Closed book and notes. No calculators. No cell phones. Show all work. Name: Solution Problem 1: /25 Problem 2: /25 Problem 3: /25 Problem 4: /25 Total: /100 Miscellaneous: b b 2 4 ac 2 a , 1 . 8 n , 4 . 6 , exp parenleftBigg radicalbig 1 2 parenrightBigg , atan([0.1 1/3 0.5 2/3 1 2 5])*180/pi = [5.7 18.4 26.6 33.7 45 63.4 78.7] acos([0.5 0.7071 0.866])*180/pi =[60 45 30] asin([0.5 0.7071 0.866])*180/pi =[30 45 60] When m 11 , m 12 , m 21 and m 22 are scalars: bracketleftBigg m 11 m 12 m 21 m 22 bracketrightBigg 1 = bracketleftBigg m 22 m 12 m 21 m 11 bracketrightBigg m 11 m 22 m 21 m 12 . When M 11 and M 22 are invertible matrices: bracketleftBigg M 11 M 12 M 22 bracketrightBigg 1 = bracketleftBigg M 1 11 M 1 11 M 12 M 1 22 M 1 22 bracketrightBigg . In turn, when the number of columns of C 1 agrees with the size of M 11 and the number of rows of B 2 agrees with the size of M 22 , bracketleftBig C 1 bracketrightBig bracketleftBigg M 11 M 12 M 22 bracketrightBigg 1 bracketleftBigg B 2 bracketrightBigg = C 1 M 1 11 M 12 M 1 22 B 2 . Name: Problem 1 A particular levitatedball system has dynamics given by the nonlinear equations = 1 z 2 1 + 2 z = z + tildewide u tildewide y = . 1. Find an operating point corresponding to a constant output equal to 3. Is there one or more than one such operating point? We take the state to be x = ( x 1 ,x 2 ,x 3 ) = ( , ,z ) . This gives the equations x 1 = x 2 x 2 = 1 x 2 3 1 + x 2 1 x 3 = x 3 + tildewide u tildewide y = x 1 Operating points with output equal to 3 must satisfy x * 1 = 3 , x * 2 = 0 , ( x * 3 ) 2 1 + 9 = 1 , u * = x * 3 . So there are two operating points, corresponding to x * 3 = 10 and x * 3 = 10 . 2. Compute the transfer function from tildewide u u * to tildewide y 3 where u * is an input that makes the output a constant equal to 3. (You may find helpful the expressions on the first page of the exam.) We consider the operating point with x * 3 = 10 . The transfer function for x * 3 = 10 is the same except for an extra minus sign as a factor. We compute d = 0 , C = bracketleftBig 1 0 0 bracketrightBig , B = bracketleftBig 0 0 1 bracketrightBig T , A = 1 2 x * 1 ( x * 3 ) 2 (1 + ( x * 1 ) 2 ) 2 2 x * 3 1 + ( x * 1 ) 2 1 = 1 3 5 radicalbigg 2 5 1 ....
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This note was uploaded on 08/03/2010 for the course ECE PROF. VOLK taught by Professor Volkanrodoplu during the Spring '10 term at UCSB.
 Spring '10
 VolkanRodoplu

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