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midterm_solution

# midterm_solution - ECE 147A FEEDBACK CONTROL SYSTEMS THEORY...

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ECE 147A FEEDBACK CONTROL SYSTEMS - THEORY AND DESIGN F09 Midterm Exam Closed book and notes. No calculators or cell phones. Show all work. Name: Solution Problem 1: /15 Problem 2: /15 Problem 3: /15 Problem 4: /20 Total: /65 Miscellaneous: tan 1 (1 / 3) 18 degrees, tan 1 (1 / 2) 26 degrees, tan 1 (1) = 45 degrees. tan 1 (2) 63 degrees, tan 1 (3) 72 degrees, tan 1 (5) 78 degrees, tan 1 (10) 84 degrees, tan 1 (20) 87 degrees, sin(45 ) = 2 / 2 0 . 7, sin(50 ) 0 . 77, sin(55 ) 0 . 82, sin(60 ) = 3 / 2 0 . 87. 4 . 6 ζω n , 1 . 8 ω n , exp parenleftbigg πζ/ radicalBig 1 ζ 2 parenrightbigg .

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Name: Problem 1 Consider the nonlinear system ¨ z = u 2 + z ˙ z + sin( z ) , y = z. 1. Which values of the output can be obtained through valid operating points? Express all valid operating points as a function of the operating output value y . Operating points must satisfy ( u ) 2 = sin( y ) . When this equation is satisfied, the operating point is given by ( y , 0 ,u ) when taking the state to be x = ( z, ˙ z ) . The equation has a solution as long as sin( y ) 0 . This condition holds for y [(2 i 1) π, 2 ] for each integer i . 2. Find the transfer function as a function of the output component of the operating point y and identify those values of y for which the transfer function is stable. Hint: use the fact that bracketleftBigg a b c d bracketrightBigg 1 = bracketleftBigg d b c a bracketrightBigg ad bc . The equations of motion are ˙ x 1 = x 2 , ˙ x 2 = u 2 + x 1 x 2 + sin( x 1 ) , y = x 1 We get A = bracketleftBigg 0 1 x 2 + cos( x 1 ) x 1 bracketrightBigg , B = bracketleftBigg 0 2 u bracketrightBigg , C = bracketleftBig 1 0 bracketrightBig , D = 0 . In turn, using that x 2 = 0 , the transfer function is C ( sI A ) 1 B + D = bracketleftBig 1 0 bracketrightBig bracketleftBigg s 1 cos( x 1 ) s x 1 bracketrightBigg 1 bracketleftBigg 0 2 u bracketrightBigg = bracketleftBig 1 0 bracketrightBig bracketleftBigg s x 1 1 cos( x 1 ) s bracketrightBigg s 2 x 1 s cos( x 1 ) bracketleftBigg 0 2 u bracketrightBigg = 2 u s 2 x 1 s cos( x 1 ) . The transfer function is stable when the operating output x 1 , which must satisfy sin( x 1 ) < 0 , also satisfies x 1 < 0 and cos( x 1 ) < 0 . These conditions hold for ξ 1 [(2 i 1) π, (2 i 1 / 2) π ] for each nonpositive integer i . 3. Suppose a controller C has been designed to control the system above near the operating point given above, using the negative unity feedback configuration with the reference prefilter F . Draw the corresponding block diagram implementation of the control system.
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midterm_solution - ECE 147A FEEDBACK CONTROL SYSTEMS THEORY...

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