Quiz_6_sol - Solution: The part between-2 and 0. (2.5 pts)...

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ECE 483 Quiz # 6 Solution Suppose a feedback sampled-data control system has the following characteristic equation: 1 + Kz ( z + 2)( z 2 + 0 . 25) = 0 , where K 0 is an adjustable parameter. (10 pts) Find the range of K 0 so that the closed-loop system is stable. Solution: We use the Jury’s test. The characteristic equation is equivalent to Q ( z ) = ( z + 2)( z 2 + 0 . 25) + Kz = z 3 + 2 z 2 + ( K + 0 . 25) z + 0 . 5 = 0 . We apply the Jury’s test by constructing the array: z 0 z 1 z 2 z 3 0.5 K + 0 . 25 2 1 1 2 K + 0 . 25 0.5 -0.75 0.5K-1.875 0.75-K For the system to be stable, we must have: Q (1) = K + 3 . 75 > 0 K > - 3 . 75 ( - 1) 3 Q ( - 1) = K - 1 . 25 > 0 K > 1 . 25 | a 0 | = 0 . 5 < a 3 = 1 | b 0 | = 0 . 75 > | b 2 | = | 0 . 75 - K | ⇒ 0 < K < 1 . 5 . To sum up, for the closed-loop system to be stable, we must have 1 . 25 < K < 1 . 5. (2.5 pts) Find the open-loop zeros and poles of the system and plot them on the complex plane. Solution: The only open-loop zero is 0 and the open-loop poles are - 2 and ± j 0 . 5. (2.5 pts) Find the parts of the real axis on the root locus, and mark them on the above plot.
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Unformatted text preview: Solution: The part between-2 and 0. (2.5 pts) Find the asymptotes of the root locus (their center and angles) and mark them on the above plot. Solution: There are two asymptotes diverging to innity along 90 angles, with the center [(-2 + j . 5-j . 5)-0] / (3-1) =-1. (2.5 pts) Determine if z = j 2 is on the root locus. If not, what is the angle of deciency ? Solution: The angle of diciency is = 180 -n G ( j 2) =-45 , which is not 0. Therefore, j 2 is not on the root locus. See the plot next page for the root locus by Matlab. 1-3-2-1 1 2 3-2-1.5-1-0.5 0.5 1 1.5 2 2.5 Root Locus Real Axis Imaginary Axis 2...
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This note was uploaded on 04/15/2011 for the course ECE 483 taught by Professor Evens during the Spring '08 term at Purdue University-West Lafayette.

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Quiz_6_sol - Solution: The part between-2 and 0. (2.5 pts)...

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