hw5soln_F2009 - MAE170 Homework 5 Solution Fall 2009 E7.15...

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MAE170 Homework 5 Solution Fall 2009 E7.15 (a) The characteristic equation is 1 + GH ( s ) = 1 + K ( s + 1)( s + 3) s 3 = 0 and the root locus for this system is shown in figure 1. Some of the computations to determine the root locus by hand are the following. We have n = 3 with open-loop poles at 0 , 0 , 0, and m = 2 with open-loop zeros at - 1 and - 3. The break-away/break-in points are found among the solutions of 0 = q ( s ) dp ( s ) ds - p ( s ) dq ( s ) ds s 3 (2 s + 4) - 3 s 2 ( s + 1)( s + 3) = (1) = - s 2 ( s 2 + 8 s + 9) s = 0 , 0 , - 1 . 35 , - 6 . 64 . (2) Of these only 0 and - 6 . 64 are on the root locus and they are actual break- away/break-in points (the root at s = 0 is double indicating the break- away of the 3 open-loop poles at s = 0). The break-away angles at s = 0 form a symmetric pattern spaced by 360 o / 3 = 120 o . The break-in/break- away angles at s = - 6 . 64 form a symmetric pattern spaced by 360 o / 4 = 90 o . Crossings of the root locus with the imaginary axis are found by constructing the Routh array as in part (b); the critical value of gain is K = 3 / 4. For this value of K the closed-loop characteristic polynomial becomes: p cl ( s ) = s 3 + 3 4 s 2 + 3 s + 9 4 and it has zeros (closed-loop poles for K = 3 / 4) at s = - 3 / 4 , ± j 3, as it can be found using MATLAB or by substituting s = in the char. polynomial. Thus, axis crossings occur at ± j 3 for K = 3 / 4. (b) The closed-loop characteristic equation is equivalent to s 3 + Ks 2 + 4 Ks + 3 K = 0 Form the Routh array s 3 : 1 4 K s 2 : K 3 K s 1 : b s 0 : 3 K 1
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-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 -4 -3 -2 -1 0 1 2 3 4 Real Axis Imag Axis Figure 1: Root Locus for problem E7.15 where b = 4 K - 3. For stability, we require that all elements in the first column be positive. Therefore, K > 3 4 . (c) When K > 3 / 4, we have e ss = lim s 0 sE ( s ) = lim s 0 s ± 1 1 + GH ( s ) 1 s 2 = lim s 0 s 2 s 3 + K ( s + 1)( s + 3) = 0 Therefore, the steady-state error to a ramp input is e ss = 0. P7.1 Figures 2 and 3 show the root locus plots for the given systems. Some of the computations to determine the root loci by hand are the following. a) KG ( s ) = K s ( s +10) 2 . We have n = 3 with open-loop poles at 0 , - 10 , - 10, and m = 0 (no open-loop zeros). The break-away/break-in points are found among the solutions of 0 = q ( s ) dp ( s ) ds - p ( s ) dq ( s ) ds ≡ - s ( s + 10) 2 · 0 - (3 s 2 + 40 s + 10) · 1 = = - (3 s 2 + 40 s + 10) s = - 3 . 33 , - 10 . 2
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-12 -10 -8 -6 -4 -2 0 2 4 -20 -15 -10 -5 0 5 10 15 20 Real Axis Imag Axis (a) -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 Real Axis (b) Figure 2: (a) Root Locus for problem P7.1a (b) Root Locus for problem P7.1b. 3
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-7 -6 -5 -4 -3 -2 -1 0 1 2 3 -8 -6 -4 -2 0 2 4 6 8 Real Axis Imag Axis (a) -6 -5 -4 -3 -2 -1 0 1 2 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 Real Axis (b) Figure 3: (a) Root Locus for problem P7.1c (b) Root Locus for problem P7.1d. 4
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This note was uploaded on 08/31/2010 for the course ENGRMAE 170 taught by Professor Staff during the Fall '08 term at UC Irvine.

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hw5soln_F2009 - MAE170 Homework 5 Solution Fall 2009 E7.15...

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