solution hw6

solution hw6 - Problem 1 controller valve tank 1 1 0.2s +...

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Problem 1 Figure 1 The characteristic equation for the closed loop in figure 1 can be written as follows: C OL 2 2k (1 3 s) 1G ( s )1 0.2s 0.4s 1 + += + ++ (S1.1) Setting (S1.1) equal to zero and substituting k C =2 yields: 32 0.2s 0.4s 5s 12 0 + = (S1.2) Test 1: All the coefficients of (S1.2) are positive. Therefore, one can move to test 2. Test 2: (S1.2) is a 3 rd order polynomial. Thus, the Routh-Hurwitz array is composed of 4 rows. Column 1 Column 2 Row 1 0.2 5 Row 2 0.4 12 Row 3 [(0.4*5) (12*0.2)] 0.4 1 −= 0 Row 4 ( 1*12) 1 12 −− = 0 As all the coefficients of the first column of the Routh-Hurwitz array are not positive, the system is unstable. Moreover, as the sign changes two times (from row 2 to row 3 and row 3 to row 4), there are two roots with positive real part. Problem 2 Figure 2 For the proportional controller, the characteristic equation for the closed loop in figure 2 can be written as follows: controller valve tank 1 Measuring device SP y( s ) y(s) C G 1 3 1 (s 1) + 1 controller valve tank 1 Measuring device SP s ) y(s) C k( 1 3s ) + 1 2 1 0.2s 0.4s 1 + + 2
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C OL 3 k 1G ( s )1
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solution hw6 - Problem 1 controller valve tank 1 1 0.2s +...

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