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ch17 - 123456789 8 17.1 Using Eq 17-9 the filtered values...

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17-1 ±²³´µ¶·¸¹º¸ 17.1 Using Eq. 17-9, the filtered values of x D are shown in Table S17.1 time(min) α=1 α=1 α= α= 0.8 α= α= 0.5 0 0 0 0 1 0.495 0.396 0.248 2 0.815 0.731 0.531 3 1.374 1.245 0.953 4 0.681 0.794 0.817 5 1.889 1.670 1.353 6 2.078 1.996 1.715 7 2.668 2.534 2.192 8 2.533 2.533 2.362 9 2.908 2.833 2.635 10 3.351 3.247 2.993 11 3.336 3.318 3.165 12 3.564 3.515 3.364 13 3.419 3.438 3.392 14 3.917 3.821 3.654 15 3.884 3.871 3.769 16 3.871 3.871 3.820 17 3.924 3.913 3.872 18 4.300 4.223 4.086 19 4.252 4.246 4.169 20 4.409 4.376 4.289 Table S17.1. Unfiltered and filtered data. To obtain the analytical solution for x D , set s s F 1 ) ( = in the given transfer function, so that 5 5 1 1 ( ) ( ) 5 10 1 (10 1) 1 10 D X s F s s s s s s = = = - + + + Taking inverse Laplace transform x D ( t ) = 5 (1 - e - t /10 ) A graphical comparison is shown in Fig. S17.1 Solution Manual for Process Dynamics and Control, 2 nd edition, Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp
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17-2 0 2 4 6 8 10 12 14 16 18 20 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 time (min) noisy data alpha = 0.5 alpha = 0.8 analytical solution X D Fig S17.1. Graphical comparison for noisy data, filtered data and analytical solution. As α decreases, the filtered data give a smoother curve compared to the no-filter ( α =1) case, but this noise reduction is traded off with an increase in the deviation of the curve from the analytical solution. 17.2 The exponential filter output in Eq. 17-9 is ( ) ( ) (1 ) ( 1) F m F y k y k y k = α + - α - (1) Replacing k by k -1 in Eq. 1 gives ( 1) ( 1) (1 ) ( 2) F m F y k y k y k - = α - + - α - (2) Substituting for ( 1) F y k - from (2) into (1) gives 2 ( ) ( ) (1 ) ( 1) (1 ) ( 2) F m m F y k y k y k y k = α + - α α - + - α - Successive substitution of ( 2) F y k - , ( 3) F y k - ,… gives the final form 1 0 ( ) (1 ) ( ) (1 ) (0) k i k F m F i y k y k i y - = = - α α - + - α
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17-3 17.3 Table S17.3 lists the unfiltered output and, from Eq. 17-9, the filtered data for sampling periods of 1.0 and 0.1. Notice that for sampling period of 0.1, the unfiltered and filtered outputs were obtained at 0.1 time increments, but they are reported only at intervals of 1.0 to preserve conciseness and facilitate comparison. The results show that for each value of t , the data become smoother as α decreases, but at the expense of lagging behind the mean output y ( t )= t . Moreover, lower sampling period improves filtering by giving smoother data and less lagg for the same value of α . t =1 t =0.1 t α =1 α =0.8 α =0.5 α =0.2 α =0.8 α =0.5 α =0.2 0 0 0 0 0 0 0 0 1 1.421 1.137 0.710 0.284 1.381 1.261 0.877 2 1.622 1.525 1.166 0.552 1.636 1.678 1.647 3 3.206 2.870 2.186 1.083 3.227 3.200 2.779 4 3.856 3.659 3.021 1.637 3.916 3.973 3.684 5 4.934 4.679 3.977 2.297 4.836 4.716 4.503 6 5.504 5.339 4.741 2.938 5.574 5.688 5.544 7 6.523 6.286 5.632 3.655 6.571 6.664 6.523 8 8.460 8.025 7.046 4.616 8.297 8.044 7.637 9 8.685 8.553 7.866 5.430 8.688 8.717 8.533 10 9.747 9.508 8.806 6.293 9.741 9.749 9.544 11 11.499 11.101 10.153 7.334 11.328 11.078 10.658 12 11.754 11.624 10.954 8.218 11.770 11.778 11.556 13 12.699 12.484 11.826 9.115 12.747 12.773 12.555 14 14.470 14.073 13.148 10.186 14.284 14.051 13.649 15 14.535 14.442 13.841 11.055 14.662 14.742 14.547 16 15.500 15.289 14.671 11.944 15.642 15.773 15.544 17 16.987 16.647 15.829 12.953 16.980 16.910 16.605 18 17.798 17.568 16.813 13.922 17.816 17.808 17.567 19 19.140 18.825 17.977 14.965 19.036 18.912 18.600 20 19.575 19.425 18.776 15.887 19.655 19.726 19.540 Table S17.3. Unfiltered and filtered output for sampling periods of 1.0 and 0.1
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17-4 Graphical comparison : 0 2 4 6 8 10 12 14 16 18 20 0 2 4 6 8 10 12 14 16 18 20 time, t y(t) α=1 α=0.8 α=0.5 α=0.2 Figure S17.3a. Graphical comparison for t = 1.0 0 2 4 6 8 10 12 14 16 18 20 0 2 4 6 8 10 12 14 16 18 20 time, t y(t) α=1 α=0.8 α=0.5 α=0.2 Figure S17.3b. Graphical comparison for t = 0.1
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17-5 17.4 Using Eq. 17-9 for α = 0.2 and α = 0.5, Eq. 17-18 for N * = 4, and Eq. 17- 19 for y =0.5, the results are tabulated and plotted below.
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