hw7soln_F2009 - MAE170 Homework 7 Solution Fall 2009 E8.5...

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Unformatted text preview: MAE170 Homework 7 Solution Fall 2009 E8.5 Refer to Fig. E8.5 in the text that shows the magnitude Bode plot of G ( s ) approximated by the asymptotes. The lower portion of the plot for < 2 rad/s is 20log | G ( j ) | = 20log K and satisfies for = 8 rad/s 20log K 8 = 0 dB. Therefore, it follows that K = 8. Since the slope changes to 0 db/dec at = 2 rad/s , we have a zero at = 2 rad/s . Clearly, we have another zero at = 4 rad/s since the slope changes to 20 db/dec . The zero at = 4 rad/s yields a = 0 . 25. Then the slope changes again to 0 db/dec at = 24 rad/s and to- 20 db/dec at = 36 rad/s meaning that there are poles at = 24 rad/s and = 36 rad/s . The pole at = 24 rad/s yields b = 1 / 24. Therefore, G ( s ) = 8 ( 1 + s 2 )( 1 + s 4 ) s ( 1 + s 8 )( 1 + s 24 )( 1 + s 36 ) . P8.2 (a) The first transfer function is L ( s ) = G c ( s ) G ( s ) = 1 (1 + 0 . 5 s )(1 + 2 s ) . L ( s ) is already normalized with K b = 1 and no poles or zeros at s = 0. The break frequencies are 1 = 1 / 2 = 0 . 5 rad/s and 2 = 1 / . 5 = 2 rad/sec coming from real poles. Therefore the asymptote magnitude Bode plot starts at 20log | K b | = 0 db and with a slope of 0 db/dec , it breaks at = . 5 rad/s to a slope of- 20 db/dec and again at = 2 rad/s to a slope of- 40 db/dec . The asymptote phase Bode plot starts at K b = 0 o , breaks to- 90 o at = 0 . 5 rad/s and then again to- 180 o at = 2 rad/s . The Bode plots are shown in Figure 1. This is a type 0 system and the position error constant is K p = K b = 1. (b) The second transfer function is L ( s ) = G c ( s ) G ( s ) = 1 + 0 . 5 s s 2 . 1 10-2 10-1 10 10 1 10 2-60-50-40-30-20-10 Magnitude Bode Plot with Asymptotes (rad/sec) Magnitude in db 10-2 10-1 10 10 1 10 2-200-150-100-50 Phase Bode Plot with Asymptotes (rad/sec) Phase in degrees Figure 1: Bode P8.2a....
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hw7soln_F2009 - MAE170 Homework 7 Solution Fall 2009 E8.5...

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