Solutions_mae143b

# Solutions_mae143b - MAE143B Final Exam Solutions Tuesday...

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MAE143B Final Exam Solutions — Tuesday December 8, 2009 Question 1 Part (i): Lead compensator portion, s/ 2 .. 5+1 s/ 25+1 . For a lead compensator, the pole is larger than the zero and both are real and negative. It is designed to improve stability and bandwidth by adding phase lead. Lag compensator portion, s +0 . 1 s +0 . 02 . For a lag compensator, the zero is larger than the pole and both are real and negative. It is designed to improve steady-state error without affecting too much the bandwidth and stability, in spite of it adding phase lag. The design was performed in three parts; analysis of the open-loop plant with proportional control, lead compensator design to achieve closed-loop bandwidth, lag compensator design to achieve static accuracy speciﬁcation. Figure 2 shows that G ( s ) with unity feedback is marginally stable. It could be stabilized by reducing the proportional gain. But for 20 phase margin, this would yield a closed-loop bandwidth much less than one radian per second. In Figure 3 we see that the lead compensator has been included to achieve the 20 phase margin at the target frequency of 3.9 radians/sec. This is achieved with the compensator s/ 2 .. 5+1 s/ 25+1 and unity feedback. The static accuracy of this closed-loop system to a unit ramp is given by. lim t →∞ e t = lim s 0 sE ( s ) , = lim s 0 s × 1 1 + s/ 2 .. 5+1 s/ 25+1 - s +2 s ( s +1) × 1 s 2 , = 0 . 5 . In Figure 4 we see that a lag compensator, s +0 . 1 s +0 . 02 , has been included with the lead compensator to increase the velocity constant by a factor of ﬁve low frequencies – and thereby deal with the static accuracy speciﬁcation – while leaving the bandwidth and phase margin at 4 radians per second relatively unaltered. Part (ii):

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Solutions_mae143b - MAE143B Final Exam Solutions Tuesday...

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