ball beam project 1 chris hulbert

# ball beam project 1 chris hulbert - From the point on the...

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A1 Appendix A Design of Continuous-Time System Transfer Function and Kp/Kv Derivation: 084 . 3 1 2 55 . 11 55 . 4 2 2 715 . 0 69 . 0 ) ln( ) ln( 8 ~ 8 . 4 2 ) 1 ( * ) 1 ( * 1 * 1 1 * 2 2 2 1 2 2 2 2 = - = = = = = = + - = = = = + + + = + + + = + + + + + = Θ Θ - - Am Tm Kv Am Tm Kp Let Mp Mp e Mp AND Tm AmKp Tm AmKp Tm AmKv s s Tm AmKp AmKp s AmKv s Tm AmKp AmKv s Tm AmKp AmKv s Tm KpAm d n n n n n n ζϖ ϖ ζ π πζ Our Plant Model with the Servo system now becomes: 2 3 4 7 . 20 5 . 6 88 . 12 s s s + + Plotting this system, we get the plot:

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A2 Figure a Picking a high crossover frequency of 70, we see that the phase must be compensated about 255 degrees. Therefore 3 lead compensators at approximately 83.8 degrees each will give us an alpha of 0.029. We then compute the zero’s and poles: 1292 79 . 3 = = = = α ϖ c c p z The compensator plot is shown below as the series combination of 3 identical lead compensators.
A3 Figure b Now, we can place the 3 lead compensators in our system and plot the compensated response as well as the unity feedback step response, both shown below.

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Unformatted text preview: From the point on the step response, we see the Maximum overshoot is exactly 5% meeting the minimum requirements. A4 Figure c Figure d A5 9 . 285 10 * 3918 . 536 * 6825 . 18 1 ) 70 ( ) 70 ( 1 70 1 ) ( ) ( * 9 3 = = = = = =-j G j Dlead Kv j j s at s G s Dlead Kv ϖ Calculation of the Steady-State error is as follows: 49 1 02 . 1 % 2 11 . 184 82 . 1088 17 . 701 * 9 . 285 7 . 20 88 . 12 6 . 52 4399 . 54 7 . 306 * ) 7 . 20 5 . 6 ( 88 . 12 1292 79 . 3 0029 . 1 7 . 306 * )] ( ) ( [ 2 2 3 lim =-= ⇒ = = = = = + + + + = =-Kp ess Kv s s s s s Kv s G s KvDlead Kp s Since our Kp is greater than the required Kp, no Lag compensation is needed....
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## This note was uploaded on 01/23/2012 for the course EE 4580 taught by Professor Gu during the Fall '10 term at LSU.

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ball beam project 1 chris hulbert - From the point on the...

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