Figure 2 shows a unit step response of a second order process. It shows that there is a small overshoot and then the out

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Figure 2 Shows A Unit Step Response Of A Second Order Process It Shows That There Is A Small Overshoot And Then The Out 1 (85.92 KiB) Viewed 71 times

Figure 2 shows a unit step response of a second order process. It shows that there is a small overshoot and then the output settles at 1. At the bottom of the figure it shows the values of the pole location which are approximately at s = -1 + j and s=-1-1 a Using the step response software, by trial and error, find the values for the damping ratio and the natural frequency that produces this response (Hint See Figure 2.5 on page 18 of Book 2). Make sure you include a screenshot of your step response 2.0 =0.2 1.5 021 $=0.7 output 1.0 M S=1 0.5 =2 b. By substituting the values for the damping ratio, natural frequency and gain into the standard form of the second-order transfer function, the transfer function for this system is found to be 2 G (3) 82 + 25 +2 Using this transfer function, show that the poles are at approximately s = -1 +j and s = -1-j? 0.0 0 3 4. 2 (rad) Figure 2.5 Unit step response of second-order system with different types of damping 8 OF 1 sel . Gan PG sise 62 06/ . 952 0500 225 25 325 35037 400 300 Times Tine Pole location Pola 1-1001-9993 Pole 2 -1001-0991 Zoom and @le Reset 100% Pole locations Polet-10010999 Pole 2-1001-9.993 Zoom and an @ @ Reset 4009 Figure 2 (a) A unit step response of a second-order process, (b) zoomed in on the overshoot c. It is decided that the process in Figure 2 will be controlled using a PD controller with a derivative gain, Ko' of 0.5, so that the transfer function of the PD controller is: C(s) = Kp(1 +0.55) Using the combined transfer function of the forward path, C(s)G(s), use the root-locus software to plot the root locus with the value of Ko going from a minimum gain of 0 to a maximum of 50. Include a screenshot of your root-locus plot as your answer.
d. The root locus showed that at some value of gain the two closed-loop poles arrive at the real axis together. By zooming and shifting the root locus find the value of s at which point the two closed-loop poles meet on the real axis. Answers should be to 1 place after the decimal point. Remember to include a screenshot at an appropriate zoom level of your root-locus plot. e. The root-locus plot showed the position of the closed-loop poles as Kp was varied from 0 to 50. By reducing the value of the maximum gain, the root locus can be produced which stops when the two poles meet on the real axis and therefore the maximum gain is the value at this point where the two poles meet. Try lowering the maximum value of gain to find the value of the gain to 3 places after the decimal point where the two poles meet. Include a screenshot at an appropriate zoom level of your root-locus plot f. Finally, we would like you to show that the value of gain found in part (e) and the value of s found in part (d) are correct by using the closed-loop transfer function to find the value of Ko where the two poles meet on the real axis.