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Closed-loop system response and characteristics, Proportional gain 10 < paste transfer function Ts as output from Matlab hereStep response of open-loop system: Pole-zero map: <paste step response plot here> easte,pole-zero plot here> Pole-Zero Map StAnswer the following questions. Q: What type of response is output? <answer? Q: What is the maximum speed of rotation that th

Closed-loop system response and characteristics, Proportional gain 10 clear all: close all: ls J = 0.022R = 0.11;K = 0.02;R 1.5;L= 0.6; Closed loop Transfer function T(s) Cs-10; RRA pole (Tg) 22T zero (Tg) figure ; figure ; teS) characteristics natural frequency damping ratio Dr-abs(real (RpT (2)) ) / ettling time peak time ER忌 overshoot 032=100 rise time
Step response of open-loop system: Pole-zero map: easte,pole-zero plot here> Pole-Zero Map Step Response 06 0.5 0.4 03 02 2 0.1 .3 0.5 1.5 4 3.5325 2 1.50.5 Time (seconds)
Answer the following questions. Q: What type of response is output? : Compare the results for Kp3 to those for Kp-10. Discuss and explain the differences. : Observe the effect for other values of Ke such as 1, 15, etc. What do you conclude is the effect of increasing Ke answer
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Answer #1

kp=10

Figure 1 File Edit View Insert Tools Desktop Window Help System: Ts Step Response Peak amplitude: 0.567 Overshoot (%):3.67 At

emalatd valuss ia Under damh㎂ve otatin ocuss dtt Form the graphGien, =0.11 Ools +o6t0 1.66千 For k to -I0 1.669 p Als) ion. Jao.4S 5,518 o2 usn3.66 únulatud valuus Calculatid valurs 5.5183 5 518 0·ね41 o8a6 0.6a65 0-8964 ts 1.0千 .66 The values are almost ealkp=3

Figure 1 File Edit View Insert Tools Desktop Window Help Step Response 0.3 System: Ts Final value: 0.266 System: Ts ー System:terms 5.5182 o.ね49. o3.9 0.4 4 3340 0.9229 1.89 0.69 t 14 1.ot 36ty 0, 05 35% but MptRYm lon | 4. 416 | 6-2296 | 4-1864 | q-421S | 13.4382 0.68れ| 0,6423. | 0.508. | 0-424S | 0,29 0.24S 04 t.ol l.00 But Mp

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