HELLO, PLEASE SOLVE PART D ANSWERS FOR THE REST OF THE QUESTİON İS SHARED BELOW. KİND REGARDS THANK YOU MATLAB code for

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HELLO, PLEASE SOLVE PART D ANSWERS FOR THE REST OF THE QUESTİONİS SHARED BELOW. KİND REGARDS THANK YOU

Hello Please Solve Part D Answers For The Rest Of The Question Is Shared Below Kind Regards Thank You Matlab Code For 1 (505.85 KiB) Viewed 42 times

Hello Please Solve Part D Answers For The Rest Of The Question Is Shared Below Kind Regards Thank You Matlab Code For 2 (76.34 KiB) Viewed 42 times

Hello Please Solve Part D Answers For The Rest Of The Question Is Shared Below Kind Regards Thank You Matlab Code For 3 (144.21 KiB) Viewed 42 times

Hello Please Solve Part D Answers For The Rest Of The Question Is Shared Below Kind Regards Thank You Matlab Code For 4 (26.26 KiB) Viewed 42 times

MATLAB code for drawing Pole-zero plot (for calculating maximumovershoot and settling time)
T=0.02;z=tf('z',T);sys=(0.03798*z+0.00072)/(z^2+0.10248*z-0.93478);[poles,zeros]=pzmap(sys)pzmap(sys)zgrid

Hello Please Solve Part D Answers For The Rest Of The Question Is Shared Below Kind Regards Thank You Matlab Code For 5 (85.96 KiB) Viewed 42 times

R(s) T D(s) T X. G₂OH(S) H(s) Consider the system given above with D(s) = 9.4 (s+4.5) s (s+11.1) = G(s) C(s) 0.6 e-Tas Q2) Consider the system given above with G(s) ‚H(s) = 1 where the time-delay 0.3 s + 1 is Ta 20 ms and the sampling period is T = 20 ms. Then, answer the following questions. a) Draw the root locus plot for D (s) = K. b) Design a digital controller which makes the closed loop system steady state error zero to step inputs and the closed-loop system poles double on the real axis. c) Find the settling time and the overshoot of the digital control system with the controller you designed in (b).
d) Simulate the response of the with your designed controller for unit step input in Simulink by constructing the block diagram. Provide its screenshot and the system response plot.
Errer signal is output E(s) = ((8) = D*(^) E*(8) G₁₂0H (1) G₂ (1) where * thous sampled values of the corresponding signals. given by R(S)-((s)H() Sampled values of above equations obtained as E*(x) = R*(s)- [(C₂) H()]* H(s) = 1 (given) are E*(8) = R*(s) - (+) c*(x) = D*(x) {**(x) [G₂OH (1) G₂ (₂)] *_(2) Solving equations (1) and (2), weget (*(x) = D*(8) [ GIZOH (8) G18)]* R*() 1+ D)*(x) [G₁₂₂H (2) GL₂]]] and from equation (2) weget open loop transfer function (im pulse form * G (1) = (x1) = 0² (1) [G 2 (1 G D* (8) E* (1)
We know Zero order hold transfer (cor) function is defined as (₂017 (2) * (8) Gol = = In Gop (2) = D(2) Z) ...Gop (2) 8 D*(1) Z-transfer function forms D(3) TA 0.6 - Jas 03471 12 [ (1.273) (016 (2²) We know Given Th= гождес property of Zot with G(8) Z [ [ (azon (3). (^(x)] = (1-2¹) Z then above equation becames =D11-2¹) 2 (² схау едискот G₁ (2) s(0.38+1) hop (2) = D/2)2 (1-1) zētas s(x+12) ty (5)
pantial fraction of xcasse) 1 8 (8+10) 1 A B = + 8(x+10) (x+13) 1= A ( 2 + 10 + Bs сле 3 1... 8(8+10) : get values of A and 8 as A = 2 (3/10) - (31/10) B = -7/0 s Z -Tas s(s+10) we know. / = [7 Z ) 37 -Tas s - AS 3+² 2^²^ ] = [-_2²07 e амт e ята -ат 2-e modified z transform 3 parameter gives modified z-tensform of I sta where & is fractional delay s where m=1-A. 3z (2+10
given Id = 20msec готеес Td = A= 0.02sec m = 1-A = 0.98 Now equation (6) becomes Z Tas e ² [ ] = to 3 1(1+10) To (2-e) Z- 3 10 = 3 Go G₁₁ (²) = =0.02111 (2¹) - 72 From equation (3) = D(₂) 3 (0. 5 1-3 (0-9367 ·10 x0.98x20x1² Z-0.9355-(2-1) 0.9367 10 (2-1) (2-0-9355) Z-0.9355 Goz open loop transfer function (in Z-domain) closed loop transfer function (0.06332 +0.0012) z (2-0.9355) Gop (2) 1+ Gop (2)
3 Exp (a) Rot Locus for D(8) = K₂ K 3 (0.06332 + 0.00/2) 2 ( 2-0.9355) 5 Gop (2) = Now uring Z= Gop (W) = w-plane H+W tw K 3 = K 3 5 Put w=jv Gop (jv) = 1.0.0633 (H+W) +0.0012 0:0012) (H+W) (HW (1-w) - 0.9355 K 3 (1-W) (0.0645 + 0.062/1) 5 (1+w) (00645 +1.9355 W) it is similar to I-blane analysis - (0.0633 (HW) +0.0012 (1-W)) (FW) (1+w) ( 1tw- 0.9355 (rw)) k3 (+jv) (0.0645 + 0.06 21 j2) 5 (1+ju) (0.045 +1.9355 ju)
0.3 0.2 0.1 0 -0.1 -0.2 -0.3 I -35 -30 -25 I -20 -15 -10 -5 Activate Windows 5 Go to Settings to activate
Select D(z) = Z (2-1)(2+1) ess=0 for step inputs and two real poles of cloud loop Fransfer fromation Kp = lim hop (2) (b) = as lim 0(2) 0.6 (0.06332+0.99/2 2 (2-0-9355) 271 Kp = lim کھے x0.6 (0.06332+0.002) 2+₁ (2-1)(2+1) +(2-0.9355) kozoo ess= closed log poles ++ Gop (e) =o 7² +0.10248 Z-0.93478=o 2₁ = 0.916957 22 -1.01943
Imaginary Axis 0.8 0.6 0.4 0.2 -0.2 -0.4 -0.6 -0.8 -1 -1.5 0.9T/T 1 T/T 1 T/T -1 0.9T/T- 0.8π/T 0.8T/T 0.7.π/T 0:7T/T -0.5 Pole-Zero Map 0.6TT/T 0.6T/T Real Axis 0.5T/T 50 0.5PT/T 0.4T/T 0.4T/T 0.3 0.2 0:4 0.6 0.7 0.8 0.9 0.5 0.1 -0.3TT/T 0.3π/T I 0.5 0.2/T 0.2π/T 0.17/T 0.1/T Activate Windows Go to Settings to activate V
From the pole zero ہے = 0.95 4 Ts = settling time Gon at 9=0.95 Overshoot we obtain won = Mp= Mp = P p% Mp zero plat = given T = zomlec готее с 4 XT 0.95 ×0.05TT Is = e 11-12 -9.558 e = = 0.05π 0.00706 - 4x0.02 0.95X0.05 0.536 лес 0.95TT 7.06 X155 10 6% 1 - 6.95) edine