Question

Solve the following problems:

1 = P.3 Design a Lead compensator using the Bode diagrams approach, for a process which transfer function is G(s) to satisfy the following requirements: s(s+1)' 1. ts < 1.0 sec(+1%) for a unit step reference. 2. Overshoot < 8% Include a bode plot with the compensated and the uncompensated system (include the margins) and a unit step response of the compensated system. 400 P.1 Design a Lag compensator, for a process which transfer function is G(s) = following requirements: to satisfy the s(s+1)(s+2)' 1. ess < 5% 2. Overshoot s 20% Include a bode plot with the compensated and the uncompensated system (include the margins) and a unit step response of the compensated system.

This is all the information I got.

Answer 1

Giso = P-3 s(s+12 Design Lead compensation for 1. To < I sec. for moit step. Overshoot <80% 2. so Ts = 4 Gwn 4 cicis Elon - To Coto % Mp = e xl00% < 80% -te Coto е < solving we will get q = coso 70.6265 so Taking & = 0.7 Relation b/w damping ratio and Phase Margin is- im tant 28 (2) [1-29² + √14484 2 xo. 7 tant um 1-2x0,49 +1 1+4x6.734 - om 449.599 Calculating Phase Margin of uncompensated system - Grain cross over frequency (egel I Gejws) wge = 1 where wgc suge to w ge scoge +17 = 1 wgel + wge-150 x Let wge coge? a? + 2 - 1 + +4 2 t5 2 0.6180 Positive value of so 0.6180 = 0.786 rod/sec wge Wyc Phase at coge = -go-tan' ( wych -gu-tan (0.786) -128.167° 180° 128,1679 = 51,839 :. Phase Mangin since the system already has required Phase margir We need to adjust fole- zero of lead compensator ...at 10 times distance with each other so that do not affect phase Bet desired value of gain, and sls+1) yu KCI+) (1 fois) Controller plant RIS Characteristic equation 1 + Gres) Gly) - 0 It Klitos) (1-0,1) s(+b) + (1 +0.15) K 0.15² tot こい 3 tlos flok from eqin we have qwny 4 & wa taken % = 0.7 and 4 6.7 on 75.7143 مدا Standard equation. $*+ 2ętao ist wor? = 0 with eqciis 2 =lok Compare مکرا ( Wn > 5.3] lok (6) .', 3.6 L! 3.6 (ts) Lead control Gc= (tols) Hence required MATLAB Check!

close all
clear all
s =tf('s');
G = 1/(s*(s+1));
Gc = 3.6*(1+s)/(1+0.1*s); %Designed Lead controller
T = feedback(G*Gc,1); %Closed Loop T.F.
figure (1)
step(T)
title('Step Response of Compensated system');
figure (2)
bode(G,T)
legend('Uncompensated system','Compensated system');

Step Response of Compensated system 1.2 System: T Peak amplitude: 1.01 Overshoot (%): 0.877 System: T At time (seconds): 0.949 Settling time (seconds). U.ro 0.8 Amplitude 0.6 0.4 0.2 0 0 0.1 0.2 0.3 0.4 0.6 0.7 0.8 0.9 1 0.5 Time (seconds)

Bode Diagram 50 Magnitude (dB) -50 -100 -150 00 Uncompensated system Compensated system -45 Phase (deg) -90 System: Uncompensated system Phase Margin (deg): 51.8 Delay Margin (sec): 1.15 At frequency (rad/s): 0.786 Closed loop stable? Yes -135 -180 10-2 10-1 100 101 102 103 Frequency (rad/s)

P1 400 GLS SOS+20+2) Requirement 1. ess< 5% 2, overshoot < 20% We have to design Log controller 20% overshoot we have So for [ 20 acolo * 100 e 2 - Toto < @ 20.45 coso ६. Ę = -0,5 of taking we should have phase margin 29 Tozart wityGu tont e 3817 o Lag One inspection of Bode Slot from MATLAB we got to know that only gain adjustment required in compensator of form KCl toils) Geest (+) do have desined closed loop response, By adjusting k=0.0016 desired Specifications % overshoot = 17.6% we got error steady

Hence Gc(s) = 0.0016*(s+0.1)/(s+1)

CODE-

close all
clear all
s =tf('s');
G = 400/(s*(s+1)*(s+2));
Gc = 0.0016*(1+0.1*s)/(s+1);
T = feedback(G*Gc,1); %Closed Loop T.F.
figure (1)
subplot(1,2,1)
step(G)
title('Step Response of Unompensated system');
subplot(1,2,2)
step(T)
title('Step Response of Compensated system');
figure (2)
bode(G,T)
legend('Uncompensated system','Compensated system');

Response-

Step Response of Unompensated system Step Response of Compensated system X 104 5 1.2 4.5 System: T Peak amplitude: 1.18 Overshoot (%): 17.6 At time (seconds): 9.19 System: T Final value: 1 1 4 3.5 0.8 3 Amplitude 2.5 Amplitude 0.6 2 0.4 1.5 1 0.2 0.5 0 0 50 200 250 0 5 10 15 20 25 30 100 150 Time (seconds) Time (seconds)

Bode Diagram 100 50 0 Magnitude (dB) -50 -100 -150 1 -200 0 T 45 Uncompensated system Compensated system -90 -135 Phase (deg) -180 System: Compensated system Phase Margin (deg): 64.8 Delay Margin (sec): 2.56 At frequency (rad/s): 0.442 Closed loop stable? Yes System: Uncompensated system Phase Margin (deg): -66.7 Delay Margin (sec): 0.705 At frequency (rad/s): 7.26 Closed loop stable? No -225 -270 -315 10-2 10-1 100 101 102 103 Frequency (rad/s)