Question

It has the following transfer function:

2 Gp(s) K* (+1) *wn S*(s +2* S * wn) * (+1)

-What happens to the plant with different values of ($\delta$) (relative damping factor), also analyze how it influences if the values of $\omega n$ ,    and vary, for this implement scripts in Matlab.m and show the results in graphs
corresponding.
- Implement models of transfer functions in:
a) open loop
b) closed loop with unit feedback
b) closed loop with unit feedback and a PID controller

**DO IT IN SIMULINK

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Answer 1

A system has transfer function of Gp(s) K* ( + 1) *wn? s*(s +2*S*wn) * (+1) Where is the damping.satie. Wn is the natural damping frequency wais the location of a pole and wa is the location of zero of the system Initially, we vary the damping ratio, while fixing the other parameters as wn = 1 rad/secao = 2, wa = 5, and K = 1. To model this in Simulink, we first need to compute the transfer function for the different values of 0. Letus assume that the variation of damping ratio is [0, 0.2, 0.4, 0.6, 0.8, 1). For8 = 0, the transfer function would be 2s + 10 Gp(s) = 5s + 10s" For 8 = 0.2, the transfer function would be 2s + 10 Gp(s) = + 12s +45 5s Foro = 04, the transfer function would be Gp(s) = 2s + 10 + 14s + 8s ssa Foro = 0.6, the transfer function would be 2s + 10 Gp(s)=55* +165- + 125 For ô = 0.8, the transfer function would be 2s + 10 Gp(s)= 5s + 18s- + 165

Foro = 1, the transfer function would be 2s + 10 Gp(s) = 5s + 20° + 20s The corresponding Simulink model can be used to modelthe above system and test the same for open- loop, close-loop and PID controlled system unit step response. For the PID controlled system, we assume that Kp = 10, Ki = 1 and Kd = 0.1 and it is fixed for all the transfer functions.

File Edit View Display Diagram Simulation Analysis Code Tools Help 10.0 Normal untitled untitled Kp = 10, ki = 1 Kd = 0.1 25+10 583+1052 2s+10 583+1052 Step PID(S) Transfer Fcn Izeta = 0 2s+10 593+1052 Step1 Transfer Fcn6 Izeta = 0 Step2 PID Controller Transfer Fcn12 Izeta = 0 2s+10 593-12s2+4s Transfer Fcn1 Izeta=0.2 Scope Open-loop Response 25+10 593-12s2+45 PID(s) Scope1 close-loop Response 2s+10 593+12s2+45 Transfer Fcn7 Izeta=0.2 PID Controller 1 Transfer Fcn13 Izeta=0.2 Scope2 close-loop Response with PID control 2s+10 553-1452+85 2s+10 553-1452+8s PID(S) Transfer Fen2 zeta=0.4 2s+10 553-145248 Transfer Fen8 zeta=0.4 PID Controller2 Transfer Fcn14 Izeta=0.4 2s+10 553-16s2+125 2s+10 5s316s2+125 PID(S) Transfer Fcn3 Izeta=0.6 2s+10 593+1652+125 Transfer Feng zeta=0.6 PID Controller3 Transfer Fcn15 Izeta=0.6 2s+10 593-1892-165 Transfer Fcn4 Izeta=0.8 2s+10 5s3-1892-165 Transfer Fcn10 Izeta=0.8 PID(S) 25+10 553-1852-165 PID Controller 4 Transfer Fcn16 Izeta=0.8 2s+10 583-203220s Transfer Fcn5 Izeta = 1 2s+10 5s3-2032+20s PID(S) Transfer Fcn11 zeta = 1 2s+10 593+2052+205 Transfer Fcn17 zeta = 1 PID Controller5 Ready 100% ode45^

The corresponding responses are shown below

Open-loop response

50 45 Transfer Fon Izeta = 0 Transfer Fcnl \zeta=0.2 Transfer Fen2 \zeta=0.4 Transfer Fcn3 Izeta=0.6 Transfer Fcn4 \zeta=0.8 Transfer Fcn5 \zeta = 1 40 35 30 25 20 15 10 5 0 0 1 2 3 5 6 7 8 9 10 me offset: 0

Close-loop response

4 3.5 Transfer Fcn6 \zeta = 0 Transfer Fcn7 \zeta=0.2 Transfer Fcn8 \zeta=0.4 Transfer Fcn9 \zeta=0.6 Transfer Fcn10 \zeta=0.8 Transfer Fenli \zeta = 1 3 2.5 2 1.5 1 0.5 0 -0.5 -1 0 2 3 5 6 8 9 10 me offset: 0

PID controlled response

60 40 Transfer Fcn12 \zeta = 0 Transfer Fcn13 \zeta=0.2 Transfer Fcn14 \zeta=0.4 Transfer Fcn15 \zeta=0.6 Transfer Fcn16 \zeta=0.8 Transfer Fcn17 \zeta = 1 20 0 -20 -40 -60 -80 -100 -120 0 1 2 3 4 5 6 8 9 10 Time offset: 0

Now, we are varying the parameter wn, while assume , 8 = 1, W; = 2, wa = 5, and K = 1. To model this in Simulink, we first need to compute the transfer function for the different values of wrę Let us assume that the variation of Wn is (2, 4, 6, 8, 10, 12]. Forwn = 2, the transfer function would be 8s + 40 Gp(s)= 5s + 305* + 40s Forwn = 4, the transfer function would be 32s + 160 Gp(s)= 5s + 50s+ 80s Forwn = 6, the transfer function would be 72s + 360 Gp(s)= 5s + 70s + 120s Forwn = 8, the transfer function would be 128s + 640 Gp(s)= 5s + 90s + 160s Forwn = 10, the transfer function would be 200s + 1000 Gp(s) = 5s + 110s? + 200s Forwn = 12, the transfer function would be 288s + 1440 Gp(s) = 5s 3 + 130s + 240s The correspondingSimulink model can be used to modelthe above system and test the same for open- loop, close-loop and PID controlled system unit step response. For the PID controlled system, we assume that Kp = 10, Ki = 1 and Kd = 0.1 and it is fixed for all the transfer functions.

File Edit View Display Diagram Simulation Analysis Code Tools Help iii 10.0 Normal untitled untitled D Kp = 10. Ki = 1 Kd = 0.1 8s+40 593+30527405 8s+40 553+30s2+40s Step PID(s) Transfer Fcn wn = 2 Step1 Transfer Fcn6 wn = 2 8s+40 593+30s2+40s Transfer Fcn12 wn = 2 Step2 PID Controller 32s+160 593+5052+80s Scope Open-loop Response 32s+160 593+50s2+80s Transfer Fcn1 PID(S) Scope1 close-loop Response 32s+160 593+50s2+80s WN=4 Transfer Fcn7 wn=4 PID Controller1 Transfer Fcn13 wn=4 Scope2 close-loop Response with PID control 72s+360 553-7052+1205 Transfer Fen2 wn=6 72s+360 553 70s2+120s PID(s) 72s+360 5s 347052+120s Transfer Fen8 wn=6 PID Controller2 Transfer Fcn14 wn=6 128s+640 533-9052-160s 128s+640 583-9052-160s PID(S) Transfer Fcn3 wn=8 128s+640 583-9052-160s Transfer Fcn9 wn=8 PID Controller3 Transfer Fcn15 wn=8 2005+1000 553-11092-2005 200s+1000 553-110s2-2005 PID(S) Transfer Fcn4 wn=10 Transfer Fcn10 wn=10 200s+1000 593-110s 2-2008 Transfer Fcn16 wn=10 PID Controller 4 288s+1440 593+130s24240s Transfer Fcn5 wn = 12 288s+1440 553-13052-240s PID(S) Transfer Fcn11 wn = 12 288s+1440 58313092-240s Transfer Fcn17 wn = 12 PID Controller5 Ready 100% ode45^

Open-loop response

60 Transfer Fon wn = 2 Transfer Fcnl wn=4 Transfer Fen2 wn=6 Transfer Fcn3 wn=8 Transfer Fcn4 wn=10 Transfer Fcn5 wn = 12 50 40 30 20 10 o 0 1 2 3 4 5 6 7 8 10 me offset: 0

close-loop response

1.2 - Transfer Fcn6 wn = 2 Transfer Fcn7 wn=4 - Transfer Fcn8 wn=6 Transfer Fcn9 wn=8 Transfer Fcn10 wn=10 Transfer Fcn11 wn = 12 1 0.8 0.6 0.4 0.2 0 0 2 3 4 5 6 7 8 9 10 me offset: 0

PID controlled response

1.6 Transfer Fcn12 wn 2 Transfer Fcn13 wn=4 Transfer Fcn14 wn=6 Transfer Fcn15 wn=8 Transfer Fcn16 wn=10 Transfer Fcn17 wn = 12 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 1 2 3 6 8 9 10 ime offset: 0

Now, we are varying the parameter.wa, while assume , ô = 1, n = 10 rad/sec, wa = 5, and K=1. To model this in Simulink, we first need to compute the transfer function for the different values of wo Let us assume that the variation of wjis (2, 4, 6, 8, 10, 12]. Forw = 2, the transfer function would be

Gp II 200 S + 1000 5 S^3 + 110 s^2 + 200 S

For ω_b=4, the transfer function would be

For ω_b=6, the transfer function would be

Gp = 400 S + 2000 5 S^3 + 120 s^2 + 400 s

For ω_b=8, the transfer function would be

Gp = 600 S + 3000 5 S^3 + 130 s^2 + 600 S

For ω_b=10, the transfer function would be

Gp = 800 S + 4000 5 S^3 + 140 s^2 + 800 S

For ω_b=12, the transfer function would be

Gp = 1000 S + 5000 5 S^3 + 150 s^2 + 1000 s

Gр 11 1200 S + 6000 5 S^3 + 160 s^2 + 1200 S

The corresponding Simulink model can be used to model the above system and test the same for open-loop, close-loop and PID controlled system unit step response. For the PID controlled system, we assume that Kp = 10, Ki = 1 and Kd = 0.1 and it is fixed for all the transfer functions.

File Edit View Display Diagram Simulation Analysis Code Tools Help iii 10.0 Normal untitled untitled D Kp = 10. Ki = 1 Kd = 0.1 200s +1000 553-11092-2005 200s+1000 den(s) Step PID(s) Transfer Fcn wb = 2 200s +1000 583-110s 2005 Step1 Transfer Fcn6 wb = 2 Step2 PID Controller Transfer Fcn12 wb = 2 400s+2000 593+120s2-400s Scope Open-loop Response 400s+2000 15s 3-120s2-400s Transfer Fcn1 PID(s) Scope1 close-loop Response Wb=4 Transfer Fcn7 wb=4 400s+2000 583+12082-400s Transfer Fcn13 Wb=4 PID Controller1 Scope2 close-loop Response with PID control 600s+3000 553-130s2-6005 Transfer Fen2 wb=6 600s+3000 1553-130s2-600s PID(s) 600s+3000 553-13052-600s Transfer Fen8 wb=6 PID Controller2 Transfer Fcn14 wb=6 800s+4000 1553-14052-800S 800s+4000 1593+140s2-800S PID(S) Transfer Fcn3 wb=8 800s+4000 553+ 140s2+800s Transfer Fcn9 wb=8 PID Controller3 Transfer Fcn15 wb=8 1000s+5000 553-150s2-1000s 1000s+5000 553-150s2-1000s PID(S) Transfer Fcn4 wb-10 Transfer Fcn10 1000s+5000 553-15052-1000s Transfer Fcn16 wb=10 wb=10 PID Controller 4 1200s+6000 553-16052-12005 Transfer Fcn5 wb = 12 1200s+6000 553-16052-1200s PID(S) Transfer Fcn11 wb = 12 1200s +6000 553-160521200s Transfer Fcn17 wb = 12 PID Controller5 Ready 100% ode45^

Open-loop Response

60 Transfer Fon wb = 2 Transfer Fcnl wb=4 Transfer Fen2 wb=6 Transfer Fcn3 wb=8 Transfer Fcn4 wb=10 Transfer Fcn5 wb = 12 50 40 30 20 10 0 1 2 3 5 7 8 9 10 ime offset: 0

Close-loop Response

1.2 Transfer Fcn6 wb = 2 Transfer Fen7 wb=4 Transfer Fen8 wb=6 Transfer Fcn9 wb=8 Transfer Fcn10 wb=10 Transfer Fcn11 wb = 12 1 0.8 0.6 0.4 0.2 0 1 2 3 5 7 8 9 10 ime offset: 0

PID controlled Response

1.4 Transfer Fcn12 wb = 2 Transfer Fcn13 wb=4 Transfer Fcn14 wb=6 Transfer Fcn15 wb=8 Transfer Fcn16 wb=10 Transfer Fcn17 wb = 12 1.2 12 1 0.8 0.6 0.4 0.2 0 1 2 3 5 7 8 9 10 ime offset: 0

Now, we are varying the parameter.w a while assume, 8 = 1, wn= 10 rad/sec, w = 10, and K=1. To model this in Simulink, we first need to compute the transfer function for the different values of wa Let us assume that the variation of wais [2,4,6, 8, 10, 12]. Forwa = 2, the transfer function would be GP = 500 s + 1000 2 S^3 + 50 S^2 + 200 S Forwa = 4, the transfer function would be 500 S + 2000 4 S^3 + 100 s^2 + 400 S Forwa = 6, the transfer function would be 500 S + 3000 6 S^3 + 150 s^2 + 600 S Forwa = 8, the transfer function would be

Gp = 500 S + 4000 8 S^3 + 200 s^2 + 800 S Forwa = 19 the transfer function would be Gp = 500 S + 5000 10 S^3 + 250 s^2 + 1000 s Forwa = 12 the transfer function would be 11 500 S + 6000 12 S^3 + 300 s^2 + 1200 S The correspondingSimulink model can be used to modelthe above system and test the same for open- loop.close-loop and PID controlled system unit step response. For the PID controlled system, we assume that Kp = 10, Ki = 1 and Kd =0.1 and it is fixed for all the transfer functions.

File Edit View Display Diagram Simulation Analysis Code Tools Help 10.0 Normal untitled Puntitled L Kp = 10, ki = 1 Kd = 0.1 500s+1000 25350s2+2005 500s+1000 283-508242005 Step PID(S) Transfer Fcn wa = 2 500s+1000 283-50s2+2005 Step1 Transfer Fcn6 wa = 2 Step2 PID Controller Transfer Fcn12 wa = 2 500s+2000 483+10052-400s Scope Open-loop Response 500s+2000 483+10052-400 Transfer Fcn1 PID(s) Scope1 close-loop Response 500s+2000 453-100s 2.400 Wa=4 Transfer Fcn7 wa=4 PID Controller 1 Transfer Fcn13 Wa=4 Scope close-loop Response with PID control 500s+3000 165 3+15052-600s Transfer Fen2 wa=6 500s+3000 6s3-150s2-600s PID(S) 500s+3000 6s3-150s2-600s Transfer Fen8 wa=6 PID Controller2 Transfer Fcn14 wa=6 500s+4000 1893-200s2-800S 500s+4000 893-200s2-800s PID(S) Transfer Fcn3 wa=8 500s+4000 893-200s2+800s Transfer Feng wa=8 PID Controller3 Transfer Fcn15 wa=8 500s+5000 1053 +250 2+10005 500s +5000 103+250s2+1000s PID(S) Transfer Fcn4 wa-10 Transfer Fcn10 wa=10 500s+5000 1053+25052+1000S Transfer Fcn16 wa=10 PID Controller4 500s +6000 1293-300s2+1200s Transfer Fcn5 wa = 12 500s+6000 1253+300s2+1200s PID(S) Transfer Fcn11 wa = 12 500s+6000 1253+300s2+1200s Transfer Fcn17 wa = 12 PID Controller5 ha Ready 100% ode45^

open-loop response

60 Transfer Fun wa = 2 Transfer Fcnl wa=4 Transfer Fcn2 wa=6 Transfer Fcn3 wa=8 Transfer Fcn4 wa=10 Transfer Fcn5 wa = 12 50 40 30 20 10 0 1 2 3 5 7 8 9 10 ime offset: 0

Close-loop response

1.2 Transfer Fcn6 wa = 2 Transfer Fcn7 wa=4 Transfer Fen8 wa=6 Transfer Feng wa=8 Transfer Fen10 wa=10 Transfer Fcn11 wa = 12 1 0.8 0.6 0.4 0.2 0 1 2 3 5 7 8 9 10 ime offset: 0

PID controlled response

1.4 Transfer Fcn12 wa = 2 Transfer Fcn13 wa=4 Transfer Fcn14 wa=6 Transfer Fcn15 wa=8 Transfer Fcn16 wa=10 Transfer Fcn17 wa = 12 1.2 1 0.8 0.6 0.4 0.2 0 1 2 3 4 5 6 7 8 9 10 Time offset: 0