Modeling, investigating and design control system of the direct curent motor

Trang 1 Hanoi University of IndustryFaculty: Mechatronics---MAJOR ASSIGNMENT REPORT OF INTRODUCTION TOAUTOMATIC CONTROL SYSTEMSMODELING, INVESTIGATING AND DESIGN CONTROL SYSTEM OF THE DI

lOMoARcPSD|39211872 Hanoi University of Industry Faculty: Mechatronics - MAJOR ASSIGNMENT REPORT OF INTRODUCTION TO AUTOMATIC CONTROL SYSTEMS MODELING, INVESTIGATING AND DESIGN CONTROL SYSTEM OF THE DIRECT CURENT MOTOR Supervisor: Master Bui Huy Anh Student: Do Manh Dung Class: ME6159.TA1 - Course 17 Student ID: 20226006516 Hanoi - 2023 Downloaded by mon hon (monmon1@gmail.com) lOMoARcPSD|39211872 INDEX INTRODUCTION .2 SECTION 1: BRIEFT OVERVIEW OF THE DESIGNED SYSTEM 3 1.1: System modeling .3 1.2: Transfer function .4 1.3: The stability of the control system 5 1.4: The system respone 6 SECTION 2: THE INFLUENCE OF PARAMETERS ON SYSTEM QUALITY 8 2.1: The influence of electric inductance L on system quality 8 2.2: The influence of electric resistance R on system quality 10 2.3: The influence of Moment of inertia of the rotor J on system quality 11 SECTION 3: DESIGN THE PHASE DELAY CONTROLLER FOR THE SYSTEM 13 3.1: Phase delay controller .13 3.2: Investigate the dependence of position control quality on controller parameters 14 3.3: Investigate the influence of a to the quality of the control system 15 3.4: Investigate the influence of Kc to the quality of the control system.17 3.5: Investigate the influence of T to the quality of the control system 18 IN CONCULSION 20 1 Downloaded by mon hon (monmon1@gmail.com) lOMoARcPSD|39211872 INTRODUCTION The direct current motor is a peripheral device that is widely used because of its simple control and affordable price A direct current motor DC (DC is stand for “Direct Current Motors”) is a mortor controlled bt a current with a fixed direction of in other words, this is a type of motor that runs on a DC voltage source – a direct voltage (Different from AC alternating voltage) The ouput wires of the motor usually consist of two wires (power wire – VCC and ground wire – GND) DC motor is a direct curent motor with continuous rotation function DC motors come in many types, each with different specific functions Around us, small DC motors are used in tools, toys, and various household appliances In industry, applications of DC motors include conveyors and turntables, … using large power DC motors in applications such as brakes and reversing… We have some examples such as DC motor in fan, hydraulic pumps, direct current DC motor used in the robot manufacturing industry This article aims to simulate and design the control system through the application of DC motor 2 Downloaded by mon hon (monmon1@gmail.com) lOMoARcPSD|39211872 SECTION 1: BRIEFT OVERVIEW OF THE DESIGNED SYSTEM 1.1: System modeling In general, the torque generated by a DC (direct current) motor (động cơ) is proportional to the armature current and the strength of the magnetic field In this example we will assume that the magnetic field is constant and, therefore, that the motor torque is proportional to only the armature current i by a constant factor Kt as shown in the equation below This is referred to as an armature- controlled motor (1) T =Kt i The back emf, e, is proportional to the angular velocity of the shaft by a constant factor Ke (2) e=K e θ In SI units, the motor torque and back emf constants are equal, that is, Ke=Kt=K ; therefore, we will use K to represent both the motor torque constant and the back emf constant From the figure above, we can derive the following governing equations based on Newton's 2nd law and Kirchhoff's voltage law (3) J θ+b θ=T =Ki (2.1) 3 Downloaded by mon hon (monmon1@gmail.com) lOMoARcPSD|39211872 di (4) L + Ri=V L+V R=V −e=V −K θ (2.2) dt Applying the Laplace transform, the above modelling equations can be expressed in terms of the Laplace variable s di θ(t)⇒ Θ (s)θ(t)⇒ sΘ( s)θ(t)⇒ s2Θ( s)i(t)⇒ I (s )dt ⇒sI (s) (2.1) J θ+b θ=T =Ki ⇒ J s2Θ(s)+bsΘ(s)=KI (s ) J s2Φ ⇒ I (s )= ( s )+ bs Φ( s) (2.3) K di (2.2) L + Ri=V −e=V −K θ ⇒ LsI (s)+ RI (s)=V ( s)−KsΘ(s) (2.4) dt Replace (2.3) to (2.4): ( R+ Ls ) J s2Φ ( s)+bs Φ(s) =V (s )−Ks Φ( s) K ⇔ V ( s )=Φ( s) ( R+ Ls ) (J s2+bs )+K2 s K 1.2: Transfer function G(s)= Θ( s) V (s) We arrive at the following open loop transfer function by eliminating I(s) between the two above equations However, during this example we will be looking at the position as the output G(s)= Θ( s) V ( s) = s ((Js+ b)( Ls+R)+ K2 K ) (rad / s) 4 Downloaded by mon hon (monmon1@gmail.com) lOMoARcPSD|39211872 From the concept: 𝑅 = 1Ω, 𝐿 = 0.5𝐻, 𝑏 = 0.1𝑁𝑚𝑠, (K=Ke=Kt) = 0,01 Nm/Amp (J) = 0,01 kg.m^2/s^2 ⇒G (s)= 0.01 0.005 s +0.06 s +0.1001 s3 2 Because cubic function decline very quickly to quadratic function, so that the transfer function can be considered as follows: G (s)= 0.01 2 0.005 s + 0.06 s +0.1001 1.3: The stability of the control system Matlab is used to enter the parameters and transfer function just found, type the codes below in m-file and run in Command Window: J=0.01 b=0.1 K=0.01 R=1 L=0.5 num=K dem=[(J*L) ((b*L)+(J*R)) ((b*R)+(K*K))]; G=tf(num,dem) bode(G) margin(G) The bode plot is shown below ⇒ The Gm is infinity value, which means that the amplitude of the system does not exceed the steady stae It is an indicator of the maximum amplitude that the system can achieve before becoming unstable Pm is infinity value, due to the phase of the system never meet -180 degrees, which is positive because a phase near -180 degrees can leed to unstable 5 Downloaded by mon hon (monmon1@gmail.com) lOMoARcPSD|39211872 1.4: The system respone The following codes are used in Matlab to find the system respone diagram: num= [0.01] dem= [0.005 0.06 0.1001] G=tf (num, dem) t=0:0.1:10 step (G, t) grid on The transfer function is shown below: 6 Downloaded by mon hon (monmon1@gmail.com) lOMoARcPSD|39211872 step(G, 0:0.1:5) From the diagram below, we can derive the parameters as:  Peak response: 0.09%  Settling time: 2.07s  Rise time: 1.14s  Steady state: infinity ⇒ The motor DC can’t meet the system respone requirements because of the large steady state 7 Downloaded by mon hon (monmon1@gmail.com) lOMoARcPSD|39211872 SECTION 2: THE INFLUENCE OF PARAMETERS ON SYSTEM QUALITY 2.1: The influence of electric inductance L on system quality The following codes are used in Matlab J=0.01 ((b*R)+(K*K))] b=0.1 K=0.01 R=1 L=0.5 t=(0:0.1:10) num=K dem=[(J*L) ((b*L)+(J*R)) G=tf(num,dem) step(G,t) hold on grid on L=1 num=K dem=[(J*L) ((b*L)+(J*R)) ((b*R)+(K*K))] G=tf(num,dem) step(G,t) L=2 num=K dem=[(J*L) ((b*L)+(J*R)) ((b*R)+(K*K))] G=tf(num,dem) step(G,t) L=3 num=K dem=[(J*L) ((b*L)+(J*R)) ((b*R)+(K*K))] G=tf(num,dem) step(G,t) L=4 num=K 8 Downloaded by mon hon (monmon1@gmail.com) lOMoARcPSD|39211872 dem=[(J*L) ((b*L)+(J*R)) ((b*R)+(K*K))] G=tf(num,dem) step(G,t) L=5 num=K dem=[(J*L) ((b*L)+(J*R)) ((b*R)+(K*K))] G=tf(num,dem) step(G,t) legend('L=0.5','L=1','L=2','L=3','L=4','L=5') The diagram is created by the codes above Comment: When L increases from 0.5 to 5:  Peak respone does not change  Steady state is reduced  Settling time is increasing  Rise time is increased ⇒Increasing L reduces the error, but the system response time increases a lot 9 Downloaded by mon hon (monmon1@gmail.com) lOMoARcPSD|39211872 2.2: The influence of electric resistance R on system quality The following codes are used in Matlab J=0.01 ((b*R)+(K*K))] b=0.1 K=0.01 R=1 L=0.5 t=(0:0.1:10) TS=K MS=[(J*L) ((b*L)+(J*R)) G=tf(TS,MS) step(G,t) hold on grid on R=5 num=K dem=[(J*L) ((b*L)+(J*R)) ((b*R)+(K*K))] G=tf(num,dem) step(G,t) R=10 num=K dem=[(J*L) ((b*L)+(J*R)) ((b*R)+(K*K))] G=tf(num,dem) step(G,t) R=15 num=K dem=[(J*L) ((b*L)+(J*R)) ((b*R)+(K*K))] G=tf(num,dem) step(G,t) R=20 num=K dem=[(J*L) ((b*L)+(J*R)) ((b*R)+(K*K))] G=tf(num,dem) step(G,t) 10 Downloaded by mon hon (monmon1@gmail.com) lOMoARcPSD|39211872 R=25 num=K dem=[(J*L) ((b*L)+(J*R)) ((b*R)+(K*K))] G=tf(num,dem) step(G,t) legend('R=0.5','R=5','R=10','R=15','R=20','R=25') The diagram is created by the codes above Comment:  Peak respone is reduced  Steady state is reduced  Settling time is reduced  Rise time is reduced ⇒The system takes less time to be stable 2.3: The influence of Moment of inertia of the rotor J on system quality J=0.01; 11 Downloaded by mon hon (monmon1@gmail.com) lOMoARcPSD|39211872 b=0.1; ((b*R)+(K*K)) K=0.01 R=1 L=0.5 t=(0:0.1:10) num=K dem=[(J*L) ((b*L)+(J*R)) G=tf(num,dem) step(G,t) hold on grid on J=0.001 num=K dem=[(J*L) ((b*L)+(J*R)) ((b*R)+(K*K))] G=tf(num,dem) step(G,t) J=0.01 num=K dem=[(J*L) ((b*L)+(J*R)) ((b*R)+(K*K))] G=tf(num,dem) step(G,t) J=0.1 num=K dem=[(J*L) ((b*L)+(J*R)) ((b*R)+(K*K))] G=tf(num,dem) step(G,t) J=0.2 num=K dem=[(J*L) ((b*L)+(J*R)) ((b*R)+(K*K))] G=tf(num,dem) step(G,t) J=0.3 num=K dem=[(J*L) ((b*L)+(J*R)) ((b*R)+(K*K))] G=tf(num,dem) step(G,t) legend('J=0.001','J=0.01','J=0.1','J=0.2','J=0.3','J=0.4') 12 Downloaded by mon hon (monmon1@gmail.com) lOMoARcPSD|39211872 The diagram is created by the codes above In conculsion, we can derive the change in the moment of inertia J of the rotor does not seem to affect the peak response nor does improve the steady state of the system However, as the moment of inertia J becomes larger, the rise time and the response time is longer ⇒ The rise time and the response time of the DC motor system is proportional to the moment of inertia of the rotor SECTION 3: DESIGN THE PHASE DELAY CONTROLLER FOR THE SYSTEM 3.1: Phase delay controller The phase delay controller is a low-pass filter Using the phase delay stage will narrow the system’s bandwidth, causing the system’s gain coefficient for high frequency input signals to decrease, due to the phase delay stage having no effect on the respone 13 Downloaded by mon hon (monmon1@gmail.com) lOMoARcPSD|39211872 Because of reducing the gain coefficient in the high frequency domanin, the phase delay has the effect of filrtering high frequency noise that affects the system Because the gain coefficient in the low frequency domain is large, the phase delay stage reduces the system’s steady state The transfer function of the phase delay stage is written as: Therefore, the structure of the phase delay stage as below: The parameters of the phase delay controller that need to be investigated are Kc, T, and α The following codes are used in Matlab J=0.01 b=0.1 K=0.01 R=1 L=0.5 a=-1 T=100 Kc=8 lag = tf(Kc*[T*a 1],[T 1]) num=K dem=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2) ] G=tf(num,dem) Gk =feedback(lag*G,1) step(Gk) 14 Downloaded by mon hon (monmon1@gmail.com) lOMoARcPSD|39211872 From the diagram, the parameters are identified below:  Peak respone: -2.75 (5.58s)  Steady state: 0.44  Settling time: 38.8s 3.2: Investigate the influence of a to the quality of the control system Kc=8 and T=100 are kept the same , a is changed as follows: 0; -0.2; -0.4; -0.6; - 0.8; -1 The following codes are excuted as: J=0.01 b=0.1 K=0.01 R=1 L=0.5 a=-1 T=100 Kc=8 hold on for index = 0:-0.2:-1 a = index lag = tf(Kc*[T*a 1],[T 1]) 15 Downloaded by mon hon (monmon1@gmail.com) lOMoARcPSD|39211872 num=K dem=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2)] G =tf(num,dem) Gk =feedback(lag*G,1) step(Gk) end legend('a= 0','a= -0.2','a= -0.4','a=-0.6','a=-0.8','a=-1') Comment: a is increased from -1 to 0  Overshoot = 0  The settling value are all equal to 0.444  Settling time increased from 38.8s to 217s  Time gradually increases from 15.2s to 122s  Increasing a causes the system to quickly meet the steady state 16 Downloaded by mon hon (monmon1@gmail.com) lOMoARcPSD|39211872 3.3: Investigate the influence of Kc to the quality of the control system a=-1 and T=100 are kept the same , Kc is changed as follows: 0; 2; 4; 6; 8; 10 The following codes are excuted as: J=0.01 b=0.1 K=0.01 R=1 L=0.5 a=-1 T=100 Kc=0 hold on for index = 0:2:10 Kc = index lag = tf(Kc*[T*a 1],[T 1]) num=K dem=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2)] G=tf(num,dem) Gk =feedback(lag*G,1) step(Gk) end legend('Kc= 0','Kc= 2','Kc= 4','Kc= 6','Kc= 8','Kc= 10') 17 Downloaded by mon hon (monmon1@gmail.com) lOMoARcPSD|39211872 Comment:  When increasing Kc from 2 to 8, the settling time decreases from 263 to 38.8, but when changing Kc=10, the settling time increases to 886  The settling value increases from 0 to 0.5 when changing Kc from 0 to 10 3.4: Investigate the influence of T to the quality of the control system a=-1 and Kc=8 are kept the same, T is changed as follows: 0; 20; 40; 60; 80; 100 The following codes are excuted as: J=0.01 b=0.1 K=0.01 R=1 L=0.5 a=-1 T=0 Kc=8 hold on 18 Downloaded by mon hon (monmon1@gmail.com) lOMoARcPSD|39211872 for index = 0:20:100 T = index lag = tf(Kc*[T*a 1],[T 1]) num=K dem=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2)] G=tf(num,dem) Gk =feedback(lag*G,1) step(Gk) end legend('T= 0','T= 20','T= 40','T= 60','T= 80','T= 100') Comment: when increasing T from 0 to 100 the settling value is constant and equal to 0.444  Time increases  The system quickly meets the steady state 19 Downloaded by mon hon (monmon1@gmail.com)