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Lecture 5: Motors and Control of Motorized Processes.

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1 Lecture 5: Motors and Control of Motorized Processes

2  DC or AC (or Universal!)  All Motors Convert Electric Power to Mechanical Power via Magnetic Fields  All motor applications adhere to the Equation:  T = P / ω  DC motors and AC Synchronous motors exploit the Lorentz Force  F = I L X B  AC Induction Motors exploit Faraday’s Law of Induction  V = - N * d Ф /dt




6  Armature is compelled to rotate via the Lorentz force interaction between current in armature coil and B field of permanent magnet  But, according to Faraday, a coil rotating in a magnetic field will develop an induced voltage (emf):  Vind = - N d Ф /dt  This is called the ‘back emf’ or ‘counter emf’ and it opposes (Lenz’s Law) the forward rotation produced by the Lorentz Force  This also allows the motor to become a generator

7  Electrically (for Permanent Magnet DC): Te = KbФIa, but the flux, Ф, is constant in PM DC motor So Kt = Kb* Ф Ф =>

8  Any motor can be converted into a Servo- motor by introducing a feedback loop to control shaft position, angular velocity and angular acceleration  Shaft position, Ө, is usually the feedback variable, measured using an Encoder incremental Absolute

9  A closed loop (feedback) control system, generally with Single Input-Single Output (SISO)  A portion of the signal being fed back is:  Proportional to the signal ( P )  Proportional to integral of the signal ( I )  Proportional to the derivative of the signal ( D )

10  PID control works well on SISO systems of 2 nd Order, where a desired Set Point can be supplied to the system control input  PID control handles step changes to the Set Point especially well:  Fast Rise Times  Little or No Overshoot  Fast settling Times  Zero Steady State Error  PID controllers are often fine tuned on-site, using established guidelines

11  Consider a DC Motor turning a Load:  Shaft Position, Theta, is proportional to the input voltage


13  Mechanically:

14 Torque is Conserved: Tm = Te 1 and 2 above are a basis for the state-space description

15  So, the “plant”,G(s) = K / (s 2 + 2as + b 2 )  Where a = damping factor, b = undamped freq.  And a Feedback Control System would look like:

16  A 2 nd Order SISO System with an Reference Input that Controls Shaft Position:

17  Consider the block diagram shown:  C(s) could also be second order….(PID)


19  Consider the input error variable, e(t):  Let p(t) = Kp*e(t) {p proportional to e (mag)}  Let i(t) = Ki* ∫ e(t)dt {i integral of e (area)}  Let d(t) = Kd* de(t)/dt {d derivative of e (slope)} AND let Vdc(t) = p(t) + i(t) + d(t) Then in Laplace Domain: Vdc(s) = [Kp + 1/s Ki + s Kd] E(s)

20 Let C(s) = Vdc(s) / E(s) (transfer function) C(s) = [Kp + 1/s Ki + s Kd] = [Kp s + Ki + Kd s 2 ] / s (2 nd Order) THEN C(s)G(s) = K [Kd s 2 + Kp s + Ki] s(s 2 + 2as + b 2 ) AND Y/R = Kd s 2 + Kp s + Ki s 3 + (2a+Kd)s 2 + (b 2 +Kp) s + Ki

21  Kd has direct impact on damping  Kp has direct impact on resonant frequency In General the effects of increasing parameters is: Parameter: Rise Time Overshoot Settling Time S.S.Error Kp Decrease Increase Small Change Decrease Ki Decrease Increase Increase Eliminate Kd Small Change Decrease Decrease None

22  There is a fairly standard procedure for tuning PID controllers  A good first stop for tuning information is Wikipedia: 

23  In noisy environments or with energy intensive processes it may be desirable to make the controller unresponsive to small changes in input or feedback signals  A deadband is an area around the input signal set point, wherein no control action will occur


25  Given a continuous, linear time domain description of a process, it is possible to approximate the process with Difference Equations and implement in software  Time Step size (and/or Sampling Rate) is/are critical to the accuracy of the approximation

26  Definition of Derivative: dU = lim U(t + Δ t) – U(t) dt Δ t  0 Δ t  Algebraically Manipulate to Difference Eq: U(t + Δ t) = U(t) + Δ t*dU dt (for sufficiently small Δ t)  Apply this to Iteratively Solve First Order Linear Differential Equations (hold for applause)

27  Consider the following RC Circuit, with 5 Volts of initial charge on the capacitor:  KVL around the loop: -Vs + Ic*R + Vc = 0, Ic = C*dVc/dt OR dVc/dt = (Vs –Vc)/RC

28  Differential Equation: dVc/dt = (Vs –Vc)/RC  Difference Equation by Definition: Vc(kT+T) = Vc(kT) + T*dVc/dt  Substituting: Vc(kT+T) = Vc(kT) + T*(Vs –Vc(kT))/RC

29 R=1000 C=1e-4 Vs=10 Vo=5 //Initial Value of Difference Equation (same as Vo) Vx(1)=5 //Time Step dt=.01 //Initialize counter and time variable i=1 t=0 //While loop to calculate exact solution and difference equation while i<101, Vc(i)=Vs+(Vo-Vs)*exp(-t/(R*C)), Vx(i+1)=Vx(i)+dt*(Vs-Vx(i))/(R*C), t=t+dt, i=i+1, end


31  Definition of Integration (area under curve):  Approximation by Trapezoidal Areas

32 //Calculate and plot X=5t and integrate it with a Trapezoidal approx. //Time Step dt=.01 //Initialize time and counter t=0 i=2 //Initialize function and its trapezoidal integration function X(1)=0 Y(1)=0 //Perform time step calculation of function and trapezoidal integral while i<101,X(i)=5*t,Y(i)=Y(i-1)+dt*X(i-1)+0.5*dt*(X(i)-X(i-1)), t=t+dt, i=i+1, end //Plot the results plot(X) plot(Y)


34  Using Difference Equations, it is possible now to code the PID algorithm in a high level language p(t) = Kp*e(t)  P(kT) = Kp*E(kT) i(t) = Ki* ∫ e(t)dt  I(kT+T) = Ki*[I(kT)+T*E(kT+T)+.5(E(kT+T)-E(kT))] d(t) = Kd* de(t)/dt  D(kT+T) = Kd*[E(kT+T)-E(kT)]/T

35 State-Space Description of the DC Motor: 0. θ ’ = ω (angular frequency) 1. J θ ’’ + B θ ’ = KtIa  ω ’ = -B ω /J + KtIa/J 2. LaIa’ + RaIa = Vdc - Ka θ ’  Ia’ = -Ka ω /La –RaIa/La +Vdc/La In Matrix Form:


37 //PID position control of permanent magnet DC motor //Constants Ra=1.2;La=1.4e-3;Ka=.055;Kt=Ka;J=.0005;B=.01*J;Ref=0;Kp=5;Ki=1;Kd=1 //Initial Conditions Vdc(1)=0;Theta(1)=0;Omega(1)=0;Ia(1)=0;P(1)=0;I(1)=0;D(1)=0;E(1)=0 //Time Step (Seconds) dt=.001 //Initialize Counter and time i=1;t(1)=0 //While loop to simulate motor and PID difference equation approximation while i<1500, Theta(i+1)=Theta(i)+dt*Omega(i), Omega(i+1)=Omega(i)+dt*(-B*Omega(i)+Kt*Ia(i))/J, Ia(i+1)=Ia(i)+dt*(-Ka*Omega(i)-Ra*Ia(i)+Vdc(i))/La, E(i+1)=Ref-Theta(i+1), P(i+1)=Kp*E(i+1), I(i+1)=Ki*(I(i)+dt*E(i)+0.5*dt*(E(i+1)-E(i))), D(i+1)=Kd*(E(i+1)-E(i))/dt, Vdc(i+1)=P(i+1)+I(i+1)+D(i+1), //Check to see if Vdc has hit power supply limit if Vdc(i+1)>12 then Vdc(i+1)=12 end t(i+1)=t(i)+dt, i=i+1, //Call for a new shaft position if i>5 then Ref=10 end


39  Phillips and Nagle, Digital Control System Analysis and Design, Prentice Hall, 1995, ISBN- 10: 013309832X WAKE UP!!

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