engg4420 ‐‐ lecture

MODELS OF ELECTRIC CIRCUITSElectric circuits contain sources of electric voltage and current and other electronic elements such as resistors, capacitors, and transistors.

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An important building block for circuits is an operational amplifier (op‐amp) which is also an example of a complex feedback system.

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Passive circuits consist of interconnections of resistors, capacitors, and inductors.

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Active devices include diodes, transistors, and amplifiers.

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Electric circuit types are: passive and active•

Kirchoff's current law (KCL) states that the sum of currents leaving a junction or node equals the algebraic sum of currents entering the node.

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Kirchoff's voltage law (KVL) states that the algebraic sum of all voltages taken around a closed path in a circuit is zero.

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The basic equations of electric circuits are based on Kirchoff's laws:

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With complex circuits of many elements, it is essential to write the equations in a well organized way ‐‐ one method for doing this is based on the node analysis scheme (See Electrical circuits course).

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ENGG4420 ‐‐ LECTURE 7September‐19‐123:07 PM

CHAPTER 1 By Radu Muresan University of Guelph

MODELS OF ELECTROMECHANICAL SYSTEMS

F = B∙l∙i [Newton].i.

LAW OF MOTORS: if a current of i amperes [A] in a conductor of length l meters is arranged at right angles in a magnetic field of B Tesla [T], then there is a force on the conductor at right angles to the plane formed by i and B, with magnitude of:

a.

e(t) = B∙l∙v [Volts].i.

LAW OF GENERATORS: if a conductor of length l meters is moving in a magnetic field of B Tesla at a velocity of v [m/s] at mutually right angles, an electric voltage is established across the conductor with magnitude:

b.

Electric current and magnetic fields interact in two ways that are important in the operation of electromechanical actuators and sensors:

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EXAMPLE ‐‐MODELING A LOUDSPEAKERCommon loudspeaker structures contain a permanent magnet that establishes a radial field B in the cylindrical gap between the poles of the magnet. The force built on the bobbin (voice coil) in the gap causes movements of the bobbin (the cone) which produce sound. The effects of the air can be modeled as if the cone has equivalent mass M and a viscous friction coefficient b. Assume: B = 0.5 Tesla, and the bobbin has 20 turns at 2‐cm diameter.

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Write the equations of motion of the device.•


SOLUTION

The law of motors applies to the coil of length:•

The bobbin moves the cone•The current is at right angles to the field B, and the force of interest is at right angles to the plane i×B.

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The equation of motion follows from Newton's laws, and for a mass M and friction coefficient b, we have:

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EXAMPLE ‐‐ LOUDSPEAKER WITH CIRCUITConsider the loudspeaker to be connected to the electrical circuit (See figure below) and find the differential equations relating the input voltage va to the output core displacement x. Assume the effective circuit resistance is R and the inductance is L.

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SOLUTIONThe loudspeaker motion differential equation model:

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This motion results in an induced voltage across the coil as given by:

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Note: B×l = 0.63

We need to add the KVL (Kirchoff's Voltage Law) for the circuit to the previous analysis to link va and x:

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Equations (1) and (2) form the complete dynamic model for the loudspeaker. The final transfer function is:

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MODELING A DC MOTOR ‐‐ DC MOTOR FUNCTIONALITYA common actuator that functions in accordance with the law of motors and generators is the DC motor that is used to provide rotary motion.

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The motor equations relate the torque T on the rotor (from the law of motors) to the armature current ia and express the back emf voltage e in terms of the shaft's rotational velocity (from the law of generators).

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The non‐turning part (stator) has magnets which establish a field across the rotor ‐‐ the magnets may be electromagnets or, for small motors, permanent magnets.

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The brushes contact the rotating commutator which causes the current always to be in the proper conductor windings so as to produce maximum torque. If the direction of the current is reversed, the direction of the torque is reversed.

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The motor components are shown in figure below.•


DC MOTOR MODELING SCHEMATICThe free body diagram for the rotor, defines the positive direction and shows the two applied torques.

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Assume that the rotor has inertia Jm and viscous friction coefficient b.

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T is the external torque on the rotor that is produced by the armature current ia moving in the stator's field B;

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Application of Newton's Law (M = I∙α) to the rotor gives:•

KVL applied to the circuit loop that includes the back emf gives:

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The transfer function of the motor can be obtained from Eq. (3) and (4) by substituting s for d/dt:

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In many cases, the relative effect of the inductance is negligible compared with the mechanical motion and can be neglected in Eq. (4) as follows:

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From Eq. (6) it is clear that the effect of the back emf through Ke is indistinguishable from the friction effect through b (both coefficients are part of the θ dot term).

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AC MOTORS

Another device used for electromechanical energy conversion.

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Elementary analysis of AC motors is more complex than the analysis of the DC motors.

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Experimental curves of speed versus torque are normally used to derive motor constants that will provide the dynamic model of the AC motor.

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See Mechatronics course or mechatronics references!


LINEAR SYSTEM ANALYSIS USING MATLAB OR LABVIEWSTEP 1: write the differential equations describing the dynamic behaviour of the physical system (t‐domain).

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A ratio of two polynomials (num/den);i.As a factored zero‐pole form.ii.

The transfer function can be represented as:○

STEP 2: determine and designate inputs and outputs of the system and then compute the transfer function characterizing the input‐output behaviour of the dynamic system.

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STEP 3: analyze the transfer function to determine the quantitative and qualitative dynamic properties of the system.

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NOTES ON POLES AND ZEROS OF A TRANSFER FUNCTIONZEROS and POLES may be complex quantities and we can display their locations in a complex plane referred to as the 's‐plane', where s = σ + ω∙i;

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The system has the inherent capability to block frequencies coinciding with its zero locations;

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The zeros also have a significant effect on the transient properties of the system.

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ZEROS: the roots of the numerator are called the finite zeros of the system ‐‐ the zeros are locations in the s‐plane where the transfer function is zero.

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The poles of the transfer function determine the stability of the system;

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The poles also determine the modes of the system (such as natural or unforced behaviour).

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POLES: the roots of the denominator are the poles of the system ‐‐ these are locations in the s‐plane where the magnitude of the transfer function goes to infinity.

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The location of the poles and zeros lies at the heart of feedback control design and have significant practical implications for control system design.

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If the zeros at infinity are counted the system will have the same number of zeros and poles.

The system has (n‐m) zeros at infinity if m < n because the transfer function approaches zero as s approaches infinity;

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No physical system can have n < m, otherwise it will have an infinite response at ω = ∞;

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If zi = pi, then there are cancellations in the transfer function that can lead to undesirable system properties.

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Consider the degree of the denominator as n and the degree of the numerator as m:

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FOR REVIEW ON LAPLACE TRANSFORM CHECK CHAPTER 3 OF THE "FEEDBACK CONTROL OF DYNAMIC SYSTEMS" BOOK BY G. F. FRANKLIN ET AL. !!!


EXAMPLE ‐‐ DC MOTOR TRANSFER FUNCTION USING MATLAB

The output θm;a.The output θm ‐‐ HOMEWORK!!b.

In the DC motor example, using the transfer function given by Eq. (5) assume that: Jm = 0.01 kg∙m2; b = 0.001 N∙m∙sec; Kt = Ke = 1; Ra = 10 Ω; and La = 1 H. Find the transfer function between the input va and:

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MATLAB NOTES:>> [z, p, k] = tf2zp(numa, dena); % matlab functions >> % that converts from the polynomial form >> % numa/dena to the pole‐zero form>> [numG, denG] = zp2tf(z, p, k); % converts from the>> % zero‐pole form to the polynomial form

SOLUTION


The MATLAB results show that the zero matrix Z = [] is empty ‐‐ as a result no zeroes.

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The P matrix shows 3 poles ‐‐ as a result we can write the transfer function in the zero‐pole form as:

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CASE b ‐‐ rotational velocity is the output


SYSTEM MODELING DIAGRAMS

The transfer function of each component is placed in a box, and the input‐output relationship is indicated by lines and arrows => SYSTEM BLOCK DIAGRAM.

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In many control systems the system equations can be written so that their components do not interact except by having the input of one part be the output of another part.

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The system equation can be solved by graphical simplifications.

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EXAMPLE ‐‐ ELEMENTARY BLOCK DIAGRAMS

Figure (a) series, (b) parallel, and (c) feedback connections•

Negative feedback is usually required for system stability○The gain of a single‐loop negative feedback system is given by the forward gain divided by the sum of 1 plus the loop gain.

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When the feedback Y2(s) is subtracted => negative feedback•

SUPPLEMENTAL READING 1

ENGG4420 ‐‐ SECTION 1.3. FEEDBACK CONTROL, PID, TUNINGSeptember‐19‐126:00 PM


EXAMPLE ‐‐ BLOCK DIAGRAM ALGEBRA

NOTE. Figure(c) shows how the manipulations can be used to convert a general system (on the left) to a system without a component in the feedback path, usually referred to as "a unity feedback system".

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In all cases, the basic principle is to simplify the topology while maintaining the same relationship among the remaining variables of the block diagram.

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In relation to the algebra of the underlying linear equations, block diagram reduction is a pictorial way to solve equations by eliminating variables.

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EXAMPLES ‐‐ TRANSFER FUNCTIONS FROM BLOCK DIAGRAMS

,42

4242

1

42

)(

)()(

2

2

2

ss

s

ss

ss

sR

sYsT

EXAMPLE 1 ‐‐

EXAMPLE 2 ‐‐‐

,1

11

1

)(

)()(

42131

61521

2

65

31

421

31

21

GGGGG

GGGGG

G

GG

GGGGG

GGGG

sR

sYsT


DEFINITION OF A SYSTEM TYPEA stable system can be classified as a system type, defined to be the degree of the polynomial for which steady‐state system error is a nonzero finite constant.

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Example: if in a speed control system we use proportional control the system has a constant finite error to a step input => type 0 system; if the error to a ramp or first‐degree polynomial is a finite nonzero constant => type 1 system; and so on.

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System type can be defined with regards to either reference inputs or disturbance inputs.

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EFFECTS OF POLE LOCATIONS


TIME DOMAIN SPECIFICATIONSSpecifications for a control system design involve certain requirements associated with the time response of the system.

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Rise time tr; Settling time ts; The overshoot Mp; the peak time tp.

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The requirement for a step response are expressed in terms of:

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THE FINAL VALURE THEOREM OF THE LAPLACE TRANSFORM

If all poles are in the left half except for one at s = 0, then all terms will decay to zero except for the s = 0 term and that term corresponds to a constant in time => final value is given by the coefficient associated to the pole at s = 0.

Constant ○

If Y(s) has a pair of poles on the imaginary axis => y(t) will contain a sinusoid that persists forever => undefined final value

Undefined○

If Y(s) has poles (i.e. Denominator roots) in the right hand of the s‐plane (real part pi > 0) => y(t) will grow => unbound

Unbound○

Allows to compute the constant steady‐state value of a time function y(t) given the Laplace transform Y(s) ‐‐There are 3 possible final values:

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);(lim)(lim0

ssYtyst

THEOREM. The final value is given by the pole at s = 0. If all poles of sY(s) are in the left half of the s plane then:

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);(lim1

)(lim00

sGs

ssGDCgainss

DC gain for a unit‐step input U(s) = 1/s is:•


FEEDBACK CONTROL

Given a model, the next step in the design is formulation of control specifications; static and dynamic requirements such as:

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The permissible steady‐state error in the presence of a constant or bias disturbance signal;

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The permissible steady‐state error while tracking a polynomial reference signal such as a step or a ramp;

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The sensitivity of the system transfer function to changes in model parameters;

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The permissible transient error in response to a step in either the reference or the disturbance input.

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FEEBACK CONTROL

Control structures: open loop; and closed loop.•


CLOSED LOOP ‐‐ BASIC FEEDBACK CONTROL STRUCTURE

u

EQUATIONS FOR THE OUTPUT AND THE CONTROLIt is standard practice, especially if Hy is constant, to select equal scale factors so that Hr = Hy and the block diagram can be drawn as a unity feedback structure.

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UNITY FEEDBACK SYSTEM

,111

,111

VDG

DW

DG

DGR

DG

DU

VDG

DGW

DG

GR

DG

DGYcl

.111

1

111V

DG

DGW

DG

GR

DGV

DG

DGW

DG

GR

DG

DGREcl

NOTE: the above block diagram and equations apply for Hy = constant. If the sensor has dynamics that can't be ignored then different equations need to be used.


Ecl = R ‐ Ycl;○In the unity feedback system above the error is:•

The closed loop error Ecl equation can be simplified by defining the sensitivity function S and its complementary function T as:

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S = 1/(1+D∙G); and T = 1‐S = D∙G/(1+D∙G);As a result, Ecl = S∙R ‐ S∙G∙W + T∙V;

It is standard to define the transfer function around a loop as the loop gain L(s).

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ADVANTAGES OF FEEDBACK CONTROLSystem error to constant disturbances can be made smaller with feedback in comparison with the open loop systems by a factor of S = 1/(1 + DG(0)), where DG(0) is the loop gain at s = 0;

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In feedback control, the error in the overall transfer function gain is less sensitive to variations in the plant gain by a factor of S = 1/(1 + DG) compared with errors in the open‐loop gain;

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Feedback changes dynamic response and often makes a system both faster and less stable.

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END OF THE SUPLEMENTAL READING !!!!


THE PID CONTROLLERA well known structure for the controller is proportional + integral + derivative (i.e., PID) control.

•

,)(

)()(dt

tdekdekektu D

t

Ip

The corresponding transfer function in s domain is:•

.)( sks

kksD D

Ip

,1

1)(

sT

sTksD D

IPc

Increasing the proportional feedback gain reduces steady‐state errors, but high gains almost always destabilize the system.

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Integral control provides robust reduction in steady‐state errors, but often makes the system less stable.

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Derivative control increases damping and improves stability.•

ENGG4420 ‐‐ LECTURE 8September‐23‐1211:08 PM


In a number of important cases the reference input will not be constant but can be approximated as a polynomial in time long enough to the system to effectively reach steady state.

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TI is called the "reset rate" in seconds;○TD is called the "derivative rate" in seconds○

In Eq. (3), •

The effect of the derivative control term depends on the rate of change of the error. Because of the sharp effect of the derivative control on suddenly changing signals, the "D" term is sometimes introduced in the feedback path (Ex. a tachometer on a shaft of a motor).

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EXAMPLE ‐‐ P, PI, PID CONTROL OF A DC MOTOR SPEEDUsing MATLAB plot the DC motor speed control response to steps in disturbance and steps in reference input. Use the general transfer function or the simplified one.

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Jm = 1.13∙10‐2 N‐m‐sec2/rad; b = 0.028 N‐m‐sec/rad; •La = 10‐1 Henry; Ra = 0.45 Ω; •Kt = 0.067 N‐m/amp; Ke = 0.067 V‐sec/rad.•

Consider the DC motor speed control parameters as:•

kP = 3; kI = 15 sec‐1; kD = 0.3 sec.•Consider the controller parameters as:•

NOTE: In this case the plant transfer function must be speed/input


SOLUTION TO EXAMPLESYSTEM SOLUTION

P control does not eliminate the steady‐state error○PI control increases the oscillatory behaviour but eliminates the steady‐state error

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PID control reduces the oscillation while maintaining zero steady‐state error.

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The transfer function from the disturbance input W(s) to the error E(s) is E(s)/W(s) = Tw(s) for reference R = 0. System type with regards to disturbances relates to the polynomial expressing the disturbances that the system can reject in the stated state.

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In our case we need the transfer function between the motor input va and the output speed w as the plant transfer function G.

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NOTE: In Eq. (4) we used the general system transfer function between disturbance and error given by Eq. (3) where we substituted for G as the motor (plant) that outputs the speed as a function of va and D as a proportional controller (P).


Similarly for the PI and PID controllers the transfer function Tw(s) is given below in Eq. (5) and (6):


engg4420 ‐‐ lecture

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