ENGG4420 ‐‐ LECTURE 7
September‐21‐10
2:29 PM
•
•
•
•
MODELS OF ELECTRIC CIRCUITS
Electric circuits contain sources of electric voltage and current and other electronic elements such as resistors, capacitors, and transistors. An important building block for circuits is an operational amplifier (op‐amp) which is also an example of a complex feedback system.
Electric circuit types are: passive and active
○ Passive circuits consist of interconnections of resistors, capacitors, and inductors;
○ Active devices include diodes, transistors, and amplifiers.
The basic equations of electric circuits are based on Kirchoff's laws:
○ 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.
○ Kirchoff's voltage law (KVL) states that the algebraic sum of all voltages taken around a closed path in a circuit is zero.
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).
CHAPTER 1 BY RADU MURESAN Page 1
MODELS OF ELECTROMECHANICAL SYSTEMS
• Electric current and magnetic fields interact in two ways that are important in the operation of electromechanical actuators and sensors:
a. 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 F = B∙l∙i Newton.
b. 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 e(t) = B∙l∙v Volts.
EXAMPLE ‐‐ MODELING A LOUDSPEAKER
• Next figure shows a typical geometry for a loudspeaker. The permanent magnet establishes a radial field in the cylindrical gap between the poles of the magnet. The force on the conductor wound on the bobbin causes the voice coil to move, producing sound. The effects of the air can be modeled as if the cone had equivalent mass M and a viscous friction coefficient b. Assume that the magnet establishes a uniform field B = 0.5 Tesla and the bobbin has 20 turns at 2‐cm diameter.
• Write the equations of motion of the device.
CHAPTER 1 BY RADU MURESAN Page 2
SOLUTION
• The law of motors applies to the bobbin 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 ixB.
• The equation of motion follows from Newton's laws, and for a mass M and friction coefficient b, we have:
CHAPTER 1 BY RADU MURESAN Page 3
EXAMPLE ‐‐ LOUDSPEAKER WITH CIRCUIT
• For the loudspeaker presented above and the electrical circuit driving it (See figure below), 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.
SOLUTION
• The loudspeaker motion satisfies the differential equation:
• This motion results in an induced voltage across the coil as given by:
• We need to add the KVL (Kirchoff's Voltage Law) for the circuit to the previous analysis to link va and x:
• Equations (1) and (2) form the complete dynamic model for the loudspeaker. The final transfer function is:
CHAPTER 1 BY RADU MURESAN Page 4
•
•
•
•
•
MODELING A DC MOTOR ‐‐ DC MOTOR FUNCTIONALITY
A common actuator based on the law of motors and generators is the DC motor that is used to provide rotary motion. The motor components are shown in figure below
The motor equations give the torque T on the rotor (from the law of motors) in terms of the armature current ia and express the back emf voltage e in terms of the shaft's rotational velocity (from the law of generators).
The nonturning part (stator) has magnets which establish a field across the rotor ‐‐ the magnets may be electromagnets or, for small motors, permanent magnets.
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.
CHAPTER 1 BY RADU MURESAN Page 5
DC MOTOR MODELING SCHEMATIC
• The free body diagram for the rotor, defines the positive direction and shows the two applied torques.
• Assume that the rotor has inertia Jm and viscous friction coefficient b.
• T is the external torque on the rotor that is produced by the armature current ia moving in the stator's field B;
• Application of Newton's Law (M = I∙α) to the rotor gives:
• KVL applied to the circuit loop that includes the back emf gives:
CHAPTER 1 BY RADU MURESAN Page 6
• The transfer function of the motor can be obtained from Eq. (3) and (4) by substituting s for d/dt:
• 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:
• 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).
CHAPTER 1 BY RADU MURESAN Page 7
AC MOTORS
• Another device used for electromechanical energy conversion.
• Elementary analysis of AC motors is more complex than the analysis of the DC motors.
• Experimental curves of speed versus torque are normally used to derive motor constants that will provide the dynamic model of the AC motor.
See Mechatronics course!
CHAPTER 1 BY RADU MURESAN Page 8
LINEAR SYSTEM ANALYSIS USING MATLAB OR LABVIEW
• STEP 1: write the differential equations describing the dynamic behaviour of the physical system (t‐domain).
• 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.
○ The transfer function can be represented as:
i. A ratio of two polynomials (num/den);
ii. As a factored zero‐pole form.
• STEP 3: analyze the transfer function to determine the quantitative and qualitative dynamic properties of the system.
NOTES ON POLES AND ZEROS OF A TRANSFER FUNCTION
• ZEROS 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;
• 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.
○ The system has the inherent capability to block frequencies coinciding with its zero locations;
○ The zeros also have a significant effect on the transient properties of the system.
CHAPTER 1 BY RADU MURESAN Page 9
• 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.
○ The poles of the transfer function determine the stability of the system;
○ The poles also determine the modes of the system (such as natural or unforced behaviour).
• The location of the poles and zeros lies at the heart of feedback control design and have significant practical implications for control system design.
• Consider the degree of the denominator as n and the degree of the numerator as m:
○ The system has (n‐m) zeros at infinity if m < n because the transfer function approaches zero as s approaches infinity;
 If the zeros at infinity are counted the system will have the same number of zeros and poles.
○ No physical system can have n < m, otherwise it will have an infinite response at ω = ∞;
○ If zi = pi, then there are cancellations in the transfer function that can lead to undesirable system properties.
FOR REVIEW ON LAPLACE TRANSFORM CHECK CHAPTER 3 OF THE "FEEDBACK CONTROL OF DYNAMIC SYSTEMS" BOOK BY G. F. FRANKLIN ET AL. !!!
CHAPTER 1 BY RADU MURESAN Page 10
EXAMPLE ‐‐ DC MOTOR TRANSFER FUNCTION USING MATLAB
• 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:
a. The output θm;
b. The output θm ‐‐ HOMEWORK!!
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
CHAPTER 1 BY RADU MURESAN Page 11
• The MATLAB results show that the zero matrix Z = [] is empty ‐‐ as a result no zeroes. • The P matrix shows 3 poles ‐‐ as a result we can write the transfer function in the zero‐pole form as:
CHAPTER 1 BY RADU MURESAN Page 12
CASE b ‐‐ rotational velocity is the output
CHAPTER 1 BY RADU MURESAN Page 13