Control System Analysis, Electromechanical Systems

Solved Example 2 – Electromechanical system transfer function.

Find in generic terms, the transfer function of the unit feedback system shown in Figure P5.52 (b) of which the electromechanical system of Figure P5.52 (a) is a part.

1. System Dynamic

where:

2. Laplace Transform

3. Motor&Load Transfer Function (θm(s)/Ea(s))

4. Direct Transfer Function

For the system:

The open-loop transfer function Ga(s) is:

5. Closed-loop transfer function

The closed-loop transfer function Gc(s) is:

That is to say:

This problem is the first part of one where the transient response is requested so that the overshoot is 20% and the settling time is 2 seconds, see the complete problem in the following link: Example 1 – Transient response of an electromechanical system

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Control System Analysis, Electromechanical Systems

Solved Example 1- Electromechanical System Transfer Function

Obtain the mathematical model of the position control system of the figure. Get the block diagram and the ansfer function between the angle of the load and the reference angle θc(s)/θc(s).

null

Data:

null

1. System dynamic

null

2. Laplace Transform

null

3. Block Diagram

null

Simplifying conveniently to obtain a model whose transfer function is known:

null

4. Transfer function of each block of the previous diagram.

Starting from:

nullWe obtain the following:nullThen, using:null

and substituting, we obtain:

null

Substituting the value of the data in the previous equation, we obtain:

null

Simplifying:null

On the other hand, the gain of the amplifier is obtained using:

null

From where:

null

null

Finally, the gear constant is given by the data and is n = 1/10. We then obtain a block diagram with the following transfer functions:

null

5. System Transfer function.

The open-loop transfer function Ga(s) of the system shown in the previous diagram is:

null

From where we can easily obtain the closed-loop transfer function Gc (s), which is what the statement asked, using the unit feedback:

null

NEXT: Example 2 – Electromechanical system transfer function (English)

Written by: Larry Francis Obando – Technical Specialist – Educational Content Writer.

Mentoring Academic/ Entrepreneurs/ Business.

Copywriting, Content Marketing, White Papers (Spanish– English)

Escuela de Ingeniería Eléctrica de la Universidad Central de Venezuela, Caracas.

Escuela de Ingeniería Electrónica de la Universidad Simón Bolívar, Valle de Sartenejas.

Escuela de Turismo de la Universidad Simón Bolívar, Núcleo Litoral.

Contact: Caracas, Quito, Guayaquil, Cuenca – Telf. 00593998524011

WhatsApp: +593981478463

+593998524011

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Block Diagram, Control System Analysis, Electrical Engineer, Electromechanical Systems

Block Diagram of Electromechanical Systems – DC Motor

Introduction

Electromechanical Systems are hybrids of electrical and mechanical variables. Applications for electromechanical components range from robot control, sun and star trackers, disk-drive position control, DC machines control and central Air-Conditioning systems for residences.

The structure of an Electromechanical Drive System is given in Figure 2.1. It consist of energy/power source, reference values for the quantities to be controlled, electronic controller, gating circuit for converter, electronic converter (rectifier, inverter, power electronic controller), current sensors (shunts, current transformer, Hall sensor), voltage sensor (voltage divider, potential transformer), speed sensors (tachometers) and displacement sensors (encoders), rotating three-phase machines, mechanical gearbox, and the application-specific load (pump, fan, automobile). In Figure 2-1 all but the mechanical gear are represented by a Transfer Functions (output variables as a function of time). Meanwhile, the mechanical gear is represented by the Transfer Characteristic (output variable Xout as a function of the input variable Xin).

[1]

The armature-controlled dc servomotor is perhaps the most important component found in robotic applications. But in general, in electromechanical systems, we can use induction motors and synchronous motors as well, the two most common three-phase AC Machines. Here, we just derive the Dynamic of the DC Motor in order to represent it through block diagrams. So we will have to pass from schematic to block diagram as shown in Figure 2.35:

[2]

The Block Diagram for a DC Motor

To derive the block diagram representation for a separately excited DC machine, firstly we must derive the Dynamic of the System, the differential equations that govern the DC machine. After that, we use the Laplace Transform to build the block diagram. The motor can be controlled by field or by armature. In this case, let’s suppose that a stationary permanent magnet or a stationary electromagnet generates a constant magnetic flux Ф called the Fixed Field according to Figure 2.35. As a result, the motor is controlled by a voltage ea applied to the armature terminals. The armature is a rotating circuit through which a current ia, flows. When the armature passes through Ф at right angles, it feels a force F=BLia where B is the magnetic field strength and L is the length of the conductor. The resulting torque Tm turns the rotor, the rotating member of the motor. For linear analysis it is supposed that this torque or par is proportional to the flux Ф and the armature current ia, and from here, we obtain the first important relation:

[3]

As Ф is constant, KmФ = Ki, so the previous equation is as follows,

Where Ki is The Constant of Proportionality, also called the motor torque constant (or par constant) and is one the parameter given by the manufacturers when selling a motor. Ki also frequently called KT comes in N-m/A.

(Note: when the motor is controlled by the current in the field, in order to have a linear system the current of the armature must be considered to be constant and the motor torque is given by Tm= KmiF)

There is another phenomenon that occurs in the motor: A conductor moving at right angles to a magnetic field generates a voltage Vb at the terminals of the conductor. Since the current-carrying armature is rotating in a magnetic field, its voltage is proportional to speed. In this way we obtain the second important equation:

[2]

We call Vb the Back Electromotive Force (or back emf); Kb is a constant of proportionality called the back emf constant, another given parameter. It is also important to notice that θm is the angular displacement and the angular velocity of the motor is ωm:

Although the motor is itself an open-loop system, later we will see that The Back EFM Vb, generates a feedback-loop inside the motor, acting as an “electric friction” that tends to improve the stability of the motor.

After this and applying the Kirchhoff’s law for voltages, we find the relation between the armature current ia, the applied armature voltage ea, and the back efm Vb (or eb). Clearing conveniently we obtain:

Where La and Ra represent the inductance and the resistance of the armature respectively. Vb = eb. (La and Ra are given parameters of the motor)

Now, applying Newton’s Law for Rotational Mechanical Systems and clearing conveniently:

Where TL represents the load, JM is the inertia of the rotor, and Bm viscous friction coefficient (JM and Bm are given parameters of the motor)

Actually, the completed Dynamic of the DC Motor in open-loop operation is the following set of differential equations:

[3]

To get a block diagram of this dynamic we have to apply Laplace Transform to this set of equations while focused on the value we are concerned as the output: θm. So, we now that

After applying Laplace:

θm(s) = Ωm(s)/S =(Ωm(s))*(1/S)

We can represent this simple relationship between θm(s) and Ωm(s) using a Block Diagram by recalling the fact that every block has a Transfer Function inside. In this case, the transfer function is:

θm(s)/Ωm(s) = 1/S

The block diagram to represent this system is:

We see now that we need to know what is Ωm(s). To know that we use

We clear for dθm/dt and apply Laplace. We obtain:

Ωm(s)=(Tm(s) – TL(s))/(Jm(s) + Bm(s))

As a dog following its tail, we continue this procedure using each of the rest of the equations just one time, till all the variables are cleared, except the inputs TL(s) and Ea(s).

To get Tm(s) we use:

And we get:

Tm(s)= KiIa(S)

Finally, from:

We get:

Ia(S)= (Ea(s)-Eb(s))/(SLa+Ra)

Eb(s)=KbΩm(s)

Now, using, the interconnecting, θm(s), Ωm(s), Tm(s), Ia(S), Ea(s), Eb(s) and TL(s), with functional blocks and its transfer functions, summing points and pickoff points, we get The Block Diagram for a DC Motor operating as an open-loop:

[3]

Here we can corroborate what we pointed out before, that the back electromotive force, proportional to Ωm (s), represented in the diagram as Eb (s), generates a feedback loop that tends to stabilize the system.

The most commonly used electromechanical system is shown in Figure 2.15., operating at about constant speed without feedback control. Some of the more commonly occurring drive systems are presented using linear transfer function within each and every block of these diagrams as in Figure 2.15. Consequently, we usually need the transfer function of the motor and its load to represent it just through one block.

 The Transfer Function for a DC Motor and Load

Let’s consider the configuration shown in Figure 2.37:

[2]

The relation between the input Ea(s) and the output θm(s) is given by:

(Note: To see how this equation is derived, please see page 81 of bibliography resource [2])

If we assume that the armature inductance La is small compared to the armature resistance Ra, which is usual for the motor DC, we obtain:

The desired Transfer Function θm(s)/Ea(s) is found to be:

[2]

Where Jm is the equivalent inertia after load inertia JL is reflected back to the armature, and Dm is the equivalent damping after load damping DL is reflected back to the armature, and:

Usually, to find the factors Kt/Ra and Kb, the user counts with the following relations and a Torque-Speed Curve of the motor provided by the manufacturer, as it is shown in the following example:

[2]

As a consequence, we can now represent the motor and its load through one block in the diagram:

Closed-loop operation.

The open-loop operation is acceptable for a lot of drive application where a constant speed (a chainsaw) or position (an elevator) is sufficient. However, if a variable speed is needed (conveyor belt) or the position must be accurate (an Antenna), a closed-loop control, with negative feedback, must be chosen. But, in a closed-loop operation, others devices are needed to assess or transform signals: sensors, transducers and amplifiers. A typical closed-loop operation is the position control schema for an Antenna Azimuth shown as follows:

Which equivalent block diagram is as follows:

[2]

The potentiometer. 

A potentiometer is an electromechanical transducer that converts mechanical energy into electrical energy. The input of the device is a form of mechanical displacement that can be translational or rotational. When a voltage is applied across the fixed terminals of the potentiometer, the output voltage, which is measured between the variable terminal and the ground, is proportional to the input shift. Figure 4-29 shows the schema for a rotational potentiometer.

[3]

When the potentiometer housing is connected to the reference, the output voltage e(t) will be proportional to the position of the axis θc(t) in the case of a rotatory movement. So,

where Ks is the constant of proportionality. So E(s)/θc(s) = Ks will be the transfer function for a block representing this relationship

Potentiometers are often used to feedback the output position in motor control systems. Here it works as a mechanical sensor. Doing so, they allow comparing the actual position of the load with the reference position. The comparison generates an error signal that is then amplified to move the motor until it reaches the correct position. Hence its name, servomotor, slaves of the reference position.

The Tachometer. 

Like potentiometers, tachometers are electromechanical devices that convert mechanical energy into electrical energy. It works essentially as a voltage generator, with the voltage output proportional to the angular velocity of the input shaft. Figure 4-33 reflects the common use of a tachometer in a speed control system:

[3]

The Dynamic of the tachometer can be represented by:

Where et(t) is the output voltage, θ(t) is the displacement of the motor in radians, ω(t) is the speed of the rotor in rad/s, and Kt is the constant of the tachometer. Later, Et(s)/Ω(s) = Kt. In terms of: the displacement of the motor:

The gear train. 

Gear trains are used very frequently in electromechanical systems in order to reduce speed, amplify torque or to achieve the more efficient power transfer by matching the driving member with a given load. Consider the gear train of Figure 2-30 (for a more exhaustive analysis please refer to page 66 of the bibliography [5]):

[5]

Generally, in practice, the most commonly used procedure is to reflect the inertia and the damping of the load arrow to the motor shaft. The result is as follows:

Where J1eq is the equivalent inertia seen by the motor arrow and b1eq is the equivalent damping coefficient seen by the motor arrow. In the block diagrams, it is customary to represent the gear train by a block with a proportional transfer function, a constant Kg (which stands for Kgears) equivalent to the n1/n2.

Example:
  1. Before continuing, let’s see an example of how to apply the theory studied to a fairly common case: Obtain mathematical model of the position control system of the Figure. Obtain your block diagram and the transfer function between the angle of the load and the reference angle θc(s)/θc(s).

null

Solution:

null

To see the whole answer see:

Ejemplo 1 – Función de Transferencia de Sistema Electromecánico

DC Motor/Amplifier System Block Diagram

The CD Motor is always driven by a power amplifier that acts as an energy source. For this reason, it is more practical to present the Torque-Speed Curve of the combinación DC Motor/Amplifier. The Figure 4-51 shows a block diagram for the DC motor-Amplifier arrangement. And Figure 4-52 the Torque-Speed Curve:

[3]

The Proportional amplifier

A good example of a proportional amplifier is an Operational Amplifier with negative resistive feedback and inverting configuration such as shown in Figure 2.7a.

[1]

Operational amplifiers, often called Op Amps, are often used to amplify signals in sensor circuits. The Transfer Function of the Operational Amplifier is shown in Figure 2.7b under the name of G2:

Other configurations of this class are shown in Table 3-1 and Table 4-1 with their respective transfer functions:

Table 3-1

[4]

Table 4-1

[3]

Power Electronic

Performance of servomotors used for robotics applications highly depends on electric power amplifiers and control electronics, broadly termed Power Electronic. The actuators in robotic applications, mostly DC motors, must be controlled precisely so that desired motions of arms and legs may be attained. This requires a power amplifier to drive de desired level of voltage (or current indirectly) to the motor armature. The use of a linear amplifier, as the operational amplifier discussed in the previous section, is power-inefficient and impractical since it entails a large amount of power loss. An alternative is to control the voltage via ON-OFF switching. Pulse Width Modulation or PWM for short is the most commonly used method for varying the average voltage to the DC motor (For a detailed review of PWM see Motor Drives, p 663, bibliography [7])

Briefly, a typical motor drive system is expected to have some of the systems block diagram indicated in Figure 27.1. The load may be a conveyor system, a traction system, the rolls of a mill drive, the cutting tool of a numerically controlled machine tool, the compressor of an air conditioner, a ship propulsion system, a control valve for a boiler, a robotic arm, and so on.

[7]

The Power Electronic converter block in the previous diagram, for PWM control with inner current loop, may use diodes, MOSFETs, GTOs or IGBTs. Servo drive systems normally use the full four-quadrant converter of Figure 27.9, which allows bidirectional drives and regenerative braking capabilities.

[7]

The PWM is a technique to control an effective armature voltage by using the ON-OFF switching alone. Figure 2.3.3 illustrates the PWM signal:

[6]

PWM varies the ratio of the time length of the complete ON state to the complete OFF state. A single cycle of ON and OFF states is called the PWM period, whereas the percentage of the ON stage in a single period is called duty rate. The first PWM signal of Figure 2.3.3, is of 60% duty, and the second one is 25%. If the voltage supply is V=10 volts, the average voltage actually transmitted to the DC motor is 6 volts and 2.5 volts respectively. The PWM period is set to be shorter than the time constant associated with the mechanical motion. The PWM frequency is usually between 2 and 20 KHz, whereas the bandwidth of a motion control system is at most 100 Hz. Therefore, the discrete switching does not influence the mechanical motion in most cases.

If the electric time constant Te is much larger than PWM period, the actual current flowing to the motor armature is a smooth curve, as illustrated in Figure 2.3.4:

[6]

Design of Control Systems - Next Topic

The fundamental objective of analyzing a control system is to facilitate its design. The Block Diagram is the first step because it represents the mathematical model of the system that we want to analyze and improve. The dynamics of a controlled linear process can be represented by the block diagram of Figure 10-1:

[3]

Most control systems are built so that the output vector y (t) meets certain specifications that define what the system should do and how to do it in the desired way. These specifications are unique for each individual application. Relative stability, steady-state precision (error) and transient response are the most commonly used specifications in the market. The essential problem involves determining the control signal U (t) within a prescribed range so that the specifications required by the customer are met.

After determining the above, the engineer must design a fixed configuration of the system and the place where the controller will be placed in relation to the controlled process, which also involves designing the elements that make up the controller, that is, determining the controller parameters. Because most control efforts involve modifying or compensating the performance characteristics displayed by the system during the transient response analysis and steady state response, the design of a fixed configuration is also called compensation.

In short, the art and science of designing control systems can be summarized in three steps, which leads to our following issues:

  1. Determine what should be done and how to do it
    1. Transient Response Specifications
    2. Stability – Routh Criterion
    3. Stead-State Error
  2. Determine the driver configuration
    1. Proportional, Integral and Derivative Control Actions
    2. PD Controller
    3. PI Controller
    4. PID controller
  3. Determine the controller parameters

Source

  1. Chapter 2, Block Diagram of EM Systems, pp 21, 43(23) (Fuchs E.F., Masoum M.A.S. (2011) Block Diagrams of Electromechanical Systems. In: Power Conversion of Renewable Energy Systems. Springer, Boston, MA)
  2. Control Systems Engineering, Nise pp 79, 81
  3. Sistemas de Control Automatico Benjamin C Kuo pp 159, 203
  4. Modern_Control_Engineering, Ogata 4t pp 103,
  5. dinamica_de_sistemas p 66
  6. Actuators and Drive System – Robótica
  7. Libro Rashid – Power Electronic Handbook p 663-666

Written by: Larry Francis Obando – Technical Specialist – Educational Content Writer.

Escuela de Ingeniería Eléctrica de la Universidad Central de Venezuela, Caracas.

Escuela de Ingeniería Electrónica de la Universidad Simón Bolívar, Valle de Sartenejas.

Escuela de Turismo de la Universidad Simón Bolívar, Núcleo Litoral.

Contact: Caracas, Quito, Guayaquil, Cuenca – Telf. 00593998524011

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Attention:

If what you need is to solve urgently a problem of a “Mass-Spring-Damper System ” (find the output X (t), graphs in Matlab of the 2nd Order system and relevant parameters, etc.), or a “System of Electromechanical Control “… to deliver to your teacher in two or three days, or with greater urgency … or simply need an advisor to solve the problem and study for the next exam … send me the problem…I Will Write The Solution To any Control System Problem…

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Related:

Ejemplo 1 – Función de Transferencia de Sistema Electromecánico

The Block Diagram – Control Engineering

Dinámica de un Sistema Masa-Resorte-Amortiguador

Ejemplo 1 – Función Transferencia de Sistema masa-resorte-amortiguador

Transient Response Specifications

Control System Stability

PID – Basic Control System Actions