UNDERDAMPED SECOND-ORDER SYSTEM

Fuentes:

Control Systems Engineering, Norman Nise

    1. Introduction Chapter 4 pp 162 (162)
    2. Poles and Zeros 4.1 pp 162 –
    3. First Order System 4.3 pp 165-168
    4. Second Order System 4.4 pp 168-177
    5. Underdamped Second-Order System 4.6 pp 177-186
  1. Modern_Control_Engineering__4t
    1. Introduction Chapter 5 pp 219 (232)
    2. First Order Systems 221 (234)-224
    3. Second Order System pp 224 (237)-234

Literature Review, Martes 14 noviembre 2017, 05:07 am – Caracas, Quito, Guayaquil.

Introduction

Now that we have become familiar with second-order systems and their responses, we generalize the discussion and establish quantitative specifications defined in such a way that the response of a second-order system can be described to a designer without the need for sketching the response. We define two physically meaningful specifications for second-order systems. These quantities can be used to describe the characteristics of the second-order transient response just as time constants describe the first-order system response.

Natural Frequency, Wn

The natural frequency of a second-order system is the frequency of oscillation of the system without damping. For example, the frequency of oscillation of a series RLC circuit with the resistance shorted would be the natural frequency.

Damping Ratio,

We have already seen that a second-order system’s underdamped step response is characterized by damped oscillations. Our definition is derived from the need to quantitatively describe this damped oscillations regardless of the time scale.Thus, a system whose transient response goes through three cycles in a millisecond before reaching the steady state would have the same measure as a system that went through three cycles in a millennium before reaching the steady state. For example, the underdamped curve in Figure 4.10 has an associated measure that defines its shape. This measure remains the same even if we change the time base from seconds to microseconds or to millennia.

 A viable definition for this quantity is one that compares the exponential decay frequency of the envelope to the natural frequency. This ratio is constant regardless of the time scale of the response. Also, the reciprocal, which is proportional to the ratio of the natural period to the exponential time constant, remains the same regardless of the time base.

We define the damping ratio, , to be:

Consider the general system:

Without damping, the poles would be on the jw-axis, and the response would be an undamped sinusoid. For the poles to be purely imaginary, a = 0. Hence:

Assuming an underdamped system, the complex poles have a real part, , equal to -a/2. The magnitude of this value is then the exponential decay frequency described in Section 4.4. Hence,

from which

Our general second-order transfer function finally looks like this:

Now that we have defined and Wn, let us relate these quantities to the pole location. Solving for the poles of the transfer function in Eq. (4.22) yields:

From Eq. (4.24) we see that the various cases of second-order response:

Underdamped Second-Order System

Now that we have generalized the second-order transfer function in terms of and Wn, let us analyze the step response of an underdamped second-order system.

Not only will this response be found in terms of and Wn, but more specifications
indigenous to the underdamped case will be defined. The underdamped second order system, a common model for physical problems, displays unique behavior that
must be itemized; a detailed description of the underdamped response is necessary
for both analysis and design. Our first objective is to define transient specifications
associated with underdamped responses. Next we relate these specifications to the
pole location, drawing an association between pole location and the form of the
underdamped second-order response. Finally, we tie the pole location to system
parameters, thus closing the loop: Desired response generates required system
components.

Let us begin by finding the step response for the general second-order system of Eq. (4.22). The transform of the response, C(s), is the transform of the input times the transfer function, or:

where it is assumed that < 1 (the underdamped case). Expanding by partial fractions, using the methods described, yields:

Taking the inverse Laplace transform, which is left as an exercise for the student, produces:

where:

A plot of this response appears in Figure 4.13 for various values of , plotted along a time axis normalized to the natural frequency.

We now see the relationship between the value of and the type of response obtained: The lower the value of , the more oscillatory the response.

The natural frequency is a time-axis scale factor and does not affect the nature of the response other than to scale it in time.

Other parameters associated with the underdamped response are rise time, peak time, percent overshoot, and settling time. These specifications are defined as follows (see also Figure 4.14):

  1. Rise time, Tr. The time required for the waveform to go from 0.1 of the final value to 0.9 of the final value.
  2. Peak time, TP. The time required to reach the first, or maximum, peak.
  3. Percent overshoot, %OS. The amount that the waveform overshoots the steady-state, or final value at the peak time, expressed as a percentage of the steady-state value.
  4. Settling time, Ts. The time required for the transient’s damped oscillations to reach and stay within 2% of the steady-state value.

All definitions are also valid for systems of order higher than 2, although analytical expressions for these parameters cannot be found unless the response of the higher-order system can be approximated as a second-order system.

Rise time, peak time, and settling time yield information about the speed of the transient response. This information can help a designer determine if the speed and the nature of the response do or do not degrade the performance of the system.

For example, the speed of an entire computer system depends on the time it takes for a hard drive head to reach steady state and read data; passenger comfort depends in part on the suspension system of a car and the number of oscillations it goes through after hitting a bump.

Evaluation of Tp

Tp is found by differentiating c(t) in Eq. (4.28) and finding the first zero crossing after t = 0.

Evaluation of %OS.

From Figure 4.14 the percent overshoot, %OS, is given by:

 Evaluation of Ts

In order to find the settling time, we must find the time for which c(t) in Eq. (4.28) reaches and stays within ₎±2% of the steady-state value, C final.

 Evaluation of Tr

A precise analytical relationship between rise time and damping ratio cannot be found. However, using a computer and Eq. (4.28), the rise time can be found. Let us look at an example.

We now have expressions that relate peak time, percent overshoot, and settling time to the natural frequency and the damping ratio. Now let us relate these quantities to the location of the poles that generate these characteristics. The pole plot for a general, underdamped second-order system is reproduced in Figure 4.17.

Now, comparing Eqs. (4.34) and (4.42) with the pole location, we evaluate peak time and settling time in terms of the pole location. Thus:

where is the imaginary part of the pole and is called the damped frequency of oscillation, and is the magnitude of the real part of the pole and is the exponential damping frequency part.

At this point, we can understand the significance of Figure 4.18 by examining the actual step response of comparative systems. Depicted in Figure 4.19(a) are the step responses as the poles are moved in a vertical direction, keeping the real part the same. As the poles move in a vertical direction, the frequency increases, but the envelope remains the same since the real part of the pole is not changing.

Let us move the poles to the right or left. Since the imaginary part is now constant, movement of the poles yields the responses of Figure 4.19(b). Here the frequency is constant over the range of variation of the real part. As the poles move to the left, the response damps out more rapidly.

Moving the poles along a constant radial line yields the responses shown in Figure 4.19(c). Here the percent overshoot remains the same. Notice also that the responses look exactly alike, except for their speed. The farther the poles are from the origin, the more rapid the response.

Literature Review by: Larry Francis Obando – Technical Specialist

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: Ecuador (Quito, Guayaquil, Cuenca)

WhatsApp: 00593984950376

 

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Sistemas LDCID – Modeling – Fundamentos

Sistemas LDCID – Representación Matemática

Sábado 11 de noviembre, 4:53 am

Fuentes:

Análisis de Sistemas Lineales – Prof. Ebert Brea

    1. Análisis de Sistemas en el Dominio Continuo pp 29 – (58)
  1. Control Systems Engineering, Norman Nise
    1. First Order System 4.3 pp 165-168

 

 El modelo matemático y sus términos

Los sistemas lineales, dinámicos, causales, invariantes en el dominio y deterministas (LDCID) definidos en el dominio del tiempo continuo constituyen parte importante en el estudio de los sistemas eléctricos, debido al hecho de sus innumerables aplicaciones dentro de la ingeniería eléctrica.

En general podría decirse que los sistemas lineales son el resultado de aproximaciones en el modelaje de sistemas. No obstante, aun cuando los sistemas eléctricos forman parte de los llamados sistemas no lineales, su tratamiento como sistemas lineales permiten dar respuestas acertadas a las preguntas que pudiera requerir los profesionales del área.

Los modelos matemáticos de sistemas dinámicos definidos en el dominio continuo presentan términos asociados a operaciones de derivadas de las cantidades externas con respecto a la variable independiente, que por lo general será el tiempo. Estos modelos matemáticos se denominan ecuaciones diferenciales, y sus respectivas respuestas son totalmente definidas por las condiciones de cada sistema representado por el modelo matemático.

Un aspecto importante a estudiar la representación de un sistema a través de su modelo matemático es la identificación de los términos que son expresados en el modelo matemático de un sistema LDCID, el cual es representado por una ecuación diferencial ordinaria ( (ODE) involves derivatives of a function of only one variable) de orden m-ésimo en relación a la señal de excitación x(t), y de orden n-ésimo con respecto a su señal de respuesta y(t), es decir, en general un modelo matemático asociado a un sistema LDCID viene dado por:

donde y(t) representa la señal de respuesta también denominada señal de salida, x(t) representa

la señal de excitación o de entrada, y los coeficientes an;a1; …;a0 y bm;b1; … ;b0

representan los parámetros del sistema, que alteran respectivamente la señal de excitación y la señal de respuesta, así como sus derivadas ordinarias, y la variable independiente t, en este caso puede significar el tiempo, con el propósito de contextualizar el dominio en el cual está definido el modelo matemático.

Modelo matemático de primer orden

Un sistema LDCID en el dominio continuo de primer orden es representado mediante una ecuación diferencial dada por:

Note que el modelo debe ser de primer orden en lo que respecta a los operadores ,

es decir, en mayor orden de derivadas de la señal de respuesta y(t) debe ser n = 1.

Sin embargo, podría ser de cualquier orden con relación a los operadores de la excitación

para , debido al hecho de que las operaciones de derivadas sobre la señal de excitación no son consideradas parte del sistema.

Note que las operaciones definidas sobre la señal de excitación no forman parte del sistema, por cuanto las operaciones matemáticas definidas sobre la excitación constituyen el modelo matemático de la señal de excitación.

Por razones de simplificación en la nomenclatura y mediante la propiedad de superposición, se estudiará la solución de la ecuación diferencial:

Note que para obtener la solución del sistema debe conocerse al menos una condición de la respuesta del sistema, la cual usualmente es especificada a través de su condición inicial, y(0).

Luego de manipulaciones algebraicas convenientes (demostración en Fuente 1), se concluye que la solución a la ecuación diferencial representada por la Ecuación (2.12) viene dada por:

donde:

  1. Respuesta transitoria:

  1. Respuesta permanente

Ejemplo 2.2

La Figura 2.3 muestra un sistema compuesto por una resistencia y un capacitor, y cuyos valores son representados respectivamente por R y C. Además, la figura muestra que el sistema eléctrico es excitado por una señal x(t) = u(t) y su respuesta es medida a través de la tensión sobre el capacitor, donde u(t) representa la función escalón unitario:

El modelo matemático asociado al sistema representado por la Figura 2.3 puede obtenerse empleando elementales ecuación de redes eléctricas:

Entonces, al comparar el modelo matemático definido por la Ecuación (2.12) con el modelo obtenido, se tiene que el coeficiente a0 y la señal de excitación son:

,

Al aplicar la solución expresada por medio de la Ecuación (2.21), se puede afirmar que:

Al operar la Ecuación (2.26) se tiene que la respuesta del sistema es dada por:

Note que:

por cuanto el elemento de memoria representado por el capacitor no permite cambios bruscos y por tal motivo y(0-) = y(0) = y(0+). Además, para buscar una respuesta a la pregunta debe tomarse en cuenta que la excitación tiene un valor de cero y ella ha permanecido en cero desde mucho tiempo atrás, es decir, desde menos infinito, obviamente y(0) = 0.

FIRST ORDER SYSTEMS (Fuente 2)

We now discuss first-order systems without zeros to define a performance specification for such a system…

We now use Eqs. (4.6), (4.7), and (4.8) to define three transient response performance specifications:

 

  • Time Constant: We call 1/a the time constant of the response. From Eq. (4.7), the time constant can be described as the time for to decay to 37% of its initial value. Alternately, from Eq. (4.8) the time constant is the time it takes for the step response to rise to 63% of its final value.

The reciprocal of the time constant has the units (1/seconds), or frequency. Thus, we can call the parameter a the exponential frequency. Thus, the time constant can be considered a transient response specification for a first order system, since it is related to the speed at which the system responds to a step input.Since the pole of the transfer function is at a, we can say the pole is located at the reciprocal of the time constant, and the farther the pole from the imaginary axis, the faster the transient response.

 

  • Rise Time (Tr): Rise time is defined as the time for the waveform to go from 0.1 to 0.9 of its final value.

 

  • Settling Time (Ts): Settling time is defined as the time for the response to reach, and stay within, 2% of its final value.2

Modelo matemático de orden superior.

En este apartado se introducirá el operador p, el cual será empleado para representar el orden de la derivada que está operando en cada término de la ecuación diferencial ordinaria bajo estudio.

Definición 2.1 (Operador p) Se define el operador pn al operador diferencial que representa

la derivada n-ésima con respecto a la variable del dominio continuo. Es decir,

Por otra parte, se debe introducir dos definiciones que conforman la solución completa de una ecuación diferencial ordinaria.

Definición 2.2 (Respuesta transitoria) La respuesta transitoria o, también denominada natural o solución homogénea, es la solución de toda ecuación diferencial ordinaria cuando su señal de excitación viene definida por la función nula.

Definición 2.3 (Respuesta permanente) La respuesta permanente o, también denominada forzada o solución particular, es la solución de la ecuación diferencial ordinaria ante una señal de excitación que actúa sobre el sistema.

Observación 2.1 La respuesta transitoria, natural u homogénea es intrínseca del sistema y no de la excitación, a diferencia de que la respuesta permanente, forzada o particular, que además de depender del sistema, depende de la excitación.

Se conocen condiciones del sistema, bien sean condiciones iniciales a través del valor de la respuesta y(t) para t = 0 y sus primeras n-1 derivadas para t = 0, ó n valores conocidos de la respuesta completa y(t) en n distintos instantes de t, o combinación de lo anterior.

Se tiene que:

donde el coeficiente o también denominado parámetro an = 1 (ODE with leading coefficient equal to 1 is called standard ODE form)

Aplicando las Ecuaciones (2.46), se puede escribir el modelo matemático definido por la Ecuación (2.45) como:

donde D(p) es el ampliamente conocido polinomio característico del sistema.

Respuesta Transitoria

Existen diversos métodos para determinar la respuesta transitoria de un modelo matemático asociado a un sistema LDCID en el dominio continuo, el cual es representado por una ecuación diferencial ordinaria.

Método 2.1 (Determinación de la Respuesta Transitoria) Dada la ecuación diferencial ordinaria definida por la Ecuación (2.44),

Ejecute:

Paso 1. Asegúrese de que el término an de la ecuación diferencial sea igual a uno. Si no es así, divida toda la ecuación diferencial entre an.

Paso 2. Aplique el operador “p” a la ecuación diferencial.

Paso 3. Determine las n raíces que anulen el polinomio D(p) y denote las raíces reales como ri para cada i = 1; … ;nr, y las raíces complejas conjugadas como

para cada i = nr +1; ..;n, donde

tomando en cuenta la multiplicidad de cada una de las raíces denotada como mi.

EJEMPLO 2.5 Respuesta transitoria de un sistema de quinto orden

Suponga el modelo matemático de un sistema LDCID en tiempo continuo definido

por:

donde y(t) es la señal de respuesta del sistema, y x(t) representa la señal de excitación. Para el modelo matemático definido mediante la Ecuación (2.48), determine la solución homogénea del sistema aplicando el Método 2.1.

Solución. Debido a que el término a5 no es igual a 1, se debe dividir toda la ecuación diferencial entre a5, para luego aplicar el operador p, obteniéndose:

Al calcular las cinco raíces que anulan D(p), se tiene que sus raíces son: r1 = -2, r22= -3 y

z3 = -1 +- j. Entonces, se puede afirmar que las soluciones asociadas a cada raíz viene dada por:

Respuesta Permanente

Al despejar y(t) de la Ecuación (2.47):

se tiene que

donde la fracción N(p)/D(p) representa el operador del sistema L(p).

A fin de estudiar el caso más general de las señales de excitaciones más comúnmente presentes en los sistemas eléctricos, se analizará cuando la señal de excitación es considerada una exponencial definida por:

donde en general s es un parámetro o coeficiente complejo, y cuyo valor es

y B es un parámetro constante de la señal de excitación, que pertenece al conjunto de los números reales

Por otra parte, los casos en los cuales pueden ser aplicado el método que será descrito en este punto, corresponden a aquellos en donde D(s) no es igual a 0.

La Ecuación (2.50) permite representar diversas situaciones para la señal de excitación x(t), y cuyos casos son mostrados a continuación mediante la Tabla 2.1

Es importante hacer notar que la operación

ejecuta mediante la operación límite, es decir,

EJEMPLO 2.6 Considere un sistema LDCID con modelo matemático definido por:

Para el sistema representado por la Ecuación (2.52), determine la respuesta permanente del sistema si la señal de excitación es:

Solución. Dado que el coeficiente a3 es igual a uno, se puede aplicar el operador p a la Ecuación (2.52) obteniéndose:

Respuesta Completa

La respuesta completa del sistema se consigue sumando la respuesta transitoria u homogénea con la respuesta permanente o solución particular, es decir:

donde los coeficientes ci para todo i = 1; …. ;n se obtiene de n condiciones conocidas, en concordancia con el grado de la ecuación característica N(p), es decir, los coeficientes ci

para todo i = 1; …. ;n son determinados por el conocimiento de:

Por ejemplo, el problema ahora es hallar la respuesta completa del sistema, bajo las condiciones:

Solución. Claramente se tiene que el término a3 = 1, hecho que permite aplicar el operador p directamente a la Ecuación (2.52), arrojando el polinomio característico

D(p) = p3 +8p2 +19p+12, y cuyas raíces que lo anulan son r1 = -1, r2 = -3 y r3 = -4.

Como consecuencia del análisis hecho, se tiene que la solución homogénea está dada por:

De las Ecuaciones (2.53) y (2.57) se puede afirmar que la solución completa es:

Al resolver el sistema de ecuaciones lineales definido por la Ecuación (2.60)

se obtiene que c1 = 1/3, c2 = -3/2 y c3 = 5/3, los cuales al ser sustituido en la Ecuación (2.58) se llega a:

Literature Review by: Larry Francis Obando – Technical Specialist

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: Ecuador (Quito, Guayaquil, Cuenca)

WhatsApp: 00593984950376

email: dademuchconnection@gmail.com

Copywriting, Content Marketing, Tesis, Monografías, Paper Académicos, White Papers (Español – Inglés)

FIRST and SECOND ORDER SYSTEMS

FIRST and SECOND ORDER SYSTEMS

 

Fuentes:

  1. Control Systems Engineering, Norman Nise
    1. Introduction Chapter 4 pp 162 (162)
    2. Poles and Zeros 4.1 pp 162 –
    3. First Order System 4.3 pp 165-168
    4. Second Order System 4.4 pp 168-177
  2. Modern_Control_Engineering__4t
    1. Introduction Chapter 5 pp 219 (232)
    2. First Order Systems 221 (234)-224
    3. Second Order System pp 225(238)-229

 

 

TIME DOMAIN CONTROL SYSTEMS ANALYSIS

Analisis de sistemas de control en el dominio del tiempo

FIRST ORDER SYSTEMS

We now discuss first-order systems without zeros to define a performance specification for such a system…

We now use Eqs. (4.6), (4.7), and (4.8) to define three transient response performance specifications:

 

  • Time Constant: We call 1/a the time constant of the response. From Eq. (4.7), the time constant can be described as the time for to decay to 37% of its initial value. Alternately, from Eq. (4.8) the time constant is the time it takes for the step response to rise to 63% of its final value.

The reciprocal of the time constant has the units (1/seconds), or frequency. Thus, we can call the parameter a the exponential frequency. Thus, the time constant can be considered a transient response specification for a first order system, since it is related to the speed at which the system responds to a step input.Since the pole of the transfer function is at a, we can say the pole is located at the reciprocal of the time constant, and the farther the pole from the imaginary axis, the faster the transient response.

 

  • Rise Time (Tr): Rise time is defined as the time for the waveform to go from 0.1 to 0.9 of its final value.

 

  • Settling Time (Ts): Settling time is defined as the time for the response to reach, and stay within, 2% of its final value.2

Fuente [1]

Fuente [3]

Fuente [3]

SECOND-ORDER SYSTEMS

Literature Review by: Larry Francis Obando – Technical Specialist

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: Ecuador (Quito, Guayaquil, Cuenca)

WhatsApp: 00593984950376

email: dademuchconnection@gmail.com

Copywriting, Content Marketing, Tesis, Monografías, Paper Académicos, White Papers (Español – Inglés)

PAGE_BREAK: PageBreak

Rigid Bodies Modeling

Robotics Fundamentals

Date: August, 2017. Location: Quito, Pichincha, Ecuador.

Actividad WBS (Fundamentals)
Lunes 14, 8:31 am

  1. Robotics: Dynamics and Control
    1. Rigid Bodies
    2. Transformation example in Matlab

 

We’re gonna be talking about rigid bodies. The first thing you wanna know about rigid bodies is besides the fact that they’re rigid, you wanna think about modelling them. And in order to model them, we attach reference frames to them. So here I show reference frame A, which is essentially an origin O and a set of axes, x, y, and z. We would also need a set of unit vectors that are parallel to these axes, a1, a2, and a3. So these vectors are mutually orthogonal.

And in fact, it’s these vectors that are more important than the axes. In fact, I’m gonna get rid of those axes and just keep the unit vectors, a1, a2, and a3. You have to remember that these vectors are attached to the rigid body as is the origin O.

Likewise for frame B, we have another set of unit vectors, b1, b2,and b3. And an origin P that’s attached to the rigid body.

So, we have two sets of basis vectors. Each set of basis vectors consists of mutually orthogonal unit vectors. The as are attached to the frame A, the bs are attached to the frame B.

Now I have a different set of components, q1 prime, q2 prime and q3 prime. Clearly, they are different from q1, q2, and q3 because b1, b2, and b3 are different from a1, a2, and a3. And yet, we wanna find a relationship between q1, q2, and q3 on one side and q1 prime, q2 prime, and q3 prime on the other side. This relationship is called a rigid body transformation because you’re talking about the same point. And looking at it from the vantage point of two different frames, each frame attached to a different rigid body.

So how do we relate q1, q2, and q3 to q1 prime, q2 prime, and q3 prime? Well, you can write down the vectors pictorially, and you see that immediately that picture suggests that you can use the triangle law of vector addition. The vector from O to Q is simply the sum of the vector from O to P, and the vector from P to Q. I can write this down in terms of components, as we’ve discussed before.

So now, I have a vector equation.I could even try to write this equation in terms of 3 by 1 vectors.

But you cannot simply add these 3 by 1 vectors.

Instead, you should take the vector of components q1 prime, q2 prime, and q3 prime, pre-multiplied by suitable transformation matrix, so that the resulting set of components is along a1, a2, and a3. Well, this transformation is essentially due to a rotation. And the matrix in front is a 3 by 3 rotation matrix. It’s denoted by the boldface symbol R with subscripts A and B. Suggesting that you are transforming components from frame B into frame A.

How do you write the components of a rotation matrix? Well, it’s a 3 by 3 matrix, and if you look carefully at the vector equation and the equation with 3 by 1 vectors, you can see that the rotation matrix is simply a collection of dot products or scalar products. You’re taking all possible combinations of the basis vectors, b1, b2, and b3, with the basis vectors a1, a2, and a3. In fact, if you look at the first row, it is simply the components of the basis vector b1 written in frame A.

Now we’re gonna collapse everything into a single matrix, the homogenous transformation matrix. Again, the same equation that essentially describes the triangle law of vector addition in terms of components, components written in terms of a1, a2, and a3. Let’s use homogeneous coordinates where we append the regular xyz coordinates by the number 1 as the fourth element. So this set of four numbers essentially give you a vector which is a representation of the position vector, but in projector coordinates. To relate these two sets of 4 by 1 vectors, all you need is a homogenous transformation matrix that includes elements of the rotation matrix that we’ve just described and the translation from O to P given by the components p1, p2, and p3. The last row, which consists of 0s and 1s, is simply inserted to make sure that the matrix multiplication reflects the triangle law of vector addition.

Well, this 4 by 4 matrix, we’re gonna denote by the boldface symbol T with subscripts A and B. And again, the subscripts denote the fact that you’re transforming position vectors from the second letter, B, to the first letter, A. This is our 4 by 4 homogeneous transformation matrix.

For an example see: Transformation example in Matlab

Written by: Larry Francis Obando – Technical Specialist

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: Ecuador (Quito, Guayaquil, Cuenca)

WhatsApp: 00593984950376

email: dademuchconnection@gmail.com

Copywriting, Content Marketing, Tesis, Monografías, Paper Académicos, White Papers (Español – Inglés)

Control systems are dynamic systems.

Dynamic system definition.

The first thing you must understand in the analysis of control systems is that control systems are dynamic systems. According to Ogata (1987), a system is dynamic when its output in present time depends on its input in the past. If the system output in the present time depends just on an input in the present time, the system is called static. In a dynamic system, the output changes with the time if the system is not in its equilibrium state, while, in a static system, the output keeps constant if the input doesn’t change; i.e. the output changes only when the input changes. See an excellent introduction by Prof. Pedro Albertos from UPV: Systems and Signals Examples.

Figures 1 and 2 are examples of static systems and dynamic systems respectively. The first shows the balance relation of a lever supported over a fulcrum. The present value of y(t) depends on the present value of the input u(t). The second shows that the speed and the position of a vehicle depend on an input in the past.

 

Ejemplo de sistema estático

Figure 1. An example of a static system (Albertos, 2016).

Ejemplo de sistema dinámico

Figura 2. An example of a dynamic system (Albertos, 2016).

The artificial systems such as the Off-Shore Platform of Figure 3 and the Aircraft cockpit of Figure 4, are also examples of high-complex dynamic systems made by the human beings:

 

Ejemplo de sistema dinámico. Sistema artificial 1

Figure 3. Artificial system. (Albertos, 2016)

 

Ejemplo de sistema dinámico. Sistema artificial 2

Figure 4. Artificial system (Albertos, 2016)

 

The transient response.

Regarding the control systems, the author Nise defines dynamic systems as follows:  “A control system is dynamic: It responds to an input by undergoing a transient response before reaching a steady-state response that generally resembles the input” (Nise, 2011, p. 10). Figure 5 shows a system to control the position of an antenna. Here, the output is the angular position  (Azimuth Angle), while the input is the signal  sent by the potentiometer. Figure 6 shows the output (blue line) of the system showed in Figure 5, in terms of the transient response and the steady-state response, both for high gain and low gain.

 

Ejemplo de sistema dinámico. Sistema de control de posición.

Figure 5. Position control system for an Antenna (Nise, 2011)

 

Ejemplo de sistema dinámico. Respuesta de Sistema de control de posición ejemplo 5.

Figure 6. Transient response and steady-state response for the system of Figure 5.

 

The goal of the control system of Figure 5 is to place the antenna into the position determined by the input. That´s why the output follows the input. Analyzing Figure 6 we can observe the main characteristic of dynamic systems, which is that, the response in any time after t=0 depends on the input in the past, i.e. the input in t=o determines the output in any time in the future. The transient response is the response of the system before reaching completely the steady-state response (the final value). Looking to the Figure 6, we can find two transient responses. The first one corresponds to a high gain. It generates a lot of fluctuations before the system reaches its steady-state, but it has the advantage of being faster in getting the final value. Here we can image the antenna getting its final position with a fast moving but zigzagging around it. The second one corresponds to a low gain, where there are no fluctuations and we get a very cushioned movement, but the system takes much more time in getting the final value. The selection of one or the other kind of transient response depends on the requirements of the operation and the limits of the system in order to maintain stability.

LTI Systems.

Studying control system implies to obtain as a first step its model. Indeed, before analyzing a control system, we have to develop a mathematical model of a dynamic system.

A mathematical model is perceived as a set of equations representing the dynamic of the system in an exact or approximate way. Such a dynamic, being the system an electrical, a mechanical or a biological one, can be represented by mean of differential equations (Ogata, 2002). In general, resolving a problem requires getting a simple and simplified model in the first stage, in order to visualize the solution by mean of the most practical way. To obtain a simplified model, the control engineer must decide which variable are important and which factors can be ignored.

Once we have an approximate idea of the kind and scope of the solution, the natural or forced response of the system, the model can be optimized and be transformed in one of more complexity, which requires the application of specialized software to be simulated and analyzed, with the aim of obtaining hidden and valuable information.

Talking about making a model, the MIT professor John Sterman shares its philosophy about the efficacy of a model as follows:

“Every model is a representation of a system…But for a model to be useful, it must address a specific problem and must simplify rather than attempt to mirror an entire system in detail…the usefulness of models lies in the fact that they simplify reality, creating a representation of it, we can comprehend…Von Clausewitz famously cautioned that the map is not the territory. It´s a good thing it isn´t: A map as detailed as the territory would be of no use” (Sterman, 2000, p. 89) .

To obtain the equations which set up the models of dynamic systems, the engineers use the laws of the physics applied to the properties of the systems, always searching for the easier path when they are building the model. Among the most useful properties to accomplish this objective, we have the properties of linearity and invariance in time, basically for two main reasons. In the first place, a huge quantity of physical processes, overall those which concern to science, have both properties. In the second place, the linear and time-invariant systems (LTI Systems) are widely accessible in terms of available tools for their analysis. The science of signals and systems has reached a powerful development on these software tools for the systems analysis, allowing people from different academic fields to easily approach to the study of LTI systems (Oppenheim, 1996). That’s why the comprehension of LTI systems becomes the next task at the engineers’ training for the analysis of control systems.

  Analysis and Design definition.

Before getting deeper on the characteristics of the LTI Systems, the engineers need to strictly define the basic areas of their work as control engineers [1]: the analysis, the design and the synthesis of systems.

Analysis: it is the study of the functioning of a system at specific conditions, which mathematical model is known. Generally, they are varied, the values of the parameters involved in the mathematical models in order to observe the different responses and from there to get conclusions. As the analysis depends on the mathematical model, it is independent of the kind of the system studied, being this mechanical, electrical or hydraulic.

Design: given a specific task, it is the process whereby we can find the system which accomplishes that task.  Usually, it is not a direct process and requires essay and error. The design implies to make it clear the requirements of the system, typically given in qualitative and quantitative terms. Subsequently, the engineer uses the synthesis. Once he has a model, the engineer analyzes the system so foresee the compliance of the requirements by mean of computerized simulation. By applying essay and error, the engineer modifies the model until it approximately meets the desired result. If it is possible, the engineer builds a prototype and continues the analysis until it meets the final goal.

Synthesis: it is the use of a specific procedure to find a system which works in a specific way. In this case, the characteristics of the system are postulated at the beginning and afterward the engineer uses several mathematical techniques to come up with the right system.

There are therefore two methods for designing (Distefano et al, 1995):

  1. Design from the analysis: it is made by mean of the modification of the characteristics of a system which already exists;
  2. Design from the synthesis: it is the definition of a system starting from its specifications.

In relation to control systems, Nise defines the functions of an engineer as follows:“…we discuss three major objectives of systems analysis and design: producing the desired transient response, reducing steady-state error, and achieving stability” (Nise, 2011, p. 10).