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

EL CAPACITOR

EL CAPACITOR

Actividad – Literature Review
Martes 03 octubre, 11:05 am

Fuentes:

  1. Introduccion-al-analisis-de-circuitos-robert-l-boylestad
    1. Capacitores 375 (387)
  2. Análisis de Redes – Van Valkenburg, 1999 – Network Analysis – Universidad de Illinois.
    1. El Parámetro Capacitancia p 20 (23)
  3. Análisis de Sistemas Lineales – Prof. Ebert Brea
    1. Análisis de Sistemas en el Dominio Continuo pp 29 – (58)

 

Preliminares

Por tanto se concluye que la intensidad del campo eléctrico en cualquier punto a una distancia r de una carga puntual de Q coulombs, será directamente proporcional a la magnitud de la carga e inversamente proporcional al cuadrado de la distancia a la carga.

Capacitancia

Al instante en que el interruptor se cierra, se extraen los electrones de la placa superior y se depositan sobre la placa inferior debido a la batería, dando por resultado una carga neta positiva sobre la placa superior del capacitor y una carga negativa sobre la placa inferior…Cuando el voltaje en el capacitor es igual al de la batería, cesa la transferencia de electrones y la placa tendrá una carga neta Q=CV=CE

En este punto el capacitor asumirá las características de un circuito abierto: una caída de voltaje en las placas sin flujo de carga entre las placas.

El voltaje en un capacitor no puede cambiar de forma instantánea.

De hecho, la capacitancia en una red es también una medida de cuanto se opondrá ésta a un cambio en el voltaje de la red. Mientras mayor sea la capacitancia, mayor será la constante de tiempo y mayor el tiempo que le tomará cargar hasta su valor final

Ejemplo 2.2 (Fuente:3) 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.

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)

Introduction to differential equation and modeling

Fuente: Introduction to Differential Equations

    1. Motivation
    2. A secret function
    3. Cell division
    4. Classification of differential equations
    5. Homogeneous linear ODE
    6. Introduction to modeling
    7. Model of a savings account
    8. Application: mixing salt water solution
    9. Systems and signals
    10. Newtonian mechanics
    11. 5 step modeling process

Today’s objectives

  1. Identify linear first order differential equations.
  2. Model behavior of certain systems using first order linear differential equations.
  3. Use the input signal and system response paradigm to obtain an ODE for a physical system.
  4. Check reasonableness of models using unit analysis .

 

Definition 3.2 An initial value problem is a differential equation together with initial conditions.

 

4. Cell division

Here we will see how the differential equation for our secret function appears when modeling a natural phenomenon – the population growth of a colony of cells…In this example we’ll model the number of yeast cells in a batch of dough. As we work through this example, pay careful attention to the assumptions we make, and how the initial condition plays a role in the resulting differential equation.

For our system, we assume we have a colony of yeast cells in a batch of bread dough. The first step is to identify the variables, the units, and give them names.

y

number of cells

t

time measured in seconds

We also need to set some initial condition, y0, the number of cells that we begin with at t=0. In this system, this might be the number of yeast cells in a yeast packet.

A differential model

If y denotes the number of yeast cells, what can we say about the derivative y˙? The derivative represents the rate at which the number of cells is growing. How should it depend on the number of cells? In nature, cells given plenty of space and food tend to divide through mitosis regularly. If we assume that each cell is dividing independently of all other cells, then doubling the number of cells should double the rate at which new cells are born. In fact, multiplying the number of cells by any scalar factor should do the same to the derivative. So this directly implies that the growth rate of cells is proportional to the number of cells:

y˙∝y.

We can make this into a true equation by simply inserting a proportionality constant a, such that

y˙=ay.

We say that 1/a is a “characteristic” timescale for our problem, setting the rate at which the cells divide. A solution to the above differential equation is

where y0 is the number of yeast cells we started with at t=0. In our case, we assume that y0 is the number of yeast cells in a packet, which is about 180 billion yeast cells.

5. Classification of differential equations

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There are two kinds:

  • An ordinary differential equation (ODE) involves derivatives of a function of only one variable.
  • A partial differential equation (PDE) involves partial derivatives of a multivariable function.

When we consider ODEs, we will often regard the independent variable to be time…The dot notation y˙ should only be used to refer to a time derivative. If for example y is a function of a spacial variable y=y(x), we will only use the notation y′ to denote the derivative with respect to x.

Definition 5.1 The order of a DE is the highest n such that the nth derivative of the function appears…

The order is 5, because the highest derivative that appears is the 5th derivative, y(5).

7. Natural growth and decay equations..We’ve been introduced to a few basic forms of differential equations so far. The first equation we saw was a basic growth equation,

y˙=ay,

which, when a is a positive constant, governs systems like bank accounts and cell populations. If we put a negative sign in front of a we get the decay equation

y˙=−ay,

which can be used to describe things like radioactive decay of materials.

How would you classify the differential equations y˙=ay and y˙=−ay just discussed? Choose all descriptors that apply…
Solution:

These two equations are both first order, linear, homogeneous differential equations. To see that these equations are homogeneous, we can either check that y=0 is a solution (it is), or we can rewrite them in standard linear form:

8. Introduction to modeling

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There are two kinds of modeling. We’re not going to talk about the kind that involves fancy clothes and photographs. The other kind, mathematical modeling , is converting a real-world problem into mathematical equations.

Guidelines:

  1. Identify relevant quantities, both known and unknown, and give them symbols. Find the units for each.
  2. Identify the independent variable(s). The other quantities will be functions of them, or constants. Often time is the only independent variable.
  3. Write down equations expressing how the functions change in response to small changes in the independent variable(s). Also write down any “laws of nature” relating the variables. As a check, make sure that all summands in an equation have the same units.

Often simplifying assumptions need to be made; the challenge is to simplify the equations so that they can be solved but so that they still describe the real-world system well.

I have a savings account earning interest compounded daily, and I make frequent deposits or withdrawals into the account. Find an ODE with initial condition to model the balance.

Simplifying assumptions:

  • Daily compounding is almost the same as continuous compounding, so let’s assume that interest is paid continuously instead of at the end of each day.
  • Similarly, let’s assume that my deposits/withdrawals are frequent enough that they can be approximated by a continuous money flow at a certain rate, the net deposit rate (which is negative when I am withdrawing).

Variables and functions (with units): Define the following:

P

the initial amount that the account starts with (dollars)

t

time from the start (years)

x

balance (dollars)

I

the interest rate (year−1; for example 4%/year=0.04year−1)

q

the net deposit rate (dollars/year).

Here t is the independent variable, P is a constant, and x, I, q are functions of t.

Equations: Now we want to decide how the balance changes as time changes. We’ll estimate the change in the balance Δxas time increases from some time t to a time t+Δt. We can approximate the interest earned per dollar to be:

Note that the units in each of the three terms are dollars/year. Also, there is the initial condition x(0)=P. Thus we have an ODE with initial condition:

Now that the modeling is done, the next step might be to solve this DE, but we won’t do that yet.

Remark 9.2 The notation we chose suggested that the interest rate I depended only on time. However, I could have depended on x as well. This would not change the modeling process. If I does not depend on x, we obtain a linear differential equation. If it does, the equation is nonlinear.

Video: Application: mixing salt water solution

Systems and signals

Let’s get back to the savings account model:

x˙=I(t)x+q(t).

Maybe for financial planning I am interested in testing different saving strategies (different functions q) to see what balances x they result in. To help with this, rewrite the ODE as

In the “systems and signals” language of engineering, q is called the input signal , the bank is the system , and x is the output signal . These terms do not have a mathematical meaning dictated by the DE alone; their interpretation is guided by the system being modeled. But the general picture is this:

The system may be a mechanical system such as an automobile suspension or an electrical circuit, or an economic market. It is impacted by some external signal. We are interested in understanding how the system responds to the external stimulus.

  • The input signal is the external stimulus. It usually does not appear in as simple a way in the DE as it does in the example above. But it does always determine the right hand side of the DE (when written in standard linear form).
  • The system response (also called output signal ) is the measurable behavior of the system that we are interested in. It is always the unknown function that we write a differential equation for.
  • All differential equations have many solutions. The solution of interest is often determined by the state of the system at the beginning. This initial state is given by the initial conditions.

Newtonian mechanics

Let’s try to put this into the input/ system response paradigm we’ve just introduced. The system response is the displacement of the mass. This is what we are interested in.

What is the input signal? You could imagine that there are other forces acting on the mass, like there is a sail on the mass, and wind is blowing on the sail creating an input signal. But we are going to start by considering the case where the input signal is 0. Note that pulling the cart back and releasing it specifies the initial state of the system, that is, it gives the initial conditions.

Now we are ready to write down the differential equation . The equation is governed by Newton’s second law

We need to identify the forces acting on the mass. There is the force due to the spring. For the moment, we assume that air resistance is negligible, and there is no friction on the cart.

What is the spring force? When the displacement is positive, the spring is stretched, the force is negative. When the displacement is negative, the spring is compressed, the force is positive. Thus this force is modeled linearly by Hooke’s law:

which is a function of the displacement x away from the neutral position x=0. Note that this linear model is only valid for relatively small displacements. If we stretch the spring too far, the spring force won’t obey this linear law anymore.

The position at time t=0 is x(0)=x0 for some positive displacement x0>0. From the problem statement, we assume that we release the cart with zero initial velocity, x˙(0)=0.

Putting this all together, we get:

with initial conditions:

The last step is to write this in standard linear form . We obtain the following differential equation:

Now let’s consider the same mass/spring system as above where we’ve add a sail to the mass.

The mass now experiences an additional external force from the wind. How does this change the model?

Solution: The model is exactly the same. The only difference is that the input signal is no longer zero, rather it is now the external force due to the wind on the sail. This external force Fwind(t) depends on time in some complicated way that we will not try to write down. The differential equation for this system is:

5 step modeling process

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In the example on the previous page, we outlined a 5 step modeling process that we make explicit here.

  1. Draw a diagram of the system.
  2. Identify and give symbols for the parameters and variables of the system.
  3. Decide on the input signal and the system response. Identify any initial conditions.
  4. Write down a differential equation relating the input signal and the system response.
  5. Rewrite the equation in standard linear form with initial conditions.

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)

Señales de tiempo continuo – Definición

Primero que nada el analista de sistemas debe conocer cómo modelar señales en el dominio del tiempo continuo – LFO (04-09-2017)

Fuentes (Copia Textual – Literature Review).

  1. Fundamentos_de_Señales_y_Sistemas_usando la Web y Matlab
  2. LibroASL_Edicion_2015
  3. Analisis_de_Sistemas_Lineales
  4. Oppenheim – Señales y Sistemas
  5. Nota1Señales
  6. Contenido:
  1. Introducción
  2. Función Escalón
  3. Función Rampa
  4. Función Impulso
  5. Función Pulso Rectangular
  6. Función Pulso Triangular
  7. Función Periódica
  8. Función Exponencial

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Función Escalón

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Función Rampa

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Función Impulso

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Función Pulso Rectangular

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Función Pulso triangular

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Función Periódica

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Función Exponencial

Fuente: [5]

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Fuente: [5]

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)

Linear Algebra – Foundations

Linear Algebra

Start Line:

  1. DLM
    1. Appd Mathematics
      1. Linear Algebra…What You Will Learn:
        • Represent quantities that have a magnitude and a direction as vectors.
        • • Read, write, and interpret vector notations.
        • • Visualize vectors in R2.
        • • Perform the vector operations of scaling, addition, dot (inner) product.
        • • Reason and develop arguments about properties of vectors and operations defined on them.
        • • Compute the (Euclidean) length of a vector.
        • • Express the length of a vector in terms of the dot product of that vector with itself.
        • • Evaluate a vector function.
        • • Solve simple problems that can be represented with vectors.
        • • Create code for various vector operations and determine their cost functions in terms of the size of the vectors.
        • • Gain an awareness of how linear algebra software evolved over time and how our programming assignments fit into this
        • (enrichment).
        • • Become aware of overflow and underflow in computer arithmetic (enrichment).
        • Become practical with the use of Matlab to apply all these framework

 

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

Actividad WBS (Vector Algebra – )
Martes 15, 06:32 am

Martes 29, 04:41 am

Jueves 31, 04:41 am

Lunes 04, 5:37 am

Lunes 11, 5:37 am

Martes 12, 4:46 am

Jueves 14, 5:15 am

I keep attending:

  1. LAFF: Linear Algebra – Foundations to Frontiers
    1. Overview of the Course
    2. 0.3.3 MATLAB Basics
    3. Origins of MATLAB
    4. Vectors in Linear Algebra
      1. Notation
      2. Unit Basis Vectors
      3. Simple Vector Operations
        1. Equality, Assignment and Copy
        2. Vector Addition
        3. Scaling
        4. Subtraction
      4. Advanced Vector Operations
        1. Scaled Vector Addition (AXPY)
        2. Linear Combinations of Vectors
        3. Dot or Inner Product (DOT)
        4. Vector Length (NORM2)
        5. Vector Functions
        6. Vector Functions that map a vector to a vector
    5. The Science of NFL Football: Vectors

 

 

This course is not only designed to teach the standard topics in a typical linear algebra course, but it also investigates how to translate theory into algorithms. Like typical in the algebra courses, we will often start studying operations with small matrices. In practice, however, one often wants to perform operations with large matrices so we generalize the techniques to formulate practical algorithms and their implementations
If you want to learn more about MATLAB, here are some suggestions you may want to investigate:

  • Matlab Onramp is a free 2-hour interactive online tutorial.
  • MATLAB Central is a place where people interested in MATLAB can be part of a community. Here, you may want to check out ThingSpeak™, the open loT Platform with MATLAB Analytics that allows you to aggregate, visualize, and analyze live data streams in the cloud.

 

Definition from Vectors in Linear Algebra

Definition 1.1 We will call a one-dimensional array of n numbers a vector of size n:

denotes the set of all vectors of size n with components in R.

Unit Basis Vectors

 

 

 

Equality, Assignment and Copy

 

Now we could talk about an algorithm for setting y equal to x. We’re computing y becomes x. We’ve already seen that each of the components of y has to be set to a corresponding component of x. So psi sub i has to become chi sub i. We have to do this for all indices i from 0 to n minus 1. We start indexing at 0, and therefore, if the vectors are of length n, we have to run this until n minus 1. We create an algorithm for this assignment by now writing this as a FOR LOOP.

 

Vector Addition

Now if we wanted an algorithm for computing the result vector, z, that results from adding x to y, then we can expose the components of z, the components of x and y, and we recognize that the ith of z just equals to the some of the ith components of x and y. And we can then summarise that as a little loop that says for all components of z, for i from 0 to n minus 1, the ith component of z, zeta sub i, should just be computed as chi sub i, added to psi sub i.

Scaling

 

What if instead we want an algorithm that computes vector y as the stretched vector x stretched by a scaling factor alpha? y becomes alpha times x. We expose the components of y. We need to compute alpha times x, where here we expose the individual components of x. And all we need to do is set the appropriate element of y to the corresponding element of x scaled by alpha. If we do this as an algorithm, then what we need to do to set psi i equal to alpha times chi i.

Subtraction

 

Let’s review the parallelogram method for vector addition. You lay out your vectors as such. And then, the diagonal becomes x:

You can do the same thing for vector subtraction. You lay out your vectors. And then, the other diagonal becomes the vector x minus y. Obviously, you have to make sure that it points in the right direction.

Now, how do you compute x minus y? Well, you expose the different components of vectors x and y. And you simply subtract each component of y off the corresponding component of x.

In Summary:

Scaled Vector Addition (AXPY)

 

We’re now going to talk about an operation that is going to be very important as we start looking at more complex operations, and then the algebra later on. It’s hard to picture though. It’s known as the axpy operation, and it takes a vector, scales it, and then adds it to another vector. Given two vectors x and y of size n and a scalar alpha, the axpy operation is given by y is equal to alpha x plus y.

Specifically

 

These kinds of vector operations have been very important since the 1970s, and back then the language of choice in this area was Fortran 77. Fortran 77 had the limitation that the variables and subroutines had to be identified with at most six letters and numbers. So they had to be somewhat innovative about how to name operations, and subroutines that implemented them. And the axpy here is simply an abbreviation of alpha times x plus y. So it stands for scalar alpha, the a, times x, plus, p, y–axpy.

 

If we now want to look at an algorithm for performing this operation, notice that the i-th component of y has to be updated by scaling the i-th component of x and adding it to the i-th component of y. So psi i becomes alpha times chi i plus psi i. And as usual, we need to put a loop around that so this is done for all components zero to n minus 1.

About the AXPY operation, it is often emphasized that it is typically used in situations where the output vector overwrites the input vector y.

Linear Combinations of Vectors

If we’re given two vectors of length m, u and v, and two scalars, alpha and beta, then taking the linear combination of u and v with coefficients alpha and beta is given by alpha times u plus beta times v. So that’s the scalar times the vector u, plus the scalar times the vector v. If we expose the components of u and v, then what does this mean? It means that we scale the first vector, that mean scale each of the individual components, by alpha. And we scale the components of vector v by beta. And that gives us, this right here. So taking this linear combination of vectors u and v, using the coefficients alpha and beta, means that we take the same linear combination of each of the components of u and v.

More generally

Well, instead of writing things like this, we could write them like this. What is that? That’s an AXPY. Why? 0 is a vector. And then this is a scalar times a vector, which you add to that vector. Once you’ve computed this vector, you take a scalar times the second vector and add it to that. So now, the first AXPY, we computed this. The second AXPY computes this. And you can imagine that we can do that for all of the vectors until we’re completely done.

This then motivates the following algorithm. You start by setting w equal to 0. And then for j equals 0 to n minus 1, you perform this operation right here where you take a the scalar chi j times v j and at that to w. So for j equals 0, that’s this operation. That then is stored in w. Then this here is what you do for j equals 1 and so forth.

Shortly, this will become really important as we make the connection between linear combinations of vectors, linear transformations, and matrices.

Dot or Inner Product (DOT)

 

If we’re given vectors x and y of size n, then the dot product is defined as follows.

Now what do we have here? We multiply the first components together, and then we multiply the second components together, and add those to the first components. And then we keep doing that until we get to the last components. We multiply those together, and we add those in as well. We can write this more concisely as I equals 0 to n minus one, of Chi i times Psi i.

Now to motivate an algorithm, let’s look at this slightly differently. Let’s think of this as, take an alpha, and first assigning 0 to it. After that, we multiply the first two components together, and we add that to 0. After that, we multiply the next two components together, and we add that to what already is in alpha, and so forth…This motivates the algorithm given here. You start by setting alpha equal to zero. And then, for i equals zero to n minus 1, you take what is already accumulated in alpha, and you add to it the product of the components Chi i and Psi i.

Now often we will use a slightly different notation to denote the dot product. OK, the dot product is given by this. We will often write this as x transpose y. OK? This T here means transposition. Now what does transposition mean? If we expose the components of x and y, then the transposition means that you take x, which is a column vector, and you make it into a row vector, as such. So the column vector turns into a row vector. Transposition means taking the vector and putting it on its side. And then multiplying the row vector times a column vector means multiplying the first components together, and adding that to the second components multiplied together, and so forth.

Vector Length (NORM2)

If we take that further and we look at a vector of size n, then the length of that vector is given by the square root of the squares of the components,which we can use shorthand to write as the sum of the squares of the components.

There’s a relation between the dot product and the length of a vector.

And therefore, we conclude that the length of vector x is just a square root of the dot products of x with itself. We summarize that right here.

Vector Function

 

A vector function is a function that takes one or more scalars and/or one or more vectors as inputs and then produces a vector as an output.

Well, let’s look an an example. So here we have an example of f, which is a function of two scalars. How do we know these are scalars? Well, notice that I’m using Greek letters. We agree that the Greek lowercase letters we were going to use for scalars. So it takes two scalars as input, alpha and beta, and then produces a vector of size two as output, where the first component adds the two input scalars and the second component subtracts the second scalar from the first scalar.

If we want to evaluate f of -2, 1, then all we do is we substitute -2 in for alpha. And we substitute 1 in for beta. So this here then would be the vector -2 plus 1, -2, minus 1. And if you do the arithmetic, you get the vector -1, -3. That’s summarized right here.

Let’s do another example. Here we have a function of the scalar and a vector of size three. And the output is that vector, except that each of its components has been changed by adding the scalar to it. So if you want to evaluate this function for a specific input, -2 for the scalar and the vector 1, 2, 3, then again what we need to do is substitute the -2 in for alpha. And we need to substitute 1, 2, and 3 in for the components of the vector that’s the input.

We have seen other examples already. We saw the AXPY operation, which if you think of it as a function, is the function axpy of a scalar alpha and then vectors x and y. And the output is the vector alpha times x plus y. We also saw the DOT function. And notice that in the DOT function, you have two vectors as input, x and y. And the result is the DOT product of the two factors, which is a scalar. Now you might say a scalar is not a vector. But we’re going to think of a scalar often as a vector of size one.

What we will see in the next unit is that we can think of these vector functions as mapping one vector to another vector.

Vector Functions that map a vector to a vector

Now we’re ready to look at functions that map vectors to vectors. Next week, we’ll look at a special case of those kinds of functions called “linear transformations.” We’re going to be looking at our functions that map a vector of size n to a vector of size m.

In the previous units, we looked at a function that took two scalars as input and produced a vector as an output.

We can look at a function g that takes as input a vector with components alpha and beta and then produces the exact same vector as the function f produced.

Here was another example of a function that took as input a scalar and a vector.

We can instead look at a function g, but now stacks scalar on top of the vector creating a vector that is of size four instead of the size three vector that we had before and then evaluates in exactly the same way.

The whole point being that now we have a function that takes as input a vector and as output, produces a vector.

So in summary, this insight allows us to focus on vector functions that simply take one vector as input and produce one vector as output. What we will see next week is that there’s a special class of such functions called “linear transformations” that are of great importance to linear algebra.

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)