Control System Analysis, Sin categoría

Dynamik eines Masse-Feder-Dämpfer-Systems

Die Grundelemente eines jeden mechanischen Systems sind die Masse, die Feder und der Stoßdämpfer. Das Studium der Bewegung in mechanischen Systemen entspricht der Analyse dynamischer Systeme. In der Robotik zum Beispiel bezieht sich das Wort Vorwärtsdynamik darauf, was mit Aktuatoren passiert, wenn wir bestimmte Kräfte und Drehmomente auf sie anwenden.

Die Masse, die Feder, der Stoßdämpfer sind elementare Aktuatoren eines mechanischen Systems.

Um den Roboter zu steuern, ist es folglich notwendig, die Art der Bewegung eines Masse-Feder-Dämpfer-Systems sehr gut zu kennen.

Darüber hinaus wird dieses elementare System in vielen Anwendungsbereichen vorgestellt, daher die Wichtigkeit seiner Analyse. Wenn wir in der Robotik über Inverse Dynamic sprechen, sprechen wir darüber, wie man den Roboter auf eine gewünschte Art und Weise bewegt, welche Kräfte und Drehmomente wir auf die Aktoren anwenden müssen, damit sich unser Roboter auf eine bestimmte Art bewegt.

Bevor wir die dynamische Analyse unseres Masse-Feder-Dämpfer-Systems durchführen, müssen wir sein mathematisches Modell erhalten. Dies ist der erste Schritt für jeden, der die Dynamik eines Systems, insbesondere das Verhalten seiner mechanischen Komponenten, genau kennenlernen möchte.

Wir werden unsere Studie mit dem Modell eines Masse-Feder-Systems beginnen.

 

Dies ist aus folgendem Grund praktisch. Alle mechanischen Systeme haben eine Art in ihrer Bewegung, die sie zum Schwingen bringt, etwa wenn ein Gegenstand an einem Faden an der Decke hängt und mit der Hand, die wir drücken. Oder ein Schuh auf einer Plattform mit Federn. Es ist gut zu wissen, welche mathematische Funktion diese Bewegung am besten beschreibt.

Masse-Feder-System.
Abbildung 5

Die Dynamik eines Systems wird in erster Linie durch ein mathematisches Modell dargestellt, das aus Differentialgleichungen besteht. Im Falle des Masse-Feder-Systems ist diese Gleichung wie folgt:

Diese Gleichung ist als Bewegungsgleichung eines einfachen harmonischen Oszillators bekannt. Mal sehen, woher es stammt.

Wenn wir eine Formel erhalten wollen, die die Kraft beschreibt, die eine Feder gegen die Verschiebung ausübt, die sie dehnt oder schrumpft, ist es am besten, die potentielle Energie zu visualisieren, die in die Feder injiziert wird, wenn wir sie dehnen oder schrumpfen. Die folgende Grafik beschreibt, wie sich diese Energie als Funktion der horizontalen Verschiebung verhält:

Wenn sich die Masse m der vorhergehenden Figur, die an dem Ende der Feder angebracht ist, wie in Abbildung 5 gezeigt, von dem Federrelaxationspunkt x = 0 weg in die positive oder negative Richtung bewegt, sammelt sich die potentielle Energie U (x) an und steigt in parabolischer Form an und erreicht einen höheren Energiewert, wobei       U(x) = E, Wert, der der maximalen Dehnung oder Kompression der Feder entspricht. Die mathematische Gleichung, die in der Praxis diese Kurvenform am besten beschreibt und eine Konstante k für die physikalische Eigenschaft des Materials enthält, die die Steigung der Kurve erhöht oder verringert, ist die folgende:

Die Kraft ist auf folgende Weise mit der potentiellen Energie verbunden:

Deshalb:

Es ist sinnvoll zu sehen, dass F (x) umgekehrt proportional zur Verschiebung der Masse m ist. Denn es ist klar, dass, wenn wir die Feder dehnen oder schrumpfen, diese Kraft dieser Aktion entgegenwirkt und versucht, die Feder in ihre entspannte oder natürliche Position zurückzubringen. Aus diesem Grund heißt es Restitutionskraft. Die obige Gleichung ist in der Akademie als Hookes Gesetz oder Kraftgesetz für Federn bekannt. Das Folgende ist ein repräsentatives Diagramm dieser Kraft in Bezug auf die Energie, wie sie erwähnt wurde, ohne den Eingriff von Reibungskräften (Dämpfung), weshalb sie als der einfache harmonische Oszillator bekannt ist. Es ist wichtig, die proportionale Beziehung zwischen Verschiebung und Kraft zu betonen, aber mit einer negativen Steigung, und das ist in der Praxis komplexer, nicht linear.

Abbildung 4

Siehe: AMPLITUDE AND PHASE: SECOND ORDER II (Mathlets)

Sistema MRA

Nach Newtons zweitem Gesetz:

Diese Gleichung sagt uns, dass die vektorielle Summe aller Kräfte, die auf den Körper der Masse m einwirken, gleich dem Produkt des Wertes der Masse aufgrund ihrer Beschleunigung ist, die aufgrund der Kräfte erhalten wird. Mit Newtons zweitem Gesetz erhalten wir die folgende Gleichung:

Das ist:

Diese Gleichung repräsentiert die Dynamik eines idealen Masse-Feder-Systems.

System Masse-Feder-Stoßdämpfer

Wenn keine Reibungskraft vorhanden ist, oszilliert der einfache harmonische Oszillator unendlich. In Wirklichkeit nimmt die Amplitude der Oszillation allmählich ab, ein Prozess, der als Dämpfung bekannt ist und im folgenden graphisch beschrieben wird:

Die Verschiebung einer oszillierenden Bewegung ist gegen die Zeit aufgetragen, und ihre Amplitude wird durch eine sinusförmige Funktion dargestellt, die durch einen abnehmenden Exponentialfaktor gedämpft wird, der in dem Graphen als eine Hüllkurve erscheint. Die Reibungskraft Fv, die auf die amortisierte harmonische Bewegung einwirkt, ist in den meisten Fällen von wissenschaftlichem Interesse proportional zur Geschwindigkeit V. Diese Kraft hat die Form Fv = bV, wobei b eine positive Konstante ist, die unter anderem von den Eigenschaften des Fluids abhängt, das Reibung verursacht. Diese Reibung, auch bekannt als Viskosereibung, wird durch ein Diagramm dargestellt, das aus einem Kolben und einem mit Öl gefüllten Zylinder besteht:

Die gängigste Art, ein Masse-Feder-Dämpfer-System darzustellen, ist eine Reihenschaltung wie folgt:

Abbildung 6

 

Sowie die folgenden:

In beiden Fällen wird das gleiche Ergebnis bei Anwendung unserer Analysemethode erhalten. Wenn man Fig. 6 betrachtet, kann man sehen, dass dieselbe Konfiguration wie in Fig. 5 gezeigt ist, jedoch die Wirkung des Stoßdämpfers hinzugefügt wird. Indem wir Newtons zweites Gesetz auf dieses neue System anwenden, erhalten wir die folgende Beziehung:

Diese Gleichung repräsentiert die Dynamik eines Massen-Feder-Schock-Systems.

Das könnte dich interessieren:

  1. Beispiel 1 – System Transfer-Funktion Masse-Feder-Dämpfer
  2. Beispiel 1 – Elektromechanische Systemübertragungsfunktion

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Prof. Larry Francis Obando – Technical Specialist – Educational Content Writer 

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Control System Analysis

Example 1 – Transfer Function of a liquid-level system

We take into account the system of the Figure 3-23.

In accordance with the definitions of the previous article (Dinámica de un sistema de nivel de líquidos), we focus on the capacitances C1 and C2, as well as on the resistors R1 and R2. Thus, the dynamics of this system is determined by the following equations:

From here we can obtain the transfer function depending on what variables we define as input and output. For example, suppose the input to the system is q and the output variable is q2, so to find the transfer function of the system we execute the following operations:

So:

That is to say:

In the other hand:

That is:

Also: 

Applying Laplace to equations a, b and c, we obtain:

Clearing H1(s) of e and substituting this value in d we obtain:

An due to the latter is:

The transfer function of the system is:

 

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Control System Analysis, PID Control

PID – Study of the proportional action with Matlab

To make the study of proportional action, consider the following system:

1. Implement the locus of the roots indicating the points of interest. We execute the following commands in Matlab:

>> s=tf(‘s’)

>> s1=(1)/(0.2*s+1)

>> s2=(2)/(0.1*s+1)

>> s3=(1)/(s+1)

>> G=s1*s2*s3

The direct transfer function G(s) is:nullReshaping:nullIf we add a proportional controller with a gain Kp, the direct transfer function G(s) is:nullTh closed-loop transfer function Gce(s) is:null

nullThe characteristic equation is:nullThe characteristic equation in its form 1+G(s)H(s) is:

null

To obtain in Matlab the locus of the roots, we execute the following command:

>> rlocus(G)

We obtain:

Through this first exercise we get the locus of the roots of the system, we can observe the most immediate effect of applying a proportional controller: the displacement of the roots.

The change in the gain Kp allows us to change the value of the roots of the characteristic equation (when traveling through the blue, green and red lines of the locus in the previous graph, we see how the Kp gain changes), which is the same as changing the poles of the closed-loop transfer function. By changing these poles, we change the value of the damping coefficient ζ and the natural frequency ωn for a unit step input, thus adapting the transient response of the system to the design requirements that can be requested.

Note that the roots of the characteristic equation are in s=-10, s=-5 and s=-1, when Kp=0.

But in real terms Kp can not be zero, because in practice it means that we cancel the input to the system and thus, the output is zero as well. To represent the system operating without the proportional controller, we do Kp = 1. Under this condition, let’s see what is the value of ζ, as well as the value of three quantities of extreme importance for the designers: the overshoot Mp, the rise time Tr and the steady-state error ess. Then, suppose we want to modify this performance in terms of ζ and we vary Kp (we move the roots) until we achieve ζ = 0.5.

Kp=1

If Kp=1, then:null

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

We use the following commands in Matlab to know the values of ζ, the overshoot and the time of establishment:

> >Gce=feedback(G,1)

> >damp(Gce)

We obtain:

Pole Damping Frequency
-2.26e+00 + 2.82e+00i 6.26e-01 3.62e+00
-2.26e+00 – 2.82e+00i 6.26e-01 3.62e+00
-1.15e+01 1.00e+00   1.15e+01

Note that in the previous graph, Kp = Gain = 1. The dominant complex roots are s1 = -2.26 + j2.82 and s2 = -2.26-j2.82, so that, according to the graph, we can consider ζ = 0.626 when Kp = 1. Regarding the overshoot, the rise time and the steady-state error we use the following command:

>> stepinfo(Gce)

RiseTime: 0.5626

Overshoot: 7.5449

Peak: 0.7170

>> step(Gce)

If the input is the unit step and the output reaches the final value of c = 0.663, the steady-state error is ess = 1-0.663 = 0.337. We will see that despite varying the value of Kp and moving the roots, the proportional controller does not completely cancel out the steady-state error, it will always have a value different from zero, so an integral action is required to annul this error. On the other hand, the value of the overshoot is Mp = 7.17%, taking into account that the final value of the system is c = 0.663 and the maximum value reached is of c = 0.7170.

Find Kp to achieve ζ = 0.5.

The locus of the roots allows us to vary the value of the gain Kp until reaching the requested damping, ζ = 0.5. We move on the geometric place of the roots in Matlab by clicking on the line of the dominant poles and dragging the point until it reaches the requested damping:

The previous graph shows us that we can obtain a ζ = 0.5 when the gain Kp has an approximate value of 1.46. If Kp = 1.46, the direct transfer function and the closed-loop transfer function are:

We confirm the value of the damping by:

>> Gce2=(146)/(s^3+16*s^2+65*s+196)

> >damp(Gce2)

Pole Damping Frequency
-2.04e+00 + 3.51e+00i 5.02e-01 4.05e+00
-2.26e+00  – 3.51e+00i 5.02e-01 4.05e+00
-1.19e+01 1.00e+00   1.19e+01

> >stepinfo(Gce2)

RiseTime: 0.4356

Overshoot: 15.0397

Peak: 0.8569

It is observed that the overshoot will be higher (from 7.5449 to 15.0397) after the compensation (change the Kp value from 1 to 1.46) because the damping ζ is lower (from 0.626 to 5.02), while the rise time Tr improves slightly (from 0.5626 to 0.4356)

The answers to the unit step input of both systems (before and after the compensation), can be observed by the following Matlab command:

>> step(Gce,Gce2)

The final value of the system after the compensation (in red color) is approximately c = 0.748, so the steady-state error in this case is slightly lower, ess = 1-0.748 = 0.252. It is clearly seen in the graph that the rise time is shorter after the compensation, but at the cost of a larger overshoot due to a smaller damping.

Another more sophisticated Matlab tool to design compensators is SISO Design Tool. It can be called with rltool.

>> rltool

The graphical user interface is opened (GUI).

Once there, we can import systems fron the Matlab console through file>import>G>browse>available models>G>import>close>ok.

Supongamos el requerimiento ζ=0.5. Placing the cursor on the locus, we right click and select design requirement>new>design requirement type>damping ratio>0.5>ok. 

The lower legend is obtained by placing the course on the pink dot, a hand is formed and left click. We can vary the graph until we achieve approximately the desired damping. If we place the course on the left side of the graph (white color), the legend appears Loop gain changed to 1.47. That is to say Kp=1.47.

Although, to be more exact, the value of the gain is Kp = 1.4663. This value can be seen in the other window that opens simultaneously with the Editor: Control and estimation tools manager. There, when selecting the Compensator Editor tab, we can see that C = 1.4663. Therefore, the tool allows us to be much more specific in terms of the value of the gain.

Going back to the graphic editor (SISO design task), we select analysis>response to step command, we obtain the unit step response in a new window. Once there, right click, we select plot type>step. We can see the value of main characteristics of transient response with Kp=1.46:

characteristics>rise time

characteristics>peak response

Respuesta transitoria para diferentes valores de Kp.

To complete the study it is only necessary to assign several values to Kp and analyze the transient response as well as the steady-state error of the different systems, through the programming tools presented so far. It should be noted how sensitive the system is for very close Kp values. This is shown in the following graph where the answers to the unit step input appear simultaneously for different values of Kp:

Kp=0.2, Kp=0.5, Kp=1, Kp=1.5,  Kp=1.7, Kp=2.

null

Step response, Kp=2

 

 

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Control System Analysis, PID Control

Example 1 – PI Controller Design (Proportional-Integral) – Matlab

To better appreciate the effect of the PI controller, let’s look at the following example. Suppose we have the system of Figure 7-23.

null

The direct transfer function G(s) for this system is as follows:

null

Where K is the pre-amplifier constant.

Teh design specifications for this system are:

nullDonde:

  • ess: steady-state error due to a parabolic input
  • Mp: Maximum overshoot
  • Tr: Rise time
  • Ts: Settling time

The first thing we are going to do is analyze the steady-state error ess of the system before compensating, and see how much it meets or not with the first design requirement.

For a parabolic input we must work with the acceleration constant Ka:nullThis means an infinite error for a parabolic input:To improve the steady-state error we incorporate a PI controller in the direct path of the system, which will now have the following transfer function:
nullBy increasing the typology of the system from type 1 to type 2 we immediately improve the steady-state error. Now, the ess due to the parabolic input will be a constant:
nullThat is to say:

This exercise we already worked with the PD controller. For that case we select a value of K=181.17 (Example 1 – PD Controller Design (Proportional-Diferential) – Matlab)

We allow ourselves the freedom to consider the same value for K in this case in order to maintain the transient response under acceptable and known conditions when applying the PD of Example 1. If necessary we will adjust the value of K later. We can appreciate that to achieve an ess as specified in the design, the larger K is, the smaller will have to be Ki, which may be convenient. With the values of K and ess we can calculate a first approximate value of Ki to meet the requirements:

The next thing we will do is analyze the stability of the system because the selection of the Kp and Ki parameters depends on it. We will apply the Routh-Hurwitz criterion to calculate the limit values of the mentioned parameters in such a way that the system remains stable. For this we need the characteristic equation that arises from the transfer function of the closed loop system Gce(s):

null

The characteristic equation of the system is:

null

With this equation we apply the Routh-Hurwitz criterion. In this way we discovered that the system is stable for the following range of values:

null

This result indicates that the Ki / Kp of the controller can not be very close to zero, so a value as low as Ki = 0.002215 is not convenient. Another criterion for selecting Ki / Kp is that it is convenient to select the zero added by the controller in s = -Ki / Kp so that it is located near the origin and far from the most significant poles of the system. Through the locus of the roots in Matlab we can see which are the most significant poles of the characteristic equation, assuming an acceptable relationship between the Ki and Kp parameters from the point of view of the previous stability study but relatively close to the origin, say Ki / Kp = 10, keeping Ki constant and varying Kp. 

First, we reshape the characteristic equation in its form 1+G(s)H(s):

Note that in this way the controller PI is adding a zero in s=-10.

We apply the following command in Matlab to obtain the locus of the roots of this compensated system:

>> s=tf(‘s’)

>> sys=(815265*(s+10))/(s^3+361.2*s^2)

>> rlocus(sys)

So we get:

As it can be seen, the most significant pole of the characteristic equation is located in s=-361. Therefore, the criterion that we must use to select s=-Ki/Kp is:nullWith this result, the direct transfer function G(s):nullIt would be simplified as:null

The term Ki / Kp would be negligible compared to the magnitude of s when s assumes values along the root locus that corresponds to a convenient relative damping factor of 0.7 <ζ <0.1. Then, a zero cancels a pole at zero. The maximum overshoot must be equal to or less than 5%. This means that you want a relative damping factor ζ approximate to the following:

null

null

With the help of the locus we can locate the poles that correspond to this value of ζ:

According to the graph, the value of Kp required to obtain the desired damping factor is:

And so:

We also observe in the previous graph that if Kp = 0.0838, then the roots of the characteristic equation (the poles of the system) are in s1 = -175 + j184 and in s2 = -175-j184 If we look around these roots we can notice that the zero added by the controller at s = -10 is very close to the origin compared to the poles of s1 and s2, practically canceling a pole at the origin, thus ratifying the approach we made earlier for the direct transfer function of this system then of the compensation:

So:

We can observe the response to the step according to these partial results and the comparison of the compensated and uncompensated systems, by means of the following simulation:

>> Ga=(815265)/(s*(s+361.2))

>> Gd=(68319)/(s*(s+361.2))

>> Gce1=feedback(Ga,1)

>> Gce2=feedback(Gd,1)

>> step(Gce1,Gce2)

The previous graph, with the system after the compensation in red line, shows that the PI improves the steady-state error and reduces the overshoot, but at the expense of significantly increasing the rise time. The graph also shows that the maximum overshoot is 5%, therefore the requirement is met. It is necessary to note that another relationship can be selected for Ki and Kp that meets the requirement and still improve the overshoot, for example Ki / Kp = 5, or Ki / Kp = 2. You just have to pay attention to the issue of the stability of the system. With the calculated values, we re-calculate the overshoot, and we evaluate the rise time tr and the settlement time ts . The following command gives us the value of ζ and ωn, the relative damping factor and the natural frequency respectively.

>> damp(Gce2)

Pole: -1.81e+02 + 1.89e+02i /   -1.81e+02 + 1.89e+02i

Damping: 6.91e-01

Frequency: 2.61e+02

  •  Maximum Overshoot (MP):

  • Rise Time (Tr):

>> Gd=(68319)/(s*(s+361.2))

>> sys=feedback(Gd,1)

>> step(sys)

  • Settling time (Ts):null

Values that we can also see in the response to the step input graph generated previously:

We can conclude that the compensated system fulfills the requirements of the design, although it exceeds a little in the settling time. The latter can be reduced slightly with a Ki / Kp = 2 ratio, but at the expense of getting too close to the unstable area of the system

BEFORE: Example 1 – PD Controller Design (Proportional-Diferential) – Matlab

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

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Respuesta Transitoria de un Sistema de Control

Estabilidad de un sistema de control

Simulación de Respuesta Transitoria con Matlab – Introducción

 

Control System Analysis

Example 2 – Linearization of a Magnetic Levitation (MAGLEV) system – Sphere.

The magnetic suspension system of a sphere is shown in Figure 1.

The objective of the system is to control the position of the steel sphere by adjusting the current in the electromagnet through the input voltage e(t). The dynamics of the system is represented by the following differential equations:
Where:

It is requested to linearize the system around its equilibrium point.

Solutión

Figure 1 shows that the metallic ball is subject to two forces: Fem and G; that is, the electromagnetic force and gravity. From the dynamics of the systems we focus on the equation that relates both forces by Newton’s law:

This equation is non-linear and it is where it will be necessary to apply the linearization process. Writing this equation in another way we can see that the forces acting on the sphere have opposite sign, which means that they act in opposite directions:

Where:

It is logical to think that both forces are equal at the point of equilibrium, with coordinate xo, when the ball levitates and remains immobile. Since the displacement of the ball at this point remains constant, the speed and acceleration of the ball are equal to zero, so we know that:

Therefore, at the point of equilibrium:

Where:

Returning to the dynamics of the system, only one of the equations is non-linear:

To linearize this differential equation, we proceed as we have seen in: Linearization of non-linear systems

The excursion around the equilibrium point is represented by:

From where:

We write equation 1 again substituting its terms for those found for the equilibrium point:

Applying derivative law, taking into account that xo is a constant:

To linearize the electromagnetic force at the equilibrium point, we apply the following Taylor series:

Then:

Where:

Substituting these last results in equation 3 we obtain:

We now substitute this result in equation 2:

And so we achieve the goal of representing the nonlinear system by the following linear differential equations:

Note: The second equation (e (t) = …) can maintain its original form because it is linear, it was only necessary to change the independent variable i for that representing the excursion.

Source: Modelling and simulation of a magnetic levitation system, Valer Dolga, Lia Dolga, 2007.

PREVIOUS: Example 1 – Linearization of non-linear systems

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

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Control System Analysis, Matemática aplicada - Appd Math

Example 1 – Linearization of non-linear systems.

Linearize a function

Suppose we have a system represented by the following function:
nullOur task is to linearize f (x) around xo = π / 2. As:
nullWe find the following values and substitute them in the previous equation:
nullThen we can represent our nonlinear system by means of the following negative line equation:
null

The result of the linearization of f (x) around xo = π / 2 can be seen in Figure 2.48:

null

null

Linearize a differential equation

Suppose now that our system is represented by the following differential equation:
nullThe presence of the term cosx makes the previous one a non-linear equation. It is requested to linearize said equation for small excursions around x = π / 4.

To replace the independent variable x with the excursion δx, we take advantage of the fact that:
nullSo:nullWe proceed then to the substitution in the differential equation:nullWe now apply the derivation rules:nullAnd for the term that involves the cosx function we apply the same methodology that we have just seen in the previous example for a given function, that is, linearize f (x) around xo = π / 4:

Note that in the previous equation the excursion is zero when the function is evaluated exactly at the point xo. The same happens when the slope is evaluated in xo:So:

Therefore, we can rewrite the differential equation in a linear fashion around the point xo =π /4 as follows:

That is to say:

NEXT: Example 2 – Linearization of a Magnetic Levitation (MAGLEV) system – sphere. 

 

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

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Control System Analysis, Matemática aplicada - Appd Math

Linearization of non-linear systems.

Introduction

Many components and actuators have non-linear characteristics and the effectiveness of their action requires that they remain at the point of operation where they act approximately linearly, which can be a very limited interval. For example, the music that we all hear must be amplified by a circuit composed of electronic devices that only amplify the signal when they are acting at the point of operation in which the system is designed to act linearly; proof of this is that the output of the system as a whole is proportional to the input, that is, a linear system.

What is linearization? It is to express a non-linear function or differential equation with an approximate linear version, only valid in a very small range of values of the independent variable. Something like expressing a quadratic function by the mathematical formula of a straight line. To what end? Well, to be able to apply to the system represented by this function all the control techniques for linear systems studied up to now. Our objective is to design a strategy to generate a linear equation that represents a non-linear system in a very limited region, a strategy that we configure next.

To obtain a linear mathematical model of a non-linear system it is necessary to suppose that the variable to be controlled only deviates very slightly from an operation point A of coordinates (xo, f (xo)), where xo is the input to the system and f (xo) is the output. At point A we can place a line with a certain slope and assume that for small changes δx around xo we have the output f (xo+δx) moving along this line, as shown in Figure 2-47:null

We can use point A as a new center of coordinates where the independent variable δx corresponds to the input to the system, while the dependent variable δf (x) represents the output of the system. We make this convenient change of coordinates to use the equation of the slope ma of the line in the following way:
null

OrnullAnd so:null

In the same way that:null

The latter is a linear mathematical approximation for f (x).

This technique allows us to obtain a linear expression for f (x), around the point of operation A. Now, we are going to combine the obtained expressions for f (x) and δf (x). Another way of thinking it is to think that, around the point of operation A, f (x) has the value of f (xo) plus a small component of value maδx along a straight line of slope ma:
nullWhere (x-xo) is so small that it approaches δx. Mission accomplished, we will do this:
null

What theory allows us to do this? The Taylor series.

Taylor Series

The Taylor series are the expansion of a function f (x) in terms of the value of that function at a particular point xo, around that point and in terms of the derivatives of the function evaluated at that point:
null

When the excursion around the point xo is small, as the case that interests us, the derivatives of higher order can be ignored, so:
null

Knowing that the mx slope of a line at point xo is the derivative of the line evaluated in xo, we can adapt this last equation to our strategy and we obtain the formula that interests us:
nullWhere mx = df / dx evaluated at x = xo. Note that δx is now the independent variable, for which we use only a valid range of values around xo, so that δx is an excursion. Returning to Figure 2.47, this is the key tactic of the linearization process, we have created a coordinate system centered at point A, to replace the independent variable x with δx. We can continue using the δx notation or any other more practical notation such as:
nullLet’s see how this works through examples.

Linearize a function

Suppose we have a system represented by the following function:
nullOur task is to linearize f (x) around xo = π / 2. As:
nullWe find the following values and substitute them in the previous equation:
nullThen we can represent our nonlinear system by means of the following negative line equation:
null

The result of the linearization of f (x) around xo = π / 2 can be seen in Figure 2.48:

null

null

Linearize a differential equation

Suppose now that our system is represented by the following differential equation:
nullThe presence of the term cosx makes the previous one a non-linear equation. It is requested to linearize said equation for small excursions around x = π / 4.

To replace the independent variable x with the excursion δx, we take advantage of the fact that:
nullSo:nullWe proceed then to the substitution in the differential equation:nullWe now apply the derivation rules:nullAnd for the term that involves the cosx function we apply the same methodology that we have just seen in the previous example for a given function, that is, linearize f (x) around xo = π / 4:

Note that in the previous equation the excursion is zero when the function is evaluated exactly at the point xo. The same happens when the slope is evaluated in xo:So:

Therefore, we can rewrite the differential equation in a linear fashion around the point xo =π /4 as follows:

That is to say:

Linearization of a system with two independent variables

The Taylor series enables us to work with functions or differential equations that have two independent variables. In this regard, the Taylor series applies the following formula:

null

Where the point of operation has the coordinates ¯x1 y ¯x2. For small excursions around the equilibrium point, we can obviate the higher order derivatives. The linear mathematical model for this nonlinear system around the point of operation is obtained from:

Example. Linearization of a system with two independent variables.

Linearization of magnetic sphere levitation system.

The magnetic suspension system of a sphere is shown in Figure 1.

The objective of the system is to control the position of the steel sphere by adjusting the current in the electromagnet through the input voltage e(t). The dynamics of the system is represented by the following differential equations:
Where:

It is requested to linearize the system around its equilibrium point.

See the complete answer in the following link: Example 2 – Linearization of a Magnetic Levitation (MAGLEV) system – sphere. 

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

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

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

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Control System Analysis, Time Domain

Example 1 – Transient response of an electromechanical system.

The mechanical system shown in Figure P5.52(a) is used as part of the unity feedback system shown in Figure P5.52(b). Find the values of M and D to yield 20% overshoot and 2 seconds settling time.

 

1. System Dynamic

where:

2. Laplace Transform

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

4. Direct Transfer Function

For the system:

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

5. Closed-loop transfer function

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

That is to say:

6. Calculation of M and D

According to:

Besides:

In this way:

Meanwhile:

7. Matlab verification

We use Matlab to corroborate replacing all the values calculated in the original transfer function:

Find the values of M and D to yield 20% overshoot and 2 seconds settling time.

>> stepinfo (sys)

RiseTime: 0.3554

SettlingTime: 1.8989

SettlingMin: 0.9331

SettlingMax: 1.1999

Overshoot: 19.9890

Undershoot: 0

Peak: 1.1999

PeakTime: 0.8059

 

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Control System Analysis, Electromechanical Systems

Example 2 – Electromechanical system transfer function.

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

1. System Dynamic

where:

2. Laplace Transform

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

4. Direct Transfer Function

For the system:

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

5. Closed-loop transfer function

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

That is to say:

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

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

Contact: Caracas, Quito, Guayaquil, Cuenca. telf – 0998524011

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Control System Analysis, Electromechanical Systems

Example 1 – Electromechanical System Transfer Function

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

null

Data:

null

1. System dynamic

null

2. Laplace Transform

null

3. Block Diagram

null

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

null

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

Starting from:

nullWe obtain the following:nullThen, using:null

and substituting, we obtain:

null

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

null

Simplifying:null

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

null

From where:

null

null

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

null

5. System Transfer function.

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

null

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

null

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

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

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

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

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

WhatsApp: +593981478463

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