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Victor Manuel Hernández-Guzmán and Ramón Silva-OrtigozaAutomatic Control with ExperimentsAdvanced Textbooks in Control and Signal Processinghttps://doi.org/10.1007/978-3-319-75804-6_2

2. Physical System Modeling

Victor Manuel Hernández-Guzmán¹ and Ramón Silva-Ortigoza²

(1)

Universidad Autonoma de Queretaro, Facultad de Ingenieria, Querétaro, Querétaro, Mexico

(2)

Instituto Politécnico Nacional, CIDETEC, Mexico City, Mexico

Automatic control is interested in mathematical models describing the evolution in time of system variables as a response to excitation. Such mathematical models are also known as dynamical models and systems represented by these models are called dynamical systems. The dynamical models that are useful for classical control techniques consist of differential equations, in particular, ordinary differential equations where the independent variable is time. Partial differential equations arise when several independent variables exist, for instance, a flexible body where vibration has to be described with respect to both position in the body and time. However, this class of problem is outside the scope of classical control; hence, partial differential equations will not be used to model physical systems in this book.

On the other hand, ordinary differential equations can be linear or nonlinear, time-invariant or time-variant. Because of the mathematical complexity required to study nonlinear and time-variant differential equations, classical control constrains its study to linear invariant (also called constant coefficients) ordinary differential equations. As real physical systems are intrinsically nonlinear and time-variant, it is common to consider several simplifications when modeling physical systems in classical control. For instance, model nonlinearities are approximated by linear expressions and parameters that change with time (such as resistance in an electric motor) are assumed to be constant in time, but the design of the controller must be robust. This means that the controller must ensure good performance of the designed closed-loop control system, despite the changes in such a parameter.

It is important to remark that model simplifications are only valid if the simplified model is still accurate enough to represent the true behavior of the physical system. This idea is captured by stating that a good model must be simple enough to render possible its mathematical study, but it must also be complex enough to describe the important properties of the system. This is the approach employed in the present chapter to model physical systems, following many ideas in [1].

2.1 Mechanical Systems

In mechanical systems there are three basic phenomena: bodies, flexibility, and friction . The main idea is to model each one of these phenomena and then connect them to obtain the mathematical model of the complete mechanical system. Furthermore, there are two different classes of mechanical systems: translational and rotative. In the following, we study both of them.

2.1.1 Translational Mechanical Systems

In Fig. 2.1 a rigid body is shown (without flexibility) with mass m (a positive quantity), which moves with velocity v under the effect of a force F. It is assumed that no friction exists between the body and the environment. The directions shown for force and velocity are defined as positive for this system component. Under these conditions, the constitutive function is given by Newton’s Second Law [2], pp. 89:

$\displaystyle \begin{aligned} \begin{array}{rcl} F=ma,{} \end{array} \end{aligned}$

(2.1)

where $a=\frac {dv}{dt}$ is the body acceleration.

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig1_HTML.png — Fig. 2.1
A rigid body with mass m in a translational mechanical system

In Fig. 2.2, a spring is shown, which deforms (either compresses or stretches) with velocity v under the effect of a force F. Notice that, although force F is applied at point 1 of the spring, the same force but in the opposite direction must also be considered at point 2 to render spring deformation possible. Moreover, v = v ₁ − v ₂ where v ₁ and v ₂ are velocities of points 1 and 2 of spring respectively. The directions shown for forces and velocities are defined as positive for this system component. Notice that velocity v indicates that point 1 approaches point 2. It is assumed that the spring has no mass, whereas deformation is not permanent and does not produce heat.

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig2_HTML.png — Fig. 2.2
A spring in a translational mechanical system

The net spring deformation is given as:

$\displaystyle \begin{aligned} \begin{array}{rcl} x=\int_0^tv\ dt+x(0),{} \end{array} \end{aligned}$

(2.2)

where x = x ₁ − x ₂ with x ₁ and x ₂ the positions of points 1 and 2 on the spring respectively, whereas x(0) = 0 represents the spring deformation when it is neither compressed nor stretched. According to the direction of velocity v defined in Fig. 2.2, it is concluded that the net deformation x is positive when the spring is compressed. In the case of a linear spring, the constitutive function is given by Hooke’s Law [2], pp. 640:

$\displaystyle \begin{aligned} \begin{array}{rcl} F=k\ x, {} \end{array} \end{aligned}$

(2.3)

where k is a positive constant known as the spring stiffness constant.

Friction always appears when two bodies are in contact while a relative movement exists between them. In such a case, a force must be applied to maintain the relative movement. Kinetic energy converts into heat because of friction. The constitutive function of friction is given as:

$\displaystyle \begin{aligned} \begin{array}{rcl} F=\varphi(v), \end{array} \end{aligned}$

where v is the relative velocity between the bodies and F is the applied force. A very important specific case is viscous friction, which is represented by the linear constitutive function :

$\displaystyle \begin{aligned} \begin{array}{rcl} F=b v,\text{viscous friction}{} \end{array} \end{aligned}$

(2.4)

where b is a positive constant known as the viscous friction coefficient. This kind of friction appears, for instance, when a plate with several holes moves inside a closed compartment with air, as shown in Fig. 2.3. Because air must flow through the holes as relative movement exists with velocity v, it is necessary to apply a force F to maintain the velocity. It is defined v = v ₁ − v ₂ where v ₁ and v ₂ are velocities of bodies 1 and 2 respectively. The directions of forces and velocities shown in Fig. 2.3 are defined as positive. Although the device shown in Fig. 2.3 may not be physically placed between the two bodies, the effect of viscous friction is commonly present and must be taken into account.

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig3_HTML.png — Fig. 2.3
Viscous friction appears as a result of opposition to air flow through the holes

There are other kinds of friction that often appear in practice. Two of them are static friction and Coulomb friction. The respective constitutive functions are given as:

$\displaystyle \begin{aligned} \begin{array}{rcl} F&=&\left.\pm F_s\right|{}_{v=0}, \text{static friction},{} \end{array} \end{aligned}$

(2.5)

$\displaystyle \begin{aligned} \begin{array}{rcl} F&=& F_c\ \text{sign}(v), \text{sign}(v)=\left\{\begin{array}{ll} +1,&\text{if }v>0\\ -1,&\text{if }v<0 \end{array}\right.,\text{Coulomb friction}. {} \end{array} \end{aligned}$

(2.6)

Static friction is represented by a force that prevents movement only just before this starts. This is the reason why constant F _s is evaluated at v = 0 in (2.5). Coulomb friction, in contrast, only appears when movement has started (v ≠ 0) and it is constant for velocities having the same sign. The constitutive functions in (2.4), (2.5), and (2.6) are depicted in Fig. 2.4.

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig4_HTML.png — Fig. 2.4
Constitutive functions defined in (2.4), (2.5) and (2.6)

As the static friction force is not zero as velocity tends to zero, it is responsible for some of the problems in position control systems: the difference between the desired position and position actually reached by the mechanism is different from zero, which is known as “a nonzero steady state error.” On the other hand, static and Coulomb friction have nonlinear and discontinuous constitutive functions; hence, they can be handled using control techniques developed for neither linear systems nor “smooth” nonlinear systems. This is the reason why these kinds of friction are not often considered when modeling control systems. However, it must be remarked that the effects of these frictions are always appreciable in experimental situations. The reader is referred to [4] to know how to measure the parameters of some friction models.

Once the constitutive functions of each system component have been established, they must be connected to conform the mathematical model of more complex systems. The key to connecting components of a mechanical system is to assume that all of them are connected by means of rigid joints. This means that velocities on both sides of a joint are equal and that, invoking Newton’s Third Law , each action produces a reaction that is equal in magnitude but opposite in direction. This is clarified in the following by means of several examples.

Example 2.1

Consider the mass-spring-damper system shown in Fig. 2.5a where an external force F(t) is applied on mass m. This system is, perhaps, the simplest mechanical system from the modeling point of view, but it is very important from the control point of view because it represents the fundamental behavior of position control systems.

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig5_HTML.png — Fig. 2.5
A mass-spring-damper system

A free-body diagram is shown in Fig. 2.5b. Notice that two forces, equal in magnitude but with opposite directions, exist at each joint between two components. This is to satisfy Newton’s Third Law [2], pp. 88, which establishes that for each action (from A on B) there is one corresponding reaction (from B on A). Wheels under the mass indicate that no friction exists between the mass and the floor. This also means that gravity has no effect. From the friction modeling point of view, it is equivalent to eliminating a damper at the right and replace it by friction between the mass and the floor. The friction model between the mass and the floor is identical to the friction model due to a damper as considered in the following.

The nomenclature employed is defined as:

$v_m=\frac {dx_m}{dt}$ , where x _m is the mass position.
$v_b=\frac {dx_b}{dt}$ , where x _b is the position of the damper’s mobile side.
$v_K=\frac {dx_K}{dt}$ , where x _K is the position of the spring’s mobile side.

The following expressions are found from Fig. 2.5b, and (2.1), (2.3), (2.4):

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Mass:}\hspace{1cm} m\frac{d v_m}{dt}=F(t)+F_K-F_b.{} \end{array} \end{aligned}$

(2.7)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Spring:}\hspace{1cm} Kx_K=F_K.{} \end{array} \end{aligned}$

(2.8)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Damper:}\hspace{1cm} b v_b=F_b.{} \end{array} \end{aligned}$

(2.9)

Notice that, according to the sign definitions shown in Fig. 2.1, forces actuating on mass m having the same direction as v _m in Fig. 2.5b appear affected by a positive sign in (2.7); therefore, F _b appears affected by a negative sign. Signs on both sides of expressions in (2.8) and (2.9) are given according to the sign definitions shown in Figs. 2.2 and 2.3 respectively. Also notice that one side of both spring and damper has no movement and the mobile sides have a velocity and a force applied in the same direction as in Figs. 2.2 and 2.3. Finally:

$\displaystyle \begin{aligned} \begin{array}{rcl} v_K=\frac{dx_K}{dt},{} \end{array} \end{aligned}$

(2.10)

has been defined. Assuming that all of the system components are connected by rigid joints and according to Fig. 2.5b, the following can be written:

$\displaystyle \begin{aligned} \begin{array}{rcl} v_K=-v_m=-v_b.{} \end{array} \end{aligned}$

(2.11)

Hence, it is concluded that:

$\displaystyle \begin{aligned} \begin{array}{rcl} x_K=-x_m+c_1=-x_b+c_2, \end{array} \end{aligned}$

where c ₁ and c ₂ are two constants. If x _K = 0, x _m = 0 and x _b = 0 are defined as the positions of those points at the spring’s mobile side, the center of mass m and the damper’s mobile side respectively, when the system is at rest with F(t) = 0, then c ₁ = c ₂ = 0 and hence:

$\displaystyle \begin{aligned} \begin{array}{rcl} x_K=-x_m=-x_b. \end{array} \end{aligned}$

Using these facts, (2.7), (2.8), (2.9), can be written as:

$\displaystyle \begin{aligned} \begin{array}{rcl} m\frac{d^2 x_m}{dt^2}+b\frac{dx_m}{dt}+Kx_m =F(t).{} \end{array} \end{aligned}$

(2.12)

This expression is an ordinary, linear, second-order, constant coefficients differential equation representing the mass-spring-damper system mathematical model. Thus, assuming that the parameters m, b, and K are known, the initial conditions x _m(0), $\frac {dx_m}{dt}(0)$ are given, and the external force F(t) is a known function of time, this differential equation can be solved to find the mass position x _m(t) and the velocity $\frac {dx_m}{dt}(t)$ as functions of time. In other words, the mass position and velocity can be known at any present or future time, i.e., for all t ≥ 0. Finally, when analyzing a differential equation, it is usual to express it with an unitary coefficient for the largest order time derivative, i.e., (2.12) is written as:

$\displaystyle \begin{aligned} \begin{array}{rcl} \ddot x_m+\frac{b}{m}\dot x_m+\frac{K}{m}x_m =\frac{1}{m}F(t),{} \end{array} \end{aligned}$

(2.13)

where $\ddot x_m=\frac {d^2x_m}{dt^2}$ and $\dot x_m=\frac {dx_m}{dt}$ .

Example 2.2

A mass-spring-damper system is depicted in Fig. 2.6a. Both spring and damper are fixed to the same side of the mass m. The purpose of this example is to show that the mathematical model in this case is identical to that obtained in the previous example. The nomenclature is defined as:

$v=\frac {dx}{dt}$ , where x is the mass position.
$v_1=\frac {dx_K}{dt}=\frac {dx_b}{dt}$ , where x _K = x _b stands for the spring’s and damper’s mobile sides.

According to Fig. 2.6b and (2.1), (2.3), (2.4), the following can be written:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Mass:}\hspace{1cm} m\frac{d v}{dt}=F(t)+F_K+F_b.{}\\ &\displaystyle &\displaystyle \text{Spring:}\hspace{1cm} Kx_K=F_K.\\ &\displaystyle &\displaystyle \text{Damper:}\hspace{1cm} b v_1=F_b. \end{array} \end{aligned}$

(2.14)

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig6_HTML.png — Fig. 2.6
A mass-spring-damper system

Notice that, according to the sign definition shown in Fig. 2.1, all forces in (2.14) appear with a positive sign because all of them have the same direction as v in Fig. 2.6b. On the other hand, as the joint is rigid and according to the sign definition in Fig. 2.6b:

$\displaystyle \begin{aligned} \begin{array}{rcl} v&\displaystyle =&\displaystyle -v_1.{}\vspace{-4pt} \end{array} \end{aligned}$

(2.15)

Proceeding as in the previous example, if x = 0, x _K = 0 and x _b = 0 are defined respectively, as the center of mass m and the spring’s and damper’s mobile sides when the whole system is at rest with F(t) = 0, then:

$\displaystyle \begin{aligned} \begin{array}{rcl} x_K=x_b=-x.{} \end{array} \end{aligned}$

(2.16)

Notice that combining expressions in (2.14) and using (2.16), (2.15) yield:

$\displaystyle \begin{aligned} \begin{array}{rcl} \ddot x+\frac{b}{m}\dot x+\frac{K}{m}x =\frac{1}{m}F(t), \end{array} \end{aligned}$

where $\ddot x=\frac {d^2x}{dt^2}$ and $\dot x=\frac {dx}{dt}$ . This mathematical model is identical to that shown in (2.13) if x = x _m is assumed.

Example 2.3

Two bodies connected by a spring, with an external force F(t) applied on the body at the left, are shown in Fig. 2.7a. This situation arises in practice when body 1 transmits movement to body 2 and they are connected through a piece of some flexible material. It is remarked that flexibility is not intentionally included, but it is an undesired problem that appears because of the finite stiffness of the material in which the piece that connects bodies 1 and 2 is made. It is often very important to study the effect of this flexibility on the achievable performance of the control system; hence, it must be considered at the modeling stage. The dampers are included to take into account the effects of friction between each body and the floor.

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig7_HTML.png — Fig. 2.7
Two bodies connected by a spring

The free-body diagram is depicted in Fig. 2.7b. The nomenclature employed is defined as:

$v_{m1}=\frac {dx_{m1}}{dt}$ , where x _m1 is the position of body 1.
$v_{m2}=\frac {dx_{m2}}{dt}$ , where x _m2 is the position of body 2.
$v_{b1}=\frac {dx_{b1}}{dt}$ , where x _b1 is the position of damper 1’s mobile side.
$v_{b2}=\frac {dx_{b2}}{dt}$ , where x _b2 is the position of damper 2’s mobile side.
$v_{1}=\frac {dx_{1}}{dt}$ , where x ₁ is the position of the spring side connected to body 1.
$v_{2}=\frac {dx_{2}}{dt}$ , where x ₂ is the position of the spring side connected to body 2.

Using Fig. 2.7b, as well as (2.1), (2.3), (2.4), the following is obtained:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Mass 1:}\hspace{1cm}m_1\frac{dv_{m1}}{dt}=F(t)+F_{b1}-F_{K1}.{} \end{array} \end{aligned}$

(2.17)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Mass 2:}\hspace{1cm}m_2\frac{dv_{m2}}{dt}=F_{K2}-F_{b2}.{} \\ &\displaystyle &\displaystyle \text{Spring:}\hspace{1cm}Kx_K=F_{K1}=F_{K2}.\\ &\displaystyle &\displaystyle \text{Damper 1:}\hspace{1cm}b_1v_{b1}=F_{b1}.\\ &\displaystyle &\displaystyle \text{Damper 2:}\hspace{1cm}b_2v_{b2}=F_{b2}. \end{array} \end{aligned}$

(2.18)

Forces having the same direction as v _m1 and v _m2, actuating on each one of the bodies, appear with a positive sign, and a negative sign if this is not the case. Notice that it is also defined:

$\displaystyle \begin{aligned} \begin{array}{rcl} v_K=\frac{dx_K}{dt}=v_1-v_2. \end{array} \end{aligned}$

Assuming that all of the system components are connected by rigid joints then, according to Fig. 2.7b:

$\displaystyle \begin{aligned} \begin{array}{rcl} v_{m1}&\displaystyle =&\displaystyle -v_{b1}=v_1,{}\\ v_{m2}&\displaystyle =&\displaystyle v_{b2}=v_2. \end{array} \end{aligned}$

(2.19)

Hence, if x _m1 = 0, x _m2 = 0, x _b1 = 0, x _b2 = 0, x ₁ = 0, x ₂ = 0 are defined as those positions of the center of body 1, the center of body 2, damper 1’s mobile side, damper 2’s mobile side, the spring side fixed to body 1, and the spring side fixed to body 2 respectively, when the complete system is at rest with F(t) = 0, then:

$\displaystyle \begin{aligned} \begin{array}{rcl} x_{m1}&\displaystyle =&\displaystyle -x_{b1}=x_{1},{}\\ x_{m2}&\displaystyle =&\displaystyle x_{b2}=x_{2}. \end{array} \end{aligned}$

(2.20)

Hence, the expressions in (2.17) and (2.18) can be written as:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Mass 1:}\hspace{1cm}m_1\frac{dv_{m1}}{dt}=F(t)+b_1v_{b1}-Kx_K.\\ &\displaystyle &\displaystyle \text{Mass 2:}\hspace{1cm}m_2\frac{dv_{m2}}{dt}=Kx_K-b_2v_{b2}. \end{array} \end{aligned}$

Use of (2.19), (2.20), yields:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Mass 1:}\hspace{1cm}m_1\frac{d^2x_{m1}}{dt^2}=F(t)-b_1\frac{dx_{m1}}{dt}-Kx_K.\\ &\displaystyle &\displaystyle \text{Mass 2:}\hspace{1cm}m_2\frac{d^2x_{m2}}{dt^2}=Kx_K-b_2\frac{dx_{m2}}{dt}. \end{array} \end{aligned}$

On the other hand, x _K = x ₁ − x ₂ according to paragraph after (2.2) and, thus:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Mass 1:}\hspace{1cm}m_1\frac{d^2x_{m1}}{dt^2}=F(t)-b_1\frac{dx_{m1}}{dt}-K(x_1-x_2).\\ &\displaystyle &\displaystyle \text{Mass 2:}\hspace{1cm}m_2\frac{d^2x_{m2}}{dt^2}=K(x_1-x_2)-b_2\frac{dx_{m2}}{dt}. \end{array} \end{aligned}$

Finally, using again (2.20), it is found that the mathematical model is given by the following pair of differential equations, which have to be solved simultaneously:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Mass 1:}\hspace{1cm}\frac{d^2x_{m1}}{dt^2}+\frac{b_1}{m_1}\frac{dx_{m1}}{dt}+\frac{K}{m_1}(x_{m1}-x_{m2})=\frac{1}{m_1}F(t).\\ &\displaystyle &\displaystyle \text{Mass 2:}\hspace{1cm}\frac{d^2x_{m2}}{dt^2}+\frac{b_2}{m_2}\frac{dx_{m2}}{dt}-\frac{K}{m_2}(x_{m1}-x_{m2})=0. \end{array} \end{aligned}$

These differential equations cannot be combined into a single differential equation because variables x _m1 and x _m2 are linearly independent, i.e., there is no algebraic equation relating them.

Example 2.4

Two bodies connected to three springs and a damper are shown in Fig. 2.8a. The damper represents friction existing between the bodies and, because of that, it cannot be replaced by the possible friction appearing between each body and the floor (which would be the case if wheels were not placed under each body). The corresponding free-body diagram is depicted in Fig. 2.8b. The nomenclature employed is defined as:

$\dot x_{1}=\frac {dx_{1}}{dt}$ , where x ₁ is the position of body 1.
$\dot x_{2}=\frac {dx_{2}}{dt}$ , where x ₂ is the position of body 2.
$v_{b1}=\frac {dx_{b1}}{dt}$ , where x _b1 is the position of the damper side connected to body 1.
$v_{b2}=\frac {dx_{b2}}{dt}$ , where x _b2 is the position of the damper side fixed to body 2.
$v_{1}=\frac {dx}{dt}$ , where x is the position of the spring side connected to body 1.
$v_{2}=\frac {dy}{dt}$ , where y is the position of the spring side fixed to body 2.
$v_{K1}=\frac {dx_{K1}}{dt}$ , where x _K1 is the position of the mobile side of the spring connected only to body 1.
$v_{K3}=\frac {dx_{K3}}{dt}$ , where x _K3 is the position of the mobile side of the spring fixed only to body 2.

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig8_HTML.png — Fig. 2.8
A mechanical system with two bodies and three springs

Using Fig. 2.8b, and (2.1), (2.3), (2.4), the following is found:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Mass 1:}\hspace{1cm}m_1\ddot x_1=F(t)+F_{K1}-F_{K2}-F_{b}.{} \end{array} \end{aligned}$

(2.21)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Mass 2:}\hspace{1cm}m_2\ddot x_2=F_{K2}-F_{K3}+F_{b}.{} \\ &\displaystyle &\displaystyle \text{Spring between bodies:}\hspace{1cm}K_2x_{K2}=F_{K2}.\\ &\displaystyle &\displaystyle \text{Spring at the left:}\hspace{1cm}K_1x_{K1}=F_{K1}.\\ &\displaystyle &\displaystyle \text{Spring at the right:}\hspace{1cm}K_3x_{K3}=F_{K3}.\\ &\displaystyle &\displaystyle \text{Damper:}\hspace{1cm}b(v_{b1}-v_{b2})=F_{b}. \end{array} \end{aligned}$

(2.22)

Assuming that all of the system components are connected by rigid joints then, according to Fig. 2.8b:

$\displaystyle \begin{aligned} \begin{array}{rcl} \dot x_{1}&\displaystyle =&\displaystyle -v_{K1}=v_1=v_{b1},{}\\ \dot x_{2}&\displaystyle =&\displaystyle v_{b2}=v_2=v_{K3}. \end{array} \end{aligned}$

(2.23)

Hence, if x ₁ = 0, x ₂ = 0, x _b1 = 0, x _b2 = 0, x = 0, y = 0, x _K1 = 0, x _K3 = 0 are defined as the positions of the center of body 1, the center of body 2, the side of the damper fixed to body 1, the side of the damper fixed to body 2, the side of the central spring that is fixed to body 1, the side of the central spring fixed to body 2, the mobile side of the spring at the left and the mobile side of the spring at the right respectively, when the complete system is at rest with F(t) = 0, then:

$\displaystyle \begin{aligned} \begin{array}{rcl} x_{1}&\displaystyle =&\displaystyle -x_{K1}=x=x_{b1},{}\\ x_{2}&\displaystyle =&\displaystyle x_{b2}=y=x_{K3}. \end{array} \end{aligned}$

(2.24)

Thus, the expressions in (2.21) and (2.22) can be written as:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Mass 1:}\hspace{1cm}m_1\ddot x_1=F(t)+K_1x_{K1}-K_2x_{K2}-b(v_{b1}-v_{b2}).\\ &\displaystyle &\displaystyle \text{Mass 2:}\hspace{1cm}m_2\ddot x_2=K_2x_{K2}-K_3x_{K3}+b(v_{b1}-v_{b2}). \end{array} \end{aligned}$

Use of (2.23), (2.24), yields:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Mass 1:}\hspace{1cm}m_1\ddot x_1=F(t)-K_1x_{1}-K_2x_{K2}-b(\dot x_{1}-\dot x_{2}).\\ &\displaystyle &\displaystyle \text{Mass 2:}\hspace{1cm}m_2\ddot x_2=K_2x_{K2}-K_3x_{2}+b(\dot x_{1}-\dot x_{2}). \end{array} \end{aligned}$

On the other hand, x _K2 = x ₁ − x ₂ according to paragraph after (2.2), and hence:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Mass 1:}\hspace{1cm}m_1\ddot x_1=F(t)-K_1x_{1}-K_2(x_1-x_2)-b(\dot x_{1}-\dot x_{2}).\\ &\displaystyle &\displaystyle \text{Mass 2:}\hspace{1cm}m_2\ddot x_2=K_2(x_1-x_2)-K_3x_{2}+b(\dot x_{1}-\dot x_{2}). \end{array} \end{aligned}$

Finally, the mathematical model is given by the following pair of differential equations, which must be solved simultaneously:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Mass 1:}\hspace{1cm}\ddot x_1 +\frac{b}{m_1}(\dot x_{1}-\dot x_{2}) +\frac{K_1}{m_1}x_{1}+\frac{K_2}{m_1}(x_1-x_2) =\frac{1}{m_1}F(t).\\ &\displaystyle &\displaystyle \text{Mass 2:}\hspace{1cm}\ddot x_2-\frac{b}{m_2}(\dot x_{1}-\dot x_{2})+\frac{K_3}{m_2}x_{2} -\frac{K_2}{m_2}(x_1-x_2) =0. \end{array} \end{aligned}$

As in the previous example, these differential equations cannot be combined in a single differential equation because variables x ₁ and x ₂ are linearly independent, i.e., there is no algebraic equation relating these variables. Also notice that terms corresponding to the central spring and damper appear with the opposite sign in each one of these differential equations. This is because forces exerted by the central spring or the damper on body 1 are applied in the opposite direction to the force exerted on body 2.

2.1.2 Rotative Mechanical Systems

A rigid body (without flexibility) that rotates with angular velocity ω under the effect of a torque T is depicted in Fig. 2.9. It is assumed that no friction exists between the body and the environment. The directions shown for torque and angular velocity are defined as positive for this system component. According to Newton’s Second Law [2], pp. 122:

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig9_HTML.png — Fig. 2.9
A rotative rigid body with inertia I

$\displaystyle \begin{aligned} \begin{array}{rcl} T=I\alpha,{} \end{array} \end{aligned}$

(2.25)

where $\alpha =\frac {d\omega }{dt}$ represents the body angular acceleration and I, a positive constant, stands for the body moment of inertia.

A rotative spring that deforms at an angular velocity ω, under the effect of a torque T, is shown in Fig. 2.10. Although torque T is applied at the spring’s side 1, it must be considered that the same torque, but in the opposite direction, must be applied at the spring’s side 2, to render possible deformation. The angular velocities on sides 1 and 2 respectively are defined ω = ω ₁ − ω ₂ with ω ₁ and ω ₂

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig10_HTML.png — Fig. 2.10
A rotative spring

The directions shown for torque and velocities are defined as positive for this system component. It is assumed that the spring has no moment of inertia whereas deformation is not permanent and does not produce heat. The net spring angular deformation is given as:

$\displaystyle \begin{aligned} \begin{array}{rcl} \theta=\int_0^t\omega\ dt+\theta(0),{} \end{array} \end{aligned}$

(2.26)

where θ = θ ₁ − θ ₂ with θ ₁, and θ ₂ are the angular positions of the spring’s sides 1 and 2 respectively, whereas θ(0) = 0 represents deformation when the spring is neither compressed nor stretched. According to the direction defined for ω in Fig. 2.10, it is concluded that the net angular deformation θ is positive when θ ₁ > θ ₂. In the case of a linear rotative spring, the constitutive function is given by Hooke’s Law :

$\displaystyle \begin{aligned} \begin{array}{rcl} T=k\ \theta,{} \end{array} \end{aligned}$

(2.27)

where k is a positive constant known as the spring stiffness constant.

Friction in rotative mechanical systems (see Fig. 2.11) is modeled as in translational mechanical systems. The only difference is that, in rotative mechanical systems, friction is expressed in terms of angular velocity ω = ω ₁ − ω ₂ and torque T:

$\displaystyle \begin{aligned} \begin{array}{rcl} T&=&b\omega,\text{viscous friction},{} \\ T&=&\left.\pm T_s\right|{}_{\omega=0}, \text{static friction},\\ T&=& T_c\ \text{sign}(\omega), \text{sign}(\omega)=\left\{\begin{array}{ll} +1,&\text{if }\omega>0\\ -1,&\text{if }\omega<0 \end{array}\right.,\text{Coulomb friction.} \end{array} \end{aligned}$

(2.28)

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig11_HTML.png — Fig. 2.11
A damper in rotative mechanical systems

Example 2.5

A rotative body connected to two walls through a spring and a damper is shown in Fig. 2.12a. An external torque T(t) is applied to this body. The damper represents friction between the body and the bearings supporting it, whereas the spring represents the shaft flexibility. The free-body diagram is depicted in Fig. 2.12b. The employed nomenclature is defined as:

$\dot \theta =\frac {d\theta }{dt}$ , where θ is the body’s angular position.
$\omega _b=\frac {d\theta _{b}}{dt}$ , where θ _b is the angular position of the damper’s mobile side.
$\theta _K=\frac {d\theta _K}{dt}$ , where θ _K is the angular position of the spring’s mobile side.

Use of Fig. 2.12b, and (2.25), (2.27), (2.28), yields:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Inertia:}\hspace{1cm}I\ddot \theta=T(t)+T_b-T_K.{}\\ &\displaystyle &\displaystyle \text{Spring:}\hspace{1cm}K\theta_{K}=T_{K}.\\ &\displaystyle &\displaystyle \text{Damper:}\hspace{1cm}b\omega_b=T_{b}. \end{array} \end{aligned}$

(2.29)

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig12_HTML.png — Fig. 2.12
A rotative mass-spring-damper system

Notice that torques applied to the body with inertia I that have the same direction as angular velocity $\dot \theta$ , appear with a positive sign in (2.29). If this is not the case, then a negative sign appears (see T _K, for instance). Assuming that all the system components are connected by rigid joints, then, according to Fig. 2.12b:

$\displaystyle \begin{aligned} \begin{array}{rcl} \dot \theta&\displaystyle =&\displaystyle -\omega_b=\dot\theta_K.{} \end{array} \end{aligned}$

(2.30)

If θ = 0, θ _b = 0, θ _K = 0, are defined as the angular positions of the body, the damper’s mobile side and the spring’s mobile side respectively, when the complete system is at rest with T(t) = 0, then:

$\displaystyle \begin{aligned} \begin{array}{rcl} \theta&\displaystyle =&\displaystyle -\theta_b=\theta_K.{} \end{array} \end{aligned}$

(2.31)

Hence, the expressions in (2.29) can be written as:

$\displaystyle \begin{aligned} \begin{array}{rcl} I\ddot \theta&\displaystyle =&\displaystyle T(t)+b\omega_b-K\theta_{K},\\ &\displaystyle =&\displaystyle T(t)-b\dot\theta-K\theta. \end{array} \end{aligned}$

Finally, the mathematical model is given by the following differential equation:

$\displaystyle \begin{aligned} \begin{array}{rcl} \ddot \theta +\frac{b}{I}\dot\theta+\frac{K}{I}\theta &\displaystyle =&\displaystyle \frac{1}{I}T(t).{} \end{array} \end{aligned}$

(2.32)

Notice that this model is identical to that presented in (2.13) for a translational mass-spring-damper system if mass is replaced by inertia, angular position θ replaces position x _m, and external torque T(t) replaces the external force F(t).

Example 2.6

A simple pendulum is depicted in Fig. 2.13a, where θ stands for the pendulum angular position and $\dot \theta =\frac {d\theta }{dt}$ . It is assumed that the pendulum mass m is concentrated in a single point and pendulum rod has a length l with no mass (or it is negligible compared with mass m). The corresponding free-body diagram is depicted in Fig. 2.13b. The pendulum can be assumed to be a rotative body with inertia:

$\displaystyle \begin{aligned} \begin{array}{rcl} I=ml^2, \end{array} \end{aligned}$

which represents inertia of a particle with mass m describing a circular trajectory with a constant radius l [7], Ch. 9. It is assumed that three torques are applied to the pendulum: (1) the external torque T(t), (2) the friction torque, T _b, which appears at the pendulum pivot, and (3) the torque due to gravity, T _g. According to Figs. 2.13a and b, torque T _g is applied in the opposite direction to that defined by θ and its magnitude is given as:

$\displaystyle \begin{aligned} \begin{array}{rcl} T_g&\displaystyle =&\displaystyle mg d, d=l\sin{}(\theta),\\ &\displaystyle =&\displaystyle mgl\sin{}(\theta).{} \end{array} \end{aligned}$

(2.33)

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig13_HTML.png — Fig. 2.13
Simple pendulum

Use of Fig. 2.13b, and (2.25), (2.28), yields:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Inertia:}\hspace{1cm}I\ddot \theta=T(t)+T_b-T_g.{}\\ &\displaystyle &\displaystyle \text{Damper:}\hspace{1cm}b\omega_b=T_{b}. \end{array} \end{aligned}$

(2.34)

Notice that torques applied to I that have the same direction as angular velocity $\dot \theta$ appear affected by a positive sign in (2.34), or a negative sign if this is not the case (see for instance T _g). Assuming that all the system components are connected by rigid joints, then, according to Fig. 2.13b, it is found that:

$\displaystyle \begin{aligned} \begin{array}{rcl} \dot \theta&\displaystyle =&\displaystyle -\omega_b.{} \end{array} \end{aligned}$

(2.35)

Hence, if θ = 0, θ _b = 0 (where $\omega _b=\dot \theta _b$ ) are defined as the angular positions of the pendulum and the damper’s mobile side respectively, when the whole system is at rest with T(t) = 0, then:

$\displaystyle \begin{aligned} \begin{array}{rcl} \theta&\displaystyle =&\displaystyle -\theta_b.{} \end{array} \end{aligned}$

(2.36)

This means that (2.34) and (2.33) can be written as:

$\displaystyle \begin{aligned} \begin{array}{rcl} I\ddot \theta&\displaystyle =&\displaystyle T(t)+b\omega_b-mgl\sin{}(\theta),\\ &\displaystyle =&\displaystyle T(t)-b\dot\theta-mgl\sin{}(\theta). \end{array} \end{aligned}$

Finally, the mathematical model is given by the following differential equation:

$\displaystyle \begin{aligned} \begin{array}{rcl} ml^2\ddot \theta +b\dot\theta+mgl\sin{}(\theta) &\displaystyle =&\displaystyle T(t). \end{array} \end{aligned}$

Notice that this is a nonlinear differential equation, because of the function $\sin {}(\theta )$ . In Sect. 7.3, Chap. 7, it is explained how this kind of differential equation can be studied. Moreover, in Chaps. 13, 15 and 16 it is shown how to design controllers for systems represented by nonlinear differential equations.

Example 2.7

Two rotative bodies connected by a spring are shown in Fig. 2.14a. An external torque T(t) is applied to the body on the left. The mechanical system in Fig. 2.14a can be interpreted as a motor (body on the left), which moves a load (body on the right) through the motor shaft, exhibiting flexibility. This flexibility is not intentionally included, but is undesired shaft behavior arising from the finite stiffness of the material in which the motor shaft is made. In such a situation, it is very important to take into account this flexibility to study its effect on the closed-loop system performance. According to sign definitions shown in Figs. 2.9, 2.10, and 2.11, the corresponding free-body diagram is depicted in Fig. 2.14b. The employed nomenclature is defined as:

$\omega _{2}=\frac {d\theta _{2}}{dt}$ , where θ ₂ is the body 1 angular position.
$\omega _{5}=\frac {d\theta _{5}}{dt}$ , where θ ₅ is the body 2 angular position.
$\omega _1=\frac {d\theta _1}{dt}$ , where θ ₁ is the angular position of the mobile side of the damper connected to body 1.
$\omega _6=\frac {d\theta _6}{dt}$ , where θ ₆ is the angular position of the mobile side of the damper connected to body 2.
$\omega _3=\frac {d\theta _3}{dt}$ and $\omega _4=\frac {d\theta _4}{dt}$ , where θ ₃ and θ ₄ are the angular positions of the spring sides.

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig14_HTML.png — Fig. 2.14
Two rotative bodies connected by a spring

Use of Fig. 2.14b, and (2.25), (2.27), (2.28), yields:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Inertia 1:}\hspace{1cm}I_1\dot \omega_2=T(t)+T_1-T_{3}.{} \end{array} \end{aligned}$

(2.37)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Inertia 2:}\hspace{1cm}I_2\dot\omega_5=T_{4}-T_{6}.{} \\ &\displaystyle &\displaystyle \text{Left damper:}\hspace{1cm}b_1\omega_{1}=T_{1}.\\ &\displaystyle &\displaystyle \text{Right damper:}\hspace{1cm}b_2\omega_{6}=T_{6}.\\ &\displaystyle &\displaystyle \text{Spring:}\hspace{1cm}K x_K=T_{3}=T_4, x_K=\theta_3-\theta_4, \end{array} \end{aligned}$

(2.38)

where the last expression is obtained according to the paragraph after (2.26). Notice that in equations (2.37) and (2.38), torques that have the same direction as ω ₂ and ω ₅ appear to be affected by a positive sign and a negative sign if this is not the case. Assuming that all of the system components are connected by rigid joints then, according to Fig. 2.14b:

$\displaystyle \begin{aligned} \begin{array}{rcl} \omega_2&\displaystyle =&\displaystyle -\omega_1=\omega_3,\omega_{5}=\omega_6=\omega_{4}. {} \end{array} \end{aligned}$

(2.39)

If ω ₂ = 0, ω ₅ = 0, ω ₁ = 0, ω ₆ = 0, ω ₃ = 0, and ω ₄ = 0 are defined as the angular positions of body 1, body 2, the left damper mobile side, the right damper mobile side, and the spring sides respectively, when the whole system is at rest with T(t) = 0, then:

$\displaystyle \begin{aligned} \begin{array}{rcl} \theta_2&\displaystyle =&\displaystyle -\theta_1=\theta_3,\theta_{5}=\theta_6=\theta_{4}. {} \end{array} \end{aligned}$

(2.40)

Hence, (2.37) and (2.38) can be written as:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Inertia 1:}\hspace{1cm}I_1\dot \omega_2=T(t)+b_1\omega_{1}-K(\theta_{3}-\theta_4).\\ &\displaystyle &\displaystyle \text{Inertia 2:}\hspace{1cm}I_2\dot\omega_5=K(\theta_{3}-\theta_4)-b_2\omega_{6}. \end{array} \end{aligned}$

Finally, using (2.39) and (2.40), it is found that the mathematical model is given by the following differential equations, which have to be solved simultaneously:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Inertia 1:}\hspace{1cm}I_1\ddot \theta_2+b_1\dot\theta_{2}+K(\theta_{2}-\theta_5)=T(t).\\ &\displaystyle &\displaystyle \text{Inertia 2:}\hspace{1cm}I_2\ddot\theta_5+b_2\dot\theta_{5}-K(\theta_{2}-\theta_5)=0. \end{array} \end{aligned}$

These differential equations cannot be combined in a single differential equation because θ ₂ and θ ₅ are linearly independent, i.e., there is no algebraic equation relating these variables.

2.2 Electrical Systems

There are three basic phenomena in electrical systems: inductance, capacitance, and resistance. An electric current i flowing through an inductor under the effect of a voltage v that is applied at its terminals is depicted in Fig. 2.15. It is assumed that there are no parasitic effects, i.e., the inductor has neither internal electrical resistance nor capacitance. The directions shown for electric current and voltage are defined as positive for this system component.

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig15_HTML.png — Fig. 2.15
An inductor in electrical systems

The magnetic flux linkage of the inductor is given as the product of the electric current i and a positive constant L, known as inductance , depending on the geometric shape of the conductor:

$\displaystyle \begin{aligned} \begin{array}{rcl} \lambda=Li.{} \end{array} \end{aligned}$

(2.41)

According to Faraday’s Law , [2], pp. 606, [3], pp. 325, flux linkage determines the voltage at the inductor terminals:

$\displaystyle \begin{aligned} \begin{array}{rcl} v=\frac{d\lambda}{dt}=L\frac{di}{dt}.{} \end{array} \end{aligned}$

(2.42)

One capacitor is formed wherever two electrical conductors, with different electrical potentials and with an insulator material between them, are placed close enough to generate an electric field between them. These electrical conductors are called plates. An electric current i flowing through a capacitor under the effect of a voltage v applied at its terminals is shown in Fig. 2.16. It is assumed that no parasitic effects exist, i.e., there is no leakage current between the capacitor plates and there are no inductive effects due to capacitor plates. Because of the difference in potential between the plates, an electric charge is stored. It is common to use letter q to designate such an electric charge, which is the same, but has the opposite sign, at each plate. Capacitance C is a positive constant defined as a means of quantifying how much electric charge can be stored in a capacitor. The constitutive function of a capacitor is given as [3], pp. 121:

$\displaystyle \begin{aligned} \begin{array}{rcl} C=\frac{q}{v}.{} \end{array} \end{aligned}$

(2.43)

Capacitance C depends on the geometric form of plates, the distance between them and the dielectric properties of the insulator placed between them. Recall that the electric current is defined as:

$\displaystyle \begin{aligned} \begin{array}{rcl} i=\frac{dq}{dt}.{} \end{array} \end{aligned}$

(2.44)

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig16_HTML.png — Fig. 2.16
A capacitor in electrical systems

An electric resistance is a device requiring application of a voltage v at its terminals to maintain an electric current i flowing through it. The constitutive function of a linear resistance is defined by Ohm’s Law [2], pp. 530:

$\displaystyle \begin{aligned} \begin{array}{rcl} v=iR,{} \end{array} \end{aligned}$

(2.45)

where R is a positive constant known as the electric resistance. An electric current i flowing through an electric resistance when a voltage v is applied at its terminals is shown in Fig. 2.17. The directions shown for the variables are defined as positive.

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig17_HTML.png — Fig. 2.17
An electric resistance in electrical systems

Once the constitutive functions of each electrical system component have been defined, it is necessary to connect several of them to model more complex electrical systems. Kirchhoff’s Laws are the key to establishing connections between the different components of the electrical systems.

Property 2.1 (Kirchhoff’s Voltage Law)

The algebraic sum of voltages around a closed path (or mesh) must be zero.

Property 2.2 (Kirchhoff’s Current Law)

The algebraic sum of all currents entering and exiting a node must equal zero.

Example 2.8

Consider the series resistor–capacitor (RC) circuit shown in Fig. 2.18a. Using Kirchhoff’s voltage law (KVL):

$\displaystyle \begin{aligned} \begin{array}{rcl} v_i=Ri+v_0, v_0=\frac{q}{C}, \end{array} \end{aligned}$

As $i=\dot q$ , then $Ri=RC\frac {dv_0}{dt}$ ; hence:

$\displaystyle \begin{aligned} \begin{array}{rcl} v_i=RC\dot v_0+v_0. \end{array} \end{aligned}$

Finally, the mathematical model is given as the following differential equation:

$\displaystyle \begin{aligned} \begin{array}{rcl} \dot v_0+\frac{1}{RC}v_0=\frac{1}{RC}v_i. \end{array} \end{aligned}$

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig18_HTML.png — Fig. 2.18
Some first-order circuits

Example 2.9

Consider the series RC circuit shown in Fig. 2.18b. Using KVL:

$\displaystyle \begin{aligned} \begin{array}{rcl} v_i=\frac{q}{C}+v_0, v_0=Ri. \end{array} \end{aligned}$

Differentiating once with respect to time:

$\displaystyle \begin{aligned} \begin{array}{rcl} \dot v_i&\displaystyle =&\displaystyle \frac{1}{C}i+\dot v_0,\\ &\displaystyle =&\displaystyle \frac{1}{RC}v_0+\dot v_0, \end{array} \end{aligned}$

where $i=\dot q$ and v ₀ = Ri have been used. Hence, the dynamic model is given as:

$\displaystyle \begin{aligned} \begin{array}{rcl} \dot v_0+\frac{1}{RC}v_0=\dot v_i. \end{array} \end{aligned}$

Example 2.10

Consider the series RL circuit shown in Fig. 2.18c. Using KVL:

$\displaystyle \begin{aligned} \begin{array}{rcl} v_i=v_0+Ri, v_0=L\frac{di}{dt}. \end{array} \end{aligned}$

Differentiating once with respect to time:

$\displaystyle \begin{aligned} \begin{array}{rcl} \dot v_i&\displaystyle =&\displaystyle \dot v_0+R \frac{di}{dt},\\ &\displaystyle =&\displaystyle \dot v_0+\frac{R}{L} v_0, \end{array} \end{aligned}$

where $v_0=L\frac {di}{dt}$ has been used. Hence, the dynamic model is given as:

$\displaystyle \begin{aligned} \begin{array}{rcl} \dot v_0+\frac{R}{L} v_0=\dot v_i. \end{array} \end{aligned}$

Example 2.11

Consider the series-parallel RC circuit shown in Fig. 2.18d. Define i ₁, i ₂, and i as the electric currents through R ₁, C, and R ₂ respectively. Hence, Kirchhoff’s current law (KCL) can be used to obtain:

$\displaystyle \begin{aligned} \begin{array}{rcl} i&\displaystyle =&\displaystyle i_1+i_2,\\ &\displaystyle =&\displaystyle \frac{v_c}{R_1}+\frac{dq}{dt},\\ &\displaystyle =&\displaystyle \frac{q}{R_1C}+\frac{dq}{dt},{} \end{array} \end{aligned}$

(2.46)

On the other hand, KVL allows us to write:

$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{q}{C}+R_2i&\displaystyle =&\displaystyle v_i. \end{array} \end{aligned}$

Replacing (2.46) and arranging terms yield:

$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{dq}{dt}+\frac{1}{C}\left(\frac{R_1+R_2}{R_1R_2}\right)q&\displaystyle =&\displaystyle \frac{1}{R_2}v_i. \end{array} \end{aligned}$

Applying the Laplace transform (see (3.2) in Chap. 3):

$\displaystyle \begin{aligned} \begin{array}{rcl} Q(s)=\frac{\frac{1}{R_2}}{s+a}V_i(s)+\frac{q(0)}{s+a}, a=\frac{1}{C}\frac{R_1+R_2}{R_1R_2},{} \end{array} \end{aligned}$

(2.47)

where q(0) stands for the initial charge in the capacitor. On the other hand:

$\displaystyle \begin{aligned} \begin{array}{rcl} v_0&\displaystyle =&\displaystyle R_2i,\\ &\displaystyle =&\displaystyle R_2\left(\frac{q}{R_1C}+\frac{dq}{dt}\right), \end{array} \end{aligned}$

where (2.46) has been used. Arranging and using the Laplace transform:

$\displaystyle \begin{aligned} \begin{array}{rcl} Q(s)=\frac{\frac{1}{R_2}}{s+\frac{1}{R_1C}}V_0(s)+\frac{q(0)}{s+\frac{1}{R_1C}}.{} \end{array} \end{aligned}$

(2.48)

Equating (2.47) and (2.48) and after an algebraic procedure:

$\displaystyle \begin{aligned} \begin{array}{rcl} V_0(s)=\frac{s+b}{s+a}V_i(s)-\frac{v_c(0)}{s+a}, b=\frac{1}{R_1C},{} \end{array} \end{aligned}$

(2.49)

where $v_c(0)=\frac {q(0)}{C}$ represents the initial voltage at the capacitor. Notice that:

$\displaystyle \begin{aligned} \begin{array}{rcl} a>b>0. \end{array} \end{aligned}$

The expression in (2.49) represents the mathematical model of the circuit in Fig. 2.18d.

Example 2.12

A series RLC circuit is shown in Fig. 2.19a with a voltage v _i applied at its terminals. Connection of the elements of this circuit is established by KVL [8], pp. 67. Using sign definitions in Figs. 2.15, 2.16, 2.17 the electric diagram shown in Fig. 2.19b is obtained. According to this and (2.41), (2.42), (2.43), (2.44), (2.45), in addition to KVL, it is found that:

$\displaystyle \begin{aligned} \begin{array}{rcl} v_i&\displaystyle =&\displaystyle v_L+v_R+v_C,\\ v_C&\displaystyle =&\displaystyle \frac{q}{C}, q=\int_0^ti(r)dr, \text{ if }q(0)=0,{} \end{array} \end{aligned}$

(2.50)

$\displaystyle \begin{aligned} \begin{array}{rcl} v_L&\displaystyle =&\displaystyle \frac{d\lambda}{dt},\lambda=Li,{} \end{array} \end{aligned}$

(2.51)

$\displaystyle \begin{aligned} \begin{array}{rcl} v_R&\displaystyle =&\displaystyle iR, i=\frac{dq}{dt}.{} \end{array} \end{aligned}$

(2.52)

Notice that the electric current flowing through all of the circuit components is the same. Hence:

$\displaystyle \begin{aligned} \begin{array}{rcl} v_i&\displaystyle =&\displaystyle L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q,{} \end{array} \end{aligned}$

(2.53)

or, differentiating once with respect to time:

$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{dv_i}{dt}&\displaystyle =&\displaystyle L\frac{d^2i}{dt^2}+R\frac{di}{dt}+\frac{1}{C}i.{} \end{array} \end{aligned}$

(2.54)

The corresponding mathematical model is represented by either (2.53) or (2.54).

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig19_HTML.png — Fig. 2.19
A series *RLC* circuit

It is observed from (2.52) that there exists a (linear) algebraic relation between voltage and current in a resistance, i.e., Ohm’s Law v _R = iR. Also notice that the electric resistance R is the constant relating electric current and voltage $\frac {v_R}{i}=R$ . Because of this fact, it is natural to wonder whether similar relations might exist for electric current and voltage in an inductor and in a capacitor. However, from (2.50) and (2.51) it is clear that some mathematical artifice has to be employed, because such relations involve differentiation and integration. The mathematical artifice is the Laplace transform as it possesses the following properties:

$\displaystyle \begin{aligned} \begin{array}{rcl} \mathcal{L}\left\{\frac{d x}{dt}\right\}=sX(s),\mathcal{L}\left\{\int_0^t x(r)dr\right\}=\frac{1}{s}X(s),\text{if x(0)=0},{} \end{array} \end{aligned}$

(2.55)

where X(s) stands for the Laplace transform of x(t), i.e., $\mathcal {L}\{x(t)\}=X(s)$ . Thus, using (2.50), (2.51), and the Laplace transform, and assuming that all of the initial conditions are zero, yield:

$\displaystyle \begin{aligned} \begin{array}{rcl} V_C(s)&\displaystyle =&\displaystyle \frac{1}{sC}I(s),\\ V_L(s)&\displaystyle =&\displaystyle sLI(s), \end{array} \end{aligned}$

where $\mathcal {L}\{v_C(t)\}=V_C(s)$ , $\mathcal {L}\{v_L(t)\}=V_L(s)$ and $\mathcal {L}\{i(t)\}=I(s)$ . The above expressions represent the algebraic relations between electric current and voltage in an inductor and a capacitor. The factors relating these variables are called impedance which is represented as:

$\displaystyle \begin{aligned} \begin{array}{rcl} Z(s)=\frac{V(s)}{I(s)}.{} \end{array} \end{aligned}$

(2.56)

Hence, the impedance in an inductor is:

$\displaystyle \begin{aligned} \begin{array}{rcl} Z_L(s)&\displaystyle =&\displaystyle \frac{V_L(s)}{I(s)}=sL,{} \end{array} \end{aligned}$

(2.57)

and the impedance in a capacitor is:

$\displaystyle \begin{aligned} \begin{array}{rcl} Z_C(s)&\displaystyle =&\displaystyle \frac{V_C(s)}{I(s)}=\frac{1}{sC}.{} \end{array} \end{aligned}$

(2.58)

Then, using the Laplace transform and the respective impedances, it is possible to write the mathematical model in (2.54) as:

$\displaystyle \begin{aligned} \begin{array}{rcl} V_i(s)&\displaystyle =&\displaystyle sLI(s)+RI(s)+\frac{1}{sC}I(s),\\ V_i(s)&\displaystyle =&\displaystyle Z_L(s)I(s)+RI(s)+Z_C(s)I(s),\\ V_i(s)&\displaystyle =&\displaystyle (Z_L(s)+R+Z_C(s))I(s), \end{array} \end{aligned}$

which motivates the definition of the series circuit equivalent impedance as:

$\displaystyle \begin{aligned} \begin{array}{rcl} Z_t(s)&\displaystyle =&\displaystyle Z_L(s)+R+Z_C(s),\\ Z_t(s)&\displaystyle =&\displaystyle sL+R+\frac{1}{sC}, Z_t(s)=\frac{V_i(s)}{I(s)}.{} \end{array} \end{aligned}$

(2.59)

This result proves an important property of series electric circuits:

Property 2.3

The equivalent impedance of a series connected circuit is given as the addition of impedances of all of the series connected circuit components.

The concept of impedance is very important in electric circuits and it is a tool commonly used to model and to analyze circuits in electrical engineering. This means that expressions in (2.59) also represent the mathematical model of the circuit in Fig. 2.19a.

Example 2.13

A parallel connected electric circuit with an electric current i _i flowing through it is depicted in Fig. 2.20a. Models of the circuit components are connected, in this case using KCL [8], pp. 68. Using sign definitions presented in Figs. 2.15, 2.16, 2.17 the electric diagram shown in Fig. 2.20b is obtained. From this, using (2.41), (2.42), (2.43), (2.44), (2.45), and KCL it is found that:

$\displaystyle \begin{aligned} \begin{array}{rcl} i_i&\displaystyle =&\displaystyle i_L+i_R+i_C,\\ q&\displaystyle =&\displaystyle vC, q=\int_0^ti_C(r)dr, \text{ if }q(0)=0\Rightarrow i_C=C\frac{dv}{dt},{} \end{array} \end{aligned}$

(2.60)

$\displaystyle \begin{aligned} \begin{array}{rcl} i_L&\displaystyle =&\displaystyle \frac{1}{L}\int_0^tv(r)dr, v=L\frac{di_L}{dt},{} \end{array} \end{aligned}$

(2.61)

$\displaystyle \begin{aligned} \begin{array}{rcl} i_R&\displaystyle =&\displaystyle \frac{v}{R}.{} \end{array} \end{aligned}$

(2.62)

Notice that the voltage is the same in all the circuit components. Hence, the mathematical model is given as:

$\displaystyle \begin{aligned} \begin{array}{rcl} i_i&\displaystyle =&\displaystyle \frac{1}{L}\int_0^tv(r)dr+\frac{v}{R}+C\frac{dv}{dt}{} \end{array} \end{aligned}$

(2.63)

Using the Laplace transform in (2.60), (2.61), (see (2.55)) and assuming that all of the initial conditions are zero, yields:

$\displaystyle \begin{aligned} \begin{array}{rcl} I_C(s)&\displaystyle =&\displaystyle sCV(s),\\ I_L(s)&\displaystyle =&\displaystyle \frac{1}{sL}V(s), \end{array} \end{aligned}$

where $\mathcal {L}\{i_C(t)\}=I_C(s)$ , $\mathcal {L}\{i_L(t)\}=I_L(s)$ and $\mathcal {L}\{v(t)\}=V(s)$ . The factor relating the Laplace transforms of electric current and voltage in a circuit element is called admittance and is represented as:

$\displaystyle \begin{aligned} \begin{array}{rcl} Y(s)=\frac{I(s)}{V(s)}.{} \end{array} \end{aligned}$

(2.64)

This means that the admittance of an inductor is:

$\displaystyle \begin{aligned} \begin{array}{rcl} Y_L(s)&\displaystyle =&\displaystyle \frac{I_L(s)}{V(s)}=\frac{1}{sL},{} \end{array} \end{aligned}$

(2.65)

and the admittance of a capacitor is:

$\displaystyle \begin{aligned} \begin{array}{rcl} Y_C(s)&\displaystyle =&\displaystyle \frac{I_C(s)}{V(s)}=sC.{} \end{array} \end{aligned}$

(2.66)

Then, using the Laplace transform and the definition of admittance, the mathematical model in (2.63) can be written as:

$\displaystyle \begin{aligned} \begin{array}{rcl} I_i(s)&\displaystyle =&\displaystyle \frac{1}{sL}V(s)+\frac{1}{R}V(s)+sCV(s),\\ I_i(s)&\displaystyle =&\displaystyle Y_L(s)V(s)+\frac{1}{R}V(s)+Y_C(s)V(s),\\ I_i(s)&\displaystyle =&\displaystyle (Y_L(s)+\frac{1}{R}+Y_C(s))V(s), \end{array} \end{aligned}$

which motivates the definition of the parallel circuit equivalent admittance as:

$\displaystyle \begin{aligned} \begin{array}{rcl} Y_T(s)&\displaystyle =&\displaystyle Y_L(s)+\frac{1}{R}+Y_C(s),\\ Y_T(s)&\displaystyle =&\displaystyle \frac{1}{sL}+\frac{1}{R}+sC, Y_T(s)=\frac{I_i(s)}{V(s)}.{} \end{array} \end{aligned}$

(2.67)

This proves the following important property of parallel connected electric circuits:

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig20_HTML.png — Fig. 2.20
A parallel connected electric circuit

Property 2.4

The parallel circuit equivalent admittance is the addition of admittances of all of the parallel connected circuit components.

The admittance concept is very important as a tool to model and to analyze electric circuits in electrical engineering. This means that expressions in (2.67) also represent the mathematical model of the circuit in Fig. 2.20a. From definitions of impedance and admittance in (2.56) and (2.64), it is clear that admittance is the inverse of impedance, i.e.:

$\displaystyle \begin{aligned} \begin{array}{rcl} Y(s)=\frac{1}{Z(s)}{} \end{array} \end{aligned}$

(2.68)

This is also corroborated by comparing the definitions of impedance and admittance for an inductor and a capacitor shown in (2.57), (2.58), (2.65) and (2.66). Hence, according to (2.67) it is possible to write:

$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{1}{Z_T(s)}&\displaystyle =&\displaystyle \frac{1}{Z_L(s)}+\frac{1}{R}+\frac{1}{Z_C(s)}, Y_T(s)=\frac{1}{Z_T(s)}, \end{array} \end{aligned}$

where Z _T(s) stands for the equivalent impedance of the parallel connected circuit shown in Fig. 2.20a. It is stressed that no relation exists between Z _T(s) and Z _t(s) defined in (2.59) as the equivalent impedance of the series connected circuit shown in Fig. 2.19a. In the case when n circuit elements are connected in parallel, it is possible to write:

$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{1}{Z_T(s)}&\displaystyle =&\displaystyle \frac{1}{Z_1(s)}+\frac{1}{Z_2(s)}+\dots+\frac{1}{Z_n(s)}.{} \end{array} \end{aligned}$

(2.69)

When only two circuit elements are connected in parallel, it is easy to verify that the previous expression reduces to:

$\displaystyle \begin{aligned} \begin{array}{rcl} Z_T(s)&\displaystyle =&\displaystyle \frac{Z_1(s)Z_2(s)}{Z_1(s)+Z_2(s)}.{} \end{array} \end{aligned}$

(2.70)

Example 2.14

Consider the electric circuit shown in Fig. 2.21. If V _o(s) is the voltage at the terminals of the parallel connected circuit elements and Z _p(s) is the total impedance of these parallel connected circuit elements, it can be written as follows:

$\displaystyle \begin{aligned} \begin{array}{rcl} Z_p(s)&\displaystyle =&\displaystyle \frac{V_o(s)}{I(s)},{}\\ Z_p(s)&\displaystyle =&\displaystyle \frac{R\frac{1}{sC}}{R+\frac{1}{sC}}=\frac{R}{RCs+1}, \end{array} \end{aligned}$

(2.71)

where (2.70) has been used to compute equivalent impedance of the parallel connected circuit elements. Notice that I(s) is the total electric current flowing through the parallel connected circuit elements. On the other hand, if Z _sp(s) is the equivalent impedance of the series-parallel circuit, i.e., impedance measured at terminals where the voltage V _i(s) is applied, then it can be written as follows:

$\displaystyle \begin{aligned} \begin{array}{rcl} Z_{sp}(s)&\displaystyle =&\displaystyle \frac{V_i(s)}{I(s)},{}\\ Z_{sp}(s)&\displaystyle =&\displaystyle R+\frac{1}{sC}+\frac{R}{RCs+1}, \end{array} \end{aligned}$

(2.72)

The fact that the equivalent impedance of series connected circuit elements is computed by adding the impedances of all of the series connected elements whereas impedance of the two parallel connected elements is given in (2.71) has been taken into account. Use of (2.71) and (2.72) yields:

$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{V_o(s)}{V_i(s)}=G_T(s)=\frac{Z_p(s)}{Z_{sp}(s)}=\frac{\frac{R}{RCs+1}}{R+\frac{1}{sC}+\frac{R}{RCs+1}}, \end{array} \end{aligned}$

and, thus:

$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{V_o(s)}{V_i(s)}&\displaystyle =&\displaystyle G_T(s)=\frac{Ts}{T^2s^2+3Ts+1}, T=RC.{} \end{array} \end{aligned}$

(2.73)

Notice that G _T(s) is not an impedance as it is given as the ratio of two impedances or the ratio of two voltages.

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig21_HTML.png — Fig. 2.21
Series-parallel RC circuit

Example 2.15

Consider the electric circuit depicted in Fig. 2.22. Ratio $F(s)=\frac {V_2(s)}{V_1(s)}$ is computed in the following. Define three closed paths (or meshes) as shown in Fig. 2.22. It is assumed that the electric currents I ₁(s), I ₂(s) and I ₃(s) flow around each mesh. The KVL can be used using the following criterion:

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig22_HTML.png — Fig. 2.22
A RC phase-shift circuit

Criterion 2.1

Consider the i −th mesh. The algebraic sum of voltages at the circuit elements in the i −th mesh equals zero. Voltage at each circuit element is computed as the product of its impedance and the algebraic sum of the electric currents I ₁(s), I ₂(s) or I ₃(s), flowing through that circuit element. Let I _i(s) be the electric current defining the i −th mesh. If I ₁(s), I ₂(s) or I ₃(s) has the same direction as I _i(s) in that circuit element then I ₁(s), I ₂(s) or I ₃(s), is affected by a “+ ” whereas a “−” affects those electric currents with the opposite direction. A coefficient equal to zero affects the electric currents not flowing through that circuit element. The value of a voltage source is affected by a “+ ” if I _i(s) flows through that voltage source from terminal “+ ” to terminal “−”. A “−” is used if this is not the case.

This criterion constitutes the basis of the so-called mesh analysis method employed for circuit analysis in electrical engineering. The reader is referred to [9] and [8] for a deeper explanation of this method. The use of this criterion in each one of the meshes shown in Fig. 2.22 yields:

$\displaystyle \begin{aligned} \begin{array}{rcl} I_1(s)\frac{1}{sC}+(I_1(s)-I_2(s))R&\displaystyle =&\displaystyle V_1(s), {}\\ (I_2(s)-I_1(s))R+I_2(s)\frac{1}{sC}+(I_2(s)-I_3(s))R&\displaystyle =&\displaystyle 0,\\ (I_3(s)-I_2(s))R+I_3(s)\frac{1}{sC}+V_2(s)&\displaystyle =&\displaystyle 0, V_2(s)=RI_3(s). \end{array} \end{aligned}$

(2.74)

From the third expression in (2.74) the following is obtained:

$\displaystyle \begin{aligned} \begin{array}{rcl} I_2(s)R=\frac{1}{R}\left(R+\frac{1}{sC}\right)V_2(s)+V_2(s).{} \end{array} \end{aligned}$

(2.75)

Replacing this and $I_3(s)=\frac {V_2(s)}{R}$ in the second expression in (2.74), a new expression is obtained which allows I ₁(s) to be computed as a function of V ₂(s). Replacing this value of I ₁(s) and I ₂(s) given in (2.75) in the first expression in (2.74), V ₁(s) is obtained as a function of V ₂(s). Finally, suitably arranging terms, the following expression is obtained:

$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{V_2(s)}{V_1(s)}=\frac{R^3C^3s^3}{R^3C^3s^3+6R^2C^2s^2+5RCs+1}=F(s).{} \end{array} \end{aligned}$

(2.76)

2.3 Transformers

In this book, the term transformers is used to designate those components that manipulate the system variables in such a way that energy is conserved, i.e., they neither store nor dissipate energy; energy simply flows through them.

2.3.1 Electric Transformer

An electric transformer is composed of two inductors wound on the same ferromagnetic core (see Fig. 2.23). This means that inductors are magnetically coupled but electrically isolated.

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig23_HTML.png — Fig. 2.23
Electric transformer

Each inductor defines one port. v ₁ stands for the voltage at the terminals of port 1 and i ₁ represents the electric current flowing through port 1. v ₂ stands for the voltage at the terminals of port 2 and i ₂ represents the electric current flowing through port 2. The indicated directions are defined as positive. The dot placed at the upper side of each inductor is used to indicate that inductors are wound in the same direction. If this were not the case, dots would be placed on opposite sides of the inductors. The inductor at port 1 has n ₁ turns and the inductor at port 2 has n ₂ turns. As the inductors are magnetically coupled, flux linkage in inductor 1, λ ₁, and flux linkage in inductor 2, λ ₂, depend on the electric currents flowing through both inductors:

$\displaystyle \begin{aligned} \begin{array}{rcl} \lambda_1&\displaystyle =&\displaystyle L_{1}i_1+M_{12}i_2, {} \end{array} \end{aligned}$

(2.77)

$\displaystyle \begin{aligned} \begin{array}{rcl} \lambda_2&\displaystyle =&\displaystyle M_{21}i_1+L_{2}i_2, {} \end{array} \end{aligned}$

(2.78)

where L ₁, L ₂, are the inductances of inductors 1 and 2, whereas M ₁₂ and M ₂₁ are the mutual inductances between the two circuits. It is always true that M ₁₂ = M ₂₁ = M. The positive signs in terms with factors M ₁₂ and M ₂₁ in the previous expressions are due to the fact that the direction of electric currents defined as positives enter wounds through its sides labeled with a dot (see Sect. D.2 in Appendix D). Solving for i ₂ in (2.77) and replacing it in (2.78) yield:

$\displaystyle \begin{aligned} \begin{array}{rcl} \lambda_2=\frac{L_2}{M}\lambda_1+\left(M-\frac{L_1L_2}{M}\right)i_1.{} \end{array} \end{aligned}$

(2.79)

The coupling coefficient is defined as:

$\displaystyle \begin{aligned} \begin{array}{rcl} k=\frac{M}{\sqrt{L_1L_2}}. \end{array} \end{aligned}$

If both circuits are not coupled at all, then k = 0, but k = 1 if both circuits are completely coupled such that flux linkage in inductor 1 equals flux linkage in inductor 2. This case is well approximated in practice if the core of both inductors is the same and it is made in a ferromagnetic material with a large magnetic permeability μ. Assuming that this is the case, i.e., $M=\sqrt {L_1L_2}$ , then (2.79) becomes:

$\displaystyle \begin{aligned} \begin{array}{rcl} \lambda_2=\sqrt{\frac{L_2}{L_1}}\lambda_1.{} \end{array} \end{aligned}$

(2.80)

On the other hand, from (2.78) and (2.77), using M ₁₂ = M ₂₁ = M, and (2.80), it is found that:

$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{\lambda_2}{\lambda_1}&\displaystyle =&\displaystyle \frac{Mi_1+L_{2}i_2}{L_{1}i_1+Mi_2}=\sqrt{\frac{L_2}{L_1}}.\end{array} \end{aligned}$

Hence, solving for the indicated ratio:

$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{i_2}{i_1}&\displaystyle =&\displaystyle \frac{\sqrt{L_1L_2}-M}{L_2-\sqrt{\frac{L_2}{L_1}}M},\\ &\displaystyle =&\displaystyle \frac{\frac{\sqrt{L_1}}{\sqrt{L_2}}(\sqrt{L_1L_2}-M)}{\sqrt{L_1L_2}-M}=\sqrt{\frac{L_1}{L_2}}.{} \end{array} \end{aligned}$

(2.81)

The inductance of each circuit is given as [3]:

$\displaystyle \begin{aligned} \begin{array}{rcl} L_1=\mu \frac{n_1^2A }{l},\\ L_2=\mu \frac{n_2^2A }{l}, \end{array} \end{aligned}$

where A, l, and μ are the core area, the core mean length, and the core magnetic permeability respectively. Hence:

$\displaystyle \begin{aligned} \begin{array}{rcl} \sqrt{\frac{L_1}{L_2}}=\frac{n_1}{n_2}.{} \end{array} \end{aligned}$

(2.82)

Using this and (2.42) in (2.80) yields:

$\displaystyle \begin{aligned} \begin{array}{rcl} v_2=\frac{n_2}{n_1}v_1, \end{array} \end{aligned}$

and, from (2.81):

$\displaystyle \begin{aligned} \begin{array}{rcl} i_2&\displaystyle =&\displaystyle \frac{n_1}{n_2}i_1. \end{array} \end{aligned}$

This means that the power at both ports is the same, because:

$\displaystyle \begin{aligned} \begin{array}{rcl} w_2=v_2i_2=\frac{n_2}{n_1}v_1\frac{n_1}{n_2}i_1=v_1i_1=w_1. \end{array} \end{aligned}$

2.3.2 Gear Reducer

Two gear wheels with radii r ₁ and r ₂ are depicted in Fig. 2.24. Wheel 1 has radius r ₁, n ₁ teeth, and it rotates at an angular velocity ω ₁ under the effect of an applied torque T ₁. Wheel 2 has radius r ₂, n ₂ teeth, and it rotates at an angular velocity ω ₂ under the effect of an applied torque T ₂. The directions defined as positive for these variables are also shown. It is important to note that teeth allow wheels to rotate without slipping. Hence, the arch length s generated by an angle θ ₁ at wheel 1 is equal to the arch length generated by the angle θ ₂ at wheel 2. It is well known from basic geometry that:

$\displaystyle \begin{aligned} \begin{array}{rcl} s=\theta_{1}\; r_1, s=\theta_{2}\; r_2,{} \end{array} \end{aligned}$

(2.83)

where θ ₁ and θ ₂ must be given in radians. Thus, it is concluded that:

$\displaystyle \begin{aligned} \begin{array}{rcl} \theta_{1}\; r_1=\theta_{2}\; r_2.{} \end{array} \end{aligned}$

(2.84)

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig24_HTML.png — Fig. 2.24
A gear reducer

It is also known that the tooth width α is the same in both wheels, which is necessary for teeth to engage. This implies that:

$\displaystyle \begin{aligned} \begin{array}{rcl} p_1&\displaystyle =&\displaystyle \alpha n_1, p_2=\alpha n_2,\\ p_1&\displaystyle =&\displaystyle 2\pi r_1, p_2=2\pi r_2. \end{array} \end{aligned}$

where p ₁ and p ₂ stand for wheel circumferences or perimeters. Combining these expressions, it is found that:

$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{r_1}{r_2}&\displaystyle =&\displaystyle \frac{n_1}{n_2}.{} \end{array} \end{aligned}$

(2.85)

From (2.84) and (2.85):

$\displaystyle \begin{aligned} \begin{array}{rcl} \theta_{1}\; =\frac{n_{2}}{n_{1}}\;\theta_{2}. \end{array} \end{aligned}$

Differentiating with respect to time:

$\displaystyle \begin{aligned} \begin{array}{rcl} \omega_{1}\; =\frac{n_{2}}{n_{1}}\;\omega_{2},{} \end{array} \end{aligned}$

(2.86)

On the other hand, force F applied at the point where the wheels are in contact is the same for both wheels. This force satisfies:

$\displaystyle \begin{aligned} \begin{array}{rcl} T_1=Fr_1, T_2=Fr_2, \end{array} \end{aligned}$

Solving each equation for F and equating the resulting expressions:

$\displaystyle \begin{aligned} \begin{array}{rcl} T_1=\frac{r_1}{r_2}T_2=\frac{n_1}{n_2}T_2.{} \end{array} \end{aligned}$

(2.87)

Using (2.87) and (2.86), it is found that the mechanical power, w, is the same on both ports, i.e.:

$\displaystyle \begin{aligned} \begin{array}{rcl} w_1=T_1\omega_{1}=\frac{n_1}{n_2}T_2\frac{n_{2}}{n_{1}}\;\omega_{2}=T_2\omega_{2}=w_2. \end{array} \end{aligned}$

2.3.3 Rack and Pinion

Consider the rack and pinion system depicted in Fig. 2.25. Port 1 is defined in terms of rotative variables, angular velocity ω ₁ and torque T ₁, whereas port 2 is defined in terms of translational variables, velocity v ₂ and force F ₂. By definition, torque T ₁ and force F ₂ are related as:

$\displaystyle \begin{aligned} \begin{array}{rcl} T_1=rF_2,{} \end{array} \end{aligned}$

(2.88)

where r is the pinion radius, whereas angular velocity ω ₁ and velocity v ₂ are related as (see (2.83)):

$\displaystyle \begin{aligned} \begin{array}{rcl} v_{2}=r\omega_{1}.{} \end{array} \end{aligned}$

(2.89)

Thus, the power at both ports is the same:

$\displaystyle \begin{aligned} \begin{array}{rcl} w_1=T_1\omega_{1}=rF_2\frac{1}{r}v_{2}=F_2v_{2}=w_2. \end{array} \end{aligned}$

The directions of variables shown in Fig. 2.25 are defined as positive.

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig25_HTML.png — Fig. 2.25
A rack and pinion system

2.4 Converters

We label as converters those system components that constitute a connection between two systems having a different nature.

2.4.1 Armature of a Permanent Magnet Brushed Direct Current Motor

A square loop with area A = l ², which is placed within a magnetic field B, is depicted in Fig. 2.26. An electric current i flows through the loop, with the indicated direction, under the effect of an external voltage source E. Resistance R and inductance L, shown in Fig. 2.26a, stand for the loop resistance and inductance. The force $\bar {F}$ exerted by a magnetic field $\bar {B}$ on an electric current of length l is given as [2], pp. 541:

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig26_HTML.png — Fig. 2.26
A permanent magnet brushed direct current (DC) motor. (a) Superior view. (b) Front view

$\displaystyle \begin{aligned} \begin{array}{rcl} \bar{F}=il\bar{u}\times\bar{B},{} \end{array} \end{aligned}$

(2.90)

where $\bar {u}$ is a unitary vector defined in the same direction as the electric current. A bar is used on the top of variables to indicate that they represent vectors. Symbol “×” stands for the standard vector product. Hence, under the conditions depicted in Fig. 2.26b, i.e., when θ = 90^∘, the force exerted by the magnetic field on the left and right sides of the loop is given as:

$\displaystyle \begin{aligned} \begin{array}{rcl} F=ilB. \end{array} \end{aligned}$

This means that the following torque is produced on the loop axis:

$\displaystyle \begin{aligned} \begin{array}{rcl} T=2\frac{l}{2}F=iBl^2=iBA. \end{array} \end{aligned}$

Notice that lF∕2 is torque because of the force exerted on each loop side and there are two applied forces on the loop with the same magnitude: one is applied on the left side and another on the right side. The reader can verify that, according to (2.90), forces on the rear and the front sides of the loop do not produce any torque. This means that the loop rotates at an angular velocity $\omega =-\dot \theta$ , i.e., in the opposite direction defined for θ in Fig. 2.26b where θ is the angle between the direction of the magnetic field and a line orthogonal to the plane defined by the loop. According to Faraday’s Law [3], pp. 325, this induces a voltage v at the loop terminals that is given according to (2.42), i.e.:

$\displaystyle \begin{aligned} \begin{array}{rcl} v=\frac{d\lambda}{dt}, \end{array} \end{aligned}$

where λ is the flux linkage in the loop and is given as:

$\displaystyle \begin{aligned} \begin{array}{rcl} \lambda=BA\cos{}(\theta). \end{array} \end{aligned}$

Hence, using these expressions, the following is found:

$\displaystyle \begin{aligned} \begin{array}{rcl} v=-BA\dot\theta\sin{}(\theta)=BA\omega\sin{}(\theta). \end{array} \end{aligned}$

Notice that, according to Lenz’s Law [3], pp. 327, the induced voltage v has a polarity (see Fig. 2.26a) that is opposite to the change in electric current i. Now, assume that several loops are employed, each one of them having a different tilt angle, such that a mechanical commutator connects only that loop for which θ = 90^∘. Then, the voltage at the commutator terminals is given as:

$\displaystyle \begin{aligned} \begin{array}{rcl} v=BA\omega. \end{array} \end{aligned}$

This is a description of a permanent magnet brushed direct current (DC) motor . Port 1 is represented by electrical variables, i.e., voltage v and electric current i, whereas port 2 is represented by mechanical variables, i.e., angular velocity ω and torque T. The relationship among these port variables is given as:

$\displaystyle \begin{aligned} \begin{array}{rcl} v&\displaystyle =&\displaystyle BA\omega,{}\\ i&\displaystyle =&\displaystyle \frac{1}{BA}T. \end{array} \end{aligned}$

(2.91)

If it is assumed that a torque T is applied at port 2 at an angular velocity ω, producing a voltage v and an electric current i at port 1, then a DC generator is obtained and both expressions in (2.91) are still valid. Notice that in both cases power at both ports is the same:

$\displaystyle \begin{aligned} \begin{array}{rcl} w_1=vi=\frac{1}{BA}TBA\omega=T\omega=w_2. \end{array} \end{aligned}$

The expressions in (2.91) can be written in a more general manner as:

$\displaystyle \begin{aligned} \begin{array}{rcl} v&\displaystyle =&\displaystyle k_e\omega,{} \end{array} \end{aligned}$

(2.92)

$\displaystyle \begin{aligned} \begin{array}{rcl} T&\displaystyle =&\displaystyle k_m i,{} \end{array} \end{aligned}$

(2.93)

where k _e > 0 and k _m > 0 are known as the counter-electromotive force constant and the torque constant respectively. These constants are introduced to take into account the fact that several loops are usually series connected and some other variants in the construction of practical DC motors and generators. The fact that power at both ports is the same:

$\displaystyle \begin{aligned} \begin{array}{rcl} w_1=vi=k_e\omega\frac{1}{k_m}T=T\omega=w_2, \end{array} \end{aligned}$

can be used to show that k _e = k _m in any DC motor or generator. However, attention must be payed to the units employed for the involved variables: k _e = k _m is true if these constants are expressed in the International System of Units.

2.4.2 Electromagnet

A bar made in iron, placed at a distance x from an electromagnet is shown in Fig. 2.27. The electromagnet is composed of a conductor wire wound, with N turns, on a core made in iron. An electric current i flows through the conductor under the effect of a voltage v applied at the conductor terminals. According to Ampere’s Law [5], the magnetomotive force M is given as:

$\displaystyle \begin{aligned} \begin{array}{rcl} M=Ni.{} \end{array} \end{aligned}$

(2.94)

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig27_HTML.png — Fig. 2.27
Force due to a magnetic field

The reluctance $\mathcal {R}$ of the magnetic circuit is defined through the constitutive relation :

$\displaystyle \begin{aligned} \begin{array}{rcl} M=\mathcal{R}\lambda,{} \end{array} \end{aligned}$

(2.95)

where λ represents the magnetic flux in the magnetic circuit and the reluctance is given as:

$\displaystyle \begin{aligned} \begin{array}{rcl} \mathcal{R}=\frac{l}{\mu_iA}+\frac{2x}{\mu_aA}\approx\frac{2x}{\mu_aA}, \mu_i\gg\mu_a,{} \end{array} \end{aligned}$

(2.96)

where μ _i and μ _a stand for the magnetic permeability of iron and air respectively, whereas l and x represent the magnetic circuit length and the air gap respectively.

The flux linkage is defined as ψ = Nλ. According to Faraday’s Law $v=\frac {d(N\lambda )}{dt}$ . Using this and (2.94), the power w supplied by the electric circuit is given as:

$\displaystyle \begin{aligned} \begin{array}{rcl} w=iv=\frac{M}{N}N\dot{\lambda}=M\dot{\lambda}=\dot E, \end{array} \end{aligned}$

where E is the energy supplied by the electric circuit, which is stored in the magnetic circuit as:

$\displaystyle \begin{aligned} \begin{array}{rcl} E=\int_0^tM\frac{d\lambda}{dt}dt=\int_{\lambda(0)}^{\lambda(t)}Md\lambda. \end{array} \end{aligned}$

Using (2.95), this can be written as:

$\displaystyle \begin{aligned} \begin{array}{rcl} E=\int_{\lambda(0)}^{\lambda(t)}\mathcal{R}\lambda d\lambda=\frac{1}{2}\mathcal{R}\lambda^2(t),{} \end{array} \end{aligned}$

(2.97)

if λ(0) = 0. Notice that this represents the energy stored in the reluctance, i.e., reluctance stores energy. This must be stressed, as the common belief is that reluctance is the magnetic analog of electric resistance, which is not true because electric resistance dissipates energy and reluctance stores energy.

The stored energy E also represents the mechanical work performed by an external force F to create a gap of width x. This means that F is applied in the opposite direction to the magnetic force f, i.e., F = −f and:

$\displaystyle \begin{aligned} \begin{array}{rcl} E=\int_0^xFds=-\int_0^xfds, f=-\frac{\partial E}{\partial x}. \end{array} \end{aligned}$

Using (2.97):

$\displaystyle \begin{aligned} \begin{array}{rcl} f=-\frac{1}{2}\lambda^2\frac{\partial \mathcal{R}}{\partial x}. \end{array} \end{aligned}$

This and (2.96) imply that the magnetic force is produced by the reluctance and it is applied in the sense where the reluctance, energy, and gap width decrease.

On the other hand, the inductance L(x) depends on the bar position x, as explained in the following. According to (2.41), the flux linkage in the inductor is given as the product of inductance and the electric current, i.e., ψ = L(x)i = Nλ. Moreover, according to (2.96) and (2.95), if x decreases, flux λ increases if the electric current is kept constant. According to Nλ = L(x)i, an increase in flux λ only can be produced by an increase in inductance L(x), i.e., inductance increases if x decreases.

From (2.97) and (2.95), it follows that:

$\displaystyle \begin{aligned} \begin{array}{rcl} E=\frac{1}{2}M\lambda=\frac{1}{2}Ni\frac{\psi}{N}=\frac{1}{2}i\psi=\frac{1}{2}L(x)i^2. \end{array} \end{aligned}$

Finally, according to $f=-\frac {\partial E}{\partial x}$ :

$\displaystyle \begin{aligned} \begin{array}{rcl} f=-\frac{1}{2}\frac{\partial L(x)}{\partial x}i^2.{} \end{array} \end{aligned}$

(2.98)

It is stressed that f in (2.98) is applied in the sense where the air gap x decreases. The above-mentioned ideas are based on [10].

Example 2.16

Two rotative bodies connected by a rack are shown in Fig. 2.28a. An external torque T(t) is applied to the body at the left. Hence, it may be assumed that the body at the left is a motor that transmits movement to the body at the right. Following the sign definitions shown in Figs. 2.1, 2.9, 2.3, and 2.25, the free-body diagrams shown in Figs. 2.28b and c are obtained. Figure 2.28b shows the connection of both rotative bodies by means of two rack and pinion systems, whereas Fig. 2.28c shows the free-body diagram of each rotative body.¹ The employed nomenclature is defined as follows:

$v_{m}=\frac {dx_{m}}{dt}$ , where x _m is the position of the rack with the mass m.
$\omega _{1}=\frac {d\theta _{1}}{dt}$ , where θ ₁ is the angular position of the rotative body 1.
$\omega _2=\frac {d\theta _{2}}{dt}$ , where θ ₂ is the angular position of the rotative body 2.
v ₁ is the rack translational velocity produced as a consequence that body 1 rotates with angular velocity ω ₁.
v ₂ is the rack translational velocity produced as a consequence that body 2 rotates at angular velocity ω ₂.
$\omega _{b1}=\frac {d\theta _{b1}}{dt}$ , where θ _b1 is the angular position of the mobile side of the damper connected to body 1 (friction between body 1 and its supports).
$\omega _{b2}=\frac {d\theta _{b2}}{dt}$ , where θ _b2 is the angular position of the mobile side of the damper connected to body 2 (friction between body 2 and its supports).
$v_{b}=\frac {dx_{b}}{dt}$ , where x _b is the position of the mobile side of the damper connected to rack m. This damper represents the friction between rack and floor.

Using Figs. 2.28b, c, and (2.1), (2.25), (2.4), (2.28), (2.88), (2.89), the following is obtained:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Inertia 1:}\hspace{1cm}I_1\frac{d\omega_{1}}{dt}=T(t)+T_1-T_{b1},{} \end{array} \end{aligned}$

(2.99)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Inertia 2:}\hspace{1cm}I_2\frac{d\omega_{2}}{dt}=T_{2}-T_{b2},{} \end{array} \end{aligned}$

(2.100)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Mass:}\hspace{1cm}m\frac{dv_m}{dt}=F_1-F_2+F_b,{} \end{array} \end{aligned}$

(2.101)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Left damper:}\hspace{1cm}b_1\omega_{b1}=T_{b1},\\ &\displaystyle &\displaystyle \text{Right damper:}\hspace{1cm}b_2\omega_{b2}=T_{b2},\\ &\displaystyle &\displaystyle \text{Central damper:}\hspace{1cm}b v_{b}=F_{b},\\ &\displaystyle &\displaystyle \text{Rack and pinion at the left:}\hspace{1cm}T_1=rF_1, v_1=r\omega_1,{} \end{array} \end{aligned}$

(2.102)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Rack and pinion at the right:}\hspace{1cm}T_2=rF_2, v_2=r\omega_2.{} \end{array} \end{aligned}$

(2.103)

Notice that forces with the same direction as ω ₁, ω ₂ and v _m are affected by a positive sign in (2.99), (2.100), (2.101), or a negative sign if this is not the case. Assuming that all of the system components are connected by rigid joints then, according to Figs. 2.28b and c, it is found that:

$\displaystyle \begin{aligned} \begin{array}{rcl} v_{m}&\displaystyle =&\displaystyle -v_{1}=v_2=-v_b,{} \end{array} \end{aligned}$

(2.104)

$\displaystyle \begin{aligned} \begin{array}{rcl} \omega_{1}&\displaystyle =&\displaystyle \omega_{b1}, \omega_{2}=\omega_{b2}. {} \end{array} \end{aligned}$

(2.105)

According to this, if x _m = 0, θ ₁ = 0, θ ₂ = 0, x _b = 0, θ _b1 = 0, θ _b2 = 0 are defined as positions of the rack, body 1, body 2, and the mobile sides of the dampers connected to the rack and bodies 1 and 2 respectively, when the whole system is at rest with T(t) = 0, then it is concluded that:

$\displaystyle \begin{aligned} \begin{array}{rcl} x_{m}&\displaystyle =&\displaystyle -x_b,{}\\ \theta_{1}&\displaystyle =&\displaystyle \theta_{b1}, \theta_{2}=\theta_{b2}.\end{array} \end{aligned}$

(2.106)

Hence, expressions in (2.99), (2.100) and (2.101) can be rewritten as:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Inertia 1:}\hspace{1cm}I_1\frac{d\omega_{1}}{dt}=T(t)+rF_1-b_1\omega_{b1},{} \end{array} \end{aligned}$

(2.107)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Inertia 2:}\hspace{1cm}I_2\frac{d\omega_{2}}{dt}=rF_2-b_2\omega_{b2},{} \end{array} \end{aligned}$

(2.108)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Mass:}\hspace{1cm}m\frac{dv_m}{dt}=F_1-F_2+b v_{b}.{} \end{array} \end{aligned}$

(2.109)

Recall that v ₁ = −v ₂. Then, (2.102) and (2.103) can be used to find that ω ₁ = −ω ₂. Using this result, (2.105), and subtracting (2.108) to (2.107):

$\displaystyle \begin{aligned} \begin{array}{rcl} (I_1+I_2)\frac{d\omega_{1}}{dt}=T(t)+r(F_1-F_2)-(b_1+b_2)\omega_{1}.{} \end{array} \end{aligned}$

(2.110)

On the other hand, employing − v _m = v ₁ = rω ₁ = v _b, (2.109) can be written as:

$\displaystyle \begin{aligned} \begin{array}{rcl} -mr\frac{d\omega_1}{dt}=F_1-F_2-b v_{m}. \end{array} \end{aligned}$

Solving for F ₁ − F ₂ in this expression and replacing it in (2.110), the following is finally found:

$\displaystyle \begin{aligned} \begin{array}{rcl} (I_1+I_2+mr^2)\frac{d\omega_{1}}{dt}=T(t)-(b_1+b_2+r^2b)\omega_{1}. \end{array} \end{aligned}$

This differential equation represents the mathematical model of the system in Fig. 2.28a. Contrary to several previous examples, in this case it is possible to combine three differential equations in a single differential equation. This is because the three variables ω ₁, ω ₂ are v _m are linearly dependent as they are related through the algebraic equation − v _m = rω ₁ = −rω ₂. Also notice how the radii of the pinions modify (i) the rack mass to express its effect on the equivalent inertia on the axis of body 1, and (ii) friction between the rack and floor to express the equivalent friction on the axis of body 1.

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig28_HTML.png — Fig. 2.28
Two rotative bodies connected by a rack

Example 2.17

Two rotative bodies connected through a gear reducer are shown in Fig. 2.29a. The external torque T(t) is applied t the body at the left. This mechanical system can be seen as an electric motor (body at the left) transmitting movement to a mechanical load (body at the right) through a gear reducer. According to the sign definitions shown in Figs. 2.9, 2.11, and 2.24, the free-body diagram shown in Fig. 2.29b is obtained.² The employed nomenclature is defined as follows:

$\omega =\frac {d\theta }{dt}$ , where θ is the angular position of body 1.
$\omega _{4}=\frac {d\theta _{4}}{dt}$ , where θ ₄ is the angular position of body 2.
$\omega _{b1}=\frac {d\theta _{b1}}{dt}$ , where θ _b1 is the angular position of the mobile side of the damper connected to body 1.
$\omega _{b2}=\frac {d\theta _{b2}}{dt}$ , where θ _b2 is the angular position of the mobile side of the damper connected to body 2.
ω _n1 and ω _n2 are the angular velocities of the gear wheels that result from the movement of bodies 1 and 2.
T _n1 and T _n2 are the torques appearing on gear wheels as consequences of the movement of bodies 1 and 2.

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig29_HTML.png — Fig. 2.29
Two rotative bodies connected by a gear reducer

Using Fig. 2.29b, and (2.25), (2.28), (2.86), (2.87), the following expressions are obtained:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Body 1:}\hspace{1cm}I_1\dot \omega=T(t)-T_{b1}-T_{n1},{} \end{array} \end{aligned}$

(2.111)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Body 2:}\hspace{1cm}I_2\dot\omega_4=T_{3}+T_{b2},{} \end{array} \end{aligned}$

(2.112)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Damper at the left:}\hspace{1cm}b_1\omega_{b1}=T_{b1},\\ &\displaystyle &\displaystyle \text{Damper at the right:}\hspace{1cm}b_2\omega_{b2}=T_{b2},\\ &\displaystyle &\displaystyle \text{Gear reducer:}\hspace{1cm}T_{n1}=\frac{n_1}{n_2}T_{n2}, \omega_{n1}=\frac{n_2}{n_1}\omega_{n2}, T_{n2}=T_3,{} \end{array} \end{aligned}$

(2.113)

Last expression is due to the fact that, according to Newton’s Third Law , an action of A on B produces a reaction from B on A. Notice that in (2.111) and (2.112) torques that have the same direction as ω and ω ₄ appear with a positive sign and a negative sign if this is not the case. Assuming that all of the system elements are connected by rigid joints, then from Fig. 2.29b, it is found that:

$\displaystyle \begin{aligned} \begin{array}{rcl} \omega&\displaystyle =&\displaystyle \omega_{b1}=-\omega_{n1},\omega_{4}=-\omega_{b2}=-\omega_{n2}. {} \end{array} \end{aligned}$

(2.114)

Hence, if θ = 0, θ ₄ = 0, θ _b1 = 0, θ _b2 = 0, θ _n1 = 0, θ _n2 = 0 are defined as the angular positions of body 1, body 2, the mobile side of the damper at the left, the mobile side of the damper at the right, gear wheel 1, and gear wheel 2 when the whole system is at rest with T(t) = 0, then it is concluded that:

$\displaystyle \begin{aligned} \begin{array}{rcl} \theta&\displaystyle =&\displaystyle \theta_{b1}=-\theta_{n1},\theta_{4}=-\theta_{b2}=-\theta_{n2}. {} \end{array} \end{aligned}$

(2.115)

Thus, using (2.114), (2.115), (2.111), and (2.112), the following can be written:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Body 1:}\hspace{1cm}I_1\dot \omega=T(t)-b_1\omega-\frac{n_1}{n_2}T_{3},{} \end{array} \end{aligned}$

(2.116)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Body 2:}\hspace{1cm}I_2\dot\omega_4=T_{3}-b_2\dot\theta_{4}.{} \end{array} \end{aligned}$

(2.117)

On the other hand, from (2.113) and (2.114) it is found that $\omega =\frac {n_2}{n_1}\omega _{4}$ . Using this, solving for T ₃ in (2.117), replacing it in (2.116), and rearranging terms:

$\displaystyle \begin{aligned} \begin{array}{rcl} \left(I_2+\left(\frac{n_2}{n_1}\right)^2I_1\right)\ddot \theta_4 +\left(b_2+\left(\frac{n_2}{n_1}\right)^2b_1\right)\dot \theta_4 =\frac{n_2}{n_1}T(t).{} \end{array} \end{aligned}$

(2.118)

Notice that both differential equations in (2.116) and (2.117) have been combined in a single differential equation. This is possible because variables ω and ω ₄ are linearly dependent, i.e., they are related by the algebraic equation $\omega =\frac {n_2}{n_1}\omega _{4}$ . The above-mentioned differential equation represents the mathematical model of the system: if T(t) is known, then θ ₄, i.e., the position of body 2, can be computed solving this differential equation.

Example 2.18

In Fig. 2.30a, the mechanism in Fig. 2.28a (in Example 2.16) is shown again, but now it is assumed that the rack is flexible. This flexibility is not intentional, but it is an undesired problem that appears because of the finite stiffness of the steel of which the rack is made. The mathematical model in this case can be obtained taking advantage of the procedure in Example 2.16, i.e., to continue from (2.107), (2.108), (2.109), which are rewritten here for ease of reference:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Inertia 1:}\hspace{1cm}I_1\frac{d\omega_{1}}{dt}=T(t)+rF_1-b_1\omega_{b1},\\ &\displaystyle &\displaystyle \text{Inertia 2:}\hspace{1cm}I_2\frac{d\omega_{2}}{dt}=rF_2-b_2\omega_{b2},\\ &\displaystyle &\displaystyle \text{Mass:}\hspace{1cm}m\frac{dv_m}{dt}=F_1-F_2+b v_{b}. \end{array} \end{aligned}$

However, it must be considered now that, according to Fig. 2.30b and the sign definitions shown in Fig. 2.2 and (2.3):

$\displaystyle \begin{aligned} \begin{array}{rcl} F_1=K_1(x_A-x_B), F_2=K_2(x_C-x_D), \end{array} \end{aligned}$

with:

$v_A=\frac {dx_A}{dt}$ and $v_B=\frac {dx_B}{dt}$ , where x _A and x _B are the positions of the sides of the spring on the left.
$v_C=\frac {dx_C}{dt}$ and $v_D=\frac {dx_D}{dt}$ , where x _C and x _D are the positions of the sides of the spring on the right.

Using this and (2.104), (2.105), the following is found:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Inertia 1:}\hspace{1cm}I_1\frac{d\omega_{1}}{dt}=T(t)+rK_1(x_A-x_B)-b_1\omega_{1},{} \end{array} \end{aligned}$

(2.119)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Inertia 2:}\hspace{1cm}I_2\frac{d\omega_{2}}{dt}=rK_2(x_C-x_D)-b_2\omega_{2}.{} \end{array} \end{aligned}$

(2.120)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Mass:}\hspace{1cm}m\ddot x_m=K_1(x_A-x_B)-K_2(x_C-x_D)-b\dot x_{m}.{} \end{array} \end{aligned}$

(2.121)

On the other hand, according to (2.102), (2.103), (2.104), (2.106) and assuming that all the system components are connected by rigid joints:

$\displaystyle \begin{aligned} \begin{array}{rcl} x_B=x_C=x_{m}, x_A=-r\theta_1, x_D=r\theta_2,{} \end{array} \end{aligned}$

(2.122)

where θ ₁ and θ ₂ are angular positions of gear wheels 1 and 2 defined according to Fig. 2.28b. Replacing (2.122) in (2.119), (2.120), and (2.121):

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Inertia 1:}\hspace{1cm}I_1\ddot\theta_{1}+b_1\dot\theta_{1}-rK_1(-r\theta_1-x_{m})=T(t),\\ &\displaystyle &\displaystyle \text{Inertia 2:}\hspace{1cm}I_2\ddot\theta_{2}+b_2\dot\theta_{2}-rK_2(x_{m}-r\theta_2)=0,\\ &\displaystyle &\displaystyle \text{Mass:}\hspace{1cm}m\ddot x_m+b\dot x_{m} -K_1(-r\theta_1-x_{m})+K_2(x_{m}-r\theta_2) =0. \end{array} \end{aligned}$

The mathematical model is given by these three differential equations, which must be solved simultaneously. In this case, it is not possible to combine these three differential equations in a single differential equation because variables θ ₁, θ ₂, and x _m are not linearly dependent. This means that there is no algebraic expression relating these three variables. Notice that, from the modeling point of view, this is the effect introduced by the springs considered in this example. Compare this mathematical model with that obtained in Example 2.16.

Example 2.19

Two rotative bodies connected through a gear reducer and a spring are depicted in Fig. 2.31a. The external torque T(t) is applied to the body on the left. This example can be seen as a motor (body on the left) that transmits movement to a mechanical load (body on the right) through a gear reducer that exhibits teeth flexibility. This flexibility is an undesired behavior that, however, appears in practice because of the finite stiffness of steel, the material in which the gear reducer is made. This modeling problem can be solved taking advantage of the procedure in Example 2.17 which has been used to model the mechanical system in Fig. 2.29a. We can go back to expressions in (2.116) and (2.117), which are rewritten here for ease of reference:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Body 1:}\hspace{1cm}I_1\dot \omega=T(t)-b_1\omega-\frac{n_1}{n_2}T_{n2},{} \end{array} \end{aligned}$

(2.123)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Body 2:}\hspace{1cm}I_2\dot\omega_4=T_{n2}-b_2\dot\theta_{4},{} \end{array} \end{aligned}$

(2.124)

but now, according to Figs. 2.31b, 2.10 and (2.27):

$\displaystyle \begin{aligned} \begin{array}{rcl} T_{n2}=K(\theta_A-\theta_B), \end{array} \end{aligned}$

where $\omega _A=\dot \theta _A$ and $\omega _B=\dot \theta _B$ with θ _A and θ _B, the angular position of both sides of the spring. On the other hand, according to (2.113), it is found that $\theta =\frac {n_2}{n_1}\theta _{A}$ , because ω _A = −ω _n2 and ω _n1 = −θ, where θ and θ _A are the angular positions of body 1 and the side of the spring connected to gear wheel 2 (see Fig. 2.31b). Also notice that θ _B = θ ₄. Replacing the above-mentioned expressions in (2.123) and (2.124), it is found that:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Body 1:}\hspace{1cm}I_1\ddot \theta + b_1\dot\theta+\frac{n_1}{n_2}K\left(\frac{n_1}{n_2}\theta-\theta_4\right) =T(t),\\ &\displaystyle &\displaystyle \text{Body 2:}\hspace{1cm}I_2\ddot\theta_4 +b_2\dot\theta_{4}-K\left(\frac{n_1}{n_2}\theta-\theta_4\right) =0. \end{array} \end{aligned}$

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig31_HTML.png — Fig. 2.31
Transmission system with a flexible gear reducer

These differential equations must be solved simultaneously and they represent the corresponding mathematical model. Notice that these differential equations cannot be combined into a single differential equation because variables θ and θ ₄ are linearly independent, i.e., there is no algebraic expression relating both of these variables. This is the main difference with respect to the problem solved in Example 2.17 and it is due to the introduction of the spring in the present example.

Example 2.20 (Taken from [6], pp. 122)

An electro-mechanical system is shown in Fig. 2.32. There is a permanent magnet brushed DC motor that is used to actuate on a body with the mass M through a gear reducer and a rack and pinion system. The body M moves on a horizontal plane; hence, it is not necessary to consider the effect of gravity. According to the sign definitions shown in Figs. 2.9, 2.1, 2.11, 2.3, 2.24, 2.25, 2.26, the free-body diagrams in Fig. 2.33 are obtained. The electrical DC motor components are analyzed in Fig. 2.33a. The mechanical DC motor components and the gear reducer are analyzed in Fig. 2.33b. The rack and pinion system components are analyzed in Fig. 2.33c.³ Notice that friction at motor bearings, friction at pinion bearings, and friction between body M and its supports (or the floor) are considered. The nomenclature employed is defined as follows:

$\dot \theta _{m}=\frac {d\theta _{m}}{dt}$ , where θ _m is the motor angular position.
$\omega _{P}=\frac {d\theta _{P}}{dt}$ , where θ _P is the pinion angular position.
$\dot x=\frac {dx}{dt}$ , where x is the body M position.
$\omega _{b1}=\frac {d\theta _{b1}}{dt}$ , where θ _b1 is the angular position of the mobile side of the damper connected to the motor.
$\omega _{b2}=\frac {d\theta _{b2}}{dt}$ , where θ _b2 is the angular position of the mobile side of the damper connected to the pinion.
$v_{b3}=\frac {dx_{b3}}{dt}$ , where x _b3 is the position of the mobile side of the damper connected to body M.
ω ₁ and ω ₂ are the angular velocities of the gear wheels that result from the movement of the motor and the pinion.
u is the voltage applied at the motor terminals and i is the electric current flowing through the motor.
F, T _P, and v are the force, torque, and translational velocity produced at the rack and pinion system as a result of movement of the pinion and mass M.

Using Fig. 2.33, and (2.1), (2.25), (2.4), (2.28), (2.86), (2.87), (2.88), (2.89), the following expressions are obtained:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Motor inertia:}\hspace{1cm}I_m\ddot \theta_m=T-T_1-T_{b1},{} \end{array} \end{aligned}$

(2.125)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Pinion inertia:}\hspace{1cm}I_P\ddot\theta_P=T_{2}+T_{b2}+T_P,{} \end{array} \end{aligned}$

(2.126)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Mass }M\text{:}\hspace{1cm}M\ddot x=-F-F_{b3},{} \end{array} \end{aligned}$

(2.127)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Damper at motor:}\hspace{1cm}b_1\omega_{b1}=T_{b1},\\ &\displaystyle &\displaystyle \text{Damper at pinion:}\hspace{1cm}b_2\omega_{b2}=T_{b2},\\ &\displaystyle &\displaystyle \text{Damper at mass }M\text{:}\hspace{1cm}b_3 v_{b3}=F_{b3},\\ &\displaystyle &\displaystyle \text{Gear reducer:}\hspace{1cm}T_1=\frac{n_1}{n_2}T_2, \omega_1=\frac{n_2}{n_1}\omega_2,{} \end{array} \end{aligned}$

(2.128)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Rack and pinion:}\hspace{1cm}v=r\omega_P, T_P=rF.{} \end{array} \end{aligned}$

(2.129)

Notice that, in (2.125), (2.126), and (2.127), forces that have the same direction as $\dot \theta _{m}$ , ω _P and $\dot x$ appear affected by a positive sign and a negative sign if this is not the case. Assuming that all the system components are connected by rigid joints, then, according to Fig. 2.33, it is found that:

$\displaystyle \begin{aligned} \begin{array}{rcl} v&\displaystyle =&\displaystyle \dot x=v_{b3},{} \end{array} \end{aligned}$

(2.130)

$\displaystyle \begin{aligned} \begin{array}{rcl} \dot\theta_m&\displaystyle =&\displaystyle \omega_{b1}=-\omega_1, \omega_{2}=\omega_{b2}=-\omega_P, {} \end{array} \end{aligned}$

(2.131)

Hence, if x = 0, x _b3 = 0, θ _m = 0, θ _b1 = 0, and θ _b2 = 0 are defined as positions of the body M (the mobile side of the damper connected to mass M, the motor rotor, the mobile side of the damper connected to the motor, and the mobile side of the damper connected to the pinion respectively), when the whole system is at rest with T = 0, it is concluded that:

$\displaystyle \begin{aligned} \begin{array}{rcl} x&\displaystyle =&\displaystyle x_{b3},{} \end{array} \end{aligned}$

(2.132)

$\displaystyle \begin{aligned} \begin{array}{rcl} \theta_m&\displaystyle =&\displaystyle \theta_{b1}. {} \end{array} \end{aligned}$

(2.133)

Hence, (2.125), (2.126), and (2.127), can be written as:

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Motor inertia:}\hspace{1cm}I_m\ddot \theta_m=T-\frac{n_1}{n_2}T_2-b_1\omega_{b1},\\ &\displaystyle &\displaystyle \text{Pinion inertia:}\hspace{1cm}I_P\ddot\theta_P=T_{2}+b_2\omega_{b2}+rF,\\ &\displaystyle &\displaystyle \text{Mass }M\text{:}\hspace{1cm}M\ddot x=-F-b_3 v_{b3}. \end{array} \end{aligned}$

Using (2.130) and (2.131):

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Motor inertia:}\hspace{1cm}I_m\ddot \theta_m=T-\frac{n_1}{n_2}T_2-b_1\dot\theta_{m},{} \end{array} \end{aligned}$

(2.134)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Pinion inertia:}\hspace{1cm}I_P\ddot\theta_P=T_{2}-b_2\omega_{P}+rF,{} \end{array} \end{aligned}$

(2.135)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \text{Mass }M\text{:}\hspace{1cm}M\ddot x=-F-b_3 \dot x.{} \end{array} \end{aligned}$

(2.136)

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig32_HTML.png — Fig. 2.32
Servomechanism

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig33_HTML.png — Fig. 2.33
Free-body diagrams corresponding to the servomechanism in Fig. 2.32

On the other hand, according to (2.128), (2.129), (2.130), and (2.131), it is found that $\dot \theta _m=\frac {n_2}{n_1r}\dot x$ . Using this, solving for T ₂ from (2.135) and replacing it in (2.134):

$\displaystyle \begin{aligned} \begin{array}{rcl} I_m\frac{n_2}{n_1r}\ddot x+b_1\frac{n_2}{n_1r}\dot x=T-\frac{n_1}{n_2} \left[I_P\dot\omega_P+b_2\omega_{P}-rF\right]. {} \end{array} \end{aligned}$

(2.137)

From (2.129) and (2.130), it is obtained that $\dot x=r\omega _P$ . Using this in (2.137) and replacing F from (2.136):

$\displaystyle \begin{aligned} \begin{array}{rcl} I_m\frac{n_2}{n_1r}\ddot x+b_1\frac{n_2}{n_1r}\dot x=T-\left(\frac{n_1}{n_2r} I_P+\frac{n_1}{n_2}rM\right)\ddot x-\left(\frac{n_1}{n_2r}b_2+\frac{n_1}{n_2}rb_3\right)\dot x.\end{array} \end{aligned}$

Rearranging, the following is obtained:

$\displaystyle \begin{aligned} \begin{array}{rcl} \left( I_m\left(\frac{n_2}{n_1r}\right)^2 + I_P\frac{1}{r^2} +M\right)\ddot x+ \left( b_1\left(\frac{n_2}{n_1r}\right)^2 + b_2\frac{1}{r^2} +b_3\right)\dot x= \frac{n_2}{n_1r}T.\\ {} \end{array} \end{aligned}$

(2.138)

On the other hand, applying KVL (see Example 2.12) to the electric circuit shown in Fig. 2.33a yields:

$\displaystyle \begin{aligned} \begin{array}{rcl} L\frac{di}{dt}+Ri+e_b=u.{} \end{array} \end{aligned}$

(2.139)

From (2.92), (2.93) and $\dot \theta _m=\frac {n_2}{n_1r}\dot x$ , it is found that the induced voltage and the generated torque are given as:

$\displaystyle \begin{aligned} \begin{array}{rcl} e_b&\displaystyle =&\displaystyle k_e\dot\theta_m=k_e\frac{n_2}{n_1r}\dot x, T=k_m i. \end{array} \end{aligned}$

Using this result, (2.139) and (2.138), it is finally found that the mathematical model is given by the following two differential equations, which have to be solved simultaneously:

$\displaystyle \begin{aligned} \begin{array}{rcl} L\frac{di}{dt}+Ri+k_e\frac{n_2}{n_1r}\dot x&\displaystyle =&\displaystyle u,{} \end{array} \end{aligned}$

(2.140)

$\displaystyle \begin{aligned} \begin{array}{rcl} \left( I_m\left(\frac{n_2}{n_1r}\right)^2 + I_P\frac{1}{r^2} +M\right)\ddot x+ \left( b_1\left(\frac{n_2}{n_1r}\right)^2 + b_2\frac{1}{r^2} +b_3\right)\dot x&\displaystyle =&\displaystyle \frac{n_2}{n_1r}k_m i.\\ {} \end{array} \end{aligned}$

(2.141)

Another way of representing this mathematical model using a single differential equation is by combining (2.140) and (2.141). To this aim, differentiate (2.141) once with respect to time. This produces $\frac {d i}{dt}$ to appear on the right-hand side of the resulting equation. Then, replace $\frac {d i}{dt}$ from (2.140). Thus, a differential equation is obtained in terms of $\frac {d^3x}{dt^3}$ , $\ddot x$ , $\dot x$ and i. Solve (2.141) for i and replace it to obtain a third-order differential equation in the unknown variable x with voltage u as the excitation function. Thus, the mathematical model can be expressed by a single differential equation, which can be solved for the mass position x once the voltage applied to motor u is known. The reader is encouraged to solve this problem, proceeding as indicated above.

2.5 Case Study: A DC-to-DC High-Frequency Series Resonant Power Converter

Much electronic equipment at present employs a variety of electronic circuits that require several different levels of DC voltage. Several values of DC voltage can be obtained using a large electric transformer with several tabs in the secondary winding. Then, a rectifier filter circuit can be placed at each one of these tabs to obtain a different DC voltage level at the output of each circuit. However, this method is no longer utilized because it requires voluminous inductors and capacitors, aside from the fact that it produces large energy losses.

This problem is currently solved by using several power electronic circuits, which, using electronic commutation devices, are capable of delivering different DC voltage levels from a unique DC power supply. These circuits are known as DC-to-DC power electronic converters and they solve the above-described problem, reducing both equipment volume and energy losses. In fact, the impressive equipment miniaturization that it is known today is possible thanks to employment of power electronic converters.

A particular class of power electronic converters is that known as resonant converters which are divided into series and parallel according to the way in which their components are connected (see Fig. 2.34). The DC-to-DC series resonant power electronic converter depicted in Fig. 2.34a works as follows. The commutation network is composed of an inverter delivering a square alternating current (AC) power voltage (taking values ± E) at the input of the series resonant circuit. See switches on the left of Fig. 2.35. In the most simple operating mode, the frequency of the AC voltage delivered by the inverter is equal to the resonance frequency of the series inductor capacitor (LC) circuit. Hence, the transistors comprising the commutation network are turned on and off (Q ₁, Q ₃ and Q ₂, Q ₄) when the electric current through them is zero. This is very useful because it avoids transistor stress. On the other hand, because the series LC circuit works at resonance, the electric current i through the circuit has a sinusoidal waveform, which is rectified and filtered by the circuit on the right in Fig. 2.34a. Hence, the circuit load, represented by the resistance R, receives the DC voltage v ₀.

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig34_HTML.png — Fig. 2.34
DC-to-DC resonant power electronic converters . (a) Series. (b) Parallel

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig35_HTML.png — Fig. 2.35
A DC-to-DC series resonant power electronic converter showing devices comprising the inverter

In Fig. 2.36, the electric diagram of a DC-to-DC series resonant power electronic converter is shown, representing the inverter as a square AC voltage source E(t) taking the values ± E. Notice that, because of the presence of the rectifying diodes, the circuit operation can be divided in two cases: 1) when i > 0, Fig. 2.37a, and 2) when i < 0, Fig. 2.37b. The equations for i > 0 are obtained from Fig. 2.37a. Applying KVL to the closed path on the left, the following is obtained:

$\displaystyle \begin{aligned} \begin{array}{rcl} E(t)=L\frac{di}{dt}+v+v_0, i>0. \end{array} \end{aligned}$

It is important to stress that the direction of i and the polarity of v ₀ have been taken into account. On the other hand, according to (2.50), the voltage and electric current at the resonant capacitor are related by:

$\displaystyle \begin{aligned} \begin{array}{rcl} C\frac{dv}{dt}&\displaystyle =&\displaystyle i, i>0. \end{array} \end{aligned}$

Finally, applying KCL to the node that is common to both capacitors, it is found that:

$\displaystyle \begin{aligned} \begin{array}{rcl} i=C_0\frac{dv_0}{dt}+\frac{v_0}{R}, i>0. \end{array} \end{aligned}$

On the other hand, the equations for i < 0 are obtained from Fig. 2.37b. Applying KVL to the closed path on the left, the following is obtained:

$\displaystyle \begin{aligned} \begin{array}{rcl} E(t)=L\frac{di}{dt}+v-v_0, i<0. \end{array} \end{aligned}$

Notice that v ₀ is affected by a “−” because its polarity is opposite to the direction of electric current i. However, voltage and electric current at the resonant capacitor satisfy again:

$\displaystyle \begin{aligned} \begin{array}{rcl} C\frac{dv}{dt}&\displaystyle =&\displaystyle i, i<0. \end{array} \end{aligned}$

Finally, applying KCL to the node that is common to both capacitors yields:

$\displaystyle \begin{aligned} \begin{array}{rcl} i=-\frac{v_0}{R}-C_0\frac{dv_0}{dt}, i<0. \end{array} \end{aligned}$

Notice that the correct direction of the electric current through the capacitor C ₀ and resistance is from “+ ” to “−”, i.e., it is opposite to the direction indicated for i.

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig36_HTML.png — Fig. 2.36
A simplified circuit of a DC-to-DC series resonant power electronic converter

/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_2_Chapter/454499_1_En_2_Fig37_HTML.png — Fig. 2.37
Equivalent circuits for (a) i > 0 and (b) i < 0

Analyzing these two sets of equations, it is possible to use a unique set of equations representing both cases i > 0 and i < 0:

$\displaystyle \begin{aligned} \begin{array}{rcl} E(t)&\displaystyle =&\displaystyle L\frac{di}{dt}+v+v_0sign(i),{} \end{array} \end{aligned}$

(2.142)

$\displaystyle \begin{aligned} \begin{array}{rcl} C\frac{dv}{dt}&\displaystyle =&\displaystyle i,{} \end{array} \end{aligned}$

(2.143)

$\displaystyle \begin{aligned} \begin{array}{rcl} abs(i)&\displaystyle =&\displaystyle C_0\frac{dv_0}{dt}+\frac{v_0}{R},{} \end{array} \end{aligned}$

(2.144)

where:

$\displaystyle \begin{aligned} \begin{array}{rcl} sign(i)=\left\{\begin{array}{cc}+1,&i>0\\ -1,&i<0\end{array}\right.,{} \end{array} \end{aligned}$

(2.145)

whereas abs(i) stands for the absolute value of i, i.e., abs(i) = i if i > 0 and abs(i) = −i if i < 0. The equations in (2.142)–(2.144) constitute the DC-to-DC high-frequency series resonant power electronic converter mathematical model. The study of this class of power electronic converters is continued in Sect. 3.9, Ch. 3. Analyzing (2.142)–(2.144) will explain how this circuit works and why it receives the name high-frequency. For more information on modeling this class of power electronic converter, the reader is referred to [11].

2.6 Summary

The mathematical model of a physical system is a differential equation or a set of differential equations describing the evolution of the important variables of the physical system when this is subject to particular conditions. Such a description is obtained when solving the corresponding differential equations, as in Chap. 3.

In the remainder of this book, the mathematical model of the system to be controlled is employed to design a controller (another differential equation, in general). The objective is that, when connecting the controller in feedback with the system to be controlled, a new differential equation is obtained (the closed-loop system differential equation) whose mathematical properties (studied in Chap. 3) ensure that the variable to be controlled evolves over time as desired.

2.7 Review Questions

1.

What are the fundamental laws of physics that are useful for connecting components in mechanical and electrical systems?
2.

Why is the mathematical model in Example 2.16 given by a single first-order differential equation, whereas the mathematical model in Example 2.7 is given by two second-order differential equations? Notice that there are three bodies in Example 2.16 and two bodies are involved in Example 2.7.
3.

Why is the term corresponding to θ ₄, i.e., the position term, absent in the mathematical model in (2.118)?
4.

Why are first-order differential equations the mathematical models of RL and RC circuits, whereas the mathematical model of a RLC circuit is a second-order differential equation?
5.

It is often said that an inductor (and in general any electric circuit that is long enough) has properties that are analogous to the properties of momentum in mechanical systems. Can you explain what this statement refers to? Have you observed that a spark appears when a highly inductive circuit is disconnected? Do you remember what Newton’s First Law states?
6.

It has been shown that k _e = k _m in a permanent magnet brushed DC motor. On the other hand, it is clear that the electrical subsystem must deliver energy to the mechanical subsystem to render motion possible. Can you give an example of a situation in a DC motor where the mechanical subsystem delivers energy to the electrical subsystem? What is the role of the fact that k _e = k _m in this case?
7.

It has been shown that a gear box is the mechanical analog of an electric transformer. Can you list the relationship between electric currents and voltages at both windings of an electric transformer? Is it possible to establish some similar relationship between velocities and torques on both sides of a gear box? Can you list the advantages, disadvantages, and applications of these features in gear boxes and electric transformers?
8.

According to the previous question, why does an automobile run slower when climbing a hill, but it runs faster on flat surfaces?
9.

If an electric transformer can increase voltage just by choosing a suitable rate n ₁∕n ₂, why are transistor-based electronic circuits employed to amplify signals in communication systems instead of electric transformers?
10.

If the power is the same at both ports in electric transformers and gear boxes, where are the power losses?

2.8 Exercises

1.

Consider the mass-spring-damper system shown in Fig. 2.6a which was studied in Example 2.2. Suppose that this mechanism is rotated clockwise 90^∘. This means that, now, the effect of gravity appears to be on the direction in which both the spring and the damper move, i.e., it can be assumed that the mass hangs from the roof through a spring and a damper. Show that the corresponding mathematical model is given as:

$\displaystyle \begin{aligned} \begin{array}{rcl} \ddot y+\frac{b}{m}\dot y+\frac{K}{m}y =\frac{1}{m}F(t), \end{array} \end{aligned}$

where y = x − x ₀ with x ₀ = mg∕K and g the gravity constant. What does this mean?
2.

Consider the mechanical system in Example 2.17. Assume that an external disturbance torque T _p(t) is applied to the body on the right. Also assume that the body at the left is the rotor of a permanent magnet brushed DC motor, i.e., that the applied torque T(t) is torque generated by this electric motor, as in Example 2.20. Find the mathematical model from the voltage applied at the motor terminals to the position of the body on the right. Compare with the mathematical model in (10.9) and (10.10), chapter 10.
3.

Perform on (2.140), (2.141), the algebraic operations suggested in the last paragraph of Example 2.20.
4.

Perform the algebraic operations referred to in the previous exercise on the mathematical model obtained in Exercise 2 in this section.
5.

Consider the electric circuit shown in Fig. 2.38. Show that voltage v is given as:

$\displaystyle \begin{aligned} \begin{array}{rcl} v=\frac{1}{2}(v_1-v_2). \end{array} \end{aligned}$

Fig. 2.38
Electric circuit of Exercise 5
6.

Consider the electric circuits shown in Fig. 2.39. Show that the mathematical models are given as:

Fig. 2.39
Some electric circuits

For the circuit in Fig. 2.39a:

$\displaystyle \begin{aligned} \begin{array}{rcl} \dot v_0+\frac{R}{L}v_0=\frac{R}{L}v_i. \end{array} \end{aligned}$

For the circuit in Fig. 2.39b:

$\displaystyle \begin{aligned} \begin{array}{rcl} V_0(s)=\frac{R_2}{R_1+R_2}\frac{s+\frac{1}{R_2C}}{s+\frac{1}{(R_1+R_2)C}}V_i(s)+\frac{R_1}{R_1+R_2}\frac{1}{s+\frac{1}{(R_1+R_2)C}}v_c(0),\\ {} \end{array} \end{aligned}$

(2.146)

where v _c(0) is the initial voltage at the capacitor.

For the circuit in Fig. 2.39c:

$\displaystyle \begin{aligned} \begin{array}{rcl} \ddot v_0+\frac{R}{L}\dot v_0+\frac{1}{LC}v_0=\frac{1}{LC}v_i. \end{array} \end{aligned}$

For the circuit in Fig. 2.39d:

$\displaystyle \begin{aligned} \begin{array}{rcl} \ddot v_0+\frac{R}{L}\dot v_0+\frac{1}{LC}v_0=\frac{R}{L}\dot v_i. \end{array} \end{aligned}$
7.

An analog DC voltmeter is basically a permanent magnet brushed DC motor with a spring fixed between the rotor and a point at the stator. This means that such a DC motor is constrained to rotate no more than 180^∘. Based on this description, find the mathematical model of an analog DC voltmeter.
8.

Consider the system shown in Fig. 2.40, which represents a DC-to-DC buck power converter . This converter is also known as step-down converter, because the DC output voltage υ is less than or equal to the supplied voltage E, i.e., υ ≤ E. Show that the mathematical model is given as:

$\displaystyle \begin{aligned} \begin{array}{rcl} L\frac{di}{dt} &\displaystyle =&\displaystyle -\upsilon +u E,\\ C\frac{d\upsilon }{dt} &\displaystyle =&\displaystyle i-\frac{\upsilon }{R}, {} \end{array} \end{aligned}$

(2.147)

where i, υ stand for electric current through the inductor L and voltage at the terminals of capacitor C respectively. Constant E is the voltage of the power supply and R is the load resistance. Finally, u stands for the switch position, which is the control variable, taking only two discrete values: 0 or 1.

Fig. 2.40
Electric diagram of a DC-to-DC buck power converter. (a) Converter construction using a transistor and a diode. (b) Ideal circuit of the converter
9.
A buck power converter DC motor system , is shown in Fig. 2.41. It represents a suitable way of supplying power to a permanent magnet brushed DC motor. Given the following mathematical model for a permanent magnet brushed DC motor:

$\displaystyle \begin{aligned} \begin{array}{rcl} L_{a}\frac{di_{a}}{dt} &\displaystyle =&\displaystyle \upsilon-R_{a}i_{a}-k_{e}\omega,\\ J\frac{d\omega }{dt} &\displaystyle =&\displaystyle k_{m}i_{a}-B\omega,\end{array} \end{aligned}$

show that the complete mathematical model of the system Buck power converter-DC Motor, ideally represented in Fig. 2.41b, is given as:

$\displaystyle \begin{aligned} \begin{array}{rcl} L\frac{di}{dt} &\displaystyle =&\displaystyle -\upsilon+Eu,\\ C\frac{d\upsilon}{dt} &\displaystyle =&\displaystyle i-i_{a}-\frac{\upsilon}{R},\\ L_{a}\frac{di_{a}}{dt} &\displaystyle =&\displaystyle \upsilon-R_{a}i_{a}-k_{e}\omega,\\ J\frac{d\omega }{dt} &\displaystyle =&\displaystyle k_{m}i_{a}-B\omega,\end{array} \end{aligned}$

where:
- i is the electric current through the Buck power converter inductor.
- υ is the converter output voltage, which also is the voltage applied at the motor terminals.
- u stands for the switch position, i.e., the control variable, taking only two discrete values 0 or 1.
- L represents the converter inductance.
- C represents the converter capacitance.
- E is the DC power supply voltage.
- R is the converter load resistance.
- i _a stands for the electric current through the DC motor.
- ω stands for the angular velocity of the DC motor.
- L _a represents the armature inductance of the DC motor.
- R _a represents the armature resistance of the DC motor.
- J is the inertia of the DC motor.
- k _e is the counter-electromotive force constant.
- k _m is the torque constant.
- B stands for the viscous friction coefficient.
Fig. 2.41
Electric diagram of the buck power converter-DC Motor system. (a) Construction of the buck power converter DC motor system using a transistor and a diode. (b) Ideal representation of the buck power converter DC motor system
10.

Consider the series RLC circuit in Example 2.12, whose mathematical model is given in (2.53). The energy stored in the circuit is given as the magnetic energy stored in the inductor plus the electric energy stored in the capacitor, i.e., $E=\frac {1}{2}Li^2+\frac {1}{2C}q^2$ . Show that $\dot E=iv_i-Ri^2$ . What does each term of this expression mean?
11.

Consider the inertia-spring-damper system in Example 2.5 whose mathematical model is given in (2.32). The energy stored in the system is given as the kinetic energy plus the potential energy in the spring, i.e., $E=\frac {1}{2}I\dot \theta ^2+\frac {1}{2}K\theta ^2$ . Show that $\dot E=T\dot \theta -b\dot \theta ^2$ . What does each one of the terms in this expression mean?
12.

Consider the mathematical model in (2.140), (2.141), for the electro-mechanical system in Example 2.20. The energy stored in this system is given as the magnetic energy stored in the armature inductor plus the kinetic energy of the mechanism. Is it possible to proceed as in the two previous examples to find a similar expression for $\frac {dE}{dt}$ ? What does each one of terms appearing in $\frac {dE}{dt}$ mean? Can the reader identify the terms representing the energy exchange between the electrical and the mechanical subsystems?

References

1.

P. E. Wellstead, Physical system modelling, Academic Press, London, 1979.
2.

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