Dynamic Systems: Modeling, Simulation, and Analysis [1 ed.] 9781107179790, 1107179793

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4

Electrical Systems

Contents

4.1 Fundamentals of Electrical Systems 4.2 Concept of Impedance 4.3 Kirchhoffs Laws 4.4 Passive-Circuit Analysis 4.5 State-Space Representations and Block Diagrams 4.6 Passive Filters 4. 7 Active-Circuit Analysis 4.8 Dynamic Responses via Mathematica Chapter Summary References Problems

223 230 242 244 258 271 277 297 306 306

307

Practically every modern engineered dynamic system has electrical components such as motors, sensors, controllers, or, at the very least, power sources. Therefore, an understa nding of the physical processes occurring in the typical electrical circuits, and the ability ~o model behavior of an electrical subsystem are essential for anyone interested in dynaIIllC systems. Similarly to the treatment of mechanical systems, the three keys in modeling of electrical systems are identified as follows: • fundamental principles, which are Kirchhoffs current and voltage laws, as presented in Section 4.3; . 1 . . . . . d . Sections • basic e ements, which mclude resistors, mductors, and capacitors, as descnbe Ill 4.1 and 4.2; and . · • nted JO • two ways o f analysis, which are the loop method and the node method, as prese Section 4.4. 222

4 ·1

Fund amentals of Electrical Systems

223

By a chosen way of analysis (either the loop meth 0 d h lied to the basic elements resul t· • or t e node method), Kirchhoff s laws are app . mg m a math ema t·ica I model for the electrical system • h" h · •, in consideration,• w 1c 1s a •mixture of a set of d"f" . and algebra· 1 1erent1al t. Th. -key modehng concept is illustrat d h ic eq ua ions. is three . e t ro ugho ut this chapter A slightly · . different . way of analysis, based . on the descn·bed three-key model.mg concept 1.s presented m Section 4.5, , representat .ions and block ' . where . . alternative models (st·a te-space diagrams) of the electrical circmts are derived. furthermore, in Sections . 4.6 and 4.7 , Kirchhoff s I·aws an , d den·ved modeIs or basic · elem:nts are used to d~n:e m~thematical models of passive filters as well as active electrical circmts. and · o f eIectnca · I c1rcu1ts · · . glean an ms1ght . .mto their operations• Numeri·ca I ana Iys1s behav10r 1s presented m Section 4.8.

4.1

Fundamentals of Electrical Systems Electrical and electronic components constitute important building blocks of many dynamic systems. The most commonly encountered examples of electrical subsystems include power supplies, controllers, sensors, and motors. Understanding their time-dependent behavior becomes crucial for gaining an ability to model, analyze, design, and control a variety of dynamic systems.

4.1.1

Definitions An electrical system is often called an electrical circuit. A circuit consists of interconnected circuit elements such as, for example, resistors, capacitors, inductors, or sources of energy. There exist two main groups of circuit elements: active and passive. Passive elements, usually called loads, dissipate or temporarily store the energy supplied to the electrical circuit. T hey are represented by resistors, inductors, and capacitors. Active elements are the sources that supply an electrical circuit with energy. Any type of the energy-generating device such as chemical battery, solar/photovoltaic cells assembly, thermocouple junction, or mechanical generator can serve as an active element in the electrical circuit. Conversion of some kind of energy, for example, chemical for a battery, thermal for a thermocouple, or mechanical for a generator, into electrical charge is the common trait of all sources. The active elements are commonly modeled as ideal sources of either current or voltage. An ideal current source always supplies the specified current regardless of the voltage required by the circuit, while an ideal voltage source always provides the specified voltage regardless of the current generated in the circuit. A battery is considered an ideal voltage source even though its performance is influenced by the heat generate? as current flows through the load circuit. If an impact of the produced heat cannot be ignored, the battery can be modeled as an ideal voltage source connected to an internal resistor.

I

l

224

4 Electrical Systems

. 1ctrcm • ·t generally includes one or An electnca . more active elements. The energy prov .cte 1 d by a source is considered an input to the electncal system. . . The electrical circuit the~ry ~ses current and voltage as pnmary vanables to describe the behavior of any electrical ctrcmt. . . . . Ph · II urrent 1·s the flow of electrons inside the ctrcmt. Mathematically it is exp ys1ca y, c . ' ressect as the time rate of change of electrical charge through a specific area: .

dQ dt

l=-

(4.1.J)

where i is the current, t denotes time, and Q is the electrical charge, or number of electrons passing through the cross-section of a wire. Current is measured in amperes (A), and charge is measured in coulombs (C). A negative one coulomb is equivalent to the electrical charge in 18 6.2415 x 10 electrons. Similarly, a positive one coulomb is equivalent to the electrical 18 charge in 6.2415 x 10 protons. It can also be defined as the amount of electrical charge transported in one second by a constant current of one ampere. The charge, expressed in terms of current and time, is

Q(t)

=

I

idt

(4.1.2)

The positive direction of the current flow is defined as opposite to the actual movement of electrons inside the circuit. Hence, it can be visualized as a flow of positively charged particles. Voltage is defined as the work needed to move a unit charge between two points in a circuit: dW dQ

V=-

(4.1.3)

where vis the voltage, Q is the electrical charge, and w denotes the work. One joule of work performed moving one coulomb of charge along the electrical circuit yields the unit voltage called the volt (V). The voltage between any two points in a circuit can be also defined as the difference between electric potentials of these points as shown in Figure 4.1.1, where e and e denote 2 1 electric potentials at the entrance and the exit terminals of a circuit element respectively, i is the current flowing across this element, and vis the voltage computed as

0-

1

L ___ c_ir-cu-it_ ___.~

element

V

(4.1.4)

Figure 4.1.1 Electrical circuit

terminology

4.1

Fundamentals of Electrical Systems

225

The. voltage. at any point of an el ect nca . 1 circuit . i d • electnc potentials of this point and s represente as a difference between . some selected refi · . zero electnc potential is commonly h erence pomt. The ground, or pomt of ' c osen as that r fi · By convention, the element's high-volt . e ~rence pomt. age termmal 1s denoted b th " " · h"l h low-voltage terminal is denoted by th " ,, . Y e + sign, w . 1e t e . e - sign. The sign . of . the voltage difference indi·cat es whet her the energy wa t d d by the c1rcmt element A negat· . s genera e or consume · ive vo1tage differ generation while a positi d"f~ . . ence across an element corresponds to energy ' ve I ierence md1cates energy consumption. To illustrate that concept _ "d h d" con s1 er t e irect current (DC) electrical circuit depicted in Figure 4.1.2. Passive Battery circuit The electrons traveling across this circuit are Vb element repelled by the negative terminal of the battery and attracted by its positive one. Thus, their motion is clockwise, and, by definition, the positive flow of the current becomes counterclockwise (as shown by the arrows). Let vb and Ve be the respective voltage Figure 4.1.2 DC electrical circuit differences across the battery and across the passive circuit element. By definition, vb = b2 - b 1 and Ve = e1 - e2. The current gains energy inside the battery, flowing from the low (b2 ) to high (b1) voltage terminal, so the potential difference across the battery is negative: vb < 0. On the other hand, the current loses energy inside the passive circuit element, which makes Ve = e1 - e2 > 0, and e1 > e2 (as shown in Figure 4.1.2). Customarily Ve is referred to as the voltage drop, and vb - as the voltage gain. According to the law of conservation of energy, in a closed circuit the energy supplied by the battery must be fully consumed by the circuit elements. Hence, vb + Ve = 0. While the voltage drop and voltage gain have different signs as shown above, the opinion on which one is considered positive differs from one literature source to another. In this book, we adopt the convention that assigns a positive sign to voltage gain and a negative sign to voltage drop. In addition to voltage and current, electric circuits are characterized by power. The power applied to a passive element, or generated by an active element, is a product of voltage across that element and current flowing through it: P= vi

(4.1.5)

. · b easi"ly der1·ved using the definition of power as work performed in a This express10n can e unit time: (4.1.6)

226

4

Electrical Systems

. f ·s the watt (W) which corresponds to one joule of work p " ' er1011ned.in Th e urut o power 1 one second. . . l rgy consumed or generated by an element 1s A ccord mg y, ene I

W= JPdt

(4.1.7)

0

4.1.2

Basic Elements A set of symbols, used to indicate active and passive elements on the electric circuit diagrams is shown in Figure 4.1.3, where the symbols are as follows: (a) current source, (b) voltag; source, (c) battery, (d) resistor, (e) inductor, (f) capacitor, (g) ground, and (h) terminals. Understanding the voltage-current relations of individual circuit elements is paramount to the ability to model and analyze the behavior of any circuit.

Resistors A resistor is a passive circuit element that dissipates energy. A simple light bulb is an example of a resistor, which dissipates electrical energy by converting it into light and heat. The behavior of a common resistor is governed by the empirically derived Ohm's law, which states that the electric current flowing through a resistor is directly proportional to the voltage difference across that resistor: .

l

(4.1.8)

V

=R

where i is the current, v denotes voltage, and R is the resistance.

(a)

(b)

(c)

R

L

-JW0v-

____fTfTL

--H-

(e)

(f)

(d)

Figure 4.1.3 Basic elements of electrical circuits

C

_L..

--

(g)

---0 0--

(h)

4.1

Fundamentals of Electrical Systems

227

Ohm's law specifies the linear relation between voltage and current through a resistor, and is often written in the form: V

= iR

(4.l.9)

If a resistor obeys Ohm's aw, it is called linear resistor. The resistance of a linear resistor is a constant, the value of which depends on the resistor's material, geometry, and temperature. For example, the resistance of a nonideal wire can be expressed as: R=pL A

(4.l.10)

where L and A denote length and cross-sectional area of the wire respectively, and p is the temperature-dependent resistivity of the wire material. In this chapter all the resistors are considered linear, and all the connecting wires are assumed to be perfect conductors and their resistance is neglected, i.e. the voltage drop across an ideal wire is zero. The unit of resistance is the ohm (Q). A resistor with R = I Q experiences a voltage drop of 1 V when the current flowing through it equals 1A. The power dissipated by a linear resistor is P

= vi= v2 = FR R

(4.l.ll)

Capacitors A capacitor is a passive circuit element that stores energy in the form of electrical charge. The simplest capacitor circuit can be visud C alized as a pair of parallel plates (a capacitor) connected to a battery as shown in Figure 4.1.4. In this circuit the battery will transport charge from one plate to the other until the Figure 4.1.4 A capacitor voltage difference that results from the charge build-up on the capacitor equals battery voltage vb. A capacitor is characterized by its capacitance - a measure of how much charge can be stored for a given voltage difference across the element. The mathematical expression of capacitance is (4. l.12)

228

4

Electrical Systems

where Q is the stored charge, v denotes voltage, and C is the capacitance. The . capacitance is the farad (F), which corresponds to a one coulomb of charge stored ~It of capacitor experiencing a voltage drop of one volt. Ythe For a parallel-plate capacitor, depicted in Figure 4.1.4, the capacitance depe d . such as: n son its geometry and the matenal cA

k coA

d

d

C=-=--

(4.I.1 3)

where A is the area of the flat parallel metallic plates of the capacitor, k is the relative permittivity of the dielectric material ~e~~een t~e plat~s (k -:::. J for the air). 12 c0 = 8.854xl0- F/ m denotes the free-space perm1tt1v1ty, d 1s the distance between plates and c = c0 k is the permittivity. ' For a capacitor, the relation between a current through it and a voltage difference across it is not linear. According to Eq. (4. l . l 2), the capacitor voltage is

V=gC

(4.l.14)

Substituting the expression for charge, defined by Eq. (4.1.2), we obtain I

If. Qo I J. V = C 1dt = C + C 1dt

(4.1.15)

0

where Qo is a charge on the capacitor at the time t = O. For simplicity of computations, it is assumed that Q0 = 0. Hence, the relation between capacitor voltage and current is expressed as: I

Integral form

v

= ~j;dt

(4.1.16)

0

dv

Derivative form

i= C-

dt

The energy stored in a capacitor is derived by integrating the expression for power (Eq. (4.1.5)) as stated in Eq. (4.1.7). Zero initial conditions are assumed: I

W

Inductors

=

f

Pdt

0

I

=

f 0

I

vidt

=

fv ( c:~)dt = ~Cv (t ) 2

(4.I.1 7)

0

. . . . . tic fieldAn mductor 1s a passive Circmt element that stores energy in the form of a magne into a · I · d · · b f · wound The s1mp est m uctor Circmt can e visualized as a solenoid - a coil o wire . ctor)tight helix, which may or may not be wrapped around a ferromagnetic core (indu anied connected to a source, as shown on Figure 4.1.5. An electric current is always accornP

4.1

Fundamentals of Electrical Systems

229

by a magnetic field - flux - around the conductor. In the case of a solenoid conductor this flux is ____i__. concentrated almost entirely inside the solenoid, while the outside field is weak. The longer the solenoid, the more pronounced is this effect of a nearly uniform flux inside the solenoid and negligible one on the outside. According to Faraday's law (see Section 6.3.1 Figure 4.1.5 An inductor for more detailed information), the relation between flux and the current flowing through the inductor is (4.1.18)

¢=Li

where ¢ is the flux, i denotes current, and L is the inductance. The inductance is a constant that depends on solenoid geometry such as the length 1, cross-sectional area of a coil A, and the number of coils N, as well as on the material of the solenoid and its core, if one is present. The mathematical expression for the induction of the solenoid, shown in Figure 4.1.5, is as follows: µN 2A l

(4. 1.19)

L=--

where µ denotes the magnetic permeability of the solenoid. The unit of inductance is the henry (H). The relation between flux and the voltage is expressed in integral form:

¢=

Jvdt

(4.1.20)

Equations (4.1.18) and (4.1.20) can be used to derive the expression for the voltagecurrent relation of an inductor. I

Integral form

i

±

= Jvdt 0

(4.1.21)

di v=Ldt

Derivative form

The energy stored in an inductor is derived using the Eqs. (4.1.5), (4.1.7), and the derivative form ofEq. (4.1.21). Zero initial conditions are assumed: I

I

I

w = JPdt = Jvidt =Ji 0

0

0

(r

~!)dt = ~Li2(t)

(4.1.22)

230

4 Electrical Systems

Active Elements Active circuit elements include energy sources describe~ in Sectio~ 4.1.1 , and electronic devices such as integrated circuits, which include amphfiers, transistors, and logic gates. While a detailed discussion on integrated circuits is beyond the scope of this book, amplifiers will be presented later in this chapter (see Section 4.7).

4.2

Concept of Impedance By consuming energy, passive circuit elements "resist" the flow of current. This opposition to a current is called impedance, and is represented by a voltage drop across the element. Impedance is often viewed as a generalization of the resistFigure 4.2.1 Impedance ance concept, and as an analog of a transfer function that takes current as an input and yields voltage as an output, as shown in Figure 4.2.1. Impedance is typically expressed in the Laplace domain, and is derived by taking the Laplace transform of the time domain expression of the element's voltage-current relationship,

4.2.1

Definition

Impedance is defined as the ratio of the voltage drop across the circuit element to the current flowing across that element. Traditionally it is expressed in the Laplace domain as:

z = £{ v(t)} = V(s) E{i(t)}

I(s)

(4.2.I)

The relations between current and voltage for the described earlier passive circuit elements are used to derive their impedance expressions. A resistor's impedance is obtained by taking the Laplace transform of Eq. (4.1.9):

E{v} = V(s) =E{i R} = RI(s)

(4.2.2)

Then ZREStSTOR

= V(s) = R I(s)

(4.2,3)

Using the derivative form of Eq · (4 ·I ·16), t he capacitor's · · · d·· impedance is obtame

E{i} = I(s) = ZcAPACITOR

E{c!~} = Cs V(s) I Cs

=-

(4.2.4)

4.2

Concept of Impedance

231

For an inductor the derivative form of the Eq . (4.1. 21) is used:

.G{v}

= V(s)=.& {L~:}

ZtND UCTOR

= l s l (s)

=Ls

(4.2.5)

4.2.2 Combination of Impedances A typical electrical circuit contains interconnected + /(s) multiple passive and active elements. Knowledge of Equivalent the elements' arrangement is important for the corSource impedance rect modeling of the circuit since different ways of of of the connecting the same elements yield differently behavenergy circuit V(s) Zeq (s) ing circuits. /(s) Circuit analysis typically pursues two goals, the first one being the derivation of the voltage-current relation for the circuit as an entity, and the second one being the computation of the voltages across and Figure 4.2.2 Equivalent impedthe currents through every component of the circuit. ance of a circuit The former task is more easily fulfilled when the circuit in question is transformed into an equivalent form, as shown in Figure 4.2.2. Then, the computation of currents and voltage drops for every passive circuit element is straightforward with the use of the previously presented voltage-current relations for resistors, capacitors, and inductors. To find the impedance of a circuit we first need to derive the rules for computing equivalent impedance for two interconnected elements. Obviously, this equivalent impedance depends on the elements' arrangement. The individual impedances can be connected in series or in parallel, as shown in Figure 4.2.3.

.

Series Connection A closed electrical circuit that contains a power source and two loads connected in series, with impedances z, and Z2 respectively, is presented in Figure 4.2.3(a). According to the law of conservation of charge, the currents flowing through impedances Z1 and Z2 are the same: / 1(s) = h(s) = J(s). Also, according to the law of conservation of energy the algebraic sum of voltages in a closed circuit equals zero, i.e. all the energy generated by the source must be dissipated or stored by the loads. As illustrated in Section 4.1.1, the voltage gain across a source and the voltage drop across a load have opposite signs. Assigning a positive sign to voltage gain and a negative sign to voltage drop, the law of conservation of energy

232

4

Electrical Systems

/(s)

V1 (s) , --u'-'--,

-

J1(s)l

/(s)

Z 1(s) '----Q----'

/(s) (b)

(a)

Figure 4.2.3 (a) Series and (b) parallel connections of impedances

for the circuit shown in Figure 4.2.3(a) translates into: V(s) V1 (s) + V2(s) = V(s). Then the equivalent impedance of this circuit is derived as follows:

Z ( ) = V(s) = eq s I(s)

Vi (s) + V2(s) I(s)

= Z ( ) 1

s

+Z

2

( ) s

Vi (s) - V2(s) == O or

(4.2.6)

Consequently, a series connection is characterized by the same current passing through the connected electrical elements. The equivalent impedance fo r n circuit elements connected in series is a sum of individual impedances: n

Zeq(s)

= LZ;(s)

(4.2.7)

i=I

Parallel Connection A closed electrical circuit that contains a power source and two loads connected in parallel, with impedances Z1 and Z2 respectively, is presented in Figure 4.2.3(b). According to the Jaw of conservation of charge the currents flowing through impedances z, and Z2 are: I, (s) + /z(s) = I(s). The voltage drops across the loads z, and Z2 are the sa me·· Vi (s) = e, - e2 = V2(s) = V(s). Then the equivalent impedance is derived as follows:

Zeq(s)

= V(s) = I(s)

V(s) / 1(s) + /z(s)

I1(s) /z(s) Zeq(s) = V(s) + V(s) 1

(4.2,8)

I I = Z 1(s) + Z2 (s)

. 1s . characterized by the same voltage drop ac ross dthe C onsequent Iy, para11 eI connection ·n 1 connected electrical elements. The equivalent impedance for n circuit elements connecte parallel is computed as:

4.2

-

l

Zeq(s)

II

l

i= I

Z;(s)

Concept of Impedance

= I:-

23 3

(4.2.9)

Elements connected either in series or in parallel do not have to be of the same type for the computation of equivalent impedance. The following examples illustrate the use of Eqs. (4.2.7) and (4.2.9).

Example 4.2.1 Deriving equivalent impedances for combinations of circuit elements The connections of several elements and derivations of impedances of these assemblies are shown in Table 4.2.1, where entries (a) and (b) denote series connections, (c) and (d) denote parallel connection, and (e) shows a combination of series and parallel connections.

Table 4.2.1 Equivalent impedances

(a)

R

L

C

Z,q = z,

I Cs

+ Z2 + Z3 = R + Ls + -

(b)

I

I

I

I

I

Z,q(s)

Z 1(s)

Z2 (s)

R 1(s)

Rz(s)

--=-+--=--+-R2 (c)

Z, (s) q

=

R1(s)R 2(s) R, (s) + R2(s)

(d)

Zeq

= Zparallel + Zc RLs

(e)

Zparalle/

z, = q

= R + Ls RLs + R+Ls Cs

= RLCs2 + Ls + R (R+Ls)Cs

234

4

Electrical Systems

Example 4.2.2 Finding the current through a circuit element using equivalent impedances To illustrate how the use of equivalent impedance simplifies circuit an_alysis, consider the problern of finding the current through the inductor L and voltage v1 at termmals A and B for the c·ircu1t. shown in Figure 4.2.4. Solution Derive an expression for equivalent impedance for the elements connected in parallel: 1

1

1

Zp

R

I/Cs

-=-+-Zp

=

R

1 +RCs

According to the law of conservation of energy: V(s) - VL(s) - Vp(s)

=0

where V(s) denotes the voltage supplied by the source, VL (s) is the voltage drop across the inductor, and Vp(s) represents the voltage drop across the impedance Zp. Using the definition of impedance (Eq. (4.2.1 )) and the Laplace transform of the derivative form of the inductor's voltage-current relation, we obtain V(s)

= I(s)Ls + I(s)Zp = I(s) (Ls+

R ) I +RCs

The current through the inductor -I(s) - is then derived as I(s)

=

V(s)

I + RCs RLCs2 +Ls +R

L

A

+ R

C

V1 (s)

/(s)

+B Figure 4.2.4 Circuit for Example 4.2.2

4.2

Concept of Impedance

235

The voltage at terminals A and B equals the voltage drop across the connected m parallel component with impedance Zp and is

Vi (s) = 1(s)Zp = V(s )---R __ 2

RLCs + l s+ R

4.2.3 Divider Rules The concept of impedance proves very useful for the derivation of two basic rules: the voltage divider and current divider. The voltage divider rule is best demonstrated on the series connection of two impedances. Consider a circuit shown in Figure 4.2.5. The currents through impedances z, and Z2 respectively are: 11 (s) = fz(s) = 1(s)

(4.2.IO)

Using the definition of voltage drop as a difference between electric potentials on the element's entrance and exit terminals, we obtain

Vi

11

=

z,

= eA - es

]z

=

V2 Z2

= es -

z,

(4.2.11) ec

Z2

The terminals e 1 and C are located on the wire that is connected to the ground; hence, electric potentials of these terminals equal zero. Terminals e2 and A belong to the same wire; thus, considering the assumption of all wires being ideal, electric potentials of these terminals

/(s)

+

e2

-

l,(s) Source of energy V(s)

l,(s)

e1

! !

A

Z1(s)

B

/(s)

Figure 4.2.5 Voltage divider rule

)

236

4

Electrical Systems

are the same Also assigning a positive sign to voltage gain of the energy source as d · ' escribe in Section 4.2.2, we obtain d (4.2.12)

Then currents through impedances Z 1 and Z2 respectively are /1

V - en en () =--=/z =-=Is Z1

Z2

(4.2.13)

potential of the terminal B is (4.2.14)

and the voltage drops across impedances Z1 and Z2 are derived as Z1 Vi= V - - Z1 +Z2

V2

=

Z2 V--+Z2

(4.2.15)

z,

The voltage divider rule is expressed as:

v,

z,

V2

Z2

(4.2.16)

Hence, the voltage drop across the individual impedance - the member of a series connection - is proportional to the ratio of that impedance and equivalent impedance of the series connection as a whole. For a circuit that contains a single energy source and n impedances connected in series, the voltage drop across the i-th impedance is

V; == V

(4.2.17) Zeq

where V; denotes voltage drop across the i-th impedance, Vis voltage gain of the source, ao 0 and e3 = e 1 < 0. Thus, the current source experiences a voltage drop: e4 > e3, which means that this element absorbs energy and its voltage v ; is negative. For the circuit in Figure 4.3.4(b) the situation is reversed. The voltage source consumes energy, experiencing a voltage drop (e 1 > e2 ), while the current source generates energy, experiencing a voltage gain (e 4 < e3 ). Therefore, vis negative, and v; is positive. Note that Kirchhoffs laws can be expressed either in time or in Laplace domains. The choice of domain depends on the circuit model requirements. Nonetheless, since circuit analysis aims at a finding circuit response, the voltage-current relation expression in the time domain is customarily sought.

V;

(a)

V;

(b)

Figure 4.3.4 Circuits with two energy sources (see text for details)

4.4

Passive-Circuit Analysis . . . . . . le-looP The straightforward apphcat10n of the d1v1der rules and Kirchhoffs laws to a sing tive 1a . I c1rcmts . . y1e . Id s t he vo Itage-current relations for each circuit elemen t with redirect eIectnca ease, as demonstrated earlier in this chapter. For multiple-loop circuits, such a ed to " ·1 to d enve · t he correct model. Hence, several methods were develop approac h may 1ai . alysis 1t an · · J'f h I I · I · · h 1 · rcu s1mp I y t e comp ex mu t1- oop c1rcmts. While the detailed discussion on t e c of thtS methodology belongs to the field of electrical engineering and is beyond the scope book, an introduction to the most widely used approaches is presented.

4.4

Passive-Circuit Analysis

245

The methods explored further in this chapter include: (a) the loop method, (b) the node method, (c) the superposition theorem, (d) Thevenin's theorem, and (e) Norton 's theorem.

4.4.1

Loop Method The loop method of circuit analysis utilizes the concept of loop currents, which are the main parameters of the circuit model. A loop current is a current assigned for a specific loop in the circuit. The loop method encompasses the following steps. First, currents for every loop in the circuit are assigned, using either the marked polarity of the voltage power source, or the current direction indicated by the current power source. For interconnecting elements - elements that simultaneously belong to more than a single loop net currents are used. A net current is an algebraic sum of all the currents passing through the interconnecting element. In evaluation of the element's net current, the direction of the assigned loop current in the examined loop is defined as positive. After the currents passing through every circuit element are defined, the expressions for the voltages are derived using Ohm's law, the known voltage-current relations, and the divider rules. The last step of this method includes the application of KVL to every loop in the circuit. The derived KVL equations comprise the sought system model in terms of loop currents as independent variables. The output equation is also formulated in terms of loop currents.

Example 4.4.1 To illustrate operation of the loop method, consider the circuit shown in Figure 4.4.1. This circuit consists of four passive elements: two resistors R1 and R2, two capacitors C 1 and C , and a single source of energy. The polarity of the voltage source is as shown on the circuit 2 schematic. The system input is the voltage v;n supplied by the voltage source, and the output is the capacitor C2 voltage Vout• Zero initial conditions are assumed. As indicated in the circuit schematic, this circuit consists of two loops, for which the loop currents i, and i are defined as shown. The direction of i1 is clockwise due to the marked polarity 2 of the voltage source. Consequently, the direction of i2 is also clockwise. This circuit has a single interconnecting element: capacitor C1. Its net current is i1 - i2 for loop 1, and i2 - i 1 for loop 2. Applying KVL to loop 1 yields V;n -

VR 1

-

Vc 1

=Q

and to Loop 2: Ve,

+ VRz + Vc = 0 2

. h It entrelationsforresistor(Eq.(4.1.9))andcapacitor(Eq.(4.1.16)),the R eca11 mg t e vo age-curr KVL equations become

)

246

4

Electrical Systems

R2

R1

Z) --

C2

C1

l

Loop 2

Loop 1

Vout

Figure 4.4.1 Modeling a passive circuit using the loop method I

Loop l

v;n(t) = R1i1

+;

1J(i1 - i2)dt

(I)

0

I

Loop 2

R2i2

+;

1

I

J(i2 - i1)dt + ; 0

2

Ji2dt = 0

(2)

0

The output equation is derived as /

V0

u1(t)

=;

2

Ji2dt

(3)

0

These three equations constitute the required circuit model. If desired, this model can be transformed into a single differential equation that directly expresses the relation between the input and output voltages. The easiest way to achieve such a transformation is by taking the Laplace transform of Eqs. (I), (2), and (3), performing the necessary algebraic operations, and taking the inverse Laplace transform of the resulting equa· tion. This derivation is as follows (capital letters denote variables in Laplace domain): (4)

(5) (6)

From Eq. (6): (7)

4.4

Passive-Circuit Analysis

24 7

Then, substituting this result into Eq. (5), we obtain the expression for / 1:

11

=

C,s(R2

+ C~s + C~s)C2sV0111

(8)

Finally, substituting Eqs. (7) and (8) into Eq. (4), we derive the sought model:

V;11

=

(R1

+ Cl

) (R2

tS

1 )c1C2s2V0111 - - -C2sV0111 2S C1s

+ Cl + Cl JS

(9)

= ( R1R2C1 C2s2 + (R1 C1 + R2C2 + R1 C2)s +I) V0u, Taking the inverse Laplace transform of Eq. (9), the system model is obtained as a differential equation in terms of input and output voltages: (10)

4.4.2

Node Method The node method of circuit analysis operates with the concept of node voltages, which are the main parameters of the circuit model. A node is a point in an electric circuit at which connecting wires intersect or branch. The node voltage is an electric potential assigned to a node, and an independent variable in the equations that constitute the circuit model. The node method operates on the circuit as a whole, without subdividing it into loops, and encompasses the following steps. First, the nodes of interest and their electric potentials - node voltages - are defined. Then, the current for each circuit element is assigned, using the indicated polarity of the voltage source or marked current direction of the current source. The expressions for the current through each circuit element are written in terms of node voltages, using the known voltage-current relations, Ohm's law, and the divider rules. After all the expressions for component currents are derived, KCL is applied to each node. These KCL equations comprise the sought system model in terms of node voltages as independent variables. The output equation is also formulated in terms of node voltages.

Example 4.4.2 To illustrate the operation of the node method, consider the circuit used in the Example 4.4.1. As shown in Figure 4.4.2, three nodes are defined, with assigned node voltages e0 , e 1, and e2 respectively, and the element currents are as indicatedon the circuit schematic.

)

248

4

Electrical Systems

. (this is the Wire . po ten tial of any point on a bottom wire "' • is zero Remembering that the electnc d . the known voltage-current re1at1ons 1or the resisto rand It es are derived as· d connected to the ground) an usmg · ag vo e no of rms t · capacitor the element currents m e v;,, - e1

eo - e1

iR1

=

-

R1

e1 - Voul

e1 - e2

iR2 =

R2

-

R2 d

ic1

= C1 dt ( e1

ic2

=

-

d C2 dt (e2 -

R1

0)

= C1

0) =

dei dt de2

C2 dt

=

dvout

C2

R2

#1

dt

#2

ic1

l

C2

C1

Vout

l

Figure 4.4.2 Modeling a passive circuit using the node method

Applying KCL to nodes 1 and 2 we obtain Node 1 iR 1

-

ic 1 - iR2

Node 2 iR2

-

ic 2

=0

=0

Substituting the derived element current expressions into KCL equations, we obtain the required system model: e1 - Vout C de1 V;11 - e1 --- - 1-----=0 R2 dt R1 e1 - Vout _ Ci dvout = O dt R2

· · h two variables: e and Vour• Smee · · l equations . a system of two d·r·c Vout is an output, wit 1 1erentrn It 1s 1 . gle no separate output equation is needed. stn If desired, the intermediate variable e 1 can be eliminated, thus reducing the system to~ the . d ependent variable v • Expressing e in terms of Vour, u5ing . lem . o f a smg . 1equation f"" . d1 1erentia 0111 1 second equation of the above system model yields e1

dvout Voll/ = R2 C2 -;Jr+

Substituting this result into the first equation of the derived system model, we obtain

4.4

Vin _

R1

+ (-1R1 + _lR2 ) (R2C2 dvout dt

Vo ut

) _

Passive-Circuit Analysis

dvout) d2v C(R2C2dt2 + dt 0111

I

V 0 ut _ + R2 -

249

O

Algebraic transformations result in the following system model : 2

dV 0111 d v 0111 (R C R IR 2C I C2-d2 I 1+R2C2+ R1C2)--+v0111 = v;,, + ~ t As expected, this model is identical to that derived in the Example 4.4.1 (see Eq. (10)).

4.4.3 Circuits with Multiple Energy Sources Straightforward applications of the loop method and the node method to the analysis of a circuit with multiple energy sources may prove difficult. The Superposition theorem is one of the approaches that allow efficient dealing with the multiple energy sources within a circuit. Thevenin's theorem and Norton's theorem constitute the other popular approaches to the simplification of such a circuit. The usability of these three methods is particularly apparent for the equivalent transformation of a circuit that is the combination of resistors and sources.

Superposition Theorem Superposition theorem states that the current through any passive element of a circuit with multiple energy sources equals the algebraic sum of the currents produced by each individual source independently. To evaluate the current resulting from the i-th source, all the other sources are "turned off." This means replacing the voltage sources by short circuits (zero voltage), and the current sources, by open circuits (zero current). Then, either of the described earlier methods for circuit analysis can be applied.

Example 4.4.3 Consider the circuit shown in Figure 4.4.3(a). It contains two energy sources - a current source I(s) and a voltage source V(s). This problem can be solved either in the time domain or in the Laplace domain. Here, the Laplace domain solution is provided, which is easily inverted into the time domain when required. We want to evaluate the total current passing through the resistor R 1 • According to the superposition theorem, this current (I,) is a sum of two current components - one due to the current source ([11 ), and the other due to the voltage source (/vi): I 1 = Iv1

+ In

To derive the current Iv, resulting from the voltage source, the current source is turned off as shown in Figure 4.4.3(b). Since the right loop of the circuit depicted in Figure 4.4.3(b) is open, there is no current there. Thus, only the left loop consisting of the voltage source and two resistors R, and R3 participates in current generation. The resistors R1 and R3 are connected in series, so

250

4

Electrical Systems

currents flowing through these elements are t~e same: lvi =_lv3 . Iv. The current lvi is then derived using Ohm's law and the equivalent resistance of the c1rcmt:

V

fv1 = - - = R1 Rveq

V

+ R3

To derive the current In resulting from the current source, the voltage source is turned off as · · h shown in Figure 4.4.3(c). Here both loops of the c1rcmt ave a current. According to the law of conservation of charge, the current flowing through the resist w~ d . equal~ the current sup~li~d by the energy source. The resistors R1 an R3 ~re connected in parallel so, usmg the current d1v1der law (Eq. (4.2.22)), the current across the resistor R 1 is R3 In= I R1 +R3 t the currents In and/v1 have opposite directions. Assuming the clockwise directio no Note. that . b e positive, the total current across the resistor R 1 is

(a)

1, (s) 3

(b)

(c) F. . .. igure 4 .4.3 Modeling a passive circuit usin th g e superposition theorem (see text for details)

Thevenin's and Norton's Theorems

. h wo . wit 1 ill • f t' combm any · resistors and sources energy o a ion . termmals can be replaced b (Thevell · · Y an equ1vale n t circmt of a single voltage source Vr Thevenin's theorem states that

4.4

Passive-Circuit Analysis

251

voltage) and a single series resistor Rr (Thevenin resistance). The Thevenin voltage is the open-circuit voltage at the terminals. The Thevenin resistance is the resistance of the open circuit measured at the tenninals. Its value is obtained by evaluating the equivalent resistance of this circuit, where voltage sources are replaced by short circuits, a nd current sources are replaced by open circuits.

Norton's theorem states that any combination of energy sources and resistors with two terminals can be replaced by an equivalent circuit of a single current source IN (Norton current) and a single resistor RN (Norton resistance) in parallel with the current source. The Norton resistance is defined in the same way as the Thevenin resistance. The Norton current is the current that would have existed in the wire connecting the terminals of the open circuit. Its value is obtained by Ohm's law, dividing the open-circuit voltage at the terminals by the Norton resistance. If the chosen terminals for the Norton and Thevenin transfonnations are the same, then RN = Rr and IN= Vr/ Rr · . An illustration of the equivalent transformation of an electrical circuit using Thevenm's and Norton's theorems is presented in the example below.

Example 4.4.4 Consider the two-loop circuit shown in Figure 4.4.4(a). Thevenin's and Norton's equivalents of the right-hand side of this circuit at the terminals A and Bare constructed as shown in Figure 4.4.4(b) and Figure 4.4.4(c) respectively. As stated earlier, Rr =RN.To find its value consider the open circuit shown surrounded with a dashed line in Figure 4.4.4(a). Replacing the voltage source Vi with a short circuit, it can be easily

A

A---- 1

- -- --- - - - I I I I I I I I I I I I I I

a -- --- - -- -(a)

A- ---- - 1

I I RT I

I

I I I I I

I

+ +

I I I I I I I I I

B _ __ _ J Thevenin's equivalence (b)

RN

IN

I

I I I I I I

a -- - - -- - Norton's equivalence (c)

Figure 4.4.4 Modeling a passive circuit using Thevenin's and Norton's theorems (see text for details)

252

4

Electrical Systems

seen that resistors R 1 and R 3 are connected in parallel, and resistor R2 is in series . With th . e1r equivalence. Thus: R1R3

Rr = RN = R2

+ Ri + R 3

The Thevenin voltage is the open-circuit voltage at the terminals A and B. This voltage equ I · fl h · as on the Theven· the voltage drop across the resistor R 1• The resistor R2 as no m uence in Volta 'd . . . . . this voltage cons1 er the loop consist' Ofthge smce there 1s no current passmg through 1t. To find Ing ' . . voltage source Vi and resistors R 1 and R3 • In this loop, the resistors are connected in series, so the e voltage divider rule is used (Eq. (4.2.15)):

Then, the Norton current is IN

R1 R1R2 + R1R3

= -Vr = V i - - - - - - - Rr

+ R2R3

The polarity of the Thevenin's voltage source and the direction of the current provided by the Norton's current source are defined to maintain the direction of current in the transformed equivalent circuit as the same as it was before the transformat ion.

Example 4.4.5 This example encompasse s the application of the circuit analysis methods described in this section to a simple two-loop electrical circuit. The consistency of the derived results is demonstrated. Consider the two-loop circuit shown in Figure 4.4.5. We need to derive a circuit model and find the current and the voltage drop for the resistor R3,if the numerical values of circuit elements are as follows: R1 VJ

= s n, = 12 V,

Figure 4.4.5 Circuit for Example 4.4.5

= 3 n, V2 = 8 V

R2

R3

= 2 n,

4.4 Passive-Circuit Analysis

253

5 lution by the Loop Method T~e loops are defined and the loop currents are assigned in accordance with the indicated polarity 0f the voltage sources as shown in Figure 4.4.6. Resistor R2 is the interconnecting element and its net current is the same for both loops: iRi === ; 1 + i 2• Recalling the voltage-current relation of a resistor (Eq. (4.1.9)), apply KVL to loop I and loop 2: Loop I Loop2

i1R1 - (i1 + i2)R2 = 0 V2 - i2R3 - (i1+ i2)R2 = 0 V1 -

This system of two equations with two independent variables - ; 1 and ;2 - constitutes the system model. Since the only current that passes through R3 is ;2 , we find the required current and voltage by solving the above system of equations for i2 and then, using Ohm's law, we derive VRJ : .

_ . _ -R2v1 + (R1 + R2) v2 _ O lz - - - --'-----'-- - .9032 A R1R2 + R1R3 + R2R3

lR3 -

VR3

= iR3R) = 1.8064 V

Considering that system inputs are voltages v1 and v2 provided by the voltage sources, and the outputs are the current and the voltage drop for the resistor R3 , the system model can be written in terms of input-output as follows : .

System model

lR3 -

Output equations

-R2v1 + (R1 + R2)v2 R1R2 + R1R3 + R2R3

{iR3 , VR3

= iR3R3}

Solution Using Thevenin's Theorem This solution uses an equivalent transformation of the circuit. We define the terminals A and Bas shown in Figure 4.4.7(a), and substitute the left-hand side of the circuit shown encircled in a dashed rectangle with its Thevenin equivalent.

Loop 1

Figure 4.4.6 Solution by the loop method

Loop 2

254

4 Electrical Systems

- - - ---- ---

1

I I I I I I I

[ ____ ____ _ 8 Thevenin's equivalence (b)

(a)

Figure 4.4.7 Solution by Thevenin's theorem (see text for details)

The Thevenin resistance is computed as equivalent resistance for two resistors R, and R 2 connected in parallel: R1R2

Rr= --R, +R2 The Thevenin voltage equals the voltage drop across the resistor R2 and is derived using the voltage divider rule (Eq. (4.2.15)): R2 Vr = v1 R, +R2

The equivalent single-loop circuit is shown in Figure 4.4. 7(b). Since the resistors Rr and R 3 are connected in series, the current passing through them is the same and can be derived by direct application of Ohm's law. Considering the direction of this current to be as shown on the circuit schematic, we can state that the voltage source v2 generates energy, i.e. experiences a voltage gain, while the Thevenin voltage source Vr is absorbing energy, i.e. experiences a voltage drop. Thus, v2 is positive, and Vr is negative. Then, the system model is v2 -

Vr

= i(Rr +R3)

or in extended form: V2 -

) v, R, R2+ R2 = ·(R,R1R2 + R2 + R3 l

Then, the required current and voltage for the resistor R are found as: 3 R2 V2-V 1-. _ ._ R1 + R2 _ (R, + R2)v2 - R2v1 1Ri - l R1R2 - -=- ---- -- = 0.9032 A + R3 R1R2 + R1R3 + R2R3 R, + R2 %

= iR3R3 = 1.8064 V

Considering the specified inputs and outputs of the system, the system model can be defined in terms of input- output as follows:

4.4 Passive-Circuit Analysis

(R1 + R2) v2 - R2v1 R1R2 + R1R3 + R2R3

.

System model

255

IR - -----'----3 -

Output equations

{iR3 ,

= iR3 R3}

IIR,

Solution Using Norton's Theorem This solution also uses an equivalent transformation of the circuit. We define the terminals A and Bas shown in Figure 4.4.8(a), and substitute the part of the circuit shown encircled with a dashed line with its Norton equivalent. The equivalent circuit is shown in Figure 4.4.8(b). Even though the terminals chosen for the Norton transformation are not the same as those chosen for the Thevenin transformation in the previous solution, the Norton resistance is the same as the Thevenin resistance: R1R2

RN =

R1 +R2

To derive the Norton current, the voltage at the terminals A and B needs to be evaluated first. from the circuit schematic in Figure 4.4.8(a) it is clearly seen that this voltage is a difference between the voltage gain of the source v2 and the voltage drop across the resistor R2 , which is found using the voltage divider rule:

VAB

=

V2 - VR2

=

V2 - V1 R

R2

I +R2

Hence, Norton current is derived as .

VAB

IN

= RN =

V2 - V1

R1

(

+ R2 _

R1

+ R2 ) V2 -

R 2v1

R1R2

R1R2 R1 +R2

The resulting Norton-equivalence circuit has a single current source and two resistors connected in parallel. Then, the current through the resistor R3 is found using current divider rule:

- ---------,

I I

I+ I

B

,----,

A

B

I I I I I_ - - - - - I

+

R3

V2

I I I I :

- I I

- -- - - --- --- - -- - -- - -- - '

I I t--------. I I I ,...___

___,

I

I _ _ ___ __ _ _ _ _ _ I

(a)

Figure 4.4.8 Solution by Norton's theorem (see text for details)

Norton's equivalence (b)

Electrical Systems

4

256

. lR3

=

R1R2 + R2 _ (R1 + R2)v2 - R2v1 Ri (R 1 + R2)v2 - R2v1 RN . R - R1R2 + R1R3 + R2R3 R1R2 R1R2 lN RN+ R3 ---+ 3 R1 +R2

= 0.9032 A

and the voltage drop is computed using Ohm's law: VR3

= iR3R33 =

1.8064 V

Considering the specified inputs and outputs of the system, the system model can be defined in terms of input-output as follows: .

System model

lR3

=

(R1 + R2)v2 - R2v1 RiR2 + RiR 3 + R2 R3

Output equations Solution Using the Superposition Theorem

Since the Superposition theorem deals with currents through circuit elements, the system model will be derived in the form iR3

= i13 + i23

where iR3 is the current through the resistor R3, i 13 is the current through this resistor due to the voltage source v1, and i23 is the current through this resistor due to the voltage source v2 . Note that as discussed earlier in this section, the sum of the component currents is algebraic, which means that the signs of the component currents i 13 and i23 are assigned according to their directions. The directions of currents i13 and i23 are defined as shown in Figure 4.4.9. To find the current i13, the voltage source v2 is replaced by a short circuit as shown in Figure 4.4.9(a). Then, the equivalent resistance of this circuit is

(a)

Figure 4.4.9 Solution by the Superposition theorem

(b)

4.4

Passive-Circuit Analysis

257

and the circuit current is: .

l v1

.

Since resistors

V1

v1(R2+R3) R1R2 + R1R 3 + R2R3

= - - -- - = - -- -- - -

Ri +-R_2R_3_ R2 + R3

R2R3 . 1ence---are R1 and the eqmva in series the same current passes through them. ' R2 + R3

Then, i13 is found using the current divider rule:

v1R2 R1R2 + R1R3 + R2R3 Similarly, i23 is found as the circuit current for the circuit shown in Figure 4.4.9(b): . 113

--= l. v 1R2 R2 + R3

v2(R1 + R2) v2 ---'-----'-R1R2 + R1R3 + R2R3 RJ + R1R2 R1 +R2 The currents i13 and i23 have opposite directions, so they must have different signs in the algebraic sum that represents the current through the resistor R3 • Assume that ii3 is positive. Then, the required current and voltage for the resistor R3 are found as: .

123

.

. = = lv2

.

---=----c~ -

.

(R1 + R2)v2 - R2v1 = 0.9032 A RR RR R I 2 + I 3 + 2R3

IR3

= 123 - 113 =

VR)

= iR3R3 = 1.8064 V

The computed result for iR3 is positive, which means that the positive current component was chosen correctly. A negative result would have indicated that the original assumption of the direction of current through R3 is incorrect, and the current actually flows in the opposite direction. The derived expressions for current and voltage for the resistor R3 constitute the system model in terms of input and output: system model output equations

{iR3 ,

VR3

= iR3 R3}

As expected, the results derived by all the methods employed in this example are identical.

Electrical Systems

258

4

4.5

State-Space Representations and Block Diagrams

4.5.1

Block Diagrams As discussed in Section 2. 7.2, block diagrams and transfer functions are intimate) y related h . 'b' " d block diagram is an extremely good tool 1or escn mg t e mput- output relations for·A . 1 W . . 1 . . . variety of dynamic systems, and electnca ctrcmts m particu ar. hlle a circuit schemara . ic depicts the physical connections between the system components , a block diagram dea1s W1th variables being transformed and propagated through t he system. As such, it is a useful . between th e e1ements of a d v1sua1. . . g o f the re1at1ons . . at better understandm aimed medmm Ynam1c system. The generic block diagram is shown in Figure 4.5.1. Transfer Output Customarily , all the elements of a block diagram are in the Input function Laplace domain. The presented block diagram may refer to the dynamic Figure 4.5.1 Generic system as a whole, in which case it can be viewed as a diagramblock diagram matic extension of the system transfer function model. It may also refer to an individual system element. Then the transferfunction block describes the element's dynamics, while the input and output are the intermediate variables of the system. Consider the simple single-loop electrical circuit shown in Figure 4.5.2. Let the input be the voltage v;n, supplied by the voltage source, and the output be the voltage drop Vaur on the capacitor C. The construction of a block diagram for this circuit encompasses the following steps: (a) deriving the system model equations in time domain, (b) taking the Laplace transform of the Figure 4.5.2 Simple system equations considering zero initial conditions, (c) derivRC circuit their constructing ing the component transfer functions and ) graphical representatio n - component block diagrams, as required to comprehensively express the interactions between circuit elements, ao GRc(s)

. /GRc(Jw)/

=

RCs + 1 Cs

1

= RCs + I

(4,6.4)

I 'Vou1(icv)I V;,,(jw) =JI+ C2R2w2

and its cutoff frequency is 1

Wo

= Re·

To illustrate attenuation of signals with frequency greater than wo generate Bode plots using the Bode Plot Mathematica function (see Figure 4.6.2). For both filters, the resistance was the same: R = 1 k.Q, and the values of the inductance and capacitance varied as shown in Figure 4.6.2. As derived in Chapter 7, the amplitude of the frequency response of a dynamic system subject to a periodic input with magnitude A that equals A/G(jw)/ (see Eqs. (7.3.5) and (7.3.6)). The gain (/G(jw)I) equals I for all input frequencies less than or equal to the cutoff frequency w0 , which means that these inputs will be passed through the filter without a decrease in amplitude. For signal frequencies greater than w0 the amplitude of the response . will be decreasing rapidly, approaching zero - this signifies signal attenuation. The cutoff frequency for both filters increases with the decrease of inductance or capacith tance: the lowest wo were registered for the highest inductance L = 200 mH and for e highest capacitance C = 200 nF. 6 Mathematica code that generates the Bode plot for RL filter is shown in Figure 4. •3.

4.6.2 High-Pass Filter . . the purpose of which is to pass signals wit. h a frequencY . circmt, . passive . h-pass fil ter 1s ff Ah1g a cuto than lower with signals attenuate higher than the designed cutoff frequency, and frequencies. Two types of this filter are shown in Figure 4.6.4.

4.6 Passive Filters

1.0

·•,

I\

I\ I\ I

o.aL

I I I

~o.J

(!)

- - L= 0.02x10-3

ai

- ----

L=

········

L=200x1o- 3

"O

2 ·c

o.4 -

273

\I

l1 1

I

I I I

I

2x10- 3

I

I I

I I

I

ci 111

I I I I I

~

I I

o.2 L

I

I 0 .l'-::'=-:------_.____.,,

\

1""''"'---...,_LI,w.L,"t.."-1, ,~c.o::J1.0tL __·_-.._j~·:·. j_

J..!.,

0.001

1

1000

;.;ar···>;:,.,~.--.,.,--

C.002 106

109

C.003

Frequency, rad/sec (a)

-'

ai

"O ::::,

:g

0.4

\

....

I I

0.8

3 -~ ~o.6 (!)

I-. .. I\ I I - I \ii, I I ,, ~I I I II I. I .. ..J I I I I ..... -I I . I ··..J. I

- -

l

I I I

L.,I.

- - C=Of 11rg ----- C= 2x10- 9 ........ C= 200x10- 9

-

,-

C)

ct!

~

0.2

0 0.001

...+. .- -

t ►1I

l

I I I I I I I I I I I I I I I

r

~~ 10 0

1

1

I I

+

I : : ;;~~~

C.002

I

I

_J

I

I I

I I I

\ \

\

\~•••

T ---...~' ' "' C.003

I

9

lO

Frequency, rad/sec (b)

Figure 4.6.2 Bode plots (gain only) for low-pass passive filters: (a) RL filter, (b) RC filter

Using the same reasoning as in Section 4.6.1 , we can derive the transfer functions and generate Bode plots for these filters:

.

IGRL(;w) I =

1Va111Uw) I

LRw

Vin(Jw) = J I + L2R2w2

(4.6.5)

274

4 Electrical Systems

[~ ' s];

ode l l * s + r tfm = Tra nsf erF unc tio nH 10- 3 } '' 1800 ' l ➔ 0,82• 3 rul e! = { r ➔ 1800 l ➔ 2 * 18 . ,.,}; . 3 }, ' { . rul e2 = r ➔ { 0.0 01 , l·01e} , Pl otLayout ➔ "M ' 1., agn 1tude11 * 280 ➔ l f I rul e3} 800 , { , ' . rul e3= r ➔ l ➔ Automati c, tfm I. rul e2 , t mlac k Do tte d} }, Gridlines {B e!, ' Bo deP lot [{t fm /, rul k {Black, Dashed}' l k 12] ' , ac {Bl , ➔ Plo tSt yle tiv e(" La be l" , B ac . Bla ck, 16 l, ec" d/s Frame Ti cks Sty le ➔ D1 rec uency ' ra , L be l ➔ { Sty le ("Freq . ) I" Bla ck, 16 ]}, Frame a ' ( JW Sty le[ "M agn itu de, I G "L= 200 x1 0·J 11 1 14]} , ix1 0· 311 , 14 ], Sty le[ "L= le[ St J ➔le [" L = 0. 02x10_3 11 , 14 , Y Plo tle gen ds "Ab o lut e"} J · Pl d [ {Sty s {A uto ma tic , ➔ s ion ace nct Fu ing al Se {L eft , Center}

J'

n plot for the RL filter ~or generating the Bode gai d . e co tica Figure 4.6.3 Mathema

R C

J --

R

8

vout

l

--

(a)

L

vout

l

(b)

er ies RC filter, (b) series RL tilt

filters: (a) ser figure 4.6.4 Passive high-pass

. /GRc(;w)/

=

Cw !Vou,(Jw)I C2R2w2 + J1 = ) V;n(Jw

(4.6.6)

4.6.5. ass filters are shown in Figure h-p hig se the for ts plo de Bo e s. The sa~e Th clearly seen in these graph is s nal sig y enc qu -fre low e or capa~~ The attenuation of ses with decreasing inductanc rea inc w y 1 enc qu fre f 0 tof cu tendency is found: the ce L = 200 mH and for tan uc ind st he hig the for registered tance; and the lowest cvo was nF. 200 highest capacitance C =

4.6.3 Band-Pass Filter

uencY . s of the [req d pner nal sig ss . pa . to is . ich wh . r an u r· CUit, the purpose of !owe A band-pass filter 1s a passive Cir the are wo2 d an Wo1 ere b · wit· h'm the specific limits Wo1 .$ cv .$ Wo2, wh 6 emg it as shown in Figure 4·6· · cu cir C RL an by ed ent res cutoff frequencies. It is rep

,,,, 4.6 Passive Filters

1.0 -

l

(1)01 I rTfTl'I

I

/1 !I

o.sl

(.') 1--

::::: 0.6 ·

.a

0.4 -

········

I

/ ; / ; / / /

. I I I I I I

3

3

----- L= 2x10"

-0

'i:

L= 0.02x10·

(J)OJ

/1

:

3

ai

(J)02

J.•··············t-,--- --+-- --

1

L= 2oox10· 3

Cl

ro

~

0.2 -

I / ),/

I

I

I

I

I

I I I I I I I I

\ ....•·· o~ r~~====:::r::::::::::r::=:~~:=::;:~-= --±:::::::::=::::i_ ~ .L...,.., 0.001

1

1000

106

109

Frequency, rad/sec

(a)

0.8

3

g

O.G - - C=0.02x1o- 9 ----- C= 2x10· 9

Q)

u:::,

:g

0.4

•······· C= 2oox10-9

Cl

ro

~

0.2

(b)

Figure 4.6.5 Bode plots (gain only) for high-pass passive filters: (a) RL filter, (b) RC filter

275

276

4

Electrical Systems

L

C

Figure 4.6.6 Band-pass passive filter

The K VL and impedance method is used to derive the transfer function of th• .

.

IS CITCUJt:

V (s) =Ve+ VL m

1 LCs2 + RCs + l + VR =-I +Ls/ +RI= 1 ----0 G

RCs V01a(s) = RI => GRLc(s) = LCs2 + RCs + 1

. IVou1(icv) I JGRLc(Jcv) j = Vin(Jcv) =

cv(R/ L) (4.6.7)

The maximum value of the transfer function is JGRL(jco)j

= 1. The cutoff frequencies are

defined as those at which JGRL(jco)[ =~-Solving this equation, we obtain the values of cutoff frequencies:

C001

R = --+ 2L

R coo2 = 2L +

2

R) ( 2L R) ( 2L

1

+ LC

(4.6.8)

2

1

+ LC

The bandwidth is the difference between the cutoff frequencies:

coo1 - coo2

R

=1

(4.6.9)

2 Using the following numeric values for the system parameters: R = l kQ, C === nf, L = 2 µH , the Bode plot shown in Figure 4.6. 7 is generated. aBz, 6 This band-pass filter will pass the signals with frequencies between 0.16 MHz ao'• fr . . . . . True, fr . . ..St:y\e .. Coray. fl'".,..Tid, s ,,. ( {.Aut~ti c; , No~ ) , f AuT~tic, NoM l ) • Fr. . .Tic~Sty\e ... o ; ,..ecti ve t " ~abe\ " , ttac k, UJ, 1-.ceSi~• .. '"• Aspec tR•t:io .. l I 1 . s, Fr--Labe\-. (St'y\e["Ti ae , s e c;oncss ", 11.ac;k, 16 ) , Sty \ e ( '" v-,. , Vo l t: s ", tlac;k, 1 6 )) , Plot"'--cench ... P\,-~ C( St:y\e ( " e, ", 14 ] , Style [ "11.,.,, •, 14 J) , {RiJht:, Ce11ter ) 1 1

(e) Figure •.a.J Dynamic resJ)On ~ of the weuit Lo et>n~uant .u1.d periodic input vo lt:tgc:s· ( i:1 } ' 0 pc,rt>< to 11 •1t'P i11put.. wo ; wo 1s a cutoff frequenc y in radian 8 Per second. . signals (w < w0) while or·r1enng blocks low-frequency 2. High-pass filter - a circuit that . . easy passage to high-frequency signals (ro > wo); wo 1s a cutoff frequency in radians persecond h' · I · h · · · 3. Band-pass filter - a circuit t at passes s1gna s wit m a specified frequen a dcy range ro < w < w02 while blocking signals with out-of-range frequencies·, OJ01 n (JJ 01 02 are . the cutoff frequencies for the desired bandwidth. 4. Band-stop (notch) filter - a circuit that blocks signals within a specified frequenc and OJ yrange ro 01 < w < roo2 while blocking signals with out-of-range frequencies·' w01 02arethe . . . . . frequency notch called 1s ion attenuat maximum of frequency es; cutoff frequenci Both low- and high-pass filters may experience a resonance effect at the cutoff fre ue q ncy, depending on the implementation of the filter. 4.56 The circuits are shown in Figure P4.56. L

C

L

~

~

(a)

Figure P4.56

The numerical data are as follows: (a) L = 5 µH , C = 3 nF, R = l ldl' (b) L = 2 rnH, C = 0.5 nF, R = 0.5'k.Q.

(b)

R

Problems

.57 The circuits are shown in Figure P4.57. 4

r

I

1

~i (a)

C

r

L

(c)

(d)

(e)

Figure P4.57

The numerical data are as follows: (a) C=3nF, R= 1 kn; (b) L = 5 µH, R = 1 kn, R1 = 30 n , R2 = 10 n, R3 = 20 n , R4 = 15 n; (c) L = 7 µH, C = 1.5 nF, R = 1 kn; (d) C1 = 1.5 nF, C2 = 1.0 nF, R1 = 0.5 kn, R2 = 1 kn; (e) L = 5 µH , C = 3 nF, R = 1 kn.

4.58 The circuits are shown in Figure P4.58.

The numerical data are as follows: (a) L = 5 µH, C = 3 nF, R = 20 n; (b) C1 = 1 nF, C2 = 3 nF, R1 = 100 n, R2 = 1 kn; (c) L1 = 20 µH, L2 = 3 µH, R1 = 30 n, R2 = 1 kn; (d) L1=50mH, L2 =50mH, C=0.lµF, R=lkn.

323

324

4 Electrical Systems

V1,

L

C

v..,

l

~

R2

Ci

(b)

(a)

Li

Li

Li

(d)

(c)

Figure P4.58

Section 4.7 Active-Circuit Analysis

4.59 For every active circuit shown in Figure P4.59, find the transfer function G(s) = V®I V and derive the time-domain relationship between the input (v;n) and output (v0;,j

voltages. L C

(b)

(a) C

R

L

v.,.,,

(c)

Figure P4.59

(d)

p Problems

325

4.60 for the differential amplifier circuit shown in Figure P4.60, derive the time-domain relationship between the input (v;ni , v;ni) and output (v0 11, ) voltages. How will this relationship change when R1 = R2 and R3 = R4?

+

Figure P4.60

4.61 Derive the expression for the current io through the load for the Howland current source - an active circuit, shown in Figure P4.6l. Derive the balanced bridge condition a relationship between the resistances R1 , R2 , R3 , and R4 for io to be independent of the voltage drop VL across the load.

!

io

Rz Load

Figure P4.61

4.62 The proportional- integral-derivative (PIO) control - one of the commonly used linear controls - can be implemented as an active circuit with inverting op-amps, as shown in Figure 4.62.

326

4 Electrical Systems

shown control - and . Ident,·ry each op-amp circuit -. the component . . of the . PID control law as a time-domam relat10nsh1p between the input (v1n) and derive I~

~..

output

~

~

Figure P4.62

4.63 For the active circuit shown in Figure P4.63, derive the time-domain relationship between the input (v1n1 , V;n2 , V;nJ , Vin4) and output (vou1) voltages. What does this circuit do? How will the relationship between the circuit input and output voltages change when R, = R2 = R3 = R4 = Rj? +

R1

R1

V;n1

V;n2

+ R2

V;.i

v••,

+ Ri

V;,,.

-=-

+ R•

Figure P4.63

.s • elationshh1Pe . . . Figure . . shown m • C!Tcmts 4.64 For the active P4.64 derive the t1me-domam r th oft es ' b~tw~en the input (v;ni , v;.2, v;n3) and output (vou,) voltages. To show t~at bo p _64(a), 4

c1rcmts are averagers, assume that, for the circuit in Figure R, = R2 = R3 = R, while for the circuit in Figure P4.64(b), R1 = R2 == R3== ~R• express1ons. · ,s theorem for deriving the reqmred · Thevenm . · cons1'der usmg H mt.

Problems

+

+

327

+

v,., v,.2 v,.)

-w

-UJ

(b)

(a)

Figure P4.64

4.65 The active circuits shown in Figure P4.65 represent different filters. For each of these circuits, find the filter transfer function G(s) = Vout, generate a Bode plot IG(jw) I vs.

Vin

frequency (use either Wolfram Mathematica or MATLAB to obtain the required plots), and indicate which filter type (low-pass, high-pass, band-pass, or band-stop) is represented by the given circuit.

+

(b)

(a)

(c)

Figure P4.65

328

4

Electrical Systems

The numerical (a) C = 100 µF , (b) C = 100 µF , (c) C = 120 µF ,

data are as follows: R1 = IO .0, R2 = 9 .0, R3 = I .0; R1 = 10 n, R2 = 9 n, R3 = I n; R1 = 2 .0, R2 = 9 .0, R3 = 10 .0, R4

= l n.

4.66 A non-inverting op-amp can be used as an isolation amplifier in cert . . . ain 1rn 1 mentations of the first-order low-pass and high-pass filters, with and . Pe. · · sh own m · F.1gure P4.66 derive thWithout amplification. For each of the c1rcmts V. e~ transfer function G(s) =~.generate a Bode plot IG(Jco )I vs. frequency (use . h ~

ffl~

Wolfram Mathematica or MATLAB to. obt.ain the required plots), and indicate which filter type is represented by the shown circmts. For both circuits the numerical data are as follows: C = 3 nF, R = I k.Q .

+ Vin

1-

.re

CT

V ;n

VOUI

R

(a)

(b)

Figure P4.66

4.67 A non-inverting op-amp circuit with dual capacitors is often used to obtain second·

order low-pass and high-pass filters, with and without amplification. For each of the Vout . . sh own m . F 1gure . p 4.67 denve . the filter transfer function G(s) = -v, c1rcmts . , generatea tn

Bode plot IG(Jco)I vs. frequency (use either Wolfram Mathematica or MATLAB to obtain the required plots), and indicate which filter type is represented by the shown circuits . The numerical data are as follows: (a) C1 (b) C1

= C2 = 2 µF , = 0.5 µF , C2

R1 = 20 n , R2 = 5 n , R3 = 100 n, R4 = 100 Q ; - 10 n. = 1.5 µF , R1 = 20 n, R2 = I n, R3 = 100 n, R4 -

Problems

329

c. vow +

(a)

(b)

Figure P4.67

4.68 A band-pass filter can be implemented with either inverting or non-inverting op-amps, as shown in Figure P4.68. For each of the circuits shown in Figure P4.68 derive the filter transfer function G(s)

= Vow, and generate a Bode plot IG(jro)I vs. frequency (use either Vin

Wolfram Mathematica or MATLAB to obtain the required plots) to demonstrate the bandwidths passed by these filters. Which one of them passed the wider bandwidth? Numerical data are as follows: (a) C1 = 1.5 µF, C2 = 1.2 µF, R1 = 20 n, R2 = 1 n, R3 = 35 n, R4 = 2 n; (b) C1 = 1.5 µF, C2 = 1.3 µF , R1 = 0.4 n, R2 = 0.75 n.

\ 330

4 Electrical Systems

(a)

(b)

Figure P4.68

the time-domain relationship 4.69 For the active circuit shown in Figure P4.69, derive t does this circuit do? between the input (V;nJ, v;n2) and output (v0111 ) voltages. Wha t voltages change when How will the relationship between the circuit input and outpu R1=R2=R3?

+

Figure P4.69

+

Problems

331

4,70 The inverting op-amp active circuit, shown in Figure P4.70, represents a second-order low-pass filter. Derive the filter transfer function G(s) = V0111 , and generate a Bode plot

V;,, IG(Jw)I vs. frequency (use either Wolfram Mathematica or MATLAB to obta in the

required plots), considering the following numerical values:

C1

= 15 µF ,

C2 = 10 µF , R1 = 2.5 n , R2 = 0.75 n , R3 = In.

Will this circuit pass an input signal with a frequency of 0.318 MHz? How about an input signal with a frequency of 0.159 kHz?

Figure P4. 70

4.71 One of the possible implementations of a band-stop filter (notch filter) is shown in

Figure P4.71. Derive the filter transfer function G(s)

= ~;_u', and generate a Bode plot Ym

IG(Jw)I vs. frequency (use either Wolfram Mathematica or MATLAB to obtain the required plots), considering the following numerical values: C1

= 1.5 µF,

C2 = 100 nF, R1 = 7.5 n, R2

= 0.5 n,

R3

= 6 n,

R4

= 7.5 n,

Which frequency (in Hz) will be mostly attenuated by this circuit?

R

Figure P4. 71

R

R

= 15 n.

332

4

Electrical Systems

Section 4.8 Dynamic Responses via Mathematica where system input . 4.72 Consider the passive circuit shown in Figure .P4.72, is the . h h h , d outputs are the current zo t roug t e resistor R I c1n V in , and the required the Voltage Voltage drop v0 across the inductor L2.

R1

!io

Vo

L2

1----_____....._--------Vl Figure P4.72

Derive the governing equations for this circuit using the loop method and plot the dynamic response of this system to a step input, considering the following numerical values for the system parameters and input voltage:

L1

= 2 .5 H,

Li

= 1.2 H,

R1

= 15 il, Ri = 25 il,

v;n

=

IOu(t) V , simulation time 2.5 s.

Assume zero initial conditions. Hint: this system has a closed-form solution for loop currents; hence, the analytical differential equation solver DSol ve of Mathematica is applicable. 4. 73 Consider the passive circuit described in Problem 4.24. Derive the governing equations

for this circuit using the node method. Assuming zero initial conditions, plot the dynamic response of this system to: (a) a step input V;n = 5u(t) V , (b) a ramp input V;n = 5tu(t) V, . I considering the following numerical values for the system parameters and mpu voltage:

Li = 2 .5 H , Li= 1.2 H , R = 15 n , C = 0 .65 F , simulation time 115 s. tant but . . ' (c) For the case of the step mput, consider that capacitance is no longer cons instead is dependent on time as follows: c = 3(t + l)e- 1/ 1i . How does the introd~ction of the time-dependent capacitance change the character of system respons~-ut (d) For the case of the step input, consider that capacitance is no longer con5lant1, 5. as fol. owce. · · d . d epend ent on time . and the voltage drop on the capacitance 1s mstea C = (vc + e)e- ,f •i , where e = 10-6 • How does the introduction of this capacitan change the character of the system response?

p Problems

333

If the largest current the capacitor can safely handle is IO A, will the circuits for case (c) and case (d) be safe to operate? Explain your conclusion using the simulation results. Hint: for this system, the numerical differential equation solver NDSol ve of Mathematica is applicable.

4.74 Consider the passive circuit described in Problem 4.44. Derive the state-space model of this system, considering zero initial conditions and the following numerical values of the system parameters: R1 = 5 n, R2 = 1.5 n, L = 2 H, C1 = 5 mF, C2 = 2 mF. Plot the dynamic response of this system to the following inputs: (a) unit step input, simulation duration of 5 s; (b) sinusoidal input 2 sin(t), simulation duration of 25 s; { () for < 2 1 1 (c) rectangular pulse input of a duration of 2 sec., described as is(t) = ~ fort ; 2 ·

Hint: for this system use the Mathematica functions StateSpaceModel and OutputResponse.

4.75 Consider the circuit described in Problem 4.57(a). Plot the dynamic response of this · circuit to the sinusoidal input v;n(t) = sin(IOt). If the input frequency is increased to I 000 rad/sec, would the periodic output still be expected? What if the input frequency is increased to 109 rad/sec?

'i

Thermal and Fluid Systems

5

Contents

Fund amen tal Principles of Therm al Systems Basic Therm al Elements Dyna mic Modeling of Therm al Systems Fund amen tal Principles of Fluid Systems Basic Elements of Liquid-Level Systems Dyna mic Modeling of Liquid-Level Systems Fluid Inerta nce The Bernoulli Equa tion Pneu matic Systems link 5.10 Dyna mic Response via MAT LAB and Simu Chap ter Summ ary Refer ences Prob lems

5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9

335 339 346 353 356 363 368 370 374 378 384 384 384

eerin g appli catio ns, including Ther mal and fluid systems are found in a varie ty of engin indus trial processes. Thenna~ mach ines, devices, automobiles, equip ment , buildings, and fer, and fluid d:namics, e;~e and fluid syste ms are in the areas of therm odyn amics , heat trans there is a vast literature. . 11 of whic h is the subject of a comp lete text, and for whic h press ure, and flow, are spaua prope rties of therm al and fluid systems, such as temp eratu re, " n · Becau · at 10 . seare · usual ly arise in their chara ctenz · natur e, and non1·mean·ttes . "buted m d 1stn al diffe rentia l equatio~_s1 ear this, distri buted -para mete r models gover ned by nonli near parti tical solut ions for non :arequi red for an in-de pth study of these systems. Howe ver, analy le cases . Ther efore , approxi parti al differ entia l equat ions are only limited to a few simp rnall . ling or durin g solut ion. in as . tions have to be made eithe r in mode . bles chan ging . ar or Very often , a therm al or flmd system opera tes with its varia · bles, a hne , d"tions · point . By using incre ment al vana "tic opera tmg f d h hb . 1 spec1 a o oo ne1g or con · syste m can be devel oped . Also , unde r certaI·D · d mod e I o f t h e dynam ic · nze Imea

:r

334

5.1

Fundamental Principles of Thermal Systems

335

perties of thermal and fluid components can be lumped at a point, resulting in tbe pr~-parameter models. This lumping process is similar to that in the modeling of :::nical systems by using lumped elements, such as point masses, s~rings, ~nd dam~ers. . lumped-parameter models are governed by linear ordinary differential equations, Linear . h . hi b can be systematically solved by well-developed analytical and numerical tee mques. ~d~d, linear lumped-parameter m~dels have been used in many thermal and fluid problems and satisfactory results can be obtamed. . In this chapter, we shall cover the basic concepts and important aspects of physics arding thermal and fluid systems. Due to the limitations of the mathematical tools used Rg · I in this undergraduate-level textbook, linear lumped-parameter models are mam y covere d for certain problems. The chapter is divided into three parts: • modeling of thermal systems (Sections 5.1 to 5.3) • modeling of fluid systems (Sections 5.4 to 5.9) • numerical simulations of system responses by MATLAB and Simulink (Section 5.10) As in dealing with mechanical systems, three keys are adopted in the modeling of thermal and fluid systems: • fundamental principles • basic elements • ways of analysis which shall be detailed in the subsequent sections. Here, for the first time, free-body diagrams as a way of analysis for thermal and fluid systems are introduced. The concept of the threekey modeling technique is useful for general thermal and fluid systems, regardless of the level of complexity of the model to be developed. Once the governing differential equations of a lumped-parameter model are derived, they can be converted to different forms of model representation for analysis and solution, such as transfer function formulations, block diagrams, and state-space representations. Transfer function formulations and state-space representations have been introduced to mechanical systems; see Sections 3.2 and 3.7. For details on block diagrams and the use ofSimulink, refer to Sections 6.2 and 6.6. If these sections have not been covered in class or self-learning, the reader may skip the relevant contents in this chapter and revisit them later after mastering these topics. In this text, the International System of Units (SI) is used. Refer to Tables Al and A4 in Appendix A for commonly used quantities for thermal and systems, and Table AS for unit conversion between the SI and the US customary system.

S.1 Fundamental Principles of Thermal Systems A thermal system is one in which thermal energy is transferred and stored. Examples of thermal systems include heaters, air conditioners, refrigerators, ovens, and cooling systems

336

5 Thermal and Fluid Systems

for automobiles and buildings. Thermal energy is also known as heat Thermal energy transfer takes place due to temperature differences, as hea~ ~ heat energy object (location) of the higher temperature to an object (location) of the low ows frollJ a· er tem n The storage of heat energy in an object depends on the substance and the tern Perature Peratu · object. re of the Accordingly, the processes of heat transfer (thermal energy transfer) and h (thermal energy storage) are described by the following two important variable/at st0rage T - temperature, in kelvins (K) or degrees in Celsius (°C), with °C = K _ 273. 15 qh - heat flow rate, in joules per second (J/s) or watts (W), with 1 W = 11/s

Here, temperature is a measure of the potential (heat energy) of an object and h flow describes the heat energy transfer between two objects or locations. These eat variables are analogous to electric potential (e) and current (i) in an electric circuit·two , see Section 1.4.

5.1.1

Three Ways of Heat Transfer Heat energy can be transferred in three ways: conduction, convection, and radiation, which are illustrated in Figure 5.1.1.

Conduction Heat transfer by conduction occurs when two objects (locations) of different temperatures are in physical contact, which leads to diffusion of heat through a substance (either solid or fluid). By Fourier's Jaw of heat conduction, steady-state heat conduction in a onedimensional slab or a flat plate can be described as follows: qh

dT

= -k.A -dx

(5.J.l)

. . l where qh 1s the heat flow rate; k is the thermal conductivity of the matena con tained infl thew body, in W/(m · K) or W/ (m·°C); A is the cross-sectional area normal to the heat h:at th direction; and x is a spatial coordinate with its positive direction in that of edrop fl ow. The negative · sign · m · the previous · · md1cates · · · a temperature gha equation that there 1s in the direction of heat flow. For a lumped-parameter model of heat conduction throu slab, as shown in Figure 5.1.l(a), Eq. (5.1.1) is approximated by liT kA qh ;::;; - kA - = - (T1 - T2 ) & L

(5.J,2)

. . . assumed to where L 1s the thickness of the slab in the direction of the heat flow; T1 > T2 is in line with the direction of qh.

be

Fundamental Principles of Thermal Systems

5.1

qh

qh

-+

Solid

r,

,,, ,,, ,,,

I·

L (a)

337

~ Fluid

flow

qh

• r, Surface area A

•I

(b)

(c)

Figure 5.1.1 Heat transfer in three ways: (a) conduction, T1 > Ti; (b) convection, Ts> Too; and (c) radiation, T, > T2

Convection Heat transfer by convection involves the flow of thermal energy through the movement of a fluid (liquid or gas). Heat convection occurs within a fluid where warmer areas of the fluid rise to cooler areas of the fluid, as seen in water boiling in a pot. Heat convection also takes place when a moving fluid is in touch with the surface of a solid body whose temperature is different from that of the fluid, as seen in a heat exchanger or a fan blowing cool air through an object. For a fluid flowing over the surface of a solid body, as shown in Figure 5.1 .1(b), the heat flow rate is given by Newton's law of cooling, (5.1.3) where his a convective coefficient, with units of W/(m 2 · K) and W/(m 2 -°C); A is the surface area involved in heat transfer; Ts is the temperature of the solid surface; and T1 is the temperature of the fluid at some distance away from the surface.

Radiation Heat transfer by radiation occurs through electromagnetic waves. Unlike conduction and convection, radiation transfers heat from one body to another without any physical contact in between. In other words, radiation can take place in a vacuum. Examples of radiation include heating by solar radiation and heating by heat lamps. According to the StefanBoltzmann law, the heat flow rate by radiation between two separated bodies of temperatures T1 and T2 , as shown in Figure 5.1.l(c), is given by (5.1.4) 98 where ais the Stefan- Boltzmann constant with the value a = 5.6704 x 10- kg/ (s3 . K4 ) , FE is effective emissivity, FA is a shape factor, and A is the surface area of heat transfer.

338

5 Thermal and Fluid Systems As can be seen from the previous discussion, conduction problems and sorne · R a d'iatlon · · problems can be described by linear equations. pro blems in gener convec hon 1 · mak . d'ffi l · · and so]uf a have the nonlinearity as shown in Eq. (5.1.4), which es it i cu t m modelmg . . . . d chapter, we shall mainly consider lmear or lmeanze pro blems of heat transfer. ion. In th·is

5.1.2 Heat Energy For an object of mass m and specific heat c, its heat energy Eis given by

E = me (T- To)

(5.1.5)

where To is an arbitrarily chosen reference temperature. It is assumed that the temperature throughout the object is the same, without spatial distribution. (Refer to Section 5.2.3 for the justification of this uniform temperature assumption.) Note that the absolute value of Eis not very meaningful. It is the change in heat energy that affects the dynamics of a thermal system, which is similar to the potential energy of a mass m under gravity, such as mg(h - h0). Specific heat, which is also known as specific heat capacity, is the amount of heat per unit mass required to raise the temperature of an object by one unit. In the SI, specific heat has units of J/(kg·K) or J/(kg·°C). The value of the specific heat for an object depends on the thermodynam ic process involved. Specific heat is usually measured either at constant volume in an isochoric process, in which c is denoted by cv, or at constant pressure in an isobaric process, in which c is denoted by cp. For an incompressible fluid or solid whose volume is unchanged, c = cv. For a constant-pressure process, as seen in the boiling of water to steam or in the freezing of water to ice, c = cp. Either way, Eq. (5. 1.5) is valid for the calculation of heat energy.

5.1.3 Conservation of Energy All thermal systems operate under the law of conservation of energy. For a closed th~nna~

system with a well-defined boundary, the first law of thermodynam ics, which is a version° the law of conservation of energy, applies l:::.U= Q- W

(5.1,6)

h w er~ t::.U denotes the change in the energy of the system, Q represents the he~t energY s rhe supphed to the system, and W is the work done by the system on its surrounding · change in the energy is expressed by (5.t ,7) t::.U

= M+t::.ITME

· tbe

15 h M w ere . .is t h e change in the heat energy (also called the internal energy), and h,Oi11£ . l energY)· change m the mechanical energy of the system (a sum of kinetic energy and potenua

p 5.2

Basic Thermal Elements

339

Consider a thermal . system with negligible cha nge in mechanical energy (!iITME = 0). Assume that no work is done by the system on its surroundings ( W = O). The hea t energy supplied to the system during time interval tit can be written as

Ii£ = [q;,, (t) - q0111(t)]!it

(5. 1.8)

where q;,, is the rate of heat flow into the system and q0111 is the rate of heat flow out of the system. Thus, the law of conservation of energy for the system reads

dE

= q;,, -

~

qout

(5.1.9)

q0111) dt

(5. l.10)

or in integral form I

E

= J(q;,, to

where to is an initial time.

5.2

Basic Thermal Elements A lumped-parameter model of thermal systems has two types of basic elements: thermal capacitance and thermal resistance. As shown in Section 1.4, these thermal elements are analogous to capacitors and resistors in an electric circuit or springs and dampers in a mechanical system. With the law of conservation of energy, the use of thermal capacitance and thermal resistance leads to the establishment of dynamic models for thermal systems (Section 5.3).

5.2.1 Thermal Capacitance The thermal capacitance of an object is the ability of the material of the object to store heat energy. The thermal capacitance C of an object is defined as the rate of change of the heat energy stored in the object with respect to the object's temperature; that is,

dE

C= dT

(5.2.l)

where Eis the stored heat energy as described by Eq. (5.1.5), and Tis the temperature of the object. The SI units of thermal capacitance are J/K or J/0 C. If the specific heat c of an object is not a function of temperature, the thermal capacitance of the object by Eqs. (5.1.5) and (5.2.1) is given by

C = mc=pVc

(5.2.2)

340

5 Thermal and Fluid Systems

T, C

qout

Q

Figure 5.2.1 A body with input and output heat flow rates

where p and V are the density and volume of the mass m. Consider a body n of temperatu re T and thermal capacitan ce C. Assume a neg1·. • • • IgJble change in the mechanica l energy m the body and zero work by the body to its surroundings. By Eq. (5.2.1),

dE _ dEdT _ CdT dt - dT dt dt

It follows from the law of conservati on of energy as presented in Eq. (5.l.9) that the governing equation of the body is obtained as follows:

dT

C dt

= qin -

qout

(5.2.3)

Here, q;n - q 0 ur is the net heat flow rate into the body, as shown in Figure 5.2.1 . So, with ~ven input and output heat flow rates, the time-depe ndent temperatu re of the body can be determine d by solving the differentia l equation (5.2.3). Comparis on of Eq. (5.2.3) and Eq. (4.1.16) shows that the temperatu re T of a the~al capacitanc e is analogous to the voltage v (potential drop) of an electrical capacitorf Furthermo re, because Eq. (5.2.3) is a first-order differentia l equation, the temperatures 0 . . . . d lopmeol the thermal capacitanc es can be convemen tly selected as state variables m the eve of a state-space model of thermal systems, as shown in Section 5.3.

5.2.2

Thermal Resistance 1·1 is a · · about a flow of Therma1 resistance 1s heat or heat transfer through a bodY, and J111~ . . 1~~ property of the material of the body that resists heat transfer. As a resu t d pio . resistance, the temperatu re drops as heat flows through a body. This temperature ro c~ . . an elec10 ath_erma1-resistance element plays a similar role as a potential drop (voltage) 10 resistor. In other words, thermal resistance is analogous to electrical resistance. The thermal resistance in heat transfer is defined as

R= dT dqh

5.2 Basic Thermal Elements

341

which in the SI has units of K/W (or K. s/J) and °C/W (or °C . s/J). As presented in Section 5.1.1, there are three ways of heat transfer: conduction, convection, and radiation. For a linear relationship between the temperature and heat flow rate, as seen in conduction models and some convection models, Newton's law of cooling applies: (5.2.5) where the thermal resistance is a constant for any given temperature, and !:lT is the temperature difference.

If the relationship between the temperature and heat flow rate is nonlinear, as seen in radiation, the corresponding thermal resistance can be determined by a linearized model. To this end, we assume that the temperature varies in a small neighborhood of a specific temperature r•. By the Taylor series, (5.2.6) with 1 R. = - - - -

(dqh/dT)r=r.

(5.2.7)

Thermal resistances in the three forms of heat transfer are obtained in sequel. For heat conduction, as shown in Figure 5.1.l(a), Eqs. (5.1.2) and (5.2.4) give R=.!::_

kA

(5.2.8)

For heat convection between a fluid and a solid body, as shown in Figure 5.1.l(b), Eqs. (5.1.3) and (5.2.4) yield (5.2.9) For heat transfer by radiation, as shown in Figure 5.1.l(c), the relationship (5.1.4) between the heat flow rate and the temperatures is nonlinear. Assume that the temperature T2 of the radiated body or position is constant. Consider a small variation of the temperature of the radiating body; that is, T1 = T. + !:lT. With the linearized model given by Eq. (5.2.6) and through the use of Eqs. (5.1.4) and (5.2.7), the thermal resistance is obtained as follows:

1 4a FEFA AT!

R. = - - - -

(5.2.10)

Comparison of Eq. (4.1.8) and Eq. (5.2.5) reveals that the heat flow rate qh in a thermal system is analogous to the current i in an electrical system. Thus, in developing lumpedparameter models, combinations of several thermal resistances can be replaced by a single

342

Thermal and Fluid Systems

5

equivalent thermal resistance, like combination s of resistors in electrical sy,, rn thermal resistances connected in series, as shown in Figure 5.2.2(a), the equal on:· For t\Vo qh

l

= Ri (Ti - Tc),

q1,

1

= Re

(Tc - T2)

where Tc is the temperature at the interface between the thermal resistances , can be replaced by (5.2. i ))

where the equivalent thermal resistance is given by Req =R1 +R2

(5.2.12)

For two thermal resistances in parallel combination , as shown m Figure 5·2·2(b), the equations

can be reduced to Eq. (5.2.11), with the equivalent thermal resistance given by 1

1

I

(5.2.13)

+-=R1 R2 Req

For complicated thermal systems, series combination s and parallel combinations can be further combined, as in the modeling of electrical systems. The series combination of thermal resistances can be used to model heat conduction through a multilayer composite slab. The parallel combination of thermal resistances can be used to treat heat conduction through a thermal system consisting of multiple parts, as shown in the following example.

R2 (a) .

.

.

(b)

(b) parallel cot11 . . comb.mat1on; Figure 5.2.2 Combmat10ns of thermal res,·stance s.. (a ) senes

. binattOll

5.2

Basic Thermal Elements

343

Example 5.2.1

Figure 5.2.3 shows a cl~sed and hollow cy_lindrical vessel, with thickness o, length L, and outer diameter D. The vessel is made of a matenal with thermal conductivity k . Let the temperatures inside and outside the vessel be T; and To, respectively. Determine the equivalent thermal resistance of the entire vessel. Thickness o

L

Figure 5.2.3 Heat conduction in a closed cylindrical vessel

Solution The vessel has three parts: the side of the cylinder and the two circular ends. Denote the thermal resistance of the cylinder side by Re, Denote the thermal resistance of an end by Re , The equivalent thermal resistance of the vessel is a parallel combination of three resistances, as shown in Figure 5.2.4. Therefore, the equivalent thermal resistance of the entire vessel is Req =

1 )1 1 ( Re + Re + Re

1

ReRe

= 2Re + Re

(a)

with the resistances Re and Re yet to be determined.

Figure 5.2.4 Parallel combination of thermal resistances for the vessel

There are two cases of wall thickness of the vessel. For a vessel of thin wall (o/ D < < 1 and 0IL P. Derive a model of the · a_ po Iytrop1c valve s-filling process in the cy1·m d er 1s tic system in terms of the cylmder pressure P. ga pneurna f . . . " a hqu1d-level . ·1ar tot h at 1or . sim1 sotutioD system. From the free- body diagram o lotion procedure 1s . . equa tio ns o f f t he valve shown in Figure 5.9.3, the governing I ·1· so The Poto and the aux11ary the cylinder the pneumatic elements are as follows : V

dP

(a)

1

( b)

Cylinder: nRgTdt = qm Valve:

qm

= R(Ps - P)

where Eqs. (5.9.6) and (5.9.7) have been used. It follows from Eqs. (a) and (b) that the governing differential equation of the pneumatic system is

R

P,V

~

Valve Rigid cylinder

Air-supply tank Figure 5.9.2 Pneumatic system for Example 5.9.1

~

p

R

[>arnic Whil e the abov e proce dure of sel_ecting sta~e varia bles ap~li es 8 bles as parameter., l hat d Ysteni, it is useful to think abou t the physical mean mg of state varia discu ssed in Chap tr•; 3 escribe the syste m's energy. For example, for mech anica l syste ms ' such ) d l ·t (l"mear or angu lar) f h . state varia bles are displacement (linear or angular an ve oci Y dispJaee. arisin g due to Jin,. _· ore tor · ment is used to describe the poten tial energy of the syste m, siona1 . . . . 1 1 ). The velocity descr ibes the kmet lc energy Jf th e systern sprin gs (V = -kx2 or v = -kre2 2 2 ' 1 ·2 1 2 . calelem tlcctri or F ). 1a0 T= 2 arisin g due to the moti on of inerti a elements (T = 2mx or only for the elen,e nts that store ents, discussed in Chap ter 4, the state varia bles are assig ned . h are both used to describe h energ y. Thes e state variables are curre nt and volta ge, whic te 1 1 2 electrical energ y (E = 2 cJ or E = 2Li ). differ entia l equations, state For comp ound dyna mic systems that are descr ibed by a set of vecto r will be a union of the varia bles are assigned per subsystem, and the resul ting state -spac e representation of a subsy stem s' state vectors. The proce dure of deriv ing a state 6.2.3. The formulation of such comp ound system is similar to that illust rated by Exam ple in this chapter. state- space repre senta tions will be discussed in more detai l later

6.2.2

Transfer Function Formulations ion of a time-invariant linear As intro duce d in Chap ter 2 (see Section 2.7.1), the trans fer funct inpu t in the Laplace domain, system is defined as the ratio of the system outp ut to syste m assum ing zero initial cond itions (Eq. (2.7.1)). Cons ider a dyna mic system described by a gove rning equa tion dn- lx(t) dnx (t) a n ~ + an- I dtn-l dmu(t) _ - bm dim

dx(t)

+ · ••+ a1 ---;fr + aox(t )

dm- lu(t) + bm-1 dim - 1

-

du(t)

+ ... + b1 -;ft + bou(t)

(6.2- II)

output. . . . · . d h , and x(t) 1s the system w ere a, an bi are know n coefficients, u(t) is the syste m mput of Eq. (6.2.11 ): Assuming zero initial conditions, take the Lapla ce trans form

+ an- lSn- 1 + ... + a1s + ao)X (s) = = (bmsm + bm- 1Sm- l + .. . + b 1s + bo) U(s)

(ansn

Then the soug ht system trans fer function is G(s)

= X(s ) = U(s)

bmsm + bm- JSm-l + ... +bis + bo ansn + an_ 1sn- l + ... + a 1s + ao

. . . . . are ion funct fer trans this of poles the 2.4.2, on ecall~ng the d1scuss1on m Secti the solutions of the equat ion R

(6.2,12)

(6.2,1 3J

das defiJle

6.2 System Modeling Techniques

anSn + an- 1Sn- 1+·· •+ a1 s+ ao = O

409

(6.2.14)

ation and its zeros are the solutions of the equ 0 bms"' + bm- 1s"'- I + ·· ·+ bis + bo =

(6.2. l 5)

n has n poles and m zero s. Obviously, the derived transfer functio insight into system tran sfer function gives a good Knowledge of the poles and zeros of the er of a tran sfer widely used in control design . The ord system behavior; these parameters are (6.2.13) is of ce, the tran sfer function described by Eq. hen es; pol its of ber num the is n ctio fun n-th order. a zero-pole-gain form such as A transfer function can be expressed in - Pn ) G(s) = X(s) = k (s - zi)(s - z2 ) •. . (s (s - pi) (s - P2 ) · · · (s - zm ) U(s)

(6.2.16)

is the gain. A zeros respectively, and k = bm and es pol n ctio fun r sfe tran an the are zi wberep; and design . ely used in system analysis and control nsiv exte also is tion ula form ain ro-g pole-ze put for the ends on the choice of input and out dep form n ctio fun sfer tran the e Sinc dynamic system, it is not unique. ained from system, its transfer function can be obt For any linear time-invariant dynamic cast in matrix form: the system state-space representation 1 (6.2.17) G(s) = C( sl- Ar B+ D defined as shown by Eq. (6.2.4), where the matrices A, B, C, and Da re trix A. matrix of the same size as the state ma

and I is the identity

Example 6.2.4

functions into and zeros, and cast the obtained transfer Derive transfer functions, find their poles ing equations: tems described by the following govern sys ic am dyn the for form in o-ga -zer pole (a) x(t) +ax(t) = u(t) (b) i(t) +a1.i-(t) + aoX(t)

= u(t)

j

+ bou(t) (c) i(t) +a1x(t) + a0 x(r)dr = b1u(t) 0

Solution d by taking the x(t), the transfer functions ar~ obtain~ put out and u(t) ut inp em syst ows: the [;r owed by the algebraic operations as foll foll ns, atio equ ing ern gov the of m Place transfor (a)

-l:[i:(t) +ax(t)]

=E[u(t)]

410

6 Combined Systems and System Modeling Techniques

(s + a)X(s)

= U(s)

X(s)

G(s)

1

= U(s) = s + a

This is the first-order transfer function (n = 1, m = 0), with a single pole Pt It is already in the pole-zero-gain form (an = 1, bm = I). (b) .8[.x(t) + at.i(t)

=- a

.

i

no zeros.

+ aoX(t)] = .8[u(t)]

(s2 + a 1s + ao)X(s) = U(s) 1 X(s) ----G(s)- U(s) - s 2 + ats + ao

This is the second-order transfer function (n = 2 , m = 0), with two poles Pt and P'J. that are the roots of the equation s2 + ats + a0 = Oand no zeros. The pole-zero-gam form of transfer function is 1

G(s) (c)

.8

r(,) +

a,x( t) + a, ix(

(s2 + a 1 + :

0

= (s -

P1 )(s - pz)

r)dr] - .8 [b, u(t) + bou(t)l

)x(s) = (bis+ bo)U(s) 2

G(s) = X(s) = bis+ bo = b 1s + b0s 3 U(s) s2 + ai + ao s + a1s + ao s

This is the third-orde r transfer function (n = 3, m = 2), with three polesp 1 , p 2 , andp3 that are the roots of the equation s 3 + a 1s + ao = 0 and two zeros z 1 = -b0 /b 1 and z2 = 0. Note that any transfer function must be a ratio of two polynomials. Hence, the fonn G(s) =

bis+ bo s + a1 + ao/s 2

is not a proper transfer function since its denomina tor is not a polynomial. Multiplyi~~ t~ numerator and denominator by s yields the sought transfer function, but with an adm bi n e d

S y st e ms

a n ~ .:>y~-..-==• ••

•• •~- --=- •• •• ~

•

- -··· ••--.- - . .,3,

>nents Field - ce>ntre >lled DC:-me >te>r ble>ck d i a g r a m ce>mpe

Table 6 - 4 - 3

B

E, QU Ellt l e>n

VEJ r n l n g

( J:52 -+- c:,0 s)E> ( s )

=

le>c l< dla.g ra. m

-.. I,. . .,~~

T{.s )

T { s)

v-, ,.,( s )

V-,,_.( s )

=

mpe> n e, n t

JC

•

..

T:n.s

w h i c h m a y be: ca.st i:n. m a t r i x :f'rm

.x(r) { .v(r)

~ ( r ) -+- .1:166(.t") C-..:,ic(r) -+- L'66(r)

_!_ -v;~(r)

L .f

(6_4.33

6 -1 n ,,, t Current t.: '1t/)f5

where

u7

matrices are as follows:

-

AJ xJ -

0

O

I

O

~ ~

0

0

, B3r 1 =

RJ

-L1

( ~I )

. C .

f,1(1

[J

()

Lr

,(t) = ~:) , u(t) = v;,(t)

DC Motor with a Load rms useful work by rotating In its numerous applications, a DC motor never runs idle. It perfo s. Any mechanism that rotates different mechanical devices such as winches, pumps, or fan blade the interaction between the using the torque generated by a DC motor is called load. While ction is that of a load exerting an motor and its load may be quite complex, the most typical intera on applies an assumption that additional torque that opposes the motor torque. This simplificati and undamped. In this case, the the shaft connecting the motor and the load is rigid, massless, body, the inertia of which is the load and the motor can be lumped together into a single rigid the load rotate with the same sum of inertias of the motor and the load. While the motor and to the motor torque. angular velocity, the load torque acts in the opposite direction the same direction. For In some cases, the load torque and motor torque may have cart. When the cart is moving example, consider a DC motor used to rotate wheels of a nd; thus, the load torque is downhiII, the torque generated by gravity makes the wheels desce to gravity is still trying to make assisting the motor. If the cart moves uphiII, the torque due n and opposes the motor. cart wheels descend; hence, the load torque resists the motio The presence of a load changes the mechanical subsystem as shown in Figure 6.4.12. As before, 0 is the rotation angle of the armature assembly, and Tb is the motion-resisting torque generated in the shaft bearings in response to the angular velocity. Let TL denote the load torque. Then, the governing equaFigure 6.4.12 DC motor tion for this subsystem becomes with a load: mechanical (6.4.34) or

J0+c/J= T-T L

(6.4.35)

subsystem

Techniques 6 Combined Systems and System Modeling 448

m and the coupling rela tion ' 'ay ati·on for the electrical subsyste . . unchan~ e-controlled rr ' ,or w· g equ . 1th The gove, mm the set of equations that descnbes an armatur ore, ere1 a~ Th

~~

J(} + c/J == T - TL dia . V;n - l:b == Ra la + La dt == Kb0 T == Kia

l:b

racteristics of the load , s, 1 the cont The load torque depends on the physical cha ex1of independent input. DC-motor performance it is considered an onds to the first line of Ee; (6.4 6 1. The block diagram component that corresp .3 ) s !hen modified as shown in Figure 6.4.13. armature-controlled DC motor w· h Then, the expanded block diagram of an It a load ~ 14. 6.4. re Figu in d trate illus in Figure 6.4.15, where two transfier fiunctio. ns: This block diagram can be reduced as shown 0(s)

(b)

(a)

equation in the Figure 6.4.13 DC motor with a load: (a)

Laplace domain·, (b) block d'1agram component

0(s)

2

Js + cbs

~ ~

0(s)

.h t diagram Figure 6.4.14 Armature-controlled DC mo or wit a load : block

Q(s)

DC motor

Figure 6.4.15 Annature-controlled DC mo tor wi.th a load: reduced block diagram

d

6.4 Direct Current Motors

449

and Q(s) Grl(s) = TL(s)

are clearly distinguishable. from the diagram in Figure 6.4.15, the transfer functions are derived as follows:

= Q(s) K - TL(s)] 2 s [(V;n(s) - Kb s0(s)) Ra + Js + CbS Las Vin(s) (R

a

_ = (1 + KKb )n(s) + LaS~(J.S + Cb ) - TL(s)J._I Ra + Las s + Cb

=

GV;. (s)

=

Grr (s)

= O(s) = _

n(s) Vin(s)

TL(s)

K (Ra+ Las)(Js +Cb)+ KKb

(6.4.37)

Ra+ L s (Ra+ Las)(Js + cb) + KKb 0

Sometimes the current in the armature circuit is considered to be the second output of this dynamic system (in addition to the angular velocity). Then, the additional two transfer functions are derived as l L = l 0 (s) Ra+ aS s { (Ia(s)K - TL(s)) J. 2 = O(s) S + CbS

(V;n(s) - Kbs0(s))

Substituting the value for Q(s)

= s0(s) of the second equation into the first one, we obtain

which derives the sought transfer functions: Js+cb Ia(s) +Cb)+ KKb Las)(Js =(Ra+ Vin(s) = Gvin1.(s) Kb Ia(s) GrLi. (s) = TL(s) = (Ra+ Las)(Js + cb) + KKb

(6.4.38)

450

6

Combined Systems and System Modeling Techniques

The selection of state variables for the case oflumped. motor-load inet· . . r las is by the addition of the load. The state-space representation of this dynam· not int1 ic system . llen~ as isobt . d a,tted Xt = X2 State equations

Output equations y(t)

= x2

(6-4.39)

and, in the matrix form: x(t) { y(t)

= Ax(t) + Bu(t) = Cx(t) + Du(t)

where matrices are 0 A3 x 3

=

0 0

x(t) =

1 Cb

K

J

J

Kb

_Ra La

--

La

C' -0(t)) x2

X3

0 , B3 x2

=

0 0 1 La

0

1 J

, Ci x 3

=

(0

1

0 ), D

=O

0

c

(t)) = ~(t) , u(t) = ~(t) = la(t)

Similarly, for the field-controlled DC motor with a load, the governing equations are

J(} + cb0 = T - TL . di1 Vin= Rjlj +L1dt { T=Ki1

(6.4.40)

The modified block diagram is depicted in Figure 6.4.16.

71,(s) 1

T(s)

+ -

1

Figure 6.4.16 Field-controlled DC motor with a load: block diagram

0(s)

~

b

6.4 Direct Current Motors

451

sfer functions are derived as fbe tran

Gv;.(s) = O(s) = K V;11(s) (Rj + Lp)(Js + c1, )

(6.4.41)

O(s) I Gr,_(s) = - = - - -

Js + CIJ

TL(s )

If the current in the field circuit is considered an additional output, the corresponding transfer function is I1(s) V;11(s)

(6.4.42)

G~lj~)=--=-- -

R1 + L1s

Note that due to the absence of a feedback loop (unlike the armature-controlled motor the field-controlled motor does not have a back emf in the field circuit) the load torque has no influence on the current in the field circuit, so Gri11 (s) does not exist. The state-space representation of a field-controlled DC motor with a load is as follows: x, =x2

State equations

' X2

K I = - CbJX2 +1 x3 1 fi(t)

· X3

I () = - R1 LJ X3 + LJ V;n t

Output equations y(t) = x2

(6.4.43)

and, in matrix form:

x(t) = Ax(t) +Bu(t) { y(t) = Cx(t) +Du(t) where the matrices are as follows: 0 A3 x 3

=

0 0

0 Cb

K

J

J

0

, B3 x 2 =

- RJ

L1 x(t) = (:: : :~:~) u(t) = (V;n(t)) . ' TL(t) X3 = la(t)

0 0

L1

0

1

J

0

,

c, x 3 = (0

1 0) , D = 0

452

dsystem Modeling Techniques 6 Combined Systems an

, d DC motor with a load shown in Figure 6·4 · fl 'bl ped e controll . e ected to the motor via a .ex1 e dam ~ r Consider an armatur · bl , the load is .conn The load is supported with the bearings , h· With to . . In this assem Y. I d·1mping Cb rsro~ le the inertia of the load isJl-· 1I • 10aveada 11 whi ·sJ tor ' iona tors d n kha ness stiff 1 mo 1 ' Th · (ao resist,, ''. r of lhelli~inl 1 bft1e permanent magnet. Ra and Ladenote the a d coefficient ~L· e mer current in the a, rr and ind Uqruv",,, esente Y . e1Y, while ;a indicates the DC motor 1s repr circ1i' a¾ · · I nts respectiv constant is Kb, The voltage inp,, nlire · lo th earrt. ti·~ of the circuit e em~ . K d the back emf motor torque constant is ' an llra1u d h I d ri circuit is denoted v;n(t). em inputs are V;n(t) an t e oa t~rque TL (t), and ,. ·~ syste Considering that the syst out IJJ . of the load c,. d P% . f the motor w and the angular velocity • enve: O are the angular veIoc1t y

Example 6.4.1

system (a) the governing equations of this dynamic (b) a block diagram (c) a state-space representation

. Solution ystem of this dynamic system is represented As seen in Figure 6.4.17, the mechanical subs tll'o WJ!h . the load cannot be lumped together into a · inertia elements instead of one : the motor and sin~e that connects them. Thus, the mechanic I element because of the flexible damped shaft s1em Subsy a ns described by two differential equatio has two degrees of freedom and needs to be em starts with draw1.n the mechanical subsyst . The derivation of the governing. equations ford . d' . II the actm g torques. The fi ga an m 1catmg a free-body diagram for each inertia element, rhee-body and the load rotate intes e, that both the motor anie 1· d diagrams are illustrated in Figure 6.4.18. Not d' . h 1er. ear . sse 1scu as . . tion emo dtrect1on , while the load torque opposes t . g torques: T(t) - the generated motor tor The inertial elementJ is subject to three actin ~Tias a torsional spring; andi,e shaft flexibility that is modeled b the the restoring torque due to the · Tb .. it . veloc . torque due to the shaft tors1.0naIdampmg ansmg m response to angular res1. stmg a:· la; ment 0(t), while the torque Tb opposes the torque Tk works_against the angular displace gu 0. = w(t) city velo · the motor is greater than that of the load Assuming that the angular displacement of 1.e. ' (see Chapter 3) 0(t) > 0L(t), the expressions for Tk and Tb are - - -- - - - - , ,- - - - - - - ,- - - - - - - - - - - -

I

I + I

_

- -- - - - - -- 7

I

1 1 - --

,,,,,, ,,,- ~

11 II J ob I I II - 11 T /I ......J ____,,.,_ _

' 1

+

.___ I - - -- - - - --- - -- - - -

Electrical subsystem

/

T L dt - - - Mechanical subsystem

s/§=W

ple 6.4.1

bly for Exam Figure 6.4.17 DC motor with a load assem

_d0L

WL

- df

I I I .J

6.4

(a)

Direct Current Motors

453

(b)

Figure 6.4.18 Free-body diagrams of (a) the motor and (b) the load

Tk = kb(0- 0L) Tb = cb(e - fh)

Applying Newton's second law for a rotational system, the equation of motion for the motor is JO= T(t) - Tk - Tb= T(t) - kb(0- 0L) - Cb(iJ- 0L)

Similarly, the governing equation for the load is derived as JLOL = -TL(t) + Tk + Tb - Tc = -TL(t) + kb(0- 0L) + cb(iJ- 0L) - cliJL

( I)

(2)

Equations (I) and (2) represent the governing equations for the mechanical subsystem. The governing equation for the electrical subsystem, the mechanism of generating the back emf in response to the rotation of the armature assembly, and the coupling relations are the same as for the armature-controlled DC motor: .

V;n -

eb

dia

= Ra la + Ladt

(3)

d0

tb

= Kb dt

T=Kia

Hence, the governing equations for this dynamic system are JO+ kb(0- 0L) + Cb(iJ - iJL ) = T(t) JLOL - kb(0- 0L) - Cb(iJ - 0L) + CL0l .

V;n -

eb

dia

= Ra la + Ladt

= -TL(t) (4)

d0

tb

= Kb dt

T=Kia

To construct a block diagram, take the Laplace transform of Eq. (4) considering zero initial conditions:

454

6

Combined Systems and System Modeling Techniques

(Js2 + CbS + kb)E>(s) - (cbs + kb)E>i(s) = T(s) (Jis 2 + (cb + ci)s + kb)E>i(s) - (qs + kb)E>(s)

= -TL(s)

Vin(s) - Eb(s) = (Ra+ Las)la(s) Eb(s) = Kbs0(s) T(s) = Kla(s)

(S)

The last three equati ons in the set of Eq. (5) are easy to transfo rm intl > .,,1ock ct· . compo nents , as for the armature-controlled D C motor d 1scussed above and , t1.-llt, .'"s I . tao.a "' rn ponen ts are shown in Table 6.4.2. u ting corn. The constr uction of the block diagra m compo nents that repres ent the relatit, ns be torque s T(s) and Ti(s) and the rotatio n speeds 0(s) and E>i(s) requir es a nurrtb".,r ' o ftween al the transfo rmatio ns of the first two equati ons of the set ofEq. (5) to obtain the expres sions tn . the gebraic form V

0(s)

= G1 (s)T(s) + G2(s)TL(s)

E>L(s)

= G3(s)T(s) + G4(s)TL(s)

(6)

and (7)

To simplify the derivation, let the following be denote d as Js 2 + CbS + kb = g1 JLs2 + (cb + cL)s +kb= g2 cbs+k b =g3 Then, the first two equati ons of Eq. (5) are rewritten as

= T(s) g30(s) = -TL(s)

g10(s) - g30i(s ) g2E>L(s) -

(8) (9)

From Eq. (9) 0L (s )

1 g2

g3 = - -Ti(s ) +-0( s)

(10)

g2

Pluggi ng Eq. (10) into Eq. (8), we obtain

which yields the desired form of Eq. (6): (I I)

1

~

6.!!

Kbs

1◄◄a---

i)irect turrent Motors

455

l )/ 1i)

l6-1~~{ ~l O(s1+{ s ~n(s1-.

Figure 6.4.19 DC motor with a load assembly: block diagram

·m1·tar manner, the expression for E>i(s) is derived:

Jn a s1

(12) The expanded block diagram of this dynamic system is shown in Figure 6.4.19, where

This system has four transfer functions: n(s) n(s) Gv;.(s) = Vin(s)' Gri (s) = Ti(s)' The derivation of these transfer functions from the constructed block diagram or from the derived governing equations (Eq. (5)) is straightforw ard, using the approaches discussed in Sections 6.2.2 and 6.2.3. The state-space representatio n of this dynamic system is easily obtained from that derived earlier by adding two more state variables x4 = 0i and xs = (Ji.

xi = x2 . kb Cb K x2 = --:,xi -7x2 + Jx3

State equations

.

X3

.

Output equations y(t)

Cb

x4 + 7xs

Kb 1 = - -x2 - -RaX3 +-L V;n(t) La

.x4 =xs xs

kb

+7

La

a

Cb kb Cb + CL 1 xi + - x2 - -x4 - - -- xs - - Ti(t) JL JL JL Ji JL kb

=-

= (:: )

LI

---= 456

6 Combined Systems an

d System Modeling Techniques

And, in the matr ix form:

x(t)

wher e the matr ices are 0

As xs

=

kb

Cb

J

J

= Ax(t ) + Bu(t)

{ y(t)

= Cx(t) + Du(t)

0

0

0

K

-

kb J

Cb J

0

0

-

0 0

Kb La

J Ra La

0

0

0

0

kb

Cb

0

kb JL

Cb +cL JL

= (~

1 0 0 0

0 0

0

Ji

-

Ji C2xs

XJ

x(t) =

=

1

= 0(t) = B(t)

X2 X3 = ia(t) X4 = 0L(t) s=B L(t)

_]

, Bs x2

La 0 0

~), D

()=

' u t

=

0 0 0 0

1

JL

0

(Vin (t)) TL(t)

Simplification of a DC- Mot or Model roJle d DC moto r is a second-order As can be seen from Eq. (6.4.14), the arma ture- cont for the analy sis of such systems is dyna mic system. Whil e the math emat ical appa ratus syste m to a first- orde r one is desirable. exten sivel y deve loped , in many cases redu ction of this indu ctan ce of the armature circuit Such a redu ction is reaso nable since, typicaJly, the in· La < < I. Then , the expr essio n (Ra +La s) in the denom great ly excee ds the resis tance , i.e. Ra mes ator of the syste m trans fer funct ion (Eq. (6.4.14)) beco

Ra

+ Las = Ra (

1;: + I) ~ Ra

Las. . smce - 1s very small. Ra Then , the trans fer funct ion reduc es as follows: K G(s) = Q(s) = + L 0 s) a Cb)(R ~n(s) (Js +

_

+ K.Kb

- (Js

K

+ Cb)Ra + KKb

6 5 Blod: IJ1i11 rn .-s

t.t,r. nm11 Oorr .in

457

. g coefficients setttn Ku = K/ IRil ,,

1,..} • ,

and

presents the transfer function in the familiar form of a first-order system:

G(s)

= n (s) = Vin(s)

KM rs+ I

(6.4.44)

s in the Time Domain 6.5 Block Diagram Transfer function formulations (Section 6.2.2) and block diagrams in the s-domain (Section 6.2.3) are valid for linear time-invariant dynamic systems, for which the governing differential equations are linear with constant coefficients. These techniques are not applicable to linear time-variant systems and nonlinear systems, for which the governing differential equations have time-dependent coefficients and/or nonlinearities. For these systems, statespace representations (Section 6.2.1) are widely used in modeling and simulation. In this section, another modeling technique, which is called block diagrams in the time domain or time-domain block diagrams, is introduced. This technique is applicable to general dynamic systems, including linear time-invariant systems, linear time-variant systems, and nonlinear systems. One benefit of learning time-domain block diagrams is to prepare for the use of the Simulink software in the modeling and simulation of complicated systems. As will be seen in Section 6.6. l, the structure of time-domain block diagrams and that of Simulink models are quite similar. Like an s-domain block diagram, a time-domain block diagram is an assembly of basic elements or components. Table 6.5.1 lists some basic components for the construction of time-domain block diagrams. From the table, the signal, summing point, pick-off point, and constant gain follow the same rules as their counterparts in the s-domain; a differential operator D was introduced in Section 2.5.1; and integrator can be symbolically expressed by Jdt = n-1• However, three new components (the last three in the table) are not seen in the s-domain block diagrams: the time-varying gain, multiplier, and function block, which are used to describe time-variant properties and nonlinear characteristics. A multiplier can be used to describe the powers of a variable, such as x7- and .x3. A time varying gain g(t) or nonlinear function ~(t) can be used to describe the time-variant or nonlinear property of a dynamic system.

)

458

6 Combined Systems and System Modeling Techniques

Table 6.5.1 Basic components of time-domain block diagram -

Rule

Symbol

Component

~

u Signal

V

Summing point

v:=J y

Integrator

X

Constant gain

X

Time-varying gain

Multiplier

u V

►

W=U-V

V= U

1

:wu

►0

z

·8> .8

:0

W=U

dU.d x D U=- X=dt dt'

overdo!

D,

Differentiator

w

~?,

u

Pick-off point

Variable, input or output

◄

►

.. y

z(t) =J1 y(r)dr /0

..

y=kx

=g(t)x

y ..

y

..

W= UXV

w

x ,X ~

X2

Y= X1XX2XX:3

X3

Function block

t

►,

¢(/)

I

y►

y= (t)

-

. blex(t) governed by an n-th-orde . vana . characten.stlc struc·r . system wit. h its Cons1.der a dynam1c for con differential equation. Denote the system input by r(t). A systematic procedure st eps. tion of a time-domain block diagram for the system takes the following five (1) Rewrite the differential equation in the following form:

6.5

D''x = J(x , Dx, . . ,[Y

(Z) (J)

(4)

(5)

Block Diagram~ in the Time Domain

1.x

459

(6.5 1)

r,

here/is a function of variable x , its dcrivativc.J up to order,, - J. and input r f(Jr a I neu ;ynarnic system,/is a li~ear function ; for a nnnlm, ,tr ;>
-

0.1 0 .05 0 0

2

4

6

8

10

12

14

16

18

20

ple 6.6.1 Figure 6.6.2 Resp onse (x and y) of the system in Exam

Example 6.6. 2

by Eqs. (a) and (b) in Exam ple 6.5.4, has An arma ture- cont rolle d DC moto r, whic h is gove rned the follo wing para mete r value s:

La J

3 H, Ra = 0.56 n, K = Kb = 0.06 N · m/ A = l .9 x 105 = 9 x 10- kg · m2 , cb = l.2 x 10- 4 N . m • s/rad

(a)

~~

In a Simu link mod el of the moto r, Let the apph ed arma ture volta ge be v;n = lO V. Build s. I O < · t ~ O . nt ia(t) of the moto r for the rotat ion spee d cv(t) = 0(t) and the arma ture curre simu latio n, the follo wing two cases are cons idere d: .

I

6.6

Modeling and Simulation by Simulink

. the motor idles with no load torque, TL 1

46 7

=o

~ the motor idles initially and, at t = 0.04 s, a consta nt load to rque, Tt_ = 0 .6 N -m is ~d apphe

501uti00 . g Figure 6.5.6, a Simulink model of this electromechanic al system is built; see Figure BY referencinodel has two inputs, which are the load torque h and applied voltage \ ';n; a nd two rn are the rotation · speed wan d armature current ia of the motor. · . 1·tn k 6.6•3· The hich W ith the S1mu outputs, rotation speed and armature current of the motor are plotted aga inst time for the .,,odel, t 11• entioned two cases. above-rn e I as shown in Figure 6.6.4, the rotation speed of the idling motor reaches a In . h ad cas -state ' value _of 163.6 rad/s (1562 rpm) ._ The_armature c~rrent initially goes up as h1g_ ste Y A which 1s needed to overcome the mertla and damping effects of the motor, a nd it as I3.19 ' 33 A • tuallY settles at 0. even . t h e same way initially. At time t = 0.04 s when h In case 2, the motor b e h aves m t e load torque Ti is applied, the pattern of the system response starts to change. As shown in

:e

d(theta)/dl omega

c_b T K

i_a

-c 0 .5 6 1 4 - - - - - - - - - - - - - '

L..,__ _

Ra

K_b

Figure 6.6.3 Simulink model of the motor in Example 6.6.2

468

6

Combined Systems and System Modeling Techniques

1

180

J

160

1

140

7

120 U)

:ci

~

a

100

80 60 40

20 0 -20

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

0.08

0.09

0.1

14

12

- - -·- - - - - - - ---- -

10

g... ._

8 6 4 2 0 0

0.01

0.02

0.03

0.04

0 .05

0.06

0 .07

t(s)

Figure 6.6.4 Response of the motor without a load (case 1)

. kly climbs Figure 6.6.5, the motor speed significantly drops and the armature current qmc A. up. The steady-state values of the speed and current are 72.0 rad/s (688 rpm~ ao
}, Plo tRa nge Al l, Plo tSt yle {B lac k, Th ick Gr idL ine s }, Tru e, Ta ble [8. 8l• i, {i, 18} ] J, (8, se, 1ee , 158 }}, As pec tRa tio 1/3 {{{ 8, 58, 1ee , 158 }, , No ne} , {Jo in[ {8 l, Ta ble [8. 8l• i, {i, 18 })1 FrameLabel ➔ {Style["time", 12) , No ne} }, 1 Sty le[ "w (ra d/s )", 12} , No ne, Non e}}

➔ {Join[{&}, FrameTicks ➔

(• 6 + b1¢>4 + b3¢>1 where a1, a2, b1 , b2, b3 are empirical constants (Vaughan, Gamble, The Modeling and Simulation of a Proportional Solenoid Valve, ASME Journal of Dynamic Systems, Measurement, and Control, Vol. 118, March 1996). Assume that the return spring is at its free length and the electromagnetic force is zero when x = 0. (a) Derive the governing equations for this system. (b) Derive the state-space representation, considering that the system input is the voltage e;n(t), and the output is the displacement of the valve. Section 6.4 Direct Current Motors

6.17 An armature-controlled DC motor is used to rotate the load JL as shown in Figure P6. l 7. The load is connected to the motor with a very stiff massless shaft and is supported by bearings with a damping coefficient bL, The system input is the voltage e;n(t) supplied to the armature circuit, while the output is the rotational velocity of the load WL,

:

:

~: :

Armature circuit - - - - - - - - - - - - - - - - ..-- -- b - - - - - - - -- b - - - 7

! •••

I~~ R•::;/' ,: i i O JJ O

I ' - - - -- - - ~ I I I__ _ __ __ __ ____ _ __I ~ __

Electrical subsystem

!.ro __Wm

J,

___ _ __ _ _ WL

i

]

_

I I

Mechanical subsystem

Figure P6.17 Armature-controlled DC motor with a load

(a) Derive the governing equations for this system. . . f2L(s) (b) Denve the transfer function E,n(s)'

6.18 An armature-controlled DC motor is used to rotate the load JL as shown in Figure P6. I8. The load is connected to the motor through a converter Ka and is supported by bearings with a damping coefficient bl. The converter transforms the rotational velocity Wm of the motor into torque TL applied to the rotor in the following manner: TL = Kawm, where Ka is a known converter constant. The converter and the connector shafts are considered massless.

Problems

6·16

495

Consider the solenoid actuator described in Problem 6.15. led to d eve Iopment of the folJowing expressions for the soleno1.d · · Experunentat1on current and the electromagnetic force:

ir(¢>) = ai¢> 3 + a2¢> Fe,,,(¢>) = bi ¢,6 + b2 ¢4

+ b3 ¢,2

where ai , a2, bi, b2, b3 are empirical constants (Vaughan, Gamble, The Modeling and Simulation of a Proportional Solenoid Valve, ASME Journal of Dynamic Systems, Measurement, and Control, Vol. 118, March 1996). Assume that the return spring is at its free length and the electromagnetic force is zero when x = 0. (a) Derive the governing equations for this system. (b) Derive the state-space representation, considering that the system input 1s the voltage e;n(t), and the output is the displacement of the valve. section 6.4 Direct Current Motors 6.17 An armature-controlled DC motor is used to rotate the load Jr as shown in Figure

P6. l 7. The load is connected to the motor with a very stiff massless shaft and is supported by bearings with a damping coefficient bl . The system input is the voltage e;n(t) supplied to the armature circuit, while the output is the rotational velocity of the load wr. Armature circuit 1-----

I I I I

r -- --- --------- -----7

I I

I

:+

I

~~~

R a ~ La

+:

I I I I

18

~

~

I

[OJ JJ) 0:) i

I WL Wm Tm -- - - - - - - - - - - - - - - I~---------------- --Electrical subsystem

Mechanical subsystem

Figure P6.17 Armature-controlled DC motor with a load

(a) Derive the governing equations for this system.

.

.

nr(s)

(b) Denve the transfer function E;n (s) ·

6.18 An armature-controlled DC motor is used to rotate the load Jr as shown in Figure P6.18. The load is connected to the motor through a converter Ka and is supported by bearings with a damping coefficient br. The converter transforms the rotational velocity Wm of the motor into torque Tr applied to the rotor in the following manner: Tl= Kawm, where Ka is a known converter constant. The converter and the connector

shafts are considered massless.

496

6

Combined Systems and System Modeling Techniques

Armature circuit

Ra~La

~

~--------"

+

- - - - 6",_ - - - __

bm

- - - - - - - - - - - - - - - -I : - -

,......~--"T"I

I I I I

--"'t-==~~

, ·

I I I I

I I

I

-

I I

I I

~-t----.1)

'-'---+----1-"

I

_ ____ _ - - - - - __ ___ I I__ - ~ - _ ~"2_ _____ ___

Electrical subsystem

!~ _ WL

Mechanical subsystem

- - - - -

I

I I

I 1

Figure P6.18 Armature-controlled DC motor with a load and converter

The system input is the voltage e;n(t) supplied to the armature c· 1 . ' \,Ult Wh"1 output is the rotational velocity of the load wi. ' le the (a) Derive the governing equations for this system. .

.

Ci~)

(b) Denve the transfer function E;n(s)" (c) Construct a modular block diagram and indicate all the relevant sipnals s h ~ uc as th armature current ia, back emf sb, motor torque Tm, motor velot i!.y w be . e m, anng torques Tbm and Tbi, load torque Ti, and load velocity wi. 6.19 An armature-controlled DC motor is used to rotate the load Ji as shown in Fi P6. l 9. The load is supported by bearings with a damping coefficient bl , and is::~ nected to the motor with a flexible shaft, modeled as a torsional spring with a stiffness coefficient kr. The system input is the voltage e;n(t) supplied to the armature circuit, while output is the rotational velocity of the load wi. Armature circuit

,-----------------------I ~ ~

I 1

I I I I I I I I I - I I - _'-_-_-_-_-_-_-_-_-_-_-_--'_ - _I I

[8] ~..,..==--....,__

[8] ~-==-~

I

.__,,,,,--,,'-_.__

J

m

'-'---+---+-'

'-"----

I I I I I

I

I_ _ _

Tm

_ Wm _ _ _ _ _ _ _ _ _ _ _ WL _ _ _

Electrical subsystem

1

Mechanical subsystem

Figure P6.19 Armature-controlled DC motor with a load and a flexible shaft

(a) Derive the governing equations for this system. (b) Derive the transfer function nr((s)) _ E;n

s

(c) Derive the state-space representation. h 11 5 (d) Construct a modular block diagram and indicate all the relevant signals sucbaari ·t OJ e armature current ia, back emf eb, motor torque Tm, motor veloct Y m, torques Tbm and TbL, spring torque Tk , and load velocity wr.

.J. Problems

497

field-controlled DC motor is used to rotate the load Jl as shown in Figure P6.20. The 6,20 ioad is supported by bearings with a damping coefficient bl, and is con nected to the JJlO

tor through a gear pair with a gear ratio N

= nn,2 . The shafts connecting the gear pair

•th the motor and the load are short, stiff, and considered massless. . h . he The system input ts t e voI tage e;,,(t) supplied to the field circ uit, while output 1st rotational velocity of the load w l.

Wl

Field circuit

----------7

bm

I I I Rr I I I I

I

I I I

n1

I I

I I

L------------

I

Electrical subsystem

I I I

Mechanical subsystem

Figure P6.20 Field-controlled DC motor with a load and gear pair

(a} Derive the governing equations for this system. • &: f . Ql(s) (b} Denve the trans1er unction ( ). E;,, s (c) Derive the state-space representation. 6.21 A field-controlled DC motor is used to rotate the load JL as shown in Figure P6.2 l. The load is supported by bearings with a damping coefficient bl, and is connected to the motor with a flexible shaft, modeled as a torsional spring with a stiffness coefficient kr. The load is subject to an unknown drag torque TD. The system inputs are the voltage e;,,(t) supplied to the field circuit and the drag torque TD, while output is the rotational velocity of the load W£. Field circuit r - - - - - - - -

i~ 3 I

L

-

-

-

-

-

!~: !~: UJP~l JJ

71- - - - - - - - - - - - - - - - - --, 11

Rf!! -

-

-

Electrical subsystem

_,

I

-

-

Tm am 9L - - - - - - - - - - - Mechanical subsystem

-

-

-

Figure P6.21 Field-controlled DC motor with a load, flexible shaft, and drag torque

I

i -

I

498

6

Combined Systems and System Modeling Techniques

(a) Derive the governing equations for this system. QL(s)

Ql(s)

.

nd To(s) · (b) Derive the transfer functions E;n(s) a (c) Derive the state-space representation. (d) Construct a modular block diagram and indicate all the relevant sig 1 na s ~Uch . I field current 1/ motor torque Tm, motor d 1sp acement 0m, bearing t orq lies r as the I d l . ' b,,. and rbl spring torque Tk, drag torque To, oa ve oc1ty wl, and displace

merit 0

'

6.22 A field-controlled DC motor is used to rotate the loadJL a~ shown in Figure p 6 22 l is connected to the motor through a converter Ka and 1s supported by be- · The load c:1'ngs . . h '" damping coefficient bl. The converter trans1orms t. e rotational velocity OJ ,,, c·f the With a rnotor , into the torque Tl applied to the rotor in the followmg manner: TL = Ka OJm , \r,,1ereK . a ts a · rn consider~d are shafts known converter constant. The converter and the connector assJess. The load is subject to an unknown drag torque TD· The voltage supplied to the field circuit of the DC motor is driven b y _ c- generator · · h h I · · I . · Th1s vo tage 1s proportlona tot e current mt e generator Circuit as followt. .,_e1 :::: Kgig, where a Kg is a known constant. The system inputs are the voltage e;n(t) supplied to the generator and the d rag torque To, while output is the rotational velocity of the load W£ . Field circuit

Generator circuit

r-----------7r------- -- --7 I I I 1 I I

I

Lg

:

Rg::_

~

....__ _ _ __,

1I I1

.____ _ _ _.....J

0

L ____________

I

R,:

V

-7

l&l

I I I I I I I -

-

TL

Wm

Tm -

-

-

-

-

-

-

_

Electrical subsystem 1

-----

bm

II

L __________ _

Electrical subsystem 2

L -

I I I

Lr

~

:_ II

+

-

-

-

-

-

-

-

-

-

~L -

-

I I I I I I I - _ I

Mechanical subsystem

Figure P6.22 Field-controlled DC motor with a load and a generator

(a) Derive the governing equations for this system. (b) Derive the transfer functions .QL(s) and .QL(s) To(s) · E;n(s) . (c) D enve the state-space representation. (d) Chonstruct a modular block diagram and indicate all the relevant signals such as torque . · • mput · fleId crrcmt t e generator current d voltage e1 , field current iJ, motor lg, . Joa nd T.m, motor velocity w b · m, eanng torques Tbm and TbL drag torque TD, a . • ve1oc1ty wl,

Problems

499

sedion 6.5 Block Diagrams in the Time Domain

23

6· _ 6 24 _ 6 25 _ 6 26 6. 27

Construc t time-dom ain block diagram s for the dynamic systems represen ted by equations in Problem 6.8. Construc t a t~me-doma~n block d~agram for the DC-mot or system in Problem 6.18. Construc t a t~me-dom a~n block d~agram for the DC-mot or system in Problem 6. 19. Construc t a t~me-doma~n block d~agram for the DC-mot or system in Problem 6.20. Construc t a t1me-do mam block diagram for the DC-mot or system in Problem 6.21 .

Sedion 6.6 Modeling and Simulation by Simulink

conside r that the transduc er describe d in Problem 6. 11 has the followin g paramet ers: 6_28 Lo = 0.003 H, Ro = 0.8 fl, R = 1.2 n, h = 0.03 m, B = l .25 T. The input displace ment of the magnet is x(t) = t. Plot the voltage drop on the resistor R (conside r a simulati on time O< t ~ 0.02 s), find the steady-s tate value if it exists, and compute the settling time for 0.5% alJowed error. 6.29 Conside r that the transduc er describe d in Problem 6.12 has the folJowing paramet ers: Lo= 0.003 H, Ro= 0.8 fl, R

m

= 0 .25kg,

c

= 7.5

= l .2 n , h = 0.03m,

x 10- 4

-

B

= 1.25T,

N

-

A·m

The external force applied to the magnet is constant :/M(t) = 3.1 N. Plot the voltage drop on the resistor R (conside r a simulati on time O < t < 200s), find the steady-s tate value if it exists, and compute the settling time for 1.0% allowed error. 6.30 Conside r that the armature -control led DC motor with a load and converte r system described in Problem 6.18 has the following paramet ers: (N-m-s) 3 La = 0.004H, Ra = 0.75 n, Jm = 8.0 x 10- kg-m2, bm = 0.005 -'----rad (N·m•s) V-s N-m Km = 0.3 - - , Kb = 0.2 - d , Ka = 0.4 rad ,

A

JL

= 12.0 x 10-

ra

3 kg-m 2 ,

bL

= 0.006 N-m-s rad

(a) The voltage applied to the armatur e circuit of the DC motor is constan t: ein(t) = 0.1 V. Plot the velocity of the load CO£ (conside r a simulati on time 0 < t < 20 s), and find the steady-s tate value if it exists. (b) The voltage applied to the armature circuit of the DC motor is a positive half-sine periodic function ein(t) = 0.5jsin(t)I V (lsin(t)I indicates the absolute value of the sine function). Plot the velocity of the load WL (conside r simulati on time O < t < 50 s), and find the largest velocity value after the amplitud e of the oscillati ons settles to a constant sinusoid al pattern.

500

6

Combined Systems and System Modeling Techniques

. 31 Consider that the field-controlled DC motor with a load and gear pair d . escnb d 1n . . e Problem 6.20 has the followmg parameters. 4 2 L = 0.006H , R = 0.56 n, Jm = 9.1 x 10- kg•m , bm = 0.0012 (N•m-~ 1

1

rad Km= 0.06 N-m N = ni = 2.5, Ji= 15 .0 x 10- kg-m , bi= 0.002 ~ :!.. ran1 A ' (a) The voltage applied to the field circuit_ of the ~C m~tor ~s constant: ei,, ( :::: 5 ·2 V. t < 20 ~ Plot the velocity of the load roi (consider a simulation time .0 < fi nd and ", . . steady-state value if exists; plot the field current (consider simul" 1 1 . n tune 0 < t ~ 0.2 s), and find the steady-state value if it exists. (b) The voltage applied to the field circuit of the DC motor is - PU1se· . e;n(t) = 2u(t) - 2u(t - 5) V. Plot the velocity of the load wi (consider a .~·rnuJ . · ation d d. 1 . . . time 0 < t ~ 20 s), find the maxrmum ve ocity an its stea y-state value if it .., .. iSls· 1 ot 20 s), find the maxim1·rn cu•rPrent the field current (consider a simulation time 0 -< t < and its steady-state value if exists. (c) The voltage applied to the field circuit of the DC motor is a trapezoidal p1..: ,e, shown in Figure P6.31. 4

2

3.0 2.5

-

2.0

~ 1.5

1.0 0.5 0.0 0

5

10

15

20

25

30

t Figure P6.31 Input voltage

th Plot the velocity of the load roi ( consider a simulation time 0 ~ t ~ 30 s), fio 0 sgn(OL) is the signum or sign function, defined as sgn(x = - I w . hen x < 0 . { 0 whenx = 0 Do both systems converge to a steady state? Qualitatively describe the influence of the drag torque on the system response. {b) Plot the velocity of the load wL (consider simulation time 0 ~ t ~ 15 s) for the case when drag torque is applied with 5-s delay: Tv = 0.005 0Lsgn(fh)u(t - 5). Does the system converge to a steady state? Qualitatively describe influence of the delayed drag torque on system response.

Section 6.7 Modeling and Simulation by Mathematica 6.33 6.34 6.35 6.36 6.37

Solve Problem 6.28 by Mathematica. Solve Problem 6.29 by Mathematica. Solve Problem 6.30 by Mathematica. Solve Problem 6.31 by Mathematica. Solve Problem 6.32 by Mathematica.

7

System Response Analysis

Contents

e Domain 7.1 System Response Analysis in the Tim 7.2 Stability Analysis Frequency Domain 7.3 System Response Analysis in the arying and Nonlinear Systems 7.4 Time Response of Linear Time-V Chapter Summary References Problems

501 544

55! 5)1 5))

5)! 51i

been models of various dynamic systems have In the previous chapters, mathematical e. g differential equations by the first laws of natur developed; several modeling tools, includin ebeen rams, and state-space representations, hav transfer function fonnulations, block diag deterthe dynamic response of a system can be introduced. With these models and tools, aviors can be gained through analysis and beh its of ing and erst und the and ed, min simulation. physical behaviors of dynamic systems5tare ing ard reg es issu n mai e thre , pter cha In this sy em the time domain, stability analysis, and addressed: system response analysis in In the presentation, analytical method~ are response analysis in the frequency domain. ulink are used for general linear and nonlinear used for simple systems; MATLAB and Sim systems.

7.1

main System Response Analysis in the Time Do

· dorna~

· the tune sfoJ111 · ns of the response oflinear time-invariant systems m IutJo · · In th1s section, so L~~

. Refer to Section 2.6 for the ap are detennmed by Laplace transfonn. method.

502

7.1 System Response Analysis in the Time Domain

503

General Concepts

7,1.1

consider a general dynamic syS tem governed by an n-th-order linear ordinary differential equation

A(D)y(t)

= B(D)r(t)

(7.1. I)

where D == d/dt, A(D) and B(D) are polynomials of operator D n

m

k=O

l=O

A(D) = [akd , B(D) = [b,d with ak and b1 being constants, and r(t) and y(t) are the forcing function (system input) and response (system output), respectively. Refer to Section 2.5.1 for details on the operator polynomials. Let the initial conditions of Eq. (7.1.1) be y(O)

= ao,o,

Dy(O)

= ao,1, . .. , .D1'- 1y(O) = ao,n-1

(7.1.2)

where ao,o, ao,1, .. •, ao,n-1 are prescribed values. The basic problem of system response analysis in the time domain is to find the solution of the differential equation (7 .1.1) subject to the initial conditions (7.1.2).

Free Response and Forced Response The solution of the previous differential equation can be obtained by Laplace transforms. To this end, performing Laplace transform of Eq. (7.1.1) yields A(s)Y(s)

= J(s) + B(s)R(s)

(7 .1.3)

whereR(s) and Y(s) are the Laplace transforms of the input r(t) and outputy(t), respectively, and J(s) is a polynomial of s, which is a linear combination of the initial values a0,0, a0,1, • •• , ao,n-l • It can be shown that I(s) = 0 if all the initial values of system response are set to zero. Following Section 2.6, the time response of the system is given by

y(t) = Y1(t) +YF(t)

(7.1.4)

.e-

(7. l.5a)

with ()=

YI t

I { / (S)

A(s)

}

YF(t) =.8- 1{G(s)R(s)}

(7.1.5b)

1 where.e- is the inverse Laplace transform operator,y1(t) andyF(t) are the free response and forced response, respectively, and G(s) is a transfer function of the system given by

504

7

System Response Analysis

. G(s)

=

B(s) A(s)

=

bms"' + bm- ts"'- 1 +···+b is+ bo ansn + an - isn- l + · · · + a1s + ao

(7_ I-6)

For inverse Laplace transform s, refer to Section 2.4.4.

Example 7.1.1 A spring- mass-da mper system is governed by

mx + cx(t)

+ kx(t) = f(t)

x(0) b xo , .x(0) = vo Laplace transfor m of the previous diffet ntial equation yields (ms 2 +cs+ k) X(s)

= -.._.., F(s) + m(sxo + vo) +

A(s)

B(s)= I

cxo

/ (s)

By Eq. (7 .1. Sa, b ), the free response and forced response of the system are given by

Yi(t)

= .E-I {m(sxo + vo) + cxo} ms2 +cs+ k

YF(t) = .E- 1{

l

ms2 +cs+

kF(s)}

From a physical viewpoin t, y 1 (t) is the system response excited by initial disturbances and it is describe d by the different ial equation

A(D)y1(t)

=0

(7. l.7a)

subject to initial conditio ns

Y1(0)

= ao,o,

Dy1(0) = ao,1, ... , Dn-lY1(0)

= ao,n-1

(7. l.7b)

The YF(t) , on the other hand, is the system response excited by the input r(t) and it is governe d by

A(D)yF(t) = B(D)r(t)

(7. l.8a)

YF(0) = 0, DyF(0) = 0, . .. , Dn-lYF(0) = 0

(7J.8b)

with zero initial conditio ns

Because Eq. (7 .1. 7a) is a homoge neous different ial equation , its solution can be expressed by (7 .1 .9)

d

~

1

µ

I

7 .1

1

where Ak

System Response Analysis in the Time Doma in

505

are the roots of the chara cteris tic equat ion

itions (7. 1. 7b). Here , are const ants that are event ually deter mine d by the initial cond A f h o root • (a }. roots ted and k characteristic repea 0 / has A(J) If ed. assum roots ave been . tinct f, dJS . t roots J1 , A2 , . . . , Am, with l+m = n and AQ#Ak or d m d'1stmc /multiplicity) an j == I, 2, . . . , m, the free respo nse is of the form

y1(t)

1 2 =A1if 11 +A2 lf21 + .. . +AmEfml + (B1t'- I +B2t '- + ... +B1) flo

(7.1.JO)

us equa tion. Refer to Section 2.6 for more on the comp leme ntary homo geneo the conv oluti on theor em The solution of Eq. (7. l.8a), throu gh the appli catio n n integ ral (Section 2.4.3) to Eq. (7. l.5b) , can be expressed by the conv olutio I

YF(t) = Jg(t--r)r(-r)d-r

(7.1.1 1)

0

h is the solut ion of Here g(t) is the impu lse respo nse funct ion of the system, whic

= B(D)t5(t) 1 g(0) = 0, Dg(0) = 0, ... , v - g(0) = 0

A(D) g(t)

respo nse of the syste m unde r with t5(t) being the Dirac delta funct ion. Physically, g(t) is the collision, impa ct, and shock . It an impulse of unit ampl itude , whic h is often used to describe to the syste m trans fer funct ion is easy to show that the impu lse respo nse funct ion is relate d by

G(s)

= .8{g( t)}

(7.1. 12)

where ./J is the Lapla ce trans form opera tor. be obtai ned eithe r by direc t Thus, the free and force respo nses of a dyna mic syste m can b), or by the homo gene ous inverse Laplace trans form , as show n in Eqs. (7.l.5 a) and (7.l.5 1). Thes e analy tical meth ods solution ofEq . (7.1.9) and the conv oluti on integ ral of Eq. (7.1.1 shall be used altern ative ly for conv enien ce in analy sis.

Transient Response and Stea dy-S tate Resp onse d as a sum of a trans ient The response (outp ut) y(t) of a dyna mic syste m can be viewe response Yrr(t) descr ibes the response Yrr(t) and a stead y-sta te respo nse Yss• The transient it is a funct ion of time t. If change and trans ition of the syste m with respe ct to time and

506

7

System Response Analysis

a system is stable and is under a bounded input, its transient response disappe . . . d. ars after a I enough time. (The stability of a dynamic system 1s examme m Section 7 .2.) The ste ong response Yss is the remaining part of the system response after all transients have die:dY·state it is the time limit outand Yss

=

lim y(t)

t-+OO

(7.1.JJ)

The Yss is also known as a final value of the output. A transient response can be excited b initial disturbanc es or by the applicatio n of an input. A steady-sta te response, however, ca~ only be excited by an input as it physically represents a balance between input energy and output energy. In other words, a steady-sta te response is contained in the forced response YF(t) . It should be pointed out that a steady-sta te response may not exist if the system is unstable or if the input is unbounde d. Figure 7 .1 .1 shows three cases of system responses: (a) response y settles to a steady-sta te value Yss after the transients die out; (b) response y has no steady state because the limit lim y(t) does not exist; and (c) response y , being unbounded, has no (-+ 00 steady state.

y Transient response

Yss

~ - - - -- - - - - - - - - t (a)

y

y

L __

(b)

_ __ _ _ _ _ _

(c)

Figure 7.1.1 Three cases of system response (see text for details)

f

7.1

System Response Analysis in the Time Domain

507

mple 7.1.2 Exa 'd a system with response y governed by CoOSI

er

.Y + oy = ro , with y(O) = Yo > oand ro is a constant. The solution of the differential equation by a ny method given in where 0 and a =I- T , the system response by Eq. (7 .1.29) 1s t

I _:.=!_ T ro(I - e- ur)dr

y(t) = Te

J0

1

i=

r0 - -t [ -r T -(l -aT)r =-e T TeT---eTI T l -aT

r=O

aT - -t I -ut =ro+ro - - e T ---e ) l - aT ( I - aT On the right-hand side of the previous equation, the first term is the steady-state response,Yss = r0 ; thesecond term is the transient response, which vanishes as time goes by.

Estimation of the Time Constant As shown in Figures 7.1.2 to 7.1.5, the time constant Tof a first-order system characterizes its free and forced responses. Hence, modeling and analysis of a first-order system requires the knowledge of its time constant. The time constant of a first-order system can be estimated

514

7

System Response Analysis

from its response curves. For instance, consider the step response shown in Fig Ure 7 t = T, the system response by Eq . (7.l.24) isy(T) = ro(I - e- ') = 0.632Yss - In othe · 1·6·At Tis the time at which the step response is 63.2% of the steady-state value. Also th r Words, ,

the step response at the initial time is j,(O)

= r0 ..!_, indicating that T = T

e st 0 Pe

_Y ss . Addi ti of y(O) onalJy, the

steady-state error shown in Figure 7. 1.5 can be used to estimate the time constant · 1ndeed ' Eq. (7.1.27) leads to T = ess/a .

y(t)

LYss

------0.632y55 slope= -'o T

- - - -- --1""- - T~ - - - - - - - - - t Figure 7.1.6 Step response at t

= T and slope at t = 0

Example 7.1.5 The output of a first-order system subject to a step input has a final value Yss system response is 1.0. Determine the time constant of the system. Solution By Eq. (7.1.24),

where Yss

= ro has been used. Thus, 2

-e T

=

It follows that the time constant of the system is -2 T

I

l - 2.5

=

0.6

= ln(0.6 ) = 3.915

= 2.5. At t =-:

he

7.1

f

System Response Analysis in the Time Domain

515

Time Responses of Second-Order Systems 7,1 .3

nd-order systems are seen in many applications, including spring- mass-damper sys. . .h f Seco tating sha ts wit . tors1ona1 sprmgs and dampers ' RCL circuits' direct current (DC) terns, ro motors, liquid-level systems, th~rmal systems, and first-order systems with feedback control. A typical second-order system 1s governed by the differential equa tion

ay

+ by(t) + cy(t ) = f(t)

(7.1.30)

with a > 0, b 2::: 0, c > 0, where/(t) and y(t) are the system input and output, respectively. for convenience in analysis, Eq. (7.1.30) is converted to a standard form

y + 2?wny (t) + w;y(t) = w~r(t)

(7. 1.31)

where wn and 0) under a sinusoidal input is described by

y + 2i;m,,y(t) + m~(t) = m~ r0 sin mt y(0) = 0, y(0) = 0

(7. l. 77)

From Section 2.6, it can be shown that the forced response is of the form

y(t) = roH sin(mt - 0)

+ e-ur----

/ I

I

II : I /" : I I f I I : I ; I : I

0.6 0.4 0.2

/

:

/ ,.

- - - - a= 0.6 ············· a= 1.2 - · - · - a=6 - - - - a= 12

f, I : I

: I

//

- - a=oo

I

0

0

2

3

4

5

6

7

8

9

10

:~:: 7.1.21 Step response of the third-ord er system in Example 7.1.15: a= 0.6, 1.2, 6, 12,

7 .2

Stability Analysis . . and design of dynamic sys tems, especial In analysis · ly feedback control systems stability · · ,s· . .. d importan t ·. The sta b1Tity of a system is related to the bounde dness of its response to essentia lly ' m1t1a1 an external disturbances· A n unstable system generally has an unboun ded response.

7.2 Stability Analysis

545

unbounded response makes it difficult to achieve desired system performance, can f h d f: ·1 . an uch S atfuncuon an a, ure O t e system, and in a worst-case scenario even leads to . . I d' use 111phic consequences, me u mg property damage and loss of human life. Therefore, m ca . fd . tastro ynamtc systems, assurance of stability is a must. ca . nd operation

°

design a

stability Definitions

1.2.1

this text, we shall only consider stability analysis of linear time-invariant systems. :ccording to Section 7. I, the total response of a linear time-invariant system is the sum of its free response and forced response; namely,

y(t)

=y,(t) +yF(t)

. Stability can be defin~~ by eith~~ the free response or the forced response. ts first, consider stab1hty defimttons by forced responses. A system is stable if its response bounded in magnitude to every bounded input. A system is unstable if there is a bounded input that yields an unbounded output. These are stability definitions in the sense of bounded input-bounded output (BIBO). for a bounded input, say lr(t)l -5: M < oo, Eq. (7.1.11) gives I

I

I

lYF(t)l -5: jg(t - r)llr(r)ldr 5,

MI lg(t - r)ldr 0

0

where g(t) is the impulse response function of the system. This implies that the system is co

BIBO stable if f lg(t)jdr is finite. As an example, consider an underdamped second-order 0

system with the impulse response

which is given in Eq. (7.l.52b). Because

I

lg(t)ldr 5,

0

bfe-u co

co

1

1-