Ministry of Transport and Telecommunications of Ukraine
State Department of Communications
A.S. Popov Odessa National Academy of Telecommunications
Electric circuits department
ELECTRIC CIRCUITS AND SIGNALS
Unit 1
DC and AC circuit analysis
Parts 1 and 2
Learner’s guide for bachelors
Telecommunications
Telecommunication Systems
Information Systems
Odessa 2010
2
УДК 621372
План УМИ 2009 г.
N.F. Arbuznikova, A.A. Novikov, A.U. Kalashnikov, A.V. Shkulipa AC and DC
circuit analysis: Learner’s guide to “Electric Circuits and Signals”. – Odessa: PC
ONAT, 2009. – P. 1 and pp. 2 – 108 р.
The learner’s guide consists of two parts. First one gives a brief description of
theoretical material included in module 1, such as basis of DC and AC circuit
analysis. Second part contains guidelines for the laboratory works. Appendixes
provided also contain test questions and examples of problem solutions. Translated
from the latest edition that includes changes made in 2008.
Translated by A.O. Moga and V.N. Isarev
Editor of the translation E.A. Radius
APPROVED
to be published
at Department Meeting
Protocol № 9
17.04.09
3
TABLE OF CONTENTS
Introduction. Course summary……………………………………………………...
PART 1 DC and AC circuit analysis…………………………………………….....
Contents of unit 1…………………………………………………………………...
1 Basic concepts and laws of electric circuit theory……………………………...
1.1
Characteristics of electric current………………………………………
1.2
Classification of electric circuits……………………………………….
1.3
Ideal non-autonomous two-terminal networks…………………………
1.4
Ideal autonomous networks…………………………………………....
1.5
Kirchhoff’s Laws………………………………………………………
1.6
Substitution circuits…………………………………………………....
1.7
Substitution circuits with dependent sources…………………………..
1.8
Circuit topology………………………………………………………...
1.9
Basic concepts and theorems of circuit theory………………………....
1.10 Equivalent transformations of single-type elements……………………
2 Analysis of linear DC circuits……………………………………………….....
2.1
Equivalent circuits……………………………………………………...
2.2
Kirchhoff’s rules approach………………………………………….....
2.3
Node-voltage method…………………………………………………...
2.4
Superposition method…………………………………………………..
2.5
Thevenin’s and Norton’s theorems……………………………………..
2.6
Mesh-current method…………………………………………………...
3 AC circuit analysis……………………………………………………………..
3.1
Classification of electric signals………………………………………..
3.2
Harmonic oscillations. Basic concepts and definitions…………………
3.3
Spectral (frequency) representation of the harmonic oscillation…….....
3.4
Passive elements in AC circuits………………………………………...
3.5
Passive elements in AC circuits………………………………………...
4 Symbolic AC analysis………………………………………………………….
4.1
Representation of harmonic oscillations via complex numbers………..
4.2
Equations of passive elements in complex form……………………….
4.3
Basic laws of circuit theory in complex form…………………………..
4.4
Nodal equations for complex amplitudes……………………………….
4.5
Matrix methods of circuit analysis……………………………………...
4.6
Analysis of circuits with coupled inductors…………………………….
4.6.1 Magnetically coupled circuits…………………………………...
4.6.2 Series inductors with magnetic………………………………….
4.6.3 Some features of analysis of magnetically coupled circuits.
Decoupling inductors…………………………………………………...
4.7
Power balance…………………………………………………………..
4.8
Source operating modes and conditions for maximum power transfer...
4.9
Signal energy characteristics……………………………………………
4.10 Procedure of symbolic analysis of linear electric circuits………………
5
6
6
7
7
9
10
13
14
15
16
16
18
22
25
25
25
25
28
30
32
33
33
35
37
38
42
45
45
46
48
49
49
55
55
57
59
60
62
64
65
4
5
Non-linear DC circuit analysis………………………………………….
5.1
Elements and their characteristics………………………………………
5.2
VAC approximation…………………………………………………….
5.3
Methods of non-linear circuit analysis………………………………….
References…………………………………………………………………………..
PART 2 Guidelines for the laboratory works………………………………………
Laboratory work № 1.1 Analysis of autonomous two-terminal networks…………
Laboratory work № 1.2 Analysis of resistive circuits with two sources…………...
Laboratory work № 1.3 Analysis of linear AC circuits…………………………….
Laboratory work № 1.4 Input function analysis (RL, RC, RLC circuits)………….
APPENDIX A Test questions to laboratory works…. ……………………………..
APPENDIX B Examples of solutions of module 1 tasks…………………………..
67
67
68
69
72
73
73
75
77
79
85
90
5
INTRODUCTION
Course summary
Discipline Electric Circuits and Signals (ECS) is a base course given to all
technical specialty departments in Odessa National Academy of Telecommunication
named after O.С. Popov. Purpose of the course is to lay down the foundations for the
subsequent special courses, related to systems of telecommunications, networks,
radio communications.
For the successful mastering of this course one needs to have knowledge of
higher mathematics, physics and informatics. Purpose of the course is to give
understanding of basic ideas, principles and methods of analysis and synthesis of
electric circuits and signals.
Development of modern information technologies requires considering analog
circuits and signals along with digital ones, which is why foundations of discrete
systems and signals are included in the discipline.
Actual content of the course has been slightly changed and supplemented. It
takes into account a new credit-module system of education. The course is divided
into four units. Each unit presents a few topics including lectures and laboratorypractical work. Each unit ends with a test or exam.
First Unit: DC and AC Circuit Analysis contains basic concepts and laws of
electric circuits, analysis of linear DC and AC electrical circuits, analysis of nonlinear
DC electric circuits.
Second Unit: Circuit Analysis in Frequency Domain is devoted to circuit
description via complex functions, phenomena of resonance in electric circuits,
analysis and synthesis of reactive two-terminal networks, frequency analysis (Fourier
transform).
Third Unit: Laplace Transform in Circuit Analysis and Circuit Analysis in Time
Domain contains transient circuit analysis, operator (Laplace) description of circuits,
Laplace transform and its properties, relationship between frequency and time
characteristics of circuits.
Fourth Unit: Analog and Digital Signal Transforms contains nonlinear AC
circuit analysis, frequency-selective circuits description, description of linear
distortions of signals, analog-to digital conversion and synthesis of digital circuits.
6
PART 1
DC and AC circuit analysis
Contents of unit 1
Type of lessons
Number of hours
Lc
16
Pr
8
Lb
8
Self
43
Total
75
Introduction. The unit contains history of development of electrical
engineering, radio engineering, telecommunications as well as scientists who made a
considerable contribution into development of circuit theory. Also the purpose of the
course, its role in basic technical courses; its connection to other technical disciplines.
1 The basic concepts and laws of electric circuit theory
Basic definitions: circuit, network, equivalent circuit.
Classification of electrical circuits. Independent, dependent sources.
Components, their description. Ohm’s and Kirchhoff’s Laws.
Basic principles and theorems of electric circuits.
Network topology. Equivalent transformations of single-type elements.
2 Linear DC circuit analysis. Approaches.
3 Linear AC electric circuits. Properties of R, L, C elements. Time-domain
diagrams, vector diagrams, frequency dependences.
4 Symbolic AC circuit analysis. Ohm’s and Kirchhoff’s Laws for complex
amplitudes. Analysis of circuits with coupled inductors. Concept of complex power.
Maximum power transfer.
5 Nonlinear DC circuit analysis.
Laboratory works.
1.1 Analysis of autonomous two-terminal networks.
1.2 Analysis of resistive circuits with two sources.
1.3 Linear AC circuits analysis.
1.4 Input function analysis (RL, RC, RLC circuits).
Complex task.
1 Analysis of resistive circuits with two sources (superposition method, nodevoltage method).
2 RLC circuit analysis. Symbolic method.
3 Analysis of circuits with nonlinear resistive elements
The offered manual contains brief lecture notes, guideline for the laboratory
works, examples of tasks and theirs solutions, quiz for each topic, references.
7
1 BASIC CONCEPTS AND LAWS
OF ELECTRICAL CIRCUIT THEORY
An electric circuit is a set of electrical devices which, under certain conditions,
allow current to flow. These are resistors, inductors, transformers, capacitors,
machine generators, batteries, electron tubes, transistors, wires, aerials, etc.
1.1 Characteristics of electric current
Electric current is a flow of electric charge. To describe it the following values
are used in circuit theory: current (strength), voltage, power and energy.
Current strength. Current strength i is total amount of electric charge which
has passed through a wire cross-section per unit of time.
According to SI current is measured in Amps (Ampere, A), electric charge in
coulombs (C) or ampere-seconds (A-s), time in seconds (s).
Current at some moment t can be defined as
i = dq / dt,
(1.1)
where dq is infinitesimally small amount of charge
1
passing through a wire cross-section per infinitesimally
+
i
short time interval lasting from t to t + dt. In other words,
current is time derivative of charge.
u
Current direction is conventionally considered to be
direction of positive charge moving. Despite being a
scalar value the current is also denoted by an arrow
– 2
(Fig. 1.1).
Unidirectional (usually constant) current is known as Figure 1.1 – An element
of a circuit
direct current (DC); it’s denoted by letter I; the current
that periodically reverses its direction is alternating current (AC). Value of current at
any given moment is instantaneous current denoted as a function i(t).
Voltage. Voltage between points 1 and 2 is a value u that equals to energy
required to transmit single positive charge between points 1 and 2.
According to SI voltage is measured in Volts (V), work or energy in Joules (J).
Voltage is defined as charge derivative of energy, i.e.
u = dw / dq.
(1.2)
Voltage can possess positive values as well as negative ones. Being measured
from point 1 to point 2 (u12), the voltage is assumed as a positive value in case
positive charge transmission in that direction consumes energy.
If voltage is constant, then it is called direct voltage and denoted by U. Harmonic
voltage is known as alternating voltage which is characterized by instantaneous
values u(t).
8
We assume that all the values which do not depend on time are denoted by
capital letters (U, I), whereas all the time-dependent ones are denoted by small letters
[u(t), i(t)].
Conventionally positive directions of current and voltage. Before starting
analyzing circuits you should choose and denote conventionally positive directions of
currents and voltages being calculated. It can be done randomly. Usually voltage
across all elements (except sources) is chosen co-directional with current.
Positive current direction is shown by an arrow, positive voltage direction is
shown by an arrow pointing to “minus”.
Having calculated desired values, you can determine actual directions of voltage
and current. In case obtained value is positive, then its actual direction coincides with
chosen one.
If you get a negative value, than “real” current direction is the opposite to the
one chosen. The same is true for voltage.
Instantaneous power and energy. Power is a physical value, which equals to
the work performed per some unit of time. In the international system of SI electric
power is measured in Watts (W). In general, if at regular time intervals unequal
amount of work is performed, the concept of instantaneous power is used.
Instantaneous power р is defined as a rate of change of energy in time, that is to say
time derivative energy w:
p = dw / dt.
(1.3)
DC power can be easily determined basing on the following considerations: to
transfer a single charge, a circuit performs work equal to voltage U. The amount of
charge transferred during one second is equal to I. To transfer it the energy of UI is
needed. Thus, power
P = UI .
(1.4)
Thus, DC power is defined as product of voltage and current. Accordingly,
energy for the period of time from t1 to t2 can be defined as
w = UI (t 2 − t1 ).
(1.5)
As a unit of electric energy watt-second (W-s) is used. It is equal to one joule (J).
AC instantaneous power is also simply defined by multiplying equations (1.1) and
(1.2) and also taking into account formula (1.3):
p = ui.
Thus, instantaneous power is product of instantaneous voltage and current.
During time interval lasting from t1 to t2 total amount of energy is
(1.6)
9
t2
t2
w = ∫ pdt = ∫ uidt.
t1
(1.7)
t1
Note that the values of p and w can be both positive and negative. According to
formulae (1.4)...(1.7), power and energy are positive in case voltage and current have
the same signs and negative otherwise. Positive energy (power) means that the circuit
section being considered consumes energy, whereas negative value means that the
section is producing it.
1.2 Classification of electric components
When analyzed, real electric circuits are replaced with a combination of ideal
elements, each of which is attributed to certain properties. Ideal elements are
connected to each other by terminals, also known as nodes. Minimum number of
terminals is two; such elements are called two-terminal networks and the current
entering one node of such an element equal to the current leaving the other node.
Equivalents of real resistors, capacitors, inductors, one-port generators etc. are twoterminal networks. Graphically a two-terminal network is denoted in the same way as
a circuit section (see Fig. 1.1).
There are circuits which connect to other ones with three, four etc. terminals.
They are respectively called three-terminal, four-terminal networks etc. For example
electronic tubes, transistors can be classified as three-terminal networks. It is also
possible to consider them to be quadripoles (four-terminal networks) by representing
a common terminal as two, connected together.
Generally speaking, circuits which connects to other ones with N terminals are
N-terminal or multiterminal networks.
Obviously, the simplest N-terminal circuit is a two-terminal one.
It is fully described by the relationship between the voltage across its terminals
and the current flowing through it. The voltage-current relations defines the
component as linear or non-linear one.
Linear is one with a linear relationship between voltage and current. Otherwise
the element is non-linear.
Two-terminal networks can also by classified by a number of features. Here we
shall use energy criteria only. Thus, there are autonomous and non-autonomous
networks distinguished.
Non-autonomous is a network which, when disconnected from a power source,
does not produce voltage across its terminals, if open, or current through them, if
closed.
Autonomous is a network which can independently from outer sources produce
voltage or current. In other words, an autonomous network is a circuit containing
independent power sources.
Non-autonomous networks, in turn, are divided into passive and active ones.
Passive is a non-autonomous network whose total energy over time from t = 0 to
t → ∞ is non-negative:
10
∞
w = ∫ u (t ) i (t )dt ≥ 0.
(1.8)
0
Thus, a passive network consumes energy received from outer circuits (if any)
but does not produce it.
Active is a network whose total amount of energy over time of existence from
t = 0 to t → ∞ is negativeо, i.e.
∞
w = ∫ u (t ) i (t )dt < 0.
(1.9)
0
Such an element gives energy to a circuit in the same way as the autonomous
networks do. Still, there is a significant difference between them. The autonomous
networks produce voltage across theirs terminals, even if extracted from a circuit,
whereas an active network will generate voltage only if connected to the rest of the
circuit.
Active network could be an amplifier with common input and output, so that
amount of energy given to the outer circuit is greater than the amount of energy
consumed (due to some external power source).
1.3 Ideal non-autonomous elements
They are:
1) Resistance and Conductance;
2) Inductance and Inverse Inductance;
3) Capacitance and Elastance (reverse capacitance).
Resistance and conductance. Resistive element or, simply, resistance is a twoterminal network which dissipates electromagnetic energy and can be described by
equation
u (t ) = R i (t ),
(1.10)
the value R does not depend on voltage or current. The value R is itself
resistance.
The real component approaching this description is so called resistor. The device
turns electromagnetic energy into heat.
Resistive element is denoted by a symbol in Fig. 1.2, a. The arrows show
positive direction of current and voltage, which are the same for both values.
Equation (1.10) can be rewritten in another form:
i (t ) = G u (t ),
(1.11)
G = 1/ R.
(1.12)
where
11
By characterizing the network with expression (1.11) instead of (1.10) we define
conductive element or conductance. Value G is called conductance.
The SI unit of electrical resistance is ohm, symbol Ω, the SI unit of conductance
is siemens (S).
Considering that conductance and
i
i
i
R
L
С
resistance reflect the same properties of
u
u
u
some physical object, the conductive
G
Г
S
element is denoted with the same symbol as
resistive one (see Fig. 1.2, a).
б)
в)
а)
For example, over the time interval
Figure 1.2 – Passive elements of
from 0 to t resistive (conductive) element
electric circuits
receives or emits energy equal to
t
t
t
w = ∫ u (t ) ⋅ i (t )dt = R ∫ i (t )dt = G ∫ u 2 (t )dt.
2
0
0
(1.13)
0
In case of DC current u(t) = U, i(t) = I, energy is
w = R I 2 t = G U 2 t.
(1.14)
Resistance R can be either positive or negative (the same is true for
conductance).
As you may see from expression (1.13), the elements with positive resistance
(conductance) are characterized by positive sign of energy. Such elements are, thus,
passive, receiving energy from outer circuit. As these elements are obtained by the
idealization of resistor, they are used to estimate the irreversible energy loss (such as
heat dissipation, emission of radiation, etc).
The elements with the same negative resistance or conductance are characterized
by negative sign of energy. Thus, these elements are active, giving energy to outer
circuit.
Inductance and inverse inductance. An inductive element is a two-terminal
network whose voltage across and current through are related by the expression
u L (t ) = L
diL
,
dt
(1.15)
here L depends neither on current, nor on voltage. Inductance L is an element
that can store magnetic energy.
At present day there are methods invented to produce inductive elements using
transistors, resistors and capacitors only.
Graphically the inductive element is denoted by the symbol in Fig. 1.12, b. The
arrows show positive directions of current and voltage, which are assumed to be the
same for both values.
12
By rewriting expression (1.15) in the following way
iL (t ) = Г ∫ u L (t )dt ,
where
Г = 1 / L,
(1.16)
we derive an element known as inverse inductance. It is denoted by the symbol Г.
The SI unit for inductance is Henry (H), for inverse inductance is 1/Henry (1/H).
Given that the inductive and inverse inductive elements reflect the same
properties of some real object, the elements are denoted by the same symbol
(see Fig. 1.2, b).
Over the time from – ∞ to t an inductive (inverse inductive) element either
receives, or returns energy
Li 2 (t ) i 2 (t )
=
.
w = ∫ u (t ) i (t )dt =
2
2
Г
−∞
t
(1.17)*
It is assumed that when t → – ∞ current is zero. This formula indicates that total
energy of an inductive element depends on instantaneous value of current only and
does not depend on its previous states. Inductance (inverse inductance) can be either
positive, or negative.
Capacitance and elastance. A capacitive element (capacitance) is a network
whose voltage and current are related by formula below
iC (t ) = C
du C
,
dt
(1.18)
the value С does not depend on both voltage and current. The value C is known
as capacitance. Capacitance is the element that can store electrical energy.
A capacitor would be a good example of a device maximally approaching the
properties of capacitance. In a capacitor energy is stored mainly in the form of
electric field.
Graphically a capacitive element is denoted by the symbol shown in Fig. 1.2, c.
Here, as in above, the positive directions of current and voltage coincide.
By transforming equation (1.18) into expression below
uC (t ) = S ∫ iC (t )dt ,
(1.19)
where
S = 1 / C,
(1.20)
we obtain an element known as elastance (inverse capacitance). It is denoted by the
letter S.
* Prove (1.17) using equation (1.15)
13
The SI unit for capacitance is Farad (F), for elastance is 1/Farad (1/F).
As the capacitance and elastance describe the same properties of some real
object, they share the same graphical symbol (see Fig. 1.2, c).
Total amount of energy that a capacitive element receives or returns over the
time from – ∞ to t is
Cu 2 (t ) u 2 (t )
=
w = ∫ u (t ) i (t )dt =
.
2
2S
−∞
t
(1.21)**
It is assumed that when t → – ∞ the voltage is zero. As you may see, total
energy stored in capacitance depends on the instantaneous voltage only and does not
depend on its previous states. Capacitance (elastance) can be either positive, or
negative.
1.4 Ideal autonomous elements
There are two types of ideal autonomous sources distinguished: ideal voltage
source and ideal current source.
Ideal voltage source. The ideal voltage source is an autonomous two-terminal
network whose voltage across its terminals does not depend on the current through
them
u (t ) = e(t ).
(1.22)
The value e(t) is called driving voltage or source voltage. If to connect a resistive
element to a voltage source, then the current through the resistor can be estimated as
i (t ) = e (t ) / R ,
e(t)
expression above is derived from (1.10). While
resistance is decreasing, the current will be
increasing. In the extreme case of near zero
Figure 1.3 – Ideal voltage
resistance the current becomes infinitely great. Yet,
source
the voltage across the resistor remains equal to е(t).
This is why the source must not operate in such a
mode. Graphically voltage source is denoted by a symbol in Fig. 1.3.
Real device approaching properties of an ideal voltage source is a starter acid
accumulator, whose inner resistance can reach 0,01 Ω.
Ideal current source. An ideal current source is an autonomous network,
whose current does not depend on the voltage across its terminals, i.e.
** Prove (1.21) using equation (1.18)
14
i(t ) = j (t ).
(1.23)
The value j(t) is called driving current or source current. If to connect a
conductive element G to a current source j(t), then the voltage across the load can be
expressed as
u (t ) = j (t ) / G,
formula above is derived from (1.11). The lower conductance, the higher voltage
across it. In the extreme case of near zero conductance, the voltage tends to become
infinitely high. Yet, the current in such a circuit remains unchanged: j(t). It is
essential to prevent the source from running in such a mode.
The real devices approaching the properties of an
ideal current source are, for example, alkaline
accumulator and electrostatic generator. Ideal current
j(t)
source as well as ideal voltage source are used to
estimate the amount of energy delivered to a circuit.
Graphically a current source is denoted by the symbol
Figure 1.4 – Ideal current
in Fig. 1.4.
source
1.5 Kirchhoff’s Laws
If a node connects only two elements, then the connection is called series.
If a few elements are joined to each other by matching terminals, then the
connection is called parallel.
Branch is an element or set of elements between two nodes in case the current
leaving one nodes equal to the current entering the other one. Set of elements forming
a closed path for current to flow is called loop.
If a circuit consists of linear elements only, then it is linear. It is non-linear
otherwise.
Kirchhoff’s circuit laws. Kirchhoff’s Current Law states: the algebraic sum of
current into any node of a circuit is zero
∑ i(t ) = 0.
(1.24)
Obviously, the expression includes driving currents into the node (if any). When
writing this equation, the currents leaving the node are considered to be negative
while the currents entering it are positive.
Kirchhoff’s Voltage Law states: the algebraic sum of voltages in any loop must
be zero. The statement is true for both linear and non-linear circuits
∑ u (t ) = 0.
(1.25)
15
Obviously, source voltages are also included into the equation (if there are
some). To write the equation correctly, one needs to take into account voltage
polarity: if it coincides with chosen path-tracing, then the voltage is considered to be
positive, it is negative otherwise.
1.6 Substitution circuits
ICS
Circuit theory deals with idealized circuits, not real ones. Thus, to analyze a real
circuit it is necessary to replace its components with some sets of ideal elements
approaching characteristics of the original devices. Such a substitution is called a
substitution circuit of a device.
When choosing a substitution circuit, it is essential to obtain the relationships
between current and voltage at the circuit terminals meeting the real ones, obtained
by measurements, with the required degree of accuracy. The higher accuracy of a
substitution circuit is required, the more complex the circuit would be. Thus, the
same real object can be replaced by different equivalents depending on the
requirements given. All in all, the validity of a substitution can be proven by the
experience only.
We shall consider possible substitution for
u
some electrical components often used in practice.
IVS
E
Substitutions for autonomous two-terminal
networks (non-ideal sources). Figure 1.5 shows a
volt-amp characteristic of a non-ideal source,
obtained by measurements. There E is the source
voltage in idling mode, J is the current flowing
i
J
0
through the source’s short-circuited terminals. The
Figure 1.5 – Volt-ampere
dashed lines show supposed VAC of ideal sources
characteristics
(horizontal line corresponds to an ideal voltage
source, vertical one to an ideal current source).
The line through points E and J (see Fig. 1.5) can be described by equation
below
U 12 = E − R i.
(1.26)
The formula corresponds to circuit in Fig. 1.6, a.
Current i can be expressed from equation (1.26) as
i=
E − u12 E U12
u
= −
= J − 12 = J − u12G ,
R
R
R
R
E
1
, G = . Thus deduced, the equation corresponds to another possible
R
R
substitution circuit (Fig. 1.6, b). Good examples of non-ideal sources would be an
here J =
16
accumulator battery, a DC generator, a galvanic cell. R and G in Fig. 1.6 are load
elements, as current does not flow in an open circuit.
1
R E
+
i
i
2
1
u12
RL
J
G = 1/R U12
GL
i
2
a)
b)
Figure 1.6 – Substitutions of non-ideal sources:
а – voltage source; b – current source
1.7 Substitutions of dependent sources
Dependent (controllable) sources. Until now we have considered substitution
circuits made of two-terminal elements. We have also assumed that a source is a
device whose voltage or current do not depend on other voltages and currents in a
circuit, containing the source. Such a source is called independent.
There are, however, devices, such as electronic tubes or transistors, whose
substitution circuits cannot be two-terminal. They need to have ideal components
with three and more terminals. Among them the dependent (controllable) sources are
most widely used. These are four-terminal networks whose first port is either shortcircuit, or open-circuit, whereas the other port contains either voltage, or current
source. The driving voltage (current)
i1
+
+
u1
µu1
ri1 depends on voltage (current) at first port.
According to what has been said,
there are four types of dependent
a)
b)
sources: VCVS – voltage-controlled
voltage source (Fig. 1.7, a); CCVS –
i1
u1
gu1
αi1 current-controlled voltage source (Fig.
1.7, b); VCCS – voltage-controlled
c)
d)
current source (Fig. 1.7, c); CCCS –
current-controlled current source (Fig.
Figure 1.7 – Controllable sources:
1.7, d).
а – VCVS; b – CCVS;
Proportionality coefficients r and q
c – VCCS; d – CCCS
are respectively measured in Ω and S,
whereas coefficients α and µ are unitless values.
1.8 Circuit topology
Topology is a part of mathematics that deals with properties of figures and
geometric images remaining unchanged while the figures are being deformed.
17
By ignoring the nature of elements of a circuit and replacing them by line
segments, we obtain some geometric figure.
L
R3
N1
N2
R1
+
e(t )
C2
C1
a)
R2
C1
R2 C2
R1
R4
R5
N3
b)
Figure 1.8 – A circuit and its graph:
а – electric circuit; b – graph of the electric circuit
The figure is known as circuit graph (Fig. 1.8, а). We shall define the following
terms.
Graph node is an end point of a line segment or a separated point.
Rib is a line segment ending by a node.
Graph is a set of nodes and ribs, where ribs join at nodes. Each rib must end at a
node.
The graph is directed in case each rib is assigned to some direction shown by an
arrow (current graph in Fig. 1.8, b). Undirected otherwise.
Subgraph. It is a part of a graph, where each node and rib corresponds to
respective nodes and ribs of the original graph. For example, a single separated node
or a single rib, starting and ending by nodes, could be a subgraph.
Graph path is a sequence of m ribs and m + 1 nodes, in which each node, except
the first and the last ones, is common for two neighboring ribs. For example you can
find the following paths between В1 and В3 in graph in Fig. 1.8, b: R1, R3 or just R4,
etc. Nodes В1 and В3 are the first and the last nodes for whichever the path you have
chosen.
Cycle is a closed-loop path. For instance you can find the following cycles in
Fig. 1.8: N1–N2, N2–N3–N4, etc, which are shortly denoted as C1, C2, etc.
In directed graphs cycles like ribs are assigned to direction (clockwise or
counterclockwise), which is indicated by a round arrow (Fig. 1.8)
Planar graph is one which can be drawn on a plane or sphere in such a way that
no two ribs intersect each other, except in the nodes. Otherwise the graph is
nonplanar. For example graphs in Fig. 1.8, b and 1.9, a are planar, whereas
Fig. 1.9, b represents nonplanar one.
A graph is called connected if there is a path between any couple of nodes.
Tree is a connected subgraph containing all node but no loops (Fig. 1.9, c).
18
R6
1
N1
R4
R1
R2
N4
R3
N2
2
4
R5
N3
а)
3
b)
c)
Figure 1.9 – Network graphs:
a – planar graph; b – nonplanar graph; c – graph tree
1.9 Basic principles and theorems of circuit theory
Superposition principle. The following statement is valid for linear circuits:
The complex response at a given part of the circuit caused by two and more sources
is the sum of the responses which would have been caused by each source
individually. In other words, linear circuit’s response to individual stimulus can be
added independently.
The validness of this principle follows from the properties of systems of linear
equations: if an equation set contains a few different functions, then general solution
of the system can be determined by summing partial solutions for each function,
assuming that only one function exists while the others are zeroes.
Duality theorem. The conception of duality of actions and phenomena in nature
was developed long ago. At the end of XIX century, when it was only the beginning
of electricity, this conception drew attention of English physicist O. Heaviside. This
had led to development of duality principle. When applied to an electric circuit, the
principle uses Kirchhoff’s laws (Table 1.1). Significant development of the principle,
however, was only in mid-twentieth century. A great contribution to its development
was made by scientists of Odessa National Academy of Telecommunications
E.V. Zeliah, A.M. Ivanitskiy.
The duality principle can be stated as: for any theorem there is a dual theorem in
which all the values replace original ones in such a manner that the theorem remains
valid. The correct substitution can be found in Table 1.1, where couples of dual
values are given. For example, current trough one circuit (element) corresponds to
voltage across dual circuit (element) and vice versa.
According to duality principle there is always some dual law, method
corresponding to another law or method. The last raw in Table 1.1 shows couples of
dual networks (series network ↔ parallel network) containing dual elements (L, C, R
↔ C, L, G) and dual connections (loop ↔ node).
19
Table 1.1 – Dual correspondence
Original statement
Dual statement
∑ ik (t ) = 0
∑ u k (t ) = 0
Node
i =G⋅u
Conductance G
du
iC = C C
dt
Capacitance С
1
iL = ∫ u L dt
L
Inductive admittance YL
I&C = jωC ⋅ U& C
Loop
u = R ⋅i
Resistance R
di
uL = L L
dt
Inductance L
1
uC = ∫ iC dt
C
1 &
U& C =
IC
jωC
Capacitive impedance ZC
U& L = jωL ⋅ I&L
Capacitive admittance YС
Inductive impedance ZL
1 &
I&L =
UL
jω L
∑ I&k
∑ U& k
=0
L
G
1
2
3
R
C
L
C
=0
j(t)
1
4
2
3
+
4
e(t)
Dual circuits. Two circuits are called dual if, by replacing values in terms of
dual ones, equations describing one circuit become equations describing the other
one.
There is the following problem of great interest arises: how to obtain a circuit
dual to the original one. It is not difficult to transform a circuit consisting of series
and parallel connections only into a dual one (Fig. 1.10, a). In this case dual circuit
can be obtained by replacing series networks by parallel ones, parallel networks by
series ones, resistances by conductances, inductances by capacitances, capacitances
by inductances, current sources by voltage sources and vice versa.
As an example Fig. 1.10 demonstrates dual circuits
L1 i1 1i2 L2
R1
+
e(t)
1
C
2
a)
1
R2
j(t)
u1 /
u 3/
/
/
L
G1 /
C1
1
/
b)
Figure 1.10 – Dual circuits: а – original; b – dual
2/
C2 /
G2
/
u 2/
20
Substitution theorem. It can be stated as:
If voltage across or current through any bilateral branch of a circuit are known,
this branch can be replaced by any combination of elements that will maintain the
same voltage across and current through the chosen branch. For example
1 Two-terminal network (TT) connected to the circuit C (Fig. 1.11, а) can be
replaced by a voltage source without changing currents and voltages in the circuit.
Chosen source must generate voltage of the same value and polarity as the voltage
across the original branch (Fig. 1.11, b).
2 Two-terminal network (TT) can also be replaced by a current source, which
generates current of the same value and direction as the current through the original
branch (Fig. 1.11, c).
ITT
TT
UTT
+
C
UTT
C
C
ITT
ITT
a)
b)
c)
Figure 1.11 – Substitution of two-terminal network TT:
а – original two-terminal network; b – substitution by a voltage source;
c – substitution by a current source
Proving. Between the two-terminal network TT and the circuit C we shall insert
two sources with voltages of the same value as the voltage across TT but with
different polarity (Fig. 1.12, a). Obviously, the currents and voltages in the circuit
will remain unchanged. The voltage between points 1 and 1/ equals to zero. These
points, therefore, can be short-circuited without causing any change in the circuit.
Redrawing the circuit, as it is shown in Fig. 1.12, b and applying KCL to node 1
we can see that the current through branch 1-1/ is zero, which means that the branch
can be disconnected leaving the circuit parameters unchanged. As a result we obtain
circuit in Fig. 1.11, b, which proves validness of the first part of the theorem.
Sometimes you can come across this theorem as a compensation theorem.
uTT
uTT i
+ TT
1
iTT+
+
TT uTT u = 0
C
iTT 1/
iTT
a)
uTT
uTT
+
iTT
/
1
iTT
+
iTT
TT
1
iTT
b)
Figure 1.12 – Proving substitution theorem:
а – two identical opposite-polarity sources in one branch;
b – two autonomous circuits
C
21
Thevenin’s theorem. Norton’s theorem.
1 Thevenin’s theorem states that any linear circuit (Fig. 1.13, a), no matter
how complex it is, can be simplified to an equivalent circuit with just a single voltage
source and a series two-terminal network (Thevenin equivalent) Fig. 1.13, b,
connected to a load L. Source voltage must be equal to open-circuit voltage of the
simplified circuit.
2 Norton’s theorem states that any linear circuit (Fig. 1.13, a), no matter how
complex it is, can be simplified to an equivalent circuit with just a single current
source and a parallel two-terminal network (Norton equivalent) Fig. 1.13, c,
connected to a load L. Source current must be equal to short-circuit current of the
simplified circuit.
uoc
+
i
АTT
u
а)
L
TT1
b)
isc
TT1
c)
Figure 1.13 – Thevenin’s and Norton’s theorems:
а – autonomous two-terminal network with a load;
b – equivalent voltage source; c – equivalent current source
In accordance with what has been said above, Thevenin or Norton equivalents
are non-autonomous two-terminal networks which can be obtained from the original
circuit by short-circuiting voltage sources and open-circuiting current sources.
Open-circuit voltage of an autonomous two-terminal network is voltage across
its terminals when a load is disconnected, i.e. the current trough the terminals is zero.
Short-circuit current of an autonomous two-terminal network is the current
through its short-circuited terminals.
Reciprocity theorem. Reciprocity is the feature that all the passive linear
elements possess. It is proven by Reciprocity Theorem. The proof itself you will find
in respective books [4, 5, 6].
There are two ways to formulate the theorem.
The first one concerns voltage source: if a voltage source E acting in one branch
of a passive circuit (Fig. 1.14, a) causes a current I to flow in another branch of the
circuit, then the same voltage source acting in the second branch (Fig. 1.14, b) would
cause an identical current I to flow in the first branch.
Since the currents I2 = I1, the circuit is reciprocal. Nonreciprocal otherwise.
The second formulation is dual and concerns a current source rather than a
voltage one. Instead of current, the open-circuit voltage across another branch is
considered. This makes the circuit analysis easier. For further information look over
the chapter Matrix methods of circuit analysis.
22
+
Е1
PEC
I2
I1
+
PEC
Е2
a)
b)
Figure 1.14 – Passive electric circuit with a voltage source:
а – in the first branch; b – in the second branch
1.10 Equivalent transformations of single-type elements
А Series connection of single-type elements
Series resistances. Connection of elements shown in Fig. 1.14 is called series
(all elements are connected via simple nodes).
Uab
R1
a+
R2
i
Rn
R3
_
b
u1
u2
u3
un
Figure 1.14 – Series resistances R
Voltage across each resistance in Fig. 1.14 is defined by Ohm’s law:
u3 = R3i etc;
u1 = R1i; u2 = R2i;
un = Rni.
Total voltage across points a and b is an arithmetical sum of individual voltages:
uab = R1i + R2i + R3i +…+ Rni = i(R1+ R2 + R3+…+ Rn) = iRt ,
а
i
Re
where Rt is total (equivalent) resistance of the
“ab” section (Fig. 1.15), which is equal to
arithmetical sum of all resistances between
terminals a and b.
Series inductances are shown in Fig. 1.16.
b
uab
Figure 1.15 – Total resistance
+
a
L1
u1
i
L2
L3
Ln
u2
u3
un
Figure 1.16 – Series inductances
Voltage of each inductance can be determined as
Uab
_
b
23
u1 = L1
а
di
di etc.
; u2 = L2
dt
dt
+
Сe
i
–
b
uab
Figure 1.19 – Equivalent capacitance
Thus, total voltage across ab section
u ab = L1
di
di
di
di di
di
+ L2 + L3 + K + Ln
= (L1 + L2 + L3 + K + Ln ) = Le ,
dt
dt
dt
dt dt
dt
Le
_
here Le – total (equivalent) inductance
+ i
b
а
of аb section (Fig. 1.17), which equals to
uab
arithmetical sum of all inductances between
Figure 1.17 – Equivalent inductance
terminals a and b.
Series capacitances are shown in Fig. 1.18.
Uab
C1
+ i
a
u1
C2
u2
C3
u3
Cn
_
b
un
Figure 1.18 – Series capacitances
Voltage of each capacitance can be determined as
u1 =
1
1
idt ; u 2 =
∫
∫ idt etc.
C1
C2
Total voltage across ab section is described by expression
uab =
1 1 1
1
1
1
1
1 1
idt + ∫ idt + ∫ idt + K+ ∫ idt = ∫ idt + + + K+  = ∫ idt,
∫
C1
C2
C3
Cn
Cn  CE
 C1 C2 C3
Here 1/Сe is value reciprocal if equivalent capacitance of ab section (Fig. 1.19):
1
1
1
1
1 .
=
+
+
+K+
Ce С1 С2 С3
Сn
Ideal series voltage sources. When series connection of independent voltage
sources takes place in a circuit, they can be replaced by a single equivalent source
whose voltage Ee equals to arithmetical sum of individual voltages of sources. One
should be careful while choosing signs: "+" is chosen for sources of the same polarity
as the equivalent one, "–" is chosen otherwise (Fig. 1.20).
24
Example of connection of three sources is shown in Fig. 1.20.
u12
+
1
+ +
Е1
Е2
u12
2
Е3
а)
1
+
2
Ее
b)
Figure 1.20 – Equivalent transformation of ideal voltage sources:
а – series voltage sources; b – equivalent source
According to KVL
u12 = Ee = − E1 − E 2 + E3 .
B Parallel connection of single-type elements
By employing duality principle one can easily obtain expressions for equivalents
of parallel connections of single-type elements. This task is set to be solved yourself.
Key questions
1 What are current and voltage?
2 Which directions of current or voltage are assumed as positive ones?
3 What are power and energy?
4 What are two-, three-, four-terminal networks?
5 What are autonomous and non-autonomous two-terminal networks?
6 What are active and passive two-terminal networks?
7 What are ideal resistance, inductance and capacitance?
8 What are ideal voltage and current sources?
9 Statements of Kirchhoff’s laws
10 Which possible substitutions for two-terminal networks do you know?
11 Which and how many dependent sources could you name?
12 What are node, branch and network?
13 Which principles and theorems are used in electric network theory?
14 What are dual circuits?
15 What is Substitution theorem about?
16 What are Thevenin’s and Norton’s theorems about?
17 How would you determine equivalent of series resistances, inductances,
capacitances?
18 How to determine the equivalent of parallel resistances? Inductances?
Capacitances?
19 Which element can be used to replace series voltage sources?
20 Which element can be used to replace parallel current sources?
25
2 LINEAR DC ANALYSIS
2.1 Equivalent DC circuits
Linear DC analysis is aimed at determining voltage, current and power values at
any section of a circuit given. There are several methods of circuit analysis exist.
When considering DC circuits, it is recommended to replace all the elements by
respective equivalents.
Table 2.1 represents electrical elements and theirs equivalents.
2.2 Kirchhoff’s rules approach
The method applies equations obtained by Kirchoff’s Laws. Branch currents are
the variables in such equations.
Let some circuit consists of nb branches and nn nodes.
We shall denote each branch of the circuit with the assumed current. Then it is
possible to set (nn – 1) equations describing respective nodes in accordance with
KCL. In the same way KVL can be used to set (nb – nn + 1) linearly independent
equations describing voltages across circuit’s branches.
Consequently, combination of (nn – 1) equations based on KCL and (nb – nn + 1)
equations based on KVL form a system of nn – 1+ nb – nn + 1= nb linearly
independent algebraic equations with respect to the same variables.
Such an equation set is known to have a single solution that allows to find all the
branch currents and therefore, voltage values between any couple of nodes.
2.3 Node Voltage Method
R1
The method is based on Kirchhoff’s
I1
and Ohm’s Laws. It is one of the most
widely used approaches in circuit analysis.
E1
The method solves for unknown
R4
R3
I4
I
3
+
voltages at (nn − 1) nodes with respect to a
reference node, whose potential is assumed
u2 0
u30
to be zero. Then, in accordance with Ohm’s E2,
R5
J
u10
Law, branch currents are calculated.
R2
+
Figure 2.1 represents an electric circuit
I5
to be analyzed via the Node Voltage
I2
Method.
Figure 2.1 – Electric circuit with
First, it is necessary to choose an
four nodes
arbitrary node as a reference one. It is
denoted as a zero node. Then the nodes remaining are arbitrarily numbered as 1, 2, 3
so that there are (nn – 1) = 3 independent nodes in the circuit.
Nodal voltages are voltage values between these nodes and the reference one.
They are denoted as u10, u20 and u30 respectively.
Table 2.1 – Equivalents of electrical elements
Non-DC equivalents: u(t), i(t)
Name
Ideal voltage
source
Ideal current
source
Symbol
e(t)
+
i
u
i j(t)
u
Relationships between i
and u
Name
u = e(t ) does not
depend on the value and
direction of the current i
Ideal voltage
source
i = j (t ) does not
depend on the value and
polarity of the voltage u
Ideal current
source
Resistance
u = iR
L
i
u=L
ii
C
C
i=C
Capacitance
uu
i1
L1
1
i 2(∗) L2
3
*
u1
(∗) 2
М
+
–
+
U
I
J
U
R
Resistance
U
u
Coupled
inductances
I E
I
u
Relationships
between I and U
U = E does not
depend on the
value and
direction of the
current I
I = J does not
depend on the
value and polarity
of the voltage U
U = I ⋅R
U
R=
I
Symbol
R
i
Inductance
DC equivalents: u(t)= U, i(t) = I
u2
4
di
dt
Short circuit (SC)
du
dt
Open circuit (OC)
di1
di
±M 2
dt
dt
di1
di
+ L2 2
u2 = ± M
dt
dt
I UL = 0
R=
U
I
IC = 0
R=
U
1
u1 = L1
Short circuit (SC)
R1
2
R2
3
4
U
=0
I
U
→∞
I
R1 = 0
R2 = 0
27
Now we choose arbitrarily directions of branch currents, except from the one
containing the current source, and set up KCL equations describing the independent
nodes (nodes 1, 2, 3):
Node I 3 − I1 − I 2 = 0 ;
Node J − I 3 − I 4 = 0 ;
Node I 4 + I1 − I 5 = 0 .
Defining each of the currents by means of Ohm’s Law:
I1 =
u10 − u 30 − E1
= (u10 − u 30 − E1 )G1 ;
R1
I2 =
u10 + E 2
= (u10 + E 2 )G 2 ;
R2
I3 =
I4 =
u 20 − u10
= (u 20 − u10 )G3 ;
R3
u 20 − u 30
= (u 20 − u 30 )G 4 ;
R4
u
I 5 = 30 = u30G5 ,
R5
where
G1 =
1
1
1
1
1
; G2 =
; G3 =
; G4 =
; G5 =
R1
R2
R3
R4
R5
are conductances of the respective branches.
The currents expressed in terms of nodal voltages and branch conductances now
can be substituted into equation set (2.1)
(u 20 − u10 )G3 − (u10 − u30 )G1 + E1G1 − u10G2 − E2G2 = 0;

 J − (u 20 − u10 )G3 − (u 20 − u30 )G4 = 0;
(u − u )G + (u − u )G − E G − u G = 0.
30
4
10
30
1
1 1
30 5
 20
(2.2)
Transforming equation set (2.2) by grouping summands with the common nodal
voltages, inverting signs and shifting those do not containing voltage variables to the
right part of the equations:
u10G11 − u 20G12 − u30G13 = I11;

− u10G21 + u 20G22 − u30G23 = I 22 ;
− u G − u G + u G = I ,
20 32
30 33
33
 10 31
(2.3)
28
where G11 , G22 and G33 are self-conductances of the respective nodes (first,
second, third one) equal to sum of conductances of those branches joined in the node:
G11 = G1 + G2 + G3 ; G22 = G3 + G4 ; G33 = G1 + G4 + G5 ;
G12 , G 21 , G13 , G31 , G 23 , G32 are mutual conductances between the respective
nodes, i.e. total conductance of branches connecting the nodes:
G12 = G3 , G13 = G1 , G 23 = G 4 , G12 = G21 , G13 = G31 and G23 = G32 ;
I11, I22 and I33 are currents in the respective nodes:
I 11 = E1G1 − E 2 G2 , I 22 = J , I 33 = − E1G1 .
When choosing signs of the components in expressions above pay attention to
the polarity of sources connected to a node. The signs need to be the same with the
source’s potential entering the node.
Generally, if a circuit consists of n independent nodes, then it can be described
by the following equation set
u10G11 − u 20 G12 − u30G13 − K − u n 0 G1n = I11 ;
− u G + u G − u G − K − u G = I ;
 10 21
20 22
30 23
n0 2n
22

KKKKKKKKKKKKKKKKKKK.
− u10Gn1 − u 20Gn 2 − u30 Gn 3 − K + u n 0 Gnn = I nn .
(2.4)
System (2.4) meets certain rules and requirements and is known as canonical
(normal) system.
Thus, by solving system (2.3) with respect to three independent variables, nodal
voltages u10, u20 and u30 are determined.
Current value in each branch is calculated by Ohm’s Law.
In case a current value is negative, then its real direction is opposite to the one
chosen.
2.4 Method of Superposition
The method implies analysis of a complex linear multisource circuit by
considering individual influence of each source.
Example. Given the following values Е1, J, R1, R2, R3 and R4 (Fig. 2.2, а). The
task is to determine all the currents flowing through the resistors.
29
I1 R1
R3
J
+
E1
I1′ R1
I3
I2
R4 E
1
R3
+
R4 sc
R2
а)
R3
I 3′′
J
OC
R2
I1′′ R1
I 3′
I 2′′
R4
R2
b)
c)
Figure 2.2 – Method of Superposition:
a – original circuit; b – current source extracted; c – voltage source extracted.
Solution
We choose conventionally positive current directions I1, I2, I3.
1 Estimate currents caused by the voltage source E1 only (i.e. with the current
source extracted). The current source then must be replaced by its inner resistance
(R→ ∞). In this case the circuit transforms into one in Fig. 2.2, b.
The circuit in Fig. 2.2, b has single voltage source. The currents at resistors R1, R3
and R4 are calculated by Ohm’ Law:
I'2 = 0, I 1′ = I 3′ = I 4′ =
E1
.
R1 + R3 + R4
2 Estimate individual influence of the current source. To do this, replace the
voltage source by its inner resistance (R = 0).
Thus, the circuits becomes one in Fig. 2.2, c.
Evidently, current I''2 = J. Currents I''1 and I''3 can be calculated according to
Current Divider Rule:
R3 + R4
,
R1 + R3 + R4
R1
I 3′′ = J
.
R1 + R3 + R4
I1′′ = J
3 Now, taking into account current directions in Fig. 2.2, b and c along with their
values, it is possible to define real values and directions of currents in the original
circuit (Fig. 2.2, a):
I1=I'1 – I''1 and it is co-directed with the greater of the partial currents;
I4 = I3 = I'3 + I''3 ; I2 = I'2 + I''2 = J since the partial currents do not oppose each
another, the real current shares their direction.
Thus, Method of Superposition leads to analysis of bunch of single-source
circuits.
30
Thevenin’s and Norton’s theorems
The theorems are advisable to apply in cases current or voltage values in one
branch only are required.
These theorems employ the same principle of replacement of some part of a
circuit by an equivalent source and load.
1) Thevenin’s theorem states that ideal voltage source can be used as an
equivalent one.
2) Norton’s theorem implies use of ideal current source.
More information about the theorems. Thévenin's theorem for linear electric
circuits states that any combination of voltage sources, current sources and resistors
with two terminals is electrically equivalent to a single voltage source V and a single
series resistor R0.
This can be used in order to define unknown current or voltage values in some
branch. According to this method, the branch analyzed is replaced by open circuit.
Next steps are to determine voltage across terminals remaining after extracting branch
and equivalent resistance connected to them. When dealing with resistance, all the
sources are replaced by their inner resistances. The current then can be calculated by
Ohm’s Law.
Example. There is a circuit shown in Fig. 2.3, а.
R3
R3
а
E2
R1
+
+
R2
E1
E2
R1
R4
а
Uoc аb
+
+
R2
E1
b
b
а)
R3
b)
а
Re
а
R1
Rаb = Re
I4
+ Uoc ab
R4
SC
R2
b
b
c)
d)
Figure 2.3 – Solution via Thevenin’s Theorem:
a – original circuit; b – open-circuit voltage determination;
c – equivalent resistance determination; d – determination of unknown current
31
Given the values E1, E2, R1, R2, R3 and R4. The current flowing through
resistor R4 is unknown.
Solution
1 Extract the branch we are searching current in and determine the open-circuit
voltage Uoc ab across its terminals (points a and b). Fig. 2.3, b
Uoc аb = –E2 + I⋅R2,
where
I=
E1 + E 2
.
R1 + R2 + R3
2 Determine the equivalent resistance connected to the terminals a and b
(Fig. 2.3, c):
R (R + R3 )
Re = 2 1
R1 + R2 + R3
3 The wanted current equal to (Fig. 2.3, d):
I4 =
U oc аb
.
Re + R4
Norton’s theorem for a linear electric circuit states that any combination of
resistances and DC sources is electrically equivalent to an ideal current source in
parallel with a single resistor R. The value of R is the same as that in the Thevenin’s
equivalent. The value of Isc can be obtained by short-circuiting the branch being
considered and thereby calculating current flowing through it.
Considering the previous example,
а
the voltage value U4 is determined
according to expression below:
1
1
ISC Geq =
G4 =
U4
I sc
Req
U4 =
R4
Ge + G 4
b
where Isc – short-circuit current
1
1
Ge =
, G4 =
(Fig. 2.4).
Re
R4
Figure 2.4 – Voltage determination via
Norton’s Theorem
32
2.6 The Mesh Current Method
Applying Duality Principle (see 1.9) to the Node Voltage Method (see 2.2), we
obtain equations on which the Mesh Current Method is based.
Description of the method is recommended to do yourself.
Key questions
1 What do Kirchhoff’s Laws state?
2 How many equations is it necessary to describe an electric circuit when
applying Kirchhoff’s Analysis?
3 How many equations is it necessary when applying the Node Voltage
Analysis?
4 How to determine self-conductance of a node?
5 How to determine mutual conductance between two nodes?
6 How to determine node-currents?
7 What is a reference node?
8 What is the principle the Method of Superposition is based on? What does it
state?
9 What are the Thevenin’s and Norton’s Theorems?
10 When is it reasonable to apply these theorems?
11 How many equations is it necessary when applying the Mesh Current
Analysis?
12 How to determine self-resistance of a loop?
13 How to determine mutual resistance between two loops?
14 How to determine loop voltage?
15 How to determine branch currents via the Mesh Current Method?
33
3 AC NETWORK ANALYSIS
3.1 Classification of electric signals
Signal is a physical message carrier, e. g. light, colour, sound, temperature,
electromagnetic waves. The very word “signal” comes from the Latin “signum”,
which means sign. Electric signals (further signals) can be represented by means of
mathematical expressions, vector diagrams, graphs, etc. The same signal can be
described in more than one way including time-domain and frequency-domain
representations.
There are several signal classifications based upon different criteria. For instance
all the signals can be divided into two groups:
– deterministic signals are ones that can be described by an analytical expression
at any moment of time (this includes periodic and non-periodic signals);
– random signals can not be described as such in closed analytical form but are
only statistically describable (speaking, music, noise).
Random signal carry information. Deterministic signals, however, are often used
as testing signals, for adjustments and measurements.
Further considered are deterministic signals.
The signals can be distinguished upon:
1. Signal waveform:
– continuous (analogous, analog) signal is a varying quantity whose domain
(time) is continuum [6] (Fig. 3.1, а and b);
– discrete signal is time series consisting of sequence of quantities (samples) or
steps remaining on the same level over time intervals (Fig. 3.1, c).
f(t)
f(t)
f(t)
t
t
t
b)
c)
Figure 3.1 – Signals:
a – continuous signals; b – quantized signal; c– sequence of samples
a)
Continuous signals are:
– periodic, which means they repeat themselves after certain amount if time
multiple of a period Т f(t + kT) (Fig. 3.2, а and b);
– aperiodic, which means they never repeat, that is to say, do not meet the
requirement (Fig. 3.2, c and d).
34
f(t)
f(t)
f(t)
0 Т
t
0
0
t
tp
f(t)
t
0
tp
t
Т
а)
b)
c)
Figure 3.2 – Periodic (a, b) and aperiodic (c, d) signals
d)
Fig. 3.3 shows some of the most frequently appearing continuous signals.
f(t)
f(t)
1
1
e
δ(t)
1(t)
f(t)
1 cos t
1
f(t)
sin t
– αt
0
t 0
a)
t
0
t
1
1
t
− tp0
tp
2 d) 2
c)
b)
Figure 3.3 – The simplest pulse signals:
a – decreasing exponent; b – unit-step function;
c – delta pulse; d – cosine pulse; e – sine pulse
tp
0
t
e)
Fig. 3.3, а shows decreasing exponent which is represented analytically as:
 0, t ≤ 0;
f (t ) =  −αt
1e , t > 0,
where е is base of natural logarithm; α is a real number.
In case α tends to 0, then the function transforms into function 1(t) which looks
like single step or unit step (see Fig. 3.3, b). This is also known as Heaviside
function. It is written analytically:
0, t ≤ 0;
f (t ) = 
1, t > 0.
Another rarely used in mathematics function is one shown in Fig. 3.3, с. It is
called delta pulse δ(t) and described by expression below
 0, t ≠ 0;
δ (t ) = 
∞, t = 0.
35
Practically this function is remarkably important. For example its properties are:
∞
∞
−∞
−∞
∫ δ(t )dt = 1, ∫ f (t )δ(t )dt =
∞
f (0), or
∫ f (t )δ (t − t
d
)dt = f (t d ).
−∞
In these formulae f(t) is function of some continuous signal; δ (t − t d ) delayed by
td delta pulse.
Fig. 3.3 shows harmonic pulses such as cosine (d) and sine (e) pulses. Their
properties are left for further consideration.
3.2 Harmonic oscillations.
Basic concepts and definitions
Along with DC energy sources in electric circuits are often used alternators
producing current which changes its direction or voltage possessing polarity which
switches in time in some particular manner, often periodically. Any periodic
oscillation meets the equation
i(t) = i(t + kT),
where T is period of the oscillation, that is to say, the smallest time interval over
which a function repeats itself.
The most common type of a periodic oscillation is harmonic wave. Its
instantaneous value is described by a sine or cosine function:
i(t) =Im sin (ωt + φi )=Im sin ψi (t),
(3.1)
where Im is amplitude, in other words, the maximal value possessed by the
function;
φi is initial phase (radians or degrees);
ψi(t) = ωt + φi total phase;
ω = dψi /dt angular frequency, which is the rate of change of argument (rad/s).
Period Т and angular frequency ω are related with one another as
Т = 2π/ω.
Number of periods per time unit
(usually 1 s) is called cyclic
frequency f = 1/T which is measured
in Hz. It is officially adopted that
standard frequency of alternating
current in European countries is
50 Hz.
i(t)
Im
ωt
t
0
ϕi
ω
Т
Figure 3.4 – Time diagram of AC current
over time
36
It is clear that ω = 2πf = 2π/Т rad/s.
Expression (3.1) can be transformed into one using cosine wave:
i(t) = Im cos (ωt + φi);
(3.2)
φί΄ = φί + π/2.
Figure 3.2 shows time diagram of AC current.
Vast frequency range of harmonic waves is used in telecommunication systems.
It varies from a few hertz to hundreds of gigahertz.
Apart from current amplitude Im, there are effective (root mean square, RMS)
value I and average value Iav used. The effective current can be determined according
to expression below
I=
1T 2
∫ i dt.
T0
(3.3)
Considering current to be
i (t ) = I m cos ωt
T
∫i
0
T
2
dt = ∫
I m2
0
(3.4)
I m2 T
I m2
cos ω tdt =
∫ (1 + cos 2ω t )dt = 2 T .
2 0
2
In accordance with (3.3),
I=
Im
≈ 0,707I m .
2
(3.5)
In the same way effective voltage:
U m ≈ 0,707U m .
If to raise expression (3.3) to square and multiply both parts of the obtained
equation by RT, then total amount of heat dissipated on a resistor R over one period T
can be determined:
T
T
wR = ∫ pdt = ∫ Ri 2 dt = TRI 2 .
0
(3.6)
0
Hence, the effective value of AC current is equivalent to some DC current that
while flowing through the resistance R over the time interval T dissipates the same
amount of heat as AC current does.
Naturally, the average of AC current (voltage) over a period
37
I av
1T
= ∫ idt
T0
(3.7)
is zero, since the positive part of the wave is compensated by the negative one
(see Fig. 3.4). In this regard, half-period average value is introduced:
′ =
I av
T
4
T
4
−
T
−
4
2I
2
idt = m
∫
T T
T
4
∫ cos ωt dt =
T /4 2
2I m
sin ωt
= I m ≈ 0,637 I m .
−T /4 π
ωT
(3.8)
Same for AC voltage:
U av ≈ 0,637U m .
3.3 Spectral (frequency) representation
of harmonic oscillations
Alternating current i(t) is shown in Fig. 3.4. Description of harmonic oscillation
in time-domain is clear but far too complicated to use when solving problems, since it
requires cumbersome trigonometric transformations. Other ways to represent
harmonic oscillations depending on the type of a task are to be introduced further.
A harmonic oscillation i(t) (see Fig. 3.4) can be fully described by the basic
parameters such as amplitude Im, angular frequency ω and initial phase ϕi . These are
used to form amplitude and phase spectrums of the oscillation considered. The
spectrums are formed by means of vertical line segments, which correspond to
amplitude and initial phase values in Cartesian coordinates.
Fig. 3.5 shows amplitude (a) and phase (b) spectrums of AC current i1(t). Where
i1 (t ) = I m1 cos(ω1t + ϕ1 )
Such a frequency analysis
of harmonic oscillation in some
cases is much more convenient
than analysis in time-domain.
Spectral approach to harmonic
oscillations and subsequent
method of circuit analysis are
considered further.
Im
(3.9)
φ
Im1
φ1
ω1
ω
ω1
ω
а)
b)
Figure 3.5 – Spectrums of a harmonic oscillation:
а – amplitude spectrum; b – phase spectrum
38
3.4 Passive elements in AC circuits
Resistance
If AC voltage is applied to some resistive element R, then it causes current flow
through it, which can be easily calculated by Ohm’s Law.
AC voltage
u = U m cos(ωt + ϕu ) ,
(3.10)
the current, then
u U
i = = m cos(ωt + ϕu ) = I m cos(ωt + ϕi ),
(3.11)
R
R
here amplitude of the current is
Im =
Um
;
R
and initial phase is
ϕi = ϕu = ϕ .
Thus, voltage across a pure resistive element and current flowing through it are
in phase: ϕi − ϕu = 0 .
Instantaneous power in a pure resistive circuit
p = ui = U m I m cos 2 (ωt + ϕ) = UI [1 + cos 2(ωt + ϕ)] = UI + UI cos 2(ωt + ϕ) (3.12)
It follows from (3.12) that instantaneous power p has two components:
Direct level UI and harmonic level U I ⋅ cos 2(ωt + ϕ) (function of doubled
frequency 2ω). Since current and voltage are in phase, i.e. they are simultaneously
positive or negative, the instantaneous power is always positive. It means that
resistance always dissipates power releasing it in the form of heat.
Fig. 3.6 shows time-diagrams of current, voltage and instantaneous power for
some resistance R.
i, u, p
p
i
R
u
UI
i
u
0
t
a)
ϕi
>0
ω
b)
Figure 3.6 – Time diagrams of u, i, p for a restive element:
а – resistive element; b – graphs of i(t), u(t), p(t)
39
Period average value of power dissipated by a resistive element,
1T
1T
Pav = ∫ pdt = ∫ u i dt
(3.13)
T0
T0
It is known as true power and measured in Watts (W). It follows from expression
(3.12) and Fig. 3.6 that
Pav = UI = I 2 R = U 2 G.
(3.14)
Inductance
If there is some current flowing through an inductive element
i = I m cos(ωt + ϕi ),
then it induces some voltage across the element, which can be estimated as:
u=L
di
π

= −ωLI m sin (ωt + ϕi ) = U m cos ωt + ϕi +  = U m cos(ωt + ϕu ) , (3.15)
dt
2

where
π
π
ϕu = ϕi + , ϕu − ϕi = ,
2
2
U m= I m ω L , U = I ω L .
The value of ωL is expressed in Ohms, just like resistance is, except that its
mathematical symbol is XL instead of R and it is called inductive reactance. The
reciprocal of it is
1
1
BL =
=
,
X L ωL
known as inductive susceptance. Hence
I m = BLU m , I = BLU .
It follows from (3.15) that initial phases of current and voltage in an inductive
element differ by angle of π/2, in other words current lags voltage by π/2.
Instantaneous power delivered to a pure inductive element can be expressed as
follows:
p = ui = UI sin (2ωt + 2ϕi ) = ωLI 2 sin (2ωt + 2ϕi )
(3.16)
40
So it is described by a sine wave with doubled frequency (voltage and current are
graphed in Fig. 3.7).
i, u, p
p
u
i
i
L
u
а)
0
t
b)
Figure 3.7 – Time diagrams of u, i, p for an inductive element:
а – inductive element; b – graphs of i(t), u(t), p(t)
As it is shown in the diagram above, instantaneous power possesses negative
values as well as positive ones, which means that pure inductance alternately absorbs
power from and returns it to an AC source. The average power, thus, is zero.
Energy of the magnetic field caused by inductance is
Li 2 LI m2
LI 2
2
[1 + cos 2(ωt + ϕi )]
wL =
=
cos (ωt + ϕi ) =
2
2
2
(3.17)
LI m2
It varies with frequency of 2ω from 0 to
. It is equal to zero when current is
2
zero and reaches its maximum at the peak value of current. Thus, there is an exchange
of energy between the source and the inductive element. The power delivered from
the source is stored in the form of the magnetic field around the inductor and then it is
returned to the source.
Capacitance
If there is some voltage applied to a capacitive element
u = U m cos(ωt + ϕu ).
Then it creates some current flowing through the element
i=C
where
du
π

= −ωCU m sin (ωt + ϕu ) = I m cos ωt + ϕu +  = I m cos(ωt + ϕi ), (3.18)
dt
2

41
π
π
ϕ i = ϕu + , ϕ u − ϕ i = − .
2
2
Amplitude and effective values of the current are:
I m = ωCU m ; I = ωCU .
The value
BC = ωC
is expressed in Siemens in the same way as conductance and known as capacitive
susceptance. The reciprocal of it is capacitive reactance
XC =
1
.
ωC
The initial phases of current and voltage in a capacitive element differ by the
angle of π/2 or, to be specific, voltage lags current by π/2. Time diagrams of the
current and voltage are shown in Figure below.
i, u, p
u
p
i
i
C
u
а)
t
0
b)
Figure 3.8 – Time diagrams of u, i, p for a capacitive element:
а – capacitive element; b – graphs of i(t), u(t), p(t)
Instantaneous power delivered to a capacitive element has the same character as
in the case of inductance:
p = ui = UI sin (2ωt + 2ϕu ) = ωCU 2 sin (2ωt + 2ϕu ).
(3.19)
The average power over a period is also equal to zero. Energy of electrical field
generated by capacitance can be estimated as:
42
Cu 2
wC = ∫ pdt =
.
(3.20)
2
0
Just like in inductive circuits, there is an energy exchange in capacitive ones.
Capacitance absorbs the energy delivered from a source. It is stored in the form of
electric field between plates and then returned to the source with the electric field
diminishing.
In conclusion, it should be noted that expressions (3.3) and (3.13) defining
respectively RMS and period average values are correct for periodic current or
voltage of any waveform.
The bottom line of what was considered about harmonic oscillations in linear
electric circuits is that any linear circuit containing R, L, C supplied with a harmonic
current with frequency of ω will be characterized by currents and voltages of the same
frequency.
Table 3.1 represents expressions describing instantaneous and effective values of
harmonic oscillations in time-domain.
Table 3.1 – Equations of the elements
Harmonic mode
Time-domain
Element
representation
Instantaneous values
RMS values
(general)
Resistive
u R = RI m cos(ωt + ϕ)
u R = Ri
i G
U = RI
i R = Gu
i R = GU m cos(ωt + ϕ)
I = GU
uR
R
T
Inductive
i L
uL
Capacitive
i C
uС
uL = L
di
dt
1 t
i L = ∫ udt
L −∞
1 t
∫ idt
C −∞
u L = ωLI m cos(ωt + ϕ)
iL =
1
π

U m cos ωt + ϕ − 
ωL
2

U = ωLI
I=
1
U
ωL
1
I m cos(ωt + ϕ)
1
ωC
U=
I
ωC
π

du
iC = ωCU m cos ωt + ϕ + 
I = ωLU
iC = C
2

dt
3.5 Vector representation of single-frequency
harmonic oscillations
uC =
uC =
When analyzing AC circuits, vector representation of harmonic oscillations is
much more convenient than their description in time-domain.
As it was mentioned above, in a linear electric circuit with an AC source of
single frequency ωs all the currents and voltages are harmonic functions with of the
same frequency. This lets us omit ωs, as it carries no new information about variables,
so that we can characterize the variables by two parameters only. They are amplitude
43
and initial phase. In this case each function corresponds to some rotating vector of
fixed length and initial phase.
These vectors are graphed in polar coordinate system and all the operations with
them are performed in accordance with vector algebra.
Application of vector representation
L
R
31 C
2
4
of harmonic oscillations is demonstrated 1 i
by an example in Fig. 3.6.
The harmonic current i = I m cos ωt , u
uc
uR
uL
when flowing through the circuit, creates
harmonic voltage across circuit terminals,
which in accordance with KVL equals to
Figure 3.6 – Series RLC circuit
u = u R + u L + uC .
According to Table 3.1
π
1
π


u = RI m cos ωt + ωLI m cos ωt +  +
I m cos ωt − .
2  ωC
2


Sum of these harmonic oscillations corresponds to vector sum:
π
1
π
RI m ∠0 + ωLI m ∠ +
I m ∠ − = U m ∠ϕ ,
2 ωC
2
or
(3.21)
π
π
U mR ∠0 + U m L ∠ + U m C ∠ − = U m ∠ϕ .
2
2
UmL
UmR
1
3
2
ϕ
UmC
4
Um
Figure 3.7 – RLC voltage
vector diagram
Vector sum is represented in Fig. 3.7. Any
vector start where the previous one ends. Such a
vector diagram is known as a topographic
diagram (XL < XC).
It appears from Fig. 3.7 that amplitude of the
applied voltage is:
2
U m = U mR
+ (U mL − U mC ) 2 ,
and its initial phase is
ϕ = arctg
U mL − U mC
.
U mR
Knowing these parameters of the voltage applied to the circuit in Fig. 3.6, we can
determine its instantaneous value
44
u = U m cos(ωt + ϕ).
Analysis of electric circuits in harmonic mode by means of vector diagrams is
simple and also convenient when checking results. The task to graph current vector
diagrams in a tank is a set to be accomplished yourself.
However, vector analysis of electric circuits is usually applied to rather simple
circuits.
Another way to describe complex circuits in harmonic mode algebraically was
developed by Arthur Kennelly and Joseph Steinmetz in 1893 – 1894. This method is
known as method of complex amplitudes or symbolic method. All the further
description of electrical network theory as well as specialized disciplines is based on
this method. Symbolic method combines analytical approach with geometrical
representation.
Key questions
1 What is harmonic current (voltage)?
2 What are instantaneous, mean and effective values of harmonic current
(voltage)?
3 Which parameters is harmonic oscillation characterized with?
4 What is vector diagram of current (voltage)?
5 What is the difference between initial phases of current and voltage in
inductance (capacitance)?
6 Which ways to represent harmonic oscillation do you know?
7 What are inductive reactance and susceptance?
8 What are capacitive reactance and susceptance?
45
4 METHOD OF COMPLEX AMPLITUDES
(SYMBOLIC METHOD)
4.1 Representation of harmonic oscillations via complex numbers
It’s known that any point on a complex plane is defined by its radius-vector. It is
a vector which begins at the origin of coordinates while its terminus coincides with
the point corresponding to the complex number given (Fig. 4.1).
When represented in polar (exponential) form,
a complex number A& is described by expressions: + j Im
A& = A∠α = Ae jα ,
А
where А is modulus, α is argument, j = − 1 is А2
imaginary unit.
By applying Euler’s formula, we obtain
trigonometric representation of a complex number:
A& = A cos α + jA sin α or its respective algebraic 0
representation:
A& = A1 + jA2 , where A1 = A cos α ;
A1 = A sin α .
It’s obvious that:
α
Re
+
А1
Figure 4.1 – Point А on
complex plane (Re – real axis,
Im – imaginary axis)
A2
.
A1
If some vector rotates counter-clockwise with angular rate of ω, then at any
moment of time its position is determined by complex value
A = A12 + A22 ; α = arctg
Im
jωt
A& е
ωt
α
0
A& (t=0)
+Re
Ae j (ωt +α ) = A& e jωt ,
(4.1)
where
A& = Ae jα
is a complex amplitude representing the
vector at the moment t = 0 (Fig. 4.2), i.e. A& is
a complex number which does not depend on
time. Modulus and argument are respectively
&
equal to amplitude and initial phase of a
Figure 4.2 – Rotating vector A
harmonic oscillation.
on complex plane
If to multiply complex amplitude A& by
rotation operator e jωt , then vector A& will
rotate by angle of ωt in a positive direction. From trigonometric representation of
complex function (4.1)
46
Ae j (ωt +α) = A cos(ωt + α) + jAsin(ωt + α) it follows that harmonic function
A cos(ωt + α) can be considered as real part of complex function (4.1), that is to say a
projection of the rotating vector on real axis. Conventionally it can be written as
follows:
(
A cos(ωt + α) = Re A& e jωt
)
(4.2)
π

For example, if current is i = 14,1 cos ω t +  , then its complex amplitude is
6

j
π
/6
(Im=14,1; ϕ = π/6) I& m = 14,1e
and its effective complex value is
π
&
j
&I = I m = 10 e 6 .
2
4.2 Equations of passive elements in complex form
In fact, complex currents and voltages are symbols (images) of real harmonic
currents and voltages.
We shall define complex impedance Z of a two-terminal network as
Z=
U& m
,
I&m
where I&m , U& m are complex current flowing through the network and complex voltage
across it respectively.
Resistance
If there is a harmonic current flowing through some resistance R
i = I m cos(ωt + ϕ),
with a complex amplitude
I&m = I m e jϕ ,
Then, in accordance with the equation of this element u = Ri, the voltage appears
u = R / I m cos(ωt + ϕ),
and its complex amplitude is
U& m = RI m e jϕ .
47
Hence we obtain
RI m e jϕ
ZR =
=R .
I m e jϕ
(4.3)
Inductance
If there is a current flowing through an inductance
i = I m cos(ωt + ϕ) ⇔ I m e jϕ ,
then in accordance with the equation
u=L
di
,
dt
the voltage appears

π
j  ϕ+ 
π

u = ωLI m cos ωt + ϕ +  ⇔ ωLI m e  2  .
2

Hence
π
j ( ϕ+ )
2
U&
ωLI m e
Z L = mL =
I&
I e jϕ
mL
= ωL ⋅ e
j
π
2.
m
Regarding Euler’s formula
e
j
π
2
= j
we obtain
Z L = j ωL
(4.4)
or
Z L = jX L .
Capacitance
Similarly, capacitance can be described by the expressions
ZC =
1
1
=−j
jωC
ωC
or
Z C = − jX C .
(4.5)
48
4.3 Basic laws of circuit theory in complex form
The equations of passive elements above are identical to each other:
U& mR = I&mR R; U& mL = I&mL Z L ; U& mC = I&mC Z C ;
This is why the expressions
U& m = I&m Z ;
(4.6a)
U& = I& Z , a I& = U Y ,
(4.6b)
and
where
Y=
1
Z
are called Generalized Ohm’s Law. It should be noticed that relationships between
complex values remain the same as between their instantaneous values. According to
Kirchhoff’s Current Law (KCL) at any node of a circuit
l
l
k =1
k =1
∑ ik = ∑ I mk cos(ωt + ϕik ) = 0 .
By substituting complex amplitudes I& mk for instantaneous current values ik, we
obtain:
– amplitude values:
l
∑ I&mk
=0
(4.7а)
= 0.
(4.7b)
k =1
– effective values
l
∑ I&k
k =1
Similarly, in accordance with Kirchhoff’s Voltage Law for any closed-loop
circuit
n
∑U& mk
=0
(4.8а)
= 0.
(4.8б)
k =1
and
n
∑U& k
k =1
It follows that there is a complete analogy in description DC and AC circuits. We
can say therefore that all the methods of DC analysis are valid for AC circuits as well.
49
4.4 Nodal equations for complex amplitudes
Let us analyze an AC circuit by means of Node Voltage Method.
When in canonical form, equation set for complex amplitudes differs from one
for DC circuit only in symbols used. To make it simpler, we usually use complex
effective values instead of complex amplitude values
U&
U&
U& = m = m e jϕu = Ue jϕu ,
2
2
I&
I&
I& = m = m e jϕu = Ue jϕu .
2
2
Often the terms “complex current” or “complex voltage” are used instead of
“complex effective value of current or voltage”.
Thus, the equation set of complex nodal voltages is:
1
2
N
Y11U& 1 − Y12U& 2 − ... − Y1N U& N = J&11 ;
− Y21U& 1 + Y22U& 2 − Y23U& 3 − ... − Y2 N U& N = J& 22 ;
........................................................
− YN 1U& 1 − YN 2U& 2 − YN 3U& 3 − ... + YNN U& N = J& NN ,
(4.9)
where Ykk is sum of admittances of branches joined at node k; Ykl = Ylk is equivalent
admittance of the branch between k-th and l-th nodes. An example of a subcircuit is
given in Fig. 4.3. Admittance of the branch
Lkl
connecting nodes k and l equals to:
1
1
Ykl =
+
+ jωC kl .
jωLkl Rkl
k
Rkl (Gkl)
l
Ckl
Work out the equations for other
methods on your own.
Figure 4.3 – Parallel RLC circuit
When using computers for harmonic
analysis of circuits, some programs can be employed to estimate impedances and
admittances of typical passive two-terminal networks which these circuits consist of.
4.5 Matrix method of circuit analysis
The mathematical basis of this chapter is the theory of matrices. Therefore,
before proceeding further, it is strongly recommended to have a look through the
relevant section of the mathematics.
50
J&1
U& 1
(2×2)
I&1
J&1
U& 1′
J&2
U& 2
а)
U& 2′
(2×2)
b)
U& 1′′
We introduce the concept of pass
quadripole
(two-port
network),
considering the relationships between
its input and output currents and
voltages ( I&1 ,U& 1 , I&2 , U& 2 ). Apparently,
the coefficients describing these
values are either unitless (İ1↔İ2,
U& 1 ↔ U& 2 ) or measured in Ω, S
( I& ↔ U& ). Thus, there are six different
ways of setting equations describing a
quadripole
(employing
different
groups of quadripole parameters).
Different cases require different
parameters to use. When writing
equations, we assume that currents I&1
and I&2 enter the quadripole and
(2×2)
I&2
U& 2′′
J&2
c)
Figure 4.4 – Quadripoles:
а – with two current sources;
b – with a single current source on the left;
c – with a single current source on the right
polarity of the voltages U& 1 and U& 2 is
as shown in Fig. 4.4, a.
As an example, we shall deduce one of the systems possible. Imagine that a
quadripole is connected to some circuit. There is a power supply source in the circuit,
which means there is some current and voltage at the quadripole’s input. We shall
write equations describing relationships between voltages U& 1 , U& 2 and currents İ1 , İ2.
Applying Compensation Theorem to the circuit being analysed, we replace external
network by two current sources J&1 and J& 2 (Fig. 4.4, а). Then we use Superposition
Principle (Fig. 4.4, b, c):
U& 1 = U& 1′ + U& 1′′ 
(4.10)

U& 2 = U& 2′ + U& 2′′ 
For the circuit in Fig. 4.4, b we can write:
U& 1′ = Z11 I&1 ;
U& 2′ = Z 21 I&1 ,
(4.11)
Similarly for the circuit in Fig. 4.4, c:
U&1′′ = Z12 I&2 ;
U& ′′ = Z I& .
2
(4.12)
22 2
From equations (4.11), (4.12), (4.10) we easily deduce the following system:
51
U& 1 = Z11 I&1 + Z12 I&2 ; 

U& 2 = Z 21 I&1 + Z 22 I&2 .
Z coefficients in this system are known as impedance parameters or Zparameters. Generally speaking, they are complex and depend on circuit
configuration and its components only. If solved with respect to currents İ1, İ2, these
equations produce another type of parameters, etc.
Table 4.1 contains all six types of parameters and their physical meanings.
Table 4.1 – Matrix parameters of quadripoles
Systems of equations
Physical meaning of the coefficients
I&2
I&1
U& 1
I&1=Y 11U&1 +Y 12U& 2
I& =Y U& +Y U&
2
21 1
22
U& 1= Z 11I&1 + Z 12I&2
U& = Z I& + Z I&
21 1
I& 2
U& 2 U& 1 = 0
I&
= 1 &
U& 2 U 1 = 0
Y22 =
Y12
I&1
OC U& 2
I&2
U& 1 OC
U& 1
I&1 I&2 = 0
U&
Z 21= 2 &
I& I 2= 0
Z 11=
22 2
U& 2
SC
I&1
U& 1 U& 2 = 0
I&
Y 21= 2 &
U& 1 U 2 = 0
U& 1
2
SC
Y 11=
2
I&2
I&1
U& 1
U& 1= A11 U& 2+ A12 I&2
I& = A U& + A I&
1
21
2
OC U& 2
Z 22
2
U& 1
U& 1
U& 2 I&2 = 0
I&
A 21 = 2 &
I& I = 0
A11 =
22 2
1
I&1 = F 11U& 2+ F 12I&2
U& 2= F 21U& 2+ F 22I&2
F 11=
SC
A22
2
OC U& 2
I&1
U& 1 I& 2= 0
I&2
I&1
U& 1
− I&2 U& 2 = 0
I&
= 1 &
− I&2 U 2 = 0
A12 =
I&1
U& 1
U& 1
I&2 I&1 = 0
U&
= 2 &
I& I 1= 0
Z12 =
1
I&1
U& 2
I&1
I&2
U& 2
SC
F12 =
I&1
− I& 2 U& 1 = 0
52
F 21=
U& 2
U& 1 I&2 = 0
F22 =
I&2
I&1
U& 1
U& 1= H 11I&1 + H 12U& 2
I& = H I& + H U&
2
21 1
22
2
SC
U& 2= B11 U& 1+ B12 I&1
I& = B U& + B I&
2
21
1
22 1
I&2
U& 1 OC
U& 1
I&1 U& 2 = 0
I&
= 2 &
I&1 U 2 = 0
U& 2
U& 1
U& 2 I&1 = 0
I&
= 2 &
U& 2 I 1= 0
H 11 =
H 12=
H 21
H 22
I&2
U& 1 OC
U& 2
U& 2
U& 1 I&1 = 0
I&
= 2 &
U& I = 0
B11 =
B21
1
1
U& 2
− I&2 U& 1 = 0
I&2
I&1
SC
U& 2
U& 2
I&1 U& 1 = 0
I&
= 2 &
I& U = 0
B12 =
B22
1
1
It is often required to use various parameters, as long as we describe the same
circuit. Thus, it is necessary to define the relations between them. Table 4.2 contains
exactly that type of expressions.
Table 4.2 – Relations between matrix parameters
Y
Z
Y
Z 22
Z 12
Y11 Y12
Y21 Y22
Z
A
H
F
Y 22
Y
− 12
∆Y
∆Y
Y 21 Y 11
−
∆Y
∆Y
Y
1
− 22
Y 21 Y 21
∆Y Y 11
−
Y 21 Y 21
1
Y 11
Y 21
Y 11
∆Y
Y 22
Y
− 21
Y 22
∆Z
Z
− 21
∆Z
−
Z11
Z12
Z21
Z22
Z 11
Z 21
1
Z 21
Y 12
Y 11
∆y
Y 11
∆Z
Z 22
− Z 21
Z 22
Y 12
Y 22
1
Y 22
1
Z 11
Z 21
Z 11
−
∆Z
Z11
∆Z
∆Z
Z 21
Z
− 22
Z 21
−
A
A 22
A12
1
A12
A11
A 21
1
A 21
A11
H
∆A
A12
A
− 11
A12
∆A
−
A 21
A
− 22
A 21
−
A12
A21 A22
Z 12
Z 22
1
Z 22
A12
A 22
1
A 22
∆A
A 22
A
− 21
A 22
Z 12
Z 11
∆Z
Z 11
A 21
A11
1
A11
∆A
A11
A
− 12
A11
−
1
H 11
H 21
H 11
∆H
H 22
H 21
H 22
∆H
−
H 21
H
− 22
H 21
F
H 12
H 11
∆H
H 11
H
− 12
H 22
1
H 22
H 11
H 21
1
H 21
−
H11 H12
H21 H22
H 22
∆H
H
− 21
∆H
H 12
∆H
H 11
∆H
−
F 12
∆F
F 22
F 22
F 21
1
−
F 22 F 22
F
1
− 12
F 11
F 11
F 21
∆F
F 11
F 11
F 22
1
−
F 21 F 21
F
∆F
− 11
F 21 F 21
F 22
∆F
F
− 21
∆F
F11
F21
F 12
∆F
F 11
∆F
−
F12
F22
54
Just as two-terminal networks, quadripoles can be connected in different ways.
There are 5 types of their connections. They are shown in Table 4.3 alongside with
the respective matrix parameters used.
Table 4.3 – Types of quadripole connection
Parallel-parallel
Series-series
I&1
I&2
&I 2
I&1
(1)
(1)
U& 1
U& 2
(2)
U& 1
U& 2
(2)
(n)
n
[ Z ] = [ Z (1) ] + [ Z ( 2) ] + K + [ Z ( n ) ]
[Y ] = [Y (1) ] + [Y ( 2) ] + K + [Y ( n ) ]
Series – parallel
I&1
I&2
(1)
U& 1
Parallel – series
I&1
(1)
U& 2
(2)
U& 1
[H ] = [H ] + [H
( 2)
U& 2
(2)
(n)
(1)
I&2
(n)
] + K + [H
( n)
]
(1)
[ F ] = [F ] + [F
(2 )
] + K + [F
( n)
]
Cascade
U&1
I&1
I&2
(1)
(2)
(n)
U& 2
[ A] = [ A(1) ] [ A( 2 ) ] K [ A( n ) ]
There are a few examples of determination of a circuit matrix parameters (Yparameters are considered).
Y2
I&2
Y2 I&2
I&1
I&1
Y2
I&2
I&1
U& 1
[Y] =  11
Y
Y21
U& 2
Y1
Y12  (Y1 + Y2 )
=
Y22   −Y2
а)
U& 1
Y
−Y2 
[Y] =  11

Y2 
Y21
Y3
Y12   Y2
=
Y22  −Y2
U& 2
U& 1
Y1
Y3
−Y2 
(Y +Y )
Y
Y
[Y] =  11 12 =  1 2

(Y2 +Y3 )
Y21 Y22  −Y2
b)
Figure 4.5 –Examples of quadripoles
c)
U& 2
−Y2 
(Y2 +Y3 )
55
4.6 Analysis of circuitswith coupled inductors
Previously considered circuits did not contain inductors mutually coupled by a
common magnetic field. If there were a few coupled inductors, then voltage across
each of them would depend on its own current as well as the currents flowing through
the inductors nearby. This phenomenon is known as mutual inductance.
4.6.1 Magnetically coupled circuits
Let us consider some special features of setting up equations for linear circuits
containing coupled inductors. We also assume that there is no energy loss in
inductances.
M
M
M
i1
i2
L1
L2
а)
L1
L2
i1
i2
L1
b)
L2
c)
Figure 4.4 – Coupled inductances:
а – current i1 flows only; b – current i2 flows only;
c – both currents i1 and i2 flow
First, we consider the case when there is a current i1 flowing trough inductor L1
while terminals of the second inductance are open (Fig. 4.4, а). Some part of the
magnetic flux generated by current i1 crosses coils of the second inductor.
Flux linkage (i.e. total magnetic flux coupled to all the coils of an inductor) of the
first inductor we denote with ψ1 and flux linkage of the second one ψ12. Both values
are proportional to i1 and respectively equal to:
ψ1 = L1i1
and
ψ12 = Mi1,
where М is mutual inductance. M is measured in Henry (H).
Mutual inductance is defined by the expression
M = k L1 L2 ,
where coupling coefficient k possesses values of 0 ≤ k ≤ 1. In case the whole magnetic
flux generated by an inductor crosses coils of another one, k = 1. In absence of any
connection between inductors k = 0.
56
Similarly, in case current i2 is flowing through inductance L2 (Fig. 4.4, b), while
terminals of the first inductance are open, ψ2 = L2i2 and ψ21 = Мi2.
In case currents are flowing through both inductances (Fig. 4.4, c), total flux
linkage of the first and second inductances will be:
ψ11 = ψ1 ± ψ21
and
ψ22 =ψ2 ± ψ12
or
(4.10)
ψ11 = L1i1 ± Mi2
and
ψ22 = ± Mi1 + L2i2.
Signs of summands in (4.10) depend on directions of magnetic fluxes in
inductances, whereas directions of magnetic fluxes themselves are defined by
directions of currents flowing through the inductances. If inductances are connected in
such a way that flux linkages ψ1 and ψ21 (and consequently, ψ2 and ψ12) add, then it is
called cumulative connection (“positive” sign in 4.10), otherwise it is differential
connection (“negative” sign in 4.10).
For convenience, when charting electric circuits containing coupled inductances,
the concept of corresponding terminals is employed. Corresponding terminals are
nodes in respect of which currents with the same directions generate additive
magnetic fluxes.
In Fig. 4.5 represents corresponding terminals for the cases of aiding (Fig. 4.5, а)
and opposing (Fig. 4.5, b) connection of inductances L1 and L2.
M
M
i1
L1
L2
а)
і2
i1
i2
L1
L2
b)
Figure 4.5 – Connections of inductances: а –aiding; b – opposing
By deriving equation (4.10) with respect to the variable t, we obtain expressions
describing relationships between currents and voltages at terminals of two coupled
inductances:
dψ11
di
di
= u1 = L1 1 ± M 2 ;
dt
dt
dt
(4.11)
dψ 22
di1
di2
= u2 = ±M
+ L2
.
dt
dt
dt
57
Specifically for AC circuits, expressions above can by rewritten by means of
complex numbers:
U& 1 = jωL1 I&1 ± jωMI&2
(4.12)
U& = ± jωMI& + jωL I& .
2
1
2 2
Circuits containing three and more coupled inductors can be treated in the same
way:
U& 1 = jωL1 I&1 ± jωM 12 I&2 ± jωM 13 I&3 ± ... ± jωM 1N I&N ;
U& 2 = ± jωM 21 I&1 + jωL2 I&2 ± jωM 23 I&3 ± ... ± jωM 2 N I&N ;
U& 3 = ± jωM 31 I&1 ± jωM 32 I&2 + jωL3 I&3 ± ... ± jωM 3 N I&N ;
M
U& N = ± jωM N 1 I&1 ± jωM N 2 I&2 ± jωM N 3 I&3 ± ... + jωLN I&N .
i
u
4.6.2 Series inductors with magnetic coupling
R1
R1
L1
i
R2
M u
L2
R2
L1
M
L2
а)
b)
Figure 4.8 – Series connection of two coupled inductors:
а – cumulatively coupled inductors; b – differentially coupled inductors
Figure 4.8, a shows cumulative coupling of two series inductors. Energy loss
there is represented by resistances R. According to KVL, voltage across the terminals
of the network being considered is
U& = ( R1 + R2 ) I& + jω( L1 + L2 + 2 M ) I& .
(4.16)
Z ec = Re + jωLec ,
(4.17)
Hence, total impedance
where
R е = R1 + R2 ;
Lec = L1 + L2 + 2 M .
Expression (4.16) now can be written as:
U& = Z ec I& = Ue jϕ .
u
Phase difference between current and voltage is:
(4.18)
58
ϕec = ϕu − ϕi = arctg
ωLec
.
Re
(4.19)
Complex voltage across each inductor can be expressed as
U& 1 = [R1 + jω(L1 + M )] I& and U& 2 = [R2 + jω(L2 + M )] I& .
Total impedance of differentially coupled series inductors (Fig 4.8 b) is
Z ed = Rе + jωLed ,
(4.20)
where
Led = L1 + L2 − 2 M .
Phase difference between current and voltage is
ϕed = ϕu − ϕi = arctg
ωLed
.
Rе
(4.21)
Finally, complex voltage across each inductor
U& 1 = [R1 + jω(L1 − M )] I& and U& 2 = [R2 + jω(L2 − M )] I& .
Vector diagrams beneath represent voltage across cumulatively and differentially
coupled inductors L1 and L2.
jω M I&
U&
R2 I&
-jω MI&
jω L2 I&
U&
-jωM I&
R2 I&
jωM I&
jωL1 I&
ϕec
0
I&
R1 I&
a)
jωL2I&
jωL1 I&
ϕed
0
R1 I&
b)
Figure 4.9 – Voltage vector diagrams:
а – cumulative coupling; b – differential coupling
I&
59
4.6.3 Some features of analysis of magnetically coupled circuits.
Decoupling inductors
When analyzing magnetically coupled circuits, the Mesh Current Method is
normally used, since other methods, such as the Node Voltage Method or Thevenin’s
(Norton’s) Theorem, are invalid due to magnetic coupling.
Yet to apply all the known methods of analysis, the circuits being considered
needs to be decoupled.
In Figure below (Fig. 4.10, a) there are two coupled inductors joined at a node by
corresponding terminals. Figure 4.10, b shows their equivalent, obtained by
decoupling the original circuit.
M
i2
i1
L1
i3
i1
L2
i2
L1– M
L2 – M
M
i3
а)
b)
Figure 4.10 – Cumulatively coupled inductors decoupling:
а – original circuit; b – decoupled circuit
In case the inductors were joined in ‘opposing’ way (Fig. 4.11 a), the equivalent
would be as in Fig. 4.11, b.
M
i2
i1
L1
i3
L2
i1
i2
L1 + M
L2 + M
–M
а)
i3
b)
Figure 4.11 – Differentially coupled inductors decoupling:
а – original circuit; b – decoupled circuit
That such a transformation is valid is offered to check in the recommended
literature.
60
4.7 Power balance
If to supply some electric circuit
(Fig. 4.12) with AC voltage
Electric
circuit
u = Um cos (ωt+ϕu).
Figure 4.12 – A circuit with a source
plugged in
It will cause current flow
i = Imcos (ωt + ϕi).
The period average value of power produced by the source can be estimated as
1T
1T
1T
1
(
)
(
)
ui
dt
=
U
I
cos
t
+
cos
t
+
dt
=
U m I m cos(ϕ u + ϕ i ) dt +
ω
ϕ
ω
ϕ
m m
u
i
∫
∫
∫
T0
T0
T0
2
T
T
U I
U I
1
1
1
+ ∫ U m I m cos(2ωt + ϕ u + ϕ i ) dt = m m cos(ϕ u − ϕ i ) ∫ dt = m m cosϕ = U I cosϕ ,
T0
2
2
T0
2
where ϕ = ϕu – ϕi.
Pav = UI cosϕ = zI ⋅ I cos ϕ = I 2 z cosϕ
(4.22)
Pav =
The period average power is also known as consumed power. It is defined in
terms of RMS values of voltage U, current I and cosine of angle between them.
Taking into account that Z = R + jX = z⋅cosφ + jz⋅sinϕ, it follows that
z=
R
cos ϕ
and equation (4.22) transforms into:
Pav = P = I 2 R = U 2 G
(4.23)
Thus, period average power is equal to energy dissipated at resistive elements,
which leads us to another definition of this value as true power, that is to say, actual
amount of power used or dissipated in a circuit. It is measured in units of Watts (W).
Along with the true power in the AC circuits the concept of reactive power is used.
Q = UI sin ϕI 2 X = U 2 B
(4.24)
Hence, total amount of power can be expressed as complex numbed
∗
~
S = P + jQ= U& ⋅ I ,
(4.25)
61
∗
*
Where I is complex conjugate of the current, i.e. if I& = I e jϕ , then I = I e − jϕ .
Modulus of complex power is called apparent power.
S = P2 + Q2 .
(4.26)
Reactive power is measured in Volt-Amps-Reactive (VAR); apparent power – in
Volt-Amps (VA). In accordance with (4.25)…(4.23)
~
S = UI cos ϕ + jUI sin ϕ.
(4.27)
In other words:
∗
~
P = Re[U& I ] = Re[ S ];
∗
~
Q = Im[U& I ] = Im[S ];
(4.28)
~
S = UI = S ;
cos ϕ =
P
.
S
(4.29)
Cosine of ϕ is called power factor. The greater cosϕ, the more efficient circuit.
Power factor possesses maximum possible value of cosϕ = 1 in case of a purely
resistive circuit, where P = S, Q = 0.
Power balance in the AC circuits can be defined by means of Tellegen’s
Theorem in complex form. Since the Kirchhoff’s Laws are valid for complex currents
and voltages as well as for their complex conjugates, the following expressions are
obtained:
nв
∑U& k I&k = 0 .
(4.30)
k =1
nв
*
nв
~
∑U& k I k = ∑ S k = 0,
k =1
(4.31)
k =1
It, therefore, can be stated that total amount of complex power consumed by all
the branches of a circuit is zero. In other words, this equation represents power
balance. Definition of the latter can be reformulated in the following way: total
amount of complex power produced by all sources equals to the amount of power
consumed in all branches:
nв
nв
~
~
(4.32)
∑ S k pr = ∑ S k cons .
k =1
k =1
In accordance with (4.32), the conditions for true and reactive power balance are
derived:
62
nв
nв
k =1
k =1
∑ Pk pr = ∑ Pk cons ;
nв
nв
k =1
k =1
(4.33)
∑ Qk pr =∑ Qk cons .
(4.34)
4.8 Source operating modes and conditions for maximum power transfer
Active two-terminal network generating AC voltage (current) can be replaced by
an equivalent network (in accordance with Thevenin’s or Norton’s Theorems, see
Fig. 4.11).
Z0
U&0
+
I&0
Z0
U&0 = Z0 I&0
a)
b)
Figure 4.13 – Equivalent sources: а – voltage source; b – current source
Where U& 0 equals to complex voltage across the source’s open-circuit terminals
(Fig. 4.13, а) and I&0 is complex current flowing through the source’s short-circuit
terminals
(Fig.
4.13,
b).
The
impedance
of
a
two-terminal
element
.
Z0 =
U0
.
= R0 + jX 0 is called source’s inner impedance. If to connect some load
I0
impedance to a source
Z l = Rl + jX l ,
I&
U&0
+
Z 0 = R0 + jX 0
Z н = Rн + jX н
then question of maximum power transfer
arises.
According to Ohm’s Law the complex
current in circuit in Fig. 4.14 is
I& =
Figure 4.14 – Voltage source
with a load connected
U& 0
U& 0
=
.
Z 0 + Z L R0 + RL + j ( X 0 + X L )
The true power dissipated by the load is
U 02 RL
.
P = I RL =
( R0 + RL ) 2 + ( X 0 + X L ) 2
2
63
It is obvious that power consumed by the load reaches its maximum in case
X0 + XL = 0,
P η
which means that
Pmax
X L = −Х0,
then
1
η
U 02RL
. (4.35)
P=
(R0 + RL )2
0,5
Figure 4.15 shows
diagram Р(Rl) (continuous
curve).
Its
maximum
function Р(Rl) reaches
when
RL /R0
0
1
2
3
4
5
Figure 4.15 – True power Р and efficiency η graphs
with respect to load impedance
RL = R0.
(4.36)
Thus, amount of power consumed by load in AC circuit is maximal if the load
impedance is complex conjugate of the generator’s inner impedance:
Zl = R0 – jX0.
The maximal amount of power consumed is
P max =
U 02
.
4R0
(4.37)
The source’s efficiency is only 50 %, since inner impedance of the generator
consumes the same amount of power as the load does.
While increasing Rl, average power in the load decreases. The efficiency,
however, is growing. The latter statement is represented in Fig. 4.15 by the dashed
line.
It follows from (4.35) that maximum power transfer can be ensured only at one
frequency or, at least, narrow frequency band.
In communication systems the load impedance value is usually chosen to be
equal to generator’s inner impedance:
ZL = Z0 = R0 + jX0 .
This is known as matched load. It is shown in Figure 4.16.
(4.38)
64
Z0
.
.
U L = U 0/ 2
+
U0
Z0=ZL
It means that independently of
frequency, the voltage in load is always
half of the generator voltage. Further it
is proved that it allows avoiding signal
distortion when it is being transmitted
by the circuit.
Figure 4.16 – Matched load
4.9 Signal energy characteristics
Previously a few criteria of classification of electrical signals were introduced.
Here some energy characteristics are given.
A signal can be estimated by its instantaneous power:
u 2 (t ) 2
p(t) = s (t) =
= i (t ) R , W.
R
2
(4.39)
where s(t) is function describing signal in time-domain.
If to assume that s(t) is an instantaneous value of u(t) or i(t), then p(t) would be
the amount of instantaneous power dissipated by the resistor of 1 Ω. The value of p(t)
is measured in А2 Ω = V2 S = W.
Signal average power over some time interval:
1 t
Pav =
∫ p(t)dt , W.
t 2 − t1 t
2
(4.40)
1
Often signal average power is considered over the whole signal lifetime:
T
2
1
Pav = lim ∫ p(t)dt , W.
T →∞
T−
(4.41)
T
2
Signal energy over a time interval is defined by expression below:
W = Pav (t 2 − t1 ) = PavT , W·s.
(4.42)
It should be noted that “energy” is a concept which can be applied to signals of
finite duration only.
Example. Signal u(t) is described by system:
65
 0, t ≤ 0;
U t

u (t ) = 
, 0 < t ≤ tp ;
t
p

 0, t > t p .
u(t)
U
Its diagram is shown in Fig. 4.17.
The
instantaneous
power
accordance with (4.39) is:
0
tp
t
Figure 4.17 – u(t) signal waveform
in
U2 2
p (t ) = u (t ) = 2 t , W.
tp
2
The average power is:
1
Pav =
tp
tp
∫
0
t
1 U2 2
U 2 t 3 tp U 2
p (t ) dt =
∫ t dt = t 3 3 0 = 3 , W.
t p t p2 0
p
p
The signal energy over the interval [0; tp] is:
W = Pav t p =
U 2t p
3
, W·s.
4.10 Procedure of symbolic analysis of linear electric circuits
1 Replace all the elements of the initial circuit by their complex equivalents in
accordance with Table. 4.1.
2 Calculate impedances (admittances) of passive elements in the circuit.
3 Choose optimal analysis method and set up a system of equations by means
of complex numbers.
4 Solve of the equations with respect to complex currents or voltages unknown.
5 Verify the results by checking power balance in complex form or plotting
vector diagrams.
6 Transform obtained vectors (complex numbers) into time-domain.
Key questions
1
2
3
4
5
6
7
How to write alternating voltage or current in terms of symbolic method?
What is impedance (admittance) of a two-terminal network?
Write the Kirchhoff’s and nodal equations for complex amplitudes?
What are coupled inductors?
How to define currents and voltages for cumulative and differential coupling?
Write the equations for power balance in an AC circuit.
What is the condition of maximum power transfer from a source to a load?
66
Table 4.1 – Complex equivalents of elements
Elements of equivalent circuits for complex amplitudes U& m and I&m
Equivalents for arbitrary u(t), i(t)
Relationship
Schematic
Relationship between
Name
Name
Schematic symbol
between I&m and U& m
symbol
i and u
u = e(t ) does not
U& m = E& m does not
E& m
e(t)
&
i
I
Ideal voltage
m
+
+
depend on value and Complex voltage
depend on I&m ,
source
source
u
direction of current i
U& m
U& = U e jϕ u
m
i j(t)
Ideal current
source
u
i = j (t ) does not
depend on value and
direction of voltage u
Complex current
source
u = iR
Resistance
R
i
Resistance
u
L
i
Inductance
di
u=L
dt
u
i
C
i=C
Capacitance
u
Coupled
inductances
1
3
i1
L1
u1
(*)2
i 2 (∗) L2
*
Inductive
impedance
+
_
М
4
u2
u1 = L1
du
dt
di1
di
±M 2
dt
dt
u2 = ± M
di1
di
+ L2 2
dt
dt
Capacitive
impedance
Complex coupled
inductances
J& m
I&m
I&m = I m e jϕ i
U& m = Z R I&m
I&m ZR
U& m
ZL
ZR = R
U& m = Z Z I&m
Z L = jωL
U& m
I&m
I&m = J& m does not
depend on U& .
m
U& m
I&m
m
ZC
U& m
Z U& m 1
I&m 1
(∗ )
I&m (∗) Z Z
2
U& m 2
U& m = Z C I&m
1
ZC =
jωC
U& m1 = Z 1 I&m1 ± Z М I&m 2
U& m2 = ±ZМ I&m1 + Z2 I&m2
Z 1 = jωL1 Z 2 = jωL2
Z М = jωМ
67
5 NON-LINEAR DC CIRCUIT ANALYSIS
5.1 Elements and their characteristics
Non-linear element is one whose parameters, such as resistance (impedance),
depend on the value of current flowing through voltage or across it. It can be
described by functions below
i = f(u) or u = f –1(i).
(5.1)
Any electric circuit with at least one non-linear element in it is non-linear
circuit (NLC).
Non-linear elements along with linear ones can be divided on resistive, inductive,
and capacitive elements.
Non-linear circuit which does not contain
u
reactive elements is called resistive
(memoryless) circuit. Schematic symbol for
a non-linear element is shown in Fig. 5.1.
i NLE
Graphical representation of NLE
function in Cartesian coordinates is known as
Figure 5.1 – Schematic symbol
volt-ampere characteristic (VAC, also i-u
for NLE
diagram).
An example of NLE is a semiconductor diode. Its designation and approximate
VAC are shown in Fig. 5.2, a and b. The ideal diode (valve) would be described by
VAC in Fig. 5.2, c.
і
і
0
u
-
+
i
u
0
a)
и
b)
c)
Figure 5.2 – Schematic symbol (а), VAC of non-ideal (b) and ideal (c) diodes
VAC can also be represented as a table of corresponding values of current and
voltage.
Classification of NLE can be performed in different ways:
– upon VAC symmetry in respect of origin of coordinates;
– upon VAC monotony;
– upon number of terminals: two-terminals (diodes), three-terminals (triodes),
etc.
We suggest looking into NLE classification on your own.
68
To define NLE properties completely, the whole VAC is necessary, however its
points can be individually described by static and differential resistances.
Static resistance of NLE is voltage-to-current ratio at some point of VAC
(Fig. 5.3, а):
Rs =
u1
.
i1
(5.1)
Differential resistance (also incremental resistance or slope resistance) is
ratio of infinitesimal increment of voltage at a VAC point to corresponding increment
of current:
∆u du
= .
∆i→0 ∆i
di
Rd = lim
(5.2)
Evidently, differential resistance can possess negative values across decreasing
sections of VAC.
Often VAC steepness is used instead of Rd. Steepness is a reciprocal of
differential equation. Thus, it equals to:
S =
di
.
du
(5.3)
Fig. 5.3, a illustrates how to calculate the NLE parameters. Also a VAC of a
tunnel diode is shown in Fig. 5.3, b. Differential equation across section (и1 – u2) is
negative. It should be noticed that there are 3 points in Fig. 5.3, b: 1, 2, 3, which are
different in values of static and slope resistances.
i
i
i1
i1
u1
1
u
а)
3
2
u1
u2
u
b)
Figure 5.3 – Volt-ampere characteristics of NLE:
а – single operating point on VAC;
b – three operating points at the same current i1
5.2 VAC approximation
The VAC of non-linear elements are obtained experimentally. To perform
calculations, however, function (i.e. analytical expression) needs to be defined. This
raises a problem of selecting such a function that would be consistent with a
69
reasonable degree of accuracy to represent experimental characteristic. This problem
is called approximation.
Piece-wise linear approximation. The method consists in replacing real
characteristic by straight lines of different slopes of i = i0 + iu.
Polynomial approximation. The most common math tool to use is polynomial
of n-th power:
i = a0 + a1u + a2u2 + a3u3 +….+ anun,
where coefficients a0, a1, a2, a3 … are real numbers which can be determined
according to expression
1 d ni
an =
,
n! du n u = U 0
here U0 is an operating point.
di
Easy to notice that a1 =
is VAC steepness at point u = U0, а2 is first-order
du
derivative of steepness (taken with a coefficient of 1/2!), а3 is second-order derivative
of steepness (with a coefficient 1/3!), etc. Number of terms is determined by accuracy
of calculation required.
There are also other ways of approximation, such as or exponential, either
trigonometric polynomial, either hyperbolic function approximation.
5.3 Methods of non-linear circuit analysis
Analytic method consists in employing Kirchhoff’s equations. The difficulty of
this approach is that equations would necessarily contain functions describing NLE.
This would make such non-linear equations, which leads to certain difficulties when
solving them.
Example. Circuit in Fig. 5.4 contains 2 nodes and 3 branches. If the circuit was
linear, then there would be only 3 equations necessary to calculate it:
1L. u1 + u2 – E = 0,
2L. u3 + uL – u2 = 0,
1 i1 – i2 – iL = 0.
In addition to resistors, there is a NLE in the circuit. The approximation function
iL = f(uL) needs to be defined for it. For most complicated circuits analytic solution
offers difficulties, so much more effective approaches appear to be numerical methods
of solving systems of non-linear equations.
Graph-analytic method is employed for respectively simple circuits containing
only one NLE. In this case optimal method is based on Thevenin’s or Norton’s
theorems.
70
+
R2
NLE ul
2c
1c
Е
For example circuit in Fig. 5.4 can
be replaced by an equivalent
(Fig. 5.5, a), where
R3
R1 i 1 1
U eg = U oc = i2 o R2 =
il
i2
E ⋅ R2
R1 + R2
in accordance with Fig. 5.5, b;
Figure 5.4 – Two-loop circuit
containing single NLE.
Reg =
R1 ⋅ R2
+ R3
R1 + R2
in accordance with Fig. 5.5, c.
Reg
R3
R1
il
R3
R1
+
+
NLE
Uevs
+
ul
-
Е
а)
R2
UOC
R2
i2o
b)
Revs
c)
Figure 5.5 – Equivalent transformation of circuit in Fig. 5.4:
а – equivalent voltage source; b – open-circuit voltage Uoc;
c – thevenin’s equivalent
To determine operating point of NLE one can use a load characteristic which is
defined by two points:
E
1) ul = 0, isc = eg
Reg
2) il = 0, U evs = U oc .
i
ics
il
VAC
OP
ul
u
U EVS = UOC
Figure 5.6 – Determination of
operating point by using load
characteristic
The line is shown in Fig. 5.6. Intersection
of two characteristics defines operating point
(ul and il).
Now you can verify the result by
Kirchhoff’s Law:
u L + iL ⋅ Reg = U eg .
Having determined values of uL and iL you
can calculate other variables for the circuit in
Fig. 5.4.
The expressions are recommended to derive by your own.
71
Graphical method of NLC equivalent transformation consists in replacing VAC
sections which describe series or parallel connection of elements by a single
equivalent VAC. Such a substitution is performed by summing individual
characteristics of NLEs with respect to current axis (in case of parallel connection) or
voltage axis (series connection).
There are characteristics of non-linear elements in Fig. 5.7, а.
NLE
R
i
i
∑
∑
NLE
R
i1+i2
i
u
u1 u2
а)
i2
i1
u1 + u2
u
u
b)
Figure 5.7 – VAC transformation:
a – series connection of elements;
b – parallel connection of elements.
When a series connection of elements, these characteristics should be summed
with respect to fixed values of current, as it is shown in Fig. 5.7, a and thus the
equivalent VAC is obtained.
When a parallel connection of elements, currents are added with respect to fixed
values of voltage, as it is shown in Fig. 5.7, b. Advantage of this method is that such
an equivalent substitution exposes influence of individual elements on general
solution, whereas original characteristic shape is of no importance as well as solving
non-linear equations is no longer required. A series drawback of the method is high
inaccuracy of the results (15…20 %).
Key questions
1
2
3
4
5
6
7
8
What is a non-linear circuit?
What are static and differential resistances of NLE?
What is VAC steepness?
What is VAC approximation?
What is the idea of piece-wise linear approximation?
What is the idea of analytic method of NLC calculation?
What is the idea of graphical method of NLC calculation?
What does the accuracy of graphical method depend on?
72
References
1 Бакалов В.П., Воробиенко П.П., Крук Б.И. Теория электрических цепей.
Радио и связь, 1998.
2 Белецкий А.Ф. Теория линейных электрических цепей. – М: Энергия,
1986.
3 Зелях Э.В. Теория линейных электрических цепей: Учеб. пособие. Разд.
первый – Одесса: ОЭИС им. О.С. Попова, 1978.
4 Атабеков Г.И. Теоретические основы электромеханики. Ч. 1 – М.:
Энергия, 1978.
5 Матханов П.Н. Основы анализа электрических цепей. Линейные цепи –
М.: Высшая школа, 1990.
6 Зевеке Г.В. и др. Основы теории цепей: Учебник для вузов.– М.,
Энергия, 1975.
7 Воробиенко П.П. Теория линейных электрических цепей. Сборник задач
и упражнений: Учеб. пособие для вузов. – М.: Радио и связь, 1989.
8 Шебес М.Р., Каблукова М.В. Задачник по теории линейных
электрических цепей: Учеб. пособие для электро- и радиотехнических спец.
вузов. – М.: Высшая школа, 1990.
73
PART 2
Guidelines for the laboratory works
Laboratory work № 1
ANALYSIS OF AUTONOMOUS TWO-TREMINAL NETWORKS
1 Purpose of work
1.1 To analyze a few autonomous two-terminal networks (ATTN) by running
their VAC.
1.2 To determine inner resistances and possible substitutions.
2 References
2.1 Бакалов В.П., Воробиенко П.П., Крук Б.И. Теория электрических цепей:
Учебник для вузов. – М.: Радио и связь, 1998.
2.2 Воробиенко П.П. Теория линейных электрических цепей. – М.: Радио и
связь, 1989. – № 2.11, 2.12, 2.13.
3 Quiz
3.1 Questions to check your knowledge on the topic are given in Appendix A.
4 Home task
4.1 To answer test questions.
4.2 To look over laboratory work description.
4.3 Prepare tables for VAC of ATTN.
5 Laboratory task
5.1 Measure ATTN characteristics (Fig. 5.1). The results write in respective table.
5.2 Plot respective VAC.
5.3 Determine inner resistances of each ATTN and show possible substitution
circuits.
Figure 5.1 – Virtual model for running volt-ampere characteristics of ATTN
74
6 Procedure of performance of laboratory work
6.1 To determine an ATTN characteristic, you need to connect respective
ATTN to the circuit. To do that, point mouse cursor at the switch connected in series
with the ATTN given and click left mouse button. This closes the switch.
6.2 Each ATTN is analyzed in three modes: open-circuit (o.c., switch on
position “2”), with a load (switch on position „3”), and short-circuit (s.c., switch on
position “4”). To choose the mode, use mouse. Ammeter A and Voltmeter V readings
need to be registered in s.c. and o.c. modes.
6.3 Use your mouse to connect analyzed ATTN to load (switch on position “3”).
While varying resistance value RL (to do that, use buttons “up” and “down”, next to
RL), register ammeter and voltmeter readings for each value of load resistance.
6.4 Write the readings in respective table.
6.5 Similarly, analyze the characteristics of other ATTN.
6.6 According to the results you have obtained plot volt-ampere characteristics
for each ATTN.
6.7 According to VAC determine inner resistances (conductances) of each
ATTN and give possible substitution circuits.
6.8 In accordance with VAC determine which type of energy source is it.
7 The contents of the protocol
7.1
7.2
7.3
7.4
7.5
7.6
Topic and purpose of work.
Analyzed circuit.
Result table and VAC diagrams.
Calculation results.
Substitution circuits for ATTN.
Conclusion.
75
Laboratory work № 2
ANALYSIS OF RESISTIVE CIRCUITS
WITH TWO SOURCES
1 Purpose of work
1.1 To master application of Ohm’s and Kirchhoff’s Laws.
1.2 To master superposition method.
1.3 To check calculation results experimentally.
2 References
2.1 Белецкий А.Ф. Теория линейных цепей. – М.: Радио и связь, 1986.– С.
49…62, 31…39.
2.2 Воробиенко П.П. Теория линейных электрических цепей. – М.: Радио и
связь, 1989.– задачи № 1.13, 1.18; № 2.1, 2.2; № 3.1, 3.2, 3.4, 3.5, 3.8, 3.
3 Quiz
3.1 Questions to check your knowledge on the topic are given in Appendix A.
4 Home task
4.1 Look over the theoretical material and answer the test questions.
4.2 Draw the circuit with two sources (given by an instructor).
4.3 Draw s circuit with a single voltage source (current sources need to be opencircuited).
4.4 Determine equivalent resistance Req with respect to remaining source.
4.5 Determine currents in all branches and show their conventionally positive
directions.
4.6 Verify obtained results according to Kirchhoff’s Laws.
4.7 Draw a circuit with a single current source (voltage sources need to be shortcircuited).
4.8 Apply paragraphs 4.4...4.6 to the circuit.
4.9 Determine real currents in the original circuit (with two sources) by
superposition method. Use calculation data from paragraphs 4.5, 4.8.
4.10 Check compliance with Kirchhoff's laws.
5 Laboratory task
5.1 Measure all the currents in the circuit with a voltage source only. Compare
with calculated data.
5.2 Measure all the currents in the circuit with the current source only. Compare
with calculated data.
5.3 Measure all the currents in the original circuit. Compare with calculated
data.
5.4 Reverse direction of the current source. Measure all the currents and
compare them to the results from paragraph 5.3.
5.5 Make conclusions.
76
6 Procedure of performance of laboratory work
6.1 Assemble the circuit with two (three) energy sources. To do this, point your
mouse cursor to each of two-terminal networks Z1…Z6 (Fig. 6.1) and click left mouse
button. The symbol of chosen element will turn into symbols R, E or J (when clicked
repeatedly, the symbol is replaced by next option). While choosing sources, pay
special attention to their polarity. Having finished assembling the circuit, confirm
your choice by pressing button „схема выбрана”.
Figure 6.1 – Virtual model to analyse multi-source DC circuits
6.2 There are entry fields in the upper right corner. Enter values of the elements
(in SI) there and then confirm your choice by pressing button „Значения элементов
введены”.
6.3 Automatically opens first partial circuit (with a single source), indicating
directions of the partial currents whose values are shown on the right of the circuit.
6.4 To view remaining partial and general (original) circuits and to determine
corresponding currents, press buttons from „2” to „6” (depending on the number of
partial circuits) or „Р”. The buttons are in the lower right corner of the window.
6.5 For each circuit compare obtained results with calculations and derive
conclusions.
6.6 To change source’s polarity or analyze new circuit press button „Новая
схема” in the lower right corner of the window and repeat paragraphs 6.1…6.5.
7 The contents of the protocol
7.1 Topic and purpose of work.
7.2 Analyzed circuit.
77
7.3 Calculation data for partial and general circuits.
7.4 Experimental data.
7.5 Conclusions.
Laboratory work № 3
ANALYSIS OF LINEAR AC CIRCUITS
1 Purpose of work
1.1 To analyze properties of RLC AC series and parallel circuits.
2 References
2.1 Бакалов В.П., Воробиенко П.П., Крук Б.И. Теория электрических
цепей: Учебник для вузов. – М.: Радио и связь, 1998.
2.2 Воробиенко П.П. Теория линейных электрических цепей. – М.: Радио и
связь, 1989. – С.120, зад. 7.1.
3 Quiz
3.1 Questions to check your knowledge on the topic are given in Appendix A.
4 Home task
4.1 Look over the theoretical material in recommended books.
4.2 Answer the test-questions.
4.3 Learn how to determine basic parameters of harmonic oscillation
(amplitude, frequency, initial phase) by its time diagram.
4.4 Learn how to determine difference in initial phases of two harmonic
oscillations by their diagrams.
4.6 Know how to represent a harmonic oscillation on a complex plane or in
frequency-domain.
4.7 Know how to derive formula from time diagram.
5 laboratory task
5.1 Analyze series connection of R, L, C elements (canonical series-oscillating
circuit):
5.1.1 Determine resonant frequency ω0, amplitude and effective values of
voltage across elements, maximal values of reactive powers, draw time diagrams of
voltages and powers.
5.1.2 Repeat paragraph 5.1.1 for frequencies ω1 = ω0/2 and ω2 = 2ω0.
5.1.3 Using time-diagrams, determine formulae and draw vector diagrams of
voltages.
5.1.4 Determine impedances of the reactive elements at each frequency.
5.2 Analyze parallel connection of R, L, C elements (canonical paralleloscillating circuit, or tank):
78
5.2.1 Repeat paragraphs 5.1.1…5.1.4 choosing currents and active power as
analyzed values.
5.3 For an arbitrary frequency different from ω1, ω0 and ω2 determine difference
in initial phases of voltages across the elements (for series circuit) and currents
through the elements (for parallel circuit), draw vector diagrams.
6 Procedure of performance of the laboratory work
6.1 Select tab „Последовательный контур” (Fig. 6.1). Now you must see a
model for series-oscillating circuit analysis. There are circuit itself and formula of
input current in the upper right corner of the program window. In the upper left corner
there are buttons to increase (clockwise direction) and decrease (counterclockwise
direction) frequency; there are also time diagrams of voltages and powers beneath
(curve’s colour corresponds to element’s colour).
6.2 Varying frequency by respective buttons, you can determine resonant
frequency of the circuit (voltages across the reactive elements must be equal in
magnitude and opposite in sign).
6.3 Resonant frequency and corresponding diagrams should be noted in your
protocol.
6.4 Set frequencies ω1 and ω2 (to do that, use buttons for varying frequency).
Note corresponding diagrams of voltages and powers in your protocol.
6.5 For each curve define formula and draw vector diagrams at all 3
frequencies.
6.6 To analyze parallel-oscillating circuit, choose tab „параллельный контур”.
A model similar to previous one appears (Fig. 6.2), where time diagrams present
currents and active power.
6.7 Similarly analyze the circuit in respect of currents and instantaneous power
in the resistor.
7.1
7.2
7.3
7.4
7 The contents of the protocol
Topic and purpose of work.
Results of your home task.
Circuits, graphs, vector diagrams, formulae.
Conclusions.
79
Figure 6.1 – Virtual model to analyze series-oscillating circuit
АД
Figure 6.2 – Virtual model to analyze parallel-oscillating circuit
Laboratory work № 4
INPUT FUNCTION ANALYSIS
(RL, RC, RLC)
1 Purpose of work
1.1 Analyze frequency dependence of linear passive two-terminal networks
containing R, L, C elements.
80
2 References
2.1 Бакалов В.П., Воробиенко П.П., Круг Б.Н. Теория электрических
цепей: Учебник для вузов. – М.: Радио и связь, 1998.
2.2 Белецкий А.Ф. Теория линейных электрических цепей. – М.: Радио и
связь, 1986.
3 Quiz
3.1 Questions to check your knowledge on the topic are given in Appendix A.
4 home task
4.1 Learn properties of passive R, L, C elements in AC circuits.
4.2 Calculate values of the elements using the following expressions
R = n , Ohm; L =
n
10
, mH; C =
, µF,
m+n
n( m + n)
where m is a number of the academic group (for example TE-02, m = 2), n is a
student’s serial number in the group list.
4.3 Calculate circuits in Fig. 4.1.
Ug
Ug
L
R
C
R
1
Ug
L
R
2
C
3
Figure 4.1– Analyzed circuits
Formulas describing circuit 1 (Fig. 4.1):
Z RL = R + jX L = z RL e jϕRL ;
X L = ωL; z RL = R 2 + X L2 ; I RL =
Ug
z RL
; ϕ RL = arctg
XL
.
R
When carrying out the calculation, assume frequency to be equal to: 0; 0,25ω11;
0,5ω11; ω11; 2ω11; 3ω11; ∞,
here ω11 is a reference frequency derived from the equation XL = R, hence
ω11 =
R rad
,
; U Г = 10 V.
L s
81
Tabulate the results (Table 4.1). Plot the frequency responses using the
calculation results.
Table 4.1
ω, rad/s
XL, Ω
zRL, Ω
IRL, mA
ϕ°RL
Use these formulas to describe circuit 2 (Fig. 4.1): Z RC = R − jX C = z RC e jϕRC ;
U
X
1
; z RC = R 2 + X C2 ; I RC = g ; ϕ RC = −arctg C .
z RC
R
ωC
Calculations needs to be carried out at the frequencies of: 0; 0,25ω12; 0,5ω12;
ω12; 2ω12; 3ω12; ∞,
here ω12 is a reference frequency derived from the expression XC = R, hence
XC =
ω12 =
1 rad
,
; U g = 10 V.
RC s
Tabulate the calculation results. Plot the frequency responses using the
calculation results.
Table 4.2
ω, rad/s
XС, Ω
z RС , Ω
IRС, mA
ϕ°RС
Use these formulas to describe circuit 3 (Fig 4.1): Z = R + jX LC = z RLC e jϕRLC ;
2
X LC = X L − X C ; z RLC = R 2 + X LC
; I RLC =
Ug
z RLC
; ϕ RLC = arctg
X LC
.
R
Calculations needs to be carried out at the frequencies of: 0; 0,25ω0; 0,5ω0; ω0;
2ω0; 3ω0; ∞,
Here ω0 is a reference frequency derived from the expression:
ω0 =
1
, rad/s.
LC
Tabulate the calculation results. Plot the frequency responses using the
calculation results.
Table 4.3
XL, Ω
ω, rad/s
XC, Ω
XLC , Ω
zRLС , Ω
IRLС , mA
ϕ°RLС
82
5 Laboratory task
5.1 Assemble the circuit shown in Fig. 4.1. Choose nominal values of the
elements, generator voltage.
5.2 Vary AC source frequency f [Hz], measure and note voltmeters readings.
Fill them in Table 4.1. Carry out measurements at the frequencies of: 0,25f11; 0,5f11;
ω 

f11; 2f11; 3f11  f11 = 11  .
2π 

Table 5.1
f, Hz
UR, V
UL, V
IRL, mA
XL, Ω
zRL, Ω
ϕ°RL
5.3 Calculate the following parameters of the circuit using the expressions
U
U
U
U
I RL = R ; X L = L ; z RL = g ; ϕ RL = arctg L ,
R
I RL
I RL
UR
Obtained results fill in the Table 5.1.
5.4 Repeat paragraphs 5.1…5.3 for circuit 2 (Fig 4.1). Use the following
formulas:
U
U
U
U
I RC = R ; X C = C ; z RC = g ; ϕ RC = −arctg C .
R
I RC
I RC
UR
Tabulate the results.
Table 5.2
f, Hz
UR, V
UС , V
IRС , mA
XС, Ω
z RС , Ω
ϕ°RС
5.5 Repeat paragraphs 5.1…5.3 for circuit 3 (Fig 4.1). These are the formulas to
use:
I RLC =
U
U
U − UC
UR
U
; X L = L ; X C = C ; z RLC = g ; ϕ RLC = arctg L
.
R
I RLC
I RLC
I RLC
UR
Tabulate the results.
Table 5.3
f, Hz UR , V UL, V
UС, V
ULС , V IRLС, mA
XL, Ω
XС , Ω zRLС, Ω ϕ°RLС
Using the tables plot corresponding diagrams.
6 Procedure of performance of the laboratory work
6.1 Assemble analyzed circuit (RL, RC, RLC). Click button „Выбор
элементов” at the top of the computer model (Fig .1). Replace networks Z1, Z2, Z3
with corresponding components (R, L, or C). When analyzing networks RL and RC
replace Z3 by short circuit. Figure 6.2 shows an example of model for RL circuit
analysis.
83
Figure 6.1 – Computer model of analysed circuit
Figure 6.2 – Computer model for RL circuit analysis
6.2 Choose nominal values of R, L, C components in accordance with the
variant given. To do this, click „Значения элементов” button. Enter values of the
elements in the window appeared. Press “OK”.
Note. Use the floating-point representation of input values. For instance 5 µF
becomes 5е-6 F.
6.3 Set parameters of the generator. Press button „Параметры генератора”. In
opened window enter generator voltage (set one time only), frequency, initial phase
(usually 0 degrees). Press “OK”.
Note. Frequency of 0 or ∞ Hz cannot be set.
6.4 Take readings. To do this, click the voltmeter (ammeter) with left button of
mouse. The readings will appear in opened window.
6.5 To plot the frequency responses of the circuit, carry out paragraphs 6.3 and
6.4 using frequencies from the range offered.
6.6 To modify the circuit, repeat paragraphs 6.1 and 6.2.
6.7 An approximate view of frequency responses can be observed, if pressed
„Результаты измерений”. To return to the menu of circuit assembling, press
„Изменить схему”.
84
7.1
7.2
7.3
7.4
7 The contents of the protocol
Topic and purpose of work.
Results of the home task.
Analyzed circuit, formulas related to it, diagrams, tables.
Conclusions.
85
APPENDIX A
Test questions to laboratory works
Laboratory work № 1
ANALYSIS OF AUTONOMOUS TWO-TREMINAL NETWORKS
1 What is an electric circuit?
2 What is an equivalent circuit?
3 What is an autonomous network?
4 What is an active element? Which electric elements are active?
5 What is instantaneous power?
6 What is an electric network?
7 What is a linear element?
8 What is an ideal voltage source?
9 What is an ideal current source?
10 What states Kirchhoff Current Law?
11 What states Kirchhoff Voltage Law?
12 What is a series connection of electric components?
13 What is a parallel connection of electric components?
14 Explain the substitution circuit for a non-ideal voltage source.
15 Explain the substitution circuit to a non-ideal current source.
16 What is a dependent source?
17 How many dependent sources do you know? What are they?
18 What is a passive electric component?
19 What is a non-autonomous electric component?
20 For a given node write an equation using Kirchhoff Current Law.
21 Write an equation describing a capacitive element С.
22 Write an equation describing an inductive element L.
23 Plot VAC of a non-ideal voltage source.
24 Plot VAC of a non-ideal current source.
25 How to calculate the amount of energy?
26 What is a non-linear element?
27 What units is conductance measured in?
28 What is the difference between voltage source and current source?
86
Laboratory work № 2
ANALYSIS OF RESISTIVE CIRCUITS
WITH TWO SOURCES
1 What are the relationships between values of current and conductance in a
resistive current divider?
2 What are the relationships between values of current and capacitance in a
capacitive current divider?
3 What are the relationships between values of current and inductance in an
inductive current divider?
4 What are the relationships between values of voltage and resistance in a
resistive voltage divider?
5 What are the relationships between values of voltage and capacitance in a
capacitive voltage divider?
6 What are the relationships between values of voltage and inductance in an
inductive voltage divider?
7 What is the essence of the Superposition Method in circuit analysis?
8 What is the essence of the Node Voltage Method in circuit analysis?
9 How do we apply Thevenin’s theorem in circuit analysis?
10 How do we apply Norton’s theorem in circuit analysis?
11 What is a current divider?
12 What is a voltage divider?
13 Write the equation governing a voltage divider?
14 Write the equation governing a current divider?
15 What is self-conductance of a node?
16 Is the following statement correct: components connected in series are joined
in a common node?
17 Is the following statement correct: the algebraic sum of the voltages around
any closed circuit is zero?
18 Is the following statement correct: the algebraic sum of current flowing
around any closed circuit is zero?
19 For a given node write Kirchhoff’s Current Law.
20 Write the equation describing a capacitive component С.
21 Write the equation describing an inductive component L.
22 Plot VAC of a non-ideal voltage source.
23 What is the graph of an electric circuit?
24 What is a graph node?
25 What is a graph rib?
26 What a cycle of a graph?
27 What is a graph tree?
28 What is a directed graph?
29 What is an undirected graph?
30 What is a planar graph?
87
Laboratory work № 3
ANALYSIS OF LINEAR AC CIRCUITS
1 What does the difference in initial phases of AC voltage and current in a
resistor equal to?
2 What does the difference in initial phases of AC voltage and current in an
inductor equal to?
3 What does the difference in initial phases of AC voltage and current in a
capacitor equal to?
4 What does the difference in initial phases of AC voltages in L and C
components equal to, in case they are connected in series?
5 What does the difference in initial phases of AC voltages in R and L, in case
they are connected in series?
6 What does the difference in initial phases of AC voltages in R and C
components equal to, in case they are connected in series?
7 What does the difference in initial phases of AC voltages in a RLC circuit
equal to, in case UR = UС = UL = 1 V? What about currents?
8 What does the difference in initial phases of AC currents in L and C
components equal to, in case they are connected in parallel?
9 What does the difference in initial phases of AC currents in R and C
components equal to, in case they are connected in parallel?
10 What does the difference in initial phases of AC currents in R and L
components equal to, in case they are connected in parallel?
11 What are harmonic oscillations?
12 Which units is angular frequency ω measured in?
13 How to derive voltage amplitude value from its RMS value?
14 Which units is frequency f measured in?
15 What is period of a harmonic oscillation?
16 What are the relationships between angular frequency ω and period T?
17 What is the average value of a harmonic oscillation?
18 How to calculate the amount of energy dissipated by a resistor?
19 How to determine inductive impedance?
20 How to determine inductive admittance?
21 How to determine capacitive impedance?
22 How to determine capacitive admittance?
23 What is electric resonance in a circuit?
24 How to determine circuit impedance?
25 How to determine circuit conductance?
88
Laboratory work № 4
INPUT FUNCTION ANALYSIS
1 Which way of complex numbers representation is convenient to use in order
to determine total impedance of a series circuit?
2 Which way of complex numbers representation is convenient to use in order
multiply them?
3 What does total voltage across a series connection of L and R components
equal to, in case individual voltage across each is 2 V?
4 What does the angle between individual voltage vectors in a series RL circuit
equal to, in case UL = UR = 10 V?
5 What does the angle between vectors of total voltage and voltage across the
resistor in a series RL circuit equal to, in case UL = UR = 1 V?
6 What does total impedance of a series RC circuit equal to, in case R = 3 Ω,
ХС = 4 Ω?
7 What is an amplitude frequency characteristic?
8 What is a phase frequency characteristic?
9 How does inductive impedance depend on frequency?
10 How does capacitive impedance depend on frequency?
11 How different are currents in a series RL circuit?
12 How different are currents in a series RC circuit?
13 How does inductive admittance depend on frequency?
14 How does capacitive admittance depend on frequency?
15 Which way of complex numbers representation is the most convenient for
subtracting two values of impedance?
16 Which way of complex number numbers representation is the most
convenient to use in order to determine total impedance or admittance of a RLC
circuit?
17 What does the total current of parallel RC circuit equal to, in case IR = IC =
= 1 A?
18 What does the angle between vectors of current in a parallel RC circuit equal
to, in case IR = IC = 1 A?
19 What does the angle between vectors of total current and current trough a
resistor in a parallel RC circuit equal to, in case IR = IC = 1 A?
20 What does the total current in a parallel RC circuit equal to, in case IG = IC =
= 1 A?
21 What does the total current in a series RC circuit equal to, in case UR = UC =
= 2 V?
22 What does the total current in a parallel RL circuit equal to, in case IR = 3 А,
IL = 4 А?
23 What does the total voltage in a series RC circuit equal to, in case UR = 2 V,
UC = 1 V?
24 What does the frequency ω equal to, in case XL = R = 100 Ω, L = 0,1 H?
89
25 What does the frequency ω equal to, in case XС = R = 100 Ω, С = 10 µF?
26 What does the current through a series RL circuit equal to, in case there is DC
voltage of 10 V applied to it and R = 10 Ω, L = 10 mH?
27 What does the current through a series RC circuit equal to, in case there is
DC voltage of 10 V applied to it and R = 10 Ω, C = 10 µF?
28 What does the phase shift between voltage across an inductor and current
through a series RL circuit equal to, in case frequency is zero?
29 What does the phase shift between voltage across an inductor and current
through a series RL circuit equal to, in case R = XL = 10 Ω?
30 What does the phase shift between voltage across a capacitor and current
through a series RC circuit equal to, in case R = XС = 10 Ω?
31 What does the impedance magnitude of a series RL circuit equal to, in case
R = XL = 10 Ω?
32 What does the impedance magnitude of a series RLC circuit equal to, in case
R = 1 Ω, XL = 2 Ω, XС = 1 Ω?
33 What does the total voltage of a series LC circuit equal to, in case UL = UC =
= 5 V?
90
APPENDIX B
Examples of problem solution for unit 1
Kirchhoff’s Laws
Problem 1
A few branches are joined in a node of an electric circuit. The currents through
them are shown in Figure below.
Determine current I5.
I1
5A
Solution
I2
8A
According to Kirchhoff’s Current Law:
I3
11A
I5
I1 − I 2 − I 3 + I 4 − I 5 = 0 ,
I4
3A
Hence
I 5 = I1 − I 2 − I 3 + I 4 =
Figure 1.1 – The node being
analysed
= 5 − 8 − 11 + 3 = −11A.
Verification:
5 – 8 – 11 + 3 – (– 11) = 0,
0 ≡ 0.
Problem 2
There is an electric circuit in Fig. 2.1: R1 = 10 Ω, R2 = 20 Ω, Е = 15 V, J = 1 A.
Find uj.
E +
R1
R2
u1
u2
Solution
uj
J
I
Figure 2.1 - The circuit
being analysed
Current I flows in the same direction as
source current J does. Voltage drops U1 and
U2, therefore, are directed as it is shown in
the Figure and respectively equal to:
U 1 = I ⋅ R1 = 1 ⋅ 10 = 10 V ;
U 2 = I ⋅ R2 = 1 ⋅ 20 = 20 V .
In accordance with Kirchhoff’s Voltage Law:
91
U1 + U 2 + U j − E = 0;
U j = E − U 1 − U 2 = 10 − 10 − 20 = −10 V.
Sign “–” indicates that the source polarity is opposite to that in the Figure.
Problem 3
Find branch currents applying Kirchhoff’s Laws to a circuit in Fig. 3.1.
Initial data: R1 = 10 Ω, R2 = 20 Ω, R3 = 30 Ω, Е1 =20 V, Е2 =10 V.
For the circuit given you can set up one equation by Kirchhoff’s Current Law
and two more by Kirchhoff’s Voltage Law. To do that, choose arbitrarily directions of
currents I1, I2, I3, as it is shown in Fig. 3.1, and
R1 I1
directions of loop-tracing (shown by dashed
I2 R2
arrows)
I3
E1 + u1
E
2
u
Thus, in accordance with Kirchhoff’s
R3 u3 2
Laws you obtain an equation set
l1
l2
Node: I 1 − I 2 − I 3 = 0;
+
l1:
Figure 3.1 – The circuit
being analysed
l2:
U 1 + U 3 − E1 = 0;
U 2 − E 2 − U 3 = 0.
Expressing voltages U1, U2, U3 in terms of currents and resistances:
I1 − I 2 − I 3 = 0;


 I1 ⋅ R1 + 0 + I 3 ⋅ R3 = E1;
0 + I ⋅ R − I ⋅ R = E .
2
2
3
3
2

Now currents I1, I2, I3 can be calculated by determinants:
I1 =
∆
∆1
∆
; I2 = 2 ; I3 = 3 ,
∆
∆
∆
where
1
∆ = R1
0
0
∆1 = E1
−1
0
E2
R2
−1
0
R2
−1
1
R3 = 10
− R3
0
−1
0
20
−1
30 = −600 − [10 ⋅ (30 + 20)] = −1100 ;
− 30
−1
R3 = − E1 [R3 − R2 ] + E2 [− R3 ] = −20 (50) + 10 (−30) = −1000 − 300 = −1300;
− R3
92
1
0
−1
∆ 2 = R1
E1
R3 = − E1 [− R3 ] − E2 [R3 + R1 ] = 20 (−30) − 10 (40) = −600 − 400 = −1000;
0
E2
− R3
1
−1
0
∆3 = R1
0
0
R2
E1 = −E1 ⋅ R2 + E2 ⋅ R1 = −20⋅ 20+10⋅10 = −400+100= −300;
E2
I1 =
1300
1000
300
= 1,182 A , I 2 =
= 0,91A , I 3 =
= 0,273 A .
1100
1100
1100
Verification of the obtained values is performed by checking power balance:
∑ Pk = 0 , or Рs = Рcons.
− I1 ⋅ E1 − I 2 ⋅ E2 + I12 ⋅ R1 + I 22 ⋅ R2 + I 32 ⋅ R3 = 0.
When substituted, currents will fit the last equation. This operation is leaved out
in order to be accomplished yourself.
Analysis of single-source circuits
Problem 4
I1
R1
I3
I2
U1
E
R3
R2
U3
U2
R4
U4
Find currents and voltages in a circuit
shown in Fig. 4.1. Values of components:
Е = 6 V, R1 = 5 Ω, R2 = 30 Ω, R3 = 10 Ω,
R4 = 20 Ω.
Solution
Figure 4.1 – The circuit
being analysed
1 Set directions of the currents and
voltages, as it is shown in Fig. 4.1.
2 Since the circuit contains mixed-type connections, transform it into more
appropriate form to deal with. Resistors R3 and R4 in Fig. 4.1 are connected in series,
so they can be replaced by an equivalent resistor R34
R34 = R3 + R4 = 10 + 20 = 30 Ω.
Fig. 4.2 demonstrates the circuit after this transformation.
In the latter circuit resistors R2 and R34 are connected in parallel, so they can be
replaced by an equivalent R234:
93
R2 ⋅ R34
30 ⋅ 30
=
= 15 Ω
R2 + R34 30 + 30
= U 3 + U 4 . Currents in branches can be found in
R234 =
The voltage U 2 = U 34
accordance
with
current
divider
equations:
R34
R2
; I 3 = I1
.
I 2 = I1
R2 + R34
R2 + R34
Now the circuit becomes one in Fig.
4.3.
It contains voltage source and two
resistors: all connected in series. The
current in the circuit and voltages across
elements can be found by Ohm’s Law:
I1
R1
I3
I2
U1
E
R2
E
6
=
= 0,3 A;
R1 + R234 5 + 15
R1
U1
E
R234
U 1 = I1 ⋅ R1 = 0,3 ⋅ 5 = 1,5 V;
U 2 = I1 ⋅ R234 = 0,3 ⋅ 15 = 4,5 V.
Currents I2 and I3 (see above)
30
= 0,15A;
30 + 30
30
I 3= 0,3
= 0,15A.
30 + 30
Voltages by Ohm’s Law
I 2 = 0,3
Figure 4.3 –
Equivalent circuit
U 3 = I 3⋅R3 = 0,15 ⋅ 10 = 1,5 V ;
U 4 = I 3⋅R4 = 0,15 ⋅ 20 = 3 V.
3 Result can be verified by checking power balance of the original circuit:
− E ⋅ I1 + I 1 ⋅ U 1 + I 2 ⋅ U 2 + I 3 ⋅ U 3 + I 4 ⋅ U 4 = 0.
Substituting calculated values we obtain:
− 6 ⋅ 0,3 + 0,3 ⋅ 1,5 + 0,15 ⋅ 4,5 + 0,15 ⋅ 1,5 + 0,15 ⋅ 3 = 0;
− 1,8 + 0,45 + 0,675 + 0,225 + 0,45 = 0;
− 1,8 + 1,8 = 0; 0 ≡ 0.
Thus, validity of the results is confirmed.
U34
Figure 4.2 – Equivalent
transformation of the circuit
I1
I1 =
R34
U2
U234
94
Problem 5 Superposition method
R1
I1
1 I3
I
R3
I4
J
R2
E
2
R4
3
Figure 5.1 – The circuit
being analysed
I1′
I3′
R1
R3
I2′
E
R2
OC R4
Figure 5.2 – First partial
circuit
Find branch currents in a circuit in Fig.
5.1 employing superposition method. There
are the following values known:
Е = 6 V; I = 0,6 A; R1 = R2 = 20 Ω;
R3 = R4 = 10 Ω.
Solution
1 Choose
conventionally
positive
directions of currents, as it is shown in Fig.
5.1.
2 Assume that the source current is zero
(J = 0). The source needs to be open-circuited
then. Now we obtain first partial single-source
circuit (Fig. 5.2).
To find partial currents I1′ − I 3′ you can
use results from previous task:
R34 = R3 + R 4 = 10 + 10 = 20 Ω;
R234 =
R2 ⋅ R34
20 ⋅ 20
=
= 10 Ω;
R2 + R34 20 + 20
E
6
=
= 0,2 A;
R1 + R234 20 + 10
R34
20
I 2′ = I1′
= 0,2
= 0,1A;
R
+
R
20
+
20
′
′
I
R
2
34
R1 I ′′ 1
3
3 2 I 4′′
1
R2
20
′
′
I
=
I
=
0
,
2
= 0,1A.
3
1
′
′
I2
R2 + R34
20 + 20
J
R4
SC
R2
3 Assume that the source voltage is zero
(Е = 0). The source needs to be short-circuited
then. Thus we obtain another partial single0
source circuit (Fig. 5.3).
Figure 5.3 – Second
Transform circuit in Fig. 5.3 by replacing
partial circuit
R1, R2, R3 parallel-series connection with an
1
equivalent resistor R123:
′
′
I
′
′
I3
4
R ⋅R
R123 = 1 2 + R3 =
R1 + R 2
R4
R123
J
20 ⋅ 20
=
+ 10 = 20 Ω.
20 + 20
0
Figure 5.4 – Second
partial circuit
I1′ =
95
The current can be determined in accordance with current divider equation:
I 3′′ = I
R4
10
= 0,6
= 0,2 A;
R13 + R4
20 + 10
R13
20
= 0,6
= 0,4 A;
R13 + R4
20 + 10
R2
20
I1′′ = I 3′′
= 0,2
= 0,1A;
R1 + R2
20 + 20
I 4′′ = I
I 2′′ = I 3′′
R1
20
= 0,2
= 0,1A.
R1 + R2
20 + 20
4 Now you can find resulting values by employing superposition principle. We
assume that all the currents are directed as it is shown in the original circuit (Fig. 5.1).
Partial currents of the same direction with those in the original circuit we suppose to
be positive (“+” sign), negative otherwise (“–” sign). Thus, real currents equal to:
I 1 = I 1′ − I 1′′ = 0, 2 − 0,1 = 0,1 A;
I 2 = I 2′ + I 2′′ = 0,1 + 0,1 = 0,2 A;
I 3 = I 3′ − I 3′′ = 0,1 − 0, 2 = −0,1 A;
I 4 = I 4′ + I 4′′ = 0,1 + 0, 4 = 0,5 A.
Negative values mean that real current direction is opposite to the one chosen.
5 Verifying the results by checking KCL (node 3):
− I1 + I 2 − J + I 4 = 0.
Substituting current values:
− 0 ,1 + 0 , 2 − 0 ,6 + 0 ,5 = 0 .
That confirms validity of the results.
1
R1 I1 I3 R3
Problem 6 Node voltage
analysis
Find currents in a circuit in Fig.
6.1 by node voltage approach. Use the
same initial data as in problem 5.
2
I4
I2
+
E
R2
U10 U20
J
R4
0
Figure 6.1 – The circuit being
analysed
96
Solution
Choose conventionally positive current directions (Fig. 6.1).
1 The circuit contains three principal nodes: 1, 2, 0.
Let node 0 be a reference node. Remaining nodes can be described by the nodal
equation set with respect to voltages U10 and U20:
1
G11U 10 + G12U 20 = I11 ;
2
G 21U 10 + G 22U 20 = I 22 ,
where U10, U20 are desired nodal voltages;
G11, G22 are self-conductances of nodes 1 and 2;
G12 = G21 is mutual conductance between nodes 1 and 2;
I11, I22 are nodal currents of respectively 1st and 2nd nodes;
1
1
1
1
1
1
+
+
=
+
+ = 0,2 S;
R1 R2 R3 20 20 10
1
1
1 1
G22 =
+
= + = 0,2 S;
R3 R4 10 10
1
1
G12 = G21 = −
= − = −0,1 S;
R3
10
E G
I11 =
=
= 0,3 А;
R1 20
G11 =
I 22 = J = 0,6 А.
Substitute obtained values in the equation set and solve it with respect to U10 and
U20:
 0,2U 10 − 0,1U 20 = 0,3;

− 0,1U 10 + 0,2U 20 = 0,6.
Solving by determinant method:
∆=
∆1 =
0,2
− 0,1
− 0,1
0,2
0,3
− 0,1
0,6
0,2
= 0,2 ⋅ 0,2 − (−0,1)(−0,1) = 0,04 − 0,01 = 0,03 ;
= 0,3 ⋅ 0,2 − (−0,1) 0,6 = 0,06 + 0,06 = 0,12 ;
97
∆2 =
0,2
0,3
− 0,1
0,6
= 0,2 ⋅ 0,6 − 0,3(−0,1) = 0,12 + 0,03 = 0,15 ;
U 10 =
∆1 0,12
=
= 4 V;
∆ 0,03
U 20 =
∆ 2 0,15
=
= 5 V.
∆ 0,03
2 Finding branch currents:
E − U 10 6 − 4
=
= 0,1 А;
R1
20
U
4
I 2 = 10 =
= 0,2 А;
R2 20
U − U 10 5 − 4
U
I 3 = 21 = 20
=
= 0,1 А;
R3
R3
10
U
5
I 4 = 20 =
= 0,5 А.
R4 10
I1 =
3 The results can be confirmed by checking power balance:
− E ⋅ I1 + I12 R1 + I 22 R2 + I 32 R3 − U 20 ⋅ J + I 42 ⋅ R4 = 0;
− 6 ⋅ 0,1 + 0,12 ⋅ 20 + 0,2 2 ⋅ 20 + 0,12 ⋅ 10 − 5 ⋅ 0,6 + 0,5 2 ⋅ 10 = 0;
− 0,6 + 0,2 + 0,8 + 0,1 − 3 + 2,5 = 0;
− 3,6 + 3,6 = 0;
0 ≡ 0.
As you may see the results are valid.
Equivalent source (Thevenin’s and Norton’s theorems)
Problem 7
Find current І3 in a circuit shown in Figure 7.1, а using Thevenin’s theorem, in
case Е1 = 20 V, Е2 = 10 V, R1 = 10 Ω, R2 = 20 Ω, R3 = 30 Ω.
98
1
R1
I3
+
E1
R3
R2
Rg
∗
1 I3
I1 R1
1 R2
1
+
U
R3 E
Uoc
1
R
Lp1
R=
2
2
+
E2
Eg
+
2
а)
b)
E2
+
c)
Figure 7.1 – The circuit being analysed (а), equivalent voltage source (b),
open-circuit voltage determination (c)
Solution
Consider branch containing resistor R3 to be a load and all the rest part of the
circuit to be an equivalent voltage source Eg with a series resistor Rg. Thus, you
transform the circuit into its equivalent shown in Figure 7.1 b. In this circuit current
Eg
I3 =
(*).
Rg + R3
To determine Eg, you need to disconnect resistor R3 in the original circuit and
calculate the open-circuit voltage between nodes 1 and 2 (Fig. 7.1, c).
First, find current I1 :
E + E2 30
I1 = 1
=
=1A .
R1 + R2 30
R1
1
R2
2
Figure 7.2 – Rg determination
Then U 1 = I 1 ⋅ R1 = 1 ⋅ 10 = 10 V . Loop 1
thus can be described in accordance with
KVL:
U 1 + U oc − E1 = 0 ;
U oc = E1 − U 1 = 20 − 10 = 10 V ;
U oc = Eg = 10 V .
To find Rg = Rin12, assume values E1
and E2 to be zero. Hence, you obtain the
circuit in Fig. 7.2.
Resistance Rin12 equals to
RR
10 ⋅ 20
Rin 12 = 1 2 =
= 6,666 Ω .
R1 + R2 10 + 20
Substituting Uoc = 10 V and Rg = Rin12
= = 6,666 Ω in expression (*) you obtain
5
C3
4
R1
e(t)
+
C2
1
L4
2
C5
3
Figure 8.1 – The cicruit being
analysed
R6
99
I3 =
10
= 0,272 A .
6,666 + 30
As you may have noticed application of Thevenin’s theorem gives the same
results as other methods (see Problem 3).
Dual transformations
Problem 8
Describe a circuit dual to the one in Figure 8.1.
Solution
To transform the circuit into dual one, you need to replace all the components
and sections in accordance with Table 8.1
Finding the dual circuit would be a good idea to do in stages:
а) In each closed loop choose a node and number it;
b) Then connect the nodes with lines so that they cross all the components of the
original circuit (see Fig. 8.1). Now you have a graph of dual circuit (dashed lines);
Table 8.1 – Dual values
Node
Loop
u
i
1
R
L
G1′
Node
i
u
G
C
L
L5′
2
C 4′
C
j(t)
Loop
L2′
L3′
4
3
G 6′
5
Figure 8.2 – Circuit dual to the
one in Fig. 8.1
c) Now in each branch of the dual graph
insert respective components and thus, the
circuit obtained in the end would be dual to the original one.
The result is shown in Figure 8.2.
Complex amplitudes
Problem 9
L1
R1
R2
L2
C1
Figure 9.1 – The circuit being analysed
There is a two-terminal network
in Figure 9.1. Find its impedance
Z ( jω) and admittance Y ( jω) .
100
Solution
ZL1 = jωL1 R1
Z C1 =
If to describe the circuit in terms
of impedances of its components,
then it can be transformed into the
one shown in Figure 9.2. All the
impedances in circuit below are
measured in [Ohm].
Here ZL1 and R1, ZL2 and R2 are
connected in series.
Thus
R2 ZL2 = jωL2
1
jωC1
Figure 9.2 – Symbolic equivalent of the
original circuit
Z 1 = Z L1 + R1 = jωL1 + R1 ,
Z 2 = Z L 2 + R2 = jωL2 + R2 .
Z1
Z2
Z С1
Figure 9.3 – Further transformation
of circuit in Fig. 9.2
Z ( jω ) =
=
Z e1 ⋅ Z C1
Z e1 + Z C1
In the latter circuit (Fig. 9.3)
components Z1 and ZС1 are connected in
parallel and these two together are in series
with Z2. Then total impedance of the
circuit given can be expressed as:
1
jωC1
+ Z e2 =
+ jωL2 + R2 =
1
jωL1 + R1 +
jωC1
( j ωL
1
+ R1 )
jωL1 + R1 + ( jωL2 + R2 )[( jωL1 + R1 ) jωC1 + 1]
.
( jωL1 + R1 ) jωC1 + 1
Its admittance
Y ( jω) =
( jωL1 + R1 ) jωC1 + 1
1
==
.
Z ( jω)
jωL1 + R1 + ( jωL2 + R2 )[( jωL1 + R1 ) jωC1 + 1]
The expression obtained can be further transformed to a standard form
Z=
a + jb
.
c + jd
This is recommended to accomplish on your own.
101
Problem 10
There is a two-terminal network in Fig. 10.1. The components it contains are:
R = 50 Ω, С = 20 µF, L = 0,1 H. Find its
L
impedance at frequencies of ω = 0 (DC), ω → ∞,
R
and ω = 103 rad/s.
С
Figure 10.1 – Analysed circuit
Solution
In terms of its impedances the circuit can be described as it is shown in Fig. 10.2.
The total impedance is:
Z Z
Z ( jω) = C L + R =
Z C+Z L
1
jωL
jωL
jωC
=
+R=
+ R.
2
1
−
ω
LC
1
1
+ jωL
Z
=
C
jωC
jω C
R
Thus we obtain
Z L = jωL
Figure 10.2 – Circuit impedances
Z (0) = R = 50 Ω ;
Z (∞ ) = R = 50 Ω .
Z (1000) =
j ⋅ 1000 ⋅ 0,1
j ⋅ 100
−
50
=
+ 50 = − j ⋅ 100 + 50 ;
1− 2
1 − 10 6 ⋅ 0,1 ⋅ 20 ⋅ 10 −6
Z = 502 + 1002 = 111,8 Ом ; θ Z = arctg
− 100
= −63,43о ;
50
Z ( jω) = 111,8 ∠ − 63,43 o .
Problem 11
Voltage u1 (t ) is applied to a network in
Fig. 11.1
u1 (t ) = 80 sin(1000 t + 40 о ), V
The current through the network is
i1(t) (2·1)
u1(t)
Figure 11.1 – Analysed
network
102
i1 (t ) = 10 sin(1000 t + 85 о ), mA .
Synthesize a circuit that meets the requirements given.
Solution
Harmonic current and voltage can be
represented by their complex amplitudes
o
o
U& m1 = 80e j 40 ; I&m1 = 0,01e + j 85
Network impedance can therefore be obtained
I&m1
Z(jω)
U& m1
Figure 11.2 – Impedance of
the network
as
o
o
U& m1
80 e j 40
Z ( jω) =
=
= 8000 e − j 45 = 8000 ⋅ cos(−45o ) − j ⋅ 8000 ⋅ sin(+45o ) =
o
I&m1 0,01e + j 85
= 5656 − j 5656.
Z = R − jX . Consequently R = 5656 Ω ; X = 5656 Ω ; X =
1
;
ω⋅С
1
= 0,17568 ⋅ 10 −6 = 0,1768 µF .
ωX
Thus, we may use a circuit in Fig. 11.3.
R
С
It provides the required current at frequency
ω = 1000 rad/s.
Figure 11.3 – The circuit required
C=
Problem 12
Find currents and voltages in a circuit in Figure 12.1, in case
π

e(t ) = 6 cos10 6 t +  ; R1 = 1000 Ω; R2 = 2000 Ω; L2 = 2 mH.
4

The answer represent in time-domain. Draw current and voltage vector diagrams.
103
i1(t) R1
1
i3(t)
İ1
i2(t)
e(t)
Figure 12.1 – Analysed circuit
1
U& 1
E&
L2
R2
R1
Lp1
İ3
İ2
R2
Lp2
U& 2
jωL2
U& 3
Figure 12.2 – Impedances of the
circuit analyzed
Solution
Each parameter of the circuit whether it is impedance of some component or
current (voltage) value can be represented with respective complex value (see Figure
6 j⋅45о
12.2). The source RMS voltage is E& =
e
= 3 + 3 j . Choose arbitrarily current and
2
voltage directions. The circuit contains a single source. Total impedance of the circuit
j ωL2 R2
with respect to the source is Z = R1 +
.
j ωL2 + R2
E&
According to Ohm’s Law I&1 = , U& 1 = I&1 R1 . Currents leaving node 1 can be
Z
expressed as
jωL2
R2
I&2 = I&1
;
I&3 = I&1
.
jωL2 + R2
R2 + jωL2
By Ohm’s Law, U& 2 = I&2 R2 ; U& 3 = I&3 jωL2 .
6
−3
Substituting values given, you obtain: ω L2 = 10 ⋅ 2 ⋅ 10 = 2000 Ω.
j ⋅ 2000 ⋅ 2000
j ⋅ 2000
= 1000 +
=
j ⋅ 2000 + 2000
1+ j
2000 j ⋅45 о
= 1000 +
e
= 2000 + j ⋅ 1000 = 1000 5 e
2
Z = 1000 +
I&1 =
6 j⋅45о
e
2
1000 5e j⋅26,57
j ⋅ 26 ,57 о
о
о
= 1,9 ⋅ 10 −3 e j⋅18, 43 = (1,8 + j 0,6)10 −3 , A;
о
о
U& 1 = 1,9 ⋅ 10 −3 e j⋅18, 43 ⋅ 1000 = 1,9e j⋅18,.43 = 1,8 + j 0,6 , V;
,Ω
104
о
I&2 = 1,9 ⋅ 10 −3 e j⋅18, 43
о
j ⋅ 2000
= 1,34 ⋅ 10 −3 e j⋅63,63 = (0,6 + j1,2)10 −3 , A;
j ⋅ 2000 + 2000
о
о
U& 2 = 1,34 ⋅ 10 − 3 e j ⋅63 , 63 2000 = 2 , 68 e j ⋅63 , 63 = 1, 2 + j 2 , 4 , V ;
о
I&3 = 1,9 ⋅ 10 − 3 e j ⋅18 , 43
о
2000
= 1,34 ⋅ 10 − 3 e − j ⋅25 , 55 =
j ⋅ 2000 + 2000
= (1, 2 − j 0 ,6 ) 10 − 3 , A ;
о
о
U& 3 = 1,34 ⋅ 10 − 3 e − j ⋅63 , 63 j 2000 = 2 , 68 e j ⋅63 , 63 = 1, 2 + j 2 , 4 , V .
Confirm the results by checking Kirchhoff’s Laws. For node 1
− I&1 + I&2 + I&3 = 0 , or I&2 + I&3 = I&1 .
Substituting values:
o
o
o
− 1,9 ⋅ 10−3 e j⋅18, 43 + 1,34 ⋅ 10−3 e j⋅63,63 + 1,34 ⋅ 10−3 e − j⋅25,55 =
= (−1,8 − j ⋅ 0,6) + (0,6 + j ⋅ 1,2) + (1,2 − j ⋅ 0,6)10−3 = 0.
Current vector diagram is shown in Figure 12.3.
Applying KVL to loop Lp1, you obtain:
Im
İ2
− E& + U& 1 +U& 2= 0 , or U& 1 +U& 2= E& .
İ1
Re
0
İ3
Figure 12.3 – Current vector
diagram
Im
i1 (t ) = 1,9 2 ⋅ 10 −3 cos(10 6 t + 18,43o ), A;
U& 2
E&
i2 (t ) = 1,34 2 ⋅ 10 −3 cos(10 6 t + 63,63o ), A;
Re
0
Substituting values:
o
o
6 j⋅45o
−
e
+ 1,9 ⋅ e j⋅18, 43 + 2,68 ⋅ e j⋅63, 63 =
2
= ( −3 − j ⋅ 3) + (1,8 + j ⋅ 0,6) + (1,2 + j ⋅ 2,4) = 0.
These operations confirm the results.
Transform obtained complex values into
time-domain functions:
U& 1
Figure 12.4 – Voltage
vector diagram
i3 (t ) = 1,34 2 ⋅ 10 −3 cos(10 6 t − 25,55 o ), A;
u1 (t ) = 1,9 2 cos(10 6 t + 18,43o ), V;
u 2 (t ) = u 3 (t ) = 2,68 2 cos(10 6 t + 63,63o ), V.
Voltage vector diagram is in Figure 12.4.
105
Checking power balance:
~
~
S pr = S cons ;
~
S pr = Ps + jQs = S s ⋅ e jϕ ;
~
S cons = Pcons + jQcons = S cons ⋅ e jϕ ;
∗
o
6 j 45o
~
S s = E& I 1 =
e 1,9 ⋅ 10 −3 e − j18, 43 = 7,2 ⋅ 10 −3 + j 3,6 ⋅ 10 −3 ;
2
S s = E I1 = Ps2 + Qs2 = 8,06 ⋅10 −3 , VA ;
Pcons = I 12 R1 + I 22 R2 = (1,9 ⋅ 10 −3 ) 2 1000 + (1,34 ⋅ 10 −3 ) 2 2000 = 7,2 ⋅ 10 −3 , W ;
Qcons = I 32 X L = (1,34 ⋅ 10 −3 ) 2 2000 = 3,6 ⋅ 10 −3 , VAR ;
2
2
S cons = Pcons
+ Qcons
= 8,05 ⋅ 10 −3 , VA .
When comparing results, you might see
Ps = Pcons = 7,2 ⋅ 10 −3 , W and Qs = Qcons = 3,6 ⋅ 10 −3 , VAR ;
S s = S cons = 8,05 ⋅ 10 −3 , VA .
The task is solved correctly.
Nonlinear circuit analysis
Problem 13
Find branch currents of a circuit shown
in Fig. 13.1. The components are: E = 10 V,
R1 = R2 = 10 Ω, R3 = 5 Ω, VAC of the
nonlinear resistor is defined by function
i = 0,1u2, А.
R 1 I 1 I 3 R3
1
I2
+
E
R2
INLE
UNLE
NLE
Solution
2
Figure 13.1 – Analysed circuit
Rg
+
Eg
INLE
NLE
UNLE
Figure 13.2 – Equivalent voltage
source in the original circuit
1 Use Thevenin’s Theorem, thus
replacing the circuit section containing
resistors R1, R2, R3 by an equivalent
voltage source (Fig 13.2).
Variables Eg and Rg can be
determined in the following way: the
source’s voltage is open-circuit voltage
106
UOC with the nonlinear resistor extracted (Fig. 13.3).
Resistors R1 and R2 form voltage divider. Hence
R1
R3
1
U oc = U R = E
2
+
E
R2
Uoc
E = U oc = 5 V.
Resistance RG is an input resistance of the
circuit with respect to the NLE under the
condition that the source is short-circuited (Fig.
13.4):
R R
Rg = RIN = R3 + 1 2 =
R1 + R2
2
Figure 13.3 – Circuit with
nonlinear resistor extracted
R1
SC
1
R3
=5+
RIN
R2
0
0
10 ⋅ 10
= 10 Ω.
10 + 10
2 Voltage and current in NLE (INLE and
UNLE) can be found by graphic method.
Using expression i = 0,1u2 А we can plot
VAC of the NLE.
2
Figure 13.4 – Input resistance
determination
U, B
I, A
i, А
R2
10
= 10
= 5 V;
R1 + R 2
10 + 10
1
2
3
0,1 0,4 0,9
4
1,6
5
2,5
1,6
VAC
1,4
1,2
1,0
0,8
0,6
Isc
Load line
OP
0,4
INLE
0,2
0
Eg
1
UNLE 2
3
4
5
Figure 13.5 – The nonlinear element’s VAC
u, V
107
Thevenin’s source voltage Eg = 5 V is marked on the voltage-axis (Figure 13.5)
E
5
and maximal possible current I sc = g = = 0,5 А in the circuit is marked on the
Rg 10
current-axis. By joining these two points with a line you obtain so called load line,
which crosses VAC at an operating point (OP). The OP coordinates are actual values
of current and voltage in the NLE (INLE and UNLE). Thus, the values required are
I NLE ≈ 0,3 А, U NLE ≈ 1,7 V.
3 Back to the original circuit (Fig. 13.1). It is obvious that I NLE = I 3 = 0,3 А.
Applying Ohms’ and Kirchhoff’s Voltage Laws to the circuit section containing
R2, R3, NLE you find current I2:
I 3 R3 + U NLE − I 2 R2 = 0 ,
I2 =
I 3 R3 + U NLE 0,3 ⋅ 5 + 1,7
=
= 0,32 А.
R2
10
In accordance with KVL, current I1 is:
I1 − I 2 − I 3 = 0 ;
I1 = I 2 + I 3 = 0,32 + 0,3 = 0,62 А.
Accuracy of such a method to a considerable extent depends on graphical part
accuracy.
Checking power balance:
− E ⋅ I 1 + I 12 ⋅ R12 + I 22 ⋅ R22 + I 32 ⋅ R32 + I NLE ⋅ U NLE = 0 ;
− 10 ⋅ 0,62 + (0,62) 2 ⋅ 10 + (0,32) 2 ⋅ 10 + (0,3) 2 ⋅ 5 + 0,3 ⋅ 1,7 = 0 ;
− 6,2 + 3,844 + 0,1024 ⋅ 10 + 0,09 ⋅ 5 + 0,51 = 0 ;
− 6,2 + 3,844 + 1,024 + 0,45 + 0,51 = 0 ;
6,2 = 3,844 + 1,024 + 0,45 + 0,51;
6,2 ≈ 5,83 .
6,2 − 5,82
Calculation error is
100 % = 6 %, which is acceptable for graph6,2
analytic method.
108
N.F. Arbuznikova
A.A. Novikov
A.U. Kalashnikov
A.V. Shkulipa
AC and DC circuit analysis:
Learner’s guide to
“Electric Circuits and Signals”