asp08

Analog Signal Processing

Dr.-Ing. Tuan Do-HongDepartment of Telecommunications Engineering

Ho Chi Minh City University of [email protected]

September, 2008

Course Overview

0.1 Course Description

Introduction to analog signal processing, with an emphasis on underly-ing concepts from circuit and system analysis: linear systems, review ofelementary circuit analysis, dierential equation models of linear circuitsand systems, Laplace transform, convolution, stability, phasors, frequencyresponse, Fourier series, Fourier transform, active lters and AM radio.

0.2 Prerequisite

Calculus. Dierential equations. Physics-bases treatment of electricity andmagnetism. Introductory exposure to circuit analysis.

0.3 Goals

To introduce fundamentals of analog signal processing, with major empha-sis on circuit analysis, dierential equations, convolutions, Fourier methods,and applications in ltering and AM radio.

0.4 Topics

Examples of signals and signal processing systems Analog linear shift-invariant systems Circuits and linear systems Review of DC circuit analysis: KCL, KVL, dependent sources Capacitors and inductors as circuit elements Op-amp circuits

2

0.5. ASSESSMENT 3

Characterization and solution of LSI systems via linear, constant-coecient dierential equations

Transient response of RLC circuits Complex numbers and functions of a complex variable One-sided Laplace transform Impedance Laplace transform solution of dierential equations General form of solution to a dierential equation Transfer function and block diagrams Impulse as a generalized function Convolution Stability Sinusoidal steady-state analysis and phasors Steady-state analysis of circuits Frequency response Fourier series Fourier transform Design of active lters AM radio Sampling theorem and overview of digital signal processing

0.5 Assessment

Home assignments: 20% Mid-term exam: 20% Final exam: 60%

40.6 Texts

B. P. Lathi, Signal Processing and Linear Systems, Berkeley-CambridgePress, 1998.

David E. Johnson, John L. Hilburn, Johnny R. Johnson, Basic ElectricCircuit Analysis, 3rd edition, Prentice-Hall, 1986.

0.7 Course Content

1. Fundamentals of Circuit Analysis

Circuit introductory. Circuit reduction techniques. Circuit analysis techniques. Real sources, controlled sources and maximum power transfer. Operational ampliers. Time-dependent circuits. First-order circuits. Second-order circuits.

2. Introduction to Signals and Systems

Classication of signals. Signal operations and models. Classication of systems. System model: Input-output Description.

3. Time-Domain Analysis of Continuous-Time Systems

System response to internal conditions: Zero-input response. Unit impulse response h(t). System response to external conditions: Zero-state response. Classical solution of dierential equation. System stability.

4. Signal Representation by Fourier Series

0.7. COURSE CONTENT 5

Signals and vectors. Signal representation by orthogonal signal set. Trigonometric Fourier series. Exponential Fourier series. LTIC system response to periodic inputs.

5. Continuous-Time Signal Analysis: Fourier Transform

Aperiodic signal representation by Fourier integral. Properties of Fourier transform. Signal transmission through LTIC systems. Ideal and practical lters. Application to communications: Amplitude Modulation (AM).

6. Continuous-Time Signal Analysis Using Laplace Transform

Laplace transform and properties. Solution of dierential and integro-dierential equations. Analysis of electrical networks: Transformed network. Block diagram. Application to feedback and controls.

7. Frequency Response and Analog Filters

Frequency response of an LTIC system. Bode plots. Filter design by placement of poles and zeros of H(s). Butterworth lters. Chebyshev lters.

8. Sampling

Sampling theorem. Numerical computation of Fourier transform: Discrete Fourier

Transform (DFT).

Fast Fourier Transform (FFT).

Chapter 1

Fundamentals of Circuit Analysis

1.1 Circuit Introductory

1.1.1 Circuit Variables

Scientic investigation of static electricity was done in late 1700s andCoulomb is credited with most of the discoveries. He found that elec-tric charges have two attributes: amount and polarity. There are two typeof charges opposite charges attract and similar polarity ones repel eachother. Charge polarity is indicated by positive and negative signs becausepositive and negative charges cancel each other when brought together. Asa results, the electric charge can be described by an algebraic number, q,with units of Coulomb (C).

Because opposite charges attract each other, energy is expanded to sepa-rate them from each other. This energy is stored in the electric eld betweenthe two reservoir of separated charges and is recovered when the chargesare allowed to come together. The stored energy per unit charge is calledthe voltage or potential dierence between the two reservoir of charges:

v =dW

dq(1.1)

where the unit is Volt (V = JC ).

Note that we need two reservoir of charges. So voltage is between twopoints. We also use + and - signs to indicate the direction for measuringv. From denition of voltage above, w is energy needed to move a positivecharge from - reservoir to + reservoir.

One can dene a reference point for measuring voltages (typically shown

6

1.1. CIRCUIT INTRODUCTORY 7

as ground). The voltage between any point and this reference point is callthe potential of that point. It is always assumed that + is at the point andthe - sign is at the reference point. Therefore, there is no need to indicate+ and - signs for potential. Voltage between two points is the dierencebetween the potential of the two points (see Figure 1.1).

Voltage between two charge reservoirs is analogous to height dierencebetween two uid reservoir and the same way, the potential of each pointis analogous to its elevation compared to some reference (e.g., sea level).

++

--

vv

--

++

vv

VV11

VV22

vv == vv1122 == VV11 -- VV22

vv == vv2211 == VV22 -- VV11vv == --vv

00

PPootteennttiiaall

hh11

hh22

00 SSeeaa lleevveell

EElleevvaattiioonn

++

--

hh

hh == hh11 hh22

Figure 1.1: Voltage between two charge reservoirs.

If we connect the charge reservoirs, electric charges ow from one toother. The rate of the charge ow through a specic area is called theelectric current:

i =dq

dt(1.2)

with the current owing in the direction of the charge ow (it means thata positive current is associated with the ow of positive charge). The unitof current is Ampere (A = C/s).

In principle, electric charges generate an electric eld and motion of thecharged particles (current) generates a magnetic eld. This electromagneticeld interacts with all charges and aect them. The behavior of such asystem is described by Maxwells equation. Solution of Maxwells equations,

8 CHAPTER 1. FUNDAMENTALS OF CIRCUIT ANALYSIS

however, is dicult and not needed expect for some cases (propagation ofelectromagnetic wave and light, antennas, etc.).

In most relevant engineering cases the problem can be greatly simpliedby noting that electric charges preferentially ow through a conductor (ora semiconductor) as opposed to vacuum, air, or any insulator. In this case,the system can be described as a circuit containing circuit elements andconnecting ideal wires. Circuit theory is the scientic discipline that de-scribes behavior of circuits be and is built upon the following assumptions:

All of the electromagnetic phenomena occurs inside each circuit ele-ment. They communicate with the outside world only through thevoltage across and current that particular element.

Circuit elements are connected to each other with ideal wires that donot impede ow of charge. They can be stretched (making them longeror shorter, for example) without any eect on the circuit.

Net amount of charge cannot be accumulated in any circuit elementor any location in the circuit. If a net charge of q enters a circuitelement, the same amount of charge should leave the element. Thismeans: (a) a circuit element should have at least two terminals, (b)Because current travels through the system at a good fraction of speedof light, we can safely assume that the total current entering a circuitelement is exactly equal to the current leaving that element at anyinstant time.

++

--

vv

ii

ii

11

22

CCiirrccuuiitt eelleemmeenntt

IIddeeaall wwiirree

Figure 1.2: Circuit elementwith two terminals.


In the context of circuit denition above, the important physical quan-tities are current and voltages (and electric power). These are the circuitvariables that form the basis for communication between circuit elements.

The value of voltage between two points, v, is incomplete unless the +and - signs are assigned to the two points. Similarly, the value of currentis incomplete unless a reference current direction is assigned. However, asv and i are algebraic numbers, the assigned position of + and - for voltageand direction for the current is arbitrary. The sign of v and i ips accordingto this choice.

Internal of each circuit element impose a relationship between the cur-rent owing in the element and the voltage across that element. This re-lationship is called the element law or i-v characteristics. While the choiceof reference directions for current through and voltage across an elementis arbitrary, the element law or i-v characteristics of an element dependson the reference directions. To see this point, note that there exists fourpossible choices for voltage and current directions in a two-terminal circuitelement:

++

--

vv

ii

11

22

ii

11

22

++

--

vv

((aa))

++

--

vv

ii

11

22

ii

11

22

++

--

vv

((bb))

Figure 1.3: (a) Passive sign convention. (b) Active sign convention.

In two left cases, the current direction is marked such that it ows from+ to - signs. This case is called the passive sign convention. In the tworight cases, current ows from - to + signs. This case is referred to as activesign convention. Obviously, the i-v characteristic will be dierent in eachcase as the sign of current or voltage is switched. To resolve this confusion,we follow this convention:


i-v characteristics of two-terminal element are written assuming pas-sive sign convention.

Use passive sign convention when marking current and voltages in acircuit as much as possible.

So, in marking references for voltages and currents in the circuit, it isbest to arbitrarily choose either the voltage or current reference directionand then use passive sign convention to mark the other variable. Note thatit may not be possible or practical to mark every element using passivesign convention. In this case, a good rule is to mark every element exceptvoltage and current sources using passive sign convention.

Electric power produced or absorbed in an element

The power, denoted by P , is dened as

P =dW

dt(1.3)

where the unit is Watt (W = J/s). Since v = dW/dq and i = dq/dt (bothtotal derivatives), then:

P =dW

dt=

dW

dq dq

dt= i v (1.4)

The quantities v and i are generally function of time. Therefore P is atime-varying quantity, sometimes called the instantaneous power.

Using the denition of the voltage, one nds that if we use passive signconvention:

P < 0: Element is producing power.

P > 0: Element is absorbing power.

Obviously, if we use active sign convention:

P < 0: Element is absorbing power.

P > 0: Element is producing power.


AA

++

--

ii

vv BB

Figure 1.4: Absorbedand supplied powers.

Example

Find the power absorbed or supplied by elements A and B if i = 25A andv = 120V.

For element A,

P = vi = 25 120 = 3, 000W = 3KW

Since element A reference directions follow passive sign convention andP > 0, element A is absorbing 3 KW of power.

For element B,

P = vi = 25 120 = 3, 000W = 3KW

Since element B reference directions follow active sign convention and P > 0,element B is supplying 3 KW of power.

1.1.2 Kirchho Laws

A circuit consists of circuit elements attached to each other with idealwires or connectors.

Node: A node is a point in the circuit that is connected to two or morecircuit elements with ideal wires.

Loop: A loop is a closed path in a circuit through at least two circuitelements which return to starting node without passing through any nodetwice.


Kirchho Current Law (KCL): KCL follows from our assumptionthat no net charge can be accumulated at any point in the circuit (or in anode): Sum of currents entering a node is equal to sum of currents leavinga node.

Example: Write KCL for the marked node.

+i1 i2 i3 + i4 = 0

ii11 ii44ii22 ii33

Figure 1.5: Example for KCL.

Kirchho Voltage Law (KVL): KVL follows from the denition ofthe voltage between points. Voltage is dened as the amount energy tomove unit charge from one point to another. Zero net energy should beexpanded if a charge q is moved around a closed loop and is returned backto its original position, i.e.,

W = q

v = 0

v = 0 (1.5)

Algebraic sum of all voltages around any closed path in a circuit is zero.

Example: Write KVL for the marked loop.

+vb + vc va = 0


++

--

vvaa

++ --vvbb

++

--

vvcc

Figure 1.6: Example for KVL.

KVLs and KCLs are constraints on circuit variables which arise becauseof the circuit arrangement (attachment of connection wires). In addition,internal of each circuit element impose a relationship between the currentowing in the element and the voltage across that element (element lawsor i-v characteristics). Combination of these two set of constraints resultsin a unique set of values for the circuit variables (currents and voltages).

In a circuit with E elements, there are 2E circuit variables (i and v foreach element). We need 2E equations to nd these circuit variables. If thecircuit has N nodes, we can write (N 1) KCL equations (KCL on theNth node is exactly the sum of KCL on the other (N 1) nodes). We canalso write EN +1 independent KVLs and E i-v characteristic equations.Totally, it is 2E equations.

If the i-v characteristic is a linear relationship between i and v, theelement is a linear element. If a circuit is made of linear element, theresulting set of 2E equations in 2E variables for a linear algebraic set ofequations. In this course, we only use linear elements.

1.1.3 Linear Circuit Elements

Resistor

i-v characteristic: v = Ri.

Resistance: R. Unit: Ohm ().

Conductance: G = 1/R. Unit: Siemens (S).

P = vi = Ri2 (1.6)


P > 0: Resistor always absorbs power.

ii ++ --vv

RR

ii

vv

ii == vv//RR

Figure 1.7: Resistor: Circuit symbol and i-v characteristic.

Independent Voltage Source (IVS)i-v characteristic: v = vs for any i.

ii++

--vvss

ii

vv

++

--

vvvvss

vv == vvss

Figure 1.8: Independent voltage source: Circuit symbol andi-v characteristic.

Independent Current Source (ICS)i-v characteristic: i = is for any v.

Short Circuiti-v characteristic: v = 0 for any i.Note that a short-circuit element is a special case of a resistor (with

R = 0) or an ideal voltage source (with vs = 0).

Open Circuiti-v characteristic: i = 0 for any v.Note that a open-circuit element is a special case of a resistor (with

R ) or an ideal current source (with is = 0).


ii

iiss

ii

vv

++

--

vv

iiss ii == iiss

Figure 1.9: Independent current source: Circuit symbol andi-v characteristic.

iiii

vv

++

--

vv

Figure 1.10: Short circuit: Circuit symbol and i-v character-istic.

iiii

vv

++

--

vv

Figure 1.11: Open circuit: Circuit symbol and i-v character-istic.


Switch

ii++

--

vv

i-v characteristic:

Switch open: Open circuit i = 0 for any v.

Switch close: Short circuit v = 0 for any i.

1.1.4 Example

Consider the circuit as in Figure 1.12. Find i1

++

--

2255

55 1100

7700 88118800VV

((aa))

ii11

ii00

++

--

2255

55 1100

7700 88118800VV

((bb))

ii11

ii00++ --vv00

++ --vv22++

--

vv11

++ --vv33++

--

vv44

ii44

ii33ii22

ii55

Figure 1.12: Example of circuit analysis using KVL and KCL. (a) Original circuit. (b) Circuitwith nodes and loops.

We rst mark the circuit variables consisting of 11 unknowns:

v0, v1, v2, v3, v4, i0, i1, i2, i3, i4, i5


identify the nodes, and write the i-v characteristics equations:

v0 = 25i0v1 = 70i1v2 = 5i2v3 = 10i3v4 = 8i4

Note that we have 6 elements so we have 12 circuit variables. However,the IVS element law species the voltage across its terminal, so we have 11circuit variables which have to be found.

The circuit has four nodes. Since we need to write KCL only in N 1nodes, we choose not write KCL in the bottom node. The KCLs are:

i0 + i2 + i5 = 0i3 + i1 i2 = 0i4 i3 i0 = 0

We need to write NKV L = 11 5 3 = 3 KVLs. Choosing 3 loops withsmallest number elements, we get:

+v0 v3 v2 = 0+v2 + v1 (180) = 0+v3 + v4 v1 = 0

Above are eleven equations in eleven unknowns that can be solved. Thenumbers of equations to be solved can be halved by using i-v characteristicsequations to substitute in KVLs and KCLs. For example, if we substitutefor voltages from i-v characteristics equations in KVLs, we get 6 equationsin 6 unknown currents:

i0 + i2 + i5 = 0i3 + i1 i2 = 0i4 i3 + i0 = 025i0 10i3 5i2 = 05i2 + 70i1 = 18010i3 + 8i4 70i1 = 0

Note: As can be seen, even relatively small circuits results in a largenumber equations to be solved. Several techniques for reducing the numberof equation to be solved are introduced next.


1.2 Circuit Reduction Techniques

Combination of KVLs, KCLs, and i-v characteristics equations result ina set of linear equations for the circuit variables. While the above set ofequation is complete and contains all necessary information, even small cir-cuits require a large number of simultaneous equations to be solved as seenpreviously. We learned that we can use i-v characteristics equations whenwe are marking the circuit variables and reduce the number of equationsto be solved to only number of KCLs and KVLs. We will learn later twomethods, node-voltage and mesh-current, which reduce the number of equa-tions to be solved further to either number of KCLs or number of KVLs.This is the best we can do in this direction.

As it is easier to solve smaller sets of equations (e.g., it is easier tosolve two sets of two equations in two unknown as compared to a set of 4equations in 4 unknowns), one can break up the circuit into smaller piecesand solve each individually and assemble back the whole circuit. We canuse this principle to combine circuit elements and make a much smallercircuit. These techniques are described below.

Recall that we are using lumped circuit elements, i.e., circuit elementscommunicate to outside world and other circuit elements only through iand v. Conversely, the outside world (the rest of the circuit) communicatewith the circuit element through i and v. This means, for example, thata resistor in a circuit is viewed by the rest of the circuit as a black boxwith an i-v characteristics of v = Ri. The rest of the circuit does not knowsome elements what is inside the box. In fact, we can replace the resistorwith any black box (containing whatever) with the same i-v characteristicsof v = Ri and the rest of the circuit behaves exactly the same.

Alternatively, if a black box containing many circuit elements is attachedto a circuit and has an i-v characteristics of v = 5i, we can replace this blackbox with a 5 resistor with no change in the circuit behavior.

This observation allows the circuit to be divided into two or many partsand each solved independently. We dene a box containing several elementa subcircuit or a device. The above gure shows a two-terminal device orsubcircuit.

1.2. CIRCUIT REDUCTION TECHNIQUES 19

SSuubbcciirrccuuiittRReesstt ooff

tthhee cciirrccuuiitt

++

--

vv

ii

AA bbooxx wwiitthh ssoommee

cciirrccuuiitt eelleemmeennttss

RReesstt ooff

tthhee cciirrccuuiitt

++

--

vv

ii

EEqquuiivvaalleenntt

ssuubbcciirrccuuiitt

Figure 1.13: Subcircuit and rest of the circuit.


Subcircuits play an important role in linear circuit theory. Thevenintheorem states that any subcircuit containing linear circuit element has ani-v characteristics of Av+Bi = C (where A, B, and C are constants) and itcan be reduced to a subcircuit containing at most two linear circuit elements(Thevenin and Norton Forms). We will discuss Thevenin Theorem later.Below, we explore subcircuits in the context of elements that are in seriesor in parallel. In each case, we nd the i-v characteristics of the subcircuitand use that to nd the equivalent element.

Two Resistors in Series

KCL : i = i1 = i2KVL : v + v1 + v2 = 0i-v : v1 = R1i1 = R1i

v2 = R2i2 = R2i

++

--

vv

++ --vv11

ii11RR11

RR22

++

--

vv22

ii22

ii

Figure 1.14: Two resistorsin series.

Substituting from i-v characteristics equations in KVL, we get

v = R1i + R2i= (R1 + R2)i= Reqi

Req = R1 + R2

So, a subcircuit containing two resistors in series has an i-v characteris-tics of the form v = Reqi and is equivalent to a resistor, Req = R1 + R2.

The above can be easily extended: k resistors in series are equivalent to


one resistor with

Req =k

j=1

Rj (1.7)

A Resistor in Series with a Current Source

KCL : i = i1 = isKVL : v + v1 + v2 = 0 v = R1is = v2

++

--

vv

++ --vv11

ii11RR

++

--

vv22 iiss

ii

Figure 1.15: A resistors inseries with a current source.

The i-v characteristics of ICS states that its current is is, independentof its voltage (v2). Above equations show that the i-v characteristics ofsubcircuit is i = is and is independent of voltage v = v1 + v2 (as value ofv1 can be anything). The equivalent subcircuit is an independent currentsource with strength is.

++

--

vv

ii RR

++

--

vvss

Figure 1.16: Thevenin form.

A Resistor in Series with a Voltage Source

This is the Thevenin form and cannot be reduced further.


A Voltage Source in Series with a Voltage Source

KVL : v + vs1 + vs2 = 0v = vs1 + vs2

++

--

vv

ii

++

--

vvss22

++ --vvss11

Figure 1.17: A voltagesource in series with a volt-age source.

The i-v characteristics of IVS states that their voltages are vs1 and vs2,respectively, independent of their currents. Above equation shows that thei-v characteristics of subcircuit is v = vs1+vs2 and is independent of currenti. The equivalent subcircuit is an independent voltage source with strengthvs1 + vs2.

A Voltage Source in Series with a Current Source

++

--

vv

ii

++

--

vvss22

++ --vv11

iiss11

ii22

Figure 1.18: A voltagesource in series with a cur-rent source.


KCL : i = is1 = is2KCL : v + v1 + vs2 = 0 v = v1 + vs2

The i-v characteristics of ICS states that its current is is1, independentof its (v1). Above equations show that the i-v characteristics of subcircuitis i = is1 and is independent of voltage v = v1 + vs2 (as value of v1 can beanything). The equivalent subcircuit is an independent current source withstrength is1.

Note: A current source in series with any element reduced to a currentsource. Series element all have the same current and the current sourcerequires the current through to be equal to its strength.

A Current Source in Series with a Current Source

KCL : i = is1 = is2

++

--

vv

ii

iiss22

iiss11

Figure 1.19: Two currentsources in series.

A current source in series with any element reduces to a current source.However, in the case of two current sources in series, KCL requires is1 = is2.Thus, two current sources can be attached in series only if is1 = is2. If so,the equivalent subcircuit is an independent current source with strengthis = is1 = is2.


Two Resistors in Parallel

KCL : i + i1 + i2 = 0KVL : v = v1 = vs2i-v : v1 = R1i1

v2 = R2i2

Substituting from i-v characteristics equations in KCL, and using v =v1 = v2, we get

i =v

Req 1

Req=

1

R1+

1

R2

So, a subcircuit containing two resistors in parallel has an i-v charac-teristics of the form v = Reqi and is equivalent to a resistor, 1/Req =1/R1 + 1/R2.

The above can be easily extended: k resistors in series are equivalent toone resistor with

1

Req=

kj=1

1

Rj(1.8)

A Resistor in Parallel with a Current Source

++

--

vv

ii

iissRR11

++

--

vv11

ii11++

--vv22

Figure 1.20: Norton form.

This is the Norton form and cannot be reduced further. We will showlater that Norton and Thevenin forms are equivalent.

A Resistor in Parallel with a Voltage Source

KCL : i + i1 + i2 = 0KVL : v = v1 = vs


++

--

vv

ii

vvssRR11

++

--

vv11

ii11++

--

ii22

Figure 1.21: A resistorin parallel with a voltagesource.

The i-v characteristics of IVS states that its voltage is vs, independentof its current, i2. Above equations show that the i-v characteristics ofsubcircuit is v = vs independent of current i. The equivalent subcircuit isan independent voltage source with strength vs.

A Current Source in Parallel with a Current Source

KCL : i + is1 + is2 = 0 i = is1 + is2

++

--

vv

ii

iiss22

iiss11++

--vv11

++

--vv22

Figure 1.22: Two currentsources in parallel.

The i-v characteristics of ICSs state that their currents are is1 and is2,respectively, independent of their voltage, v. Above equations show thatthe i-v characteristics of subcircuit is i = is1 + is2 independent of value ofvoltage v. The equivalent subcircuit is an independent voltage source withstrength is = is1 + is2.


A Current Source in Parallel with a Voltage Source

KCL : i + i1 + i2 = 0KVL : v = vs = v2

++

--

vv

ii

iiss

ii11++

--vvss

++

--

vv22

Figure 1.23: A currentsource in parallel with avoltage source.

The i-v characteristics of IVS states that its voltage is vs, independentof its current, i1. Above equations show that the i-v characteristics ofsubcircuit is v = vs and is independent of current i = i1 + is (as value ofi1 can be anything). The equivalent subcircuit is an independent voltagesource with strength vs.

Note: A voltage source in parallel with any element reduces to a voltagesource. Parallel elements all have the same voltage and the voltage sourcerequires its voltage across to be equal to its strength.

A Voltage Source in Parallel with a Voltage Source

KVL : v = vs1 = vs2

A voltage source in parallel with any element reduces to a voltage source.However, in the case of two voltage sources in parallel, KVL requires vs1 =vs2. Thus, two voltage sources can be attached in parallel only if vs1 = vs2.If so, the equivalent subcircuit is an independent voltage source with strengthvs = vs1 = vs2. circuit conguration arises because we are dealing witidealized circuit element. We will discuss real sources later.


++

--

vv

ii

ii22ii11++

--vvss11

++

--vvss22

Figure 1.24: Two voltagesources in parallel.

Thevenin and Norton Forms and Source Transformation

RRTT

++

--

vvTT RRNN iiNN

Figure 1.25: Thevenin and Norton forms are equivalent ifRT = RN and vT = iNRT .

Thevenin and Norton forms are equivalent if RT = RN and vT = iNRT .One can replace one with the other. This is called source transformationand is helpful in rearranging other elements in the circuit and sometimearriving at more element being in series and parallel.

Voltage Divider

The two resistors can be replaced by an equivalent resistor, Req = R1 +R2.Thus

vs = Reqi i = vsReq

v1 = iR1 =R1Req

vs

v2 = iR2 =R2Req

vs


RR11++

--

vvss

RR22

++

--

vv11

++

--

vv22

ii

Figure 1.26: Voltage di-vider.

Also,

v1v2

=R1R2

This circuit is called a voltage divider as the two resistors divide thevoltage of the IVS between them proportional to their values. This circuitcan be extended by adding more resistors to the circuit and get more ref-erence voltages. This circuit is used extensively in electronic circuits. Thebasic reason is that power supplies are bulky and/or expensive. Typically,one power supply with one voltage is provided. On the other hand, morethan one voltage may be needed for the circuit to operate properly.

Example: A battery operated radio has a 9V battery. Part of radiocircuits require a 6V supply. Design a voltage divider circuit to supply 6Vvoltage to these circuits. The desired circuit is the voltage divider circuitabove with vs = 9V and v2 = 6V. Then,

v2 =R2Req

vs 6 = R2R1 + R2

9 R2R1 + R2

=6

9

This is one equation in two unknowns and one is free to choose oneparameter. For example, choosing R1 = 1K, we get R2 = 2K. Voltagedividers are aected by the load current drawn from them (see gure). Thevoltage divider formula can only be used for the circuit if il i.


RR11++

--

vvss

RR22

++

--

vv11

++

--

vv22

ii

LLooaadd

iiLL

Figure 1.27: Voltage divider with load.

Current Divider

The two resistors can be replaced by an equivalent resistor,

1/Req = 1/R1 + 1/R2

++

--

vv

iissvv

RR11

++

--

vv

ii11++

--

ii22RR22

Figure 1.28: Current divider.

Thusv = Reqi

v = i1R1 i1 = vR1

=ReqR1

is

v = i2R2 i2 = vR2

=ReqR2

is

Also,i1i2

=R2R1


This circuit is called a current divider as the two resistors divide thecurrent of the ICS between them (inversely proportional to their values).This circuit can be extended by adding more resistors to the circuit andget more reference currents.

Wheatstone Bridge

++

--vvss

vvBB

RR11

++

--vv22

--

RRAA

vvmm

++

++ --AA BB

RRBBRR22

Figure 1.29: Wheatstone bridge.

A typical Ohm-meter measures the resistance of a resistor by using theOhms Law. It applies a known voltage of vs across the resistor, measuresthe current owing through the resistor, and its dial are set to convertthe measured value of current into the value of resistance by using R =vs/imeasured. (This is why one cannot measure the value of a resistor whileit is attached in a circuit, Ohm-meter works only if the resistor is notattached to anything but the meter.)

A typical digital multi-meter measure resistance within an accuracy ofabout 1%. In some cases, higher accuracy is needed. Resistor bridges areused for this purpose and they are made of two voltage divider circuits putin parallel with each other. The bridges operate based on the fact thatwhile it is dicult to measure the dierence between 1 and 1.01 V or 1and 1.01 A distinctly (they are only 1% apart and within the accuracy ofmeter), it is easy to measure 0.01 V. A most widely used bridge is theWheatstone bridge shown. It consists of two voltage divider circuits with avoltmeter measuring the voltage between points A and B (denoted by vm).

1.3. CIRCUIT ANALYSIS TECHNIQUES 31

We note from voltage divider formulas

v2 =R2

R1 + R2vs

vB =RB

RA + RBvs

KVL : v2 + vm + vB = 0 vm = v2 vB = vs(

R2R1 + R2

RBRA + RB

)

Measuring resistance accurately: Suppose the unknown resistance is RB.Two known and accurate resistors R1 and RA are chosen. An accurate butvariable resistor is used for R2. The variable resistor R2 is varied until themeter read zero voltage for vm. Then

vm = vs

(R2

R1 + R2 RB

RA + RB

)= 0 RB = RAR2

R1

Note that we do not need to know value of vs to nd RB. This way, RBis measure within the accuracy of resistors, R1, R2, and RA.

1.3 Circuit Analysis Techniques

In this section, we learn two methods, node-voltage and mesh-current meth-ods, which reduce the number of equations to either number of KCLs ornumber of KVLs.

1.3.1 Node-Voltage Method

The node-voltage method is based on following idea. Instead of solving forcircuit variables, i and v of each element, we solve for a dierent set ofparameters, node voltages in this case, which automatically satisfy KVLs.As such, we do not need to write KVLs and only need to solve KCLs.

Recall denition of potential. Potential was dened as the voltage be-tween any point and the reference point (usually called ground or common).Take a circuit and identify the nodes on the circuit. Dene the node voltageas the potential of that node (voltage between that node and the reference).In the circuit shown, the node voltages are denoted as v1, v2, and v3. Thevoltage across each element is simply the voltage of the node attached to


++

--

vvaa

++ --vvbb

++

--

vvcc

vv11 vv22

vv33

RReeffeerreennccee nnooddee

((ggrroouunndd))

Figure 1.30: Node voltage: voltage be-tween that node and the reference.

+ terminal of the element minus the voltage of the node attached to - ter-minal of the element:

va = v1 v2vb = v1 v3vc = v2 v3

Note that using node voltage denitions, KVLs will be automaticallysatised:

KVL : vb + va + vc = 0(v1 v3) + (v1 v2) + (v2 v3) = 0 0 = 0

As the choice of the reference point (ground) is arbitrary, we can choosethe reference point to be any node on the circuit. For example, we couldchoose node 3 in the above circuit as the reference mode (v3 = 0). Thenumber of node voltages is reduced to 2. The voltages across the threeelements are

va = v1 v2vb = v1vc = v2

and KVL is still automatically satised. With this simplication, we haveN1 node voltages in each circuit where N is the number of nodes. Recall


that number of independent KCLs is N 1. If we can write the currentin each element in terms of the node voltages, then the circuit reduced toN 1 KCLs in N 1 node voltages.

++ --vv1122

RR

ii11 vv11

vv22ii22

++

--vvssiiss vv22vv11

ii11 ii22

++-- vv2211

Figure 1.31: Node voltages of current source and resistor.

Current Source: The current owing through current source is inde-pendent of voltage across it and, therefore, is independent of node voltages.From the gure, it is obvious that the current leaving node 1 is i1 = is andcurrent leaving node 2 is i2 = is.

Voltage Source: The voltage across a voltage source is constant andis independent of current owing through it. The i-v characteristics ofthe voltage source cannot be used to nd the current owing through it.Voltage sources should be treated in a special manner that will be discussedlater.

Resistor: Resistors follow Ohms Law. We need to calculate the voltageacross the resistor rst and nd the current using the Ohms Law. Forexample, if we want to nd current leaving node 1, i1, we rst nd voltagev12:

v12 = v1 v2 i1 = v12R

=v1 v2

RAlternatively, if we want to nd the current leaving node 2, i2, we calculatev21 rst:

v21 = v2 v1 i2 = v21R

=v2 v1

R

In summary, we have found that we can dene node voltages as circuitvariables and:

Voltage across each element can be written in terms of node voltages.


Current through each element (resistor and ICS) can be written interms of node voltages using the i-v characteristics of that element.

KVLs are automatically satised. By using node voltages, we have reduced the number of variables to

N1 node voltages (one node is the reference node with zero voltage)and N 1 KCLs to solve.

v1 v2

is1 is2

i2

i3i1

i2 R2

R1 R3

Figure 1.32: Example for node-voltage method.

Example: The circuit in Figure 1.32 has three nodes. We choose thenode at the bottom as the reference (ground).The voltages at the othernodes are denoted by v1 and v2. We write KCL at these two nodes:

KCL 1 : i2 + i1 is1 = 0 v1 v2R2

+v1 0R1

= is1

KCL 2 : is2 + i3 + i2 = 0

v2 0R3

+v2 v1

R2= is2

The above are two equations in two unknowns (node voltages v1 andv2). Note that a circuit with 5 elements which, in principle, has 10 circuitvariables and 10 equations to be solved is reduced to 2 equations in 2unknowns. The previous equations can be rearranged as:

v1

(1

R1+

1

R2

)+ v2

(1

R2

)= is1

v1(

1

R2

)+ v2

(1

R2+

1

R3

)= is2


or using the denition of conductance, G = 1/R, we have:

(G1 + G2)v1 G2v2 = is1 G2v1 + (G2 + G3)v2 = is2In the matrix form, the equations are:(

(G1 + G2) G2G2 G2 + G3

)(v1v2

)=

(i1s1is2

)

This can be generalized to any circuit with N nodes (let n = N 1).The node-voltage equation can be written as

Gv = is

where

G =

G11 G12 . . . G1nG21 G22 . . . G2n

......

......

Gn1 Gn2 . . . Gnn

, v =

v1v2...vn

, is =

is1is2...

isn

Matrix G is called the conductance matrix. If the KCLs are written assum of current exiting each node, the diagonal elements of this matrix willbe positive and the o-diagonal elements will be negative. The diagonalelements Gii are equal to the sum of all conductances connected to node i.The matrix is symmetric, i.e. Gij = Gji. These o-diagonal elements, Gij,are sum of conductances directly connecting nodes i and j. Vector v is thearray of node voltages. Vector is is the array of current sources. Elementisi is net equivalent of source currents entering the node.

Because of the above features of the node-voltage equations, it is possibleto directly write the node voltage equations in the matrix form of Gv = isand follow the above rules to construct matrix G and vectors v and is. Nodevoltage method is a powerful and simple method.

Example: Consider the circuit in Figure 1.33. Find i.

The circuit has four nodes. Assigning the node at the bottom as thereference node. Using the rules to construct matrix G and vectors v and


17A

v1 v3

7A

i12

4 4

6

12 3

5A

v2

Figure 1.33: Another example for node-voltage method.

is, we obtain

G =1

12

4 1 01 6 2

0 2 7

, v =

v1v2

v3

, is =

2517

From

Gv = is

we obtain the elements of vector v as v1 = 12V, v2 = 24V, and v3 = 36V.Using Ohms law, we get

i =v2 v3

6= 2A

Circuits with Voltage Sources

Node voltage method as outlined above has to be modied for circuitswith voltage sources as the i-v characteristics of IVS does not specify thecurrent through it. This modications can be explained in the context ofthe example below:

Example: Consider the circuit in Figure 1.34. Find v.

The circuit has four nodes. We will assign the node at the bottom asthe reference node and we have three node voltages as unknowns. If we didnot have voltage sources, we would write 3 KCL at the 3 nodes to arrive at


v1 v2

20V 6mA

3V

2K 4K

v3

+

-

v

6K+ -

-

+

Figure 1.34: Circuit with voltage sources.

3 equations in 3 unknowns. For this circuit, writing KCLs is not useful asthe i-v characteristics of IVS does not specify the current through it. Wenote that there are two possible conguration for voltage sources:

a) Voltage source is connected to the reference node (such as the 20-V IVSin the circuit above). If we write the voltage across the IVS in terms of thenode voltages:

v1 0 = 20 v1 = 20VIn this case, we do not need to write any KCL. The value of node voltage

can be derived directly from the i-v characteristics of the IVS.

b) Voltage source is connected between two nodes that are not referencenodes (such as the 3-V IVS in the above circuit). We need to write twoequations for nodes 2 and 3 (one for each). KCLs for those nodes (twoequations) cannot be used as both contain the current through IVS that isunknown. One equation is the voltage across the IVS in terms of the nodevoltages:

v2 v3 = 3The second equation can be found by noting that as charge cannot beaccumulated in any part in the circuit, KCL should apply to any portionof the circuit. Consider a portion of the circuit including nodes 2, 3 andthe 3 V voltage source. We call this portion a supernode (see Figure 1.35).KCL should be valid for a supernode. This gives

KCL for supernode 2 & 3 : 6 103 + v34 103 +

v22 103 +

v2 v16 103 = 0

72 + 3v3 + 6v2 + 2(v2 v1) = 0 2v1 + 8v2 + 3v3 = 72


We now have three equations in three unknowns:

v1 = 20v2 v3 = 32v1 + 8v2 + 3v3 = 72

which gives v1 = 20V, v2 = 11V, and v3 = 8V. Therefore problem unknownv = v3 0 = v3 = 8V.

v1 =20V v2

20V 6mA

3V

2K 4K

v3=v2-3

+

-

v

6K+ -

-

+

i

Supernode

Figure 1.35: Circuit with supernode.

Noting v1 = 20, we can write the value of the node voltage on the circuit.Also, using v2 v3 = 3, we can arrive at v3 = v2 3 as shown in the gure.

Now, we only have one unknown, v2, which can be found by writing KCLon the supernode:

KCL for supernode 2 & 3 : 6 103 + v3 34 103 +

v22 103 +

v2 206 103 = 0

The above is one equations in one unknown which can be solved rapidly.


Notes:

1. Voltage sources actually simplify the node voltage equations. Byusing the i-v characteristic equations of IVS on the circuit diagram, we canreduce the number of unknown node voltages (thus, number of equationsto be solved) to N 1NIV S, where NIV S is the number of IVS.

2. The situation is much simpler for voltage sources that are connectedbetween a node and ground (such as the 20-V source in the above circuit).As the location of the reference node is arbitrary, a good rule is: Choosethe reference node as the node with maximum number of voltage sourcesattached to it.

3. While we do not need to write KCL at the nodes connected to avoltage node in order to nd the node voltages, we need to use KCL if wewant to nd the current through a voltage source. For example, in thecircuit above, current i can be found by writing KCL at node 3 (after wehave found all node voltages):

i 6 103 + v2 34 103 = 0

i = 4mA

Summary for Node Voltage Method

1. Identify nodes and supernodes.

a. Choose the reference node as the one with maximum number ofvoltage sources attached to it.

b. Use i-v characteristic equations of IVS to nd node voltage valuesand reduce the number of unknowns.

2. Write KCL at each node or supernode.

3. Solve node-voltage equations.

4. Calculate problem unknowns from node voltages. If you need to calcu-late the current in a voltage source you may have to write KCL at anode connected to that voltage source.


1.3.2 Mesh-Current Method

The mesh-current method is analog of the node-voltage method. We solvefor a new set of variables, called mesh currents, that automatically satisfyKCLs. As such, mesh-current method reduces circuit solution to writing abunch of KVLs.

Note: Mesh-current method only works for planar circuits: circuits thatcan be drawn on a plane without any elements or connecting wires crossingeach other. Note that in some cases a circuit that looks non-planar can bemade into a planar circuit by moving some of the connecting wires.

vvss11 vvss22++

RR11

--

++

ii22

--

ii33

ii11

ii11 ii22

RR22

RR33

ii11 ii22

ii33

ii11

Figure 1.36: Example of mesh-current method.

Mesh-current method is best explained in the context of example circuitin Figure 1.36.

A mesh is dened as a closed path (a loop) that contains no closed pathwithin it.

Mesh current is the current that circulates in the mesh i.e.,

a. If an element is located on a single mesh (such as R1, R2, vs1, and vs2)it carries the same current as the mesh current,

b. If an element is located on the boundary of two meshes (such as R3),it will carry a current that is the algebraic sum of the the two meshcurrents:

i3 = i1 i2i3 = i2 i1


In this way, KCLs are automatically satised. In addition, as we canwrite current in each element in terms of mesh currents, we can use i-vcharacteristics of element to write voltage across each element in terms onmesh currents. Therefore, we need only to write KVLs in terms of meshcurrents. In the circuit above, KVLs give:

Mesh 1 : R1i1 + R3(i1 i2) vs1 = 0 (R1 + R3)i1 R3i2 = vs1Mesh 2 : R3(i2 i1) + R2i2 + vs2 = 0 R3i1 (R2 + R3)i2 = vs2

or in matrix form,(R1 + R3 R3R3 R2 + R3

)(i1i2

)=

(vs1vs2

) Ri = vs

which is similar in form to matrix equation found for node-voltage method.i is the vector of mesh currents (unknowns), vs is the vector of independentvoltage sources, and R is the resistance matrix and is symmetric. Thediagonal element, Rjj, is the sum of resistance around mesh j and the o-diagonal elements, Rjk, are the sum of resistance shared by meshes j andk.

Example: Consider example in Figure 1.37. Find i and v.

vv

1166VV 66VV

66

++

--

22

++ --

--

++

ii22

++

--

99VV

33

ii

ii11

ii11 ii22

Figure 1.37: Another example of mesh-currentmethod.

Using mesh-current method:

Mesh 1 : 2i1 + 9 + 3(i1 i2) 16 = 0Mesh 2 : 6i2 + 6 + 3(i2 i1) 9 = 0

we get, i1 = 2A, i2 = 1A.


The problem unknowns, i and v can now be found from the mesh cur-rents:

i = i1 i2 = 1Av = 2i1 = 4A

Mesh-Currents Method for Circuits with Current Sources

1100VV66

2222

--

++22AA

ii22 ii33

11AA

ii11

Figure 1.38: Mesh-current method for circuitwith current sources.

Because of i-v characteristics of a current source does not specify itsvoltage, we have to modify mesh-current method. This is best seen in theexample in Figure 1.38.

From the circuit, we note:

If a current source is located on only one mesh (1-A ICS i1 in thecircuit), the mesh current can be directly found from the current sourceand we do not need to write any KVL: i1 = 1A

If a current source is located on the boundary between two meshes (2-A ICS in the circuit), KVL on these meshes (mesh 2 or 3 in the abovecircuit) contain the voltage across the 2-A ICS which is unknown. Weneed two equations to substitute for the two KVLs on meshes 2 and 3that are not useful now.

The rst equation is found from the i-v characteristics of the cur-rent source (its current should be 2 A): i3 i2 = 2A


The second equation can be found by noting that KVL can be writ-ten over any closed loop. We dene a supermesh as the combina-tion of two meshes which have a current source on their boundaryas shown in the Figure 1.39. While KVL on mesh 2 or on mesh3 both include the voltage across the 2-A current source that isunknown, KVL on the supermesh does not include that, we have:

Supermesh 2 and 3 : 2(i2 i1) + 2(i3 i1) + 6i3 10 = 0 4i1 + 2i2 + 8i3 = 10

1100VV66

2222

--

++22AA

ii22==ii33--22 ii33

11AA

ii11== --11

Figure 1.39: Supermesh for circuit with currentsources.

Totally, we have

i1 = 1i3 i2 = 24i1 + 2i2 + 8i3 = 10

which results in i1 = 1A, i2 = 1A, and i3 = 1A.

Summary for Mesh-Current Method

1. Check if circuit is planar.

2. Identify meshes, mesh currents, and supermeshes.

a. Rearrange the circuit if possible to position current source on asingle mesh.


b. Use i-v characteristic equations of ICS to nd mesh currents andreduce the number of unknowns.

3. Write KVL at each mesh and supermesh.

4. Solve for mesh currents.

5. Calculate problem unknowns from mesh currents. If you need to calcu-late the voltage across a current source you may have to write KVLaround a mesh containing the current source.

6. For consistency and elimination of errors, always mark all mesh currentsin clockwise direction and write down KVLs in the same direction.

Comparison of Node-Voltage and Mesh-Current Methods

Node-voltage and mesh-current are powerful methods that simplify circuitanalysis substantially. They are methods of choice in almost all cases (ex-cept for very simple circuits or special circuits). Examination of the circuitcan also tell us which of the two methods are best suited for the circuitat hand. We always want to reduce the circuit equations into the small-est number of equations in smallest number of unknowns. The number ofequations from node-voltage method, NNV and mesh-current method, NMCare given by:

NNV = Nnode 1NV SNMC = Nmesh NCS

where Nnode and Nmesh are number of nodes and number of meshes, respec-tively; NV S and NCS are numbers of voltage and current sources, respec-tively. Thus, always inspect the circuit, nd NV S and NCS, and proceedwith the method that results in the smallest number of equations to solve.

Note: We need to check to ensure that the circuit is a planar circuit. Ifit is not one cannot use mesh-current method and should use node-voltagemethod.

1.3.3 Principle of Superposition

If a linear circuit is driven by more than one independent source, the re-sponse of the circuit can be written as the sum of the responses of thecircuit to individual sources with all other sources having their strength setto zero. Note that a source is set to zero does not mean removing it.


Example: Consider the circuit in Figure 1.40(a). Find v by superposi-tion.

Because we have two independent sources, we rst set the current sourceto zero to obtain the circuit in Figure 1.40(b), and then set the voltagesource to zero to arrive at the circuit in Figure 1.40(c). By superposition,v = va + vb.

Figure 1.40(b) is a voltage divider circuit and va can be written downdirectly as

va =5

5 + 10 15 = 5V

Figure 1.40(c) is a current divider circuit and current i can be writtendown directly as

i =10

5 + 10 3 = 2A

vb = 5i = 10VThus,

v = va + vb = 5 10 = 5V

Note: Using superposition results in slightly simpler circuits (one el-ement is replaced with either a short or open circuit) but more circuits.In general superposition requires more work than node-voltage or mesh-current methods. Superposition is used:

a. If sources are fundamentally dierent (e.g., DC and AC sources). In thiscase superposition may be the only choice.

b. If circuit is repetitive such that circuits resulting from applying super-position look identical and, thus, we need only to solve one circuit.

1.3.4 Reduction of Two-Terminal Subcircuits to Thevenin Form

We used the equivalency of Norton and Thevenin forms in circuit reduction.Recall our discussion of equivalent elements and subcircuits. We can replaceany two-terminal subcircuit with another one as long as they have thesame i-v characteristics. It is shown that the i-v characteristics of anytwo-terminal element containing linear elements is in the Thevenin form of

v = vT RT i


1155VV

1100

55--

++

33AA++

--

vv

1155VV

1100

55--

++

11AA

++

--

vv

((aa))

((bb))

1100

55

++

--

vv

((cc))

33AA

Figure 1.40: Example of superposition. (a) Originalcircuit. (b) Setting the current source to zero: opencircuit. (c) Setting the voltage source to zero: shortcircuit.


Consider the two-terminal subcircuit and its Thevenin equivalent (theyhave exactly same i-v characteristics) in Figure 1.41. Let the current i = 0and calculate v, i.e., calculate the voltage across the terminals of the sub-circuit, while the terminal are open circuit. This voltage is called the opencircuit voltage, voc. Examination of the Thevenin form shows that if i = 0,then vT = voc.

RRTT

--

++

vvTT

++

--vv == vvoocc

ii == 00

SSuubbcciirrccuuiitt

++

--

vv == vvoocc

ii == 00

((aa))

RRNN

--

++

iiNN vv == 00SSuubbcciirrccuuiitt

++

--

vv == 00

ii == iisscc

((bb))

ii == iisscc

Figure 1.41: Two-terminal subcircuit and its Thevenin/Norton equivalentcircuits. (a) Thevenin equivalent circuit. (b) Norton equivalent circuit.

Next, consider the two-terminal subcircuit and its Norton equivalent(they have exactly same i-v characteristics). Let the voltage v = 0 andcalculate i, i.e., calculate the current, while the subcircuit terminals areshorted. This current is called the short circuit current, isc. Examinationof the Norton form shows that if v = 0, iN = isc.

Lastly, if one set all of the sources in the subcircuit to zero, the remainingcircuit should be equivalent to the Thevenins resistance, RT .


Finding Equivalent Thevenin/Norton Forms

Compute two of the following quantities by solving the appropriate circuits:vT = voc, iN = isc, and RT by setting zero the sources. The third parameteris found from vT = RT iN .

Example: Find Thevenin equivalent of the subcircuit in Figure 1.42.

44

33AA++

--

2200

55

2255VV

Figure 1.42: Example of Thevenin equiv-alent.

Thevenins Theorem: Find two of the following three: RT , voc, and isc.

a) Find RT by setting the sources to zero (see Figure 1.43)

44

2255VV 33AA++

--

2200

55 44

33AA2200

55

44

55////2200 == 44 RRTT == 88

Figure 1.43: Circuit for nding RT .


b) Find vT = voc (set i = 0)

4

3A+

-

20

5

25V

25V voc

+

-

voc

i = 0

v

Figure 1.44: Circuit for nding vT = voc.

Using node-voltage method (note that the voltage drop across the 4resistor is zero).

voc 255

+voc 0

20 3 = 0

vT = voc = 32V

c) Find iN = isc (set v = 0)

4

3A+

-

20

5

25V

25V v2

+

-

v=0

isc

Figure 1.45: Circuit for nding iN = isc.

Using node-voltage method:

v2 255

+v2 020

3 + v2 04

= 0

v2 = 16V iN = isc = v2

4= 4A


1.3.5 Real Sources

ii

vvvvss

RReeaall

ssoouurrccee

IIddeeaall ssoouurrccee

vv == vvss

((aa))

++

--

vvss

RRss

++

--

vv

ii

iiss RRss vv

--

++

ii

((bb))

Figure 1.46: Real source. (a) i-v characteristic of real source with a straight line. (b)Circuit model of a real source.

In an ideal voltage source, the voltage is constant no matter what currentis drawn from the source. In a real, practical voltage source (like a battery),however, the output voltage typically decreases as more and more currentis drawn, as is shown in the Figure 1.46. Typically, a real source is ratedfor currents below a current i which corresponds to a voltage v 95%vs(region near vs in the gure). For this region, it is a good approximationto model the i-v characteristics of a real source with a straight line. Theequation of this line is (using active sign convention):

v = vs RsiThe above approximate i-v characteristics of a real source is a Thevenin

form and, therefore, a real source can be modeled with ideal voltage source,vs, and a resistance Rs. Rs is called the internal resistance of the source(note that it is not a real resistor inside the real source)and is typicallysmall (an ideal voltage source has Rs = 0). The same arguments can beapplied to real current sources. An approximate model for a real currentsource is in Norton form. Rs is again the internal resistance of the source(and again, it is not a real resistor inside the real source). For a realcurrent source, Rs is typically large (an ideal current source has Rs ).

1.4 Dependent or Controlled Sources

Most analog electronic devices include ampliers. These are four-terminaldevices (two input and two output terminals). The voltage or current in

1.4. DEPENDENT OR CONTROLLED SOURCES 51

the output terminals are proportional to voltage or current of the inputterminals. We need a new circuit element in order to model ampliers.These elements are controlled or dependent sources. There are four type ofcontrolled sources as shown in Figure 1.47.

++

--

++

--

vv11

mmvv11

((aa))

++

--

nnii11

((bb))

ii11

++

--

vv11

ppvv11

((cc))

qqii11

((dd))

ii11

Figure 1.47: Controlled Sources. (a) Voltage-controlled voltage source.(b) Current-controlled voltage source. (c) Voltage-controlled currentsource. (d) Current-controlled current source.

Note that the element located in the input with the controlling currentor voltage can be any element: a short circuit, an open circuit, or a resistor.

Controlled sources behave similar to ideal (or independent) sources. Forexample, in the voltage-controlled voltage source in the above gure, theoutput voltage is v1 no matter what current is drawn from the circuit. Allanalysis method developed so far (KVL and KCL, node-voltage and mesh-current methods, superposition, etc.) can be used for circuits containingcontrolled sources, and by treating the controlled source similar to an idealsource. In node-voltage and mesh-current methods, however, we need towrite an auxiliary equation which relates the controlling parameter tonode-voltage or mesh-current methods as is seen in the examples below.


++

--++

--

vvss

RRss

RRpp

iixx

rriixx

RRcc

RRLL

++

--

vvoo

iioo

Figure 1.48: Example of circuit with a current-controlled voltage source.

Example: Find vo in Figure 1.48, using KVL and KCL:

KVL : RSix + RP ix vs = 0 ix = vsRS + RP

KVL : RCio + RLio + rix = 0 io = rixRC + RL

Substituting for ix from the rst equation in the second and notingvo = RLio, we get:

vo = RLio =

[ rRL(RS + RP )(RC + RL)

]vs

Example: Consider the circuit in Figure 1.49. Find v1 using node-voltagemethod.

++

--

1166VV

44 22

ii11

88ii11

11..2255AA

44

++ --

00..7755vv11

22

-- ++

++

vv11

Figure 1.49: Example of node-voltage method with con-trolled sources.


Circuit has 5 nodes and 2 voltage sources (one independent and onecontrolled voltage source). Thus, the number of node voltage equations isNNV = 5 1 2 = 2. We choose the reference node to be the one withmost voltage sources attached to it. Then, we get directly

vA = 16

vC = 8i1

00..7755vv11

++

--1166VV

44 22

ii11

88ii11

11..2255AA44

++ --

22

-- ++

++

vv11

vvAA== 1166VV vvBBvvCC== 88ii11

vvDD

Figure 1.50: Node voltages of the circuit in Figure 1.49.

We then proceed with writing KCL at the other two nodes:

Node vB :vB 8i1

2+

vB vD4

1.25 = 0 3vB vD = 5 + 16i1

Node vD :vD 0

4+

vD 162

+vD vB

4 0.75v1 = 0

vB + 4vD = 32 + 3v1

Two above equations are two equations in two unknowns (vB and vD).But they also contain the control parameters i1 and v1. We need to writetwo auxiliary equations relating these control parameters to our nodevoltages:

i1 =vD 16

2, v1 = vB 8i1 v1 = +vB 4vD + 64


We now substitute for control parameters i1 and v1 in our node-voltageequations to get:

3vB vD = 5 + 8(vD 16) 3vB 9vD = 123vB + 4vD = 32 + 3(+vB 4vD + 64) 4vB + 16vD = 224

which can be solved to nd the node voltages vB = 4V and vD = 15V. Thecontrol parameters are: i1 = 0.5A and v1 = 8V.

Thevenin Equivalent of Subcircuits with Controlled Sources

Two-terminal subcircuits containing controlled sources can be reduced toThevenin form by the method discussed in previous section. The methodwas to nd two of three parameters: RT (by setting independent sourcesto zero), vT = voc and iN = isc.

Example: Find the Thevenin equivalent of this subcircuit in Figure 1.51.

++

--

3322VV

22KK

44ii

66KK

++

--

vv

ii

Figure 1.51: Example of Thevenin equivalent ofsubcircuit with controlled source.

Finding voc:

Since the circuit is simple, we proceed to solve it with KVL and KCL(noting that i = 0) :

KCL : i1 + i + 4i = 0 i1 = 0KCL : i2 4i + i1 = 0 i2 = 0KVL : 32 + 2 103i2 + 6 103i1 + voc = 0

vT = voc = 32V


++

--

3322VV

22KK

44ii

66KK

++

--

vvoocc

ii==00

ii22ii11

Figure 1.52: Circuit for nding voc.

Finding isc

Since the circuit is simple, we proceed to solve it with KVL and KCL:

KCL : i1 + i + 4i = 0 i1 = 5iscKCL : i2 4i + i1 = 0 i2 = iscKVL : 32 + 2 103isc + 6 103isc = 0

iN = isc = 4mA

++

--

3322VV

22KK

44ii

66KK

++

--

vv==00

ii==iisscc

ii22ii11

Figure 1.53: Circuit for nding isc.

Therefore, vT = 32 V, iN = 4mA, and RT = vT/iN = 8K.

While nding voc and isc is preferred method for most circuits, in somecases, the Thevenin equivalent of the subcircuit is only a resistor (you willnd voc = 0 and isc = 0) , or only a voltage source (you will nd voc = 0but nding isc leads to inconsistent or illegal circuits), or only a currentsource (you will nd isc = 0 but nding voc leads to inconsistent or illegal


circuits). For these cases, one has to either nd RT directly and/or directlynd i-v characteristics of the subcircuits as is shown below for the circuitof previous example.

Finding RT

22KK

44ii

66KK

++

--

vv

ii

ii22ii11

++

--

vv

Figure 1.54: Circuit for nding RT .

To nd RT , we set all independent sources in the circuit to zero. Theresulting circuit cannot be reduced to a simple resistor by series/parallelformulas. This is why nding voc and isc is the preferred choices for subcir-cuits containing controlled sources. We can nd RT by attaching an idealvoltage source with a known voltage of v and calculate i. Since the subcir-cuit should be reduced to a resistor (RT ), we should get i = v/(constant)where the constant is RT . (Negative sign comes from active sign conventionused for Thevenin subcircuit).

Since the circuit is simple, we solve it with KVL and KCL:

KCL : i1 + i + 4i = 0 i1 = 5iKCL : i2 4i + i1 = 0 i2 = iKVL : 0 + 2 103i + 6 103i + v = 0

i = v8 103

Therefore, RT = 8 103 = 8K.Note that we could have attached an ideal current source with strength

of i to the problem, proceeded to calculate v, and obtain v = 8 103i.

1.5. LINEAR AMPLIFIERS AND OPERATIONAL AMPLIFIERS 57

Finding i-v Characteristics Equation:

As mentioned above, in some cases, we have to directly nd the i-v char-acteristics equation in order to nd the Thevenin equivalent of a subcircuit.The procedure is similar to nding RT .

22KK

44ii

66KK

++

--

vv

ii

ii22ii11 ii

++

--

3322VV

Figure 1.55: Circuit for nding i-v characteristicsequation.

Attach an ideal voltage source to the circuit. Assume v is known andproceed to calculate i in terms of v. (Alternatively, one can attach anideal current source, assume i is known and nd v in terms of i). Thenal expression should look like v = vT iRT and vT and RT can be readdirectly.

Since the circuit is simple, we proceed to solve it with KVL and KCL:

KCL : i1 + i + 4i = 0 i1 = 5iKCL : i2 4i + i1 = 0 i2 = iKVL : 32 + 2 103i + 6 103i + v = 0

v = 32 8 103iwhich is the characteristics equation for the subcircuit and leads to vT = 32V,RT = 8 103 = 8K, and iN = vT/RT = 4mA.

1.5 Linear Ampliers and Operational Ampliers

Ampliers are two-port networks in which the output voltage or current isdirectly proportional to either input voltage or current. Four dierent kindof ampliers exits:

Voltage amplier: Av = Vo/Vi: voltage gain (constant)


Current amplier: Ai = Io/Ii: current gain (constant) Transconductance amplier: Gm = Io/Vi: constant Transresistance amplier: Rm = Vo/Ii: constant

Figure 1.56: Voltage amplier model.

In this section, we will focus on voltage ampliers. Voltage amplierscan be accurately modelled with three circuit elements as in Figure 1.56.Ri and Ro are, respectively, input and output resistances. A good voltageamplier has a large input resistance, and a small output resistance. Anideal voltage amplier has Ri and Ro 0.

1.5.1 Practical Ampliers

Amplier Saturation

Ampliers do not create power. Rather, they act as a valve adjustingthe power ow from the power supply into the load according to the inputsignal. As such, the output voltage amplier cannot exceed the powersupply voltage (it is usually lower because of voltage drop across some activeelements). The fact that the output voltage of a practical amplier cannotexceed certain threshold value is called saturation. A voltage amplierbehaves linearly, (i.e., Vo/Vi = Av: constant), as long as the output voltageremains below the saturation voltage, Vsat

Vsat < Vo < VsatNote that the saturation voltage, in general, is not symmetric, i.e.,

Vsat,1 < Vo < Vsat,2. For an amplier with a given gain, Av, the aboverange of Vo translate into a certain range for Vi

Vsat < Vo < Vsat


since Vo = AvVi, then

VsatAv

< Vi 1 10M), andlow output resistance (< 100). They are constructed as a dierenceamplier, i.e., the output signal is proportional to the dierence betweenthe two input signals (see Figure 1.60)

Vo = A0Vd = A0(Vp Vn)In Figure 1.60, Vs and Vs are power supply attachments. They set the

saturation voltages for the OpAmp circuit. Power supply ground shouldalso be connected to the OpAmp ground. + and - terminals of the OpAmpare called, respectively, non-inverting and inverting terminals.

OpAmp Models

Figure 1.61: OpAmp models.

Because Ri is very large and Ro is very small, ideal model of the OpAmpassumes Ri and Ro 0. Ideal model is usually a very good modelfor OpAmp circuits. Very large input resistance also means that the inputcurrent into an OpAmp is very small:

Ip In 0An important feature of OpAmp is that because the gain is very high,

the OpAmp will be in the saturation region without negative feedback. Forexample, take an OpAmp with a gain of 105 and Vsat = 15V. Then, forOpAmp to be in linear region, Vi 15 105 = 150V (a very smallvalue). OpAmps are never used by themselves. They are always part of acircuit which employ either negative feedback (e.g., linear ampliers, activelters) or positive feedback (e.g., comparators). Examples below showsseveral OpAmp circuits with negative feedback.


1.5.4 Inverting Amplier

Figure 1.62: Inverting amplier.

The rst step in solving OpAmp circuits is to replace the OpAmp withits circuit model (ideal model is usually used).

Vp = 0, Vo = A0Vd = A0(Vp Vn) = A0VnUsing node-voltage method and noting In 0:

Vn ViR1

+Vn Vo

R2= 0

Substituting for Vn = Vo/A0 and multiplying the equation by R2, wehave:

R2VoA0R1

R2ViR1

VoAo Vo = 0

VoVi

= (R2R1

)(1

1 + 1/A0 + R2/(A0R1)

)

Since OpAmp gain is very large, 1/A0 1. Also if R2 and R1 are chosensuch that their ratio is not very large, R2/R1 A0 or R2/(A0R1) 1,then the voltage gain (voltage transfer function) of the OpAmp is

VoVi

= R2R1

The circuit is called an inverting amplier because the voltage gain isnegative. The negative sign means that there is 180 phase shift betweeninput and output signals.

Note that the voltage gain of the inverting amplier is independentof the OpAmp gain, A0, and is only set by the values of the resistors R1


and R2. While A0 is quite sensitive to environmental and manufacturingconditions, the resistor values are quite insensitive and, thus, the gain ofthe system is quite stable. This stability is achieved by negative feedback,the output of the OpAmp is connected via R2 to the inverting terminalof OpAmp. If Vo increases, this resistor forces Vn to increase, reducingVd = Vp Vn and Vo = A0Vd, and stabilizes the OpAmp output.

An important rule of OpAmp circuits with negative feedback is thatbecause the OpAmp is NOT saturated, Vd = Vo/A0 is very small (becauseA0 is very large). As a result, Vd 0 Vn Vp

Note: For OpAmps circuits with negative feedback, the OpAmp adjustsits output voltage such that Vd 0 or Vn Vp. This rule is derived byassuming A . Thus, Vo cannot be found from Vo = A0Vd = 0 =indenite value.

Input and Output Resistances of Inverting Amplier

From the circuit,

Ii =Vi 0R1

Ri = ViIi

= R1

The input impedance of the inverting amplier circuit is R1 (althoughinput impedance of OpAmp is innite). The output impedance of thecircuit is zero because Vo is independent of RL (Vo does not change whenRL is changed).

1.5.5 Non-inverting Amplier

Vp = Vi

Negative Feedback Vn Vp = ViVn 0R1

+Vn Vo

R2= 0

Substituting for Vn = Vi, we get

R2R1

Vi + Vi Vo = 0Finally,

Av =VoVi

= 1 +R2R1


VVii

VVoo

RR22

RR11

IIii++

--

++

++

--

--

++

--

VVdd

VVpp

VVnn

VViiIIpp

AAVVdd

VVoo

RR22RR11

Figure 1.63: Non-inverting amplier.

Input Resistance

Since Ii = Ip = 0. Therefore, Ri .

Output Resistance

Vo is independent of RL, so Ro = 0. Note that Ri and Ro = 0 should betaken in the context that we are using an ideal OpAmp model. In reality,the above circuit will have input and output resistances equal to that ofthe OpAmp itself.

1.5.6 Voltage Follower

VVii

VVoo++

--

Figure 1.64: Voltage follower.


In some cases, we have two-terminal networks which do not match well,i.e, the input impedance of the later stage is not very large, or the outputimpedance of preceding stage is not low enough. A buer circuit isusually used in between these two circuits to solve the matching problem.These buer circuit typically have a gain of 1 but have a very large inputimpedance and a very small output impedance. Because their gains are 1,they are also called voltage followers.

The non-inverting amplier above has Ri and Ro = 0 and, there-fore, can be turned into a voltage follower (buer) by adjusting R1 and R2such that the gain is 1

VoVi

= 1 +R2R1

= 1 R2 = 0So by setting R2 = 0, we have Vo = Vi or a gain of unity. We note thatthis expression is valid for any value of R1. As we want to minimize thenumber of components in a circuit, we can remove R1 from the circuit.

1.5.7 Inverting Summer

VV22

VVoo

++

--

++

--AAVVdd

VV11

RR22

RR11

VVnn

VVpp

RRff

Figure 1.65: Inverting summer.

Vp = 0

Negative Feedback Vn Vp = 0Vn V1

R1+

Vn V2R2

+Vn Vo

Rf= 0


Finally,

Vo = RfR1

V1 RfR2

V2

So, this circuit adds (sums) two signals. An example of the use of thiscircuit is to add a DC oset to a sinusoidal signal.

1.5.8 Non-Inverting Summer

VV22

VVoo++

--

++

--AAVVdd

VV11

RR22

RR11 VVnn

VVpp

RRff

RRss

Figure 1.66: Non-inverting summer.

Negative Feedback: Vn VpVp V1

R1+

Vp V2R2

= 0 Vp(

1

R1+

1

R2

)=

V1R1

+V2R2

Vn 0R1

+Vn Vo

Rf= 0 Vo =

(1 +

RfRs

)Vn

Substituting for Vn in the second equation from the rst (noting Vp = Vn),we obtain

Vo =

(1 + Rf/Rs

1/R1 + 1/R2

)(V1R1

+V2R2

)

So, this circuit also adds (sums) two signals. It does not, however, invertsthe signals.


1.5.9 Dierence Amplier

VV22

VVoo

++

--

++

--AAVVdd

VV11

RR22

RR11 VVnn

VVpp

RRff

RR33

Figure 1.67: Dierence amplier.

Negative Feedback: Vn VpVp V2

R2+

Vp 0R3

= 0 Vn Vp = R3R2 + R3

V2

Vn V1R1

+Vn Vo

Rf= 0

If one choose the resistors such that R3/R2 = Rf/R1, then

Vo =RfR1

(V2 V1)

1.5.10 Current Source

Negative Feedback: Vn Vp = VsiL =

VnR

=VsR

: constant

For a xed value of Vs, the current IL is independent of value of RL andoutput voltage Vo. As such, this circuit is an independent current source.

The value of the current can be adjusted by changing Vs (for a xed RL.Therefore, this circuit is also a voltage to current converter or a transcon-ductance amplier.

1.6. DYNAMIC OR TIME-DEPENDENT CIRCUITS 69

Vo

+

-

Vs

RL

Vn

Vp

R

+

-

IL

Figure 1.68: Current source.

Grounded Current Source

+

-Vs

R2

R1 Vn

Vp

Rf

RL

R3

IL+

-

Vo

Figure 1.69: Grounded current source.

The problem with the above current source is that the load is notgrounded. This may not be desirable in some cases. This circuit hereis also a current source with a grounded load if Rf/R1 = R3/R2.

1.6 Dynamic or Time-Dependent Circuits

In this section we discuss circuits that include capacitors and inductors.The i-v characteristics of these two elements include derivatives or integral


of either i or v resulting in time-dependent circuits.

Capacitor

A typical capacitor is made of two parallel conducting plates separated byan insulator. If we connect a capacitor to a voltage source, electrons travelfrom the voltage source to the capacitor and charge one of the plates withexcess electrons. The electric eld from these electrons repel electrons fromthe opposing plate and equal but positive charge develops on that plate.The repelled electrons from the positive plate travel to the source. As such,the total charge stored in a capacitor is zero but consists of two separatedamount of equal charges with dierent polarity. In this manner, a capacitorcan store energy in the form of electric eld between the two plates, whichis proportional to the voltage between the two plates. For most parallelplate capacitors, the voltage across the capacitor is directly proportional tothe charge accumulated in the capacitor. The constant of proportionality isthe capacitance. Note that the total charge stored in a capacitor is zero, thecharge mentioned here is the charge on one of the plates. The capacitance,C is

C =q

vUnit: Farad (F)

Most commercial capacitors are in F (106F), nF (109F), and pF(1012 F) range.

As the number of electrons owing into the negative plate is equal to thenumber of electrons owing out of the positive plate, the current owinginto one plate of capacitor is exactly equal to the current owing out of theother plate. The i-v characteristics equation for a capacitor (using passivesign convention) is

q = Cv i = dqdt

= Cdv

dt

Note that the current i ows as long as the voltage (and charge, q = Cv)changes in time. At the DC steady-state condition, dv/dt = 0 and capacitorcurrent, i = 0, and capacitor acts as an open circuit.

Energy stored in a capacitor, W , can be found from the denition ofelectric power:

W (t) =1

2Cv2(t)


Time dependent circuits either have a time-dependent source (we willexamine circuit with sinusoidal sources later) and/or have a source whichis switched on or o at some time. In all switching circuits, we assumethat switch is thrown instantaneously. This means that for these switchedcircuit, the voltage or current waveform will have a discontinuity in time.

For example, consider the circuit shown in Figure 1.70. The switch isinitially closed and is opened at time t0. For t < t0, the switch is closed.By KCL, i = is and a voltage of v = Ris appears across the resistor. Fort > t0, i = 0 and v = 0. There is a discontinuity in voltage and current attime t = t0 when the switch is thrown.

t = t0

RIs

+

-

v

i

t < t0

RIs

+

-

v

i

t > t0

RIs

+

-

v

i

Figure 1.70: Switched resistive circuit.

Values of current and voltage at the exact switching time is undened.The values are known just prior to the throwing the switch and just afterthat. We denote that time just prior to throwing the switch as t0 (t

0 = t0

with being innitesimally small). We denote the that time just afterthrowing the switch as t+0 (t

+0 = t0 + ). In this sense, current and voltage

are well dened at t0 and t+0 . In the circuit shown, both current and voltage

have a discontinuity at t = t0 as i(t0 ) = i(t

+0 ) and v(t

0 ) = v(t

+0 ).

This behavior is in general true for all resistive circuit. There is a discon-tinuity in both voltage and current waveforms at the time that the switch


is thrown. This is not true for circuits with capacitors (or inductors). Forcapacitors, the i-v characteristics equation i = Cdv/dt implies that if therewere a discontinuity in the voltage, an innite amount of current shouldow through the capacitor. Therefore, the voltage waveform across a ca-pacitor should be continuous. This means that if a switch is thrown at timet0, we should have v(t

0 ) = v(t

+0 ). Note the capacitor current waveform can

have a discontinuity.

Inductor

A typical inductor is made of a a wire wrapped around a core of magneticmaterial. As current ows through the wire, a magnetic eld is produced.As such, an inductor can store energy in the form of magnetic eld. Thei-v characteristics equation for an ideal inductor is given by Faradays Law(wire in an ideal inductor has no resistance):

v = Ldi

dtInductance, L. Unit: Henry (H)

using passive sign convention. Energy stored in an inductor, W , can befound as

W (t) =1

2Li2(t)

For inductors, the i v characteristics equation v = Ldi/dt implies thatif there were a discontinuity in the current waveform, an innite amountof voltage should appear across the inductor. Therefore, the current wave-form in an inductor should be continuous. This means that if a switch isthrown at time t0, we should have i(t

0 ) = i(t

+0 ). Note the inductor voltage

waveform can have a discontinuity.

Comments

Circuits containing one or more capacitors and/or inductors are calleddynamic circuits. Voltage and currents in dynamic circuits are, ingeneral, functions of time, i.e., their value change as time progresses.

All analysis method (KVL, KCL, circuit reduction, node-voltage andmesh current methods) can be applied to dynamic circuits with nomodication.

Application of the analysis methods to a dynamic circuit results ina set of linear dierential equations (as opposed to a set of algebraic


equations for resistive circuits). Solution of these equations requiresinitial conditionsone for each capacitor or inductor.

Two-terminal subcircuits containing capacitors and/or inductors can-not be reduced to Thevenin or Norton forms.

Switched Circuits

Time dependent circuits either have a time-dependent source and/or have asource which is switched on or o at some time. Most circuits have both.The voltage source in the circuit is a time-dependent voltage (sinusoidalvoltage with a frequency of 60 Hz). When the switch is o, the voltagesand currents in the is zero. After the switch is thrown, the voltages andcurrents rise from zero and after some time start to follow the source voltagewaveform in a steady manner.

The evolution of circuit voltages and current in time, thus, can be di-vided into two distinct time scales. 1) A transient response starting withthe throw of the switch and 2) A steady response to the time dependentsources. We will see that the transient response of the circuit is completelyset by the circuit elements themselves and not by the source. This tran-sient response also stops after some prescribed time. Because of thesetwo distinct responses of a time-dependent circuit, transient and steadyresponses can be studied separately.

As an example, consider a switched circuit as in Figure 1.71. The switchis open at t = t0. Is is DC current source. It is usually assumed that thecircuit has been in the rst state (t < t0) for a long time and the responseof the circuit for t > t0 is needed.

The solution to the circuit for t > t0 leads to a set of linear dierentialequations. These equations require a set of initial conditions (one for eachcapacitor and inductor). These initial conditions are found by solving thecircuit for t < t0 and using the no-discontinuity conditions for voltagesacross capacitors and currents in inductors at the switching time. So, theprocedure for solving switched circuits with DC sources is:

Solve DC steady-state case for t < t0. Determine vc(t0 ) and iL(t0 ). Use no-discontinuity conditions to nd the initial condition for the

second circuit: vc(t+0 ) = vc(t

0 ) and iL(t

+0 ) = iL(t

0 ).


t = t0

R CIs

t < t0

R CIs

t > t0

R CIs

Figure 1.71: Switched circuit with capacitor.

Solve the time dependent circuit for t > t0.

DC Steady-State Analysis

In a dynamic circuit with DC sources, voltages and currents reach constantvalues after long time. This state of the circuit is called DC SteadyState. By denition, all time derivatives vanish at DC steady state condi-tion:

For a capacitor, i = Cdv

dt(for any v). Thus, a capacitor acts as an open

circuit DC steady state.

For an inductor, v = Ldi

dt(for any i). Thus, an inductor acts as a short

circuit in DC steady state.

Therefore, to analyze a circuit in DC steady state condition, we replacecapacitors with open circuit and inductor with short circuit and proceed tosolve the resulting resistive circuit.

Example

This circuit has been in DC steady state and t = t0 the switch is opened attime t = t0. Find the initial conditions for the dynamic circuit for t > t0.

First, we mark vC and iL on the original circuit (those are our initial


t = t0

200 C1A

+

-

vc

L iL

50

1A

iL

50200

+

-

vc

vc

Figure 1.72: Example of switched circuit with capacitorand inductor.


conditions). We draw the circuit for t0 < t (closed switch) and replace thecapacitor with an open circuit and the inductor with a short circuit. Wethen proceed to solve the resistive DC steady state circuit.

The resulting circuit is current divider and iL and vC can be found readilyfrom current-divider formulas. Alternatively, using node-voltage method,we have

vc 050

+vc 0200

1 = 0 vc = 40V

iL =vc 050

= 0.8A

Note that since the circuit is in DC steady state, values of vC and iL areconstant in time. Using the no-discontinuity conditions for vC and iL, wecan nd the initial conditions for the time-dependent circuit as

vc(t = t+0 ) = 40V iL(t = t+0 ) = 0.8A.

Chapter 2

Introduction to Signals and Systems

2.1 Classication of Signals

A signal is a set of information or data. In this lecture, we deal almostw

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system analysis

circuit analysis techniques

time systems system

signal transmission

signal operations

classication of systems

fourier methods

analog signal processingdr