ece 303 notes final

ECE 303 - Sum10 Notes Set 1:Transmission Lines and the Wave Equation 1

INTRODUCTION

One purpose of a transmission line is to transport electromagnetic energy from one locationto another. For example, we might wish to “send” a digital bitfrom CPU to a memorychip in a computer or “send” an analog television signal downa coaxial cable. We canunderstand many applications by using the concepts of current and voltage from basiccircuits. This approach is the main topic of this set of Notes, which leads to a very importantmathematical result called the Wave Equation.

This set of Notes shows how circuit theory, which you studiedin ECE 200 and ECE 211,can be used to derive a mathematical characterization of transission lines. For future refer-ence, here are some areas where transmission line analysis is sometimes used:

Coaxial CablesCoaxial cables provide connection media in the Cable TV and Cable Modem net-works. Coax can also connect RF (Radio Frequency) devices like antennas to displaydevices, like televisions.

Twisted-Pair Wire PairsTwisted pairs of insulated wires, usually copper, is used toconnect wireline tele-phones to an access point in the Public Switched Telephone Network. Twisted-paircopper is also used in Digital Subscriber Line (DSL) internet access. Twisted-paircan also be used in short-length high-speed computer networks.

Computer BusAs computer systems become “faster”, the pulse widths become shorter and the con-ducting paths in a computer system become transmission lines. You can think of a16-bit bus as being 16 parallel transmission lines, locatedvery close together.

We will first derive the general transmission lines equations. After that we will studysome of the mathematical properties of simple lossless transmission lines. Properties andcharacteristics of other lines will be the topics of later sets of Notes.


TRANSMISSION LINES

Most metallic transmission lines have two conductors in proximity, which are separated byan insulator. Here are physical forms for some common transmission lines (for clarity theinsulator material is not shown):

TWISTED−PAIR

COPPERx

COAXIAL CABLE

x = 0 x + x∆

There is insulation between the conductors in both examplesshown above. The twisted-pair of lines do not touch in the diagram; the view is like we are looking down on thewire-pair. The length fromx to x+∆x is called aninfinitestimalsection. You can think ofthe line as being composed of a large number of these infinitesimal sections.

QUESTION:

In order to anlayze and design many devices, a property we need to understand is howvoltagev(x; t) and currenti(x; t) are related for the transmission line. For example, for theresistor we havev(t) = i(t) R. We need a similar result for the transmission line.

APPROACH:) Use Kirchoff’s laws to relate voltage and current over the infinitesimal section∆x.) Extrapolate results to find the differential equation relating voltage and current for anyvalue ofx.


CIRCUIT MODEL

In this course, we will consider transmission lines having two metallic conductors in prox-imity. One useful circuit model of a∆x section for this type of transmission line is shownbelow. We will discuss in class the physical origin of the parameters in the figure:

i(x,t)

v(x, t)

x

+ +

−−

L

G

R∆

∆ C ∆

i(x + , t)∆

x +

v(x + , t)∆

∆ x

∆x

x

x

x x

x

The above circuit models the transmission line as having resistance. inductance, capaci-tance, and conductance for any infinitesimal segment of the line. In this circuit model theparameters are:

R = RESISTANCE per unit length ( likeΩ/m )L = INDUCTANCE per unit length ( like H/m )C = CAPACITANCE per unit length ( like F/m )G = CONDUCTANCE per unit length ( like S/m )

These parameters quantify theseriesandparallel properties of the line: You can think ofR andL as series, or in-line, parameters of the conductors. Forexample, a wire has a series resistance. You can think ofC andG as parallel parameters between the two conductors. Forexample, two conductors close together form a capacitor.

We next use Kirchoff’s Laws to obtain two equations in the twounknownsv(x; t) andi(x; t).


KIRCHOFF’S LAWS

LOOP EQUATION:

To derive one of the needed equations, use Kirchoff’s Voltage Law around the outer loopof the∆x section. This procedure obtains the following equation:v(x; t) + R∆x i(x; t) + L∆x ∂

∂ti(x; t)+v(x+∆x; t) = 0 (1)

Note that we have to use a Partial Derivative in equation (1) to quantify the voltage acrossthe inductor. This is because we have two independent variables, x and t, and we arekeepingx constant while we take the derivative with respect tot. Next, divide both sides of(1) by ∆x and rearrange the resulting terms:

v(x+∆x; t)v(x; t)∆x

=Ri(x; t)L∂∂t

i(x; t) (2)Now take the limit of both sides of (2) as∆x! 0. The left-hand side of (2) becomes apartial derivative:

∂∂x

v(x; t) =Ri(x; t)L∂∂t

i(x; t) (3)Equation (3) is a Partial Differential Equation (PDE). It tells us the voltage (and current)are functions of both time and distance on the line.

NODE EQUATION:

To derive the other equation relatingv(x; t) and i(x; t), use Kirchoff’s Current Law at theNODE in the∆x-section and follow the above limit procedure. This processwill eventuallygive

∂∂x

i(x; t) =Gv(x; t)C∂∂t

v(x; t) (4)Problem: 1-1


TRANSMISSION LINE EQUATIONS

The two equations (3) and (4) derived on the previous page aresometimes collectivelycalled the Transmission Line Equations. Althoughv(x; t) and i(x; t) are strictly functionsof both x and t, there is a simpler notation we can use for our course. To simplify thenotation, letv(x; t) be replaced by simplyv and leti(x; t) be replaced by simplyi. Usingthese simplifications, the two partial differential equations in (3) and (4) then become

∂v∂x

= Ri L∂i∂t

(5a)∂i∂x

= Gv C∂v∂t

(5b)We will call the partial differential equations in (5) the Transmission Line Equations. Theyare the first of many partial differential equations we will be using in this course. Thesolutionsv = v(x; t) and i = i(x; t) from (5) give the manner in which the voltage andcurrent vary as a function of timet and distancex on the transmission line.

In these partial differential equations, one position on the line may have a different voltageand/or different current than another position on the line.You may have seen this typeof behavior before: think of a string vibrating between two fixed end points (like a guitarstring). The parts of the string near the middle, away from the fixed ends, vibrate with themaximum amplitude, while the portions of the string near thefixed ends vibrate very little.Thus, the amplitudeA= A(x; t) of the vibrating string becomes a function of both positionx on the string as well as timet.

This vibrating string example is mentioned for familiarity. The electrical transmission linehas a different mathematical form for its solution, which iswhat we will discuss next.


WAVE EQUATION SOLUTIONS

In general, none ofR;L;G, or C in (5) are zero, and the solution process for (5) to obtainv(x; t) andi(x; t) is quite involved. There is a set of Supplementary Notes accessible fromthe course Home Page called “Solution of Transmission Line Equations” which investigatesactually solving the equations in (5). If you are interestedin this solution method theseSupplementary Notes should be useful. However, in the work which follows we will justuse the solutions as needed to understand properties of transmission lines.

Whereas the solution of (5) for generalR;L;G, andC is quite complicated, there is onechoice forR and G which leads to a very simple math solution and which describes asituation common to many applications of transmission lines. Such a situation arises forthe line havingR= 0 andG = 0 in (5). (We will discuss in class situations in whichthis is a valid approximation.) This line is called the Lossless Line and the PDE resultingfrom R= 0 andG= 0 is sometimes called the Wave Equation. To develop this equation,substituteR = G = 0 in equations (5a) and (5b), giving

∂v∂x

= L∂ i∂t

(6a)∂i∂x

= C∂v∂t

(6b)It is straightforward to show that the two first-order partial differential equations in (6) areequivalent to one second-order partial differential equation given by

∂2v∂x2 = LC

∂2v∂t2 (7)

Equation (7) for voltagev = v(x; t) is sometimes called the Wave Equation. (There is also asecond-order wave equation for currenti.) On the next page we discuss the solution to theWave Equation.

Problems: 1-2, 1-3


SOLUTION PROPERTIES

The solutionv(x; t) to (7) has a very simple form. Letg(t) be any voltage waveshape whichcan be successfully introduced to the pointx= 0+ on the infinite length transmission line.Then the solution for voltage and current at any pointx and at any timet is

v = v(x; t) = g

t xvp

(8a)i = i(x; t) = v(x; t)

R0(8b)

where the parametervp in (8a) is called the Phase Velocity (in m/sec) and the parameterR0

in (8b) is called the Characteristic Resistance. These parameters are computed by

vp = 1pLC

; R0 =rLC

(9)You can see the Supplementary Notes on Solution of Transmission Line Equations for thespecific solution method, as well as the derivation ofvp andR0 in (9).

Useful Solutions:

It is straightforward (but somewhat lengthy) to show that the following functions are solu-tions to the Wave Equation :

v1(x; t) = e j ( ω tβx )v2(x; t) = cos(ω tβx+φ)

v3(x; t) = sin(ω tβx)where the parameterβ in these equations is called the phase constant and is definedby

β = ωvp

(10)We will frequently use these solutions in our work since theyare easily differentiable withrespect to both timet and spacex.

Problems: 1-4 through 1-8


TRANSFORMATION OF VARIABLES

We have now seen that transmission lines techniques introduce the “t xvp

” transforma-

tion into the argument of functions. Let’s now examine in detail the implications of thistransformation.

QUESTION:

What does the transformation term “t xvp

” mean in the argument of the “g”

function in equation (8a) ?

ANSWER:As t increases, this term moves the “g” shape in the direction of propagation, witha velocityvp.

We can show this property by choosing a simple “g” function and mathematically inves-

tigating the effect oft xvp

. Let a simple rectangular pulse be defined by

g(α) =A; 0 α T0; else

(11)To getg

t x

vp

, substituteα = t x

vpin equation (11):

g

t xvp

=(A; 0 t x

vp T

0; else(12)


PROPAGATION PROPERTIES

Now let’s locate the non-zero portion of the “g” function on thex-axis.

QUESTION: Where isg

t xvp

in (12) non-zero on thex-axis?

ANSWER: Solve the inequality on the right side of equation (12) such that onlyx is inbetween the inequality signs, as shown below:

0 t xvp T (13). &

Lower : 0 t xvp

t xvp

< T : Uppert xvp

xvp

< T t

vp t x x vp tvpT& Combine: .vp(tT) x vp t (14)

Equation (14) shows that the non-zero portion of the “g” function is localized to a particularregion inx. Note also from (14) that timet is in the upper and lower bounds of this region.This means that the region, and hence the “g” function, moves as timet changes. We canthink of this as a pulse moving through space as time increases.


PROPAGATION EXAMPLE I

Thus we see that the inequality on the right side of (12) is equivalent to the inequality in(14). Substituting (14) into (12) then gives

g

t xvp

=A; vp(tT) x vp t0; else

(15)The argumentt x

vpmoves the “g” waveform to the right ast increases. The movement

of the waveform in space is like a “wave” moving. We next examine two examples whichdemonstrate how this movement occurs.

EXAMPLE: Let vp = 2(10)8 m/sec. and let the duration of the pulseT = 1 µsec. Find theposition of the pulse in (15) above att = 0.

APPROACH: At t = 0 we havevpt = 0 andvp(tT) = 2(10)8(106) = -200. Thus,

g

t xvp

= A; 200 x 00; else

(16)A representation of this pulse is shown below.

x = xLphysical cable

region

A

x = −200 x = 0

At time t=0, the pulse is “still in the source” ... the pulse has not yetentered the PHYSICALLINE. The next example will show how the pulse travels down the line.

Problems: 1-9, 1-10, 1-11


PROPAGATION EXAMPLE II

As timet increases, the pulse travels down the transmission line. The example below willshow how this occurs.

EXAMPLE:As in the previous example, letvp = 2(10)8 m/sec,T = 1 µsec. Find the position on theline of the pulse att = 2 µsec.

APPROACH:At t = 2(10)6, we have the values given below:

vpt = 2(10)82(10)6 = 400; vp(tT) = 2(10)8h2(10)6106i = 200

Substitute these values into equation (29), giving

g

t xvp

= A; 200 x 4000; else

(17)This pulse is shown below.

x = 0 x = 200 x = 400

A

As time t increases, the rectangular pulse travels farther to the right (increasingx) until itreaches the end of the line.

Problems: 1-12, 1-13, 1-14

ECE 303 - Sum10 Notes Set 2:Propagation on Transmission Lines 1

INTRODUCTION

In the previous set of Notes we developed some of the mathematics quantifying the currentand voltage on a transmission line. In some cases, the line may be approximated as theLossless Line, which led to the Wave Equation. We then solvedthe Wave Equation andbriefly examined some of the solution properties.

With the introduction of transmission lines, we now find we have two seemingly differentmethods for analyzing electrical systems: (i) Circuit Theory, which you learned about inECE 211, and (ii) Transmission Line techniques (like wave theory), which we are currentlystudying. One of the things we will learn in this set of Notes is when Circuit Theory isadmissable and when we must apply transmission line techniques.

At some level, every circuit has transmission line characteristics. However, often the layoutof the circuit, the size of the circuit, and the frequency of operation are such that CircuitTheory techniques are sufficiently accurate for successfulanalysis and design. In othercases, such as very high frequencies and/or very large circuits, we may have to use trans-mission line techniques. We will learn in this set of Notes how to understand some ofthese considerations and make the choice between circuit techniques and transmission linetechniques.

In this set of Notes we will also develop an understanding of the physical meaning of thesolution to the Wave Equation. We will see that the solution has a very satisfying physicaland geometrical interpretation which can sometimes save ussubtantial mathematical effort.


THE WAVE EQUATION

Understanding when to use transmission line techniques requires us to examine in detailthe solution to the wave equation. From our previous work we derived the Wave Equationfor the lossless transmission line. We found that this equation is a second-order PDE forthe voltagev = v(x; t) given by

∂2v∂x2 = LC

∂2v∂t2 (1)

whereL andC are the per unit Inductance and Capacitance, respectively.In Notes Set 1 weshowed that the solution to the wave equation forv in (1) is given by

v = v(x; t) = g

t xvp

(2)where the parametervp in (2) is called the Phase Velocity (in m/sec). Recall thatvp and theCharacteristic Resistance,R0, of the lossless line are computed by

vp = 1pLC

; R0 =rLC

(3a; 3b)The function “g” in (2) is any physically realizable function, like a pulse or a sinusoid,which can be successfully coupled to the line. It is straightforward to show that among thesolutions to the wave equation are the following:

v1(x; t) = e j ( ω tβx )v2(x; t) = cos(ω tβx+φ)

v3(x; t) = sin(ω tβx)The parameterβ in these equations is called the phase constant, defined by

β = ωvp

= 2πλ

(4a)whereω = 2π f in (4a) is the radian frequency of the sinusoid andλ is the wavelength,defined by

λ = vp

f(4b)

We will frequently use these solutions in our work since theyare differentiable with respectto both timet and spacex.


CIRCUIT THEORY OR WAVE TECHNIQUES ?

In ECE 211 you developed the ability to perform circuit analysis. However, understandingthe current material requires us to learn about waves and propagation. At first glance theseseem to be two very different approaches to finding currents and voltages. For the followingdiscussion let’s adopt the following informal definitions:

Circuit Techniques :

Circuit Techniques are the application of Kirchoff’s Laws,ordinary derivatives, andtheRLC voltage/current relations likev(t) = i(t)R, v(t) = Ldi=dt, etc. to find volt-ages and currents in a device or system.

Transmission Line Techniques :

Transmission Line Techniques are the application of the wave equation, partial deriva-tives, and reflections (which will be developed in a later Notes Set) to find voltagesand currents in a device or system.

Under some constraints, transmission line techniques and circuit techniques give the sameanswers to problems. Consequently, circuit theory is widely used in situations where theseconstraints are met. But there remains the questions of (i) what these constraints are, and (ii)when these constraints are met. Understanding the implications of the solution to the waveequation, given by equation (2) on the previous page, will help us answer the followingfundamental question concerning the analysis and design ofsystems and circuits :

Given a circuit layout of a particular size and operating frequency, when canwe analyze/design the circuit using Circuit Techniques andwhen must we useTransmission Line techniques ?

The answer to this question depends on the relation between the physical size of the layoutand the wavelength of the propagating voltage/current wave. We can understand this byexamining an example which we begin on the following page.


SOURCE-LINE-LOAD SYSTEM

The figure below shows a lossless transmission line connecting an ideal source, with sourceresistanceRs, to a loadRL = R0, whereR0 is the characteristic resistance given by (3b).(As we will learn later, settingRL = R0 will eliminate reflections, butRL = R0 will stilldemonstrate the need for transmission line techniques.) The source-line connection is atx = 0 and the line-load connection is atx = xL :

+−

x = 0

sourceregion

line region loadregion

( )

x = xL

=R L

R s

R 0

v (t)sR

0

The figure above looks very much like a circuit schematic. Therefore, let’s redraw thesource-line-load system and attempt to analyze it using circuit theory. In the figure below,the “X” represent locations atx = 0+ andx = xL . Measuring the voltages at these pointsprovides the voltagesv0(t) at x = 0+ andv1(t) at x = xL .

+−

X

X

X

X

+ +

− −

x = 0

=

x = x

x = x−L

L

R L

R s

x = 0+

v (t)s

R 0

v (t)0 v (t)1

The pointx = xL is the last point on the transmission line before the connection with theload. On the next page we will apply circuit theory to obtainv0(t) andv1(t) .


CIRCUITS-BASED SOLUTION

Let’s analyze the bottom figure on the previous page using thecircuits skills you learned inECE 211. Let the transmission line be lossless withvp = 2(10)8 m/sec. Since the line is alossless conductor, then circuit-theory suggests the points x = 0+ andx = xL should havethe same voltage, that is,

v0(t) = v1(t) (5)Supposevs(t) is one-period (Tp seconds) of a sine wave at frequencyf Hertz. Let thissource produce the initial voltage on the line given by

v0(t) = A sin(2π f t); 0 t Tp

0; else(6)

Combining(5) and(6) then gives the circuits-based result


0; else(7)

Note that the result in(7) is independent of the actual Hz frequencyf , the propagationvelocityvp, or the length of the line.

Therefore, the circuits-based approach, represented by the results in equations (5) - (7),suggests the voltage on the two ends of the line is the same. This means the voltages arethe same regardless of whether the line is 1 m long or 0.01 m long or 10 m long. However,let’s see what happens when we use the solutions required by the wave equation (that is,solutions of the type given in equation (2) ). This is done on the next page.


WAVE-BASED SOLUTION

We can see the limitation of the circuits-based result by relating the voltages in the previousfigures to the solutions to the wave-equation. The solution from equation (2) for the losslesswave equation is repeated here for easy reference:

v(x; t) = g

t xvp

(8)whereg is the wave shape appearing on the first point of the line, thatis, atx = 0+. Butv0(t) has been defined to be the voltage atx = 0+. Therefore, equation(8) thus relates thevoltages at each end of the line to solutions of the wave equation. At x = 0+ we have

v0(t) = v(x; t)x=0+ = g

t x

vp

x=0+ = g(t) (9)

At the load end of the line we havex = xL and therefore equation(8) gives

v1(t) = v(x; t)x=xL = g

t x

vp

x=xL = g

t xL

vp

(10)Comparing(10) to (9) we see an equivalent way of expressingv1(t) is

v1(t) = v0

t xL

vp

(11)Notice how the waveform at the end of the line is a time-delayed replica of the waveformat the beginning of the line. “Time” is a very appealing dimension for humans, but thequantity of Phase (in radians) is frequently more useful forunderstanding sinusoidal results.We will explore the implications of this phase on the following page.


IMPLICATION OF PHASE

To see the implication of phase, let’s examine what occurs atthe load end of the line. At theload end of the linex = xL . Therefore, replacet by t (xL=vp) in the sinusoid waveformin equation(6). Then using equation(11) provides an alternate expression forv1(t) :

v1(t) = v0

t xL

vp

=8<: A sin

2π f (t xL

vp); 0 t xL

vp Tp

0; else(12)

Use (4b) to simplify the sinusoidal and the inequality on theright side of(12), obtaining

v1(t) =8<: A sin

2π f t2πxL

λ

; xL

vp t Tp+ xL

vp0; else

(13)Equation(13) can be simplified using two substitutions. Let the phase angle θL and thetime delaytL be defined by

θL = 2πxL

λ; tL = xL

vp(14; 15)

Substituting(14) and(15) into (13) then provides the more compact form

v1(t) =( A sin

2π f tθL

; tL t Tp+ tL0; else

(16)Now recall (6) forv0(t), the signal at the beginning of the line, repeated here for reference:


0; else(17)

Comparing (16) and (17), we see thatv1(t), the signal atx = xL , is not the same asv0(t),the signal atx = 0+. There is a time-delay between the two signals, as expressedin thetime range on the right side of (16). The non-zero portion ofv0(t) begins at timet = 0, butthe non-zero portion ofv1(t) does not begin until timetL.


IMPLICATION OF PHASE (cont.)

The time displacement between the two signalsv0(t) and v1(t) on the previous page isinteresting, but it does not entirely answer the question ofwhether we need to use circuittheory or transmission line techniques. For answering thisquestion, it turns out that thephase delayθL, given by the negative phase component in the sinusoid in (16), is the moreuseful quantity. This phase delay is the quantity which helps us answer the question ofwhen circuit theory is valid.

From equation(14), we see thatθL becomes non-negligible as the length of the linexL

becomes an appreciable fraction of the wavelengthλ. We will work two examples on thefollowing pages which will help us understand the relation betweenθL andλ. In each ofthese examples you can refer to the line diagrams on page 4 fordefinition of parameters.Also for simplicity, we will assume that there are no reflections from the load and that thepropagation velocity is independent of frequency.


EXAMPLE: LOW FREQUENCY

Let’s first examine an example which connects a low-frequency source to a load. Let thelength of the lossless transmission line bexL = 1 meter and let the propagation velocityvp

bevp = 2(10)8 m/sec. Using (15), this gives the time length of the line as

tL = 12(10)8 = 5 nanosec= :000005 ms (18)

ExpressingtL in milliseconds (ms) will be useful in the final result. Let the frequency ofthe low-frequency sinusoidal source bef = 200 Hz, in which case the time periodTp is

Tp = 1f

= 1200

= 5 ms (19)This is a signal in the “audio” range. From (4b) the wavelength of this sinusoid is

λ = vp

f= 2(10)8

200= 106 meters (20)

The ratio of the length of the line (1 m) to the wavelength for this case is therefore

xL

λ= 1

106 = 106 (21)Equation(21) shows that the size of the entire source/line/load system isvery small whenmeasured in units of wavelengths. Substituting (21) into (14) gives the phase term

θL = 2π106 rad 3:6(10)4 degrees (22)Finally, substitute (18), (19) and (22) into (16) and (17), obtaining the expression for thesignals at the beginning and the end of the line:

v0(t) =A sin(2π200t); 0 t 5 ms0; else

(23)v1(t) =(

A sin

2π200t0:00036Æ ; :000005 ms t 5:000005 ms

0; else(24)


EXAMPLE: LOW FREQUENCY (cont.)

The phase valueθL = 2π106 rad 3:6(10)4 degrees in (24) is very small comparedto one sinusoidal cycle, which is 2π radians or 360 degrees. This means that the signalv1(t), at the load end of the line, is essentially the same asv0(t), the signal at the beginningof the line. Their plots “overlap in time”, as discussed in class. You can see this by plottingv0(t) andv1(t) from (23) and (24) on the same graph, as shown below:

t = 5

A

t (ms)

t = 0

Lx = 1, f = 200

differencephaseo0.00036

v (t)0

1v (t)

The plot above showsv0(t), and thenv1(t) displaced by 3:6(10)4 degrees. The sinusoidsare virtually indistinguishable. In fact, the displacement of v1(t) in the figure is actuallyexaggerated for illustration purposes.

Plotting the two graphs shows easily that the two waveforms are essentially the same. Sincecircuit theory predicted thatv0(t) = v1(t), this is equivalent to saying that circuit theory isvalid for analyzing this system.


EXAMPLE: HIGH FREQUENCY

Now let’s examine an example which connects a high-frequency source to a load usingthe samexL = 1 meter line. This means thattL , the time length of the line, is stilltL =5 nanosec, as given by (18). However, now let the frequency ofthe sinusoidal source bef = 100 MHz, in which case the periodTp in seconds is

Tp = 1f

= 1108 = 10 nanosec (25)

This is an electrical signal in the “FM-radio” range. The electrical wavelength of thissinusoid is

λ = vp

f= 2(10)8

100(10)6 = 2 meters (26)The ratio of the line length (1 m) to the wavelength for this high-frequency case is thus

xL

λ= 1

2= 0:5 (27)

Equation(27) shows that now the size of the source/line/load system is non-negligiblewhen measured in units of wavelengths. Substituting (27) into (14) gives the phase term

θL = 2π(0:5) = π rad = 180 degrees (28)Finally, substitute (18), (25) and (28) into (16) and (17), obtaining the expressions for thesignals at the beginning and the end of the line:

v0(t) =A sin(2π108 t); 0 t 10 nanosec0; else

(29)v1(t) =(

A sin

2π108 t180Æ; 5 nanosec t 15 nanosec

0; else(30)


EXAMPLE: HIGH FREQUENCY (cont.)

The results in (29) and (30) mean that the signal at the end of the line, v1(t), is quitedifferent fromv0(t), the signal at the beginning of the line. They do not “overlapin time”like for the low frequency case. You can see this by plottingv0(t) andv1(t) from (29) and(30) on the same graph, as shown below:

t (nanosec)

t = 15t = 10t = 5

A

t = 0

v (t)0

1v (t)L

x = 1, f = 100 MHz

From the figure above you can see that the waveforms are not thesame. For example,observe the waveform values att = 7:5 nanosec. At the beginning of the line we havev0(t = 7:5 ns) = A. However, at the load end we havev1(t = 7:5 ns) = A, the negativeof the voltage at the beginning of the line. This means a design based uponv0(t) = v1(t)would be in error.

Therefore, circuit theory like you learned in ECE 211 is not acompletely accurate approachfor analyzing the currents and voltages in this “high-frequency” distributed system. How-ever, if the line were substantially “shorter” in physical lengthxL, then circuit theory mightonce again be valid.

The preceeding material does not mean circuit theory is not auseful technique. Providedthere are no reflections from the load back onto the line, circuit theory can be applied verywell to analyze and design systems whose sizes are “small” with respect to a wavelength.The problems listed below give you some practice in decidingwhether circuit theory ortransmission line methods are appropriate.


ECE 303 - Sum10 Notes Set 3:Lossless Lines - Pulses 1

INTRODUCTION

In previous work we have learned that the general transmission line has losses due to thein-line resistance of the conductors and the conductance between the conductors. Theconductance is due to a current path existing between the twoconductors.

However, there are many applications in which, at least to a first approximation, we canassume that the lines are lossless. Some applications whichuse this assumption will bediscussed in class. The lossless approximation means the per unit resistance,R, and theper unit conductance,G, may be assumed to be zero. In fact, these were the assumptionswhich led to the wave equation.

In this set of Notes we examine some properties of pulse propagation on lossless lines.For example, a high speed computer might use rectangular pulses to transmit informationbetween CPU and main memory. Additionally, a local area network (LAN) might use rect-angular pulses to transmit information between clients andservers. In these applicationsthe short transmission lines may often be assumed as lossless.

However, we will see that simply using a lossless line for transmission does not necessarilyremove all distortion from the transmission system. We willsee that there may be reflec-tions created by the impedance mismatches among the transmitter, the transmission lineitself, and the receiver. In this set of Notes we will study the effects of these reflections onthe transmission environment for pulses. In a future set of Notes we will study what occursif the transmitted signals are sinusoids.


LOSSLESS TRANSMISSION LINES

In a common application, we will use a Transmission Line to connect a Source to a Load.A Lossless Line is shown below:

R L

R s

R 0

v (t)s

+−

x = 0

sourceregion

line region loadregion

( )

x = xL

In the figure above, the parameters are given by:

vs(t) = Source Voltage, Rs = Source ResistanceRL = Load Resistance, R0 = Characteristic Resistance of Line

We encountered the characteristic resistance before in Notes Set 1. The characteristic resis-tanceR0 is not a series resistance. As we saw in previous notes, the characteristic resistanceis the ratio of voltage to current on the lossless line when there are no reflections. To under-stand this, let a voltage pulse enter the transmission line at x= 0 andt = 0. Then a physicalinterpretation ofR0 is given by

R0 = v(x= 0+; t = 0+)i(x= 0+; t = 0+) (1)

For a lossless line, the characteristic resistanceR0 and the velocity of propagationvp (phasevelocity) may be calculated by

R0 = rLC

; vp = 1pLC

(2)These relations were developed in Notes Set 1.


COMPUTING LINE PROPERTIES

A 5 meter lossless transmission line has per unit parametersL = 106 H/cm andC =5(10)9 F/m. The line is terminated aty= 5 m withRL = 0 (a short circuit ). Compute theproperties requested in (a), (b), (c), and (d) below:

(a) Computevp, the velocity of propagation in m/sec;

ANSWER:

Usevp = 1pLC

) vp = 1:414(10)6 m/sec

(b) Compute the characteristic resistance,R0, of the line in Ohms.

ANSWER:

UseR0 =rLC

) 141:2 Ω

(c) Compute the total series resistance of this 5 meter line in Ohms.

ANSWER:

The total series resistance of any lossless line isZERO.

(d) Let the source resistanceRs = 100Ω. A rectangular pulse on the line with amplitude6 volts and duration 1µsec. begins propagating towardRL at timet = 0. Plot the voltageacrossRL over the range 0 t 10µsec.

ANSWER:

The voltage drop across the short-circuit (RL = 0) is ZERO.


PULSE PROPAGATION ON LOSSLESS LINE

Let’s first consider the time period when there are no reflections from the load. The ampli-tude of the ideal pulse generated by the source is scaled as itenters the transmission line.When there are no reflections from the load, an Equivalent Circuit is as shown below:

R s

v (t)sR

0+−

x = 0

sourceregion

line region

Definev0(t) = v(x= 0+; t) as the initial voltage pulsev0(t) propagated on the line. Thesource pulsevs(t) andv0(t) are related as shown below:

v (t)sv (t)0

t = 0 t = Tt = 0t = T

A

K

t t

Note that the shape ofvs(t) has not changed as the pulse enters the line. Finding theamplitude ofv0(t) is similar to voltage division from circuits courses. The valuesK andAare related by

A= R0

R0+Rs

K (3)On the following page we will begin to examine the line as having dimensions in both soace(meters) and time (seconds).


PULSE AND LINE PROPERTIES

QUESTION: What is the distance “length” of a pulse on a line?

Answer: Supposev0(t) is a 1 V pulse which is 4µsec long in time:

v (t)0

t = 0t

1

t = 4 sµ

When this pulse is on the line it occupies a span of

dp = vpT = 2(10)8 m/sec 4(10)6 sec = 800 m

QUESTION: How many seconds “long” in time is a transmission line which has physicallengthxL meters?

Answer: Electromagnetic energy propagates at a velocity ofvp m/sec on the line.

Therefore it takes energy a time oftL = xL

vpseconds to travel from one end of the

line to the other. For example, ifxL = 1 mile 1609 meters andvp = 2(10)8 m/sec,

then the “time” length of the line istL = 1609 m2(10)8 m/sec

= 8.05 µsec.

In this course, it will be useful to think of pulses and lines as having dimensions in bothspatial and time units. We will next consider what happens tothe energy when it reachesthe end of the transmission line where the load is located.

Problem: 3-1


REFLECTION AND TRANSMISSION

Consider again the source/line/load transmission system on page 2. A source with sourceresistanceRs is connected to a lossless transmission line with characteristic resistanceR0.The transmission line is then connected atx= xL to a load having resistanceRL. When theenergy in the pulse propagates to the load then some energy inthe pulse will be transferedinto the load, but some energy might be reflected from the load. The reflection from theload is quantified by the load reflection coefficient and the transmission into the load isquantified by the load transmission coefficient.

LOAD REFLECTION COEFFICIENT:

Let A be the voltage amplitude of a pulse on a line which reaches theload. ThenVre f l, thevoltage amplitude of the pulse which is reflected back onto the line, is

Vre f l = ΓL A (4)whereΓL is the Load Reflection Coefficient, computed by the formula

ΓL = RLR0

RL+R0(5)

LOAD TRANSMISSION COEFFICIENT:

Let A be the voltage amplitude of a pulse on a line which reaches theload. ThenVload, thevoltage amplitude of the pulse which enters the load, is

Vload = τL A (6)whereτL is the Load Transmission Coefficient, computed by the formula

τL = 2 RL

RL+R0(7)

Using the preceding definitions, you can show thatΓL andτL at the load are related by

τL = 1+ΓL (8)


REFLECTION AND TRANSMISSION (cont.)

If the load is mismatched to the line, then load reflections will travel back toward thesource. If the source is also mismatched to the line, then reflections will be created at theline-source boundary, and additional reflections will be created. These source reflectionsare then sent back toward the load. The amplitude of the source relections are determinedby the Source Reflection Coefficient.

SOURCE REFLECTION COEFFICIENT:

LetVin be the voltage amplitude of a pulse on a line which reaches thesource from the lineside. ThenVback, the voltage amplitude of the pulse which is reflected back onto the line, is

Vback= Γs Vin (9)whereΓs is the Source Reflection Coefficient, computed by the formula

Γs= RsR0

Rs+R0(10)

Some computation examples for relflection coeffients are given on the following page.


REFLECTION AND TRANSMISSION EXAMPLES

The reflection and transmission coefficients defined on the previous pages quantify how thevoltage of the pulse is changed as energy is either reflected back onto the line or transmittedinto the load. Here are some examples which demonstrate computations.

EXAMPLE 1:

Let Rs= 150; R0 = 100; RL = 75. Use equation (4) to compute the reflection and transmis-sion coefficients.

Γs= 150100150+100

= 0.20 ; ΓL = 7510075+100

= - 0.143 ; τL = 2(75)75+100

= 0.857

EXAMPLE 2:

Let Rs= 100; R0 = 100; RL = 100. Compute the reflection and transmission coefficients.

Γs= 100100100+100

= 0 ; ΓL = 100100100+100

= 0 ; τL = 2(100)100+100

= 1.0

In the second example above there are no reflections. We say the source is match to the lineand the line is matched to the load.

Problems: 3-2, 3-3


PULSE AND REFLECTION EXAMPLE

At t = 0 an ideal voltage source havingRs = 100Ω produces a rectangular pulse with ampli-tude 6 volts and duration 10 nsec. The amplitude of the pulse which initially travels towardthe load is 4 volts. The amplitude of the first reflected pulse which travels back toward thesource is 2 volts. Compute the load resistanceRL and the characteristic resistance of theline R0.

APPROACH:

Solution for R0:

The amplitude of the source pulse isK = 6 andA = 4 volts initially propagates on theline. Use the relation betweenA andK given by euation (3) to solve forR0:

A= R0

Rs+R0

K ) 4= R0

100+R0

6

Solve this last equation forR0 and you will obtain

R0 = 200 Ω

Solution for RL:

To solve forRL, use the relation betweenA and the reflected voltageΓL A:

ΓL A = 2 ) RLR0

RL+R0A= 2 ) RL200

RL+2004= 2

Solve this last equation forRL and you will obtain

RL = 600 Ω.


BOUNCE DIAGRAMS

Some important properties of pulse voltages are conveniently expressed in a Bounce Dia-gram. The Bounce Diagram displays propagation and reflection using the Two-Dimensionalrepresentation shown below:

SOURCE LOAD

x = xL

ΓLAΓs

ΓLA

ΓLAΓs

2

t = 2tL

t = 3tL

t = tL

τL

ΓLAΓs

τLA

x = 0

At = 0

Some energy of the original pulse (with amplitudeA) moving to the right is transmit-ted into the load. This amount of voltage isτLA. Some energy of the original pulse (with amplitudeA) moving to the right undergoesa reflection given byΓL. Therefore, the amplitude of the first reflected pulse movingback to the left isΓLA. The pulse moving to the left with amplitudeΓLA then encounters the source andundergoes a reflectionΓs. The amplitude of the pulse reflected from the source isΓsΓLA


BOUNCE DIAGRAM EXAMPLE

Let A = 3 volts,T = 10 ns,vp = 2(10)8 m/sec and line length equal 20 m. Also letΓL =0.429 andΓs= 0:200. Plot the location and amplitude of the pulse att = 20 ns, 120 ns,and 240 ns using the bounce diagram.

APPROACH: Compute the necessary parameters:

tL = 202(10)8 = 100 ns, dp = 2(10)8 10(10)9 = 2 m

ΓLA = 1.29, ΓsΓLA = -0.26

t = 20 ns: d = 2(10)8 20(10)9 = 4

t = 120 ns d = 2(10)8 120(10)9 = 24 m 4 m toward source

t = 240 ns: d = 2(10)8 240(10)9 = 48 m 8 m toward load

SOURCE LOAD

t = 100 ns

t = 300 ns

t = 200 ns

x = 20x = 0

t = 0

1.29

3.00 (volts)

−.26

Given the bounce diagram, we can now plot the voltage on the transmission line for selectedtimes. This is shown on the next page.


LINE “SNAPSHOTS”

Plot the voltage on the physical line (a “snapshot”) corresponding to the bounce diagramin the previous example.

SOURCE LOAD

x = 20x = 02 4

1.29

3

−.26

6 8

16 18

t = 120 ns

t = 240 ns

t = 20 ns

The above plot shows the pulse changing polarity and reflecting off of both the load and thesource. We can also fix the position at any value of distancex and compute the time-domainvoltage at this position. This corresponds to the physical situation of using an oscilloscopeat x= x1 to acquire the time voltage tracev(x1; t).


WORKED LINE EXAMPLE - 1

The following pages are some worked examples for transmission lines and pulses. You canstudy these to help you understand some of the concepts developed for voltages, pulses,and reflections.

EXAMPLE: An ideal signal generator with impedanceRs = 70Ω is at x= 0 on a loss-less transmission line (R0 = 70Ω, vp = 2(10)8 m/sec). This signal generator produces thewaveform

vs(t) = 5; t 00; else

The line is terminated atx= 30 m with a 50Ω load. Plot the voltage on the line att = 200nanoseconds.

ANSWER:

This source signalvs(t) is a step, which is infinite in time length. Therefore, it is alsoinfinite in spatial length. The amplitude which first appearson the line is

A= R0

Rs+R0

K = 70

70+70

5 = 2:5 volts

The time length of the line is

tL = xL

vp= 30

2(10)8 = 150 nsec

At t = 200 nsec, the leading edge of the step has travelled

d = vp t = 2(10)8200(10)9 = 40 meters

The load reflection coefficient is

ΓL = RLR0

RL+R0= 5070

50+70= 20

120= 0:167

With this information you can obtain the following answer for the problem:

v(x) = 2:500; 0< x 202:082; 20< x< 30


WORKED LINE EXAMPLE - 2

An ideal signal source with impedanceRs = 70Ω is at x= 0 on a lossless transmissionline (R0 = 70Ω, vp = 2(10)8 m/sec). This ideal source (atx= 0) produces the waveform

vs(t) = 3; t 00; else

The line is terminated atx= 50 m with a 110Ω load. Plot the voltage on the line att = 400nanoseconds.

ANSWER:

This source signalvs(t) is a step, which is infinite in time length. Therefore, it is alsoinfinite in spatiallength. The amplitude which first appears on the line is

A= R0

Rs+R0

K = 70

70+70

3 = 1:5 volts

The time length of the line is tL = xL

vp= 250 nsec.

At t = 400 nsec, the leading edge of the step has travelled

d = vp t = 2(10)8400(10)9 = 80 meters

The load reflection coefficient is

ΓL = RLR0

RL+R0= 11070

110+70= 40

180= 0:222

With this information you can obtain the following answer for the problem:

v(x) = 1:500; 0< x 201:833; 20< x< 50


ECE 303 - Sum10 Notes Set 4:Lossless Lines - Sinusoids 1

INTRODUCTION

In an earlier set of Notes we studied the behavior of pulses transmitted on lossless lines.We learned that for short pulses we could actually “see” the propagation and reflectionprocesses using the bounce diagram. In the current set of Notes we will extend theseconcepts to the case of transmitting sinusoids onto lossless lines. The propagation andreflection processes are still present but we will see that they now take a different form.

Many information systems can be analyzed and designed by considering the input to be asingle frequency sinusoid. For example, an analog cable television system might transmita television signal by modulating a single high frequency sinusoid called a carrier. Thetime-domain variation of the video information signal is slower (i.e., at a lower frequency)than the rapid time-domain variations of the sinusoidal carrier. Therefore, the physicalcable tends to propagate the modulated signal very similarly to how it would propagate thecarrier by itself.

Even “digital” communication systems, like digital cable TV or digital cable modem, fre-quently use a carrier modulation method to transmit the “digital” data. For example, alogical 1 might be 100 cycles of a positive cosine at the carrier frequency and a logical 0might be 100 cycles of a negative cosine at the carrier frequency. The transmitted signallooks very much like a single frequency sinusoid except the these “phase transition” times.Consequently, design and analysis of the physical system can often be accomplished byconsidering the input as a single frequency sinusoid.

In this set of Notes we will see how the sinusoidal nature of the propagating signal requiresa different analysis than did pulse propagation. Many of thetools needed in this analysisare complex-valued and therefore we will be using quite a bitof complex functions in thisset of Notes.


SINUSOIDS ON LOSSLESS LINES

If the source connected to a transmission line is a sinusoidal source, the wave propagatedon the line has some specific properties:

WAVELENGTH:

The velocity of propagationvp (in m/sec), the Hertz frequencyf (in Hz), and the wave-lengthλ (in m) of a wave on a transmission line are related by

vp = f λ (1)EXAMPLE: A sinusoid with frequency off = 400 MHz is found to have a velocity com-puted to bevp = 2(10)8 m/sec on a transmission line. The wavelength on the line is thus

λ = vp

f= 2(10)8

4(10)8 = 0:5 meters

PHASE CONSTANT:

The phase constantβ (in rad/m), is given by the equivalent expressions

β = ωvp

= 2πλ

(2)The parameterβ is sometimes called thewavenumber, since it computes the number ofradians in one meter of line.

COMPLEX-EXPONENTIAL (PHASOR) NOTATION:

In this course, a tilde will sometimes represent a complex-exponential function (frequentlycalled a phasor). Here is one example:

v(α) = Aejω α (3)Here is another example. Replaceα in (3) with α = t x

vp:

g = g(x; t) = Aejω ( t x

vp) = Aej(ω tβx) (4)

Problems: 4-1, 4-2


SINUSOIDAL WAVES

Energy is not localized in position on a transmission line when the source signal is a sinu-soid. This requires a different mathematical approach for understanding the properties ofthe signals on the line. One difference is that the load is placed at positionx= 0, such thatthe line extends back into negativex values. This signifies we reference voltage, current,and impedance phenomena with respect to the load, like we sometimes reference oscillo-scope time measurements with reference to a triggering waveform. This change of loadposition will also simplify some of the resulting mathematical formulas.

Previously we found that some energy on a line will propagatetoward the load resistanceRL and will be dissipated as productive “work” across the load resistance. We also foundthat some energy will be reflected from the load ifRL is not equal to the line characteristicresistanceR0. This situation is represented for sinusoids in the figure below:

Ro

RL

x = 0x = − d ( )

f vb~v~

In the figure above we use the “wave packet” type of symbol to represent the sinusoidalenergy on the line. The wave packet symbol ˜vf pointing toward the load in the above figurerepresents the voltage in the Forward Wave (traveling toward the load). The wave packet ˜vb

symbol pointing toward the source represents the voltage inthe reflected Backward Wave(traveling back toward the source). The total voltage on theline, the total current on theline, and the line impedance at any line position can be computed using these concepts ofForward and Backward waves. We begin a mathematical analysis of these properties onthe following page.


FORWARD AND BACKWARD VOLTAGE WAVES

Let’s consider first the voltage wave on the line when the input is a sinusoid. Let ˜vf be the“forward” voltage wave (due to a source) which is propagating to the right on the line:

vf = Ae j ( ω tβ x) (5)Let vb be the “backward” voltage wave (due to a reflection) which is propagating back tothe left on the line:

vb = AΓL e j ( ω t+β x) (6)Note in (6) that the reflection at the load has changed the amplitude of the reflected waveto AΓL. The reflection has also changed the direction of propagation and this is signifiedby the changed sign ofβx in the argument of the backward wave exponential. The totalphasor voltage wave is therefore the sum of the backward and forward waves. Adding (5)and (6) then gives

v = vf + vb = Ae j ( ω tβ x)+AΓL e j ( ω t+β x) (7)Equation (7) can be simplified to provide

v = Ah

e j β x+ΓL e j β xi

e j ω t (8)Note that the form in (8) has been factored into a spatially varying functionV(x) and thetime-varying phasore j ω t to give the simple form

v = V(x) e j ω t (9)where the spatial functionV(x) in (9) is seen to be given by

V(x) = Ah

e j β x+ΓL e j β xi (10)

This form in (10) for the spatial voltage dependence will be very useful when we solve forthe current on the line. This is done on the next page.


FORWARD AND BACKWARD CURRENT WAVES

To compute the current wave we need to consider again the PDEsfor the lossless transmis-sion line. From Notes Set 1 we derived the following for the lossless line, modified herefor phasor ˜v andi:

∂v∂x

= L∂i∂t

; ∂i∂x

= C∂v∂t

(11a;11b)Equation (11a) requires a spatial partial derivative of ˜v. We can compute this partial deriva-tive using (9):

∂v∂x

= e j ω t ∂∂x

V(x) (12)Compute the derivative ofV(x) in (12) using the form forV(x) given in (10):

∂∂x

V(x) = ∂∂x

hAe j β x+AΓL e j β x

i (13)Operating with the partial derivatives on the right of (13) then gives

∂∂x

V(x) = A

∂∂x

e j β x+AΓL

∂∂x

e j β x (14)

Computing the required partial derivatives in (14) produces

∂∂x

V(x) = j βAe j β x+ j βAΓL e j β x (15)Substituting (15) into (12) and simplifying then gives the expression for the right side of(11a):

∂v∂x

= j βAe j ω th

e j β xΓL e j β xi (16)

We will assume that the phasor current, like the phasor voltage, can also be factored into aspatially varying functionI(x) and a time-varying phasor. This provides the simple form

i = I(x) e j ω t (17)where the spatial functionI(x) in (17) is still unknown. On the following page we willcomplete the solution fori by solving forI(x).


CURRENT WAVE SOLUTION

The right side of (11a) requires a partial derivative ofi with respect tot. Using (17) thisderivative is given by

∂i∂t

= j ω I(x)e j ω t (18)Substitute (16) into the left side of (11a) and (18) into the right side of (11a). The PDEfrom (11a) then becomes j βAe j ω t

he j β xΓL e j β x

i = j ω L I(x)e j ω t (19)Divide both sides of (19) by j ω Le j ω t . This gives a preliminary form forI(x):

I(x) = βAω L


i (20)The constant multiplying the spatial exponentials in (20) can be simplified recalling fromprevious work the definitions of phase constantβ, propagation velocityvp, and character-istic resistanceR0:

β = ωvp

; vp = 1pLC

; R0 =rLC

(21a;21b;21c)Using (21a) first gives

βω L

= ωvpω L

= 1vpL

(22)Now use (21b) in the far right term of (22):

βω L

= 1vpL

= pLCL

= pCpL

(23)Then use (21c) in the far right term of (23) to obtain the desired simplification:

βω L

= pCpL= 1q

LC

= 1R0

(24)


CURRENT WAVE SOLUTION (cont.)

Substituting (24) into (20) gives the following form for thespatial function for current:

I(x) = AR0


i (25)The desired solution for the current on the lossless line with reflections is now obtained bysubstituting (25) into (17):

i = AR0

he j βxΓL e j βx

ie j ω t (26)

Comparing (26) with (8), we see the forward and bacward current and voltage wave arerelated by

i = i f + ib = vf

R0 vb

R0(27)

Note that the backward current has a minus sign when written in terms of the voltage. Thisis because the backward current is flowing in a direction opposite to the forward current.Given the phasor form of a voltage or current, the real-valued (i. e., measured) quantity isobtained by taking the real part of (26) and (27):

v(x; t) = Refvg; i(x; t) = Refig (28; 29)To see the time-domain structure of the voltage, substitute(8) into (28), giving the follow-ing for the real-valued voltage:

v(x; t) = Ren

Ae j ( ω t β x )+AΓL e j ( ω t+ β x ) o (30)For our course, allA andΓL are real. Therefore, (30) becomes

v(x; t) = Ah

cos(ω t β x)+ΓL cos(ω t+β x) i (31)An expression for the real valued current may be obtained in asimilar manner.

Problems: 4-3, 4-4, 4-5


STANDING WAVE RATIO

Equation (31) for the real-valued voltage on the loaded lineis quite complicated. When youwork Problem 4-5 you will find that the corresponding currentis also quite complicated.However, in many problems we may proceed with analysis and design without having toknow the detailed structure of voltage and current waves. The high level information inthese expressions is often summarized by the Voltage Standing Wave Ratio (VSWR). TheVSWR is defined as the following ratio:

VSWR= 1+ jΓL j1jΓL j (32)

The VSWR is essentially the ratio of maximum voltage magnitude to minimum voltagemagnitude on the transmission line. Recall that the load reflection coefficient for voltage isgiven by

ΓL = RLR0

RL+R0(33)

LIMITING CASES:

Matched Load:For the Matched Load we have the conditionRL = R0 and consequentlyΓL = 0. Equation(32) then gives VSWR = 1. We will later see that this is the caseof total power flow intothe load.

“Open-Circuit” Load:For an “open-circuit” load, we haveRL ! ∞ and consequentlyΓL = 1. Equation (32) thengives VSWR! ∞. We will later see that this is the case of zero power flow into the load.This is equivalent to a standing wave on the line.

“Short-Circuit” Load:For a “short-circuit” load, we haveRL = 0 and consequentlyΓL =1. Equation (32) againgives VSWR! ∞.

Problem: 4-6

ECE 303 - Sum10 Notes Set 5:Line Impedance 1

INTRODUCTION

In Notes Set 3 we studied the problem of reflection and transmission of pulses on trans-mission lines. We found that when the reflection coefficientΓL was non-zero then voltagepulses could reflect back onto the physical transmission line. In a similar manner, in NotesSet 4 we extended the concept of transmission and reflection to include sinusoidal signalson the line. The reflection/transmission environment for sinusoidal voltages and currentson the line was seen to be very complicated.

This set of Notes provides an alternate approach to quantifying the system-level propertiesof the reflected and transmitted waves on the transmission line. Remember from basiccircuits that the ratio of voltage to current is, in the general case, called Impedance. NotesSet 4 has just introduced us to the concept of reflected/transmitted voltage and currentwaves. Thus, we naturally wonder if there is some ratio of “reflected/transmitted voltage”to “reflected/transmitted current” which has the units of impedance and which can be usedto simplify problems.

Such a ratio is called the Line Impedance, which is the topic of the present set of Notes.This quantity might also be called the “Line-Load Impedance” since it incorporates theeffects of the load at the end of the line, as well as the lengthof the transmission lineitself. The line impedance will be seen to be a much simpler computation than the voltagesand currents of Notes Set 4. However, accurately using the line impedance requires moreunderstanding of the physical properties of the transmission line having a load. One of thegoals of the present Notes Set is to aid in this understanding.


LINE IMPEDANCE DEFINITION

The definition of the line impedance uses the formula for the transmitted/reflected voltageand current waves from Notes Set 4. These formulas are repereated here for easy reference.The phasor voltage ˜v= v(x; t) was seen to be given by

v(x; t) = Ah

e j β x+ΓL e j β xi

e j ω t (1)Similarly, the phasor currenti = i(x; t) is given by

i(x; t) = AR0

he j βxΓL e j βx

ie j ω t (2)

We can now determine the line impedanceZ(x), which is defined as the ratio

Z(x) = v(x; t)i(x; t) (3)

You can think ofZ(x) as the “looking in” impedance (toward the load) at the pointx.Substituting (1) and (2) into (3) this ratio becomes:

Z(x) = R0

"e j βx+ΓL e j βx

e j βxΓL e j βx

# (4)Now we need a useful formula for computing the looking-in line impedance. One suchform is obtained by substituting forΓL in the numerator and denominator in (4):

Z(x) = R0

2664 e j βx+ RLR0

RL+R0e j βx

e j βx RLR0

RL+R0e j βx

3775 (5)


LINE IMPEDANCE COMPUTATIONS

Now multiply numerator and denominator of the right side of (5) by RL+R0 to obtain

Z(x) = R0

" (RL+R0)e j βx+ (RLR0) e j βx(RL+R0)e j βx (RLR0) e j βx

# (6a)Grouping theR0 and theRL terms in the right side of (6a) then gives

Z(x) = R0

"RL (e j βx+e j βx )R0(e j βxe j βx )R0(e j βx+e j βx)RL (e j βxe j βx ) # (6b)

Euler’s formulas for cosine and sine may be manipulated to give the following:

2 cosβx= e j βx+e j βx; 2 j sinβx= e j βxe j βx (6c)Substituting (6c) into (6b) then gives

Z(x) = R0

2RL cosβx j 2R0 sinβx2R0 cosβx j 2RL sinβx

(7)Divide numerator and denominator of (7) by cosβx, remember that tanβx= sinβx

cosβx, and

simplify the result to obtain the useful form

Z(x) = R0

RL j R0 tanβxR0 j RL tanβx

(8)Although equation (8) is mathematically valid, a change in the way of positioning the trans-mission line with respect to thex-axis will simplify future computations. When consideringa line having a load, it will be convenient to place the load atx= 0 and let the transmissionline extend in thex region. This change in positioning is explored on the next page.


LINES WITH LOADS

When considering a line having a load, it will be convenient to place the load atx= 0 andlet the transmission line extend in thex region. The figure below shows this geometryand shows a forward and reflected wave.

Ro

RL

x = 0x = − d ( )

f vb~v~

In the above figure the input terminals to the transmission line at located atx=d. Definethe looking-in impedanceZin asZin = Z(x=d). Therefore, substitutex=d in equation(7) for Z(x) on the previous page, and recognize that tan(βd) = tan(βd). Performingthese operations then provides

Zin = R0

RL+ j R0 tanβdR0+ j RL tanβd

(9)Equation (9) shows that the looking-in impedance to the transmission line can have a dif-ferent impedance depending on the value ofx. One implication is that even though theload may be fixed, a source will see a varying impedance which depends on the line length.Another implication is thatZ(x) may be complex-valued, even if the load resistanceRL andcharacteristic resistance of the lineR0 are real-valued. We will frequently use equation (9)to find the impedance of a lengthd line connected to a load.


LINE IMPEDANCE EXAMPLES

EXAMPLE 1: A lossless transmission line operating atf = 10 MHz hasvp = 2:36(10)8m/s andR0 = 141Ω. The line is terminated atx= 0 m withRL = 100Ω. ComputeZin, the“looking in” impedance, on the line at the pointx=50 m from the load.

APPROACH:Use equation (9), repeated here, to compute this impedance:

Zin = Z(x=d) = R0


whered = 50 for this problem. Compute the phase constantβ:

β = ωvp

= 2π107

2:36(10)8 = 0:266 rad/m

Thus,βd = 0.266 (50) = 13.3 rad 762o. This gives tanβd = 0:90 and the lookingin impedance becomes

Zin = Z(50) = 141

100+ j 141(:90)141+ j 100(:90) = 141

161:6e j 51:8o

167:6e j 32:6o

In polar and rectangular we thus have

Zin = 136:2e j 19:2o = 128:6+ j 44:8One important point to note is that even though the loadRL is a real-valued load, the line-plus-load combination seen by the source is a complex value.


LINE IMPEDANCE EXAMPLES (cont.)

EXAMPLE 2: A transmission line (x< 0) operating atf = 50 MHz hasvp= 2(10)8 m/secandR0 = 100Ω. The line is terminated atx= 0 m with a 50Ω load. What non-zero lengthmust the line have such that the looking in impedance isZin = RL = 50 Ω? (NOTE: Thetrivial line lengthd = 0 is not allowed.)

APPROACH:

Zin is given by equation (9): Zin = R0


.

We wantZin = RL = 50. This occurs if tanβd = 0:

Zin = R0

RL

R0

= RL

But tanβd = 0 for βd = nπ, wheren is a positive integer. Thus,

βd= nπ ) d = nπβ

= nλ2π

= nλ2

Thus we need to compute the wavelength for the frequencyf = 50 MHz:

λ = vp

f= 2(10)8

5(10)7 = 4 ) d = 2n

From this last result we see an example of the periodicity of the line impedance. Specifi-cally, the line impedance equalsRL every 2 meters.



LIMITING CASES

ZERO-LENGTH LINE: d = 0

The general formula forZin should reduce to just the load impedance when the linelengthd = 0. Does it? Whend = 0, we haveβd = 0, and thus

tanβd = tan(0) = 0 (10)Using (10) in equation (9) then gives

Zin = R0

RL+ jR0 tanβdR0+ jRL tanβd

= R0

RL+ jR0 (0)R0+ jRL (0) = R0

RL

R0

= RL (11)Yes, the general line impedance formula gives the intuitively correct answer for the“very short” line.

LOW FREQUENCIES:

For “low” frequencies, wave theory should reduce to circuittheory; that is, two loss-less conductors across an impedanceRL should only have an impedanceRL. To checkthis case, note that

β = ωvp

= 2π ff λ

= 2πλ

Thus we haveβ d = 2πdλ

. Therefore, whend << λ, we haveβd 0. This again

givestanβd tan(0) = 0

from which we again obtain

Zin = R0


= R0

RL+ j R0 (0)R0+ j RL (0) = R0

RL

R0

= RL

Yes, the general line impedance formula gives the intuitively correct circuit theoryanswer for the “low” frequency line.


SHORT-CIRCUITED LINES

Let a lengthd line be terminated in ashort circuit. This is sometimes called a stub. Thisline hasRL = 0, as shown in the figure below:

( )x = − d x = 0oR

inZ

Using (9), the input impedanceZin is given by

Zin = Z(d) = R0

RL+ jR0 tanβdR0+ jRL tanβd

= R0

0+ jR0 tanβd

R0+0

= jR0 tanβd (12)EXAMPLE:

Let λ= 1 m, f = 200 MHz, andR0 = 50Ω. ComputeZin for the lengthd transmissionline terminated in a short-circuit.

APPROACH:We haveβ = 2π=λ = 2π and thusZin = j50tan2πd. If d = 0.1 m, then

Zin = j 50 tan(2π0:1) = j 50(:726) = j 36:3 ( INDUCTIVE

If d = 0.35 m, then

Zin = j 50 tan(2π0:35) = j 50(1:376) = j 68:8 ( CAPACITIVE

Problem: 5-9


EXAMPLE FOR INDUCTOR

EXAMPLE:

It is desired to create anL = 106 H inductor using a short-circuit lossless transmissionline which has a length less than 1 m. The line characteristicresistanceR0 = 100Ω, theoperating frequency isf = 100 MHz, and the velocity of propagation isvp = 2(10)8 m/sec.Compute the length of line which meets the specification.

APPROACH: We wantZin = jωL

ComputeZin:

Zin = j R0 tanβd ( β = ωvp

= 2π108

2(10)8 = π

This givesZin = j 100 tanπd

ComputejωL:

jωL = j 2π108(10)6 = j200πEquateZin and jωL:

j100tanπd = j200πtanπd = 2ππd = tan1(2π)d =

1π

tan1(2π) =1π(1:413) = 0.45 m.

Remember to find tan1(2π) in radians.

Problem: 5-10


OPEN-CIRCUITED LINES

Let a lengthd line be terminated in aopen circuit. This line hasRL ! ∞.

( )x = − d x = 0oR

inZ

To computeZin, divide numerator and denominator ofZin by RL:

Zin = R0

264 RL

RL+ j

R0

RLtanβd

R0

RL+ j

RL

RLtanβd

375 = R0

264 1+ jR0

RLtanβd

R0

RL+ j tanβd

375 (13)Take the limit asRL ! ∞:

Zin = R0

1

j tanβd

= jR0

tanβd(14)

EXAMPLE:Let λ = 1 m, f = 200 MHz, andR0 = 50Ω. ComputeZin for the lengthd transmission lineterminated in an open-circuit.

APPROACH:

For these values we haveβ = 2π=λ = 2π and thusZin = j50

tan2πd. If d = 0.1 m, then

Zin = j500:726

= - j 68.9 ( CAPACITIVE

Problem: 5-11


EXAMPLE FOR CAPACITOR

EXAMPLE: It is desired to create aC= 1010 F capacitor using an open-circuit transmis-sion line which has a length less than 1 m. The line characteristic resistanceR0 = 50Ω, theoperating frequency isf = 100 MHz, and the velocity of propagation isvp = 2(10)8 m/sec.Compute the length of line which meets the specification.

ANSWER:

We wantZin = 1jωC

= j1

ωC

ComputeZin:

Zin = jR0

tanβd( β = ω

vp= 2π108

2(10)8 = π

This givesZin = j50

tanπd

Compute j1

ωC: j

1ωC

= j1

2π108(10)10 = j1002π

EquateZin and j1

ωC: j50

tanπd= j100

2π) tanπd

50=

2π100

) tanπd = π

Therefore,

πd = tan1(π)d =

1π

tan1(π) =1π(1:263) = 0.402 m.

Once again, remember to find tan1(π) in radians.


IMPEDANCE FOR PARALLEL COMBINATIONS

Lines can be connected in parallel. For example, letZ1 be the input impedance of line 1and letZ2 be the input impedance of line 2. The physical picture and equivalent circuit isshown below:

line

1

line 2

Physical Lines Equivalent Circuit

RA

BRZin ZinZ1

Z2

The input impedancesZ1 and Z2 are functions ofZA, ZB, and other line parameters asshown on the previous page. Using circuit theory, the input impedance for the parallelcombination of lines is given by

Zin = Z1 jj Z2 = Z1Z2

Z1+Z2(15)

EXAMPLE:An open-circuit line with input impedanceZ1 = j68:9 Ω and a short-circuit line withinput impedanceZ2 = j36:3 Ω are connected in parallel. ComputeZin.

APPROACH: Zin = Z1Z2

Z1+Z2= ( j68:9)( j36:3) j68:9+ j36:3 = j76:7


ECE 303 - Sum10 Notes Set 6:Lossy Lines 1

INTRODUCTION

In previous work we have studied lossless transmission lines. We saw that this was anaccurate model for some transmission lines which were smallin length. In this set of Noteswe extend our study of transmission lines to include lossy transmission lines. We mightthink of these as lines which are long enough in length such that the total line resistanceand conductance values become appreciable.

Examples of these lossy lines are common, especially in manycommunication systemswhich use twisted-pair or coaxial cable for transmission over substantial distances. Twoexamples of these lossy line systems are as follows:

Digital Subscriber Line (DSL):

Some DSL systems use the entire length of the twisted-pair copper telephone wireswhich connect the subscriber with the remote Central Office.This line is the sameline which carries your voice conversations, and the line may be quite long (on the or-der of miles for many lines). Due to the lossy nature of the twisted-pair line, DSL cannot be implemented on all of these regular telephone lines. The lossy nature createsa maximum length of line over which high-speed DSL services can be sucessfullyimplemented.

Cable TV/Cable Modem:

Many modern cable TV/cable modem systems have fiber-optic networks which carryinformation from the cable service provider to a central neighborhood location. How-ever, many of the connections from the central neighborhoodlocation to the sub-scriber premises is still a coaxial cable connection. The length of this last cableconnection is frequently long enough that the line must be analyzed as a lossy line.

We will see that analyzing a lossy line is more challenging than understanding the losslessline. Mathematically, this is because the solution for the lossy line PDE is not as simple asthe Wave Equation for the lossless line. We will denote the origin of these challenges whenwe encounter them in our development.


LOSSY LINE PDEs

A transmission line is called a lossy line whenR 6= 0 andG 6= 0 in the transmission linecircuit model. To understand the lossy line we need to returnto the general transmissionline equations, which were given by equations (5a) and (5b) in Notes Set 1. They arerepeated here for easy reference:

∂v∂x

=RiL∂i∂t

(1a)∂i∂x

=GvC∂v∂t

(1b)Our studies are simplified if we assume the input current and voltage for the line are pha-sors. This is an accurate mathematical condition for many communications systems whichuse what is called “carrier modulation” for sending information. Many modern analog anddigital communication systems use various types of carriermodulation methods.

The boundary conditions for the PDEs in (1) will be the conditions imposed at the transmit-ter end (x= 0+) of the transmission line. Using the tilde notation for a complex “rotating”phasor, suppose the following are the boundary conditions:

v(x= 0+; t) = v0(t) = Ae j ω t (2a)i(x= 0+; t) = v0(t)

Z0= A

Z0e j ω t (2b)

whereZ0 is an unknown, possibly complex, scaling factor relating phasor current and volt-age. We have previously seen a scaling factor relating current and voltage on the losslessline. It was the characteristic resistanceR0, which related the forward lossless current wave,i f = i f (x; t), to the forward lossless voltage wave ˜vf = vf (x; t). That relation was

i f = vf

R0(3)

Notice the similarity between equation (3) and Ohm’s Law from circuits. Therefore, wecan think of (2b) as an assumption that “Ohm’s Law holds for Transmission Lines”. Thisis a reasonable assumption. If this assumption is incorrect, then our solution method willexpose an inconsistency.


PROPAGATION ON LOSSY LINES

The boundary conditions atx= 0, given by equation (2), represent the transmitter’s inputsignal on the line. The solution we obtain for the lossy line PDE will be the voltage andcurrent along the entire line for any value of timet. It is straightforward to show that thereal part of the boundary condition in (2a) is equivalent to the real-valued condition

v(x= 0+; t) = v0(t) = A cosω t (4)Think of equation (4) as giving us the known, real-valued transmitter voltage. The solutionfor all x will then be the propagation of the transmitter voltage and current “down the line”.

Whereas the real-valued boundary condition of (4) is understandable from a physical stand-point, it is actually easier to mathematically solve the lossy PDE using the complex-valuedphasor representation given in equation (2). Subject to theboundary conditions of (2),we will next show that the voltage and current phasors given in equations (5) below solvethe lossy transmission line PDEs given in equation (1). We will assume the form of thesolutions can be written as

v = Aeγx e j ωt (5a)i = v

Z0= A

Z0eγxe j ωt (5b)

whereγ andZ0 are (possibly complex) constants which must be determined.For simplicityof notation we have used the simplified forms ˜v = v(x; t) andi = i(x; t).We will next verify that the forms given in (5) are solutions of the lossy PDEs given by(1). We will do this on the following pages by substituting (5a) and (5b) into (1) and thentesting the results for consistency. If the forms in (5) are not valid solutions for (1), then aninconsistent or inadmissable condition will result. If no inconsistent conditions arise, thenthe forms in (5) are solutions. Additionally, this approachwill display the required formsfor the unknown constantsγ andZ0. We will see that they are easily calculable from theline parametersR, L, G, andC.


SOLUTION PROCESS

Examining equations (1a) and (1b), we see we need to compute four partial derivatives.Using the solution forms from (5a) and (5b) these partial derivatives are given below:

∂v∂x

= γ A eγxejω t = γ v (6a)

∂i∂x

= γAZ0

eγxejω t = γZ0

v (6b)

∂v∂t

= j ω A eγxejω t = j ω v (6c)

∂i∂t

= j ωAZ0

eγxejω t = j ωZ0

v (6d)

Now substitute the partial derivative results from (6a) - (6d) into the equations (1a) and (1b)and simplifiy: γ v = R

Z0v j ωL

Z0v = R+ jωL

Z0v (7a) γ

Z0v = G v j ωC v = (G+ jωC) v (7b)

The equations in (7a) and (7b) are consistent if the coefficients multiplyingv on each sideare equivalent. Therefore, equating the coefficients on both sides gives

γ = R+ j ωLZ0

(8a)γ

Z0= G+ j ωC (8b)

One the next page we solve (8) for the constantsγ and Z0. We will see that they aregeneralizations of parameters we have already seen in the context of the lossless line.


LINE PARAMETERS

Solving equations (8a) and (8b) forγ andZ0 gives the expressions in equations (9) and(10) below. These two quantities are very important parameters of the lossy line. They arecalled thePropagation Constantand theCharacteristic Impedance:

Propagation Constant:

γ = p (R+ j ωL) (G+ j ωC) (9)Characteristic Impedance:

Z0 =s R+ j ωLG+ j ωC

(10)The propagation constant basically determines how voltage(or current) decreases alongthe length of the line. The characteristic impedance determines the magnitude and phase ofthe current with respect to the voltage. We will illustrate these properties on the followingpages. Note also the results for the lossless line are special cases of (9) and (10) obtainedby settingR= 0 andG= 0.

Equations (9) and (10) show that computation of the propagation constant and characteristicimpedance requires multiplications, divisions, and square roots for complex numbers. Thiswill require you to use some of your complex function skills you developed previously. Tohelp you with some of these methods a computational example appears on the next page.


COMPUTATION EXAMPLES

EXAMPLE: A lossy line operating atf = 10 KHz has the following per meter parameters:R= 0:1 Ω; G= 10µS; L = 0:2 µ H, C = 100 pF . Compute the propagation constantγand characteristic impedanceZ0.

APPROACH:

Refer to equations (9) and (10) for the this problem. Using the relationω = 2π f , we have:

R+ j ωL = 0:1+ j 2π1040:2(10)6 = 0:1+ j 0:0125

= 0:101e j 7:1o

G+ j ωC = 105+ j 2π1041010 = 105h

1+ j0:628i

= 1:18(10)5 e j 32:1o

To find γ :

γ = q (0:101e j 7:1o) (1:18(10)5e j 32:1o) = q1:189(10)6e j 39:12o= 0.00119e j 19:6o

= 0.0011 + j 0.0004polar rectangular

To find Z0 :

Z0 = s0:101e j 7:1o

1:18(10)5e j 32:1o = p8559e j 25o = 92.5e j 12:5o

It was previously stated that the propagation constant determines how voltage decreasesalong the length of the line. We will illustrate this property on the following page.

Problems: 6-1, 6-2, 6-3


PROPAGATION CONSTANT

Now that we know how to compute the propagation constant, let’s see how it affects theproperties of a lossy line. Recall from equation (9) the formula for the propagation constantγ, repeated here for reference:

γ = p (R+ j ωL) (G+ j ωC) (11)In general,γ is a complex-valued parameter, with a real partα and imaginary partβ:

γ = α+ j β (12)In equation (12),α is called theattenutatonconstant andβ is called thephaseconstant.Onceγ is computed from (11),α andβ may be computed by

α = Refγg; β = Imfγg; (13)To see the effect of these constants, substitute equation (12) for γ into the transmission linesolution forv from equation (5a):

v = Aeγxe j ω t = Ae(α+ j β)x e j ω t = Aeαx e j (ω tβx) (14)Sincev(x; t), the real voltage at timet and positionx, is given byv(x; t) = Refvg, take thereal part of both sides of (14), giving

v(x; t) = Aeαx cos(ω tβx) (15)Sinceα > 0 for a lossy line, equation (15) shows that the amplitude of the wave decreasesas a function of distance along the line. As a practical matter, make sure you use theappropriate units for distancex. For example, ifα andβ are given in per meter units, thenx also has to be in meters. Ifα andβ are given in per mile units, thenx also has to be inmiles, etc.



PROPAGATION CONSTANT EXAMPLES

EXAMPLE 1: A lossy line with no reflections hasγ = 3+ j30 (in per kilometer units). Avoltagev0(t) = 10cos(ω t) enters the transmission line atx= 0+ and flows to the right. Atpositionx= x1 suppose we measurev(x1; t)=A1 cos(ω t15). Assumingx1 > 0, computethe value ofA1.

APPROACH:Sinceγ = 3+ j30 we haveα = 3 andβ = 30. The general real-valued solution is

v(x; t) = Aeαxcos(ω tβx)At x= x1, we see that the phase is given by

βx1 = 15 ) x1 = 15β= 15

30= 0:5

At x= x1 = 0:5, we see that the magnitude is given by

A1 = Aeαx1 = 10e3(0:5) = 2.23

EXAMPLE 2: At positionx= x2 on this line e we measurev(x2; t) = 5 cos(ω t30x2).Assumingx2 > 0, compute the value ofx2.

APPROACH:Sinceγ = 3+ j30 we haveα = 3 andβ = 30. Also, since

v(x; t) = Aeαxej(ω tβx) = 10e3xej(ω t30x)we have jv(x; t)j = 10e3x

At x= x2, this last result gives 5= 10e3x2, from which e3x2 = 0:5. Taking ln of both sides

of this last result givesx2 =13

ln(0:5) = 0.23 m


CHARACTERISTIC IMPEDANCE

We can compute the current waveform on the lossy line by usingthe voltage solution andthe characteristic impedance. Let’s see how the characteristic impedance affects the currentproperties on the lossy line. Recall from equation (10), thecharacteristic impedanceZ0,repeated here:

Z0 =s R+ jωLG+ jωC

(16)In general,Z0 is a complex-valued parameter, having a magnitudejZ0j and a phaseφ:

Z0 = jZ0j ejφ (17)To see the effect of the characteristic impedance magnitudeand phase on the line current,substitute equation (17) forZ0 into the transmission line solution fori from equation (5b):

i = AZ0

eγxe j ωt = AjZ0j e j φ e(α+ jβ)x e j ωt = AjZ0jeαx e j (ωtβxφ) (18)Sincei(x; t), the real voltage at timet and positionx, is given byi(x; t) = Refig, take thereal part of both sides of equation (18), giving

i(x; t) = AjZ0j eαx cos(ω tβxφ) (19)The result in (19) shows thatjZ0j scales the current magnitude, andφ causes the currentand voltage to be out of phase on the line. This is different from the lossless infinite lengthline. In that case, voltage and current were in phase.


CHARACTERISTIC IMPEDANCE EXAMPLE

At ω = 106 rad/sec a lossy transmission line with no reflections has permeter parametersR = 0.6 Ω, G = 0, C = 1010 F, andL = UNKNOWN. The characteristic impedance isZ0 = 80:0 j37:3. Compute the per meter inductanceL.

APPROACH:

From the definition ofZ0 and the problem statement we have

Z0 =sR+ jωLjωC

= 80:0 j37:3= 88:27e j 25o

Squaring both sides gives

R+ jωLjωC

= 7791:6e j 50o

Multiply both sides byjωC and substitute forω, R, andC:

0:6+ j106L = j1061010h7791:6e j 50oi= 0:78ej 40o

Compute the Cartesian form for the term on the right:

0:60+ j106L = 0:78cos(40o)+ j0:78sin(40o) = 0:60+ j0:50

The real parts cancel, giving

j106L = j0:50 ) L = 0:50106 = 5(10)7


ECE 303 - Sum10 Notes Set 7:Power Flow on Transmission Lines 1

INTRODUCTION

One of the main purposes of transmission lines is to transport an information signal fromone place to another. For analog signals or digital information impressed on a carrier sinu-soid, the power transported is an important measure of performance.

Computing the power transported by a transmission line willextend some of the propertiesof power you learned from basic circuits. We will start this set with a review of the familiarrelations from circuits and then see how the power relationschange for transmission lines.We will study three important transmission line cases in this Notes Set: Lossless Lines with no reflections Lossy Lines with no reflections Lossless Lines with reflections

Lossless lines with no reflections will produce results quite similar to the circuit expressionsyou learned earlier. However, we will see that the conditions of lossy lines and reflectionson lossless lines require changes in the computation of power as compared to the circuitexpressions. We will see that lossy lines produce a loss in propagated power as the poweris dissipated in the line. In lossless lines, reflections from the load cause energy to travelbackward on the line toward the source. This represents energy not efficiently transportedfrom source to load.


INSTANTANEOUS POWER

Let’s first review what we learned about power from basic circuits. Consider the case ofcurrenti(t) flowing through a resistor ofR ohms, causing a voltagev(t) across the resistor.From basic circuits the Instantaneous Power dissipated across the resistor is

p(t) = v(t) i(t) (1)Alternate forms ofp(t) are given by

p(t) = v2(t)R

; p(t) = i2(t) R (2a; 2b)If we think of R as the load of a device, we frequently say thatp(t) is the instantaneouspowerdelivered to the load.

EXAMPLE:

As an example, let the voltage across a resistorR bev(t) = 2cos(2π500t) and let the currentbe i(t) = 0:01cos(2π500t). Equation (1) provides a straightforward method to computetheinstantaneous power:

p(t) = v(t) i(t) = 2 cos(2π500t) 0:01 cos(2π500t)= 0:02 cos2(2π500t) = 0.02

h 12+ 1

2cos(2π1000t) i

= 0.01 + 0.01 cos(2π1000t) watts

Note thatp(t) is still a function of timet. In fact, notice thatp(t) has a frequency which istwice the frequency of the original voltage sinusoid.


TIME-AVERAGE POWER

Another useful measure of power is the Time-Average Power for a sinusoid. This is theaverage power delivered to the load over one period of the sinusoid. In this course we willuse the symbolPavg for this quantity. The definition of the time-average power is given by

Pavg = 1Tp

Z Tp

t=0p(t) dt (3)

This average power is similar to the “rms” measures you calculated in your circuits courses.

EXAMPLE:

As an example, let’s compute the time-average power for the example on the previous page.For reference, the voltagev(t) across the resistorR and the currenti(t) through the resistorare given below:

v(t) = 2 cos(2π500t) ; i(t) = 0:01 cos(2π500t)The frequencyf of the voltage sinusoid is seen to be 500 Hz and the period is therefore givenby Tp = 1= f = 0.002 sec. Equation (3) provides a staightforward method to compute thetime-average power:

Pavg = 10:002

Z :002

t=0

h0:01 + 0:01 cos(2π1000t) i dt

= 500Z :002

t=00:01dt + 500

Z :002

t=00:01 cos(2π1000t) dt

= 0.01 + 0 = 0.01 watt

Note that there is no spatial argument (likex or y) in any of the preceding computations.This signIfies we are computing power for a circuit which is very small compared to thewavelength of the carrier sinusoid. However, transmissionlines are frequently larger thanthe wavelength, and thus the voltage and current in a transmission line DO have spatialdependence. How is power now computed for a spatially-distributed transmission line?This concept is studied next.

Problem: 7-1


COMPLEX-VALUED VOLTAGE AND CURRENT

Power computations for the spatially-distributed transmission line can be simplified ifwe use the complex phasor notation. Purely time-domain computations are sometimescomplicated because trigonometric identities and evaluations of triginometric integrals arefrequently necessary. There is an alternative computationfor time-average power usingcomplex-valued phasors which is often much simpler.

To develop these phasor relations, letv(t) = A cos(ω t) be the voltage across a resistorR.Ohm’s law gives the current through the resistor as

i(t) = v(t)R

= AR

cos(ω t) (4)The complex phasor voltage and current corresponding to thequantities in (4) are therefore

v(t) = Ae j ω t ; i(t) = v(t)R

(5a; 5b)The real quantitiesv(t) andi(t) are related to the phasor ˜v(t) andi(t) by taking the real partof the corresponding complex quantity:

v(t) = Ref v(t)g= RefAe j ω t g= RefAcosω t+ j A sinω t g= A cosω t (6a)i(t) = Ref i(t)g= Ref v(t)

Rg= Ref A

Rcosω t+ j

AR

sinω t g= AR

cosω t (6b)With the previous definitions of complex-valued voltage andcurrent, the time-averagepowerPavg may be computed by the formula

Pavg = 12

Refv(t) i (t)g (7)wherei (t) is the complex conjugate of the complex current. Using equation (7) to com-putePavg is frequently easier than using equation (3).


COMPLEX-VALUED POWER COMPUTATIONS

We will do an example which shows that the complex-valued computation gives the sameanswer forPavg as does the time-domain integral in equation (3).

EXAMPLE: Let the voltage across a resistor bev(t) = 2 cos(2π500t) and let the current throughthe resistor bei(t) = 0:01 cos(2π500t). Then the complex voltage and current are:

v(t) = 2e j2π500t ; i(t) = 0:01e j2π500t

Using v(t) andi(t) in equation (7) then gives

Pavg = 12

Ref2e j2π500t 0:01e j2π500t g= 12

Ref0:02g = 0:01 watt

Note that 0.01 watts is the same answer previously obtained using the time-domain methodfor time-average power.

Computation ofPavg using the preceding complex-valued approach is frequentlysimplerthan the corresponding integral of trigonometric functions. In this course we will considerPavg for the following three cases: Lossless line with no reflections. Lossy line with no reflections. Lossless line with reflections.

For some transmission lines the time-average power will notbe a function of position.However, for other transmission lines the average power will be a function of position onthe transmission line. One implication is that all the powerintroduced to the transmissionline may not be successfully transported to the load. We willnext consider each of the threecases above individually.

Problem: 7-2


LOSSLESS LINE WITH NO REFLECTIONS

If a transmission line is lossless, thenR = 0, G = 0, and thus

β = ωvp

= ωp

LC; Z0 = R0 =rLC

(8a;8b)With no reflections, there is only a forward voltage wave and aforward current wave:

v(x; t) = v f (x; t) = Ae j(ωtβx); i(x; t) = i f (x; t) = AR0

e j(ωtβx) (9a;9b)Thus using equations (9a) and (9b) in equation (7) shows thatPavg(x), the time-averagepower at pointx on the line, is given by

Pavg(x) = 12

Re

v(x; t) i(x; t)= 12

Re

Ae j(ωtβx) A

R0e j(ωtβx)= A2

2R0(10)

Note in this case the power flow is independent of positionx.

EXAMPLE: Let the voltage wave bev(x; t) = 5 cos(ω tβx) on a lossless transmission line withR0 = 50Ω. Assuming no reflections, computePavg(x).APPROACH: The phasors are ˜v(x; t) = 5e j (ωtβx) and i(x; t) = 0:1e j (ωtβx). Thus, using (10)gives

Pavg(x) = 12

Ref5e j(ωtβx) 0:1e j (ωtβx)g = 0:25

Note from the givenv(x; t) andR0 thatA2

2R0= 52

2(50) = 0:25

Problems: 7-3, 7-4


LOSSY LINE WITH NO REFLECTIONS

Assuming no reflections on the lossy line, the voltage and current phasors are given by theforward waves:

v(x; t) = A eγ x e j ω t ; i(x; t) = AZ0

e γ x e j ω t (11a;11b)whereγ=α+ j β is the propagation constant andZ0 = jZ0je j φ is the characteristic impedanceof the lossy line.

Again using equation (7) showsPavg(x), the time-average power at pointx on the line, isgiven by

Pavg(x) = 12

Re

v(x; t) i(x; t) (12)Substituting the phasors from (11) into (12) then gives

v(x; t)i(x; t) = A e αx e j ( ωtβ x) AjZ0j e j φ e ( α+ j β )x e j ωt (13)

Completing the computations in (13) shows thatPavg(x) becomes

Pavg(x) = 12

Re

A2jZ0j e 2 α xe j φ

(14)Since Euler’s formula givese jφ = cosφ+ j sinφ then equation (14) becomes

Pavg(x) = A2

2jZ0j e 2 α x cosφ (15)Note in (15) the power is a function of positionx. Since the functione 2 α x decreases asa function ofx, we see that the power transmitted by the lossy line descreases as the linegets longer.


LOSSY LINE EXAMPLE

A lossy transmission line with no reflections has per kilometer parametersR = 300 Ω,G = 0, C = 5(10)8 F, andL = 6(10)4 H. At y = 0+, the input waveform to the line isv(t) = 5cos(π106 t). Compute the time average power flowing on the line aty = 500 m.

APPROACH:

Compute some of the necessary terms:

R+ j ωL = 300+ j π1066(10)4 = 300+ j1885 = 1908e j81o

G+ j ωC = 0+ j π1065(10)8 = j0:157 = 0:157e j90o

To find α = Refγg first computeγ :

γ = q (1908e j81o) 0:157e j90o) = p300e j171o = 17:3 e j85:5o

Now take the real part ofγ to find α :

α = Refγg = 17:3cos(85:5o) = 1.38

Now find Z0 in polar form :

Z0 = r1908e j81o

0:157e j90o = p12;153e j9o = 110.2e j4:5o

Thus,Pavg becomes

Pavg = A2

2 jZ0j e2αx cos(φ) = 252(110:2) e2(1:36)(0:5) cos(4:5o) = 0.029 watts

Notex = 0:5 in the last computation. Since theR;L;G; andC parameters are given in perkilometer units, we must therefore convertx into kilometers:x = 500 m = 0.5 km.



LOSSLESS LINE WITH REFLECTIONS

In this case both ˜v = v(x; t) andi = i(x; t) have forward and backward waves:

v = Ah

e j βx+ΓL e j βxi

e j ω t ; i = AR0

he j βx ΓL e j βx

ie j ω t (16a;16b)

The time-average power at pointx on the line is again given by equation (7)

Pavg(x) = 12

Re

v i (17)Compute the phasor product required in equation (17):

v i = A2

R0

he j βx+ΓL e j βx

ihe j βxΓL e j βx

i=

A2

R0

h1ΓL e j 2βx +ΓL e j 2βx Γ2

i=

A2

R0

h1Γ2

L +ΓL

e j 2βx e j 2βx

iSimplifying this last result using Euler’s formula gives

v i = A2

R0

1Γ2

L + j 2ΓL sin2βx (18)

Using the result from (18) in equation (17) for the time average power formula then gives

Pavg(x) = A2

2R0

1Γ2

L

(19)In equation (19) the parameterA is the amplitude of the forward voltage wave which ispropagated on the line. You can see the reflection (represented by the parameterΓL ) hasreduced the power which successfully enters the load.


EXAMPLE WITH REFLECTIONS I

A forward voltage wave ˜v f (t) = 5 e j 2π(10)8 t propagates fromx = 0 on a lossless trans-mission line. The line has characteristic impedanceR0 = 50 Ω, vp = 2(10)8 m/sec, and isterminated atx = 0:25 m with a 100Ω load. Compute the time-average power delivered tothe load.

APPROACH:

Compute the reflection coefficientΓL:

ΓL = RLR0

RL+R0= 10050

100+50= 0:333

Thus, the time average power is

Pavg(y) = A2

2R0

1Γ2

L

= 52

2(50) 10:3332Completing the computations gives

Pavg(x) = 0.222 w


EXAMPLE WITH REFLECTIONS II

Suppose the voltage phasor and current phasor on a transmission line are given by

v = 2e j (ωtβx)+ e j (ωt+βx ); i = 0:06e j (ωtβx )0:03e j (ωt+βx )In these equationsω = 108 and β = 0:5. Compute the real-valued time-average powerflowing on this line.

APPROACH:

Note that the voltage and current expressions above includea forward wave and a backward(reflected) wave. To find the power we can use

Pavg(x) = 12

Re

v iCompute the product ˜v i

v i = h2e j (ωtβx )+ e j (ωt+βx ) ih

0:06e j (ωtβx)0:03e j (ωt+βx ) i= 0:12 :06e j 2βx+ :06e j 2βx :03= 0:09+ j 0:12sin(2βx)Taking the real part then provides

Pavg(x) = 12

Ref0:09+ j0:12 sin(2βx)g = 0.045 w

Problems: 7-11, 7-12, 7-13


LINE AS CIRCUIT ELEMENT

If we know that a line plus load has an input impedanceZin, then we may use circuittheory to determine the time-average power flowing on the line. As an example, let a signalsourcevs(t) have source resistanceRs. Note this implies a phasor source ˜vs(t). The sourceis connected to a lossless line/load havingZin, as shown below:

.

.

x = 0

sv (t)

SR

~ inZL

R

Since we know the input impedanceZin we do not need specific properties of the line likelength or characteristic resistance. To determine the power flowing on the line we can usecircuit theory to determine the power across the “lumped” impedanceZin. To compute thisaverage power, apply equation (7) to obtain

Pavg = 12

RefVin I ing (20)In equation (20),Vin is the phasor voltage acrossZin, obtained by voltage division,

Vin = Zin

Zin+Rs

vs(t) (21)

Also in equation (20),Iin is the phasor current throughZin, obtained by Ohm’s law

Iin = vs(t)Zin+Rs

(22)


LINE AS CIRCUIT ELEMENT : EXAMPLE

Here is an example of this approach. Let a signal sourcevs(t) = 5 cos(1000t) have sourceresistanceRs = 70Ω, The source is connected to a lossless line/load having “looking-in”impedanceZin = 50+ j50 Ω, as shown in the figure on the previous page. Compute thetime-average power flowing on the transmission line.

SOLUTION:Replace the transmission line and load in the preceding figure with the discrete impedanceZin. Also, work the problem using phasors. This simplifies the transmission line circuitdown to a voltage supply (given by ˜vs(t) = 5e j 1000t ) in series withRs = 70Ω andZin.Since the line is lossless, all power flowing into the line (i.e., flowing acrossZin) will bedelivered to the load. Thus, we need only to find thePavg acrossZin. DefineVin as thevoltage acrossZin andIin as the current throughZin. This current and voltage are related by

Iin = Vin

Zin(23)

Using (23) thePavg formula becomes

Pavg = 12

ReVin Iin= 1

2Re

Vin V

in

Zin = 12

Re

jVinj2Zin (24)

Use voltage division to computeVin:

Vin = Zin

Zin+Rs vs(t) = 50+ j50

50+ j50+705e j 1000t = 2:715e j (1000t+0:391) (25)

Substitute (25) andZin into the right side of (24), obtaining

Pavg = 12

Re

2:7152

50 j50

= 12

Re

7:371

50 j50

= 3:686Re

1

50 j50

(26)On the following page we will provide two methods for finding the real part of the complexexpression in (26). Being able to use both methods will be very helpful in working withcomplex functions


EXAMPLE (cont.)

On this page we w provide two methods for finding the real part of the complex expressionin (26). One method uses the polar form of a complex number andthe other method usesthe rectangular form.

Polar Form:

Write the denominator expression in (26) in polar form:

Pavg = 12

Re

7:371

50 j50

= 12

Re

7:371

70:7 e j 0:7853

= 12

Ren

0:1043e j 0:7853o (27)

Rewrite (27) by factoring out the constant inside the “Re” argument and expanding thecomplex exponenetial using Euler’s formula:

Pavg = 0:0521Ren

cos(0:7853)+ j sin(0:7853)o (28)Extracting the real part in (28) then gives the desired result:

Pavg = 0:0521 cos(0:7853) = 0:0521(0:707) = 36:9 mW (29)Rectangular Form:

Multiply the numerator and denominator expressions in (26)by the complex conjugate ofthe denominator:

Pavg = 12

Re

7:371

50 j50

= 3:686Re

50+ j50(50 j50)(50+ j50) (30)

Perform the multiplication inside the “Re” argument in equation (30):

Pavg = 3:686Re

50+ j50

2500+2500

= 3:686Re

50

5000+ j

505000

(31)Extracting the real part of the expression in equation (31) then gives the desired answer:

Pavg = 3:68650

5000= 36:9 mW (32)


ECE 303 - Sum10 Notes Set 8 :Electric Charges and Fields 1

INTRODUCTION

An awareness of many important physical phenomena can be achieved by assuming theexistence of electric charges in fixed positions. This is sometimes calledElectrostatics,denoting the stationary location of the charges. A fundamental property of these stationaryelectric charges is that they effect each other through a property called theElectric Field.

The effect of this electric field appears as a force exerted onone charge by another charge.This force is not like a mechanical force, in which particlesor objects must be in physicalcontact to transmit the force. The charges themselves can begreat distances apart and stillproduce forces on each other. However, the specific type of effects depend on the motionof the charges.

Here is a hierarchy of forces and effects produced by electric charges:

Stationary ChargesIf charges are stationary, then they can only effect each other by the Electric Field.Each charge exerts a force on the other charges. This type of force is sometimescalled the Coulomb force.

Constant Velocity ChargesCharges in constant velocity create a magnetic field. This magnetic field causes aforce which can effect other charges having non-zero velocity. A DC current in aconductor is an example of constant velocity charges.

Accelerating ChargesAccelerating charges radiate energy away from their location. This energy is some-times called the Electromagnetic Wave. Antennas use this principle to send televisionand radio signals. An AC current is an example of accelerating charges.

In this set of Notes we will discuss the effects of stationarycharges on each other. We willalso discuss the motion of charges under the influence of the electric field. In future sets ofnotes we will discuss magnetic fields and electromagnetic wave propagation.


EXAMPLES OF CHARGE PHENOMENA

Below are some examples of how the concepts of electric charge and electric field candescribe some physical problems.

Semiconductor DevicesSome diodes use the important semiconductor structure called a PN-junction. Nearthis junction positive and negative ions are held fixed in thelattice of the semiconduc-tor crystal. These fixed ions carry a charge, which creates anelectric field allowinga preferential direction of current flow. This is sometimes quantified as the thresholdvoltage, which is about 0.7 volts for silicon.

Biological StructuresMany important biological structures have electrical charges. For example, someamino acids (which are the building blocks of proteins) are positively charged andsome amino acids are negatively charged. The attracting andrepelling forces pro-duced by these charged structures cause the resulting protein molecule to twist, turn,and fold to assume a specific structural shape. In many cases,it is this structuralshape which determines the function of the protein.

Bio-ElectricityPositive ions such as sodium and potassium and negative ionssuch as chlorine aresignificant for the electrical properties of biological cells. These (and other) ions areresident in the fluids inside and outside the cell, being separated by the cell mem-brane. However, their concentration imbalances across thecell membrane causes apotential difference on the order of 50 - 100 millivolts to begenerated across the cellmembrane. Therefore, we can think of each cell as a tiny battery, storing potentialenergy and available for doing work. Cells such as nerves andneurons can generateand propagate pulses using these ions. In this way nerves canprovide electrical com-munication pathways between sensory locations in the body and command centersin the brain.

In the above examples, the importance of the charges is theirparticulate nature, rather thanthe wave energy radiated by the charges as they move. We will discuss the wave nature ofpropagated energy in future sets of Notes.


ELECTRIC FIELDS

We will first consider stationary electric charges; the study of these charges is sometimescalled Electrostatics. An electric charge produces an electric field E which affects othercharges. As a first example, let chargeQ exist at the origin. At pointP the electric field dueto chargeQ is given by

E = 14πε

Qjrj2 ur (1)where the new quantities in equation (1) are defined as follows: r is the vector between the location ofQ and the pointP, ur is a unit vector in the direction ofr (pointing radially away from the origin), ε is the electrical permittivity (in Farads/m) of the medium.

The units of the electric field arevolts/meter. If the charges are located in free-space, thenthe permittivity (having units offarads/meter) is

ε = ε0 8:85(10)12 (2)The following evaluation, obtained using equation (2), often simplifies computations:

14πε0

9(10)9 (3)Equation (1) shows that for a charge at the origin the electric field E has magnitudejE jand directionur pointing directly away from the charge (which is at the origin). Let’s referto the direction of a generalE-field vector asuE . Thus, theE-field in equation (1) can alsobe written in magnitude-direction form as

E = jE j uE ) jE j= 14πε

Qjrj2 (4a;4b)The electric fieldE is also a linear field. This means that the fields due to each of multiplecharges add independently. For example, let chargeQ1 produce fieldE1 at pointP and letchargeQ2 produce fieldE2 at pointP. The total field,ET is the vector sum

ET = E1+E2 (5)


ELECTRIC FIELD EXAMPLE

Let’s work an example which shows how to compute theE-field. A charge ofQ = 1 nC islocated at(x;y) = (0;0) in free space. Find theE field at the point(x;y) = (4;3) and writeit in the formE = E uE . This geometry is shown below:

Ey = 3

x = 4

Qr

The vectorr = 4ux+3uy. Thusjrj=p42+32 =p25= 5. The unit vector is therefore

uE = rjrj = 4ux+3uy

5= 0.8ux + 0.6uy

The magnitude of theE-field is computed by

E = 14πε0

Qjrj2 = 9(10)9 1(10)9

52 = 0.36

Two forms for theE-field will be useful:

Magnitude-Direction: E = E uE = 0:36h

0:8ux+0:6uy

iCartesian: E = 0:288ux+0:216uy

The last line is the Cartesian form. To obtain the Cartesian form, multiply the magnitude,E, times the unit vector,uE .

Problems: 8-1, 8-2


LINEAR FIELD EXAMPLE

The electric fieldE is a linear field. This means that the fields due to each of multiplecharges add independently, as shown below:

1Q

2Q E1

E2

ET E1 E2

.

x

y= +

EXAMPLE: Let Q1 = 2/9 nCoul be located at(x;y) = (1;0) in free space and letQ2 = 1/3nCoul be located at(x;y) = (0;3). Compute the totalE-field at the pointP = (x;y) = (5;4).APPROACH: Draw the location of the charges and the pointP:

Q1

Q2

1r

= ( 5, 4)P

r2

y

x

.

SinceE1 andE2 are in the directions ofr1 andr2, you should now be able to sketch thedirections of the individual field. The specific math computations using the figure aboveare carried out on the next page.


LINEAR FIELD EXAMPLE (cont.)

Substitute the expressions for the individual fields and factor out common terms:

ET = E1+E2 = Q1

4πε0jr1j2 u1 + Q2

4πε0jr2j2 u2= 14πε0

Q1jr1j2 u1+ Q2jr2j2 u2

From the problem statement we can find the unit vectors and lengths:

r1 = 4ux+4uy ) u1 = 0:71ux+0:71uy and jr1j2 = 32r2 = 5ux+uy ) u2 = 0:98ux+0:20uy and jr2j2 = 26

Substituting the above and equation (7) into the expressionfor ET then gives

ET = 9(10)9" 29(10)9

32u1+ 3

9(10)9

26u2

#=

232

h0:71ux+0:71uy

i+ 326

h0:98ux+0:20uy

i= 0:157ux+0:067uy

Another form forET is the magnitude-direction form. You should be able to show that theboxed result above is equivalent to

ET = 0.171uE , whereuE = 0:918ux+0:392uy



ELECTRIC (COULOMB) FORCE

Once an Electric Field is established, there will be a FORCE on any other charge in thatfield. This is sometimes called the Coulomb force.

Fixed and Mobile Charges

It will simplify our study to think of one charge,Q, as being in a fixed position and anothercharge,q, as being free to move. In this way,Q produces the fieldE andq moves due tothis field. The force, in Newtons, exerted onq is

F = qE (9)Here is an example of how to compute the force.

EXAMPLE:Let an existing electric field beE = 0:36

h0:8ux+0:6uy

i. Suppose a chargeq = -5

nC is placed at the point(x;y) = (3;4). Find the force which is exerted on chargeq.

APPROACH:Using (9), the forceF onq is given by

F = qE =5(10)90:36h

0:8ux+0:6uy

i=1:8(10)9uE (10)whereuE = 0:8ux+0:6uy.

The negative sign in equation (9) signifies the force is an ATTRACTION. We thus have thefollowing very useful result: Same polarity chargesREPEL . Opposite polarity chargesATTRACT.

Problems: 8-7, 8-8

ECE 303 - Sum10 Notes Set 9 :Voltage 1

INTRODUCTION

Voltage is a type ofPotential energy, which is the ability to do work. In fact, voltage isoften referred to as the electric potential. This set of Notes examines the definition andorigin of voltage using geometries of charges.

It is instructive to use some mechanical analogies to understand electric voltage. Below aretwo mechanical analogies which will be very useful for us during this course:

PendulumOne type of potential energy is the gravitational potentialenergy stored in a mass.Raising the mass to a higher location increases the gravitational potential energy,which then gives the mass the ability to do work. For example,upward displacementof the mass at the end of a pendulum increases the gravitational potential of the mass.Releasing the mass then causes the pendulum to move downwardthrough its motiontrajectory.

SpringAnother type of potential energy is compressing or stretching a spring. Assume oneend of the spring is fixed and a load is attached to the other end. Then compressingthe spring “stores” potential energy in the spring. When thecompressing force isremoved, the potential energy is transformed into the motion of the load.

Electrical voltage is somewhat more abstract since we don’tactually see the charges beingmoved into the configuration which produces the voltage. However, we can sometimes“see” the effect of the voltage. For example, the voltage of aflashlight battery moveselectrons through a filament, which in turn glows, producingthe light.


DOT PRODUCT

The definition of voltage uses the vector concept of a dot product. It is easy to use Cartesiancoordinates to demonstrate some properties of the dot product. However, keep in mind thatthe principles also apply to vectors in cylindrical and spherical coordinate systems.

Let W =Wx ux +Wy uy+Wz uz andV = Vx ux +Vy uy+Vz uz be vectors in Cartesian coor-dinates. TheDot Product of these two vectors is given by

WV =WxVx+WyVy+Wz Vz (2)An alternate formula for the dot product is

WV = jWj jVj cosθwv (3)whereθwv is the angle between the vectorsW andV. An example will demonstrate usingthe dot product.

EXAMPLE:

Let W = 3ux+4uy andV = 2ux+5uy. Drawing these vectors in thexy-plane showsthat the angle between them is small. To show this mathematically, equating (2) and(3) above gives

WxVx+WyVy+Wz Vz = jWj jVj cosθwv (4)Now solve (4) for cosθwv, producing

cosθwv = WxVx+WyVy+WzVzjWj jVj = (3)(2)+(4)(5)+(0)(0)p32+42p22+52

= 0:966

Thus, the angleθwv may thus be found as

θwv = cos1(0:966) = 15:1o = 0.263 rad.

It is frequently a good idea to draw the vectors (such asW andV above) to validate yourmathematical angle computation.

Problems: 9-1, 9-2, 9-3


CHARGES AND WORK

Now that we have reviewed the dot product, let’s examine the definition of voltage. We willsee that a configuration of charges creates the voltage “field”. For now, let’s consider thesimple charge configuration shown below. Assume both charges are positive and assumechargeQ is fixed in position. Chargeqt is allowed to move.

qt

r = rA

r = rB

AB

Q

X

It is instructive to think of the interaction betweenQ andqt in three steps: The fixed chargeQ produces a fieldE at pointA. Thus, there is a force onqt , determined byF = qtE. This forceF (or, equivalently, the fieldE) does workW in movingqt from pointA topointB.

From basic physics, the work (in Joules) done by the field (i.e., the force) is given by

W = Zl

Fdl = qt

Zl

Edl (1)The l subscript on the integral in (1) signifies aLine integral anddl is theline differential.The “” in the integrand of (1) signifies the Dot Product.


DEFINITION OF VOLTAGE

Now consider the situation of using an external source of energy to move a chargeqt againstan electric fieldE. Denote the work doneby this external source asWex. Since this workis done against the Force (or against the Field) this work is thus the negative of the workintegral in equation (1):

Wex =Zl

F dl (5)Since the forceF on qt is due to the existing fieldE, we haveF = qtE and therefore (5)becomes

Wex =Zl

qtE dl (6)The common engineering unit ofVoltage, having units of Joules/Coulomb, is defined as thework per unit positive charge done by the external source:

V = Wex

qt= Z

lqtEdl

qt(7)

We will use the symbolV (or v) to denote voltage in this course. In evaluating (7) we seethat the chargeqt cancels, giving the following definition of voltage in termsof the electricfield:

V =Zl

Edl (8)Voltage is sometimes called aPotential Difference, which signifies voltage is a differencein potential energy between two points. In the definition of equation (8) these two pointsare the endpoints on the line integral.

You can check the sign of a voltage computation using physical arguments: If the directionon the pathl is Against the field, then work must be done in moving a positive test charge.This should produce aPositive sign on the resulting computation ofV in (8).


VOLTAGE FIELD FOR POINT CHARGE

We can use our prior results to find the scalar Voltage Field due to a point chargeQ at theorigin. Due to the spherical symmetry of theE-field in this problem, it is easiest to workin spherical coordinates. The electric field in spherical coordinates due to this charge hasbeen given in Notes Set 7 by

E = Q4πεr2 ur (9)

Theuθ anduφ components ofE in (9) are zero and have been omitted for simplicity. Theline differentialdl in spherical coordinates is given by :

dl = dr ur + r dθ uθ+ r sinθdφ uφ (10)Substituting (9) and (10) into (8) then gives

V =Zl

Q4πεr2 ur dr ur + r dθ uθ+ r sinθdφ uφ

(11)To show the relation between this problem for computing a line integral (in spherical co-ordinates) and the previous example (in Cartesian), let’s divide the computation into twosteps: (a) Computation of the dot product, and (b) Evaluation of the scalar integral:

(a) Dot Product:

Note theuθ anduφ components of theE-field are zero. Therefore, theuθ anduφ com-ponents of the line differentialdl do not affect the dot product and thus equation (11)simplifies to the following:

Edl = Q4πεr2 ur dr ur = Q

4πεr2 dr (12)Substitute (12) into (11) for the voltage, giving the scalarintegral

V =Zl

Q4πεr2 dr (13)

We now need to choose a path “l” and evaluate the line integral in (13). This is done on thenext page.


VOLTAGE FIELD FOR POINT CHARGE (cont.)

(b) Evaluation of Line Integral:

To evaluate the line integral in (16), we first need to choose apath “l” . Since the integrandin (16) is only a function ofr, we can therefore use a path which only varies as a functionof r, that is, a radial line intersecting the origin. This is a simple geometry which is shownbelow:

r0

radius = circle,

Q. . .x

rpath l

E = Er u r at point x

Let the path start atr!∞, and then move inward along a radial direction such thatdθ = 0anddφ = 0 along the path. Let the stopping point on the path ber = r0. Enforcing theseconditions on the integral in (13) thus gives

V (r0) = Z r=r0

r=∞

Q4πεr2 dr (14)

Evaluating the integral in (14) then gives

V (r0) = Q4πε

Z r=r0

∞

drr2 = Q

4πε

1r

r0

∞(15)

This example is completed on the following page.


VOLTAGE FIELD FOR POINT CHARGE (cont.)

Now we need to evaluate the limits in (15). Asr ! ∞, we have1r! 0 and thus the lower

limit in (15) goes to zero. Evaluation at the upper limit in (15) then gives

V (r0) = Q4πε

1r0

= Q4πεr0

(16)Sincer0 was completely arbitrary we may replacer0 in (16) with the spherical coordinateradial variabler, giving the desired result:

V (r) = Q4πεr

(17)Equation (17) is very useful and very flexible for applications. Specifically, (17) may beused even when the point charge is not located at the origin. For example, letQ be locatedat a general pointP. Then the voltage at a distanced from P is given by

V (d) = Q4πεd

(18)In another application, suppose we know the value ofQ but not its location. If we know thevalue of voltage at pointP is VP, then equation (18) allows us to determine how far pointPis from the location ofQ:

VP = Q4πεd

) d = Q4πεVp

(19)An extension of these results shows that knowing the voltageat two pointsP1 andP2 allowsus to compute information aboutQ and/or the location ofQ. This concept is illustrated inthe detailed problem solutions.



VOLTAGE FOR MULTIPLE CHARGES

Voltage is also a linear field, which means the voltages due toseparate charges add at apoint in space. For example, suppose we have two chargesQ1 andQ2 which are fixed atthe locationsx = x1 andx = x2, and we want to compute the voltage at a point in betweenthe charges. This is shown in the figure below:

x 2x = x

2d1d

Q2Q1

1x = x

To compute the voltage at the positionx shown in the figure, we have

V (x) =V1(x)+V2(x) (20)whereV1(x) andV2(x) are the voltages due toQ1 andQ2 independently. Using equation(21) for each of these voltages then gives

V (x) = Q1

4πεd1+ Q2

4πεd2(21)

whered1 andd2 are the distances to each of the charge from the pointx. Sinced1 andd2

must be non-negative, equation (21) becomes

V (x) = Q1

4πε(x x1) + Q2

4πε(x2 x) (22)This same approach can be used if the pointx is outside the rangex1 x x2. If this werethe case, draw a figure like the one above to help you compute the distancesd1 andd2 interms of the charge locations. (Note: In either of the preceeding cases, the voltage at acharge location is undefined, because the distance to the charge would then be zero.)

Problems: 9-9, 9-10

ECE 303 - Sum10 Notes Set 10 :Voltage Surfaces 1

INTRODUCTION

Now that we have studied voltage, we can begin to use visualization techniques for under-standing many properties in electromagnetics. Specifically, voltage is a scalar field, whichis much simpler to understand than a vector field, such as the electric field. A scalar fieldis also much simpler to draw. For example, we will see that a voltage which is a functionof two spatial variablesx andy can be represented as a surface above thexy-plane.

However, a voltage “surface” is much more important than just providing us with a picture.The voltage surface has geometrical properites, such as height, steepness, smoothness, andcurvature. We will see that there is a definite relation between the geometrical propertiesof a voltage surface at a point and the corresponding electric field at that point. We can usethese geometrical properties to visualize the direction and magnitude of charge movementwithout doing complicated mathematics.

In addition, there does exist a specific mathematical relation between the voltage surfaceand the electric field. This means that we can rigorously compute the electric field atany point if we know the mathematical expression for the voltage. The vector operatorlinking the scalar voltage and the vector electric field is the gradient. We will discovera very intuitive meaning for the gradient operator and use this property to enhance ourunderstanding of the electric field.


PLOTTING A SURFACE

To help us eventually understand voltage surfaces, let’s first examine a simpler surface.Plotting a simple three-dimensional surface will aid in developing visualization skills. Sup-pose we have a surface defined by the equation

f (x;y) = x2+y2 (1)For now, we don’t have to think off (x;y) as a voltage. Just think off (x;y) as a surfacehaving varying height above thexy-plane. You could use Matlab andmesh to plot thissurface very accurately. However, there are some simple ways for obtaining an approximatesketch of the surface which will be accurate enough for our purposes.

In the left-hand figure below we have plotted the height of thesurface f (x;y) above thex-axis over the range2 x 2 and the height of the surface above they-axis over therange2 y 2. These lines trace out two parabolas above thexy-plane.

f = 4

f = 4

1 2 x

f(x,y)

f = 1 f = 1

y

f = 4

f = 2.5

x

f = 1

y

f(x,y)

f = 4

Then in the right-hand figure above we have plotted the heightof the surface above thexy-plane for three values for height:f (x;y) = 1, f (x;y) = 2:5, and f (x;y) = 4. You can thinkof all these lines as “stripes” on the surfacef (x;y). With some imagination, you shouldbe able to “see” the parabolic surfacef (x;y) rising “above” thexy-plane. This parabolicsurface resembles a “bowl”.


PARTICLE MOTION

Now let’s perform a thought experiment to relate the function f (x;y) to energy and motion.Think of the surface on the previous page as a physical surface, like a bowl, rising abovethe flat xy-plane. Suppose we held a small ball on the surface directly above the point(x;y) = (1;1), or directly above(x;y) = (1;1), as shown in the figure below. Whatwould happen at either place if the restraining force on the ball were removed? Obviously,the ball would “roll down the surface”, as shown in the figure.Notice that the direction ofthe roll is dependent on the initial position of the ball. Also note that the balls are rollingtoward the “bottom of the bowl”.

x

y

Original Position

Later Position

(x,y) = (1,1)

f(x,y) = 2

Later Position

Original Position

(x,y) = (−1,−1)

Path of Movementon Surface

In this example, the ball has been released in the scalar “gravitational potential energyfield”, and the ball, having positive mass, receives a force of movement due to the field.

This is exactly what happens when a chargeq is released in a scalar voltage fieldV(x;y;z).The charge moves under the influence of the voltage field in a manner similar to the ballrolling down the surface. A positive charge will move from locations of higher voltage(potential) in space to points of lower voltage (potential), much as the ball would movefrom location of greater height to lower height.


CONTOUR PLOT

Although the three-dimensional surfaces examined on the previous pages are very instruc-tive, they are very cumbersome to reproduce every time we want to investigate chargemovement in a voltage field. Consequently, we desire some picture which is easier to drawbut which still has the same information as the three-dimensional surfaces. Such a reduceddimension plot does exist and it is called a Contour Plot.

A contour plot is created by projecting the location of properties of the three-dimensionsurfacef (x;y) onto thexy-plane. To see how this is done, consider again the example ofthe balls rolling down thef (x;y) surface on the previous page. The three-dimension pictureof this is shown again in the figure below:

x

y

(x,y) = (1,1)

(x,y) = (−1,−1)

f(x,y) = 2

f(x,y) = 0.5

In the above figure the balls’ path on the three-dimensionalf (x;y) surface is projecteddown onto the two-dimensionalxy-plane. As the balls roll toward the bottom of the bowl,their movement is directed toward the origin, which is the bottom of the bowl. In thetwo-dimensionalxy-plane this is represented by the two arrows pointing inwardtoward theorigin. We will next show how to extract this same information in a simpler manner usingthe contour plot.


CONTOUR PLOT (cont.)

Important properites in the current example are the initiallocations of the balls, their initialdirections of roll, and some measure relating to the initialvelocities of their roll. Thecontour plot provides much of this information. Referring to the three-dimensionalf (x;y)surface on the previous page, consider the locus of all points in thexy-plane which give asurface heightf (x;y) = 2. Also consider the locus of all points in thexy-plane which givea surface heightf (x;y) = 0:5. In thexy-plane these loci are the circlesx2+ y2 = 2 andx2+y2 = 0:5, which have been drawn in the figure below:

.

.x

y

(x,y) = (1,1)

(x,y) = (−1,−1)x + y = 0.52 2

Initial Position

Later Position

.

x + y = 22 2

In terms of the above contour plot, the balls were initially released at their respective pointson the circlex2+ y2 = 2 and they begin moving toward the origin. Suppose that at thelater time the balls have now rolled down to a height off (x;y) = 0:5. Therefore, the balls’xy-positions on the contour plot have now reached the circlex2+y2 = 0:5. This situation isrepresented by the arrows in the above figure. The arrows trace the movement of the ballsprojected onto thexy-plane.

One the following pages, we will study more about contours, how to plot them, and howthey are useful in voltage problems.


PLOTTING CONTOURS

If we know a two-dimensional functiong(x;y) we can plot the contours. For example,suppose we have the two-dimensional function

g(x;y) = x2+9y2 (2)Suppose we want to plot the contours in thexy-plane corresponding to the surface heights

g(x;y) = K (3)whereK is a constant. We can solve for the entire family of contours by substitutingequation (2) into equation(3):

x2+9y2 = K (4)Equation (4) is recognized to be the equation for a family of ellipses whose size dependson the value ofK. For example, if we want the contour in thexy-plane corresponding tothe height ofK = 50 on theg(x;y) surface, then we substituteK = 50 into equation (4):

x2+9y2 = 50 (5)We can draw this ellipse by solving for the points where it crosses thex-axis and the pointswhere it crosses they-axis. When it crosses thex-axis, we havey= 0 in (5), and thereforeequation (5) becomes

x2 = 50 ) x= p50 = 7:07 (6)When the ellipse crosses they-axis, we havex= 0 in (5), and therefore equation (5) be-comes

9x2 = 50 ) x= r509

= 2:36 (7)When the ellipse crosses the liney= 0:5x, we havey= 0:5x in (5), and therefore equation(5) becomes

x2+9(0:5x)2 = 50 ) x = 3:92 (8)These are enough points to sketch theK = 50 ellipse, which is done on the next page.


PLOTTING CONTOURS (cont.)

From our work on the previous page, we now have enough points to sketch theK = 50ellipse, which is shown in the figure below:

x = 3.92

y = 2.36

y = 0.5xy

x

y = −2.36

x = 7.07x = −7.07

ellipse

22x + 9y = 50

Although plotting a single contour is sometimes useful, thereal value of the contour plotarises when when plot the contours for several heights on thesurface. The figure belowplots the elliptical contours for the surface of equation (4) whenK = 25; 50; 75; and 100.

100

K = 25ellipses

22x + 9y = K

x

x = 10

x = 7.07x = −7.07x = −10

7550y



E-FIELD USING GRADIENT

The previous material in this Notes Set discussed geometrical properties of voltage sur-faces and contour plots. A natural question is how a voltage surface (which is a scalarfield) relates to the electric field (which is a vector field). Another question concerns themathematical operators which can give us quantitative properties of these fields. This nextsection of Notes addresses these questions.

We found earlier that if we know the vector electric field thenwe can find the scalar voltageusing the line integral :

V = Zl

E dl (9)Now we are interested in the reverse question: if we know a voltage functionV(x;y;z),can we find the electric fieldE(x;y;z) ? The answer is “Yes”, by using the mathematicaloperation of theGradientof V. The computational formula is

E =∇V (10)where “∇” in (10) is the gradient operator. The math term for “∇V” is “gradient ofV”.Note thatV is a scalar, but∇V is a vector. Once again,∇V is coordinate-specific and thereare three forms:

Cartesian: ∇V = ∂V∂x

ux+ ∂V∂y

uy+ ∂V∂z

uz (11a)

Cylindrical: ∇V = ∂V∂r

ur + 1r

∂V∂φ

uφ+ ∂V∂z

uz (11b)

Spherical: ∇V = ∂V∂r

ur + 1r

∂V∂θ

uθ+ 1r sinθ

∂V∂φ

uφ (11c)

In this course we will primarily study the Cartesian form forthe gradient. We will examinean example on the next page.


GRADIENT EXAMPLE

Let’s work an example which shows the mechanics of computingthe gradient of a scalarfunction. For simplicity, the first example will use Cartesian coordinates.

EXAMPLE:

Let a scalar function ofx andy in Cartesian coordinates be given by

F = F(x;y) = 4xy2+z3

Compute the gradient ofF, which is denoted as∇F.

APPROACH:

From equation (11a) we have

∇F = ∂F∂x

ux+ ∂F∂y

uy+ ∂F∂z

uz

Compute the partial derivatives required above, giving

∇F = 4y2 ux+8xyuy+3z2uz

Note that∇F is a vector. Notice also that∇F has a different magnitude and direction fordifferent values of(x;y;z).


E-FIELD EXAMPLE

As another example of using the gradient, let’s compute the vectorE-field due to a pointcharge by starting with the scalar voltage field surroundingthe charge. From previouswork, the voltage field due to a chargeQ at the origin is given by

V(r) = Q4πεr

(12)Computing the gradient in spherical coordinates will keep us from having to rewrite (12) interms of Cartesianx, y, andz. Now operate on both sides of (12) with the gradient operatorin spherical coordinates from equation (11c):

E = ∇V = ∂V∂r

ur + 1r

∂V∂θ

uθ+ 1r sinθ

∂V∂φ

uφ

(13)But notice from (12) thatV(r) is only a function of radial variabler. Therefore, theφ andθ partial derivatives in equation (13) vanish:

∂V∂φ

= 0; ∂V∂θ

= 0 (14)Substitute the results from (14) into the expression forE in (13):

E = ∂V∂r

ur = ∂∂r

Q

4πεr

ur = Q

4πε∂∂r

1r

ur (15)

Computing ther partial derivative in (15) then provides the desired result:

E = Q4πεr2 ur (16)

Equation (16) is the same result as given by Coulomb’s law. However, notice that it wascomputationally simpler to start withV and differentiate to obtainE (this example) that itwas to start withE and integrate to obtainV .



VOLTAGE FOR CHARGE PAIR

Some interesting physical voltage phenomena can occur whenthere are multiple fixedcharges. Suppose we have two chargesQ1 andQ2 which are fixed at the locationsx= x1

andx= x2, as shown in the figure below:

x 2x = x

2d1d

Q2Q1

1x = x

Let’s compute the voltage in the regionx1 < x< x2. For anyx-value in this range we have

V(x) = Q1

4πεd1+ Q2

4πεd2(17)

whered1 andd2 are the distances to each of the charge from the pointx. Sinced1 andd2

must be non-negative, equation (17) becomes

V(x) = Q1

4πε(xx1) + Q2

4πε(x2x) (18)There is some value ofx in the rangex1 < x < x2 which gives a minimum for voltage forthisx-range. Call this valuexmin and denote the corresponding voltage asVmin. We can findthis valuexmin by solving the differential equation

ddx

V(x) = 0 (19)for the value ofxmin. We will do an example on the next page which demonstrates thisprocedure.


VOLTAGE MINIMUM EXAMPLE

Consider again the charge configuration shown in the figure onthe preceding page. Thereis a voltage minimum somewhere in the regionx1 < x < x2. To show this let’s assumefree-space and work a specific example having the following parameters:

x1 = 5 ; Q1 = 1010 Coul ; x2 = 10 ; Q2 = 3(10)10 Coul

Substituting these values into (18) and remembering that 1=4πε0 = 9(10)9, we get thefollowing equation:

V(x) = 0:9(x5) + 2:7(10x) (20)Now take the first derivative of (20) and set it equal to zero:

2:7(10x)2 0:9(x5)2 = 0 (21)Now add the terms in (21), and simplify, obtaining

2:7 (x5)20:9 (10x)2 = 0 (22)Multiply out the squares in (22) and then collect terms, producing

1:8 x29x22:5= 0 (23)You can use the quadratic formula on equation (23) to find the roots are

x=1:83 ; x= 6:83 (24)The positive root in (24) is in the range 5< x< 10. Thus,

xmin= 6:83 (25)The voltage minimumVmin is obtained by using (20) to evaluateV(x) at x= xmin= 6:83:

Vmin= 0:9(6:835) + 2:7(106:83) = 1:344 (26)Problems: 10-10, 10-11, 10-12

ECE 303 - Sum10 Notes Set 11 :Gauss’ Law 1

INTRODUCTION

A fundamental property we have seen is that charges create electric fields. Sometimesthe charges are discrete charges or point charges, as we studied previously. However, inmany physical structures it is more instructive to considerthese discrete charges as beingdistributed over a spatial extent. This creates the conceptof a distributed region in spacewhich has a particular “charge density”. An example of such acharged region is the deple-tion region near the PN-junction of a semiconductor diode.

If we begin with a knowledge of the charge density in a region,then Gauss’ Law gives usa way to compute the electric fieldE over the region. This might at first seem to be morecomplicated than just using Coulomb’s Law to compute theE-field given just a point chargeQ. However, the region which is modeled as having a charge density may actually comprisean enormous number of individual discrete charges. To use Coulomb’s Law to find theE-field component due to each of these discrete charges may not be computationally feasable.Consequently, an alternate approach is to use Gauss’ Law, starting with the charge densitydue to the individual charges.

There are two forms of Gauss’ Law: the Integral Form and the Differential Form. The in-tegral form considers theE-field existing over a surface in space. However, the differentialform gives usE-field at every point within the surface, as well as theE-field on the surface.For these reasons, the differential form is also often called the Point Form of Gauss’ Law.This set of Notes studies the differential form of Gauss’ Law. The PN-junction voltagefrom solid-state electronics is used as a motivating application to show us how Gauss’ Lawcan be applied to a useful problem.

If you take a course in solid-state electronics you may learnabout the PN-junction in muchmore detail. The diffusion equation and current density equations are often used to developvery quantitative results for hole and electron flow across the junction. In this notes Set, wewill only examine the PN-junction from the standpoint of Gauss’ Law, which is in keepingwith our study of electromagnetic phenomena.


THE P-N JUNCTION

This page describes a simple model for the hole and electron densities on the differentsides of the PN-junction, prior to their being joined to actually form the junction. Onthe following page, we will place the sides in contact and observe the hole and electronmovement.

Lattice Geometry

In the figure below, the bulk material on both sides of the dotted line is silicon. Atoms fromcolumn V in the periodic chart have been added to the silicon lattice to form the N-typeregion on the right side and atoms from column III in the periodic chart have been addedto the silicon lattice to form the P-type region on the left side. These added atoms are oftencalled the impurities.

+

+

+4

+4

+3

+3

+4

+4

_

_

+4

+4

+5

+5

+4

+4

JunctionP−type(Acceptor)

N−type(Donor)

Donors and Acceptors

At room temperature, one of the outer shell electrons from the column V impurity is heldvery loosely by its nucleus and is therefore very mobile. This material is called the N-type(or Donor) because these mobile electrons are Negative charges. Conversely, there is anelectron vacancy in the outer shell of the column III impurity, and this vacancy can acceptmobile electrons. However, it is conceptually easier to envision the electron “vacancy” asmoving through the lattice. This vacancy is called a hole, and it is a virtual Positive charge.Thus, this material is called the P-type material (or Acceptor).


DIFFUSION OF HOLES AND ELECTRONS

We can show that the diffusion of holes and electrons across the PN-junction creates anelectric field. It is this field which eventually stops the diffusion and causes the process toreach a steady-state.

Before Diffusion:

Consider the figure below, which shows a mobile hole and a mobile electron before dif-fusion begins. The +3 and +5 are the column III and column V nuclei which are heldimmobile in the semiconductor lattice:

_+5

+4

+

+3

+4 N−typeP−typeDiffusion

JunctionBEFORE Diffusion:

After Diffusion:

Now let the diffusion process occur and suppose the mobile electron diffuses to the left andfills the hole. These mobile charges cancel, but the bound charge imbalances due to the +3and +5 lattice nuclei still remain:

+3

+4 N−typeP−type

AFTER Diffusion:

+5

+4

Junction


ELECTRIC FIELD

Note from the bottom figure on the previous page that, after diffusion, the +3 lattice nucleiin the P-type material is surrounded by 4 outer shell electrons. Therefore, there is a resulting“q” charge on the left. Additionally, we see that the diffusionof the electron from theN-type material on the right results in a charge of+q remaining at the +5 location in theN-type material.

Therefore, from electrostatics we see there is anElectric Field created between the+qcharge and theq charge. Thus, diffusion generates bound stationary charges and anElectric field is created.

Since there are many mobile elections on the right, another electron will try to move to theleft by diffusion. However, note theq previously created will oppose the diffusion of thesecond electron to the left. As many electrons diffuse, the opposing electric field builds up,until no more electrons from the right can diffuse. A steady state electric fieldEx(x) thenexists, as shown in the figure below:

++

++

+

__

__

_x = x2

x = x1

E−field

N−typeP−type

Junction

This region betweenx1 andx2 in the figure is sometimes called theDepletion Regionina course on solid-state electronics. From the figure you can see that the “bound” chargesform two regions of charge density. On the following pages, we will show how we can useGauss’ Law to determine the electric field if we know these original charge densities.


GAUSS’ LAW: DIFFERENTIAL FORM

This concept of a distributed charge region (such as that in the depletion region) can bequantified by introducing the Charge Density. Charge density has the symbolρ, and hasunits of coulombs per m3. When the charge environment is given by charge densityρ ina material of permittivityε, then the electric fieldE is given by Gauss’ Law. One form ofGauss’ Law is called the differential form and is given by

∇ E = ρε

(1)The vector operation on the left side of equation (1) is called the Divergence operation. Themathematical equation resulting from Gauss’ Law in (1) is a Partial Differential Equationfor the electric fieldE, given a charge densityρ and a medium with permittivityε. The mathform of the divergence depends on the coordinate system used. For example, in Cartesiancoordinates the divergence is given by

∇ E = ∂Ex

∂x+ ∂Ey

∂y+ ∂Ez

∂z(2)

Substituting (2) in (1) then gives

∂Ex

∂x+ ∂Ey

∂y+ ∂Ez

∂z= ρ

ε(3)

You can think of the charges, represented byρ = ρ(x;y;z), as creating anE-field withcomponentsEx, Ey, andEz. To find these components, we would generally have to solveequation (3) forEx;Ey; andEz. However, specific problems can often be made simpler byusing geometrical considerations. To help our intuitive understanding of Gauss’ Law, let’sfirst examine the geometrical properties of divergence. This will in turn help us understandthe physical interpretation of Gauss’ Law.

Problems: 11-1, 11-2


DIVERGENCE - PHYSICAL INTERPRETATION

Many problems in electromagnetics can be simplified if we understand the geometicalproperties of the mathematical operators. Such an operatoris the divergence operator givenin the statement of Gauss’ Law in equation (1).

To develop our understanding, let’s consider equation (3) when the partial derivaties withrespect toy and z are equal to zero and charge densityρ is only a function ofx. Thisdoes not trivialize the problem, as will be explained in class. Using these simplifications,equation (3) becomes

dEdx

= ρ(x)ε

(4)where we have also setEx = E for simplicity. Now recall the definition of the ordinaryderivative in (4):

dEdx

= lim∆x!0

E(x)E(x∆x)∆x

(5)If ∆x in (5) is small, but non-zero, we can approximate the derivative in (5) by the right-hand expression without the limit operation:

dEdx

E(x)E(x∆x)∆x

(6)Now substitute (6) into (4) and rearrange somewhat:

E(x)E(x∆x) = ρ(x)ε

∆x (7)Now we can investigate (7) for the two possible cases: (i) theregion∆x does not con-tain charge, or (ii) the region∆x does contain charge. These regions are explored on thefollowing page.


DIVERGENCE - PHYSICAL INTERPRETATION (cont.)

ZERO CHARGE:

If the region∆x does not contain any charge, then we haveρ(x) = 0 and (7) becomes

E(x) = E(x∆x) (8)Equation (8) signifies that for no charge in the region∆x, then theE-field is the same oneither side of the∆x segment. This non-zeroE-field has been produced by a charge notlocated in the∆x segment. A figure describing this situation is shown below:

x − x∆

E(x)

x

E(x − x)∆

NON-ZERO CHARGE:

If the region∆x does contain charge, then the fields on either side of the∆x segment arenot equal. To see this, observe the situation in the figure below:

x

q

qE (x − x)∆

∆E(x − x) E(x)

∆x − x

qE (x)

The fieldsE(x) andE(x∆x) due to charges not in∆x are shown, as in the preceding figure.However, due to the presence of chargeq within ∆x, there is an additional fieldEq(x∆x)at x∆x and a fieldEq(x) at locationx. These two fields are in opposite directions, asrequired by Coulomb’s Law. This example is completed on the following page.


DIVERGENCE - PHYSICAL INTERPRETATION (cont.)

To obtain the total fields atx andx∆x we have to add the two field components existingat each boundary, as shown on the following page. Let the subscript s, for sum, representthe summation of the fields at the boundaries of the∆x segment. Visually adding theE andEq fields shown in the previous figure then gives the resulting fields shown below:

q

x∆x − x

sE (x)sE (x − x)∆

You can see in the figure that the fields are no longer equal on each side on the segment.As ∆x! 0, there is a field discontinuity at the location of the charge, or more strictly, thelocation of the charge density. With the preceding understanding, we now have a simpleinterpretation of the differential form of Gauss’ Law:

If the divergence of theE-field is non-zero at a point in space, then there ischarge (or charge density) located at that point.

When we have a volume of space which has a non-zero charge density, like the PN-junctiondepletion region, then the divergence of theE-field is non-zero over the entire volume.Similarly, if we have a region where theE-field divergence has been determined to be non-zero, then we are guaranteed that a non-zero charge density exists throughout that region.


E-FIELD DIFFERENTIAL EQUATION

We will work an example using the differential form of Gauss’Law to demonstrate how tofind the electric fields, given the charge density.

EXAMPLE: Let a region on thex-axis have the following charge density:

ρ(x) = K; w< x 0K; 0< x< w(9)

Find the differential equation forE over the regionw< x< w.

APPROACH: Start with Gauss’ Law in Cartesian coordinates from equation (3):

∇E = ∂Ex

∂x+ ∂Ey

∂y+ ∂Ez

∂z= ρ(x)

ε(10)

whereρ(x) is the charge density function is given in (9). Sinceρ(x) is only a functionof x, thenE = Ex ux is only a function ofx. Thus, only the partial derivative withrespect tox in (10) is non-zero, and equation (10) becomes

∂Ex

∂x= ρ(x)

ε= K=ε; w< x 0K=ε; 0< x< w

(11)Since there is only one independent variable (x) we may replace the partial derivativein (11) with an ordinary derivative. Equation (11) then becomes

dEx

dx=

K=ε; w< x 0K=ε; 0< x< w(12)

Equation (12) gives the two differential equations which must be solved to find theEx-field.Other properties, such as continuity and boundary values, are necessary to actually solvethese equations forEx.


BOUNDARY VALUE EXAMPLE

Computing a specificE-field requires that the solution of the ODE, like equation (12) onthe previous page, satisfies specific Boundary Conditions. An example of Boundary Con-ditions is shown in the following example.

EXAMPLE: Let the charge density in a regionw< x< w be given by

ρ(x) = K; w< x 0K; 0< x< w(13)

ComputeEx(x), subject to the Boundary ConditionsEx(w) = Ex(w) = 0.

APPROACH: Equation (13) above is actually two ODEs. Therefore, solve the ODE foreach region, and apply the boundary condition appropriate for each region.

LEFT REGION: w< x 0Sinceρ(x) = K in this region, the following differential equation holds:

dEx

dx= K

ε(14)

Integrating both sides of the differential equation in (14)gives

Ex(x) = Kε

x+C1 (15)whereC1 is a constant of integration. Apply the left-side boundary condition:

Ex(x=w) = 0 ) Kε(w)+C1 = 0 ) C1 = K

εw (16)

Substituting the value forC1 from (16) into equation (15) forEx(x) then gives

Ex(x) = Kε

x+ Kε

w (17)


BOUNDARY VALUE EXAMPLE (cont.)

RIGHT REGION: 0< x wSinceρ(x) =K in this region, the differential equation is:

dEx

dx= K

ε(18)

Integrating both sides of equation (18) gives

Ex(x) =Kε

x+C2 (19)whereC2 is a constant of integration. Apply the right-side boundarycondition to (19):

Ex(x= w) = 0 ) Kε(w)+C2 = 0 ) C2 = K

εw

Putting this value forC2 into (19) forEx(x) gives

Ex(x) =Kε

x+ Kε

w (20)A plot of Ex(x) as a function ofx is shown below

w

x = wx = 0

x = − w

xE (x)Kε_

OtherE-field problems may be solved in a similar manner.


ECE 303 Sum10 Notes Set 12 :Poisson/Laplace Equations 1

INTRODUCTION

We have previously seen how a charge density created an electric field. We have also seenhow an electric field produced a voltage difference between two points in space. Therefore,computing a voltage difference based upon an initial chargedensity required the solutionof two different equations. We naturally wonder if there is perhaps a more straightforwardway of computing a voltage based upon a specified charge density.

Indeed there is such a straightforward method and it is the topic of this Notes set. Althoughonly one new approach is needed, there has been a historical division in the names givento the resulting equations. This division is based upon whether the region in question has azero charge density or a non-zero charge density. These equations are called: Laplace’s Equation:

If the region under consideration has a zero charge density,the resulting equation iscalled Laplace’s Equation. This equation is useful in regions which are dielectrics,insulators, or free-space. We will use the region in the dielectric between the platesof a capacitor to illustrate this equation. Poisson’s Equation:If the region under consideration has a non-zero charge density, the resulting equationis called Poisson’s Equation. This equation is useful in regions which have a chargedenisty, like the depletion region near a PN junction.

Laplace’s Equation is usually easier to mathematically solve. Therefore, we will concen-trate on solving Laplace’s Equation for regions of zero charge density. However, the ex-tension is straightforward to charged regions (Poisson’s Equation), and this is explored insome of the problems.


THE EQUATIONS

There is a way to find voltage,V, directly from charge density,ρ, without having to firstfind theE-field. To derive this method, start with the point form of Gauss’ Law:

∇E = ρε

(1)Recall that theE-field is the negative gradient of voltage

E =∇V (2)Thus, substituting (2) into (1) thus gives

∇ (∇V) = ρε

(3)Multiplying both sides of (3) by minus one produces the form known as Poisson’s equation:

∇ (∇V) =ρε

(4)The form∇ (∇V) on the right-hand side of (4) is not a dot product ... it is the Divergenceof the Gradient ofV. The simplified notation

∇2V = ∇ (∇V) (5)is frequently used. Thus, Poisson’s equation is often written as

∇2V =ρε

(6)It is easy to see that if the charge densityρ equals zero, then

∇2V = 0 (7)Equation (7) is called Laplace’s Equation. It is valid wherever the medium under considera-tion has no charges. Examples of where this occurs are in free-space away from conductors,or in an insulator separating conductors.


SECOND-ORDER DIFFERENTIAL EQUATIONS

Laplace’s Equation and Poisson’s Equation are examples of second order partial differentialequations. To solve these equations, we have to consider thespecific form for∇2V. Forexample, in Cartesian coordinates we have

∇2V = ∂2V∂x2 + ∂2V

∂y2 + ∂2V∂z2 (8)

In many of our cases, there may be a spatial symmetry and consequently voltage will varyonly as a function of one variable. For example, suppose voltage varies only as a functionof x. Laplace’s equation thus becomes

∇2V = ∂2V∂x2 = ∂

∂x

∂V∂x

= 0 (9)Since (9) has only one independent variable, it is equivalent to the second order ordinarydifferential equation (ODE) given by

d2Vdx2 = d

dx

dVdx

= 0 (10)Equation (10) says thatV =V(x) is a function whosesecondderivative (with respect tox)is zero. Thus, thefirst derivative ofV(x) is a constant, or

dVdx

= K1 (11)whereK1 is the unknown constant. We have transformed the second-order ODE into asimpler first-order ODE. Geometrically, this implies that theSlopeof V(x) is constant. Tofind V(x) we integrate the first-order ODE in (11) to obtain

V =V(x) = K1 x+K2 (12)whereK2 is another constant of integration. It is easy to see thatV(x) is the equation of astraight line. The specific constantsK1 andK2 are obtained by using boundary conditions.


LAPLACE’S EQUATION EXAMPLE

A parallel plate capacitor has its upper plate atz = 0 and its lower plate atz = d. Thevoltage on the upper plate isV(z= 0) = V0 volts and the voltage on the lower plate isV(z=d) = 0 volts. Compute the voltage in the regiond < z< 0.

APPROACH:

Since there is only one independent variable (z) we may replace partial derivatives withordinary derivatives, giving

ddz

dVdz

= 0 (13)Integrating equation (13) once gives

dVdz

= K1 (14)whereK1 is a constant of integration. Integrating (14) then gives

V(z) = K1z+K2 (15)whereK2 is another constant of integration. To evaluateK1 andK2 we have to use theboundary conditions or boundary values for the specific problem.

Boundary Conditions:V(z= 0) = V0: ) K1 (0)+K2 =V0 ) K2 =V0

V(z=d) = 0: ) K1 (d)+K2 = 0 ) K1 = V0

d

Using these evaluations forK1 andK2 in (14) we see that the equation for the voltage in thedielectric is given by

V(z) = V0

dz+V0 (16)



CYLINDRICAL COORDINATES

Since many electrical structures (like many wires and coaxial cables) are cylindrical it issometimes easier to work problems using Cylindrical Coordinates. The coordinates of apoint P in cylindrical coordinates are specified by the threevaluesr, φ, andz, where r is the radial distance from the origin to the projection ofP onto thexy-plane, φ is the rotation angle from thex-axis to the radial containing the projection ofP onto

thexy-plane, z is the vertical distance from thexy-plane to the pointP,

An example of the pointP= (r;φ;z) defined in cylindrical coordinates is shown below:

.

y−axis

x−axis

r−directionangleφ

z−axis

, zP = φr, ( )

Here are the equations for mapping the Cartesian coordinatesx;y andz into the correspond-ing cylindricalr;φ andz

r =px2 + y2; φ = tan1 y

x

; z= z

We will next work an example solving Laplace’s equation in cylindrical coordinates.


CYLINDRICAL COORDINATE EXAMPLE

Let’s work an example using Laplace‘s Equation in cylindrical coordinates. Cylindricalcoordinates are very useful when the structure under consideration (like a wire or cable)has a cylindrical geometry.

EXAMPLE:Let a very long coaxial cable have inner conductor with radius r = a and outer conductorwith radiusr = b. The conductors are separated by an insulating dielectric material. Thevoltage on the outer conductor isV(r = b) = V0 and the voltage on the surface of the innerconductor isV(r = a) = 0. Compute the voltage in the regiona< r < b.

APPROACH:

In cylindrical coordinates the Laplacian ofV is given by

∇2V = 1r

∂∂r

r

∂V∂r

+ 1r2

∂2V∂φ2 + ∂2V

∂z2 (17)By symmetry considerations, the voltage in the coaxial cable dielectric varies only as afunction of r. Therefore, the partial derivatives in (17) with respect tor andφ are zero.That is,

∂V∂z

= 0; ∂V∂φ

= 0 (18a;b)Using these simplifications in (17) thus produces

∇2V = 1r

∂∂r

r

∂V∂r

(19)Using (19), Laplace’s equation in cylindrical coordinates(equation (7)) thus becomes

1r

∂∂r

r

∂V∂r

= 0 (20)This example is continued on the next page.


CYLINDRICAL COORDINATE EXAMPLE (cont.)

Multiplying both sides of (20) byr then gives

∂∂r

r

∂V∂r

= 0 (21)Finally, since there is only one independent variable (r) in (21), we can replace the par-tial derivatives in (21) with ordinary derivatives. Therefore, (21) becomes the ordinarydifferential equation

ddr

r

dVdr

= 0 (22)We next need to solve equation (22) for the unknown voltageV =V(r). To do this, integrateboth sides of (22), giving

rdVdr

= K1 (23)whereK1 is a constant of integration. Dividing both sides of (23) byr then provides

dVdr

= K1

r(24)

Integrating (24) then gives the desired result

V(r) = K1 ln r +K2 (25)whereK2 is another constant of integration independent ofr. As before,K1 andK2 arefound using the boundary conditions.


ECE 303 - Sum10 Notes Set 13 :Time-Invariant Magnetic Fields 1

INTRODUCTION

Stationary electric charges produce only an electric field in space. This phenomenon isoften calledElectrostatics. However, when these charges move to create a current, a mag-netic field is also produced in space. This magnetic field can then effect other movingcharges, or, equivalently, other currents. The inclusion of magnetic field phenomena withelectric field phenomena is often calledElectromagnetics. If the charges are moving withconstant velocity, then the magnetic fields produced are time-invariant. However, thesetime-invariant magnetic field can vary as a function of space.

When charges accelerate, that is, change their velocity, then the magnetic field becomestime-varying, as well as spatially-varying. Additionally, electromagnetic energy can alsobe radiated in this case, which means the energy can be propagated in space. Transmit-ting antennas are structures which support this charge acceleration and specifically allowelectromagnetic energy to be radiated in an effective manner. An AC current is also anexample of charges which undergo acceleration. Time-varying electric and magnetic fieldsare examined in future sets of Notes.

This set of notes studies the magnetic phenomenon which occurs when time-invariant cur-rent flows. An example of this time-invariant current is DC flowing in a wire or a conduc-tor. We will find that this time-invariant current produces atime-invariant magnetic field inspace.


MAGNETIC FORCE

Our primary interest in this course is the magnetic field produced by currents, that is, bylarge numbers of charges in motion. However, it is instructive to first see how the magneticfield affects one moving charge. Recall that an electric chargeq placed in an electric fieldE experiences a force given byF = qE. Note that this force is in the same direction astheE field. Now let the charge be moving with vector velocityv in a region where thereis a existing magnetic flux densityB, having units of Webers/m2. Then the moving chargeexperiences an additional forceFm (in newtons) given by the cross product

Fm= q(v X B) (1)This magnetic force is illustrated in the sequence of figuresbelow:

z

xB

qy

v

qy

z

x

Fm

vv

z

xB

qy

The figure on the left is the chargeq moving in thez-direction, with no magnetic fluxdensity present. The middle figure shows the appearance of the magnetic flux densityB inthe x-direction. The right figure then shows the direction of the resulting forceFm alongthey-axis. This means the velocity of the charge will now change in response to the force.Consequently, the velocity vector will no longer be alignedwith thez-axis. Thus, we seethat the presence of qB-field changes the direction of q moving charge. The directionchange is orthogonal to both the initial velocity and the direction of the magnetic field.The cross product operation quantifies this orthogonal direction. If you need a review ofthe mathematics of the cross product you can see the information in the Review Notes onVectors.



MAGNETIC FIELD UNITS

Now let’s examine the units of the fields related to magnetics. Even though equation (1) is across product, all the terms of the cross-product will have units given by the scalar relation

F = qvB (2)Let’s write equation (2) in terms of the units of the quantities:

Newtons= ( Coul ) m

sec

Webers

m2

(3)But from basic circuits we know that

Coulsec

= ( Amp ) (4)Using (4) in (3) then gives

Newtons= ( Amp ) ( m )

Webers

m2

(5)Solve (5) for the units of Webers/m2, obtaining

Webers

m2 = NewtonsAmp - m

(6)“Webers per meter-squared” seems to be the units of q density(due to the appearance of

meters-squared). Therefore, let’s examine what a Weber might be. Multiply both sides of(6) by m2, obtaining

Webers = NewtonsAmp - m

m2 = Newton - m

Amp

(7)But a Newton-meter is a Joule and therefore (7) becomes

Webers = JoulesAmp

(8)


MAGNETIC FIELD UNITS (cont.)

Equation (8) has shown that the units for a Weber are “Joules per Amp”. Remember that aJoule is one of the standard units for Work. When interpretedthis way, the magnetic fluxin Webers is the amount of work a particular magnetic field canperform on a conductorcarrying one amp of current.

To gain even more familiarization with the unit of Weber, recall from our work in NotesSet 9 that the units for the Volt are

Volts = JoulesCoul

(9)When interpreted this way, the electric potential in Volts is the amount of work an electricfield can perform on a charge of one coulomb.

Comparing equations (8) and (9) shows that the Weber in magnetics is somewhat analogousto the volt in electrostatics. The Weber is frequently called the unit for the “Magnetic Flux”.Using this terminology, the unit ofB, Webers per meter-squared, is frequently called theunit for the “Magnetic Flux Density”. One Weber per meter-squared is also related to theunits Tesla and Gauss through the following:

1Weber

m2 = 1 Tesla = 104 Gauss (10)There is another field, sometimes called the Magnetic Field “Intensity”, which is denotedby the symbolH. TheH field is related toB by the equation

B = µH (11)whereµ is called the Magnetic Permeability of the medium. For free-space,µ= µ0, whichhas the value

µ0 = 4π(10)7

Henriesm

(12)The units ofH are Amps/meter, and this may be a more familiar unit. Many of our problemswill be worked using theH-field.


AMPERE’S CIRCUITAL LAW

Ampere’s Circuital law is a fundamental result which relates current and the magnetic field.This law allows us to compute the magnetic field produced by a current-carrying wire. Tounderstand Ampere’s Law, we will use a very simple geometry.Let a uniformly distributedcurrentI flow inside a cylindrical conductor with cross-section surfaceSas shown below :

I

S

C

Let a closed contourC surround the conductor as shown in the figure. The magnetic fieldH is related to currentI by Ampere’s Circuital Law:

The Line Integral of the Magnetic Field Intensity (H) around aClosed Path is equal to the current enclosed by the path.

A mathematical statement of qmpere’s Circuital Law isIC

Hdl = Ienc (13)The left side of (13) is a line integral around the closed contourC and the right side is thecurrent enclosed byC. If the contourC encloses the entire wire thenIenc= I . The value ofH is found by applying (13) for the specific geometry of the current.


MAGNETIC FIELD DIRECTION : QUALITATIVE

One useful result of the magnetic field definition in equation(13) occurs for a very longstraight conductor. Suppose a time-invariant (DC) currentflows in a very long conductororiented with thez-axis as shown below. This current produces a Time-Invariant magneticfield both inside and outside the conductor. Unlike the electric field lines, which are in adirection radial from charges, the magnetic field lines formclosed paths. The direction ofthe magnetic field outside the conductor is given by the Right-Hand Rule, as shown in thefigure below:

I

lines of HCLOSED

Ampere’s Circuital Law also holds for the magnetic field inside the conductor, but rightnow we are primarily interested in the field outside the conductor. In the figure above,notice that theH field lines extend out into the space surrounding the conductor. Thismeans the magnetic field shown can affect current flowing in conductors which are nearby.To determine such an effect, we need quantitative expressions forH, which are examinedon the following pages.


LONG CYLINDRICAL CONDUCTOR

Ampere’s Circuital Law in equation (13) can be used to obtainthe magnetic field due tocurrent in a very long conductor. Let the uniformly distributed DC currentI be flowing ina cylindrical conductor (radiusa) aligned in thez-direction as shown below:

x

y

zI

φ

a

Let a pointP be defined in cylindrical coordinates as being at radial distancer from thecenter of the conductor and have angleφ with respect to thex-axis. Then we can useAmpere’s Law to show the magnetic field atP due to this current is given by

H =8>>><>>>: I r2πa2 uφ; 0< r a

I2π r

uφ; r > a

(14)The problem below gives a derivation of the magnetic field outside the conductor, which isthe bottom relation in (14). The top relation in (14) can be derived via a similar process.

Problem: 13-3


MAGNETIC FIELD DIRECTION: QUANTITATIVE

While in general the direction of the magnetic field follows the right-hand rule, in practicewe often have to compute this direction. Let’s examine a specific case to understand moreabout how to compute the direction of the magnetic field. Assume the following: Let a currentI = 0:2 amps be in the+y-direction (directed into thexz-plane) at the

point (x;z) = (2;1). Find the magnetic fieldH at the pointP= (x;z) = (5;2).The intercept of the current with thexz-plane is shown in the figure below:

q

pI=0.2

X

P = (x,z)=(5, 2)z=2

H

.

xx=5x=2

z=1

z

There are three basic steps to finding the direction of theH vector in the figure above: Find the vectorp linking the location of the current (at the point(x;z) = (2;1)) andthe pointP where we want to find theH field. Find the vectorq which is orthogonal to the vectorp. Use the right-hand rule to determine whetherH point “up” onq or “down” q.

In the figure we have used the right hand rule to qualitativelyshow the direction ofH. Onthe following page we will quantitatively determine theH field.


MAGNETIC FIELD DIRECTION: QUANTITATIVE (cont.)

The vectorp is the vector which begins at the location of the current and ends atP. Fromthe figure on the preceeding page this vector is

p = (52)ux+(21)uz = 3ux+uz (15)To find q, which is a vector orthogonal top, observe that ifq is orthogonal top, then thedot product ofp andq is zero:

p q = 0 (16)The general form for the vectorq is given by

q = qxux+qzuz (17)whereqx andqz are unknown scaling factors which must be determined. Substituting (15)and (17) into (16) then gives the following:

p q = 3ux+uz

qxux+qzuz

= 0 (18)Computing the dot product in (18) and solving for the relation betweenqx andqz then gives

3qx+qz = 0 ) qz = 3qx (19)Let the unit vector in the direction ofq be denoted byua. Therefore we know that

q2x+q2

z = 1 (20)Substitute (19) forqz into (20), giving

q2x+(3qx)2 = 1 ) qx = 0:316 (21)


MAGNETIC FIELD DIRECTION: QUANTITATIVE (cont.)

Note that (21) gives two possibilities forqx : one pointing to the right( +0:316) and onepointing to the left( 0:316). Examining the figure for this example, the right-hand ruleshows thatH must point “down” and “to the right”. Therefore, we choose the positive (“tothe right”) value forqx in (21):

qx = 0:316 (22)Substituting (22) forqx into (21) then shows thatqz is given by

qz = 3(0:316) = 0:948 (23)Since we know thatua is in the same direction asuH , then we also have

uH = qxux+qzuz (24)Finally, substitute (22) and (23) into (24) to obtain desired the unit vectoruH :

uH = 0:319ux0:948uz (25)Compare the quantitative result of (25) with the qualitative direction ofH shown in theearlier figure. You will see that the result in (25) does indeed “point in the correct direction”.


MAGNETIC FIELD MAGNITUDE

Now let’s find the magnitude for theH-field in the previous example. First, observe that thepoint P= (x;y) = (5;2) is outside of the current carrying conductor. Letd be the distancebetween the location of the current at(x;y) = (2;1) and the pointP= (5;2). The magnitudeof H, denoted byH, is then found using the bottom relation in (14) :

H = I2πd

(26)From the problem statement and figure we know thatI = 0:2 andd=p10= 3:16. Substi-tuting these values into (26) and simplifying then gives theresult

H = 0:01 (27)Finally, using (25) and (26) then gives the result for the vector H-field at pointP :

H = H uH = 0:01

0:319ux0:948uz

(28)

Problems: 13-4, 13-5, 13-6


MAGNETIC FIELD: MULTIPLE CURRENTS

There are many applications where multiple conductors in close proximity carry multiplecurrents at the same time. Some common applications where this occurs are: Parallel data transfer in digital systems Simultaneous telephone voice conversations in the cable bundle outside the residence Simultaneous DSL/voice transmission in the cable bundle outside the residence

We can understand the effect of the magnetic field on adjacentconductors by using a simpleexample. Let multiple currentsIA; IB; and IC be positioned parallel to they-axis, and beflowing in they-direction, as shown in the figure below.

IA

IB

ICHB

HC

HA

y

z

x

x = 1

x = −1

z = 1

x = 0

Suppose we want to find the total magnetic field at the pointP = (0;0;1) above thexy-plane. In the figure above you can use the right-hand rule to qualitatively determine thedirection of the magnetic fields at this point due to each current. If we recognize thatHA,HB, andHC are all in thexz-plane, then we can simplify the problem to a two-dimensionalfigure as shown on the following page.


MAGNETIC FIELD: MULTIPLE CURRENTS (cont.)

SinceHA, HB, andHC are all in thexz-plane, then the fields and currents from the previouspage can be simplified to the two-dimensional figure as shown below :

IC IB IA

X X X

HA

HB

HC

z = 1

x = 0 x = 1x = −1

The currentsIA, IB, andIC are all into the page, and the directions of the individual magneticfields are found using the right-hand rule. The total magnetic field atz= 1 is the sum of thethree individual magnetic fields.

Therefore, finding the magnetic field due to the multiple currents involves finding each ofHA, HB, andHC and then adding them together to find the totalH. Finding the magnitudesof the individualHi is straightforward. However, finding the direction of theseindividualHi is a little harder. You might want to review Problem 13-4 and 13-5 for how to find thesedirections.

On the next page we will complete the numerical example for computing the total magneticfield at the pointP= (0;0;1).


LINEAR FIELD EXAMPLE

EXAMPLE: Find theH-field at the point(x;y;z) = (0;0;1) if IA = IB = IC = 1:0 in thefigure on the previous page.

APPROACH: The totalH-field at pointP is H = HA+HB+HC where the subscripts referto the current elements. All currents are “into the page”, that is, in the+y-direction.

HB: Using the right-hand rule shows the direction ofHB to beux. The magnitude is

HB = IB2πr

= 12π(1) = 0:159

HA: The right-hand rule shows the direction ofHA to beuA = .707ux + .707uz.(See also Prob. 13-8 for more information.) The magnitude is

HA = IA2πr

= 1

2πp

2= 0:113

HC: The right-hand rule shows the direction ofHC to beuC = .707ux :707uz.(See also Prob. 13-8 for more information.) The magnitude is

HC = IA2π r

= 1

2πp

2= 0:113

The total magnetic field is thus

H = 0:159ux+0:113(:707ux+ :707uz)+0:113(:707ux :707uz)Adding the vector terms in the above equation then gives

H = 0.319ux


ECE 303 - Sum10 Notes Set 14 :Curl and the Magnetic Field 1

INTRODUCTION

We have previously studied the integral form of Ampere’s Circuital Law. This integral formis very descriptive and is useful for magnetic field computations in simple geometries. Forexample, the integral form lets us find the magnetic field due to a cylindrical conductor.

However, there is an equivalent differential form, or pointform, which is very useful inmore general environments and in our eventual use of Maxwell’s Equations. In theory, thisdifferential form allows us to compute the magnetic field at any point in space. Using thedifferential form requires us to understand a new quantity,called Current Density, whichhas a value at every point in space.

Stokes’ Theorem is the mathematical tool which we use to obtain the differential form ofAmpere’s Law from its integral form. Therefore, Stokes’ Theorem is similar in function tothe Divergence Theorem, which can be used to obtain the differential form of Gauss’ Lawfrom its integral form. This is similar to the manner in whichwe used Gauss’ Law to findthe time-invariant electric field at any point in space.

Another reason for studying curl relations concerns deriving computational results fromMaxwell’s equations. Expressing Maxwell’s equations in their differential form allows usto theoretically compute the time-varying electric and magnetic fields at any point in space.We will see that the spatial differential operators in Maxwell’s equations are defined by thecurl operator. Therefore, familiarization with the curl will be very valuable in later workwhen we study Maxwell’s equations.


CURRENT DENSITY

To derive the differential (point) form for Ampere’s Law, weneed to understand the CurrentDensity, which is analogous to charge density. For example,we can think of a region ofcharge (Coulombs) as being distributed over a volume (with units m3). At any point withinthe region, there is a charge densityρ having units of Coulombs/m3. We can similarlythink of the total currentI (in Amps) passing through a surface (having units m2). At anypoint on the surface, there is a current densityJ having units of Amps/m2. This allowsconsideration of currents which vary as a function of space.

The current densityJ is sometimes called the Conduction Current density, because it isdue to the actual physical movement of charge carriers (likeholes or ions or electrons).These charges are usually resident in conductors. Just as there is a “macroscopic” relationbetween current and voltage (Ohm’s Law), there is an equivalent “microscopic” relationbetween electric fieldE and the vector current densityJ. Recall that Ohm’s Law relates thescalar voltageV and scalar currentI :

I = 1R

V (1)Now consider the units on both sides of equation (1),

Amps=

1ohms

( volts) = voltsohms

(2)Divide the units of each side of (2) by m2: Amps

m2

= volts

ohms - m2

= 1

ohms - m

volts

m

(3)Observing the units in (3), we see that Amps/m2 is the unit of current density. We alsosee that 1/ohm-m is the unit of conductivity, and volts/m is the unit for electric field. Thesymbols for these quantities areJ for current density,σ for conductivity, andE for theelectric field. Therefore, equations (1) and (3) show that the relation between the vectorelectric fieldE and the vector current densityJ is given by

J = σE (4)The result from (4) is useful for quantifying properties at points within the conductor.


CURRENT DENSITY AND CURRENT

We frequently use current when we are more interested in the macroscale properties of adevice. An example is circuit analysis, in which we study a circuit to determine its voltageand current properties. In circuits the conductors carrying the current are not consideredrelevant to the problem. However, there are many applications where the dimensions andthe geometry of the conductors are very important. One application is when the currentdensity is spatially varying, as in “high-frequency” applications using metallic conductors.The general case thus requires that we consider current density, rather than current.

Obtaining the current from a current density is like obtaining the mass of an object if youknow its mass density: You would “sum up” the specific mass densities over the spatialvolume of the object. In a similar manner the total current ina conductor can be found by“summing up” all the individual current densities passing through the cross section of theconductor. To see how this is done, consider the cylindricalconductor in thez-directionshown in the figure below. The current densityJ = J(x;y) flows only inside the conductorand the surfaceS is a cross section of the conductor:

S

J

= inside conductor

J current densityJ

J

The total current in the conductor is the sum of all the current densities flowing throughthe surfaceS. “Summing up” this current density is done by a Surface Integral, which isexamined on the following pages.


SURFACE INTEGRAL

Let’s examine the surface integral using a simple Cartesianexample. Let a specific vector

field be given byG = ux+1:5uy+2uz. Evaluate the surface integralZ

SG dS over the

rectangle in thexy-plane having corners at the points(x;y) = h (0;0); (5;0); (5;3); (0;3)i.APPROACH:

Draw the geometry of the fieldG and the surfaceS:

( 5 , 3 )

Surface S

Gy

z

x( 5 , 0 )

( 0 , 3 )

Sd

For simplicity in the above figure,G anddS have only been shown at one point. However,keep in mind that that bothG anddS exist over the entire surfaceS. This means thatGdShas a value over the entire surface.

Direction of a Surface

Note that the surface elementdS is a vector and therefore has a “direction”. One way ofdenoting this direction is

dS = dSuN (5)whereuN is the unit vectorNormal to the surface element anddS is the scalar surfacedifferential element. Since the rectangle in the above figure is in thexy-plane, the surfaceelement in (5) becomes

dS = dSuz= dx dyuz (6)


SURFACE INTEGRAL (cont.)

Note that we could also have chosenuN in theuz direction. This will only change thesign of the surface integral. Now compute the dot product:

G dS = hux+1:5uy+2uz

i dxdyuz= 2dxdy (7)Substitute (7) into the expression for the surface integraland integrate over the dimensionsfor x andy shown in the previous figure:Z

SG dS = Z x=5

x=0

Z y=3

y=02dxdy= 2 x

5x=0

y3y=0

= 2(5)(3) = 30

In summary, evaluating a surface integral has two steps: Compute the scalar dot product. Evaluate the integral over the limits defining the surface.

NOTE: If G anddS are constant with respect to space over the surfaceS thenZS

GdS = ZS

G uN dS = G uN

ZS

dS (8)Evaluating the last integral on the right hand side of (24) then givesZ

SGdS = ( G uN ) A = AjG j cosθ (9)

whereA on the right hand side of (9) is the area of the surfaceSandθ is the angle betweenthe surface normal andG.. These results sometimes simplify computing surface integrals.


CIRCULAR CROSS SECTION

Consider again the current densityJ = J(x;y) flowing through the cross section surfaceS. Using the surface integral, we therefore see thatItot, the total current flowing in theconductor, is given by

Itot = ZS

J dS (10)If the integration in (10) is performed in cylindrical coordinates, thendS must be expressedin cylindrical coordinates. A circular cross section of a conductor with radiusa is shown inthe figure below:

x−axisdr

r = across−section

circular

surface S

z−axisy−axis

φr d

uN

φ angle

The unit vectoruN in the figure is orthogonal to the surfaceS. Keep in mind that a unitvectoruN exists for every point on the surface, but for simplicity only one is shown in thefigure. The computation ofdS in the figure thus becomes

dS = dSuN = rdφ dr uz (11)As shown in the figure,dr is the infinitesimal length along ther-axis andrdφ is the in-finitesimal length along theφ angle.



AMPERE’S LAW AND CURRENT DENSITY

Ampere’s Circuital law may now be restated to incorporate current density. We will usethe same integral form which we used previously. Let the current densityJ flow inside acylindrical conductor with cross-section surfaceSas shown below :

I

S

C

The current densityJ inside the conductor creates the total currentI as shown in the figure.Let a closed contourC surround the conductor as shown in the figure. The magnetic fieldH is related to currentI by Ampere’s Circuital Law:

The Line Integral of the Magnetic Field Intensity (H) around aClosed Path is equal toIenc, the current enclosed by the path.

But now we know that the current enclosed by the path is related to the surface integral.Therefore, the mathematical statement of Ampere’s Circuital Law becomesI

CHdl = Ienc = Z

SJdS (12)

The left side of (12) is a line integral around the closed contour C and the right side is asurface integral over the surface enclosed byC. The new information in (12) is the relationof magnetic fieldH to current densityJ.


STOKES’ THEOREM

Having developed the concepts of current density and surface integral, we can now apply avector identity to derive the differential (point) form forAmpere’s Law. This vector identityis called Stokes’ Theorem and it is very important in many areas of applied mathematics.

Let G be a general vector field which intersects a surfaceS. Stokes’ Theorem relates aproperty ofG occuring inside the surface with a property ofG occuring on the boundaryof the surface. This surfaceS is simply a mathematical “shape” inside a boundary curveC which encloses the surface. The mathematical shape might beflat or have curvature.Stokes’ Theorem states that the following surface integraland line integral are equal:Z

S(∇ X G) dS = I

CG dl (13)

In (13), the left side is a surface integral and the right sideis a line integral. The term“∇ X G” on the left side integral in (13) is the Curl ofG. The circle and the subscript “C”on the right side integral in (13) means this line integral isevaluated around the Contour(line) which encloses the surfaceS.

Equation (13) is a general property for any general vector field. With G replaced by thespecific magnetic fieldH, equation (13) becomesZ

S(∇ X H) dS = I

CH dl (14)

Now recall the integral form of Ampere’s circuital law from (12), repeated here for refer-ence: I

CHdl = Z

SJdS (15)

whereJ is the current density intersecting the surface. The derivation of the point form forAmpere’s Law is completed on the next page.


AMPERE‘S LAW: POINT FORM

To complete the derivation of the point form for Ampere’s Law, substitute equation (15)for theH line integral into (14). This process gives the following:Z

S(∇ XH) dS = Z

SJdS (16)

Since equation (16) must hold for any arbitrary surfaceS, limiting arguments would showthat an equivalent requirement is that the integrands of (17) must be equal. That is, we havethe condition

∇ XH = J (17)Equation (17) is the differential point form for Ampere’s Circuital Law. Since this relationholds at every point in space it is often also called the pointform of Ampere’s Law.

Observe that equation (17) is a vector partial differentialequation. If the right-hand side(the current densityJ) is given, then spatial integrations may be performed to determine themagnetic fieldH. Equivalently, if we knowH, then performing the curl (spatial derivatives)will provide theJ which produced the magnetic field.

Note that (16) and (17) have been derived independent of coordinate system, and thereforehold for Cartesian, Cylindrical, and Spherical coordinates. On the next few pages we willstudy some specific examples for Cartesian and Cylindrical coordinates.


CURL COMPUTATIONS

Since the point-form of Ampere’s Circuital Law uses the curl, let’s develop some familiaritywith this operation. The formulas for curl in Cartesian and Cylindrical coordinates aregiven below:

CARTESIAN:Let G(x;y;z) = Gx ux+Gy uy+Gz uz. In Cartesian coordinates the curl ofG is:

∇ X G = ux uy uz

∂∂x

∂∂y

∂∂z

Gx Gy Gz

= ux

∂Gz

∂y ∂Gy

∂z

+uy

∂Gx

∂z ∂Gz

∂x

+uz

∂Gy

∂x ∂Gx

∂y

(18)CYLINDRICAL:Let F(r;φ;z) = Fr ur +Fφ uθ+Fz uz. In cylindrical coordinates the curl ofF is:

∇ X F = 1r

ur r uφ uz

∂∂r

∂∂φ

∂∂z

Fr rFφ Fz

= ur

1r

∂Fz

∂φ ∂Fφ

∂z

+uφ

∂Fr

∂z ∂Fz

∂r

+uz

1r

∂∂r(rFφ) ∂Fr

∂φ

(19)We will not use the spherical curl in this course, but spherical curl is very important in thestudy of power radiated from antennas.


CURL EXAMPLE: CARTESIAN

Let’s work an example in Cartesian coordinates for computing the curl. Let the vector fieldG be given by

G(x;y;z) = 3zux+2x uy+y2 uz (20)Equation (20) shows that thex, y, andz components ofG are given by

Gx = 3z ; Gy = 2x ; Gz= y2 (21)Substitute the relations in (21) into equation (18) for the curl in Cartesian, giving

∇ X G = ux uy uz

∂∂x

∂∂y

∂∂z

3z 2x y2

(22)Evaluating the terms in the “determinant-like” structure of (22) then gives

∇ X G = ux

∂∂y

y2 ∂∂z

2x

+ uy

∂∂z

3z ∂∂x

y2 + uz

∂∂x

2x ∂∂y

3z

(23)Finally, evaluating the partial derivatives in (23) gives the desired form

∇ X G = 2y ux + 3 uy + 2 uz (24)Problems: 14-3, 14-4


CURL EXAMPLE: CYLINDRICAL

Now let’s work an example in cylindrical coordinates for computing the curl. Let the vectorfield F be given in cylindrical coordinates. by

F(r;φ;z) = 3rz ur + 2r

uφ+ r φ uz (25)Equation (13) shows that ther, φ, andz components ofG are given by

Fr = 3rz ; Fφ = 2r

; Fz= r φ (26)Substitute the relations in (26) into equation (19) for the curl in cylindrical, giving

∇ X F = 1r

ur r uφ uz

∂∂r

∂∂φ

∂∂z

3rz r ( 2r) rφ

(27)Note the term “

1r

” outside of the determinant in (27). This means do all of the partial

derivatives inside the determinant first, and then multiplythe results by1r

. Therefore,

evaluating the terms in the “determinant-like” structure of (27) then gives

∇ X F = ur

1r

∂∂φ

rφ ∂∂z

2r

+ uφ

∂∂z

3rz ∂∂r

rφ

+ uz1r

∂∂r

r ( 2r) ∂

∂φ3rz

(28)


CURL EXAMPLE: CYLINDRICAL (cont.)

Finally, evaluating the partial derivatives in (28) gives the desired form

∇ X F = ur + (3r φ) uφ + (0) uz (29)Since theuz term in (29) is zero, it is easier to just omit this term in the expression. There-fore, (29) may be written in the simpler form as

∇ X F = ur + (3r φ) uφ (30)



CURL OF H AND CURRENT DENSITY

Sometimes a simple way of computing the curl ofH is to instead compute the currentdensityJ. Recall that time-invariant current densities and magnetic fields satisfy the pointform for Ampere’s Circuital Law, repeated here for reference:

∇ X H = J (31)Equation (31) shows that knowing ofJ immediately gives the curl ofH. For example, let acylindrical conductor have current in thez-direction. A cross section of the cylinder in thez= 0 plane is shown below, in which the shaded area signifies presence of current density :

JJ = 0 = 0outside conductor :

inside

x−axis

y−axis

cylinder :conducting

Using (31) we now have the equivalent figure in terms of curl ofH :

HXHX = 0

outside conductor :inside

J =

x−axis

y−axis

cylinder :conducting

Therefore, the curl ofH is found by computingJ instead. The problems listed below giveyou some practice using this method.


ECE 303 - Sum10 Notes Set 15 :Time-Varying Magnetic Fields 1

INTRODUCTION

As we have progressed in our study of electromagnetics, we have given increasing degreesof freedom to the electric charge. Recall that first we required the charges to be stationary,in which case the only field effect was an electric field quantified by Coulomb’s Law. Next,we allowed the charges to move with constant velocity, whichproduced a time-invariantmagnetic field which was quantified by Ampere’s Circuital Law. The magnetic field pro-duced by these time-invariant currents can vary as a function of space, but can not vary asa function of time.

In this set of Notes we remove the constraint of time-invariant current. A time-varyingcurrent does indeed produce a time-varying magnetic field inspace. However, we willshow later that a time-varying current can also produce a magnetic field and electric fieldwhich can propagate like a wave through space.

In this set of Notes we concentrate on the effects of the time-varying magnetic field with-out investigating the wave nature of the propagating energy. We will study Faraday’s Law,which describes how we can transform the electromagnetic field energy back into a voltageand/or current. In a later Notes Set we will study Maxwell’s Equations, which do incorpo-rate the propagation characteristics of the electromagnetic wave.


MAGNETIC FLUX

A magnetic field changing in time can induce a voltage and current in a circuit. This isquantified by Faraday’s Law, which is the main topic of this set of notes. The mathematicsof Faraday’s Law can be simplified by using the magnetic flux density,B(t). This magneticflux density is related to the magnetic field intensity,H(t), by

B(t) = µH(t) (1)whereµ is the magnetic permeability in Henries/meter of the medium. For free space thevalue is

µ= µ0 = 4π(10)7 (2)In equation (1) we have now allowed for the possibility of a time-varying magnetic field.Units of these quantities are:

H(t)) Ampsm

; B(t)) Webersm2 ; µ) Henries

m(3)

The Magnetic Flux,ΦM(t), over a surfaceS is the surface integral given by

ΦM(t) = ZS

B(t)dS (4)The surface integral in (4) is computed using the techniquesof surface integrals, which wecovered in a previous set of notes. Evaluation forΦM(t) in equation (4) is perhaps easier ifwe divide the problem into two steps: Compute a scalar dot productB(t)dS. Evaluate an integral over the limits defining the surface.

Note that if B = B(t) is time-varying, thenΦM = ΦM(t) is time-varying. An exampleof computing a constantΦM is shown on the next page, after which we will compute atime-varyingΦM(t).


MAGNETIC FLUX EXAMPLE

Let’s work an example in which we compute the time-invariantmagnetic fluxΦM over asurface. Let a time-invariant currentI = 0:3 be flowing in thez-direction as shown in thefigure below. A metallic conductor in free space forms a closed path over the rectanglewith corners at(x;y;z) points (3, 0, 2), (6, 0, 2), (6, 0, 7), and (3, 0, 7). ComputeΦM

passing through the rectangle if the rectangle surface normal uN = uy.

uy

x=3 xx=6

z=7

z=2

z

I = 0.3

B = Binto page

In the above figure they-axis is into the page. Let’s first evaluate the dot product requiredin equation (4) for the magmetic flux.

Dot Product :

Since the surfaceS is a rectangle in thexz-plane then we have the following fordS:

dS = dSuy = dx dzuy (5)We see that the loop is outside the current-carrying wire andwe need only the formula forB outside a current carrying wire. From our previous work, we know that the cylindrical-coordinate expression for theH-field produced outside a long, straight wire is

H = I2π r

uφ (6)Comparing (6) with the figure above we see that in the plane of the loop we have

r = x ; uφ = uy (7)The example is continued on the next page.


MAGNETIC FLUX EXAMPLE (cont.)

Substituting (7) into (6) and using the free-space resultB = µ0 H then gives

B = µ0 I2πx

uy (8)Now substitute (5) and (8) into the formula for the dot product required in (4) :

BdS = µ0 I2πx

uy dx dzuy (9)Performing the dot product in (14) then gives the desired result for the dot product:

BdS = µ0 I2πx

dxdz (10)Evaluation of Integral :

From the figure, we see that the surface is the rectangle in thexz-plane spanning 3 x 6and 2 z 7. Substitute these limits into the limits of integration required by (4) and alsosubstituteBdS given by (10) into (4). This gives

ΦM = Z 7

z=2

Z 6

x=3

µ0 I2πx

dx dz= µ0 I2π

Z 7

z=2dz

Z 6

x=3

dxx

(11)UseI = 0:3 from the problem statement and evaluate the integrals in (11). This gives

ΦM = 0:75µ0

πln x

6x=3

(12a)Useµ0 = 4π(10)7 and evaluate the expression in (12a), providing the desiredresult

ΦM = 2:08(10)7 (12b)Problems: 15-1 through 15-4


FARADAY’S LAW

The previous pages have constrained the magnetic field in space to be time-invariant. How-ever, a new and extremely important phenomenon occurs when the magnetic field becomestime-varying. Faraday’s Law states that a time-varying magnetic field causes a voltage (anda current) to be induced in a closed path. A mathematical statement of Faraday’s Law is

vind(t) = ddt

ΦM(t) (13)whereΦM(t) is the magnetic flux passing through the loop. The geometry ofof Faraday’sLaw is shown below:

uN(t)B

vind (t)

+

x

y

z

−

In the example in the figure, the time-varying magnetic field intersects a circular loop inthexy-plane. Two things to keep in mind are: ΦM(t) must be time-varying; otherwise there is no induced voltage. The closed path (loop, rectangle, etc.) must have non-zero area; otherwise there is

no induced voltage.

On the following page we will work an example which demonstrates the computation of thetime-varying magnetic flux. After that we will give an example which shows how Faraday’slaw computes the voltage induced by the changing magnetic flux.


TIME-VARYING FLUX EXAMPLE

Here is an example in which the flux densityB(t) is time-varying, in which case.ΦM(t) isalso time-varying. Let a magnetic flux density given by

B(t) = 109cosω t uz (14)exist in a region of free space. A metallic conductor in this region forms a closed path overthe rectangle with corners at the four(x;y;z) points (100, 4, 0), (105, 4, 0), (105, 14, 0),and (100, 14, 0). ComputeΦM(t) passing through the rectangle if the rectangle surfacenormaluN = uz. This geometry is shown below, in which thez-axis is out of the page:

...x=105 xx=100

into pageu

y=14

y = 4

y

y=14

y = 4

z= B(t)Bout of page

x=105x

x=100

...

zu= B(t)By

Note from (14) that the direction of the magnetic field changes depending on the polarityof the cosine. The figure on the left above is for the case of cosω t negative (field into thepage) and the figure on the right above is for the case of cosω t positive (field out of thepage). However, notice the surface does not change orientation and therefore the surfaceelementdS is given by

dS = dSuz = dx dyuz (15)As before, we will first compute the dot product and then evaluate the surface integral.


TIME-VARYING FLUX EXAMPLE (cont.)

Dot Product :

Substitute (14) and (15) into the expressionB(t) dS, producing

B(t) dS = 109 cosω t uz dx dyuz (16)Computing the dot product operation on the right of (16) thengives

B(t) dS = 109 cosω t dx dy (17)Evaluation of Integral :

Substitute (17) into (4) and integrate over the dimensions of the rectangle in thexy-plane:

Φ(t) = Z 14

y=4

Z 105

x=100109 cosω t dx dy (18)

Since cosω t is not a funvtion of space it can be factored out of the integral, producing

Φ(t) = 109cosω tZ 14

y=4dy

Z 105

x=100dx (19)

Evaluating the integrals in (19) then gives the desired answer

Φ(t) = 109cosω t105= 5(10)8 cosω t (20)Problem: 15-5


VOLTAGE COMPUTATION

To compute the voltage induced in a loop, we use the mathematical statement of Faraday’sLaw given by equation (13). In the following example we will use the same geometry asin the previous example. Let’s assume that the magnetic flux is given by equation (20) andwe want to find the voltage induced in the loop shown below:

zu= B(t)B

−+..

y=14

y = 4

y

...x=105 xx=100

From equation (20) we found the time-varying flux to be

ΦM(t) = 5(10)8 cosω t (20)Therefore, substituting (20) into (13) shows the induced voltage is given by

vind(t) = 5(10)8 ddt

cosω t = 5(10)8ω sinω t (21)For example, ifω = 106 rad/sec, the induced voltage would bevind(t) = 0:5 sin106 t.



FARADAY’S LAW : DIFFERENTIAL FORM

The integral form of Faraday’s Law given by (13) is very descriptive and is useful forcomputations in simple geometries. Equation (13) is repeated here for easy reference:

vind(t) = ddt

ΦM(t) = ddt

ZS

B(t) dS (22)where we have also expressed the magnetic fluxΦM(t) as a surface integral. In equation(22), S is the surface enclosed by the closed path (the circuit) which inducesvind(t). Wemay move the derivative on the right hand side of (22) inside the integral by transformingit to a partial derivative:

ddt

ZS

B(t) dS = ZS

∂B(t)∂t

dS (23)Substituting (23) into (22) thus gives

vind(t) = ZS

∂B(t)∂t

dS (24)Now recall that a voltage, such asvind(t), is the line integral of an electric fieldE(t) overa path. Faraday’s law requires that the “line” form a closed path (or closed contourC).Therefore, since the fields are “doing the work” we have

vind(t) = IC

E(t) dl (25)where the circle on the integral in (25) signifies the closed path which encloses the surfaceS. Equating (25) and (24) then givesI

CE(t) dl = Z

S

∂B(t)∂t

dS (26)We complete the derivation of the differential form of Faraday’s Law on the following page.


DIFFERENTIAL FORM (cont.)

Stokes’ Theorem now states that the line integral ofE(t) on the left side of (26) is equiva-lent to a surface integral of the curl ofE(t) :I

CE(t) dl = Z

S

∇ X E(t) dS (27)

Substituting equation (27) in equation (26) then givesZS

∇ X E(t) dS = Z

S

∂B(t)∂t

dS (28)Note that both sides of (28) are now surface integrals over the same surfaceS. Now let thesize of the surfaceS shrink to zero. Since (28) must hold for any arbitrary surface, thenlimiting arguments would show that (28) is valid if the left side terms in the dot productsare equal; that is, if

∇ X E(t) = ∂B(t)∂t

(29)Finally, substitueB(t) = µH(t) into (29), giving

∇ X E(t) = µ∂H(t)

∂t(30)

Equation (30) is the point (or differential) form for Faraday’s Law. We will use the form(30) when we develop Maxwell’s Equations.

ECE 303 - Sum10 Notes Set 16 :Maxwell’s Equations 1

INTRODUCTION

In electromagnetics, there are four vector partial differential equations which are collec-tively known as Maxwell’s Equations. Three of them are similar to equations we havepreviously studied under other names. The derivation of thefourth equation requires alittle more investigation, which is one of the topics of thisset of notes.

To derive this fourth equation we will need another basic property relating charges andcurrent density. This property is called the continuity equation. For brevity we will notderive the continuity equation, although it is certainly within our mathematical skill levelat this time. We will, however, outline the derivation stepsin class discussions.

In this set of Notes, we will largely discuss qualitative properties of Maxwell’s equations,and will defer specific solution of the equations until laterNotes Sets. However, there is agreat deal of insight we can derive from these qualitative studies. Additionally, recallingthe relation between excitation and response from ordinarydifferential equations will helpus understand the implication of Maxwell’s equations.


DISCUSSION OF PREVIOUS RESULTS

We have previously examined several different phenomena concerning electric and mag-netic fields. For easy reference, we will list below the differential form for some of theimportant relations we have found. We will also include a brief discussion of the physicalmeaning of each law.

GAUSS’ LAW:

From Notes Set 10, we learned that the differential form, or point form, of Gauss’Law is given by

∇ E = ρε

(1)whereρ is the charge density at the point andε is the electric permittivity. In NotesSet 10 we also examined the physical interpretation of the divergence in (1) above,and learned that non-zero divergence ofE implies the existence of electric charges.Depending on the sign ofρ on the right-hand side of (1), one of three possibilitiesemerges:

ρ> 0 :If, at a specific point, we haveρ > 0, then the right hand side of (1) means there isa positive charge at that point. The left hand side side of (1)then means thatE-fieldlines are directed outward through a small closed surface surrounding the charge.Therefore, we can think of positive charges as Sources ofE-field lines.

ρ< 0 :If ρ< 0, the right hand side of (1) means there is a negative charge at that point. Theleft hand side side of (1) then means thatE-field lines are directed inward througha small closed surface surrounding the charge. Therefore, we can think of negativecharges as Sinks, or Terminations, ofE-field lines.

ρ = 0 :If ρ = 0, the right hand side of (1) means there is no charge at that point. The lefthand side side of (1) then means that allE-field lines which enter a small surfacesurrounding the point will exit that surface (perhaps on theother side).


DISCUSSION OF PREVIOUS RESULTS (cont.)

DIVERGENCE OF MAGNETIC FIELD:

Recall the results we found for the magnetic field lines due toa constant current in aconductor. Notes Set 12 showed that external to the conductor the lines of magneticfield H formed closed circles around the conductor. There was no beginning andno termination to the magnetic field lines. Therefore, for any small closed surface (asmall volume element) in the space surrounding the conductor, we have the result thatanyH-field line that enters the small volume also exits the small volume. But this isexactly the qualitative description of the mathematical condition that the divergenceof the lines must be zero. Consequently, we arrive at the second divergence equation:

∇ H = 0 (2)Although we have discussed the situation outside of a conductor (because it is rela-tively easy to envision), equation (2) also holds inside theconductor. Equation (2)states that magnetic field lines have no beginning and no end.Another implication isthat there is no magnetic counterpart to the electric charge. (Recall that electric fieldlines begin on positive electric charges and terminate on negative electric charges).Isolated magnetic “charges” in this sense do not exist.

FARADAY’S LAW:

We found in Notes Set 14 that a time-varying magnetic field could induce a voltageand current in a closed conducting path, i. e., a circuit witha load. An easy tounderstand form of Faraday’s Law for this induced voltage was given in Notes Set14, repeated here:

vind(t) = ddt

ΦM(t) (3)whereΦM(t) is the magnetic flux, which is related to the magnetic flux density (andmagnetic fieldH ) through a surface integral. Recall in Notes Set 9 that a voltage canbe expressed as a line integral involving the electric fieldE. Therefore, as shown inNotes Set 14, we see that the left side of (3) is a function of the electric fieldE andthe right side of (3) is a function of magnetic fieldH.


DISCUSSION OF PREVIOUS RESULTS (cont.)

In Notes Set 14 a vector identity called Stokes’ Theorem was then then used to pro-duce the differential for Faraday’s law:

∇ X E =µ∂H∂t

(4)The differential form in (4) is not as physically descriptive as the form in (3). How-ever, equation (4) is the form frequently given in Maxwell’sequations, especiallywhen we wish to actually compute the electric and magnetic fields at a point in space.

AMPERE’S CIRCUITAL LAW :

In Notes Set 12 we studied the time-invariant magnetic field produced by a time-invariant current. In that Notes Set we outlined the derivation of the differential formfor Ampere’s Circuital Law, repeated below:

∇ XH = J (5)In the form given in (5),J is the current density in a conductor. For this reason,J issometimes called the conduction current density. Additionally, a constraint on (5) isthatJ be time invariant.


SUMMARY OF EQUATIONS

The equations discussed in the material on the previous pages are summarized below:

∇ E = ρε

(6)∇ H = 0 (7)

∇ X E =µ∂H∂t

(8)∇ X H = J (9)

As given above, these field equations are very general and express PDEs having time andspatial variables. Recall that to obtain useful resuls we previously had to examine spe-cific applications and strict geometries. Examples of this were using Gauss’ Law in one-dimension to find the electric field for the PN-junction and using Ampere’s Circuital Lawto find the magnetic field in the vicinity of a long, straight DC-carrying conductor.

In addition, the field equations (6) - (9) above reflect the experimental nature of early elec-tromagnetic science. It was often the case that experimental effects were observed and themathematical laws governing the physical principles were discovered later.

However, there is one inconsistency in the above equations,specifically equation (9). Re-moving the inconsistency using mathematical techniques led to some very important pre-dictions concerning physical properties of electromagnetic phenomena. This is a casewhere the mathematical work preceded the experimental results. On the following pageswe will examine this inconsistency and study one approach for resolving it.


THE CONTINUITY EQUATION

When we deal with the differential form of equation (9), we see that “current” is representedby current densityJ. Current densityJ (in units of amps/m2) is related to currentI (in unitsof amps) by the following surface integral:

I = ZS

J dS (10)Therefore, you can see the formulas which involveJ are actually related to current, eventhough we specifically callJ the current density.

A very important equation called the Continuity Equation relates the current densityJ tothe charge densityρ by the following:

∇ J = ∂ρ∂t

(11)This equation has a very intuitive physical interpretation, which we will now obtain. Theoperation on the left side of (11) is the divergence of the current density. Let’s examinewhat (11) implies about the charge inside a very small closedsurface, that is, inside a verysmall volume. If there is non-zero divergence of current density (that is, if∇ J 6= 0 ),then current must be flowing across the surface of the very small volume.

Since current is flowing across the surface, this means the charge density must be changinginside the volume of the closed surface. This in turn means wemust have the time derivativeof ρ inside the surface is not equal to zero. Further, the minus sign on the right side of (11)states that if current flows outward through the closed surface (positive divergence) thenthe charge density within that closed surface must decrease(negative time derivative).

Although there are many interesting implications of (11), our current interest is how therelation in (11) affects the field equations (6) - (9). We willnext see how (11) leads toan inconsistency in Ampere’s Circuital Law listed in (9). Onthe following pages we willoutline an approach for removing the inconsistency. This will result in the set of equationscalled Maxwell’s Equations.



A VECTOR IDENTITY

Equation (11) incorporates the divergence of current density; the right hand side of Am-pere’s Circuital Law in (9) incorporates the current density. Therefore, let’s take the diver-gence of both sides of (9) and investigate the resulting equation. Taking the divergence ofboth sides of (9) produces

∇ ∇ X H= ∇ J (12)

By the continuity equation in (11), the left side of (12) should be the negative of the firstderivative of charge density. Therefore, substituting (11) into (12) gives

∇ ∇ X H

?=? ∂ρ∂t

(13)Is the relation in (13) true? (We have used the question marksin (13) to denote the tentativenature of this conclusion.) Let’s see if (13) is true.

Here is an vector operation identity from the area of mathematics called vector analysis.For any arbitrary vector fieldA the following equation is true:

∇ ∇ X A= 0 (14)

Equation (14) holds for any vector field and holds regardlessof the coordinate system.Proving (14) requires more mathematics than we have had so far in this course, but showingthat (14) is true is straightforward. Lengthy, but straightforward. (You are asked to do thisin Cartesian coordinates in Problems 16-7 through 16-9.)

If we substitute the magnetic fieldH into the relation given by (14) we get

∇ ∇ X H= 0 (15)

But the relation (15) contradicts (13). (Equation (15) turns out to be the true relation.)Therefore, we have an inconsistency which we must resolve.

Problems: 16-7, 16-8, 16-9


THE FINAL EQUATION

Here is one approach to resolving the inconsistency createdby equation (9). Rewrite thecontinuity equation of (11) as

∇ J+ ∂ρ∂t

= 0 (16)Since the right side of (15) is zero, substitute (16) into theright side of (15), obtaining

∇ ∇ X H= ∇ J+ ∂ρ

∂t

(17)Recall now Gauss’ Law relating theE-field and charge density:

∇ E = ρε

(18)Solving (18) for charge density then gives

ρ = ε

∇ E (19)

Substituting (19) forρ into (17) and rearranging then gives

∇ ∇ X H= ∇ J+ ε

∂∂t

∇ E

(20)The two divergences on the right side of (20) can be rearranged to give

∇ ∇ X H= ∇ J+ ε

∂∂t

E (21)

Using limiting arguments, it is straightforward to show that equation (21) is valid if thequantities inside the parentheses are equivalent; that is,if

∇ X H = J+ ε∂E∂t

(22)


MAXWELL’S EQUATIONS

Equation (22) is therefore the appropriate form for the curlof H. In words, equation (22)says a magnetic field (H) can be created by either a conduction current densityJ or a time-varying electric field (the first derivative ofE term). This time-varyingE-field term hassubstantial implications which we will explore in the next set of Notes.

For easy reference, the four field equations are listed below. Collectively, they constitutethe set of equations called Maxwell’s Equations:

∇ E = ρε

(23a)∇ H = 0 (23b)

∇ X E =µ∂H∂t

(23c)∇ X H = J+ ε

∂E∂t

(23d)In the next set of Notes we will solve Maxwell’s equations fora specific set of conditions.We will see that the set of conditions we choose leads to the phenomena of a propagatingelectromagnetic wave. The problems listed below give you some practice in working withcurl differential equations.


ECE 303 - Sum10 Notes Set 17:Plane Wave Propagation 1

INTRODUCTION

In the preceding set of Notes we introduced Maxwell’s equations. In the form in whichwe saw them they are a very general set of equations, and applyto Cartesian, Cylindrical,and Spherical coordinate systems. However, to gain useful results we frequently have toformulate a problem in a particular coordinate system. It isat this point we have to evaluatethe necessary partial derivatives and then solve the remaining partial differential equation.

In this set of Notes we examine one of the simpler geometries and physical situations.Specifically, we want to find the electric field in a non-conducting medium, at a pointwhich is “far” from any current sources. Although at first these constraints might seemto be somewhat limiting, this situation accurately describes the electric field geometry inbroadcast television, broadcast radio, cell phones, wireless internet access, and many othercommunications environments.

The reason for enforcing these constraints is that it greatly simplifies the mathematics ofsolving Maxwell’s equations. We will be able to follow the solution process all the wayfrom the initial statement, using vector derivative operators, to a scalar wave equation whichwe have already solved. Therefore, we will be able to use our prior knowledge of the waveequation to infer properties of the electromagnetic field ina three-dimensional space.


NON-CONDUCTING MEDIUM

Let’s investigate solving Maxwell’s curl equations for theE andH fields in a non-conductingregion (a medium) having zero charge density (ρ= 0). This accurately describes free-spaceand links used in broadcast TV, radio, cell phones, and wireless internet access.

For easy reference, here are the two curl equations from Maxwell’s equations listed in thepreceding Notes Set:

∇ X E =µ∂H∂t

(1)∇ X H = J+ ε

∂E∂t

(2)If the medium is non-conducting, then it has conductivityσ = 0. SinceJ = σE, thisimpliesJ = 0. Therefore, usingJ = 0 in (2) gives the following two equations relatingEandH:

∇ X E =µ∂H∂t

(3)∇ X H = ε

∂E∂t

(4)To find the electric and magnetic field, we have to solve equations (3) and (4) forE andH.Even though the equations look very complicated, they are actually two equations in twounknowns. Thus, our approach is to manipulate equations (3)and (4) via math and vectoroperations to produce one equation in one unknown. This one equation might be a vectorpartial differential equation, and it might be difficult to solve, but it will be one (vector)equation in one (vector) unknown.


SOLUTION APPROACH

To derive the desired equation, first take the curl of both sides of the equation (3):

∇ X ( ∇ X E ) =µ∂∂t

( ∇ X H ) (5)Substitute equation (4) into the right-hand side of equation (5):

∇ X ( ∇ X E ) =µ∂∂t( ε

∂E∂t

) =µ ε∂2E∂t2 (6)

A VECTOR IDENTITY :There is a general vector identity which will simplify equation (6). Given a generalvector fieldG, the following is true:

∇ X ( ∇ X G ) = ∇ ( ∇ G ) ∇2 G (7)where∇2 G in (7) is called thevectorLaplacian of the fieldG.

With the general vectorG in (7) replaced by the specific electric field vectorE we have

∇ X ( ∇ X E ) = ∇ ( ∇ E ) ∇2 E = ∇ ( ρε) ∇2 E (8)

But in the non-conducting medium we have assumed that the charge densityρ = 0. Sub-stitutingρ = 0 into (8) gives the following result:

∇ X ( ∇ X E ) = ∇2 E (9)Substituting (9) into (6) then gives

∇2 E = µ ε∂2E∂t2 (10)

Since∇2 E is the vector Laplacian (instead of the scalar Laplacian), equation (10) needssome further explanation.


VECTOR LAPLACIAN

Suppose the electric field in equation (10) is given in Cartesian coordinates by

E = Ex ux+Ey uy+Ez uz

Then the Vector Laplacian is given by

∇2 E = ( ∇2 Ex ) ux+( ∇2 Ey ) uy+( ∇2 Ez ) uz (11)Each of the three scalar Laplacians in equation (11) has three terms:

∇2 Ex = ∂2Ex

∂x2 + ∂2Ex

∂y2 + ∂2Ex

∂z2 (12a)∇2 Ey = ∂2Ey

∂x2 + ∂2Ey

∂y2 + ∂2Ey

∂z2 (12b)∇2 Ez = ∂2Ez

∂x2 + ∂2Ez

∂y2 + ∂2Ez

∂z2 (12c)SIMPLIFICATION 1: Suppose thatEx andEy are zero; that is

E = Ez uz= Ez(x;y;z; t) uz (13)Substituting the simplication (13) into equations (10) and(11) provides

∇2 Ez = µ ε∂2Ez

∂t2 (14)Substituting (12c) into (14) then gives the following

∂2Ez

∂x2 + ∂2Ez

∂y2 + ∂2Ez

∂z2 = µ ε∂2Ez

∂t2 (15)Equation (15) is like the wave equation we saw when we studiedtransmission lines. Wewill use another simplification to make equation (15) easierto visualize.


PLANE WAVE GEOMETRY

SIMPLIFICATION 2: Suppose thatEz has no variation in thex or thez direction; that is

∂Ez

∂x= ∂Ez

∂z= 0 (16)

The simplification in (16) leads to the field geometry below:

x

plane plane

z

y

E E

y = y2y = y1

The figure above shows thatEz may only vary as a function ofy. Oncey is fixed, theEz

is fixed, regardless of the coordinates ofx or z. This configuation for the electric field iscalled a Plane Wave.

The plane wave approximation is very widely used in communications applications. If areceiver is located many wavelengths away from the transmitting antenna, then the receivedelectromagnetic field is essentially a plane wave.


WAVE EQUATION SOLUTION

How do these simplifications affect the mathematical solution for theE-field? Substitutingequation (16) into equation (15) shows that

∂2Ez

∂y2 = µ ε∂2Ez

∂t2 (17)The partial derivative on the left hand side of (17) shows that Ez can only vary as a functionof y andt. Furthermore, the following two properties are very important: If time t is fixed, thenEz varies only as a function of the positiony. If positiony is fixed, thenEz varies only as a function of timet.

Equation (17) is the one-dimensional Wave Equation, just like we studied in lossless Trans-mission Lines. Therefore, the solution to (17) is known immediately to be

Ez(y; t) = g(t yvp) ; where vp = 1p

µ ε(18)

In equation (18),vp is the velocity of the wave and “g” is any realizable function.

EXAMPLES: Let β = ω=vp. The following are solutions of the wave equation given byequation (17): Ez(y; t) = Ez(y; t) = A ej(ω tβy) (19) Ez(y; t) = A cos(ω tβy) = RefEz(y; t)g (20)

Note that there is no attenuation of the amplitude in either of the above solutions. In actualapplications, the waveEz-field will experience an attenuation due to Spherical Spreading.We will not address that phenomenon in this course.



MAGNETIC FIELD SOLUTION

We now have a solution which shows theE-field to be a propagating wave. How do wedetermine theH-field? One method is to substitute our solution forE into Maxwell’sequations and solve forH. Let’s choose the simple one-dimensional exponential wavesolution from (19). This represents a wave having only auz-component and propagating inthey-direction:

E = Ae j (ω tβ y ) uz= Ez uz (21)SubstituteEz from equation (21) into equation (3), giving

∇ X E = ux uy uz

∂∂x

∂∂y

∂∂z

0 0 Aej(ω tβy) = µ

∂H∂t

(22)Evaluating the curl on the left side of (22) then provides

∇ X E = ux

h j βAej(ω tβy)i= j β Ez ux (23)Substituting (23) into (22) thus gives j β Ez ux =µ

∂H∂t

(24)The left side of (24) shows that the time-varying component of H has only aux component.Thus, we need only solve for thisHx component. Also, note that theH-field is orthogonalto both theE-field and the direction of propagation. The ODE for theHx-component isobtained from (24) as

dHx

dt= jβ

µEz (25)

Equation (25) says thatHx is porportional toEz. We may now solve forHx using solutiontechniques for Ordinary Differential Equations, as outlined in the problems below.



INTRINSIC IMPEDANCE

There is another approach to solving for the magnetic field which emphasizes a fundamen-tal property of the medium. Again letEz have the form from equation (21). However, nowassume that the magnetic fieldHx is related toEz by just an unknown scaling factorη:

Hx = Ez

η= A

ηej(ω tβy) (26)

Note the similarity of equation (26) to Ohms Law:I = VZ

. Compute the time-derivative of

(26) and substitute the result into equation (25), giving

j βAej(ω tβy) = j ωµAη

ej(ω tβy) (27)Solving for the unknownη then gives

j βA= j ωµ Aη

) η = ωµβ

(28)Use the definitions ofβ = ω=vp andvp = 1=pµε to simplify η in (28):

η = ωµβ

= vpµ= µpµ ε

(29)Performing a final simplification of (29) then gives the desired result:

η =rµε

(30)Since the units ofη are in Ohms,η is called theintrinsic impedanceof the medium. In freespace,ε0 = 8:85(10)12 andµ0 = 4π(10)7. These values gives the following importantresult for free-space:

η 377 Ohms (31)Problem: 17-5


FIELD DIRECTIONS

The previous example illustrates a general property: the electric field propagates with themagnetic field. The propagating electric field can not exist without a propagating magneticfield. Thus, the phenomenon is called an Electro-Magnetic wave. Here are some importantproperties for propagating electromagnetic waves in a lossless medium: TheE-field andH-field are orthogonal to each other. Both fields are orthogonal to the direction of propagation ofthe wave. The magnitude of the magnetic field is porportional to the magnitude of the electric

field: H = E=η.

The figure below illustrates the propagation of the real-valued fieldsE= Acos(ω tβy) uz

andH = Aη

cos(ω tβy) ux. These waves are propagating in the+y-direction.

x

y

z

H

E

E = A t − y uω βcos( ) z

direction ofpropagation

H ω t βyA_η

u= cos( − ) x

Problem: 17-6


POWER DENSITY

We have now found that theE andH fields of an electromagnetic wave are orthogonal tothe direction of propagation. There is another field which has a component aligned withthe direction of propagation. This field is the instantaneous power density fieldP, which iscomputed using the cross-product:

P = E X H (32)The Cross-Product operation has been studied previously. Since the result of a cross-product is orthogonal to each vector contained in the cross-product, then (32) gives theresult thatP is orthogonal to bothE andH and is therefore in the direction of the wavepropagation. Since the units ofP are in Watts/m2, the vectorP is sometimes called thePower Density vector. The figure below displays the relationbetweenP, E, andH:

E

H

P

Let the vectors in the figure above be given byP= P uP, E = E uE, andH = H uH . SinceP, E, andH are all mutually orthogonal, this gives the following useful properties:

uE X uH = uP (33a)uH X uP = uE (33b)uP X uE = uH (33c)

Equations (33a) - (33c) are very important. They state that knowledge of the directions ofany two ofP, E, or H allows us to compute the direction of the third.



POWER DENSITY EXAMPLE

As an example, consider an electromagnetic wave propagating in a lossless medium. Sup-pose the real-valued electric field associated is

E = 2cos(ω tβy) uz (34)The argument of the cosine in equation (34) tells us the EM-wave is propagating in the+y-direction. The directions ofE andP now fix the direction ofH. Using equation (33c)we see that sinceuP X uE = uH , theH-field must be in theux-direction.

To find the magnitude ofH, remember thatH = Eη

. Thus,H = 2377

0:0053. This gives

theH-field asH = 0:0053 cos(ω tβy) ux (35)

To compute the numerical value forP, use the cross-product from (32):

P = E X H = ux uy uz

0 0 2cos(ω tβy)0:0053 cos(ω tβy) 0 0

(36)Evaluating this cross-product then gives

P= 0:0106 cos2(ω tβy) uy (37)This last result shows that propagating power flows in theuy direction. Note that the waveitself propagates in this same direction. Also note thatP in (37) is still a function of time.Therefore,P is sometimes called theInstantaneouspower density.



TIME-AVERAGE POWER DENSITY

The power density vectorP in equation (37) can be written in a form which separates itsmagnitude and direction:

P = P(t) uP (38)whereuP is a unit vector in the direction of power density flow andP(t) is the scalaramplitude of the time-dependent power density. This direction uP is also the direction ofpropagation of the electromagnetic wave. In general,P(t) will also be a function of space,as well as time.

The scalarP(t) in (38) is aninstantaneousfield, meaning it varies over timet and space.We are frequently interested in the time-average power density Pavg, given by

Pavg = Pavg uP (39)Note thatPavg points in the same direction as power density. However, the scalar amplitudePavg now is not a function of time and is given by the average ofP(t) over one period ofthe sinusoid:

Pavg= 1Tp

Z Tp

t=0P(t) dt (40)

whereTp is the time period of the sinusoidal frequency. In the complex-valued phasordomain, an equivalent expression for the vectorPavg in (39) is given by the complex-domaincomputation

Pavg = 12

Ren

E X Ho (41)Equation (41) is similar to the expression we found in previous work for the time-averagepower flowing on a transmission line. However, in equation (41), Pavg is a Power Densityvector and has units of Watts/m2, instead of watts.


TIME-AVERAGE POWER DENSITY : EXAMPLE

As an example, let’s reconsider the real-valued example from the previous pages. Thecomplex-valued electric field corresponding to the real-valuedE in (34) is given by

E = 2ej(ω tβy) uz (42)The complex-valued magnetic field corresponding to the real-valuedH in (35) is given by

H = 0:0053ej(ω tβy) ux (43)Now compute the cross-product required in (41) :

E X H = ux uy uz

0 0 2e j (ω tβy)0:0053e j( ω tβy) 0 0

= 0:0106uy (44)Substituting (44) into equation (41) then gives the desiredresult :

Pavg = 12

Ren

E X Ho = 0:0053uy (45)

ECE 303 - Sum10 Notes Set 18 :Wave Propagation in Lossy Media 1

INTRODUCTION

In previous work we studied the propagation of electromagnetic waves in free-space. Wemodified the general problem to examine a wave propagating ina region where it could beconsidered a plane wave. We found that this simplified situation led to the wave equation,which showed that plane wave experienced no attenuation as it propagates.

In actual applications, we would find that there is an attenuation due to spherical spreading,in which the transmitted power is distributed over a larger surface as the wave propagates.However, this spherical spreading is not a function of the non-conducting region throughwhich the wave is propagating.

In constrast to this lossless propagation, waves propagating in conducting media do expe-rience a loss in amplitude. For this reason, these media are sometimes called lossy media.We will study this case in detail in this set of Notes. We will find that the wave propaga-tion characteristics in a lossy media are substantially different from waves propagating innon-conducting media or free-space.


MAXWELL’S EQUATIONS: Time-Domain

The beginning point is again Maxwell’s equations. However,now the medium in which thewave is propagating is a lossy medium; that is, the medium hasa non-zero conductivity.Let this non-zero conductivity beσ in units of Siemens/meter. Ifσ > 0, then the currentdensityJ in the conducting medium isJ = σE. Therefore, Maxwell’s two curl equationsare as previously given :

∇ X H = σE + ε∂E∂t

(1)∇ X E =µ

∂H∂t

(2)We do not need to solve (1) and (2) using the detailed approachwe used previously. In fact,we can use the form of the plane wave solution and propose a similar form for the solutionsto (1) and (2) above. Therefore, let’s assume the following phasor solutions forE andH:

E = F(x;y;z) e j ω t uE (3a)H = 1

ηF(x;y;z) e j ω t uH (3b)

whereF(x;y;z) is a scalar function which is not dependent on timet, andη is a possiblycomplex constant signifying a possible phase difference betweenE andH. The unit vectorsfor E andH areuE anduH . Specifiying the unit vectors separately allows for the possibilitythatE andH are in different directions.

Here is a point to realize when we ”assume a solution form”: Ifthe form of the assumedsolution is “wrong”, then a mathematical inconsistency will appear in the suceeding anal-ysis. We saw this earlier when we needed to make a modificationto Ampere’s CircuitalLaw, such that it was consistent with the Continuity Equation.


MAXWELL’S EQUATIONS: Phasor-Domain

The solution forms assumed in equations (3) have separated the solution into a time-domainpart (e j ω t) and a spatial-domain partF(x;y;z). In this approach, the time-domain part isknown once we specifiy the frequencyω. Therefore, we only have to solve Maxwell’sequations (2) for the spatial solutionF(x;y;z).Since the forms for the solutions in equation (3) have time-domain phasors, then thetime-derivatives are easy to compute. Therefore, using (3)we have the following time-derivatives:

∂E∂t

= j ω F(x;y;z) e j ω t uE = j ω E (4)∂H∂t

= j ω1η

F(x;y;z) e j ω t uH = j ω H (5)Equations (4) and (5) show one of the benefits of working with the complex-valued phasors.Specifically, time-differentiation translates into aj ω multiplier.

Now substitute (4) and (5) into (1) and (2), respectively. This gives the following equationsfor the phasor fieldsE andH:

∇ X H = (σ + j ω ε) E (6)∇ X E = j ω µH (7)

These last two equations must now be solved simultaneously for E andH. We will do thison the next page.


SOLUTIONS

We will again limit our study to plane waves propagating in a conducting medium. Let theE-field be in thez-direction and let the wave be propagating in they-direction. A solutionprocess similar to that for the non-conducting medium (likefree-space) in Notes Set 16would show that the following are the phasor solutions for the E andH fields:

E = Aeγ yejω t uz (8a)H = A

ηeγ yejω t ux (8b)

In equations (8a) and (8b) above,γ is called the propagation constant for the medium andη is called the intrinsic impedance. The solution process would show that the fields havethe relatively simple forms given in (8a) and (8b), providedγ andη are computed by thefollowing expressions:

γ =p jω µ (σ+ jω ε) (9a)η =s jω µ

σ+ jω ε(9b)

The quantities in (9) are similar to the related quantities of propagation constant and charac-teristic impedance we obtained for transmission lines. Thequantitiesγ andη are complexnumbers and are very useful when written as

γ = α+ j β (10a)η = jη j e j φ (10b)


SOLUTIONS (cont.)

Using (10a) and (10b) forγ andη in (8a) and (8b) forE andH then gives the followingphasor solutions for the fields:

E = A e α y e j ( ω tβ y ) uz (11a)H = Aj η jeα y e j ( ω tβ yφ ) ux (11b)

We notice that the solutions in (11) are somewhat different from the solutions for free-spacepropagation we obtained in Notes Set 16. Specifically, two properties are immediatelyevident from equations (11a) and (11b): The amplitude of bothE andH attenuates as the wave propagates into the medium. The phase of theH-wave differs from the phase of theE-wave by the angle of theintrinsic impedance.

These two properties directly affect the power that an incident wave can carry into a con-ducting medium. We will explore this phenomenon in the remainder of this Notes Set.

Problems: 18-1, 18-2, 18-3


E-FIELD EXAMPLES

Suppose a medium existing overy 0 has the parameters(σ;µ;ε) = (4;µ0;72ε0). Let awave with f = 5 MHz have the form

E = A e α y e j ( ω tβ y ) uz (12)as it propagates in this medium. Suppose the following boundary condition is known:

E(y= 0; t = 0) = 10uz (13)Do the following: (a) ComputeA, (b) Find the direction of propagation, (c) Computeγand write it in the formγ = α+ j β, (d) Write the expression for the real-valuedE-field.

(a) ComputeA.

To computeA, substitutey= 0 andt = 0 into (12) and equate the result to the bound-ary condition given in (13):

E(y= 0; t = 0) = A eα ( 0 ) e j ( ω ( 0 )+β ( 0 ) ) = 10 (14)The exponentials on the left side of (14) evaluate to unity and thus we find that

A= 10 (15)(b) Find the direction of propagation.

Consider the exponential in (12), given bye j ( ω tβ y ). Since the spatial argumentin this exponential is given byβy, then the propagation of the wave is in the+ydirection (rightward direction).


E-FIELD EXAMPLES (cont.)

(c) Computeγ and write it in the formγ = α+ j β.

Equation (9a) gives the expression for the propagation constant, repeated here forreference:

γ =p j ω µ ( σ+ j ω ε ) (16)Some of the computations required in (16) are shown below:

j ω µ= j π 1074π 107 = j 39:48 (17a)σ+ j ω ε = 4+ j 107 π 72(8:85)1012= 4+ j 0:02 (17b)

Substituting (17a) and (17b) into (16) then gives

γ =r j 39:48

4+ j 0:02 (18)

It is straightforward to show that (18) is equivalent to

γ = 12:58e j 45o = 8:89+ j 8:89 (19)The rectangular form forγ on the far right side of (19) then shows that

α = 8:89 and β = 8:89 (20)(d) Write the expression for theE-field.

Take the real part of (12) to obtain the expression for the real-valued field:

E = A e α y cos(ω tβ y) uz (21)Substituting the computed parameters from (15) and (20) into (21), the form formthe real-valued field becomes

E = 10e 8:89y cos(107π t8:89y) uz (22)Problems: 18-4, 18-5


H-FIELD EXAMPLES

Now let’s work some examples for determining the magnetic field. For simplicity, we willuse some of the results of the previous example for the electric field. Suppose a mediumexists overy 0 and has(σ;µ;ε) = (4;µ0;72ε0). Suppose anE-field propagating to theright in this medium.has the form

E = 10e 8:89y cos(107π t8:89y) uz (23)Do the following: (a) Compute the direction ofH, (b) Compute the value of the intrinsicimpedanceη, (c) Compute the real-valuedH field.

(a) Compute the direction ofH.

From (23), we see theE-field of the wave is in theuz direction. Equation (23) alsoshows the wave is propagating in the+y direction. Therefore power densityP alsoin the+y direction, and thus has auy unit vector. In equation form we now have

uE = uz ; uP = uy (24)The direction ofH is found using the cross product relations given in Notes Set16:

uE X uH = uP (25)Substituting (24) into (25) then gives

uz X uH = uy (26)Equation (26) is satisfied if

uH = ux (27)(b) Computeη.

Equation (9b) gives the expression for the intrinsic impedance, repeated here forreference:

η =s j ω µσ+ j ω ε

(28)


H-FIELD EXAMPLES (cont.)

Some of the computations required in (28) are shown below:

j ω µ= j π 107 4 π 107 = j 39:48 (29a)σ+ j ω ε = 4+ j 107 π 72(8:85)1012= 4+ j 0:02 (29b)

Substituting (29) into (28) then gives

η = sj ω µ

σ+ j ω ε= s

j 39:484+ j 0:02

(30)Simplifying (30) then gives the desired form forη :

η = j η j e j φ = 3:14e j 45o (31)(c) Compute the real-valuedH field.

With the direction of the magnetic field given from (27) asuH = ux, we can useequation (11b) to write the phasorH as follows:

H = Aj η je α y e j ( ω tβ y φ ) ux (32)Substitute (15) forA, (31) for j η j andφ, and (20) forα andβ into (32), giving

H = 3:18e 8:89y e j ( 107π t8:89y 45o) ux (33)Thus, the real-valuedH-field is given by taking the real part of (33):

H = 3:18e 8:89y cos(107π t8:89y 45o) ux (34)Problem: 18-6

ECE 303 - Sum10 Notes Set 19 :Wave Reflection and Transmission 1

INTRODUCTION

This set of Notes considers the case when waves propagating in one medium are incidentupon the boundary containing another medium, Some of the incident wave energy may betransmitted into the new medium and some energy may be reflected back into the originalmedium. Common examples of this are the electromagnetic waves broadcast for television,radio, cell phone, and wireless internet applications.

In many of these examples, a broadcast antenna sends electromagnetic waves out over aregion of space, but the waves do not have a great deal of directionality. Indeed, this is oneof the advantages of a broadcast medium: a user or subscriberis not limited to a specificfixed location. Think of using a cell phone in different places in a neighborhood or campus:you can get a “good” signal in many places. Think of using the radio in a vehicle: you canget a particular station over a wide area.

However, this very advantage of the broadcast application leads to the following situation.The electromagnetic waves can be reflected from nearby structures, and this can lead tointerference effects. This interference occurs because there now may be multiple paths fromthe broadcast antenna to the receiving antenna. It is therefore possible for these multiplepaths to create “destructive interference” at the receiver.

This set of Notes examines wave reflection and transmission at boundaries. We will see theintrinsic impedance of the different media determine the degrees of reflection and transmis-sion. You are probably familiar with the wave reflection process in acoustics: it is calledan echo. We will see the electromagnetic situation is very similar in concept to the acousticecho.


TWO-MEDIA BOUNDARY

This page lists the properties of the two adjoining losslessmedia which are important forunderstanding wave transmission and reflection. The figure below shows the geometry ofthe boundary, and properties of medium 1 and medium 2:

Medium 1

τΓ

Medium 2

2η

1η

2 2( µ , ε )11( µ , ε )

At the basic level, the lossless media are defined by their electric permittivity ε and mag-netic permeabilityµ. This is the top level of descriptors shown in the figure above. How-ever, we have seen that these basic descriptors define the intrinsic impedanceη of themedia. Therefore, if we know theεi andµi of theith medium, we can compute the intrinsicimpedanceηi of the ith medium using the previously obtained formula

ηi = rµi

εi(1)

This computation is represented by the downward arrow to theηi in the figure. The ar-rows down from theηi to the reflection coefficient,Γ, and the transmission coefficientτimply that Γ andτ can be computed from theηi . Reflection coefficient and transmissioncoefficient are explained on the next page.


TWO-MEDIA BOUNDARY (cont.)

The intrinsic impedanceη is a very important parameter for a medium, However, if weare interested primarily in wave transmission and reflection, then we can use theη to cal-culate even more useful parameters. These are the values of reflection coefficientΓ andtransmission coefficientτ shown in the last line of the figure on the previous page.

The theoretical computation ofΓ andτ proceeds from the basic property that the tangentialcomponent of theE-field must be continuous across a boundary. Enforcing this constraintand using limiting arguments gives the following computational formulas for reflectioncoefficientΓ and transmission coefficentτ :

Γ = η2η1

η2+η1(2)

τ = 2η2

η2+η1(3)

It is much easier to work reflection/transmission problems using reflection coefficientΓand transmission coefficentτ. On the following pages we will show how these parameterscan help us obtain the fields in the vicinity of a reflection/transmission boundary.

Suppose we have the case whereη1 andη2 are real-valued and positive. Then from (3)we see that the reflection coefficientΓ may be either positive or negative, depending onthe specific values ofη1 andη2. If Γ is negative this denotes a 180Æ phase change (a signchange) in the reflected waveER . However, note from (3) that sinceη2 can not be negative,then the transmission coefficientτ can not be negative. Therefore, there is no sign changebetweenEI andET as the electric field crosses the boundary between the media.

Problem: 19-1


NORMAL INCIDENCE

In this course we will only consider the case of Normal Incidence. This means the unitnormaluN of the reflecting boundary is orthogonal to the incidentE-field vector. For ourexample figure shown below, assume allE-fields are in theuz direction (up in the figures)and we have two media which are both lossless. Let the phasor form for the incident wavepropagating in medium 1 be given by

EI = A e j ( ω t β1 y) uz (4)The incidient wave is propagating in they-direction (to the right), as shown below:

Medium 1 Medium 2

Incident

Reflected

Transmitted

z

y

IE~

TE~

RE~

.x (out)

The reflected and transmitted waves shown in the figure are easily described using thereflection coefficientΓ and the transmission coefficientτ. These fields are given by

ER = A Γ e j ( ω t + β1 y) uz (5)ET = A τ e j ( ω t β2 y ) uz (6)

Note in (5) that the sign on the spatial phase termβ1y is now a positive sign. This means thereflected wave is propagating in they direction. Also note in (5) and (6) that the phaseconstantβ is a function of the particular medium. Specifically, let theith medium havevelocity of propagationvi = 1=pµi εi . Then we have

βi = ωvi

= ωp

µi εi (7)


TRANSMISSION EXAMPLE

Let’s work an example for finding theE field transmitted from free-space into a medium.Suppose an incident plane wave in free space propagates in the +y direction and has anelectric field given by

EI = 5e j ( 2π 109 tβ1 y )uz (8)whereβ1 is the phase constant in medium 1 (free space). This wave is incident on theboundary of a dielectric having permeabilityε = 10ε0 and permeabilityµ= µ0. Computethe real-valued fieldET transmitted into the dielectric.

APPROACH :

We will first find ET and then take the real part to findET . Here are some properties of theelectric field transmitted into the dielectric: The Hertz (or rad/sec) frequency of the electric field does not change as the wave

enters medium 2. ThusET has a Hz frequencyf = 109. The vector orientation of the electric field does not change as the wave enters themedium 2. ThusET has a unit vector in theuz direction. The direction of propagation does not change as the wave enters medium 2, but theβ constant does change. ThusET has a “β2 y” in its complex exponential. The amplitude ofET is the amplitude ofEI scaled by the transmission coefficientτ.

Using (6) and (8) and incorporating the above properties then gives the following forET :

ET = 5τe j ( 2π 109 tβ2 y )uz (9)where in (9) we will have to computeτ andβ2. We will do this on the next page.


TRANSMISSION EXAMPLE (cont.)

To computeβ2, substituteε2 = 10ε0 andµ2 = µ0 into equation (7):

β2 = ωp

µ2ε2 = 2π109q

8:85(10) 11 4π(10) 7 (10)Evaluating (10) then gives the following value forβ2 :

β2 = 66:3 (11)To computeτ we needη2. To computeη2, substituteε2 = 10ε0 andµ2 = µ0 into (1):

η2 = rµ2

ε2= s 4π(10) 7

8:85(10) 11 (12)Evaluating (12) then gives the following value forη2 :

η2 = 119:2 (13)Since medium 1 is free-space we therefore know thatη1 = 377. Substituting this value and(13) into (3) then gives the evaluation forτ:

τ = 2(119:2)119:2+377

= 0:48 (14)Substituting (11) and (14) into (9) then gives the phasor form for the transmitted field as

ET = 2:4e j ( 2π 109 t66:3 y )uz (15)Finally, taking the real part ofET in (15) then provides the real-valued field transmittedinto medium 2:

ET = 2:4 cos( 2π 109 t66:3 y )uz (16)Problem: 19-2


REFLECTION EXAMPLE

Now let’s examine the previous example and compute the electric field reflected back intomedium 1 (which is free-space in this example). Here are someproperties of the electricfield transmitted back into free-space: The Hertz (or rad/sec) frequency of the reflected electric field does not change as the

wave reflects. ThusER has a Hz frequencyf = 109. The vector orientation of the electric field does not change as the wave reflects. ThusER has a unit vector in theuz direction. The direction of propagation reverses as the wave reflects. Therefore,ER has a“+β1 y” in its complex exponential. The amplitude ofER is the amplitude ofEI scaled by the reflection coefficientΓ. IfΓ is negative, you can think of this a reversing the direction of ER. For this example,this meansER has a unit vector in theuz direction.

Using (5) and (8) and incorporating the above properties then gives the following forER :

ER = 5Γe j ( 2π 109 t + β1 y )uz (17)where in (17) we will have to computeΓ andβ1. To computeβ1, substituteε1 = ε0 andµ1 = µ0 into equation (7):

β1 = ωp

µ2ε2 = 2π109q

8:85(10) 12 4π(10) 7 (18)Evaluating (18) then gives the following value forβ1 :

β1 = 20:95 (19)This example is completed on the following page.


REFLECTION EXAMPLE (cont.)

Since medium 1 is free-space we know thatη1 = 377. Substituting this value forη1 andη2 from (13) into equation (2) then gives

Γ = 119:2377119:2+377

= 0:52 (20)Substituting (19) and (20) into (8) then gives the phasor form for the reflected field as

ER = 2:6e j ( 2π 109 t+20:95y )uz (21)Finally, taking the real part ofER in (21) then provides the real-valued field reflected backinto medium 1:

ER = 2:6 cos( 2π 109 t +20:95y )uz (22a)An alternative representation ofER in (22a) is to associate the minus amplitude with achange in direction of the unit vector. Therefore, moving the minus sign to the unit vectorlocation in (22a) thus gives

ER = 2:6 cos( 2π 109 t +20:95y ) [ uz (22b)Equation (22b) specifically denotes the “direction change”of the E-field vector as it isreflected back into medium 1. The direction has changed from “up” (+uz) for the incidentwave in medium 1 to “down” (uz) for the reflected wave. This representation will be veryhelpful in understanding the total fields throughout both media.

Problem: 19-3, 19-4


INTRINSIC IMPEDANCE: CONDUCTORS

Now let’s consider the case of finding the reflection from an ideal conductor. We will modelthis ideal conductor by letting conductivityσ!∞. (The conductivity for a good conductorlike copper isσ = 5:8(10)7 Siemens/m .) From Notes Set 17 we have seen that the intrinsicimpedance of a conductor is given by

η = s jω µσ+ jω ε

(23)Divide numerator and denominator of the intrinsic impedance in equation (23) byσ:

η = s jω µσ+ jω ε

= vuut jω µσ

1+ jω εσ

(24)Take the limit of this last form in (24) asσ! ∞ :

η = limσ!∞

vuut jω µσ

1+ jω εσ

= r0

1+0= 0 (25)

Therefore, the intrinsic impedance of a ideal conductor is zero. In practice, the conductiv-ity of actual conductors, like gold, silver, and copper, is frequently approximated as zero.However, this is not the case for semiconductors, like silicon and germanium. Semiconduc-tors have finite conductivities, and thus require that equation (23) be used for the intrinsicimpedance.

The relation in (25) is very similar to an approximation you used in circuits. When you an-alyzed circuits, you may have assumed the voltage drop alongthe wires was zero. This wasequivalent to assuming zero resistance in the wires, which is similar to assuming infiniteconductivity of the conductor at the boundary.

Problem: 19-5


REFLECTION AND TRANSMISSION: CONDUCTORS

The fact that good conductors have approximately zero intrinsic impedance has substantialimplication for the behavior of electromagnetic waves in the vicinity of conductors. Let’sdetermine how this effects the reflection and transmission coefficients for a ideal conductor.Consider again the figure on page 4 showing the boundary between two media. Let themedium 1 be free-space (η1 = 377) and medium 2 be an ideal conductor for whichη2 = 0:

η1 = 377; η2 = 0 (26)Now substitue the relations in (26) into equations (2) and (3) for the reflection coefficientΓ and the transmission coefficientτ, obtaining

Γ = 03770+377

= 1 (27)τ = 2(0)

0+377= 0 (28)

Equation (27) means all of the incidentE field energy is reflected back from the idealconductor and equation (28) means no energy is transmitted into the conductor. We willshow that the negative sign inΓ in (27) implies a phase shift ofπ-radians in the wavereflected back into the medium 1.

On the following pages we will solve for the total electric and magnetic fields at the bound-ary between free-space and the ideal conductor. Since the boundary coincides with thesurface of the ideal conductor then this is equivalent to solving for the fields on the surfaceof the conductor.

Problem: 19-6


ELECTRIC FIELD ON CONDUCTORS

You may be familiar with receiving antennas, such as those onportable televisions andradios. These antennas are frequently constructed from a good conductor (like a metal)which has many electrons in the conduction band. These electrons are very mobile withinthe metal conductor. Consequently, they may easily be made to move in response to exter-nal fields, like those in the incident electromagnetic wave.

We may understand the basics of how a current is generated in areceiving antenna bycomputing what occurs when an EM wave is incident on the surface of an ideal conductor.The material to follow uses the figure on page 4 for reference.You can refer to that figiurefor the field directions and propagation directions in the material below.

Let’s compute theE-field on the surface of the conductor. Refering to the figure on page 4,let an EM wave in free-space (medium 1) be propagating such that the incidientE-field isgiven by

EI = Ae j ( ω tβ1 y ) uz (29)Let this wave be normally incident on the surface of an ideal conductor occupying the re-giony 0. The reflectedE-field is given by equation (5), repeated here for easy reference:

ER = A Γ e j ( ω t + β1 y) uz (30)Since the boundary is the ideal conductor, substituteΓ = 1 from (27) into (30). Thisshows that the reflected electric field is given by

ER =A e j ( ω t + β1 y) uz (31)Although (31) is mathematically correct, there is another form which will later help us findthe direction of the reflected magnetic field. If we associatethe minus sign in (31) with theunit vector, we see that (31) is equivalent to

ER = A e j ( ω t + β1 y) [ uz (32)We continue this material on the following page.


ELECTRIC FIELD ON CONDUCTORS (cont.)

Now recall that any phasor electric field has the generic scalar-unit vector form given by

E = EuE (33)Comparing the reflected electric field from (32) to the generic form in (33) then gives thefollowing for the reflected electric field :

E = A e j ( ω t + β1 y) ; uE = [ uz (34a; b)Equation (34b) shows that the ideal conductor has “reversed” the vector direction of theelectric field in the reflected wave. Note also that (34a) shows the direction of propagationhas been reversed, as well.

The totalE-field in medium 1 is thus the sum of the incident and reflected fields:

E1 = EI + ER (35)Substituting (29) and (31) into (35) then gives

E1 = Ae j ( ω tβ1 y ) uz Ae j ( ω t+β1 y ) uz (36)Now evaluate (36) at the boundary (y= 0) to find the phasor electric field on the surface ofthe ideal conductor:

E1(y= 0) = Ae j ω t uzAe j ω t uz = 0 (37)Equation (37) tells us that theE-field on the surface of an ideal conductor is zero. Thisis similar to the circuits approximation where we assume there is no voltage loss on theconductors between circuit elements.


MAGNETIC FIELD ON CONDUCTORS

We have now found that the electric field on the surface of the ideal conductor is zero. Nowlet’s find the magnetic field on the surface. Like theE-field in medium 1, theH-field inmedium 1 is the sum of the incident and reflected magnetic fields:

H1 = HI + HR (38)Refering to the field directions and propagation directionsin the figure on page 4 willhelp us determine the directions ofHI and HR. Since we know the directions of theEcomponents and their propagation directions, we can use thefollowing relation from NotesSet 14 to find the directions of theH components:

uH = uP X uE (39)Equation (39) applies to both the incident and reflected components of the fields.

Direction of HI :

From equation (29) we see the the following properties for the incident waveEI :

uP = uy; uE = uz (40a; b)Substituting (40a) and (40b) into (39) then shows that the direction of the magnetic fieldfor the incident wave is given by

uH = uy X uz = ux (41)Using (41) thus gives the phasor form for the incident magnetic filed as

HI = EI

η1uH = A

η1e j ( ω tβ1 y )ux (42)

Direction of HR :

From equation (31) we see the the following properties for the reflected waveEI :

uP = [ uy ; uE = [ uz (43a; b)


MAGNETIC FIELD ON CONDUCTORS (cont.)

Substituting (43a) and (43b) into (39) then shows that the direction of the magnetic fieldfor the reflected wave is given by

uH = [ uy X [ uz = ux (44)Using (44) thus gives the phasor form for the reflected magnetic field as

HR = Aη1

e j ( ω t+β1 y )ux (45)Total Magnetic Field :

To find H1, the total magnetic field in medium 1, substitute (42) and (45) into (38), giving

H1 = Aη1

e j ( ω tβ1 y )ux + Aη1

Γe j ( ω t+β1 y ) ux (46)Evaluate (46) at the boundary (y= 0) to find theH-field on the surface of the ideal conduc-tor :

H(y= 0) = Aη1

e j ω t ux + Aη1

e j ω t ux (47)Simplifying (47) then gives the desired result for the phasor magnetic field on the surfaceof the ideal conductor :

H(y= 0) = 2Aη1

e j ω t ux (48)Taking the real part of (48) then gives the real-valued magnetic field on the surface of theideal conductor :

H(y= 0) = 2Aη1

cosω t ux (49)Equation (49) shows that the EM wave incident on the surfae ofthe ideal conductor has thuscreated a time-varying sinusoidal magnetic field on the surface of the conductor. You canthink of this time-varying magnetic field as “moving” the mobile charges available in theconduction band of the conductor. This constitutes a sinusoidal current on the conductorsurface.

If you think of theE andH plane waves as being “sheets” incident on the conductor, thenyou can think of the current as a “sheet” moving up and down in thez-direction on thesurface of the conductor.

ece 303 notes final

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