ELECTROMAGNETIC THEORY AND TRANSMISSION LINES … · 2 0 1 G[E = r' . 4 0 U & 2 0 S 1 dS E = r' . 4...

ELECTROMAGNETIC THEORY AND TRANSMISSION LINES

B.TECH IV SEM-ECE

Prepared ByDr.P.Ashok Babu,Professor

Mrs.A.Usha Rani,Assistant ProfessorMr.Murali Krishna, Assistant Professor

Dr.S.Pedda Krishna, Professor.

Introduction to Co-Ordinate System and

Electrostatics

UNIT -I

Spherical Coordinates

Cylindrical Coordinates

Cartesian Coordinates

P (x, y, z)

P (r, θ, Φ)

P (r, Φ, z)

x

y

zP(x,y,z)

Φ

z

rx

y

z

P(r, Φ, z)

θ

Φ

r

z

yx

P(r, θ, Φ)

Rectangular Coordinates

Or

X=r cos Φ,Y=r sin Φ,Z=z

X=r sin θ cos Φ,Y=r sin θ sin Φ,Z=z cos θ

• dx, dy, dz are infinitesimal displacements along X,Y,Z.

• Volume element is given by

dv = dx dy dz

• Area element is

da = dx dy or dy dz or dxdz

• Line element is

dx or dy or dz

Ex: Show that volume of a cube of edge a is a3.

P(x,y,z)

X

Y

Z

3

000

adzdydxdvV

aa

v

a

dx

dy

dz

Cartesian Coordinates

X

Y

Z

r

φ

Z

Cylindrical coordinate system

• dr is infinitesimal displacement

along r, r dφ is along φ and

dz is along z direction.


dv = dr r dφ dz

• Limits of integration of r, θ, φ

are

0<r<∞ , 0<z <∞ , o<φ <2π

Ex: Show that Volume of a

Cylinder of radius ‘R’ and

height ‘H’ is π R2H .φ is azimuth angle

X

Y

Z

rφ

r dφ

dz

dr

r dφ

dr

dφ

Cylindrical coordinate system

Differential quantities:

Length element:

Area element:

Volume element:

dzardadraldzr

ˆˆˆ

rdrdasd

drdzasd

dzrdasd

zz

rr

ˆ

ˆ

ˆ

dzddrrdv

Limits of integration of r, θ, φ are 0<r<∞ , 0<z <∞ , o<φ <2π

• dr is infinitesimal displacementalong r, r dθ is along θ and rsinθ dφ is along φ direction.


dv = dr r dθ r sinθ dφ

• Limits of integration of r, θ, φare

0<r<∞ , 0<θ <π , o<φ <2π

Ex: Show that Volume of a sphereof radius R is 4/3 π R3 .

P(r, θ, φ)

X

Y

Z

r

φ

θ

drP

r dθ

r sinθ dφ

r cos θ

r sinθ

Spherical Coordinate System

3

3

0 0

2

0

2

2

3

42.2.

3

sin

sin

RR

dddrr

dddrrdvV

R

v

Try Yourself:

1)Surface area of the sphere= 4πR2 .

Volume of a sphere of radius ‘R’

Spherical Coordinates: Volume element in space

System Coordinates dl1 dl2 dl3

Cartesian x,y,z dx dy dz

Cylindrical r, φ,z dr rdφ dz

Spherical r,θ, φ dr rdθ r sinθdφ

• Volume element : dv = dl1 dl2 dl3

• If Volume charge density ‘ρ’ depends only on ‘r’:

Ex: For Circular plate: NOTE

Area element da=r dr dφ in both the

coordinate systems (because θ=900)

drrdvQ

v l

2

4

Points to remember

6 - 12

The gradient is the closest thing to an ordinary derivative, taking a

scalar-valued function into a vector field.

The simplest geometric definition is “the derivative of a function with

respect to distance along the direction in which the function changes

most rapidly,” and the direction of the gradient vector is along that most-

rapidly changing direction.

The Gradient operator

ˆˆ ˆr e c ta n g u la r : = xx

1ˆˆ ˆc y lin d r ic a l: = r

1 1ˆ ˆˆs p h e r ic a l: = r

s in

y zy z

zr r z

r r r

g ra d ( ) ( )f r f r r r

6 - 13

Divergence in spherical coordinates

2

2

(s in )( )1 1 1

s in s in

rvvr v

d iv v vr r r r

r r

The coordinate system is orthogonal if the surfaces made by setting the

value of the respective coordinates to a constant intersect at right angles.

In the spherical example this means that a surface of constant r is a sphere.

A surface of constant θ is a half-plane starting from the z-axis. These

intersect perpendicular to each other. If you set the third coordinate, φ , to

a constant you have a cone that intersects the other two at right angles.

volumearea

6 - 14

Gauss’s Theorem

0 0

1 1lim limV V

d Vd iv v v d A

V d t V

rr r

Ñ

Fix a surface and evaluate the surface integral of over the surface:v

r

S

v d Arr

Ñ

Recall the original definition of the divergence of a vector field:

kVNow divide this volume into a lot of little volumes, with

individual bounding surfaces . If you do the surface integrals of over each of these pieces and add all of them, the result is the

original surface integral.

kS

v d A

rr

6 - 15

Stokes’ Theorem

The expression for the curl in terms of integrals is

Use exactly the same reasoning as that was used in the case of the Gauss’s theorem, this leads to

d A v

r rv r

1h A

1A

1nh

Let us first apply it to a particular volume, one that is very thin and small.

Take a tiny disk of height , with top and bottom area . Let be

the unit normal vector out of the top area. For small enough values of

these dimensions, is simply the value of the vector

inside the volume times the volume

16

Coulomb’s Law

• Coulomb‟s law is the “law of action” between charged bodies.

• Coulomb‟s law gives the electric force between two point charges in an otherwise empty universe.

• A point charge is a charge that occupies a region of space which is negligibly small compared to the distance between the point charge and any other object.

17

Coulomb’s Law

2

120

21

12

4ˆ

12

r

QQaF

R

Q1

Q212r

12F

Force due to Q1

acting on Q2

Unit vector in

direction of R12

18

Coulomb’s Law

• The force on Q1 due to Q2 is equal in

magnitude but opposite in direction to the

force on Q2 due to Q1.

1221 FF

19

Electric Field

• Consider a point chargeQ placed at the origin ofa coordinate system inan otherwise emptyuniverse.

• A test charge Qt broughtnear Q experiences aforce:

2

04ˆ

r

QQaF t

rQt

Q

Qt

r

20

Electric Field

• The existence of the force on Qt can be

attributed to an electric field produced by Q.

• The electric field produced by Q at a point in

space can be defined as the force per unit

charge acting on a test charge Qt placed at that

point.

t

Q

Q Q

FE

t

t0

lim

21

Electric Field

• The electric field describes the effect of a

stationary charge on other charges and is an

abstract “action-at-a-distance” concept, very

similar to the concept of a gravity field.

• The basic units of electric field are newtons

per coulomb.

• In practice, we usually use volts per meter.

22

Electric Field

• For a point charge at the origin, the electric

field at any point is given by

3

0

2

044

ˆr

rQ

r

QarE

r

23

Electric Field

• For a point charge located at a point P’

described by a position vector

the electric field at P is given by

rrR

rrR

R

RQrE

where

43

0

r

Q

P

r R

rO

24

Electric Field

• In electromagnetics, it is very popular to

describe the source in terms of primed

coordinates, and the observation point in terms

of unprimed coordinates.

• As we shall see, for continuous source

distributions we shall need to integrate over

the source coordinates.

Field due to Different Types of Charges

E = ?++

+

++ + ++ + +

+

Source

+

Continuous Charge Distributions

•Cut source into small (“infinitesimal”) charges dq

•Each produces

2

( )ˆ

e

d qd E k

r r

dE

r

dq

2e

d qd E k

r

or

Steps:•Draw a coordinate system on the diagram

•Choose an integration variable (e.g., x)

•Draw an infinitesimal element dx

•Write r and any other variables in terms of x

•Write dq in terms of dx

•Put limits on the integral

•Do the integral or look it up in tables.

Continuous Charge Distributions

2

0

1 λ d xE = r ' .

4 π ε r '

2

0 S

1 d SE = r ' .

4 π ε r '

2

0 V

1 d VE = r ' .

4 π ε r '

Charge distributed along a line:

Charge distributed over a surface:

Charge distributed inside a volume:

If the charge distribution is uniform, then , , and can be taken outside the

integrals.

Electric Flux Density

30

Electric Flux Density

Consider a point charge at the origin:

Q

31

Electric Flux Density of a Point Charge

(1) Assume from symmetry the form of the field

(2) Construct a family of Gaussian surfaces

rDaDrr

ˆ

spheres of radius r where

r0

spherical

symmetry

32


(3) Evaluate the total charge within the volume

enclosed by each Gaussian surface

V

evencldvqQ

33


Q

R

Gaussian surface

QQencl

34


(4) For each Gaussian surface, evaluate the

integral

DSsdD

S

24 rrDsdD

r

S

magnitude of D

on Gaussian

surface.

surface area

of Gaussian

surface.

35


(5) Solve for D on each Gaussian surface

S

QD

encl

24

ˆr

QaD

r

2

004

ˆr

Qa

DE

r

36

Electric Flux Density of a Spherical Shell of Charge

Consider a spherical shell of uniform charge density:

otherwise,0

,0

braqq

eva

b

Gauss Law, It’s Applications to

Symmetrical Charge Distributions

38

Gauss’s Law

• Gauss‟s law states that “the net electric flux

emanating from a close surface S is equal to

the total charge contained within the volume

V bounded by that surface.”

encl

S

QsdD

39

Gauss’s Law (Cont’d)

V

Sds

By convention, ds

is taken to be outward

from the volume V.

V

evencldvqQ

Since volume charge

density is the most

general, we can always write

Qencl in this way.

40

Applications of Gauss’s Law

• Gauss‟s law is an integral equation for the

unknown electric flux density resulting from a

given charge distribution.

encl

S

QsdD known

unknown

41

Applications of Gauss’s Law (Cont’d)

• In general, solutions to integral equations must

be obtained using numerical techniques.

• However, for certain symmetric charge

distributions closed form solutions to Gauss‟s

law can be obtained.

42

Applications of Gauss’s Law (Cont’d)

• Closed form solution to Gauss‟s law relies on

our ability to construct a suitable family of

Gaussian surfaces.

• A Gaussian surface is a surface to which the

electric flux density is normal and over which

equal to a constant value.

Electric Potential: Potential Field

Due To Different Types of Charges

44

Electrostatic Potential• An electric field is a force field.

• If a body being acted on by a force is moved from one point toanother, then work is done.

• The concept of scalar electric potential provides a measure of thework done in moving charged bodies in an electrostatic field.

• The work done in moving a test charge from one point to another ina region of electric field:

b

a

b

a

ba ldEqldFW

ab

q

F

ld

45

Electrostatic Potential

• In evaluating line integrals, it is customary to take the dl in the

direction of increasing coordinate value so that the manner in

which the path of integration is traversed is unambiguously

determined by the limits of integration.

3

5

ˆ dxaEqWxba

46


• The electrostatic field is conservative:

– The value of the line integral depends only on the end

points and is independent of the path taken.

– The value of the line integral around any closed path is

zero.

0C

ldE

C

47


• The work done per unit charge in moving a test charge

from point a to point b is the electrostatic potential

difference between the two points:

b

a

ba

abldE

q

WV

electrostatic potential difference

Units are volts.

48


• Since the electrostatic field is conservative we can write

aVbV

ldEldE

ldEldEldEV

a

P

b

P

b

P

P

a

b

a

ab

00

0

0

49


• Thus the electrostatic potential V is a scalar field that is

defined at every point in space.

• In particular the value of the electrostatic potential at any point

P is given by

P

P

ldErV

0 reference point

50


• The reference point (P0) is where the potential is zero

(analogous to ground in a circuit).

• Often the reference is taken to be at infinity so that the

potential of a point in space is defined as

P

ldErV

51

Electrostatic Potential and Electric Field

• The work done in moving a point charge from point a

to point b can be written as

b

a

abba

ldEQ

aVbVQVQW

52


• Along a short path of length Dl we have

lEV

lEQVQW

or

53


• Along an incremental path of length dl we have

• Recall from the definition of directional derivative:

ldEdV

ldVdV

>Thus

VE

Potential Gradient and the Dipole

field due to Dipole, Maxwell’s Two

Equations for Electrostatic Field

55

Charge Dipole

• An electric charge dipole consists of a pair of equaland opposite point charges separated by a smalldistance (i.e., much smaller than the distance atwhich we observe the resulting field).

d

+Q -Q

56

Dipole Moment

• Dipole moment p is a measure of the strength

of the dipole and indicates its direction

dQp +Q

-Q

d

p is in the direction from

the negative point charge to

the positive point charge

57

Electrostatic Potential Due to Charge Dipole

observation

point

d/2

+Q

-Q

z

d/2

P

Qdapz

ˆ

R

Rr

58

Electrostatic Potential Due to Charge Dipole (Cont’d)

R

Q

R

QrVrV

0044

,

cylindrical symmetry

59

Electrostatic Potential Due to Charge Dipole (Cont’d)

d/2

d/2

cos)2/(

cos)2/(

22

22

rddrR

rddrR

R

R

r

P

60

Electrostatic Potential Due to Charge Dipole in the Far-Field

• assume R>>d

• zeroth order approximation:

RR

RR

0V

not good

enough!

61

Electrostatic Potential Due to Charge Dipole in the Far-Field (Cont’d)

• first order approximation from geometry:

cos2

cos2

drR

drR

d/2

d/2

lines approximately

parallel

R

R

r

62


• Taylor series approximation:

cos2

111

cos2

11

cos2

11

cos2

111

r

d

rR

r

d

r

r

d

r

dr

R

1,11

:Recall

xnxxn

63


2

0

0

4

cos

2

cos1

2

cos1

4,

r

Qd

r

d

r

d

r

QrV

64


• In terms of the dipole moment:

2

0

ˆ

4

1

r

apV

r

Energy Density in Electrostatic

Field

Energy Density in Electrostatic Field

• To determine the energy that is present in an

assembly of charges

• let us first determine the amount of work

required to assemble them.

• Let us consider a number of discrete charges

Q1, Q2,......., QN are brought from infinity to

their present position one by one.

Energy Density in Electrostatic Field

• Since initially there is no field present, the

amount of work done in bring Q1 is zero.

• Q2 is brought in the presence of the field of Q1,

the work done W1= Q2V21 where V21 is the

potential at the location of Q2 due to Q1.

It takes no work to bring in first charges

1 0W

Work needed to bring in q2 is :

1 12 2 2

0 1 2 0 1 2

1 1[ ] ( )

4 4

q qW q q

R R

1 2 1 23 3 3

0 1 3 0 2 3 0 1 3 2 3

1 1 1[ ] ( )

4 4 4

q q q qW q q

R R R R



31 24 4

0 1 4 2 4 3 4

1[ ]

4

qq qW q

R R R

for q1

Total work

W=W1+ W2+ W3 +W4

1 3 2 3 3 41 2 1 4 2 4

0 1 2 1 3 2 3 1 4 2 4 3 4

1

4

q q q q q qq q q q q q

R R R R R R

0

1

1

4

nj

ii j

j

qV P

R

j i

01 1

1

4

n ni j

i ji j

q q

R

ij jiR R

01 1

1 1

2 4

n nj

ii j

i j

qq

R

0

1 1

1

8

n ni j

i ji j

q q

R

1

1

2

n

i i

i

W q V P

j i j i

Where is the energy stored? In charge or in field ?

Both are fine in ES. But,it is useful to regard the energy

as being stored in the field at a density

2

02

E Energy per unit volume

The superposition principle,not for ES energy

201 1

2W E d

202 2 τ

2W E d

201 2( )

2to tW E E d

1 2 0 1 2( )W W E E d

2 201 2 1 2( 2 )

2E E E E d

Convection and Conduction Currents, Continuity Equation

and Relaxation Time

Current (in amperes) through a given area is the electric charge

passing through the area per unit time

Current density is the amount of current flowing through a surface,

A/m2, or the current through a unit normal area at that point

Current density

Convection and Conduction Currents

Currentd Q

Id t

Where

IJ

S

.S

I J d S

Depending on how the current is produced, there are different

types of current density

• Convection current density

• Conduction current density

• Displacement current density

- Current generated by a magnetic field

Does not involve conductors and does not obey Ohm‟s law

Occurs when current flows through an insulating medium such as

liquid, gas, or vacuum


Convection current density

Where v is the velocity vector of the fluid


v v y

Q yI S S v

t t

y v y

IJ v

S

Conduction current density

Current in a conductor

Obeys Ohm‟s law

Consider a large number of free electrons travelling in a

metal with mass (m), velocity (v), and scattering time (time

between electron collisions), τ


The carrier density is determined by the number of electrons, n,

with charge, e

m vF q E

vn e

Conduction current density can then be calculated as2

v

n eJ v E E

m

This relationship between current concentration and electric field is

known as Ohm‟s Law.

Continuity Equation

Continuity Equation and Relaxation Time

Where Qin is the total charge enclosed by the closed surface.

Invoking divergence theorem

.in

o u tS

d QI J d S

d t

. .v

S

J d S Jd v But,

in v

vv v

d Q ddd v d v

d t d t d t

Due to the principle of charge conservation, the time rate of

decrease of charge within a given volume must be equal to the net

outward current flow through the closed surface of the volume.

Thus, the current coming out of the closed surface is

From the above three equations, we can write as

which is called the continuity of current equation.

The continuity equation is derived from the principle of

conservation of charge

It states that there can be no accumulation of charge at any point

For steady currents,

Hence,

The total charge leaving a volume is the same as the total charge

entering it.

v

v v

v

J d v d vt

Jt

0v

t

0J


Utilizing the continuity equation and material properties such

as permittivity and conductivity, one can derive a time

constant

We start with Ohm‟s and Gauss‟ Laws

J

v

v v

E

Et

0v v

t

v

v

t

Relaxation Time

J E

0ln ln

v v

t


0

r

t

T

v ve

rT


is the initial charge density. The relaxation time(Tr) is the time

it takes a charge placed in the interior of a material to drop by e-1

(=36.8%) of its initial value.

For good conductors Tr is approx. 2*10-19 s.

For good insulators Tr can be days

Capacitance- Parallel plate, Co-axial and Spherical Capacitor

Capacitance is an intuitive characterization of a capacitor. It tells you,

how much charge a capacitor can hold for a given voltage

The property of a capacitor to „store electricity‟ may be called its

capacitance

Generally speaking, to have a capacitor we must have two (or more)

conductors carrying equal but opposite charges

Capacitance

Figure 1. Charge carriers of conductor with opposite polarity

Suppose we give Q coulomb of charge to one of the two plates of

capacitor, the potential difference V is established between the two

plates, then its capacitance is

The capacitance C is a physical property of the capacitor and in

measured in farads (F)

The charge Q on the surface of the plate and the potential difference

Vab between the plates can be represented in terms of electric field

Capacitance

a b

QC

V

.

.

b

a b

a

E d SQC

VE d l

.E d S Q

.

b

a b a b

a

V V V E d l

Therefore, the capacitance C can be written as

• Choose a suitable coordinate system.

• Let the two conducting plates carry charges + Q and - Q

• Determine E using Coulomb's or Gauss's law and find V

• Finally, obtain C from

Capacitance

Procedure for Obtaining Capacitance

Capacitance can be determined using first method are as follows

Capacitance can be obtained for any given two-conductor

capacitance by following either of these methods:

• Assuming Q and determining V in terms of Q (involving Gauss's

law)

• Assuming V and determining Q in terms of V (involving solving

Laplace's equation)

QC

V

Parallel-Plate Capacitor

S

Q

A

the charge density is given by

The flux passing through the medium and flux density is given by

. .n S

S S

n S

Q

D d A Q d A

D

Figure 2. Parallel plate conductors

Parallel-Plate Capacitor

The charge density in terms of electric field as

D E

S

Q

A

Where,

But, we know,

SE

The above equation modifies to

Also, the relation between electric field and electric potential canbe written as

QE

A

0

.

d

QV E d l E d d

A

Q AC

Q dd

A

Thus, the parallel plate capacitor can be written as

Coaxial(Cylindrical) Capacitor

Consider length l of two coaxial conductors of inner radius a and

outer radius b (b > a) as shown in Figure 3

2

0 0

. .

.

( 2 )

S S

l

S

Q D d A E d A

Q E d A E d d l

Q E l

By applying Gauss's law to an arbitrary Gaussian cylindrical surface

of radius ρ (a < ρ < b), we obtain

Figure 3 Cylindrical conductors

Coaxial(Cylindrical) Capacitor

.2

ln2

ln ln ln ln2 2

ln2

a

b

a

b

Q dV E d r

l

Q

l

Q QV a b b a

l l

Q bV

l a

The potential difference between the inner and outer conductors

can be written as

Thus the capacitance of a coaxial cylinder is given by

2

ln ln2

Q Q lC

Q b bV

l a a

Spherical Capacitor

Consider the inner sphere of radius a and outer sphere of radius

b(b> a) separated by a dielectric medium with permittivity ε as

shown in Figure 4

22

0 0

2

. .

. s in

( 4 )

S S

S

Q D d A E d A

Q E d A E r d d

Q E r

Figure 4. Spherical conductor

By applying Gauss's law to an arbitrary Gaussian spherical surface

of radius r(a<r<b), we obtain

Spherical Capacitor

S

Q

A

Therefore, the potential difference between the inner and outersphere can be written as

2

Q.

4

Q 1

4

Q 1 1

4

a

b

a

b

d rV E d l

r

r

Va b

Thus, the capacitance of a spherical capacitor is given by4

1 1Q 1 1

4

Q QC

V

a ba b

By letting b ∞, C4πεa, which is the capacitance of aspherical capacitor whose outer plate is infinitely large

Magnetostatics and Time varying fields

UNIT II

Biot-Savart Law

The Law of Biot-Savart is

1 1 1 2

2

1 2

2

4

I dd

R

L aH

(A/m)

To get the total field resulting from a

current, you can sum the contributions

from each segment by integrating

2

4.

RId

R

L aH (A/m)

Note: The Biot-Savart law is analogous to the Coulomb‟s law

equation for the electric field resulting from a differential charge

92

Vector Magnetic Potential

• Vector identity: “the divergence of the curl of

any vector field is identically zero.”

• “If the divergence of a vector field is

identically zero, then that vector field can be

written as the curl of some vector potential

field.”

0 A

93

Vector Magnetic Potential (Cont’d)

• Since the magnetic flux density is solenoidal, it

can be written as the curl of a vector field

called the vector magnetic potential.

ABB 0

94


• The general form of the B-S law is

• Note that

V

vdR

RrJrB

3

0

4)(

3

1

R

R

R

95


• Furthermore, note that the del operator operates

only on the unprimed coordinates so that

R

rJ

rJR

RrJ

R

RrJ

1

1

3

96


• Hence, we have

vdR

rJrB

V

4

0

rA

97


• For a surface distribution of current, the vector

magnetic potential is given by

• For a line current, the vector magnetic potential is

given by

sd

R

rJrA

S

s

4)(

0

LR

ldIrA

4)(

0

98


• In some cases, it is easier to evaluate the

vector magnetic potential and then use B

= A, rather than to use the B-S law to

directly find B.

• In some ways, the vector magnetic potential A

is analogous to the scalar electric potential V.

99


• In classical physics, the vector magnetic

potential is viewed as an auxiliary function

with no physical meaning.

• However, there are phenomena in quantum

mechanics that suggest that the vector

magnetic potential is a real (i.e., measurable)

field.

Forces due to Magnetic Fields,

Ampere’s Force Law

101

Ampere’s Law of Force

• Ampere’s law of force is the “law of action”

between current carrying circuits.

• Ampere’s law of force gives the magnetic force

between two current carrying circuits in an

otherwise empty universe.

• Ampere‟s law of force involves complete

circuits since current must flow in closed

loops.

Inductances and Magnetic

Energy

103

Flux Linkage

• To discuss about inductance, first we have to

know the flux linkage

• Consider two magnetically coupled circuits

C1

I1

S1 S2 C2

I2

104

Flux Linkage (Cont’d)

• The magnetic flux produced I1 linking thesurface S2 is given by

• If the circuit C2 comprises N2 turns and thecircuit C1 comprises N1 turns, then the totalflux linkage is given by

2

2112

S

sdB

2

2121122112

S

sdBNNNN

105

Mutual Inductance

• The mutual inductance between two circuits is

the magnetic flux linkage to one circuit per

unit current in the other circuit:

1

1221

1

12

12

I

NN

IL

106

Mutual Inductance

2

2

21

1

21

21

1

21

1

1221

1

12

12

C

S

ldAI

NN

sdBI

NN

I

NN

IL

107

Mutual Inductance (Cont’d)

112

110

1

4C

R

ldIA

1 2

2

12

21210

21

1

21

12

4C C

C

R

ldldNN

ldAI

NNL

108

Mutual Inductance (Cont’d)

• The Neumann formula for mutual inductance

tells us that

– L12 = L21

– the mutual inductance depends only on the

geometry of the conductors and not on the current

109

Self Inductance

• Self inductance is a special case of mutual

inductance.

• The self inductance of a circuit is the ratio of

the self magnetic flux linkage to the current

producing it:

1

11

2

1

1

11

11

I

N

IL

110

Self Inductance (Cont’d)

• For an isolated circuit, we call the self

inductance, inductance, and evaluate it using

I

N

IL

2

111

Energy Stored in Magnetic Field

• The magnetic energy stored in a region

permeated by a magnetic field is given by

dvHdvHBW

VV

m 2

2

1

2

1

112

Energy Stored in an Inductor

• The magnetic energy stored in an inductor is

given by

2

2

1LIW

m

113

Ampere’s Law of Force (Cont’d)

Two parallel wires

carrying current in

the same direction

attract.

Two parallel wires

carrying current in

the opposite

directions repel.

I1 I2

F12F21

I1 I2

F12F21

114


A short current-

carrying wire

oriented

perpendicular to a

long current-carrying

wire experiences no

force.

I1

F12 = 0

I2

115


The magnitude of the force is inversely proportional

to the distance squared.

The magnitude of the force is proportional to the

product of the currents carried by the two wires.

116


• The direction of the force established by theexperimental facts can be mathematicallyrepresented by

1212

ˆˆˆˆ12 RF

aaaa

unit vector in

direction of force on

I2 due to I1

unit vector in direction

of I2 from I1


of current I1


of current I2

117


• The force acting on a current element I2 dl2 by

a current element I1 dl1 is given by

2

12

11220

12

12

ˆ

4 R

aldIldIF

R

Permeability of free space

0 = 4 10-7 F/m

118


• The total force acting on a circuit C2 having a

current I2 by a circuit C1 having current I1 is

given by

2 1

12

2

12

12210

12

ˆ

4C C

R

R

aldldIIF

119


• The force on C1 due to C2 is equal in

magnitude but opposite in direction to the

force on C2 due to C1.

1221FF

120

Force on a Moving Charge

• The total force exerted on a circuit C carrying

current I that is immersed in a magnetic flux

density B is given by

C

BldIF

121

Force on a Moving Charge

• A moving point charge placed in a magnetic

field experiences a force given by

BvQ

The force experienced

by the point charge is

in the direction into the

paper.

BvQFm

vQlId

122

Lorentz Force

• If a point charge is moving in a region where bothelectric and magnetic fields exist, then itexperiences a total force given by

• The Lorentz force equation is useful fordetermining the equation of motion for electronsin electromagnetic deflection systems such asCRTs.

BvEqFFFme

Faraday’s Law and Transformer EMF, Motional EMF

Steady current produces a magnetic field

In 1831, Michael Faraday discovery that a time-varying magnetic field

would produce an electric current

According to Faraday's experiments, a static magnetic field produces no

current flow, but a time-varying field produces an induced voltage

(electromotive force)

induced emf (in volts), in any closed circuit is equal to the time rale of

change of the magnetic flux linkage by the circuit(Faraday‟s law)

According to Faraday‟s, It can be expressed as

Where,N is the number of turns in the circuit is the flux through each

turn.

The negative sign shows to oppose the flux producing it( Lenz's law)

Faraday's Law of electromagnetic Induction

em f

d dV N

d t d t

Induced emf (For a single turn N=1) in terms of E and B can be written

as

The variation of flux with time may be caused in three ways

• By having a stationary loop in a time-varying B field

• By having a time-varying loop area in a static B field

• By having a time-varying loop area in a time-varying B field

Induced emf (For a single turn N=1) in

terms of E and B can be written as

e m fS

L

d dV E d l B d s

d t d t


This emf induced by the time-varying current (producing the

time-varying B field) in a stationary loop is often referred to as

transformer emf

A stationary conducting loop is in a time varying magnetic B

field

By applying Stokes's theorem to the middle term

S S

E d S B d st

BE

t

e m fS

L

d dV E d l B d s

d t d t


Moving Loop in Static B Field (Motional emf)

When a conducting loop is moving in a static B field, an emf is

induced in the loop

F q vB

We define the motional electric field Em asm

m

FE v B

Q

Moving with uniform velocity u as consisting

of a large number of free electrons, the emf

induced in the loop is


e m fL

L

dV E d l v B d l

d t

By applying Stokes's theorem

( )m

E v B

Faraday’s law of Induction:

This describes the creation of an electric field by a changing

magnetic flux

The law states that the emf, which is the line integral of the electric

field around any closed path, equals the rate of change of the

magnetic flux through any surface bounded by that path

One consequence is the current induced in a conducting loop placed

in a time-varying B

Maxwell’s equations in integral form

Modified Ampere’s law:

It describes the creation of a magnetic field by an electric field and

electric currents

The line integral of the magnetic field around any closed path is the

given sum

Bd

dd t

E s

IE

o o o

dd μ ε μ

d t

B s

Maxwell’s Equations in Point form and Integral form for Time-

Varying Fields

Maxwell equations in differential form

Maxwell’s equations in point form

,v

D

,t

BE

,0 B

,t

DJH

The two Gauss‟s laws are symmetrical, apart from the absence of the

term for magnetic monopoles in Gauss‟s law for magnetism

Faraday‟s law and the Ampere-Maxwell law are symmetrical in that

the line integrals of E and B around a closed path are related to the

rate of change of the respective fluxes


I

E A G a u s s 's la w e le c tr ic

B A 0 G a u s s 's la w in m a g n e tis m

E s F a ra d a y 's la w

B s A m p e re -M a xw e ll la w

oS

S

B

E

o o o

qd

ε

d

dd

d t

dd μ ε μ

d t

Gauss’s law (electrical):

The total electric flux through any closed surface equals the net

charge inside that surface divided by eo

This relates an electric field to the charge distribution that creates it


oS

qd

ε E A

Gauss’s law (magnetism):

The total magnetic flux through any closed surface is zero

This says the number of field lines that enter a closed volume must

equal the number that leave that volume

This implies the magnetic field lines cannot begin or end at any point

Isolated magnetic monopoles have not been observed in nature0

S

d B A

Faraday’s law of Induction:

This describes the creation of an electric field by a changing

magnetic flux

The law states that the emf, which is the line integral of the electric

field around any closed path, equals the rate of change of the

magnetic flux through any surface bounded by that path

One consequence is the current induced in a conducting loop placed

in a time-varying B


Modified Ampere’s law:

It describes the creation of a magnetic field by an electric field and

electric currents

The line integral of the magnetic field around any closed path is the

given sum

Bd

dd t

E s

IE

o o o

dd μ ε μ

d t

B s

UNIT-III

Uniform Plane Waves

Reflection and Refraction of

Uniform Plane Wave

Till now, we have studied plane waves in various medium

Let us try to explore how plane waves will behave at a media

interface

Electromagnetic waves are often at the interface of boundary may

be reflected or refracted or changes direction at the interface

When a radio wave reflects from a surface, the ratio of the two

(reflected wave/incident wave) is known as the „reflection

coefficient‟ of the surface

How much of the incident wave has been transmitted through the

material is given by another ratio(transmitted wave /incident wave)

known as „transmission coefficient‟

Reflection and Refraction of Uniform Plane Wave

Transmission and reflection ratio depends on the

• conductivity (σ),

• permittivity (ε) and

• permittivity (ε) of the material as well as material properties of

the air which the radio wave is incident

Also some part of the wave will be transmitted through the material

• how much of the incident wave has been transmitted through

the material is also dependent on the material parameters

mentioned above

• It is given by another ratio known as „transmission coefficient‟

• It is the ratio of the transmitted wave divided by the incident

wave

Reflection and Refraction of Plane Wave contd..,

In plane wave incident from media interface, consider two cases

• Normal incidence

• Obliquely incidence

The electric and magnetic field expressions

• in all the regions of interest and apply the boundary

conditions to get the values of the transmission and

reflection coefficients

• Type of boundary interfaces for the solutions of

transmission and reflection coefficients

dielectric –conductor interface (both normal and oblique

incidence)

• dielectric –dielectric interface (both normal and oblique

incidence)


At an interface between two media, the angles of incidence,

reflection, and refraction are all measured with respect to the

normal.

• Incidence angle (θi): angle at which the

incident wave makes with the normal to

the interface

• Reflection angle (θr): angle at which the

reflected wave makes with the normal to

the interface

• Transmission or refraction angle (θt):

angle at which the transmitted

(refracted) wave makes with the normal

to the interface


If a plane wave is incident obliquely upon a surface that is not a

conductor boundary, part of the wave is transmitted and part of it

reflected.

In this case the transmitted wave will be refracted; that is the

direction of propagation will be altered.

Consider a planar interface between two dielectric media. A

plane wave is incident at an angle from medium 1.

Incident

wave

x

z

z=0Medium 1

(ε1,μ1)

Medium 2

(ε2,μ2)

ani

anr ant

AA‟ B

O‟

O

Reflected

waveRefracted

wave

θtθi

θr

In the diagram incident ray 2 travels a

distance AO‟ where as the transmitted

ray 1 travels a distance OB, and reflected

ray 1 travels from O to A‟


Snell‟s Law of refraction

• The angle of reflection is equal to the angle of incident

11 1 1 22 1 1 2

1 2 2 12 2 2 2

1

s in

s in

o

o r rt

i o r r o

u n

u n

n :

refractive

index


Reflection of a Plane Wave at Normal

Incidence

-Dielectric Boundary

Normal Incidence -Dielectric Boundary

Reflected

wave

Medium 2

2 2( , )

Medium 1

1 1( , )

Hi

Ei

ani

Incident

wave

y

x

z

z=0

Hr

Era

nr

Ht

Et

ant

Transmitted

wave

Consider the boundary is an interface between two dielectrics

We will consider the case of normal incidence, when the incident

wave propagation vector is along the normal to the interface between

two media

Generally, consider a time harmonic x-polarized electric field

incident from medium 1 (µ1, ε1, σ1) to medium 2 (µ2, ε2, σ2)

For the case of perfect dielectric

• σ1= σ2=0, ε1≠ ε2

• No loss or absorption of power

We will assume plane waves with electric field vector oriented

along the x-axis and propagating along the positive z-axis

without loss of generality

For z<0 (we will refer this region as region I and it is assumed

to be a medium1)

When a plane electromagnetic wave incident on the surface of

perfect dielectric

• part of energy is transmitted

• part of it is reflected

Let the wave incident on medium 1

the incident, reflected and transmitted fields

shown in Figure

Medium 2

2 2( , )

Medium 1

1 1( , )

Hi

Ei

ani

Incident

wave

y

x

z

z=0

Hr

Era

nr H

t

Et

ant

Transmitted

wave

Normal Incidence -Dielectric Boundary contd..,

1

0ˆ( )

j z

i x iE z a E e

10

1

ˆ( )j zi

i y

EH z a e

Incident wave ( inside

medium 1 )

Reflected wave ( inside

medium 1 )1

0ˆ( )

j z

r x rE z a E e

10

1 1

1ˆ ˆ( ) ( ) ( )

j zr

r z r y

EH z a E z a e

Transmitted wave ( inside

medium 2 ) 2

0ˆ( )

j z

t x tE z a E e

20

2 2

1ˆ ˆ( ) ( )

j zt

t z t y

EH z a E z a e


The tangential components (the x-components) of the electric and

magnetic field intensities must be continuous. ( at interface z=0 )

1 ta n 2 ta nE E

1 ta n 2 ta nH H

(0 ) (0 ) (0 )i r t

E E E

io ro toE E E

(0 ) (0 ) (0 )i r t

H H H

1 2

1( )

to

io r o

EE E

The continuity of tangential components of the electric and

magnetic fields require that

Rearranging these two equations

2 1

2 1

r o ioE E

2

2 1

2

to ioE E



1 2: 0 , 1

2 1

2 1

r o

io

E

E

Reflection coefficient ( + or - ) ≤ 1

The reflection coefficient

Special cases

The transmission coefficient

2

2 1

2to

io

E

E

20 ( ) 1sh o r t E/H, E=0 perfect

conductor !!

2( ) 1o p en H(I)=0 No current !!

1

which is related to reflection coefficient

Normal Incidence - Dielectric Boundary contd..,

Further more, the reflection and transmission coefficient

1 2

2 1

r o r o

io io

H E

H E

1 1

2 2 1

2to to

io io

H E

H E

The permeabilities of all known dielectrics do not differ

appreciably from free space, so that μ1= μ2= μ0

0 0

2 1 1 2 12 1

2 1 0 0 1 2 1 2

2 1

2,

ro to

io io

E E

E E

2 1 2

1 2 1 2

2,

ro to

io io

H H

H H

Similarly,

Reflection of a Plane Wave Normal

Incidence

-Conducting Boundary

In z-direction ( x=constant )

9 0 o u t o f p h ase

1 1sin ( cos )

j t

y iE jA z e

n o is p ro p a g a te .a v

P

In x-direction ( z=constant )

1 1c o s ( c o s )

j t

x iH B z e

trave lin g w ave

1( sin )

1

ij t x

yE C e

1

1

1s in s in

x

i i

uu

1

1

s inx

i

C = f(z) , D = g(z)1

( s in )

1

ij t x

zH D e

Oblique Incidence - Conducting Boundary

Oblique Incidence -

Conducting Boundary


When EM Wave incident obliquely on the interface between two

media boundary, it can be decomposed into:

• Horizontal Polarization

• Vertical Polarization

Horizontal Polarization, or transverse electric (TE) polarization

The E Field vector is parallel to the boundary surface or

perpendicular to the plane of incidence

Vertical Polarization, or transverse magnetic (TM) polarization

The H Field vector is parallel to the boundary surface and the electric

vector is parallel to the plane of incidence

Let us consider the wave incident on a perfect conductor

• The wave is totally reflected with the angle of incidence equal to

the angle of reflection


Perpendicular Polarization

2( )

Medium 2

Perfect conductor

Medium 1

1( 0 )

Hi

Ei

ani

Hr

Er

an

r

Incident

wave

Reflected

wave

y.

x

z

z=0

i

r


Perpendicular Polarization Assume the wave is propagating obliquely along an arbitrary

direction ar and the electric field vector is along y-direction(normal

to the plane of incidence x-z plane) and direction of propagation

Using direction cosines the direction of propagation of incidence

wave can be written as

ˆ ˆ ˆs in c o sn i x i z i

a a a

Similarly the direction of propagation of reflected wave (-z

direction )can be written as


a a a


Incident wave ( inside medium 1 )

1

1ˆ( , ) ( , )

i n i iH x z a E x z

1ˆ

0ˆ( , ) n i

j a R

i y iE x z a E e

1( s in c o s )0

1

ˆ ˆc o s s in i ij x zi

x i z i

Ea a e

1( sin cos )

0ˆ i i

j x z

y ia E e

Reflected wave ( inside medium 1 )

1( s in co s )

0ˆ( , ) r r

j x z

r y rE x z a E e

Boundary condition, z = 0

1( , 0 ) ( , 0 ) ( , 0 )

i rE x E x E x


0 0&

r i i rE E

1 1sin sin

0 0ˆ ( ) 0i r

j x j x

y i ra E e E e

, for all x

Snell’s law of reflection the angle of reflection equals the angle of incidence

1( s in co s )

0ˆ( , ) r r

j x z

r y rE x z a E e

1( s in co s )

0

( sn e ll's la w o f re fle c tio n )

ˆ

i ij x z

y ia E e


Similarly, Magnetic field of the reflected wave can be written as

1( s in co s )

0ˆ( , ) r r

j x z

r y rE x z a E e

1( s in c o s )

0

( sn e ll's la w o f re fle c tio n )

ˆ

i ij x z

y ia E e

1

1ˆ( , ) ( , )

r n r rH x z a E x z

1( s in c o s )0

1

ˆ ˆ c o s s in i ij x zi

x i z i

Ea a e


Total field in medium 1 can be written as(standing wave in terms of

electric and magnetic field)

1( , ) ( , ) ( , )

i rE x z E x z E x z

1sin

0 1ˆ 2 sin ( cos ) i

j x

y i ia j E z e

1

1

s in0

1 1

1

s in

1

ˆ( , ) 2 c o s c o s ( c o s )

ˆ s in s in ( c o s )

i

i

j xi

x i i

j x

z i i

EH x y a z e

a j z e

1 1 1cos cos sin

0ˆ ( )i i i

j z j z j x

y ia E e e e


In z-direction ( x=constant )

9 0 o u t o f p h ase

1 1sin ( cos )

j t

y iE jA z e

n o is p ro p a g a te .a v

P

In x-direction ( z=constant )

1 1c o s ( c o s )

j t

x iH B z e

trave lin g w ave

1( sin )

1

ij t x

yE C e

1

1

1s in s in

x

i i

uu

1

1

s inx

i

C = f(z) , D = g(z)1

( s in )

1

ij t x

zH D e



Parallel Polarization

2( )

Medium 2

Perfect conductor

Medium 11

( 0 )

Hi

Ei

ani

Hr

Eranr

Incident

wave

Reflected

wave

y.

x

z

z=0

i

r

Parallel Polarization Assume the wave is propagating obliquely along an arbitrary

direction ar and the electric field vector is parallel to the plane of

incidence (x-z plane) and the magnetic field vector is normal to

the plane of incidence

Using direction cosines the direction of propagation of incidence

wave can be written as


a a a

Similarly the direction of propagation of reflected wave (-z

direction ) can be written as


a a a


Incident wave ( inside medium 1 )

Reflected wave ( inside medium 1 )

Boundary condition, z = 0

1ˆ

0ˆ ˆ( , ) ( cos sin ) n i

j a R

i i x i z iE x z E a a e

1( s in c o s )0

1

ˆ( , ) i ij x zi

i y

EH x z a e

1( s in c o s )

0ˆ ˆ( c o s s in ) i i

j x z

i x i zE a a e

1( sin cos )

0ˆ ˆ( , ) ( cos sin ) r r

j x z

r r x i z rE x z E a + a e

1( s in c o s )0

1

ˆ( , ) r rj x zr

r y

EH x z a e


ro io r i

E E

0 1ˆ2 [ c o s s in ( c o s )

i x i iE a j z

1s in0

1

1

ˆ 2 c o s ( c o s ) ij xi

y i

Ea z e

1s in

1ˆ s in c o s ( c o s )] i

j x

z i i+ a z e

1( , ) ( , ) ( , )

i rE x z E x z E x z

1( , ) ( , ) ( , )

i rH x z H x z H x z

1 1s in s in

( c o s ) ( c o s ) 0i rj x j x

io i r o rE e E e

( for all x)

Total field in medium 1 can be written as(standing wave in terms of

electric and magnetic field)


In z-direction : standing-wave1,

xE

1 yH

1 1/ s in

x iu u

The wave is non uniform plane wave

1,

zE

1 yH


Reflection of a Plane Wave Oblique Incidence -Interface between

dielectric media

x

z=0

Oθi

θr

iH

iE

•

The electric field phasors for the perpendicular polarization, with

reference to the system of coordinates in the figure

Let us assume that the incident wave propagates in the

first quadrant of x-z plane and propagation makes an

angle θi with the normal

1( sin cos )

( , ) i ij x z

yi ioE x z a E e

1( s in c o s )

1

( , ) ( i ij x zio

i x zi i

EH x z a c o s + a s in )e

Let us assume that the reflected wave propagates in

the second quadrant of xz plane and reflection makes

an angle θr with the normal

1( sin cos )

( , ) r rj x z

yr roE x z a E e

1( s in c o s )

1

( , ) ( i ij x zr o

r x zr r


Oblique Incidence -Interface between dielectric media

Perpendicular Polarization

θt

Medium 1

Medium2

Similarly the transmitted fields are

Equating the tangential components of electric field

• electric field has only Ey component and it is tangential at the

interface z=0)

2( sin cos )

( , ) t tj x z

yt toE x z a E e

2( s in c o s )

2

( , ) ( i ij x zto

t x zi i


Similarly the magnetic field

• magnetic field has two components: Hx and Hz , but only Hx is

tangential at the interface z=0)

( , ) ( , ) ( , )iy ry ty

E x o E x o E x o

( , ) ( , ) ( , )ix r x txH x o H x o H x o


Boundary at z=0 gives

1 21s in s ins in

i trj x j xj x

e e e

1 21s in s ins in

1 1 2

1c o s c o s c o si tr

j x j xj x

i r te e e

If Ex and Hy are to be continuous at the interface z = 0 for all x,

then, this x variation must be the same on both sides of the

equations (also known as phase matching condition)

1 1 2s in s in s in

i r t

i r Follows Snell's law


Now we can simplify above two equations by applying Snell‟s

1 1 2

1c o s c o s c o s

i r t

The above two equations has two unknowns τ and Г and it can be

solved easily as follows

1

2

1 1

1c o s c o s

c o si r

t

1


Solving the equations and rearranging, we get

For normal incidence, it is a particular case and put

2 1

2 1

c o s c o s

c o s c o s

r o i t

io i t

E

E

2

2 1

2 c o s

c o s c o s

to i

io i t

E

E

1

0i r t


In this case, electric field vector lies in the x-z plane Since the magnetic field is transversal to the plane of incidence such

waves are also called as transverse magnetic (TM) waves So let us start with H vector which will have only y component

Similar to the previous case of perpendicularpolarization, we can write the reflected magneticfield and electric field vectors as

Parallel Polarization

θr

z=0

Oθi

iH

iE

•

rH

rE

θtx

1( s in c o s )

1

( , ) i ij x zio

i y

EH x z a e

1( sin cos )

( , ) ( i ij x z

i x zio i iE x z E a cos a sin )e

1( s in c o s )

1

( , ) r rj x zr o

r y

EH x z a e

1( sin cos )

( , ) ( r rj x z

r x zro r iE x z E a cos a sin )e


Similarly the transmitted fields are

Equating the tangential components of magnetic field

• electric field has only Hy component and it is tangential at the

interface z=0

Similarly the magnetic field

• magnetic field has two components: Ex and Ez , but only Ex is

tangential at the interface z=0)

( , ) ( , ) ( , )iy ry ty

E x o E x o E x o

( , ) ( , ) ( , )ix r x txH x o H x o H x o

2( sin cos )

( , ) ( t tj x z

t x zto t tE x z E a cos a sin )e

2( s in c o s )

2

( , ) t tj x zto

t y

EH x z a e


Boundary at z=0 gives

1 21s in s ins in

c o s c o s c o si trj x j xj x

i r te e e

1 21s in s ins in

1 1 2

1i tr

j x j xj xe e e

If Ex and Hy are to be continuous at the interface z = 0 for all x,

then, this x variation must be the same on both sides of the

equations (also known as phase matching condition)

1 1 2s in s in s in

i r t

i r Follows Snell's law


Now we can simplify above two equations by applying Snell‟s

c o s c o s c o si r t

The above two equations has two unknowns τ and Г and it can be

solved easily as follows

1 2

11

co s co s

co s

i r

t

2

1

1

Therefore,

2 1

2 1

c o s c o s

c o s c o s

ro t i

io t i

E

E

2

2 1

2 c o s

c o s c o s

to i

io t i

E

E

Note that cosine terms multiplication in the above equations is

different from the previous expression for perpendicular polarization


Brewster Angle, Critical Angle

Critical Angle

When light travels from one medium to another it changes speed and

is refracted.

As the angle of incidence increases so does the angle of refraction.

When the angle of incidence reaches a value known as the critical

angle the refracted rays travel along the surface of the medium or in

other words are refracted to an angle of 90°.

The critical angle for the angle of incidence in glass is 42°.

Surface ImpedanceIt is defined as the ratio of the tangential component of the electric

field to the surface current density at the conductor surface.

It is given by

where Etan is the tangential component parallel to the surface of the

conductor.

And Js is the surface current density.

ta n

s

s

EZ

J Ω

Critical Angle

> When the angle of incidence of the light ray reaches the critical

angle (420) the angle of refraction is 900. The refracted ray travels

along the surface of the denser medium in this case the glass.

> According to the law of refraction

1 2s in s in

i tn n

1 1

2 2

s in

s in

t

i

n

n

2i c t

i f th e n

1 12 2

1 1

s in s inc

no r

n

Total Internal Reflection

>When the angle of incidence of the light ray is greater than the

critical angle then no refraction takes place. Instead, all the light is

reflected back into the denser material in this case the glass. This is

called total internal reflection.

Brewster AngleFor unpolarized waves the electric fields are in many directions as shown in figure, e.g. unpolarized light

Where as in polarized waves the electric field vector is either vertical or horizontal as shown in figure below.

Brewster Angle

We can convert unpolarized light into polarized light by passing the light through a polarizing filter.

When unpolarized wave is incident obliquely at Brewster angle θB, only the component with perpendicular polarization (Horizontal polarization) will be reflected, while component with parallel polarization will not be reflected.

Brewster AngleIt is also called as Polarizing angle.

At Brewster angle, the angle between reflected ray and refracted ray is 900.

Normal Incidence - Conducting Boundary

Consider the boundary is an interface with a perfect conductor

The incident wave travels in a lossless medium (Medium 1 )

2( )

Medium 2

Perfect

conductor

Medium

11( 0 )

Hi

Ei

an

i

Hr

Er

an

r

Incident

wave

Reflected wave

y.

x

z

z=0

1

0ˆ( )

j z

i x iE z a E e

10

1

ˆ( )j zi

i y

EH z a e

Incident wave ( inside

medium 1 )

1( 0 )

1( )

Reflected wave ( inside

medium 1 )1

0ˆ( )

j z

r x rE z a E e

1

1ˆ( ) ( )

r n r rH z a E z

1

1ˆ( ) ( )

z ra E z

1

0

1

1ˆ

j z

y ra E e

2 20, 0

E H

Normal Incidence -Conducting Boundary contd..,

2( )

Medium

2

Perfect

conductor

Mediu

m 11( 0 )

Hi

Ei

ani

Hr

Er

anr

Incident

wave

Reflected wave

y

x

z

z=0

The boundary is an interface with a perfect

conductor(medium 2) both electric and magnetic fields

vanish.

No wave is transmitted across the boundary into the z >

0 i.e.

Total wave in medium

1

1 1

1 0 0ˆ( ) ( ) ( ) ( )

j z j z

i r x i rE z E z E z a E e E e

Continuity of tangential component of the E-field at the

boundary z = 0

1 0 0 2ˆ(0 ) ( ) (0 ) 0

x i rE a E E E

0 0r iE E

Normal Incidence - Conducting Boundary contd..,

Incident and reflected field are of equal amplitude, so all incident

energy is reflected by a perfect conductor

Negative sign indicates that the reflected field is shifted in phase

by 180° relative to incident field at the boundary

The magnetic field must be reflected without phase reversal, If

both were reversed there would be no reversal of direction .

In medium 1, the total field can be written as

(α=0)


0

1 1

1

ˆ( ) ( ) ( ) 2 c o s

i

i r y

EH z H z H z a z

1 1

1 0ˆ( ) ( )

j z j z

x iE z a E e e

0 1ˆ 2 s in

x ia j E z

The space-time behavior of the total field in medium

1(multiply with )

After multiplying a time factor and take a real instantaneous

part of the fields in medium 1 as a function of z and t can be

written as

1 1 0 1ˆ( , ) R e[ ( ) ] 2 sin sin

j t

x iE z t E z e a E z t

0

1 1 1

1

ˆ( , ) R e [ ( ) ] 2 c o s c o sj t i

y

EH z t H z e a z t

From the total fields in medium 1, we conclude that there is a

standing wave in medium 1, which does not progress

The magnitude of field varies sinusoidal with distance from the

reflecting plane

Electric field has zero at the surface and multiple of half wave

length from the surface and maxima for magnetic field

1

1

1

Z e ro s o f ( , ) o c c u r a t , o r , 0 ,1, 2 , .. .

2M a x im a o f ( , )

E z tz n z n n

H z t

1

1

1

M a x im a o f ( , ) o c c u r a t ( 2 1) , o r ( 2 1) , 0 ,1, 2 , .. .

2 4 Z e ro s o f ( , )

E z tz n z n n

H z t


It has a maximum value of twice the electric field strength of the

incident wave at the distance from the odd multiple of the quarter

wave length and zero for magnetic field

The totally reflected wave combines with

the incident wave to form a standing wave

The total wave in medium 1 is not a

traveling wave

It stands and does not travel, it consists of

two traveling waves Ei and Er of equal

amplitudes but in opposite directions.

z

x

2

0t

5 / 4t

3 / 2t

/ 4 , 3 / 4t

/ 2t

z

0t

/ 4t

/ 2t

3 / 4t

t

4

3

4

E1 versus z

H1 versus z

1 1 0 1ˆ( , ) R e[ ( ) ] 2 sin sin

j t


0

1 1 1

1

ˆ( , ) R e [ ( ) ] 2 c o s c o sj t i

y

EH z t H z e a z t


1 1 0 1ˆ( , ) R e[ ( ) ] 2 sin sin

j t


The equations also shows that there is a 90° out of time

phase between the electric and magnetic fields

0 1ˆ 2 s in co s ( )

2x i

a E z t

0

1 1 1

1

ˆ( , ) R e [ ( ) ] 2 c o s c o sj t i

y

EH z t H z e a z t


Brewster AngleIt is also called as Polarizing angle.

At Brewster angle, the angle between reflected ray and refracted ray is

900.

UNIT-IV

Transmission line Characteristics

If we are familiar with low frequency circuits and the circuit consists

of lumped impedance elements (R,L,C), we treat all lines(wires)

connecting the various circuit elements as perfect wires, with no

voltage drop and no impedance associated to them.

As long as the length of the wires is much smaller than the wavelength

of the signal and at any given time, the measured voltage and current

are the same for each location on the same wire.

Let us try to explore what happens, when the signal propagates as a

wave of voltage and current along the line at sufficiently high

frequencies

For sufficiently high frequencies, wavelength is comparable to the

length of a conductor, so the positional dependence impedance

properties (position dependent voltage and current) of wire can not be

neglected, because it cannot change instantaneously at all locations.

Transmission line equivalent circuit

So our first goal is to represent the uniform transmission line as a

distributed circuit and determine the differential voltage and current

behavior of an elementary cell of the distributed circuit.

Once that is known, we can find a global differential equation that

describes the entire transmission line by considering a cascaded

network (subsections) of these equivalent models.

So a uniform transmission line is represented as a distributed

circuit shown below

L series inductance

R series resistance

C shunt capacitance

G shunt conductance


Let us assume a differential length dz of transmission line ,V(z),I(z)

are voltage and current at point P and V(z)+dV, I(z)+dI are voltage

and current at point Q

QP

The series impedance and shunt admittance determines the variation of

the voltage and current from input to output of the cell shown below


For a differential length dz of transmission line ,the series

impedance and shunt admittance of the elemental length can be

written as

( )

( )

R j L d z

G j C d z

The corresponding circuit voltage and current equation is

( )

( )

V d V V I R j L d z

I d I I V G j C d z

From which we obtain a first order differential equation

( )

( )

d VI R j L

d z

d IV G j C

d z

Transmission line equations

We have a system of coupled first order differential equations that

describe the behavior of voltage and current on the transmission line

It can be easily obtain a set of uncoupled equations by differentiating

with respect to the coordinate z2

2

2

2

( )

( )

d V d IR j L

d zd z

d I d VG j C

d zd z

Substitute , in the above equations ,we obtain a second order

differential equations2

2

2

2

( )( )

( )( )

d VG j C R j L V

d z

d IG j C R j L I

d z


γ is the complex propagation constant, which is function of

frequency

α is the attenuation constant in nepers per unit length, β is the

phase constant in radians per unit length

Let the constant term can be represented as propagation constant,

which is written as

The transmission line equations can be written as

( )( )j G j C R j L

2

2

2

2

2

2

d VV

d z

d II

d z


Where, a,b,c, and d are the constants

Above equations represent the standard solutions of the wave

equations, which are similar to the solution of uniform plane wave

equations.

The terms and can be represented as backward and

forward wave along z-direction.

The solution of the second order transmission line equation isz z

z z

V a e b e

I c e d e


ze ze

Determination of constant

terms A and B

Determination of the constant terms A and B

To determine the constants a,b,c, and d , the above equations can be

written in terms of hyperbolic functions, where substitute

Substitute above equations in the solutions of transmission line

equations V and I

z z

z z

V a e b e

I c e d e

(c o sh c o sh ) (c o sh c o sh )

(c o sh c o sh ) (c o sh c o sh )

V a z z b z z

I c z z d z z

c o sh c o sh

c o sh c o sh

z

z

e z z

e z z

Let the solutions of the transmission line wave equations can written as

The constants a+b, a-b, c+d, and c-d can be replaced by another

constant terms A,B,C, and D respectively

So, the above equations can be written as

In order to reduce the four constant terms to two constant terms, we write

the relation between V and I by considering the following basic

differential equations

( ) c o sh ( ) s in h

( ) c o sh ( ) s in h

V a b z a b z

I c d z c d z

c o s h s in h

c o s h s in h

V A z B z

I C z D z

( )

( )

d VI R j L

d z

d IV G j C

d z


( co sh s in h )( )

d A z B zI R j L

d z

Substitute V in

( s in h co sh ( )A z B z I R j L

Differentiating in terms of z, we obtain

So, the current I can be written in terms of constants A and B as

( s in h c o s h( )

A z B z IR j L

Where γ is propagation constant, which is

( )( )G j C R j L


Substitute γ in current equation, we obtain

( )( )( s in h co sh

( )

G j C R j LA z B z I

R j L

0

s

s

V A

I Z B

0

1( s in h c o s hA z B z I

Z

The above equation can be simplified and written as

Where Z0 is another constant and can be called as characteristic

impedance along the line

0

( )

( )

R j LZ

G j C


So, the voltage and current wave equation in terms of constants A and

B can be re written as

0

c o s h s in h

1( s in h c o s h )

V A z B z

I A z B zZ

Now the constants A and B can be obtained by applying initial

conditions of the transmission line at Z=0

Let Vs and Is be the source voltage and current respectively. At the

source end, Z=0 the voltage V= Vs, current I= Is, then the above

equations can be simplified as

0

c o s h ( 0 ) s in h ( 0 )

1( s in h ( 0 ) c o s h ( 0 ))

s

s

V A B

I A BZ

0

s

s

V A

I Z B


Substitute A and B values in basic voltage and current equationswe obtain

0

0

c o s h s in h

s in h c o s h

s s

s

s

V V z I Z z

VI z I z

Z

These equations are called transmission line equations. The canrepresent voltage and current at any point z from the sourcevoltage and current.

0

s

s

V A

I Z B

The constants A and B can be simplified as


Substitute A and B values in basic voltage and current equations

we obtain0

0

c o s h s in h

s in h c o s h

s s

s

s

V V z I Z z

VI z I z

Z

These equations are called transmission line equations. The can

represent voltage and current at any point z from the source voltage

and current.

0

s

s

V A

I Z B

The constants A and B can be simplified as


Input Impedance of Transmission Line from Load

Impedance

If the source voltage and source current is known to us, input

impedance of a transmission line is derived from source end as

Similar expression can be derived from terminating voltage or current

is known

This can be derived from a transmission line length l with known

terminating voltage and current

Transmission Line with Load Impedance

0

0

0

tan h

tan h

R

in

R

Z l ZZ Z

Z Z l

This can be start from basic transmission line equations with two

unknown constants A and B

0

c o s h s in h

1( s in h c o s h )

V A z B z

I A z B zZ

Now the terminating voltage and current at a distance z=l can be written

as

To derive the constants A the above voltage and current equations can

be multiplied with coshγl and sinhγl


0

c o s h s in h

1( s in h c o s h )

R

R

V A l B l

I A l B lZ

Then we get the constants A as

2

2

0

co sh co sh s in h co sh

1s in h ( s in h s in h co sh )

R

R

V l A l B l l

I l A l B l lZ

0c o sh s in h

R RV l Z I l A

Similarly, the constant B can be obtained by multiplying voltage

equation with sinhγl and current equation with coshγl , and adding

both the equations, we get B as


Then substitute A and B constants in basic equation. we obtain voltage

and current at any point from the terminating point as

0c o sh s in h

R RV l Z I l B

0

0

c o sh ( ) s in h ( )

s in h ( ) c o sh ( )

R

R

R

R

VV V l z l z

Z

VI l z I l z

Z

Where, l-z is the distance from the terminating end

Now from the above expressions, the input impedance z=0 can be

written as


0

0

c o s h ( 0 ) s in h ( 0 )

s in h ( 0 ) c o s h ( 0 )

S R R

R

S R

V V l I Z l

VI l I l

Z

0

0

c o s h s in h

s in h c o s h

S R R

RSR

V V l I Z l

VIl I l

Z

0

0

0

tan h

tan h

R

in

R

Z l ZZ Z

Z Z l

Infinite Transmission Line


>If a line of infinite length is considered then all the power fed into it

will be absorbed. The reason is as we move away from the input

terminals towards the load, the current and voltage will decrease

along the line and become zero at an infinite distance, because the

voltage drops across the inductor and current leaks through the

capacitor.

>By considering this hypothetical line of infinite line an important

terminal condition is formed.

Infinite Transmission LineLet Vs be the sending end voltage and Is be the sending end current and

Zs be the input impedance which is given by

Current at any point distance x from sending end is given by

The value of c & d can now determined by considering an infinite line.

At the sending end x=0 and I =Is.

At the receiving end, I=0 and x=∞.

Zs =Vs/Is

-PxPxde+ce=I

d+cIs

c=0 0


Therefore

If then

Therefore

The above equation gives current at any point of an infinite line

And

Similarly the above equation gives voltage at any point of an infinite line

Px

seVV

0c

0cdI

s

Px

seII

Px

seVV

Infinite line terminated in its Z0

When a finite length of line is joined with a similar kind of infinite line,

their total input impedance is same as that of infinite line itself,

because they together make one infinite line however the infinite line

alone presents an impedance Zo at its input PQ because the input

impedance of an infinite line is Zo.

It is therefore concluded that a finite line has an impedance Zo when it is

terminated in Zo.

Or A finite line terminated by its Zo behaves as an infinite line.

Let a finite length of „l‟ is terminated by its characteristic impedance Zo

and is having voltage and current VR and IR at terminating end.


Therefore

Dividing VR by IR we get Zo

R

R

I

V=Zo

have weequations, generalin I =I,V =Vl,= xPuttingRR

sinhpl Z

V - coshpl I=I

O

S

SR


Multiplying right hand side numerator and denominator by Z0, we get

Therefore

But Vs/Is is equal to the input impedance.

Thus the input impedance of a finite line terminated in its characteristicimpedance is the characteristic impedance of the line. Since theinput impedance of an infinite line is the characteristic impedance ofthe line.

Characteristic Impedance, Phase Velocity

The characteristic impedance, Z0 can be defined as:

0

R j L R j LZ

G j C

characteristic impedance (Zo) is the ratio of voltage to

current in a forward travelling wave, assuming there is no

backward wave

• Zo determines relationship between voltage and

current waves

• Zo is a function of physical dimensions and r

• Zo is usually a real impedance (e.g. 50 or 75

ohms)

Characteristic impedance, Z0 :

Characteristic impedance, Z0 :

• Voltage waveform can be expressed in time domain as:

• The factors V0+ and V0

- represent the complex quantities. The Φ± is

the phase angle of V0±. The quantity βz is called the electrical length

of line and is measured in radians.

0 0, c os c os

z zv z t V t z e V t z e

Phase Velocity

• Phase Velocity For a sinusoidal wave, or a waveform comprised of

many sinusoidal components that all propagate at the same velocity,

the waveform will move at the phase velocity of the sinusoidal

components We‟ve seen already that the phase velocity is

• vp=ω/k

Group Velocity

• When the various frequency components of a waveform have

different phase velocities

• The phase velocity of the waveform is an average of these

velocities (the phase velocity of the carrier wave), but the waveform

itself moves at a different speed than the underlying carrier wave

called the group velocity

Group vs Phase velocity

• An analogy that may be useful for understanding the difference

comes from velodrome cycling

• Riders race as a team and take turns as leader with the old leader

peeling away and going to the back of the pack

• As riders make their way from

the rear of the pack to the front

they are moving faster than

the group that they are in.

Group vs Phase velocity

• The phase velocity of a wave is

V=vp = λ T = fλ = ω/k

and comes from the change in the position of the wavefronts as a

function of time

• The waveform moves at a rate that depends on the relative position

of the component wavefronts as a function of time. This is the group

velocity and is

• vg = dω /dk

Condition for distortion less and

minimum attenuation

Condition for Distortion Less Line

• A transmission line is said to be distortion less if the attenuation

constant is frequency independent.

• The phase constant is linearly dependent of frequency.

• From the general equations of α&β, The distortion less line results,

if the line parameters are such that,

R/L=G/C

The attenuation constant and phase constant are

α= √RG and β= ω √LC

Condition for Minimum Attenuation

• From the equation the attenuation of a line is expressed by

attenuation constant DC as

α = ½

• It is observed that depends on the four primary constants in

addition to the frequency. The Value of L for minimum attenuation.

• Let us assume the three line constants C,G and R including are

constant and only L may be varied .

• Therefore, differentiating above value of , W.R.T L and equating

it to zero, we get

Condition for Minimum Attenuation:

then by solving above equation we get

R/L=G/C

This is the condition for minimum attenuation and

distortion less line.

= 0

= 0

Condition for Distortion Less Line

CONCLUSIONS:• The phase velocity is independent of frequency because, the phase

constant of linearly depends upon frequency

• Both VP and Z0 remains the same for loss less line.

• A loss less line is also a distortion less line, but a distortion less line

is not necessarily loss less.

• Loss less lines are desirable in power transmission telephone lines

are to b distortion less.

UNIT-V

UHF Transmission Lines and Applications

UHF Transmission Lines

• UHF spectrum is used worldwide for land mobile radio systems for

commercial, industrial, public safety, and military purposes.

• Many personal radio services use frequencies allocated in the UHF

band, although exact frequencies in use differ significantly between

countries.

• Major telecommunications providers have deployed voice and data

cellular networks in UHF/VHF range. This allows mobile phones

and mobile computing devices to be connected to the public

switched telephone network and public internet.

• UHF radars are said to be effective at tracking stealth fighters, if not

stealth bombers

UHF Transmission Lines• Ultra High Frequency lines commonly abbreviated as

U.H.F lines are one of the types of the transmission lines.

• Ultra high frequency lines have operational frequency

range from 300 to 3000 MHz or wavelength from 100 cm

to 10 cm.

• Under normal frequencies the transmission lines are used

as wave guides for transferring power and information

from one point to another.


• At Ultra High Frequencies, the transmission lines can be used as

circuit elements like capacitor or an inductor .

• It means they can be used in circuits like a capacitor or an inductor.

Applications:

• UHF Television Broad casting fulfilled the demand for additional

over-the-air television channels in urban areas. Today, much of the

bandwidth has been reallocated to land mobile, trunked radioand

mobile telephone use. UHF channels are still used for digital

television


• UHF spectrum is used worldwide for land mobile radio systems for

commercial, industrial, public safety, and military purposes.

• Many personal radio services use frequencies allocated in the UHF

band, although exact frequencies in use differ significantly between

countries.

• Major telecommunications providers have deployed voice and data

cellular networks in UHF/VHF range. This allows mobile phones

and mobile computing devices to be connected to the public

switched telephone network and public internet.

• UHF radars are said to be effective at tracking stealth fighters, if not

stealth bombers

Input impedance Relations; SC and OC lines

Let the voltage and current transmission line equations at any point on

the line from the source end can be written as

A transmission line, which is terminated with some load impedance

ZR at a distance „l „from the load

The voltage and current at the terminating end is VR and IR

Input Impedance Relations

0

0

c o s h s in h

s in h c o s h

s s

s

s

V V z I Z z

VI z I z

Z

At z= l, the voltage and current can be written as

0

0

c o s h s in h

s in h c o s h

R s s

s

R s

V V l I Z l

VI l I l

Z

Now the load impedance from the terminating point can be written as

By solving the above equation, we obtain


0

0

c o s h s in h

s in h c o s h

s sR

R

sRs

V l I Z lVZ

VIl I l

Z

Where ZS is the source impedance, also called input impedance

Above expression can be written in terms of hyperbolic tan

functions as

0

0

0

s in h co sh

co sh s in h

R

s

R

Z l Z lZ Z

Z l Z l

0

0

0

tan h

tan h

R

in s

R

Z l ZZ Z Z

Z Z l

For a loss less line the input impedance can be written as


0

0

0

tan h

tan h

R

in s

R

Z j l ZZ Z Z

Z Z j l

A special cases from the above general lossless line input impedance

relations

• When the line is terminated with characteristics impedance, ZR=Z0

• When the line is terminated with open circuited, ZR=infinity(∞)

• When the line is terminated with short circuited, ZR=0

0

0

0

tan

tan

R

in s

R

jZ l ZZ Z Z

Z jZ l

Lossless Transmission line(α=0)



0

0

0

tan h

tan h

R

in s

R

Z j l ZZ Z Z

Z Z j l

The line is terminated with characteristic impedance, the input

impedance is equal to the characteristic impedance

This condition is called matched load condition

There is no reflections on the line

0 0

0 0

0 0

tan

tanin

jZ l ZZ Z Z

Z jZ l

Matched Load ZR=Z0



0

0

0

tan

tan

R

in

R

jZ l ZZ Z

Z jZ l

0

0

0

ta n 1

ta n

R

in O C

R

Zj l

ZZ Z Z

Zj l

Z

Open Circuited, i.e. ZR=infinity(∞)

0c o t

in O CZ Z jZ l

Short Circuited, i.e. ZR=0

0ta n

in S CZ Z jZ l

If the line is terminated with open circuit or short circuit, the input

impedance can be purely imaginary

0

0

0

tan

tan

R

in

R

jZ l ZZ Z

Z jZ l

From the short circuit and open circuit impedance relations, the

characteristic impedance can be written as


0

0( ta n )

ta nO C S C

ZZ Z jZ l

j l

The characteristic impedance of the line can be measured from open and

short circuit Transmission line

2

0

0

O C S C

O C S C

Z Z Z

Z Z Z

Impedance Matching, Single Stub Matching

Impedance matching is one of the important aspects of high

frequency circuit analysis.

To avoid reflections and power loss from transmission line sections

impedance matching techniques can be used

• Quarter Wavelength Transformer,

• Stub Matching

Single Stub Matching

Double Stub Matching

Impedance Matching

The quarter wave transformer needs special line of characteristics

impedance for every pair of resistances to be matched

To add quarter wave transformer, cut the main line

To avoid above difficulties, either open or short circuited transmission

line attached at some position parallel to line

Impedance Matching

Adding of either open or short line in parallel to main line is call stub

The main advantage of this stub matching is

• The length and characteristic impedance remains same

• Since the stub is added in shunt, there is no need to cut the main line

• The susceptance of the stub can be adjusted for perfect matching

A short-circuited stub is less prone to leakage of electromagnetic

radiation and is somewhat easier to realize.

Open circuited stub may be more practical for certain types of

transmission lines, for example microstrips where one would have to

drill the insulating substrate to short circuit the two conductors of the

line


Based on the number of parallel stubs connected to the line stub

matching techniques can be classified into two types

• Single-Stub Matching Technique

• Double- Stub Matching Technique

A short-circuited section (stub) of a

transmission line connected in parallel to

the main transmission line is shown in

Figure

The stub is connected in parallel, so it is

easy to deal with admittance analysis

instead of impedance



There are two design parameters for single stub matching

• The location of the stub with reference to the load can be

represented as dstub

• The length of the stub line Lstub

For proper impedance matching, the admittance at the location of the

stub can be written as0

0

1(d ) Y

A s tu b s tu bY Y Y

Z

Location of the stub(dstub)


To determine the location of the short-circuited stubs, the input

impedance of a lossless transmission line and convert it to admittance

0

ta n

1 ta n

in R

in

R

Y y j ly

Y jy l

2 2

2 2

2 2

2 2 2 2

ta n ta n ta n

1 ta n

1 ta n 1 ta n

1 ta n 1 ta n

R R R

in

R

R R

R R

y y l jy l j ly

y l

y l j y l

y l y l

0

0

0

ta n

ta n

R

in

R

Y jY lY Y

Y jY l

The normalized admittance can be written as

Separating the real and imaginary parts by rationalizing


0R e [ ] 1

in inY Y o r y

2

2 2

1 ta n1

1 ta n

R s tu b

R s tu b

y d

y d

2 2 2

2

2

1

1 0

1 ta n 1 ta n

1 1ta n

1 1ta n

2ta n

R s tu b R s tu b

R

s tu b

RR R

s tu b

R

s tu b

R

y d y d

yd

yy y

d

y

Yd

Y

For no reflection, at a distance l=dstub the real part of the admittance

is unity

At l=dstub

Simplifying the above expression, we obtain


2 2

2 2 2 2

1 ta n 1 ta nIm [ ]

1 ta n 1 ta n

R R s tu b

in s

R R s tu b

y l y dy b

y l y d

2 2

2 2 2 2

2

2

2

0 0

0

0

1 ta n 1 ta n

1 ta n 1 ta n

11

1 1

1 (1 )1

1

. .

R R s tu b

s

R R s tu b

R

RR R

s

R R RR

R

R

R

s

R R

y l y db

y l y d

yyy y

b

y y yy

y

Y

Y Y Yi e b

Y Y Y

Y

At this location the imaginary part of susceptance of bs can be

written as

Therefore at length dstub the input admittance is

Length of the stub(Lstub)


To determine the length of the short-circuited stub consider the short

circuited impedance and write in terms of admittance

That is

0

c o ts

s

Bb l

Y

0

0

0

0

c o t

ta n

R

s tu b

R

R

s tu b

R

Y YL

Y Y

Y YL

Y Y

0

0

tan

co t

sc s

sc s

Z jX jZ l

Y jB jY l

The required normalized suspetance of the short circuited stub

Equating with ysc


0

0

1

ta n1 1

R

s tu b

R

Z ZL

Z Z

0

0

R

R

Z Z

Z Z

01

0

01

0

ta n2

ta n2

R

s tu b

R

R

s tu b

R

Z ZL

Z Z

Z ZL

Z Z

Converting into impedance

Therefore, the length of short circuited stub is

for

for

The drawback of this approach is that if the load is changed, the

location of insertion may have to be moved

Smith Chart

Smith Chart

It is a polar plot of the complex reflection coefficient.

It is the transformation of complex impedance into

reflection coefficient plane.

Figure 1: Transformation of Z into

Smith Chart

• Construction of smith chart:-

• It consist of r-circles and x-circles

Fig-2: r-circles Fig-3: x-circles

Smith Chart

• Points to remember regarding smith chart:

The value of r is always positive, x can be positive (for inducatance

impedance) and negative (for capacitance impedance)

Apart from the r and x circles, we can draw the VSWR-circles or (S-circles)(ALWAYS NOT SHOWN ON THE CHART)

Figure 4: Some points on smith chart

Smith Chart At point Psc on the chart r = 0 and x = 0, it represents short circuit point. Similarly

Poc on the chart r = infinity and x = infinity represents open circuit point.

A complete revolution around the smith chart represents a distance of on the

line(360 degress).

Clockwise movement on the chart is regarded as moving toward the generator.

Similarly, counter clockwise movement on the chart corresponds to moving towards

load.

There are 3 scales around periphery of the smith chart

> Outermost scale used to determine distance on the from generator end in terms

of wave length.

> The next scale determines distance from the load end interms of WL

> Outermost scale used to determine distance on the from generator end in terms

of wave length.

> The innermost scale is a protractor and is primarily used in determining angle of

reflection coefficient

Smith Chart

To the horizontal line upper part is inductive in nature and bottom part is capacitive in nature.

As shown in figure (4):

> The left most point A on the smith chart corresponds to and therefore represents ideal short-circuit load.

> The right most point B on the Smith chart corresponds to , and therefore represents ideal open circuit load.

>The upper most point C represents a pure inductive load of unity reactance and the lower most point D represents a pure capacitive load of unity reactance.

Voltage Vmax occurs where impedance is maximum i.e point A and Vminoccurs where impedance is minimum i.e point B.

ELECTROMAGNETIC THEORY AND TRANSMISSION LINES … · 2 0 1 G[E = r' . 4 0 U & 2 0 S 1 dS E = r' . 4...

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Transcript of ELECTROMAGNETIC THEORY AND TRANSMISSION LINES … · 2 0 1 G[E = r' . 4 0 U & 2 0 S 1 dS E = r' . 4...