ECE 546 Lecture 02 Review of Electromagneticsemlab.uiuc.edu/ece546/Lect_02.pdfECE 546 –Jose...

ECE 546 – Jose Schutt‐Aine 1

ECE 546 Lecture 02

Review of ElectromagneticsSpring 2018

Jose E. Schutt-AineElectrical & Computer Engineering

University of [email protected]


Printed Circuit Board


High-Density Package


Electromagnetic Quantities

E

H

D

B

Electric field (Volts/m)

Magnetic field (Amperes/m)

Electric flux density (Coulombs/m2)

Magnetic flux density (Webers/m2)

J

Current density (Amperes/m2)

Charge density (Coulombs/m2)


Faraday’s Law of Induction

Ampère’s Law

Gauss’ Law for electric field

Gauss’ Law for magnetic field

BEt

H J D

t

D

0B

Maxwell’s Equations


B H

D E

Constitutive Relations

Permittivity : Farads/m

Permeability : Henries/m

Free Space128.85 10 /o F m

74 10 /o H m


Continuity EquationDH Jt

0DH J J Dt t

D

0Jt


0E

with E

ElectrostaticsAssume no time dependence 0

t

Since 0, such thatE

where is the scalar potentialE

2we get

Poisson’s Equation 2we get

2 0 Laplace’s Equationif no charge is present


Integral Form of ME

C S

E dl B dSt

C S S

H dl J dS D dSt

S V

D dS dv

0S

B dS


Boundary Conditions

1 2ˆ 0n E E

1 2ˆ Sn H H J

1 2ˆ Sn D D

1 2ˆ 0n B B

n̂

1 1 1 1, , ,E H D B

2 2 2 2, , ,E H D B


Faraday’s Law of Induction

Ampère’s Law

Gauss’ Law for electric field

Gauss’ Law for magnetic field

oHEt

oEHt

0E

0B

Free Space Solution


oE Ht

2o o

EE Et t

Wave Equation

22

2o oEEt

Wave Equation

can show that2

22o oHHt


Wave Equation2 2 2 2

2 2 2 2o oE E E E

x y z t

separating the components

2 2 2 2

2 2 2 2x x x x

o oE E E Ex y z t

2 2 2 2

2 2 2 2y y y y

o o

E E E Ex y z t

2 2 2 2

2 2 2 2z z z z

o oE E E Ex y z t


Wave Equation Plane Wave

0x y

(a) Assume that only Ex exists Ey=Ez=0

2 2

2 2x x

o oE Ez t

(b) Only z spatial dependence

This situation leads to the plane wave solution

In addition, assume a time‐harmonic dependencej t

xE e then jt


Plane Wave Solution

j zx o

j zx o

E E e forward waveE E e backward wave

solution

where

In the time domain

22

2x

o o xE Ez

o o propagation constant

( ) cos( ) cosx o

x o

E t E t z forward traveling waveE t E t z backward traveling wave

solution


Plane Wave Characteristics

where

In free space

o o propagation constant

1 v propagation velocity

81 3 10 /o o

v c m s


HE E j Ht

Solution for Magnetic Field

ˆ j zoE xE e

If we assume that

ˆ ˆ ˆ1 1

0 0x

x y z

H Ej j x y z

E

then ˆ j zoEH y e

ˆ j zoE xE e

If we assume that then ˆ j zoEH y e

intrinsic impedance of medium


( )P t E t H t

Time-Average Poynting Vector

We can show that

Poynting vector W/m2

time‐average Poynting vector W/m2

0 0

1 1( )T T

P P t dt E t H t dtT T

*1 Re2

P E H

where and are the phasors of ( ) and ( ) respectivelyE H E t H t


: conductivity of material medium (‐1m‐1)

HEt

EH Jt

J E

Material Medium

E j H

H J j E

1H E j E E j j Ej

2 2 1E Ej

1

j

or

since then


Wave in Material Medium2 2 21E E E

j

is complex propagation constant

2 2 1j

1j jj

associated with attenuation of wave

associated with propagation of wave


Wave in Material Medium

ˆ ˆz z j zo oE xE e xE e e

1/22

1 12

decaying exponentialSolution:

1/22

1 12

ˆ z j zoEH y e e

jj

Magnetic field

Complex intrinsic impedance


Wave in Material Medium1/2

22 1 1pv

Phase Velocity:

0

0

1. Perfect dielectric

Special Cases

Wavelength:1/2

22 2 1 1f

air, free space

and


Wave in Material Medium2. Lossy dielectric

Loss tangent: 1

2

2 2

318 2

j

2

2 218

2

2 212 8 2


Wave in Material Medium

3. Good conductors

Loss tangent: 1

j j j

f f

1j f jj


attenuation

propagation

dp H, E Examples

PEC ‐ 0 0 0 supercond

Good conductor finite copper

Poor conductor finite

Ice

Perfectdielectric 0 finite

air

2

2 1f j

1f

2

/

12

j

2

/

Material Medium


Radiation - Vector Potential

E j H

H J j E

/D

0B

(1)

(2)

(3)

(4)

Assume time harmonicity ~ j te


Radiation - Vector Potential

0 such thatB A A B

A

: vector potential

0A

0E j A E j A

Using the property: 0any vector


0vector vector

where is the scalar potentialE j A

Since a vector is uniquely defined by its curl and its divergence, we can choose the divergence of A

choose such thatA

0A j Lorentz

condition

Vector Potential


B J j E

A J j j A

2 2A A J A j

2 2A j J A j

2 2A A J

D’Alembert’sequation

Vector Potential


'

, '4 '

j r reG r rr r

'

'

''

4 '

j r r

V

J r eA r dv

r r

From A, get E and H using Maxwell’s equations

Three-dimensional free-space Green’s function

Vector potential

Vector Potential


ˆ' ( ') ( ') ( ')oJ r zI dl x y z

For infinitesimal antenna, the current density is:

ˆ4

j roI dlA r z er

ˆˆˆ cos sinz r

Calculating the vector potential,

In spherical coordinates,

Vector Potential


Resolving into components,

ˆˆˆ cos sin4

j roI dlA r z r er

cosˆ4

j ror

I dlA z er

sin4

j roI dlA er

Vector Potential


Calculate E and H fields

1H A

1 sin 0sinr

AH A

1 1 0sin

rAH rAr r

E and H Fields


1 rAH rAr r

11 sin4

j roI dlH j er j r

1E Hj

1 1sinsinrE H

r j

E and H Fields


2

2cos 1 114

j ror

I dlE j er j r j

1 1E rHr r j

2 2

1 1 1sin 14

j roj I dlE j er j r r j

E and H Fields


21 1sin 1

4j roj I dlE e

r j r j r

2

2cos 1 114

j ror

I dlE j er j r j

11 sin4

j roI dlH j er j r

E and H Fields


Far field : 1 orr r 2

1then, 0r

, whereE H

0rE

sin4

j roj I dlH er

Note that:

sin4

j roj I dlE er

Far Field Approximation


• Uniform constant phase locus is a plane• Constant magnitude• Independent of • Does not decay

Characteristics of plane waves

Similarities between infinitesimal antenna far field radiated and plane wave

(a) E and H are perpendicular(b) E and H are related by (c) E is perpendicular to H

Far Field Approximation


P t E t H t

Time-average Poynting vector or TA power density

1 Re *2

P E H

E and H here are PHASORS

22 2ˆ ˆRe sin

2 2 4oI dlrP H rr

Poynting Vector


Total power radiated (time-average)

2

0 0P ds

P = 2ˆwith sinds rr d d

22 2 2

0 0sin sin

2 4oI dl

r d dr

P =

23

0

2 sin2 4

oI dl d

P =

Time-Average Power


243 4

oI dl

P =

Poynting Power DensityDirective Gain =Average Poynting Power Densityover area of sphere with radius r

2Directive Gain =/ 4

P

r

P

Time-Average Power


22

22

sin2 4

Directive Gain4 / 43 4

o

o

I dlr

I dlr

For infinitesimal antenna,

23Directive Gain sin2

Directivity


Directivity: gain in direction of maximum value

Radiation resistance:

From 212 oRIP we have: 2

2rad

o

RI

P

For infinitesimal antenna:

2

2 2

2 4 23 4 3

orad

o

I dl dlRI

Directivity


2280rad

dlR

For free space, 120

The radiation resistance of an antenna is the value of a fictitious resistance that would dissipate an amount of power equal to the radiated power Prwhen the current in the resistance is equal to the maximum current along the antenna

(for Hertzian dipole)

A high radiation resistance is a desirable property for an antenna

Radiation Resistance

Date post:	21-Apr-2018
Category:	Documents
Upload:	lytram
View:	217 times
Download:	1 times

ECE 546 Lecture 02 Review of Electromagneticsemlab.uiuc.edu/ece546/Lect_02.pdfECE 546 –Jose...

Documents