emt electromagnetic theory module iv

MODULE IV

Electromagnetic Wave Propagation

Compiled by MKP for CEC S5 batch September 2008

SyllabusElectromagnetic wave propagation: Electromagnetic waves - wave propagation in lossy dielectrics - wave equations from Maxwell’s equations - propagation constant - intrinsic impedance of the medium - complex permittivity - loss tangent - plane waves in lossless dielectrics - plane waves in free space - uniform plane wave - TEM wave - plane waves in good conductors - skin effect - Poynting vector -Poynting’s theorem. Reflection of a plane wave at normal incidence - standing waves - Reflection of plane waves at oblique incidence - parallel and perpendicular polarization -Brewster angle. Numerical methods in electromagnetics -finite difference - finite element and moment method ( only concept need to be introduced, detailed study not required)


ReferencesText Books:

1. Mathew N.O. Sadiku, Elements of Electromagnetics, Oxford University Press

2. Jordan and Balmain, Electromagnetic waves and radiating systems,Pearson Education PHI Ltd.References:

1. Kraus Fleisch, Electromagnetics with applications, McGraw Hill2. William.H.Hayt, Engineering Electromagnetics, Tata McGraw Hill3. N.Narayana Rao, Elements of Engineering Electromagnetics, Pearson

Education PHI Ltd. 4. D.Ganesh Rao, Engineering Electromagnetics, Sanguine Technical

Publishers.5. Joseph.A.Edminister, Electromagnetics, Schaum series-McGraw Hill6. K.D. Prasad, Electromagnetic fields and waves, Sathya Prakashan


Maxwell’s Equations in final forms

VD ρ∇⋅ =

0B∇⋅ =

BEt

∂∇× = −

∂

VS VD dS dVρ⋅ =∫ ∫

0SB dS⋅ =∫

L S

dE dl B dSdt

⋅ = − ⋅∫ ∫

L S

DH dl J dSt

⎛ ⎞∂⋅ = + ⋅⎜ ⎟∂⎝ ⎠

∫ ∫

Differential form Integral form Derived from

' Gauss s Law

Nonexistance of magneticMonopole

' Faraday s Law

' Ampere s Law DH J

t∂

∇× = +∂ modified by continuity eqn


Maxwell’s Equations for lossless or non conducting medium

0D∇⋅ =

0B∇⋅ =

BEt

∂∇× = −

∂

0SD dS⋅ =∫

0SB dS⋅ =∫

L S

BE dl dSt

∂⋅ = − ⋅

∂∫ ∫

L S

DH dl dSt

∂⋅ = ⋅

∂∫ ∫

Differential form Integral form

DHt

∂∇× =

∂

In lossless medium, current density J and charge density ρ are zero and Maxwell’s equations are simplified as below.

Eqn

ABC

D


Wave equations for lossless or non conducting medium

Differentiating eq (D) with respect to t

The order of differentiation on the LHS may be changed as the curl operation itself is a differentiation

Taking the curl of eq (C)

( ) (1)DHt t t

⎛ ⎞∂ ∂ ∂∇× = − − −⎜ ⎟∂ ∂ ∂⎝ ⎠

2

0 2 (2)H Et t

ε∂ ∂∇× = − − −

∂ ∂

BEt

∂∇×∇× = −∇×

∂



Substituting eq (2) in eq (3)

(3)HEt

μ⎛ ⎞∂

∇×∇× = − ∇× − − −⎜ ⎟∂⎝ ⎠

2

2 EEt

μ ε⎛ ⎞∂

∇×∇× = − ⎜ ⎟∂⎝ ⎠2

2 (4) EEt

με ∂∇×∇× = − − − −∂

( ) ( ) But A B C A C B A B C× × = ⋅ − ⋅

( ) ( ) And so E E E∇×∇× = ∇⋅ ∇ − ∇⋅∇



Substituting in eq (4)

Equation (5) represents wave equation in free space in terms of E.

( ) 2 E E E∇×∇× = ∇⋅ ∇ −∇

0 0 0 t Eu D EB ε∇⋅ = ∇⇒ = ∇⋅ =⇒⋅2 And so E E∇×∇× = −∇

22

2 EEt

με ∂−∇ = −∂

22

2 (5) EEt

με ∂∇ = − − −∂

22

2 EEt

με ∂∇ =∂



Differentiating eq (C) with respect to t

The order of differentiation on the LHS may be changed as the curl operation itself is a differentiation

Taking the curl of eq (D)

( ) (6)BEt t t

⎛ ⎞∂ ∂ ∂∇× = − − − −⎜ ⎟∂ ∂ ∂⎝ ⎠

2

2 (7)E Ht t

μ∂ ∂∇× = − − − −

∂ ∂

DHt

∂∇×∇× = ∇×

∂



Substituting eq (7) in eq (8)

(8)EHt

ε⎛ ⎞∂

∇×∇× = ∇× − − −⎜ ⎟∂⎝ ⎠

2

2 HHt

ε μ⎛ ⎞∂

∇×∇× = −⎜ ⎟∂⎝ ⎠2

2 (9) HHt

με ∂∇×∇× = − − − −∂

( ) ( ) But A B C A C B A B C× × = ⋅ − ⋅

( ) ( ) And so H H H∇×∇× = ∇⋅ ∇ − ∇⋅∇



Substituting in eq (9)

Equation (10) represents wave equation in lossless medium in terms of H.

( ) 2 H H H∇×∇× = ∇⋅ ∇ −∇

0 0 0 t Hu B HB μ∇⋅ = ∇⇒ = ∇⋅ =⇒⋅2 And so H H∇×∇× = −∇

22

2 HHt

με ∂−∇ = −∂

22

2 (10) HHt

με ∂∇ = − − −∂

22

2 HHt

με ∂∇ =∂



The wave equations for may be obtained by multiplying the wave equation for

D Band E Hε μby and equation for by

22

2 EEt

με ∂∇ =∂

( ) ( )22

2 E

Etε

ε με∂

∇ =∂

22

2 DDt

με ∂∇ =∂

22

2 DDt

με ∂∇ =∂



22

2

HHt

με ∂∇ =∂

( ) ( )22

2

HH

tμ

μ με∂

∇ =∂

22

2

BBt

με ∂∇ =∂

22

2

BBt

με ∂∇ =∂

Similarly,



In rectangular coordinate system, the wave equations assumes theform of scalar wave equations in terms of its components

22

2x

xEEt

με ∂∇ =∂

22

2y

y

EE

tμε

∂∇ =

∂2

22

zz

EEt

με ∂∇ =∂

⎫⎪⎪⎬⎪⎪⎭

For E, , For D H B we can obtain similar equations


Wave equations for lossless or non conducting media: sinusoidal time variations

If the electric field intensity is varying harmonically with time

Using these in wave equation for

j ts

E j E et

j Eω ωω∂=

∂=

j tsE E e ω=

22 22

2j t

sE j E et

Eω ωω∂=

∂= −

22

2

EEt

με ∂∇ =∂

E

2 2E Eω με∇ = − 2 2E Eω με∇ = −


Wave equations for lossless or non conducting media: sinusoidal time variations

Similarly we may obtain the following wave equations for the other harmonically varying fields

2 2H Hω με∇ = −

2 2D Dω με∇ = −

2 2B Bω με∇ = −

2 2E Eω με∇ = −2 2D Dω με∇ = −2 2H Hω με∇ = −2 2B Bω με∇ = −

Homogeneous Vector wave equations in complex time harmonic form for free space


Wave equations for free space

2 20 0E Eω μ ε∇ = −

2 20 0D Dω μ ε∇ = −

2 20 0H Hω μ ε∇ = −

2 20 0B Bω μ ε∇ = −

22

0 0 2

BBt

μ ε ∂∇ =

∂

22

0 0 2 DDt

μ ε ∂∇ =

∂2

20 0 2 HH

tμ ε ∂

∇ =∂

22

0 0 2 EEt

μ ε ∂∇ =

∂

GENERAL SINUSOIDAL TIME VARIATIONS


Maxwell’s Equations for conducting medium

D ρ∇⋅ =

0B∇⋅ =

BEt

∂∇× = −

∂

VS VD dS dVρ⋅ =∫ ∫

0SB dS⋅ =∫

L S

BE dl dSt

∂⋅ = − ⋅

∂∫ ∫

L S

DH dl J dSt

⎛ ⎞∂⋅ = + ⋅⎜ ⎟∂⎝ ⎠

∫ ∫

Differential form Integral form

DH Jt

∂∇× = +

∂

In a conducting medium with conductivity σ and charge density ρ and Maxwell’s equations are as given below below.

Eqn

ABC

D


Wave equations for lossy or conducting medium

From equation (D)

DH Jt

∂∇× = +

∂

( )H E Et

σ ε∂∇× = +

∂

(1)EH Et

σ ε ∂∇× = + − − −∂

( )2

2

E EHt t t

σ ε∂ ∂ ∂∇× = +

∂ ∂ ∂

2

2 (2)H E Et t t

σ ε∂ ∂ ∂∇× = + − − −

∂ ∂ ∂



From equation (C)

Taking curl of this equation

Putting eq (2) in (3)

BEt

∂∇× = −

∂HEt

μ ∂∇× = −∂

(3)HEt

μ⎛ ⎞∂

∇×∇× = − ∇× − − −⎜ ⎟∂⎝ ⎠

2

2 (4)E EEt t

μ σ ε⎛ ⎞∂ ∂

∇×∇× = − + − − −⎜ ⎟∂ ∂⎝ ⎠

( ) 2 (5)But E E E∇×∇× = ∇ ∇⋅ −∇ − − −



Using eq (5) in (4)

But from equation (A) we have

Putting in equation (6)

( )2

22 (6)E EE E

t tμσ με∂ ∂

∇ = ∇ ∇⋅ + + − − −∂ ∂

( )2

22 (4)E EE E

t tμ σ ε⎛ ⎞∂ ∂

∇ ∇⋅ −∇ = − + − − −⎜ ⎟∂ ∂⎝ ⎠

D ρ∇⋅ = E ρε

∇⋅ =

22

2 (7)E EEt t

ρ μσ μεε

∂ ∂⎛ ⎞∇ = ∇ + + − − −⎜ ⎟ ∂ ∂⎝ ⎠



There is no charge within a conductor, although it may be there on the surface, the charge density ρ=0. So we can rewrite equation (7) as below.

This is the wave equation for conducting medium in terms of E

22

2

E EEt t

μσ με∂ ∂∇ = +

∂ ∂

22

2

E EEt t

μσ με∂ ∂∇ = +

∂ ∂


Wave equations for lossy or conducting mediumFrom equation (D)

Taking the curl of this equation

DH Jt

∂∇× = +

∂

( )H E Et

σ ε∂∇× = +

∂

(1)EH Et

σ ε ∂∇× = + − − −∂

( ) (2)EH Et

σ ε⎛ ⎞∂

∇×∇× = ∇× + ∇× − − −⎜ ⎟∂⎝ ⎠

( ) 2 (3)But H H H∇×∇× = ∇ ∇⋅ −∇ − − −



Using eq (3) in (2)

From equation (C)

Putting in (5) and (6) in (4)

( ) ( )2 (4)EH H Et

σ ε⎛ ⎞∂

∇ ∇⋅ −∇ = ∇× + ∇× − − −⎜ ⎟∂⎝ ⎠

(5)BEt

∂∇× = − − − −

∂

( )2

22 (7)B BH H

t tσ ε⎛ ⎞ ⎛ ⎞∂ ∂

∇ ∇⋅ −∇ = − + − − − −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

( )2

2

BEt t∂ ∂

∇× = −∂ ∂

2

2 (6)E Bt t

∂ ∂∇× = − − − −

∂ ∂



This is the wave equation for conducting medium in terms of H

( )2

22 (8)H HH H

t tμσ με∂ ∂

∇ ∇⋅ −∇ = − − − − −∂ ∂

( )2

22 (9)H HH H

t tμσ με∂ ∂

∇ = ∇ ∇⋅ + + − − −∂ ∂

0 0But B H⋅ = =⇒∇ ∇⋅

22

2 H HHt t

μσ με∂ ∂∇ = +

∂ ∂

22

2 H HHt t

μσ με∂ ∂∇ = +

∂ ∂



22

2

E EEt t

μσ με∂ ∂∇ = +

∂ ∂

22

2 H HHt t

μσ με∂ ∂∇ = +

∂ ∂

22

2

D DDt t

μσ με∂ ∂∇ = +

∂ ∂

22

2 B BBt t

μσ με∂ ∂∇ = +

∂ ∂

22

0 0 2

EEt

μ ε ∂∇ =

∂

22

0 0 2 HHt

μ ε ∂∇ =

∂

22

0 0 2

DDt

μ ε ∂∇ =

∂

22

0 0 2 BBt

μ ε ∂∇ =

∂

⇒⇒⇒⇒

0 00, , σ ε ε μ μ= = =


Wave equations for lossy or conducting medium – time harmonic form

2 2E j E Eωμσ ω με∇ = −

2 2 H j H Hωμσ ω με∇ = −

2 2D j D Dωμσ ω με∇ = −

2 2 B j B Bωμσ ω με∇ = −


Uniform plane waveAn electromagnetic wave originates from a point in free space, spreads out uniformly in all directions, and it forms a spherical wave front.An observer at a large distance from the source is able to observe only a small part of the wave and the wave appears to him as a plane wave. For such a wave the electric field and the magnetic field are perpendicular to each other and to the direction of propagation.A uniform plane wave is one in which lie in a plane and have the same value everywhere in that plane at any fixed instant.

E H

E Hand


Uniform plane wave


Uniform plane wave

E

E

E

H

H

H

O

Y

X

Z


Uniform plane waveFor a uniform plane wave travelling in the z direction, the space variations of are zero over a z=constant plane. This implies the fields have neither x nor y dependence.

A plane wave is transverse in nature, that is are both perpendicular to the direction of propagation.So they are called transverse electromagnetic waves (TEM waves)

E Hand

0x y∂ ∂= =

∂ ∂ E Hand


Wave equations for uniform plane wave traveling in z direction in free space

Consider a uniform plane wave propagating in z direction.It will have Ex and Ey components but no Ez component. Ez = 0There is no variation of the field components along x and y direction

Wave equations for free space is given by

0E Ex y

∂ ∂= =

∂ ∂

22

0 0 2

EEt

μ ε ∂∇ =

∂2 2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2 2ˆ ˆ ˆx x y y z zE a E a E ax y z x y z x y z

⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂+ + + + + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

( )2

0 0 2 ˆ ˆ ˆ (1)x x y y z zE a E a E at

μ ε ∂= + + − − −

∂


Wave equations for uniform plane wave traveling in z direction in free space

0; 0yxz

EEBut Ex y

∂∂= = =

∂ ∂

22

2 2 0yx EEx y

∂∂= =

∂ ∂

( )22

2 2

2

0 0 2ˆ ˆ ˆ ˆyxx y x x y y

EE a a E a E atz z

μ ε ∂= +

∂∂∂

+∂ ∂

(1)Putting in eq

Hence2

0

2

2 0 2xx E

tEz

μ ε ∂∂=

∂∂

2

0

2

2 0 2yy E

tEz

μ ε∂∂

=∂∂

Similarly for H we can obtain2

0

2

2 0 2xx H

tHz

μ ε ∂∂=

∂∂

2

0

2

2 0 2yy H

tHz

μ ε∂∂

=∂∂


Wave equations for uniform plane wave in free space traveling in z direction

2

0

2

2 0 2xx E

tEz

μ ε ∂∂=

∂∂

2

0

2

2 0 2yy E

tEz

μ ε∂∂

=∂∂

2

0

2

2 0 2xx H

tHz

μ ε ∂∂=

∂∂

2

0

2

2 0 2yy H

tHz

μ ε∂∂

=∂∂


Wave equations for uniform plane wave traveling in z direction in free space – sinusoidal time variations

2 2

0 02 2x xE E

z tμ ε∂ ∂

=∂ ∂2 2

0 02 2y yE E

z tμ ε

∂ ∂=

∂ ∂2 2

0 02 2x xH H

z tμ ε∂ ∂

=∂ ∂2 2

0 02 2y yH H

z tμ ε

∂ ∂=

∂ ∂

22

0 02x

xE Ez

ω μ ε∂= −

∂2

20 02

yy

EE

zω μ ε

∂= −

∂

22

0 02y

y

HH

zω μ ε

∂= −

∂

22

0 02x

xH Hz

ω μ ε∂= −

∂

⇒

⇒

⇒

⇒


Wave equations for uniform plane wave traveling in z direction in conducting medium

2 2

2 2x x xE E E

z t tμσ με∂ ∂ ∂

= +∂ ∂ ∂

2 2

2 2y y yE E E

z t tμσ με

∂ ∂ ∂= +

∂ ∂ ∂

2 2

2 2x x xH H H

z t tμσ με∂ ∂ ∂

= +∂ ∂ ∂2 2

2 2y y yH H H

z t tμσ με

∂ ∂ ∂= +

∂ ∂ ∂


Wave equations for uniform plane wave traveling in z direction in conducting medium – sinusoidal time variations

22

2x

x xE j E Ez

ωμσ ω με∂= −

∂

22

2y

y y

Ej E E

zωμσ ω με

∂= −

∂2

22

xx x

H j H Hz


∂2

22

yy y

Hj H H

zωμσ ω με

∂= −

∂


Solution of wave equationsStandard partial difference equation for wave motion frequently encountered in engineering has the form

Comparing with the EM wave equation

22

2 2

1 XXv t

∂∇ =

∂ v Velocity of the wave⇒

22

0 0 2 EEt

μ ε ∂∇ =

∂

( )0 0

22

2 2

1 1 /

EEtμ ε

∂∇ =

∂

v Velocity of the wave⇒ =0 0

1μ ε

70 4 10 /Putting F mμ π −= × 12

0 8.854 10 /and H mε −= ×


Solution of wave equations

This is the velocity of light and when referred to electromagnetic wave it is denoted by c.

8

0 0

1 3 10 /v m sμ ε

= = ×

8

0 0

1 3 10 /v c m sμ ε

= = = ×


Solution of wave equation: Uniform plane wave traveling in z direction in free space

In this case the wave equations in E are

The general solution of such a differential equation has the form

2 2

0 02 2x xE E

z tμ ε∂ ∂

=∂ ∂

2 2

0 02 2y yE E

z tμ ε

∂ ∂=

∂ ∂2 2

2 2 2

1x xE Ez c t

∂ ∂=

∂ ∂

2 2

2 2 2

1y yE Ez c t

∂ ∂=

∂ ∂8

0 0

1 3 10 /Where c m sμ ε

= = ×

1 2( ) ( )E f z ct f z ct= − + +


Uniform plane wave traveling in z direction in free space

Here f1 and f2 are any arbitrary functions of (z-ct) and (z+ct)The functions of (z-ct) and (z+ct) may assume any form as,

The first function represents a wave traveling in positive z direction while the second term represents a wave traveling in negative z direction.

( )Trigonometrical A sin z ctβ⇒ −( ) z ctExponential A e β −⇒


Solution of wave equation: Uniform plane wave traveling in z direction in lossless or non conducting medium- sinusoidal time variations

In this case the wave equations in E are 2

22

xx

E Ez

ω με∂= −

∂

22

2y

y

EE

zω με

∂= −

∂2

22 0y

y

EE

zω με

∂+ =

∂2 2 0y yD E Eω με+ =

22

2 Putting Dz∂

≡∂

( )2 2 0yD Eω με+ =

Characteristic equation is ( )2 2 0m ω με+ =


Uniform plane wave traveling in z direction in lossless or non conducting medium- sinusoidal time variations

( )2 2 0m ω με+ =2 2m ω με= −

m jω με= ±jβ= ± Where β ω με=

Hence the solution is

1 2j z j z

y m mE E e E eβ β−= +

2 and are arbitrary constants mm1E E



The above equation is the phasor form of the electric field. To find the time domain form we multiply the phasor form by and then take the real part of it.

1 2j z j z

y m mE E e E eβ β−= +

j te ω

( ){ }1 2( , ) j z j zy m

tm

jE z t Re E e E e eβ β ω−= +

{ }1 2j z j j z

mt j t

mRe eE e Ee eωβ β ω−= +

{ }( ) ( )1 2

j t z j t zm mRe E e E eω β ω β− += +

1 2( ) ( )m mE cos t z E cos t zω β ω β= − + +



1 2( ) ( )y m mE E cos t z E cos t zω β ω β= − + + Forward traveling wave

Backward traveling wave

β ω με=

1 cωβ με

= = c ωβ

=



For the phase of the forward traveling wave to remain constant,

As t increases z must also increase. This means that the wave travels in the z direction with a constant phase.Similarly for the backward traveling component

As t increases z must decrease in order to keep the phase constant. This means that the wave travels in the –z direction with a constant phase.

1 2( ) ( )y m mE E cos t z E cos t zω β ω β= − + +

= t z A constantω β−

t z A constantω β+ =

1 2( ) ( ), m myE sin tO z sin t zr E Eω β ω β= − + +


Wave motion

zλλ−

t

TT−


Phase velocityDue to the variation of E with both time and space we may plot E as a function of t by keeping z constant or vice versa.The plots of E(t,z=constant) and E(z,t=constant) are shown in figure.The wave takes distance λ to repeats itself, and hence λ is called the wave length.Also the wave takes time T to repeat itself, and hence T is called the period of the wave.Since it takes time T for the wave to travel a distance λ at the speed v, λ=vTSince T=1/f we may write v f λ=

2 fω π=1 2Tf

πω

= =vωβ = 2, Substituting πβ

λ=


Phase velocity

For every wavelength of distance traveled, the wave undergoes a phase change of 2π radians.Now consider the forward waveTo prove that this wave travels with a velocity v in the z direction, consider a fixed point P on the wave.Sketch the above wave equation at times t=T/4 and t=T/2 as in figure.As the wave advances with time, point P moves along the z direction.Point P is a point of constant phase,

2 πβλ

=

1 ( )y mE E sin t zω β= −



Phase velocity

This proves that the wave is traveling in the z direction with velocity v


dz vdt

ωβ

= =


Wave motion

E

zλ 2λ

E

tT 2T


Phase velocityE

E

E

zβ

zβ

zβ

0t =

/ 4t T=

/ 2t T=

P

P

P


Relationship between E and H and Intrinsic impedance in perfect dielectric

Maxwell’s equation derived from Faraday’s law is BEt

∂∇× = −

∂

. ., Hi e Et

μ ∂∇× = −∂

ˆ ˆ ˆ

ˆ ˆ ˆ

x y z

x x y y z z

x y z

a a a

H a H a H ax y z t

E E E

μ∂ ∂ ∂ ∂ ⎡ ⎤= − + +⎣ ⎦∂ ∂ ∂ ∂



For a uniform plane wave traveling in the z direction,

0x y∂ ∂= =

∂ ∂0zH =

ˆ ˆ ˆ

ˆ ˆ0 0 0

0

x y z

x x y y

x y

a a a

H a H az t

E E

μ∂ ∂ ⎡ ⎤= − + +⎣ ⎦∂ ∂

ˆ ˆ ˆ ˆ (1)yxy x x y x y

HHE a E a a az t t

μ μ∂∂ ∂⎡ ⎤− − = − − − −⎣ ⎦∂ ∂ ∂



From equation (1) we have

From Maxwell’s curl equation for H

By proceeding in a similar way as we did for the first curl equation we get

( )y xE H az t

μ∂ ∂

= − − −∂ ∂

( )yx HE bz t

μ∂∂

= − − − −∂ ∂

DHt

∂∇× = −

∂. ., Ei e H

tε ∂∇× =∂

( )y xH E cz t

ε∂ ∂

= − − − −∂ ∂

( )yx EH dz t

ε∂∂

= − − −∂ ∂



Let the equation of the plane wave be

Differentiating the above equation wrt t

From equation (d)

11( ) yE f z ct where cμε

= − =

1 '( ) ( )yEf z ct z ct

t t∂ ∂

= − −∂ ∂

1 '( ) c f z ct= − −

yx EHz t

ε∂∂

=∂ ∂

1 '( ) x c f z ctHz

ε− −∂

=∂



Integrating both sides,

1 ( ) xH c f z ct kε= − − + yc E kε= − +1

yEεμε

= −yEμ

ε= −

y

x

EH

με

= −

The constant k is taken to be zero as it forms only a static part of the

the solution which is not importanthere

y

x

EH

με

= − y xE Hμ

ε= −

x yH Eεμ

= −



Similarly let the equation of the plane wave be

Differentiating the above equation wrt t

From equation (c)

11( ) xE f z ct where cμε

= − =

1 '( ) ( )xE f z ct z ctt t

∂ ∂= − −

∂ ∂ 1 '( ) c f z ct= − −

y xH Ez t

ε∂ ∂

= −∂ ∂

1 '( ) y c f ctHz

zε −∂

=∂



Integrating both sides,

1 ( ) yH c f z ct kε= − + xc E kε= +1

xEεμε

=xEμ

ε=

x

y

EH

με

=

The constant k is taken to be zero as it forms only a static part of the

the solution which is not importanthere

x

y

EH

με

= x yE Hμ

ε=

y xH Eεμ

=



2 2 x yE E E= +

2 2y xE H Hμ μ

ε ε= +

2 2x yE H Hμ

ε= + Hμ

ε=

x yE Hμε

=

y xH Eεμ

=

y xE Hμε

= −

x yH Eεμ

= − E Total electric field⇒

H Total magnetic field⇒

EH

μηε

= = EH

μηε

= =



This equation is similar to ohms law R=V/I by the analogy

So it is called intrinsic impedance or characteristic impedance of the medium.For free space, intrinsic impedance is For any other medium

EH

μηε

= =

V E⇔ H I⇔

00

0

=377μηε

= Ω

0

0

r

r

μ μηε ε

= 377 r

r

μηε

= Ω


Solution of wave equation: Uniform plane wave traveling in z direction in lossy or conducting medium- sinusoidal time variations

In this case the wave equations in E are

The wave equation for Ey may be written as

22

2x

x xE j E Ez


∂

22

2y

y y

Ej E E

zωμσ ω με

∂= −

∂

( )2

2y

y

Ej j E

zωμ σ ωε

∂= +

∂2

22

yy

EE

zγ

∂=

∂ ( ) jWhere jγ ωμ σ ωε= +


Uniform plane wave traveling in z direction in lossy or conducting medium- sinusoidal time variations

Here is defined as the propagation constant.

It is a complex number and can be represented as

It has a real part α called attenuation constant and an imaginary part β called phase shift constant.

( ) j jγ ωμ σ ωε= +

jγ α β= +

22

2y

y

EE

zγ

∂=

∂2 2 0y yD E Eγ− =

( )2 2 0yD Eγ− =



Characteristic equation is

Hence the solution of the partial differential equation is

2 2 0m γ− = 2 2m γ=m γ= ±

1 2z z

y m mE E e E eγ γ− += +

( ) ( )1 2

j z j zm mE e E eα β α β− + + += +

1 2z z z z

m mE e e E e eα β α β− −= +

1 2z z z z

y m mE E e e E e eα β α β− −= +



Similarly by taking the wave equation in terms of Hx and then proceeding we get

The terms are the amplitudes of the forward-traveling and backward-traveling waves.The terms cause the amplitudes of the forward and backward waves to decay as they propagate through the medium. Hence α is termed as the attenuation constant.

1 2z z z z

x m mH H e e H e eα β α β− −= +

ze α±

1 2 z zm mE e and E eα α−



The time domain fields are obtained by multiplying Ey by ejωt and taking the real part of the result.

( )1 2( , ) z z z z j ty m mE x t Re E e e E e e eα β α β ω− −= +

1 2cos( ) cos( )z zm mE e t z E e t zα αω β ω β−= − + +


Wave propagation in lossy or conducting medium:Expression for α and β

We know that

Equating (1) and (2)

Equating the real and imaginary parts of (3)

( )j jγ ωμ σ ωε= +

2 2 (1)jγ ωμσ ω με= − − − −

jα β= +

( )22 jγ α β= + 2 2 2 (2)jα β αβ= − + − − −

2 (5)ωμσ αβ= − − −

2 22 = (4)α βω με −− − − −



From eq(5) we have

Putting eq (7) in (4)

(6)2ωμσαβ

= − − − (7)2ωμσβα

= − − −

22 2 = (8)

2ωμσω με αα

⎛ ⎞− − − − −⎜ ⎟⎝ ⎠

( )22 2

2 =4ωμσ

ω με αα

− −

2 2 4 2 2 2 4 =4ω με α α ω μ σ− −



( )22 2 4 4 =4ω με α α ωμσ− −

( )24 2 24 4 0α ω με α ωμσ+ − =2

4 2 2 02

ωμσα ω με α ⎛ ⎞+ − =⎜ ⎟⎝ ⎠

2 P t pu α =2

2 2 0 (9)2

p p ωμσω με ⎛ ⎞+ − = − − −⎜ ⎟⎝ ⎠

(9 ) Solving the quadratic equation for p we get



Similarly by putting eq (6) in (4) we get another quadratic equation in β2, solving which we get

⎫⎪⎬⎪⎭

Taking the positive root only

22

1 1 (11)2

ω με σαωε

⎧ ⎫⎪ ⎪⎛ ⎞= ± + − − − −⎨ ⎬⎜ ⎟⎝ ⎠⎪ ⎪⎩ ⎭

22

1 1 2

p ω με σωε⎛ ⎞= + −⎜ ⎟⎝ ⎠

22

1 1 (12)2

ω με σβωε

⎧ ⎫⎪ ⎪⎛ ⎞= ± + + − − −⎨ ⎬⎜ ⎟⎝ ⎠⎪ ⎪⎩ ⎭



22

1 1 2

Attenuation Constant ω με σαωε

⎧ ⎫⎪ ⎪⎛ ⎞= ± + −⎨ ⎬⎜ ⎟⎝ ⎠⎪ ⎪⎩ ⎭

22

1 1 2

Phase shift Constant ω με σβωε

⎧ ⎫⎪ ⎪⎛ ⎞= ± + +⎨ ⎬⎜ ⎟⎝ ⎠⎪ ⎪⎩ ⎭


Wave propagation in lossy or conducting medium:Expression for vp and λ

Phase velocity pv ωβ

=

22

1 1 2

pv ω

ω με σωε

=⎧ ⎫⎪ ⎪⎛ ⎞+ +⎨ ⎬⎜ ⎟

⎝ ⎠⎪ ⎪⎩ ⎭

2

1

1 1 2

pvμε σ

ω ε

=⎧ ⎫⎪ ⎪⎛ ⎞+ +⎨ ⎬⎜ ⎟

⎝ ⎠⎪ ⎪⎩ ⎭


Wave propagation in lossy or conducting medium:Expression for vp and λ

Wave length2πλβ

=

22

2

1 12

πλω με σ

ωε

=⎧ ⎫⎪ ⎪⎛ ⎞+ +⎨ ⎬⎜ ⎟

⎝ ⎠⎪ ⎪⎩ ⎭


Intrinsic impedance of lossy dielectric

The phasor equation for a wave travelling in z direction in a loosydielectric is

Since the fields vary sinusoidally with time

Putting the above values in

1z z

y mE E e eα β− −= 1z

mE e γ−=

1y z

m

EE e

zγγ −∂

= −∂

yEγ= −

( )y xE H az t

μ∂ ∂

= − − −∂ ∂

jt

ω∂=

∂

y xE j Hγ ωμ− = −

y

x

E jH

ωμηγ

= =


Intrinsic impedance of lossy dielectric

jωμηγ

=

( ) But j jγ ωμ σ ωε= +

( )j

j jωμη

ωμ σ ωε=

+

jj

ωμσ ωε

=+

jj

ωμησ ωε

=+


Conductors and dielectricsElectromagnetic materials are roughly classified as conductors and dielectrics.Maxwell’s curl equation for sinusoidally varying quantities is given by

H E j Eσ ωε∇× = +

C DH J J∇× = +

C Conduction cJ urrent density⇒

DJ Displacement current density⇒

C

D

J E

J j E

σ

ωσωεε

= =

C

D

determines the natJ

The rat ure of materiioJ

alσωε

=


Conductors and dielectrics

. 1 1 D CJ J Good dielectrics σ ωε σωε

⇒ ⇒ ⇒

2. 1 D C Semi conductor J Jσ ωε σωε

≅ ≅⇒ ⇒ ⇒≅

. 3 1 D C Good conductor J Jσ ωε σωε

⇒ ⇒ ⇒

When the displacement current is much greater than conduction currentthe medium behaves like a dielectric. σIf = 0 the medium is a perfect or lossless dielectric.σ ≠If 0 the medium is a lossy dielectric.


Conductors and dielectricsWhen the displacement current is much smaller than conduction currentthe medium behaves like a conductor. σ ωεIf the medium is a good conductor.

When the displacement current is comparable to conduction currentthe medium behaves like a semi conductor.

σ

ωεThe term is called loss tangent

DJ j Eωε=

CJ Eσ=

C DJ J J= +

( )j Eσ ωε= +θ Lossta tangenn tσ

ωεθ = ⇒


Wave propagation in good dielectrics

This can be considered as a special case of the wave motion in lossy or conducting medium.All the equations derived for this case is applicable in the case of good dielectrics with appropriate modification of parameters.For a good dielectric σ/ωε<<1 so that we can approximate the value of α and β as follows


( )2 j jγ ωμ σ ωε= +2 1j j

jσγ ωμ ωεωε

⎛ ⎞= ⋅ +⎜ ⎟

⎝ ⎠2 2 2 1j j σγ ω με

ωε⎛ ⎞= −⎜ ⎟⎝ ⎠



1j j σγ ω μεωε

= −

Expanding the second radical using Binomial theorem

( ) 2 3( 1) ( 1)( 2)1 12 3

n n n n n nx nx x x− − −− = − + − + ⋅⋅ ⋅ ⋅ ⋅

1/2 21 1 1

1 2 21 12 2

j jj σ σ σωε ωε ωε

⎛ ⎞−⎜ ⎟⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎝ ⎠− = − + + ⋅⋅ ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2112 8jσ σωε ωε

⎛ ⎞= − + + ⋅⋅⋅⎜ ⎟⎝ ⎠


Wave propagation in good dielectrics211

2 8jj σγ ω μωε ω

ε σε

⎡ ⎤⎛ ⎞− + + ⋅⋅ ⋅⎢ ⎥⎜ ⎟⎝ ⎠⎢

=⎥⎣ ⎦

jα β= +

Equating the real and imaginary parts

2jj σα ω μεωε

= ⋅−2σω μεωε

= ⋅2σ μ

ε=

2σ μα

ε=211

8j jβ ω

εμε σ

ω⎡ ⎤⎛ ⎞+⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦=


Wave propagation in good dielectrics211

8σω

β ω μεε

⎡ ⎤⎛ ⎞+⎢ ⎥⎜ ⎟⎝ ⎠⎢⎣ ⎦

=⎥

β ω με= 1Since σωε

β ω με=

, , 2

For a perfect dielectric σ μα β ω μεε

= =

2 = 118

pVelocity of propagation v ωωσω με

βωε

⎡ ⎤⎛ ⎞+⎢ ⎥⎜ ⎟⎝ ⎠⎢⎣ ⎦

=

⎥



2

1= 118

pVelocity of propagation vσμεωε

⎡ ⎤⎛ ⎞+⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

1 2z z z z

y m mE E e e E e eα β α β− −= +

1 2( , ) cos( ) cos( )z zy m mE z t E e t z E e t zα αω β ω β−= − + +

:And the wave equations are


Wave propagation in good conductors

This can be considered as a special case of the wave motion in lossy or conducting medium.All the equations derived for this case is applicable in the case of good dielectrics with appropriate modification of parameters.For a good conductor σ/ωε>>1 so that we can approximate the value of α and β as follows


1 jj ωεωμσσ

⎛ ⎞= +⎜ ⎟⎝ ⎠

1 Since orσ σ ωεωε

jωμσ=



jγ ωμσ=

1 2 12

jγ ωμσ+ −⎛ ⎞= ⎜ ⎟⎝ ⎠

( )212

j ωμσ= +

( )12

j ωμσ= + jα β= +

Equating the real and imaginary parts

2ωμσα β= =


Wave propagation in good conductorsThe wave equation in this case is of the form

When we consider the forward traveling wave only

A high frequency uniform plane wave suffers attenuation as it passes through a lossy medium.Its amplitude gets multiplied by the factor, e -αx where α is the attenuation constant.

1 2 orz z z zy m mE E e e E e eα β α β− −= +

1 2( , ) cos( ) cos( )z zy m mE z t E e t z E e t zα αω β ω β−= − + +

1( , ) cos( )zy mE z t E e t zα ω β−= −



ze α−

z

1mE

yE

y


Wave propagation in good conductorsSkin depth is defined as the distance the wave must travel to have its amplitude reduced by a factor of e-1. The exponential multiplying factor is unity at z=0 and decreases to 1/e when z=1/αSo skin depth is

Thus we see that the skin depth decreases with an increase in frequency.The intrinsic impedance of a medium may be expressed in terms ofskin depth.

1 = Skin depth δα

2=δωμσ

1=fπ μσ



= jj

ωμησ ωε+

for a conductoj rωμησ

≅

45 ωμσ

= ∠1But, =

2ωμσ

δ

2

2 45 2σωμσ

= ∠ 22

2 45 σδ

= ∠

2 45 δσ

= ∠ ( )1 1 jδσ

= + Ω ( )1 1 jηδσ

= + Ω


Wave propagation in good conductorsThe phenomenon by which field intensity in a conductor rapidly decreases with increase in frequency is called skin effect.At high frequencies the fields and the associated currents are confined to a very thin layer of the conductor surface.

1 = Skin depth δα


Poynting theoremThe vector product at any point is a measure of the rate of energy flow per unit area at that point ; the direction of energy flow is perpendicular to E and H in the direction of the vector

P E H= ×

E H×

2/P E H W atts m= ×


Poynting theorem- ProofMaxwell’s first curl equation states that

This equation can be rewritten as

Pre dotting the equation (2) with we get,

We have the vector identity

DH Jt

∂∇ × = +

∂

D EJ H Ht t

ε∂ ∂= ∇ × − = ∇ × −

∂ ∂Ei

( ) (1)D E EE J E H E E Ht t

ε∂ ∂= ∇ × − = ∇ × − −−−

∂ ∂ii i i i

( ) ( ) ( ) (2)E H H E E H⋅ ∇ × = ⋅ ∇ × − ∇ ⋅ × − − −

( ) ( ) ( )E H H E E H∇⋅ × = ⋅ ∇ × − ⋅ ∇ ×


Poynting theorem - ProofPutting eq (2) in (1)

( ) ( ) EE J H E E H Et

ε ∂= ⋅ ∇ × − ∇⋅ × −

∂i i

( )B EE J H E H Et t

ε⎛ ⎞∂ ∂

= ⋅ − − ∇ ⋅ × −⎜ ⎟∂ ∂⎝ ⎠i i

( )H EH E H Et t

μ ε∂ ∂= ⋅ − − ∇ ⋅ × −

∂ ∂i

( ) (3)H EH E E Ht t

μ ε∂ ∂= − ⋅ − −∇ ⋅ × − − −

∂ ∂i

2 21 1 (4)2 2

H EBut H H and E Et t t t

∂ ∂ ∂ ∂⋅ = ⋅ = − − −∂ ∂ ∂ ∂


Poynting theorem - ProofSubstituting equation (4) in equation (3), we get

Integrating both sides of equation (5) over a volume V, we get

( )2 21 12 2

E J H E E Ht t

μ ε∂ ∂= − − − ∇⋅ ×

∂ ∂i

( )2 2

2 2H E E H

t tμ ε⎛ ⎞ ⎛ ⎞∂ ∂

= − − − ∇⋅ ×⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

( )2 2

(5)2 2H E E H

tμ ε⎛ ⎞∂

= − + − ∇⋅ × − − −⎜ ⎟∂ ⎝ ⎠

( ) ( )2 2

2 2v v

H EE J d d E H dt ν

μ εν ν ν⎛ ⎞∂

= − + − ∇⋅ ×⎜ ⎟∂ ⎝ ⎠∫ ∫ ∫i


Poynting theorem - ProofWe apply divergence theorem to the last term on the LHS, convert the volume integral into a surface integral and get

( ) ( )2 2

2 2v S

H EE J d d E H dSt ν

μ εν ν⎛ ⎞∂

= − + − ×⎜ ⎟∂ ⎝ ⎠∫ ∫ ∫i

( ) ( )2 2

2 2S v

H EE H dS E J d dt ν


− × = + +⎜ ⎟∂ ⎝ ⎠∫ ∫ ∫i

( ) ( ) 2 2S v

B H E DE H dS E J d dt ν

ν ν⎛ ⎞∂ ⋅ ⋅

− × = + +⎜ ⎟∂ ⎝ ⎠∫ ∫ ∫i


Poynting theorem

Term 1 Term 2 Term 3

( ) ( )2 2

2 2S vEE H E d

tdH dS J

ν


+⎜ ⎟× +⎝ ⎠

− =∂∫ ∫∫ i

Ingoing power flux over the surface S

Total dissipated power within the volume V at any instant due to ohmiclosses

Rate of decrease due to total electromagnetic energy stored within the volume V

Magnetic energy stored within the volume V

Electrostatic energy stored within the volume V


Poynting theorem - Interpretation

2

2E d

ν

ε ν∫

Stored electrical energy

Stored magnetic energy

Ohmic losses

Power out

Power in

( )S

E H dS×∫

( )v

E J dν∫ i

2

2H d

ν

μ ν∫



The first term represents the rate of flow of energy inward through the surface of the volume or ingoing power flux over the surface S

The second term represents total dissipated power within the volume V at any instant. For a conductor of cross sectional area A carrying a current I and having a voltage drop E per unit length, the power loss is EI watts per unit length.The power dissipated per unit volume is

Total power dissipated in a volume is

( )S

E H dS− ×∫

( )v

E J dν∫ i

( )v

E J dν∫ i

.EI EJ watts per unit volumeA

=



The first part of the third term is the stored electric energy per unit volume ( electrostatic energy density ) of the electric field.

The second part of the third term is the stored magneticenergy stored per unit volume ( magnetic energy density ) of the magnetic fieldThe volume integral of the sum represents total electromagnetic energy stored within the volume V

212

Eε

212

Hμ


Poynting theorem - InterpretationThe net inward power flux supplied by the field over the surface S must equal the time rate of increase of electromagnetic energy inside the volume V plus the total losses in volume V, assuming the volume contains no generators. If represent the total ingoing instantaneous power flux, then represents the total power flowing out of the volume.

( )S

E H dS− ×∫( )

SE H dS×∫


Poynting vectorFrom Poynting theorem it can be seen that the vector product of electric field intensity and magnetic field intensity is another vector which is denoted by

The vector P is called Poynting vector.It measures the rate of flow of energy, and its direction is thedirection of power flow and it is perpendicular to the plane containing vectors.P is the energy per unit area passing per unit time through the surface of the volume v in watts/meter square.For plane waves the direction of energy flow is the direction ofpropagation.The Poynting vector offers a useful coordinate free way of specifying the direction of propagation of plane waves.

2 /P E H Watts m= ×

E H and

E H

emt electromagnetic theory module iv

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