The wave equation

and basic plane wave solution

Energy and Power

Microwave Engineering 3rd Edition

Mina Yamashita

Portland State University

ECE531, 1/27/11

Maxwell's equations

Maxwell's equations describe how electric charges and electric currents act as sources for the electric and magnetic fields.

0=⋅∇

=⋅∇

+∂∂

=×∇

−∂∂−

=×∇

B

D

Jt

DH

Mt

BΕ

ρ

Maxwell’s equation in differential form

ρJ

M

B

D

H

E = the electric filed intensity, in V/m

= the magnetic field intensity, in A/m

= the electric flux density, in Coul/m2

= the magnetic flux density, in Wb/m2

= the (fictitious) magnetic current density, V/m2

= the electric current density, in A/m2

= the electric charge density, in Coul/m2

0=⋅∇

=⋅∇

+=×∇

−=×∇

B

D

JDjH

MBjΕ

ρ

ω

ω

Maxwell’s equation in phasor form

EjH

HjE

ωε

ωµ

−=×∇

−=×∇

Maxwell’s curl equation in phasor form

ϵ is permittivity: The real permittivity, ϵ’=ϵrϵ0µ is permeability:

εµωµη ==

k = the wave impedance

µεω=k = the wave number

sec/1031

1

8

00

mck

v

kv

p

p

×====

==

εµω

µεω

= the phase velocity

Wave Equation

x x+∆x

y

T

Tθθ ∆+

θ

y

Tension is the same for both sides .The motion is in y direction.

θθθθ ∆=∆++−= TTTFy )sin(sin

T is tensionµ is mass of unit length

Apply Newton second law.Make dx equal to zero.

µxdm ∆=

θµθ

∆=∆

∆=

Tyx

Tydm

&&

&&)(

2

2

2cos

1

tan

x

y

dx

d

x

y

∂∂

=

∂∂

=

θθ

θ

1

x

yxT

x

yx

∂∂

∆=∂∂

∆2

2

2

µ

substitute this part

x

yxT

x

yx

∂∂

∆=∂∂

∆2

2

2

µx

yxT

x

yx

∂∂

∆=∂∂

∆2

2

2

µ

2

2

2

2

x

y

t

y

T ∂∂

=∂∂µ

µT

CCtxf =→± )( C is constant and dimension of C is m/s

µT

vvtxf =→± )(

Double Derivative in time

Double Derivative in space

2

2

2

2

2

1

x

y

t

y

v ∂∂

=∂∂

Wave Equation

T is tensionµ is mass of unit length

zz

A

y

A

x

Ay

z

A

y

A

x

Ax

z

A

y

A

x

A

z

A

y

A

x

AA

AAA

zzzyyyxxx ˆ,ˆ,ˆ

)(

)()()(

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

22

∂∂

+∂∂

+∂∂

∂

∂+

∂

∂+

∂

∂

∂∂

+∂∂

+∂∂

∂∂

+∂∂

+∂∂

=∇

∇⋅∇−⋅∇∇=×∇×∇

Now using the Maxwell’s equation for E and H.

∂∂

+∂∂

+∂∂

−=∂∂

−

−∇=×∇∂∂

−=×∇×∇

2

2

2

2

2

2

2

2

00

2)()(

z

E

y

E

x

E

t

E

EHt

E

µε

∂∂

+∂∂

+∂∂

+=∂∂

+2

2

2

2

2

2

2

2

00z

E

y

E

x

E

t

Eµε

Wave equation in vacuum for E field

smcv /100.31 8

00

×===µε 2

2

00

2

t

HH

∂∂

=∇ εµ

H field

“del square”, Laplacian of vector A

The curl of curl of vector A

The Wave Equation for E & H field

Basic plane wave solution

t

EJH

t

HΕ

H

E

∂∂

==×∇

∂∂−

=×∇

=⋅∇

==⋅∇

000

0

0

0

µεµ

ερ

Maxwell’s equation

Charge density is zero� charge free medium

zy

A

x

Ay

x

A

z

Ax

z

A

y

A

AAA

zyx

zyx

A

z

A

y

A

x

AA

zz

yy

xx

zz

yy

xx

xyzxyz

zyx

zyx

ˆˆˆ

ˆˆˆ

ˆˆˆ

ˆˆˆ

∂∂

−∂

∂+

∂∂

−∂∂

+

∂

∂−

∂∂

=∂∂

∂∂

∂∂

=×∇

∂∂

+∂

∂+

∂∂

=⋅∇

∂∂

+∂∂

+∂∂

=∇

∂∂

+∂∂

+∂∂

=∇

φφφφ

(Definition of Gradient operator, del)

A cross product in the form of a determinant

x

y

z

zyx ˆˆˆ =×

0

0

)(cos0

=

=

−=

z

y

xx

E

E

kztEE ω

Traveling wave is in z directionE vector in x direction

t

Hy

z

EE

z

E

z

E

x

xx

∂∂

−=∂∂

=×∇

∂∂

=∂∂

ˆ

2

2

002

2

µε

Associated H field is in ydirection

t

HykztkE

∂∂

−=− ˆ)sin(0 ωDerivative of E

Integrate of time

)(0 kztconkE

H x −= ωω

ck 1=ω

ykztc

EH xy ˆ)cos(

0 −= ω

E and H is in phaseH is perpendicular to EBoth are traveling Z direction

µεω=kA constant, k is calledwavenumber

z

x

y

E0x

H0y

E and H are perpendicular to z directionEvery where in the plane, E value is the same.

This is called Basic Plane Wave solution� This is linearly polarized waves

c c

Left Hand Elliptical Polarization (LHEP) Animation

http://www.youtube.com/watch?v=KZz25bmTWXo&feature=related

Right Hand Circular Polarization (RHCP) Animation

http://www.youtube.com/watch?v=jY9hnDzA6Ps&feature=related

a "snapshot" of a plane-polarized wave traveling from left to right, made up of transverse electric (E-vector) and magnetic (B-vector) components.

linearly Polarized Plane Waves

Circularly Polarized Plane Waves

Think both of the superposition of an

zjk

eyExEE0

)ˆˆ( 21

−

+=

x̂ , linearly polarized wave with amplitude E1 and ŷ, linearly polarized wave with amplitude E2 are traveling in the positive direction, ẑ

(1.78)

This means that if E1≠ 0 and E2 = 0, the plane wave linearly polarized in the x̂ direction.

If E1 = 0 and E2 ≠ 0, the plane wave linearly polarized in the ŷ direction.

If E1 and E2 are both real and nonzero, a plane wave linearly polarized at the angle will be

1

21tanE

E−=φ

zjk

eyxEE0

)ˆˆ(0

−

+=

Respects an electric filed vector at a 45degree angle from the x-axis

021 EjEE ==zjk

eyjxEE0

)ˆˆ(0

−

−=

021 EEE ==

(1.79)

The time domain form of the filed

)}2

(cos(ˆ)cos(ˆ{),( 000π

ωωε −−+−= zktyzktxEtz (1.80)

This indicates that the electric field vector changes with time or distance along with z axis. For instance, if we say z=0, the equation 1.80 becomes 1.81.

)}sin(ˆ)cos(ˆ{),0( 0 tytxEt ωωε +=(1.81)

And when ωt increase from zero, the electric field vector rotates counterclockwise from the x-axis. As a results, the angle from x-axis of the electric field vector at time t, at z=0 is

tt

tω

ωω

φ =

= −cos

sintan 1

This shows that the polarization rotates at the uniform angular velocity ω.This is a right hand circularly polarized (RHCP). The fingers of the right hand point in the direction of rotation when the thumb pointsin the direction of propagation.

a left hand circularly polarized (LHCP).

zjk

eyjxEE0

)ˆˆ(0

−

+= (1.82)

The electric field vector rotates in the opposite direction.

For the magnetic field associated with a circular polarized wave can be used Maxwell’s equation or using the wave impedance applied to each component of the electric field. For example, applying (1.76) to the electric field of a RHCP (right hand circular polarization wave (1.79) yields,

zjkzjkzjkeyjx

jEexjy

Eeyjxz

EH 000 )ˆˆ()ˆˆ()ˆˆ(ˆ

0

0

0

0

0

0 −−− −=−=−×=ηηη

EneEn

eEnk

ekjEj

eEj

eEj

Ej

H

rkj

rkjrkj

rkjrkj

×=×=

×=−×−

=

∇×−

=×∇=×∇=

⋅−

⋅−⋅−

⋅−⋅−

ˆ1

ˆ1

ˆ)(

)(

0

0

0

0

0

0

0

0

00

0

0

00

ηη

ωµωµ

ωµωµωµ

(1.76)

zjk

eyjxEE0

)ˆˆ(0

−

−= (1.79)

If the rotation is clockwise looking in the direction of propagation, the sense

is called right-hand-circular (RHC). If the rotation is counterclockwise, the

sense is called left-hand-circular (LHC). A circular polarized wave radiates

energy in both the horizontal and vertical planes and all planes in between.

The difference, if any, between the maximum and the minimum peaks as the

antenna is rotated through all angles, is called the axial ratio or ellipticity and

is usually specified in decibels (dB). If the axial ratio is near 0 dB, the antenna

is said to be circular polarized.

The crossed Yagi antenna is

used in the AWS field sites for

transmission.

It has the capability to transmit

with both right hand circular

polarization (RHCP) and

left hand circular polarization

(LHCP).

Super Gain Yagi Antenna

http://www.youtube.com/watch?v=kLvkAGAUMXw&feature=related

http://www.youtube.com/watch?v=J4FEKWuXp1U

http://en.wikipedia.org/wiki/Circular_polarization

Circularly Polarized Plane Waves

With the proper phase relationship between vertically and horizontally polarized beams, the snapshot pattern is a helix and the sectional pattern is a circle.

Energy and Power

Electromagnetic

Energy and Power

• Electromagnetic waves transport throughout space the energy.

• Electromagnetic power interacts with another set of charges and currents in a receiver, information (energy) can be delivered from the sources to another location in space.

• Electromagnetic energy and power can be stored, transmitted and dissipated.

Poynting’s Theorem

• Poynting’s theorem is the tool for determine the direction the power is flowing.

• Poynting’s theorem concerns the conservation of energy for a given volume in space.

• Poynting’s theorem is a consequence of Maxwell’s equations.

Derivation of Poynting’s Theorem in the

Time Domain

Time-Domain Maxwell’s curl equations in differential form

t

DJH

t

BME

s

s

∂∂

+=×∇

∂∂

−−=×∇

Using a vector identity

( ) HEEHHE ×∇⋅−×∇⋅=×⋅∇

t

BHMHEH

t

DEJEHE

s

s

∂∂

⋅−⋅−=×∇⋅

∂∂

⋅−⋅−=×∇⋅−

( ) HEEHHE ×∇⋅−×∇⋅=×⋅∇

t

DEJE

t

BHMH ss ∂

∂⋅−⋅−

∂∂

⋅−⋅−=

Poynting’s Theorem in the Time Domain

Integrating over a volume V bounded by a closed surface S

( )

( )∫∫

∫∫∫

×⋅∇−⋅−

⋅−

∂∂

⋅+∂∂

⋅−=⋅+⋅

VV

s

V

s

VV

ss

dvHEdvMH

dvJEdvt

BH

t

DEdvMHJE

Now integrate over a volume V and use the divergence theorem

to obtain the general form of Poynting’s theorem in the Time Domain

( )

( )∫∫∫∫

∫

⋅×−⋅−⋅−

∂∂⋅+

∂∂⋅−=

⋅+⋅

SV

s

V

s

V

V

ss

sdHEdvMHdvJEdvt

BH

t

DE

dvMHJE

( )

( )∫∫

∫

⋅×−

∂∂

⋅+∂∂

⋅−=

⋅+⋅

SV

V

ss

sdHEdvt

HH

t

EE

dvMHJE

µε

Lossless media

( )22

1A

tt

AA

t

AA

∂∂

=∂∂

=∂∂

⋅

By Using the dot product & vector derivative below

A simple, lossless media in the form of Poynting’s theorem in time domain

( )

( )∫∫

∫

⋅×−

+∂∂

−=

⋅+⋅

SV

V

ss

sdHEdvHEt

dvMHJE

22

2

1

2

1µε

Derivation of Poynting’s Theorem

in the Frequency Domain

Maxwell’s curl equations in differential form

JEjH

MHjE

+=×∇

−−=×∇

ωε

ωµ

ωσ

µµµ

ωσ

εεε

mjj

jj

−′′−′=

−′′−′=

The permeability of the medium

The complex permittivity of the medium

σ Conductivity in material

(1.27a)

(1.27b)

EJJ s σ+= The total electric source current

Poynting’s Theorem in the Frequency DomainMultiplying (1.27a) by H* and multiplying the conjugate of (1.27b) by Ē

2*2*2**

*

)(

)(*

EjEJEEjJEHE

MHHjEH

s

s

ωεσωε

ωµ

−+⋅=−⋅=×∇⋅

⋅−−=×∇⋅

&

ABBABA ⋅×∇−⋅×∇=×⋅∇ )()()(

By using the vector identities (B8)

)()(

)()()(

**22*2

***

ss MHJEHEjE

HEEHHE

⋅+⋅−−+−=

×∇⋅−×∇⋅=×⋅∇

µεωσ

Now integrate over a volume V and use the divergence theoremto obtain the general form of Poynting’s theorem in the frequency Domain

dvMHJEdvHEjdvE

dsHEdvHE

sv

svv

v s

)()(

)(

**22*2

**

⋅+⋅−−+−=

⋅×=×⋅∇

∫∫∫

∫ ∫µεωσ

µµµεεε

′′−′=

′′−′=

j

j

dvHEjdvHE

dvEsdHEdvMHJE

vv

vss

vs

)(2

)(2

22

1)(

2

1

22*2"2"

2***

µεω

µεω

σ

−+−+

+⋅×=⋅+⋅−

∫∫

∫∫∫

S is a closed surface enclosing the volume V

Poynting’s theorem, a power balancing equation

Poynting Vector

A vector called Poynting Vector

*

HES ×=

The Poynting vector has the same direction as the direction of propagation.The Poynting vector at a point is equivalent to the power density of the wave at that point.

It describes the amount of energy per unit of area that is delivered by an electromagnetic filed. Like any vector, the Poynting vector has both magnitude and direction.

The Poynting vector has units of W/m2.

The sinusoidal steady –state case, the time-average stored electric energy in a volume V

dvDEWv

e ∫ ⋅= *Re41

(1.83)

Lossless media

dvEEWv

e ∫ ⋅= *4ε (1.84)

The sinusoidal steady –state case, the time-average stored magnetic energy in a volume V

dvBHWv

m ∫ ⋅= *Re41

Lossless media

dvHHWv

m ∫ ⋅= *4µ

(1.85)

(1.86)

ϵ and µ are real and constant scalar

∫∫ +V

m

V

dvHdvE 22 σσ the instantaneous power dissipated in the electric and magnetic conductivity losses, respectively, in volume V.

∫∫ ′′+′′VV

dvHdvE 22 µωεω the instantaneous power dissipated in the electric and magnetic conductivity losses, respectively, in volume V.

∫

′+′

V

dvHE 22

2

1

2

1µε the total electromagnetic energy stored in the volume V.

( )∫ ⋅×S

sdHE the flow of instantaneous power out of the volume V through the surface S

( )∫ ⋅+⋅V

ii dvKHJE the total electromagnetic energy generated by the sources in the volume V.

( ) ( )

( )∫∫∫

∫∫∫

=⋅×+++

′′+′′+

′+′+⋅+⋅

SV

m

V

VVV

sdHEdvHdvE

dvHEdvHEjdvMHJE

0

2

1

2

1

22

2222

σσ

µεωµεω

Poynting’s theorem, a power balancing equation

This means that the sum of the power generated by the sources, the imaginary power (representing the time-rate of increase) of the stored electric and magnetic energies, the power leaving, and the power dissipated in the enclosed volume is equal to zero.

dvHEjdvHE

dvEsdHE

dvMHJE

vv

vs

sv

s

)(2

)(2

22

1

)(2

1

22*2"2"

2*

**

µεω

µεω

σ

−+−+

+⋅×=

⋅+⋅−

∫∫

∫∫

∫

Poynting’s theorem, a power balancing equation

ssv

s PdvMHJE =⋅+⋅− ∫ )(21 **

*

HES ×=∫∫ =⋅=⋅× s os PsdSsdHE 21

2

1 *By using Poynting vector

σεµ ,,,HE

ss MJ ,

v

S

n̂

A volume V, enclosed by the closed surface, S containing fieldsAnd current souses,

ss MJ ,

HE ,

Ps = the power delivered by the sourcesPo = the sum of the power transmitted through the surfacePl= the power lost to heat in the volume (Joule’s law)

lvv

PdvHEdvE =−+ ∫∫ )(222"2"

2

µεωσ

Power Absorbed by a Good Conductor

x

zP

zn ˆˆ =

S

n̂

n̂

S0

εµ,

S0 = the cross-sectional surface at the interfaceS = the surface

Lossless-medium a conductor

A field is incident from z0

The real average power entering the conductor

dsnHEPss

avˆRe

2

1

0

* ⋅×= ∫ + (1.94)

n̂ = a unit vector

The closed surface, S+S0

iE

*HES ×=ẑ

S0 S

The right hand end of S is far away from z=0that it is negligible contribution – can omit.

Using the surface of conductor to calculate

power dissipation.

dszHEPs

avˆRe

2

1

0

* ⋅×= ∫

The real average power entering the conductor through S0

(1.95)

By using vector identities, (B.3) BACCBACBA ⋅×=⋅×=×⋅

*** )ˆ()(ˆ HHHEzHEz ⋅=⋅×=×⋅ η (1.96)

EnH ×= ˆ1

0η(1.76) Ω== 377000 εµη

EnH ×= ˆ1

ηGeneralized from 1.76

=η The intrinsic wave impedance of the conductor

dsHR

Ps

s

av

2

02 ∫= (1.97)

dsHR

Ps

s

av

2

02 ∫=

x

s

s jR σδσωµ

σωµ

η1

22)1(Re)Re( ==

+==

The real average power entering the conductor through S0

(1.97)

(1.98)

S0

Rs = the surface resistivity of conductor

S

z

Z=0H

H

Ht is continuous at z=0

= tangential to the conductor surface

Does not matter whether this field is evaluated Just inside or outside conductor.

Extra Note

Microwave Engineering

General Plane Wave Solution

Helmholtz equation

• The Helmholtz equation, is the elliptic partial differential equation.

• The Helmholtz equation represents the time-independent form of the original equation, results from applying the technique of separation of variables to reduce the complexity of the analysis.

wct

w∆=

∂∂ 2

2

2

2

2

2

2

yx ∂∂

+∂∂

=∆

The wave equation

models the propagation of a wave travelling through a given medium at a constant speed c.

In free space, the Helmholtz equation for Ē can be written as

0202

2

2

2

2

22

0

2 =+∂∂

+∂∂

+∂∂

=+∇ Ekz

E

y

E

x

EEkE (1.62)

This vector wave equation holds for each components of Ē:

0202

2

2

2

2

2

=+∂∂

+∂∂

+∂∂

iiii Ek

z

E

y

E

x

E (1.63)

Where the index i = x, y, or z.

By using the method of separation of variables and solve this equation. The method of separation of variables is a widely used technique for solving the wave equation and other PDEs.

νεω=k A constant k is called the wavenumber.

Linearly polarized waves

Step 1: Let’s start to with Ex.Ex as a product of three function for each of the three coordinates:

)()()(),,( zhygxfzyxEx = (1.64)

Substituting this form into (1.63) and dividing by fgh gives

020

"""

=+++ kh

h

g

g

f

f(1.65)

0202

2

2

2

2

2

=+∂∂

+∂∂

+∂∂

iiii Ek

z

E

y

E

x

E (1.63)

The double primes is the second derivative.

Step 2: Key to recognize each term in the equation (1.65) tie toa constant. We can do this because each of them are independentto each other.

020

"""

=+++ kh

h

g

g

f

f(1.65)

2"

xkf

f−=;

2"

ykg

g−=

;

2"

zkh

h−=

022

2

=+∂

∂fk

x

fx

;

022

2

=+∂

∂gk

y

gy

;

022

2

=+∂

∂hk

z

hx

(1.66)

(1.65) & (1.66) together:

2

0

222 kkkk zyx =++

;

(1.67)

(1.63) is reduced to three separated ordinary differential equations in (1.66)

Step 3: These equations would be in the form ofzjkyjkxjk zyx eee

±±±,,

Choose the traveling wave of direction, positive or negative. Either one can solve the problem.

Example for choosing the positive direction for each coordinates. “A” is an arbitrary amplitude constant.

)(),,(

zkykxkj

xzyxAezyxE

++−=(1.68)

Define a wavenumber vector,k

nkzkykxkk zyx ˆˆˆˆ 0=++= (1.69)Define a position vector

zzyyxxr ˆˆˆ ++= (1.70)0kk = n̂ : is a unit vector of propagation

(1.68) becomes

)(),,( rkjx AezyxE⋅−= )(),,( rkjy BezyxE

⋅−= )(),,( rkjz CezyxE⋅−=

(1.71) (1.72) (1.73)

Thus, the x, y, and z components of Ē in equation (1.71), (1.72), and (1.73) are the same as kx, ky, and kz.

0222

=∂

∂+

∂

∂+

∂

∂=⋅∇

z

E

y

E

x

EE z

yx

This divergence condition need to beapplied to satisfy the Maxwell’s equation.This means that Ex, Ey, and Ez must havethe same variation in x, y and z.

This condition sets the amplitude as

rkjeEE

zCyBxAE

⋅−=

++=

0

0ˆˆˆ

0)( 000 =⋅−=∇⋅=⋅∇=⋅∇⋅−⋅−

⋅⋅− rkjrkjrkj eEkjeEeEE

By using the vector identify (B7), AffAAf ∇+∇⋅=⋅⋅∇ )(

(1.74)

This means that

00 =⋅ Ek

0E

is perpendicular to the direction of propagation, k

HjE 0ωµ−=×∇Let’s find the magnetic field’s direction. From the Maxwell’s equation, the magnetic filed is (1.75)

En

eEn

eEnk

ekjEj

eEj

eEj

Ej

H

rkj

rkj

rkj

rkj

rkj

×=

×=

×=

−×−

=

∇×−

=

×∇=×∇=

⋅−

⋅−

⋅−

⋅−

⋅−

ˆ1

ˆ1

ˆ

)(

)(

0

0

0

0

0

0

0

0

00

0

0

00

η

η

ωµ

ωµ

ωµ

ωµωµ

(1.76)

From the results, the magnetic intensity vector H

is perpendicular to E

x

z yE

H

n̂

Fig. 1.8: Orientation of the nkkHE ˆ,, 0=Vector for a general plane wave

This is called Linearly polarized waves due to E field vector is fixed.

εµ

µεωωµωµ

η ===kIntrinsic impedance

0

00 ε

µη =

Intrinsic impedanceFree space