Lecture 20 Plane Waves

8/2/2019 Lecture 20 Plane Waves

1/23

EECS 117

Lecture 20: Plane Waves

Prof. Niknejad

University of California, Berkeley

Universit of California Berkele EECS 117 Lecture 20 . 1/


2/23

Maxwells Eq. in Source Free Regions

In a source free region = 0 and J = 0

D = 0

B = 0

E =

B

t

=

H

t

H = Dt

= E

t

Assume that E and H are uniform in the x-y plane sox = 0 and

y = 0

For this case the

E simplifies



3/23

Curl E for Plane Uniform Fields

Writing out the curl of E in rectangular coordinates

E = x y z

0 0 zEx Ey Ez

(E)x = Eyz = Hx

dt

(

E)y

=

Ex

z=

Hy

dt

(E)z = 0 = Hzdt



4/23

Curl of H for Plane Uniform Fields

Similarly, writing out the curl H in rectangularcoordinates

H =

E

t

Hyz

= Ex

t

Hxz

= Ey

t

0 =

Ezt

Time variation in the z direction is zero. Thus the fieldsare entirely transverse to the direction of propagation.

We call such fields TEM waves



5/23

Polarized TEM Fields

For simplicity assume Ey = 0. We say the field polarizedin the x-direction. This implies that Hx = 0 and Hy = 0

Exz

= Hyt

Hy

z =

Ex

t

2Exz2

= 2Hy

zt

2Hyzt

= 2Ext2



6/23

One Dimensional Wave Eq.

We finally have it, a one-dimensional wave equation

2Ex

z2=

2Ex

t2

Notice similarity between this equation and the waveequation we derived for voltages and currents along a

transmission lineAs before, the wave velocity is v = 1

The general solution to this equation is

Ex(z, t) = f1(t zv

) + f2(t +z

v)



7/23

Wave Solution

Lets review why this is the general solution

Ex

t= f

1+ f

2

2Ext2

= f1 + f2

Ex

z=

1

vf1

+1

vf2

2Exz2

=1

v2f1 +

1

v2f2

A point on the wavefront is defined by (t z/v) = cwhere c is a constant. The velocity of this point istherefore v

1 1v

zt

= 0

z

t= v



8/23

Wave Velocity

We have thus shown that the velocity of this wavemoves is

v = c =1

In free-space, c 3 108m/s, the measured speed oflight

In a medium with relative permittivity r and relativepermeability r, the speed moves with effective velocity

v =c

rrThis fact alone convinced Maxwell that light is an EM

wave



9/23

Sinusoidal Plane Waves

For time-harmonic fields, the equations simplify

dEx

dz

=

jHy

dHydz

= jEx

This gives a one-dimensional Helmholtz equation

d2Ex

d2z

=

2Ex



10/23

Solution of Helmholtz Eq.

The solution is now a simple exponential

Ex = C1ejkz + C2ejkz

The wave number is given by k = = vWe can recover a traveling wave solution

Ex(z, t) = ExejtEx(z, t) =

C1ej(tkz) + C2ej(t+kz)

Ex(z, t) = C1 cos(t kz) + C2 cos(t + kz)The wave has spatial variation = 2k =

2v =

vf



11/23

Magnetic Field of Plane Wave

We have the following relation

Hy =

1

j

dEx

dz

=

1

j jkC1ejkz + C2jkejkz

Hy =k

C1e

jkz C2jkejkz

By definition, k =

Hy =

C1ejkz C2ejkzThe ratio E+x and H

+y has units of impedance and is

given by the constant = /. is known as theimpedance of free space



12/23

Plane Waves

Plane waves are the simplest wave solution ofMaxwells Eq. They seem to be a grossoversimplification but they nicely approximate real

waves that are distant from their source

source of radiation

wavefront at distant points



13/23

Wave Equation in 3D

We can derive the wave equation directly in acoordinate free manner using vector analysis

E = Ht

= (H)t

Substitution from Maxwells eq.

H = Dt

= E

t

E = 2Et2



14/23

Wave Eq. in 3D (cont)

Using the identity E = 2E + ( E)Since E = 0 in charge free regions

2E = 2Et2

In Phasor form we have k

2

=

2

2E = k2E

Now its trivial to get a 1-D version of this equation

2Ex =

2Ex

t2

2Ex

x2

= 2Ex

t2



15/23

Penetration of Waves into Conductors

Inside a good conductor J = E

In the time-harmonic case, this implies the lack of freecharges

H = J + Et

= ( + j) E

Since H 0, we have( + j) E 0

Which in turn implies that = 0

For a good conductor the conductive currentscompletely outweighs the displacement current, e.g.


C d i Di l C


16/23

Conductive vs. Displacement Current

To see this, consider a good conductor with 107S/mup to very high mm-wave frequencies f 100GHzThe displacement current is still only

10111011 1

This is seven orders of magnitude smaller than theconductive current

For all practical purposes, therefore, we drop thedisplacement current in the volume of good conductors


W E ti i id C d t


17/23

Wave Equation inside Conductors

Inside of good conductors, therefore, we have

H = E

E = ( E)2E = 2E = jH2E = jE

One can immediately conclude that J satisfies the sameequation

2J = jJApplying the same logic to H, we have

H = ( H)2H = 2H = (j+) E

2H = jHUniversit of California Berkele EECS 117 Lecture 20 . 17/

Pl W i C d


18/23

Plane Waves in Conductors

Lets solve the 1D Helmholtz equation once again forthe conductor

d

2

Ezd2x = jEz = 2Ez

We define 2 = j so that

=1 + j

2

Or more simply, = (1 + j)f = 1+jThe quantity = 1

fhas units of meters and is an

important number


S l ti f Fi ld


19/23

Solution for Fields

The general solution for the plane wave is given by

Ez = C1ex + C2ex

Since Ez must remain bounded, C2 0

Ez = E0ex = Exex/ mag

ejx/ phaseSimilarly the solution for the magnetic field and currentfollow the same form

Hy = H0ex/ejx/

Jz = J0ex/

ejx/



20/23

T t l C t i C d t


21/23

Total Current in Conductor

Why do fields decay in the volume of conductors?

The induced fields cancel the incoming fields. As

, the fields decay to zero inside the conductor.

The total surface current flowing in the conductorvolume is given by

Jsz =0

Jzdx =0

J0e(1+j)x/dx

Jsz =J0

1 + j

At the surface of the conductor, Ez0 =J0


Internal Impedance of Conductors


22/23

Internal Impedance of Conductors

Thus we can define a surface impedance

Zs =Ez0

Jsz=

1 + j

Zs = Rs + jLi

The real part of the impedance is a resistance

Rs =1

=

f

The imaginary part is inductive

Li = Rs

So the phase of this impedance is always /4


Interpretation of Surface Impedance


23/23

Interpretation of Surface Impedance

The resistance term is equivalent to the resistance of aconductor of thickness

The inductance of the surface impedance represents

the internal inductance for a large plane conductor

Note that as , Li 0. The fields disappear fromthe volume of the conductor and the internal impedance

is zero

We commonly apply this surface impedance toconductors of finite width or even coaxial lines. Its

usually a pretty good approximation to make as long asthe conductor width and thickness is much larger than


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Lecture 20 Plane Waves