II.2.6 The Poynting Vector and Power in EM Waves Let us consider a parallel plate transmission line:

From the Maxwell-Ampere law we know that

xis in the same direction as E the electric

field and is there

fore orthogonal to

. .

i.e. = 0

. .

c s

c s

But D

dS

Hence

H dl J D dS

H dl J dS

We consider w>>d and neglect fringe effects The Poynting vector is defined to be the cross product, E x H. The complex Poynting vector is defined to be ½ E x H*.

Solving for the top plate 2 yH W I

Solving for the bottom plate 2 yH W I

Since the current is I on the top plate and –I on the bottom, the H field sum inside the parallel plate transmission line, while they cancel outside

Thus InsideyH

OutsideyH

Also the electric field and the voltage are related by:

.xE d V So the transmission line wave power is:

* *1 1

2 2x yV I E H Wd

The intensity of an electromagnetic wave

* 21Re

2E H Wm

This expression is known as the complex Poynting Vector and the direction of power flow is perpendicular to E and H

Since Average power: Wavefront:

For a plane wave ,0,0xE E 0, ,0yH H

Thus *1

2 x y zS E H a

Ex

Hy Sz

and

III.2.6.1 Example – Duck a la microwave

A duck with a cross-sectional area of 0.1m2 is heated in a microwave oven. If the electromagnetic wave is:

1 1Re 750 , Re 2j t z j t zx yE e Vm H e Am

What power is delivered to the duck ?

Power = *1Re

2E H Area

II.3.1 Reflection of Incident wave normal to the plane of reflection This is a special case, which could be derived by analogy with reflection of V and I. More satisfactory and relatively simple is to obtain the same results by matching boundary conditions: At the boundary between two media the electric and magnetic fields are continuous the total field in medium 1 is equal to the total field in medium 2.

Hence: xi xr xtE E E

and yi yr ytH H H

By definition of in medium 1 and 2 we have:

1 xi xr

yi yr

E E

H H

and

2 Eliminating H using gives:

1 1 2/ / /xi xr xtE E E

Eliminating Ext gives:

2 1

2 1

xrr

xi

E

E

(6.1a)

Eliminating Exr gives: (6.1b)

2

2 1

2xtt

xi

E

E

Incident power density

Transmitted power density

Reflected power density

2

1

P =2

xii

E

2

21 22

1 2( )r i r iP P P

21 22

1 2

4(1 )

( )t i r iP P P

II.3.2 Reflection from a dielectric boundary of wave at oblique incidence The behaviour of the reflected and transmitted waves will depend on the orientation of E with respect to the boundary. We therefore split the incident waves into two polarised parts.

1) Perpendicularly polarised - electric field at right angles to incident plane

2) Parallel polarised - electric field in incident plane

II.3.2.1 Snell’s Law of Refraction

Before calculating the reflection and refraction coefficients we need to know the angles at which the reflected and refracted waves will be travelling

The wave travels in medium 1 from C to B in the same time as it does from A to D in medium 2.

Hence 1

2

CB v

AD v but sin and sini tCB AB AD AB

1

2

sin

sini

t

v

v

We know that 0

1

r

v

and that

Snell’s Law of Refraction (6.2)

For dielectrics we can assume 1=2=0

12

2 1

1 2

sin

sini

t

2 2 2

11 1

sin

sini

t

n

n

II.3.2.2 Snell’s Law of Reflection

By a similar argument it can be shown that the angle of reflection is equal to the angle of incidence. AE=CB, but CB AE

sin sini r

I.e. r i Angle of reflection is equal to angle of incidence

II.3.2.3 Incident and Reflected Power The next step is to consider the power striking the surface AB and to equate that to the power leaving that surface.

i

A

CE

DB

t

t

i

r

Medium 1

Medium 2

The incident power density is (remembering that is the intrinsic impedance)

2

1

cos2

ii

E (The cosi term comes from the angle of incidence)

Hence: 2 2 2

1 1 2

cos cos cosi r ti r t

E E E

Remembering that

12

2 1

1 2

and that cos cosr i we get:

12 22

22 2

1

cos1

cosr t t

i i i

E E

E E

(6.3)

II.3.3 Perpendicularly polarised waves In perpendicularly polarised waves the electric field is perpendicular to the plane of incidence i.e. parallel to the boundary between the two media

E (perp to plane of incidence)

H(in plane of incidence)

Plane of incidence

Reflecting Plane

E

Summing the electric fields we get: i r tE E E

Combining this with equation 6.3:

12 22 2

22 2 2

1

cos1 1

cost t tr

i i i i

E EEk

E E E

Where

12

2

1

cos

cost

i

k

We get the following:

22

2 2

( )1 r ir

i i

E EEk

E E

2

(1 ) (2 ) ( 1) 0r r

i i

E Ek k k

E E

Hence:

Thus

1/2 1/21 21/2 1/2

1 2

cos cos

cos cosi tr

i i t

E

E

This expression contains the angle of the transmitted wave, however we can use Snell’s law to obtain a more useful expression which contains only the angle of incidence.

Snell’s Law

12

2

1

sin

sini

t

21/2

2 121 2

211/2 2

22 11 2

1 2

cos 1 sincos sin

cos sin cos 1 sin

i ii i

r

i

i i i i

nn n

nE

E nn n

n

(6.4)

1 2

1 2

cos cos

cos cosi tr

i i t

n nE

E n n

II.3.4 Parallel Polarised Waves

H (perp to plane of incidence)

E (in plane of incidence)

Plane of incidence

Reflecting Plane

H

In this case E is no longer parallel to the reflecting plane. Our boundary condition applies to the component of E parallel to the reflecting plane, i.e.:

cos cos cosi i r i t rE E E

Following through the algebra our expression for the ratio of reflected to incident waves becomes:

(6.5)

This is similar to equation (6.4), but in this equation the numerator can become zero i.e. no reflected wave. The angle at which this occurs is known as the Brewster Angle. Zero reflection at the Brewster Angle explains why Polarised sunglasses cut down reflections. Setting the numerator of (6.5) equal to zero we get

12

2 2

1 1

tan( )B

n

n

Where n is the refractive index Light with parallel polarization is called p-polarized (p=parallel) Light with perpendicular polarization is called s-polarized (s=senkrecht, perpendicular in German)

1/2 1/22 1 2 11/2 1/2

2 1 2 1

cos cos cos cos

cos cos cos cosi t i tr

i i t i t

n nE

E n n

21/2

2 12 22 1

21 11/2 2

22 2 12 1

1 1 2

cos 1 sincos sin

cos sin cos 1 sin

i ii i

r

i

i i i i

nn n

nE

E nn n

n

II.3.5 Comparison between reflections of parallel and perpendicularly polarised waves

Graph of r

i

E

Eversus angle of incidence

The graphs show the values of r

i

E

E for two different permittivity

ratios. As the ratio 2

1

increases three effects can be seen

1) The Brewster angle increases

2) The value of r

i

E

Efor the perpendicularly polarized wave tends

to -1 i.e. perfect antiphase at all angles

3) The value of r

i

E

Efor the parallel polarized wave tends to 1 i.e.

perfect phase at all angles

4) When2

1

total reflection occurs at all angles of

incidence.

II.3.6 Total Internal Reflection

The previous section showed graphs for 2

1

> 1. I.e. our wave is

moving from a lower refractive index medium to a higher one. If

instead 2

1

< 1 then the phenomenon known as total internal

reflection can occur.

The term 22

1

sin i

in equations (6.4) or (6.5) can be negative

and its square root is imaginary and r

i

E

E is complex

Then

2 *

1r r r

i i i

E E E

E E E

i.e. the magnitudes of the incident and reflected powers are equal. This gives rise to total internal reflection: the incident wave is reflected at the boundary and no wave emerges from the higher refractive index medium.

This is shown in the graph below at all angles of incidence where: 2 2

1Sin

The critical angle is defined as

1/2

2 2

1 1

( )cr

nSin

n

Graph of r

i

E

E versus angle of incidence

0 45 901

0

1

Permittivity Ratio of 1/2

Perpendicular ( ) , 0.5

Parallel ( ) , 0.5

.180

Total internal reflection underpins Optical fibre technology

Comparison of Transmission Line & Free Space Waves Symbols: V: Voltage Volts E Electric Field Volts m-1

I: Current Amps H Magnetic Field Amps m-1

L: Inductance Henry m-1 permeability Henry m-1

C Capacitance Farad m-1 Permittivity Farad m-1

0Z

Characteristic impedance

Ohms Intrinsic impedance Ohms

Equations: j t x j t x

F BV e V e V e j t z j t zx xF xBE e E e E e

j t x j t xF BI e I e I e j t z j t z

y yF yBH e H e H e

0F B

F B

V V LZ CI I xF xB

yF yB

E E

H H

Wave velocity 1

LC Wave Velocity

1

LC

0

0

LB

LLF

V Z Z

V Z Z

2 1

2 1

xr

rxi

E

E

Power Reflection2

L Power Reflection2

r

Wave Power *1

Re2

V I

Wave Power *1

Re2

E H

: Frequency, radians s-1, : Spatial frequency, radians m-1

.

II.3.7 Example – Reflected Power

Diamond has r = 5.84 & r = 1. What power fraction of light is reflected off an air/diamond surface? Recalling that for Transmission lines: Similarly for E-M Waves, for the case where E and H are parallel to the reflecting plane: Now

The aim of an antenna is to get signal power from the transmitter to the receiver circuit as efficiently as possible. Almost any guide carrying an electromagnetic wave will radiate part of the wave if its end is open. The ideal antenna however sends as much radiation as possible in the desired direction and with the minimum of internal reflection.

III.1 Antennae Transmitter

Receiver

III.1.1 Slot and aperture antennae The electromagnetic waves in a guide will radiate if you chop its end off (very inefficient)

Horn antennae such as these work very well but they are bulky and therefore unwieldy. A laser can be thought of as an aperture antenna and the output roughly approximates to a plane wave.

III.1.1.1 Horn Antennae More wave is radiated if the end of the guide is flared.

Bigger aperture so

- less diffraction and more gain (directed power, see III.2)

- The launched wave is more similar to a plane wave.

- There is a gradual change between the electrical wave where the characteristic impedance is Z and the radiated wave where the intrinsic impedance is η, hence less reflection.

III.1.1.2 Laser

.

III.1.2 Dipole Antennae III.1.2.1 Half-Wave Dipole If the end of a transmission line is left open circuit, the current at the end is 0 and the reflection coefficient is 1

L

if LZ then 1L

The telegraphers’ equations show that ¼ of a wavelength back from the load the voltage must be 0

20 Vb bL

ZZ

Z

so a current source is needed to drive the wave

At long wavelengths a half-wave dipole becomes impractically big. A shorter dipole still radiates, with the penalty that more of the wave is reflected from the antenna back down the waveguide.

Opening out the two bars of the transmission line has little effect on the current distribution and the exposed oscillating current radiates an electromagnetic wave.

III.1.2.2 Short Dipole

L

. Conductors reflect radio waves because E=0 at the conductor (that’s why mirrors are shiny). A single dipole placed above a conductor radiates like one half of a dipole pair. Long wave radio masts are like this, and can be formed by covering the ground with wire mesh.

III.1.2.3 Half Dipole

III.1.3 Loop Antennae The Loop antenna is like the half-wave dipole except that it behaves like a transmission line whose end is a short circuit and is therefore driven by a voltage source.

Short Circuit a transmission line..

If the end of a transmission line is a short circuit, ZL=0

L

The telegraphers’ equations show that ¼ of a wavelength back from the load the current must be 0

20 Ib bL

ZZ

Z

so a voltage source is needed to drive the wave

Short Circuit a transmission line..

then open it out..

To get a loop antenna

Inside a portable radio you will find a ferrite rod wound with copper wire. This is the long wave antenna. It has several loops and a ferrite core. The core concentrates the electromagnetic waves into the antenna and the whole arrangement is essentially half of a transformer. Another way of concentrating the electromagnetic waves is using a parabolic mirror. Placing a source (S) at the focus of a parabaloid can result in a very directive beam

Receiver

F e r r i teR o d

III.1.4 Reflector Antennae

S

. Antennae can also be joined into an array. The array must (of course) be correctly designed so that the signals combine in phase … i.e. that they add up rather than cancel out.

III.1.5 Array antennae

a b

ej ej

Exploit superposition effects to get a highly directional wave as long as the spacings (a,b) and the phase relationships (ejej) are correct.

The power is calculated by integrating over a surface far from the radiating antenna which encloses the current carrying portion. For example a sphere a distance r away from the axis of a wire carrying a current IANT . As the distance r increases E varies as Eo/r and H = Eo/r. These are termed the far field values of the electromagnetic radiation from the antenna. The intensity of power from a real antenna is not uniform. It has a 3 dimensional pattern, which is termed the ‘radiation pattern’.

III.2 Radio III.2.1 Radiation Resistance

Antennae emit power, so they can be modelled as resistors: The radiation resistance, Ra, of an antenna is that resistance which in place of the antenna would dissipate as much power as the antenna radiates.

2rmsI

antPaR (8.1)

where S

ant dSkantP I 2 (8.2)

and *2 Re2

1HEk I ant

For a half dipole antenna

236.5 ;2

antant ant rms

a

IP I I

R

The isotropic antenna is a point source which radiates with equal intensity in all directions. Its radiation pattern is a sphere. The co-ordinate system used is ,r,

III.2.2 Gain The gain (G) of an antenna is the factor by which its maximum radiated intensity exceeds that of an isotropic antenna if they emit equal power from an equal distance.

*2

2 2*

1Re ( , , ) ( , , )

21 / 4Re ( , , ) ( , , )2

ant antMax ant

a isoiso iso

E r H r kIG

R I rE r H r

(8.3)

Provided that . .antenna isotropic

s s

Int dS Int dS

An Isotropic antenna is a hypothetical device which radiates equally in all directions

The area of radio wave intercepted by a parabolic dish antenna is pretty obvious, but in principle a half wave dipole could have no area at all and yet still receive power from a radio wave. Hence we need to define an effective area.

III.2.3 Effective Area The effective area, Aeff , of an antenna is that area of wavefront whose power equals that received from the wavefront by the antenna. Aeff = Power collected by antenna Wave intensity (i.e. power/area) into antenna

III.2.4 Example – Power Transmission If two half-wave dipoles are 1 km apart and one is driven with 0.5 amps (RMS) at 300 MHz, what power is received by the other ? [ G = 1.64, Ra=73 , Aeff=0.13 m2] Answer: Intensity r metres from an isotropic antenna = Transmitted power/(4r2) Intensity r metres from this antenna

2

24ai R

Gr

Power received by receiving antenna = Intensity x Aeff

2

24a

eff

i RG A

r

2

2

0.5 731.64 0.13

4 1000

=0.3 W