8/9/2019 AP Handout 2013

1/59

Introduction to Antennas and Electromagnetic Wave Propagation

Dr. Costas Constantinou

School of Electronic, Electrical & Computer Engineering

University of Birmingham

c.constantinou@bham.ac.uk

1 Summary of Laws of Electrodynamics

Lorentz force law on a charged particle moving through an electromagnetic field:

F= q(E + v B) (1)

The law of conservation of charge:

.j + t

= 0 (2)

Maxwells equations:

.D= (Gauss law) (3)

E=

B

t

(Faradays law) (4)

.B= 0 (5) H= j +D

t (Amperes law) (6)

Constitutive relations: Valid for macroscopic, linear, isotropic, time-invariant, local media only.

Dielectrics: D= 0E + P= 0E + 0E= 0rE

Magnetic materials: B= 0H + 0M= 0H + 0mH= 0rH

Conductors: j= E

2 Solutions of Maxwells Equations

2.1 Potentials

In electrostatics, you will have come across the notion of electric potential, or voltage, and itsrelation to the electric field.

E= Bt

= 0 (7)

1

8/9/2019 AP Handout 2013

2/59

Using the vector calculus identity () 0 for any scalar differentiable function of position(r), we can infer that,

E= V (8)

In general, we need to consider the full solution to Faradays law, not just the special case of Eq. (7).

Let us begin by considering the solution to the fourth Maxwell equation, namely,.B= 0. Usingthe vector identity.( A) 0, which holds for any differentiable vector function of position,we conclude that the solution to the fourth Maxwell equation is,

B= A (9)

where A is called the magnetic vector potential. As we shall see shortly, finding the magneticvector potential is essential in determining the solution of radiation problems.

Combining Faradays law, E= B

t , with Eq. (9), gives,

E = ( A)t

=

A

t

E +A

t

= 0

This last equation is now in the same form as Eq. (7), whose solution is in terms of a static electric

potential field, giving,

E +A

t = V

E= V At

(10)

2.2 Wave Equations for Potentials

Going back to Maxwells equations, we can see that taking the curl of Faradays law yields,

E= Bt

= t

( H) ,

which assumes that is not a function of position r.

Using the the Maxwell equation for Amperes law corrected for the displacement current, gives,

E= t j +

E

t 2

8/9/2019 AP Handout 2013

3/59

Using, E= (.E) 2E, yields,

(.E) 2E= jt

2E

t2

Using Gauss law,

.D= , then gives,

1

2E= j

t

2E

t2 ,

which assumes that does not vary with position.

Rearranging finally gives,

2E 2E

t2 =

1

+ j

t (11)

Eq. (11) is a wave equation with source terms on the right hand side. The speed of propagation ofthe wave is v= ()1/2.

In free space, = 0, = 0, = 0 and j = 0, giving, v = (00)1/2 = c 3 108ms1 and

resulting in the familiar homogeneous three dimensional wave equation,

2E 2E

t2 = 0

Eq. (11) is in general difficult to solve. Following a similar approach but eliminating E instead ofHfrom Maxwells equations gives the corresponding wave equation for the magnetic field,

2H 2H

t2 = j (12)

In the presence of sources it is more convenient to form the corresponding wave equations to (11) and(12) for the electric and magnetic vector potentials. We first note that E= Bt and .B= 0are automatically satisfied (by construction) by Eqs. (10) and (9), respectively. Substituting Eq. (10)into Gauss law,

.E =

.

V At

=

2V t

(.A) =

(13)

Substituting (10) and (9) for the fields in 1 B= j + Et ,

A= j + t

V A

t

3

8/9/2019 AP Handout 2013

4/59

2A 2A

t2 (.A)

V

t

= j

2A

2A

t2

+

V

t .A=

j (14)

Neither Eq. (13) nor Eq. (14) is in the form of a standard wave equation. However, we can transformthem into wave equations because they contain an unspecified degree of freedom: .A has neverbeen specified.

In fact any vector field is fully specified by both its curl and divergence.

To observe that we are free to choose A and V up to an unspecified, arbitrary function, considerthe following question: What transformation of A and V leaves the fields E and B unaltered?This transformation will also leave the Lorentz force, F= q(E + v B), unchanged, as well as thePoynting vector, etc. (i.e. the Physics remains the same).

Using the vector identity 0 that holds for any differentiable scalar function of position ,we can see that the following transformation of the magnetic vector potential A leaves the magneticinductionBunchanged:

A = A + , (15)

since B = A = A+ = A. What is the implication on the electric scalarpotential Vthough? The two transformations for A and V must be consistent:

E = V At

= V At

V = V + (A A)

t

Using Eq. (15) the last term can be expressed in terms of,

V= V +()t

, (16)

which can only be true if,

V = V t

(17)

Eqs. (15,17) are called Gauge Transformations and allow us to specify A up to an arbitraryfunction (r, t).

The choice of gauge

.A= Vt

(18)

4

8/9/2019 AP Handout 2013

5/59

is called the Lorentz gauge and simplifies Eqs. (13) and (14) to the more familiar form of waveequations with a source term on the right hand side:

2V 2V

t2 =

(19)

2A 2A

t2 = j (20)

In free space =|j| = 0 and = 00 = c2. In this case Eqs. (11),(12),(19),(20) indicate thatany one of the Cartesian components ofE, Hand A, as well as V , satisfy the scalar, homogeneouswave equation. We use the function (r, t) to represent any arbitrary such component.

2.3 Electromagnetic waves in free-space

In free space all the above equations take the form,

2 1c2

2

t2 = 0 (21)

2.3.1 Plane wave solutions

Without loss of generality let us assume that the wave propagates along the z-direction. In thiscase Eq. (21) takes the form,

2

z2 1

c22

t2 = 0,

the general solution of which is,

(z, t) =f(z ct) + g(z+ ct),

wherefandg are arbitrary functions of a single variable representing a wave amplitude of constantshape travelling with speed c along the positive and negative z-axis, respectively.

If we are dealing with time-harmonic waves of angular frequency radians/s,

f

cos(t) and

Eq. (21) simplifies to,

2

z2 +

2

c2 = 0.

This is a simple harmonic oscillator ODE having a general solution,

= A cos

cz

+ B sin

cz

.

Choosing the origin of the coordinate system appropriately we can force B = 0, thus giving,

5

8/9/2019 AP Handout 2013

6/59

(z, t) = 0cos

cz

cos(t)

= 0

2 cos

c [z ct]

+

02

cos

c [z+ ct]

.

The wavelength of such a wave is = (2c)/; we define the wavenumber, k by,

k=

c =

2

(22)

2.3.2 Spherical waves

In the case of spherical waves the function must be a function of|r| =r =

x2 + y2 + z2, wherewe assume that the origin of the wave coincides with the origin of the coordinate system. The 3Dhomogeneous scalar wave equation in Cartesian coordinates,

2

x2+

2

y2+

2

z2 1

c22

t2

(x,y ,z ,t) = 0,

can be expressed in terms of the radial coordinate, r, of a spherical polar coordinate system usingthe chain rule to change the derivatives of with respect to r as follows:

Using

x

=

r

r

x

2

x2

=2

r2

r

x2

+

r

2r

x2

and

r

x =

x

r

2r

x2 =

1

r

1 x

2

r2

we have,

2

x2 =

x2

r22

r2 +

1

r 1 x2

r2

r

Adding the corresponding terms for 2

y2 and

2z2

similarly derived, gives,

2(r) = 2

r2 +

2

r

r

= 1

r

2

r2(r)

The wave equation then becomes,

6

8/9/2019 AP Handout 2013

7/59

1

r

2

r2(r) 1

c22

t2() = 0,

2

r2

(r)

1

c2

2

t2

(r) = 0,

which is exactly the same as the one-dimensional wave equation for plane waves, but with anargument r. The solution for a wave travelling in the radially outward (positive r) direction, isr = f(r ct), giving,

(r, t) =f(r ct)

r (23)

Spherical waves behave just like planar waves, except that their amplitude decreases as 1 /r, whichis consistent with the principle of conservation of energy.

2.3.3 Vector plane waves

If the unit vector along the direction of propagation is s, we can write,

E = E (r.s ct) (24)H = H (r.s ct) (25)

Substituting the above expressions into into Maxwells equations with = 0 and j= 0 and defineE= E/(r.s ct) (i.e. the derivative w.r.t. its argument), a simple application of the chain ruleyields,

E

t = cE

( E)x = Ez

y Ey

z

= Ez

(r.s ct)(r.s ct)

y Ey

(r.s ct)(r.s ct)

z= Ez sy Ey sz=

s Ex

Using the above two equations Faradays law (Eq. 4) and Amperes law (Eq. 6) can be simplifiedto,

s E = 0cHs

H =

0cE

7

8/9/2019 AP Handout 2013

8/59

Integrating the above equations w.r.t. (r.s ct) and setting the constant of integration to zero(since we are not interested in static, uniform fields in space) and using c = 1/

00, gives,

E=

00

s H (26)

H=

00

s E (27)

E, H ands are mutually orthogonal, i.e. Eand H are transverseto the direction of propagation.

Since the Poynting vector is given by S= E H, using Eqs. (26) and (27) gives,

S=

00

|H|2s=

00

|E|2s

The quantity,

Z0=

00

120 377 , (28)

is the impedance of free space.

Note: When sufficiently far away from any source, all waves are locally planar. Hence, all radiatedfields satisfy Eqs. (26) and (27) in the far field of antennas and scatterers.

3 Radiation

3.1 General solution to Maxwells equations

We shall soon consider the general solution of Eq. (20). Let us begin by considering spherical wavesfrom a point source. We briefly examine the following scalar problem,

2 1c2

2

t2 = s. (29)

Away from the point source taken to be situated at the origin of the chosen coordinate system,r =

x2 + y2 + z2 = 0, Eq. (29) reduces to Eq. 21, whose outward propagating solution in three-dimensions we have seen to be,

(x,y ,z ,t) =f(t r/c)

r . (30)

Now let us examine Eq. (30) near but not at the source of strength s at r = 0. the term r/cwill beso small that we can neglect it,

=

f(t)

r (r 0) (31)

8

8/9/2019 AP Handout 2013

9/59

This looks like the Coulomb field of a point charge at the origin that varies with time. Thecorresponding electrostatic problem would be,

V =Q/40

r

where

Q=

dV

2V = /0

Following the same mathematics as the electrostatic problem, we can say that in Eq. (31) satisfies,

2 =

s (r

0) (32)

where s and fare related by,

f= S

4 =

1

4

sdV. (33)

The only difference with the electrostatic problem is that s and Scan now be functions of time.

It should be stressed that Eqs. (29) and (32) are not in contradiction, since the 1/r dependence of

as r 0 makes2 1c2

2t2

. The magnitude of the time derivative does not change asr 0since we are only considering the time derivatives off(t

r/c).

To summarise, for a point source s(t), localised at the origin, the total strength is,

S(t) =

s(t)dV (34)

and

(r, t) = 1

4

S(t r/c)r

. (35)

The termr/c is the retardation time taken for electromagnetic disturbances to travel over adistance r at a speed c.

Finally, in the case of extended, vector sources it is straight-forward to argue that the generalsolution of Eq. (20) is given by,

A(r1, t) = 04

V2

j(r2, t r12/c)r12

dV2 (36)

where r12= |r1 r2| and V2 completely contains the source currents j.

In the special case of time-harmonic currents, j(r, t) =j(r)ejt , Eq. (36) becomes,

9

8/9/2019 AP Handout 2013

10/59

Figure 1: Coordinate system for current element radiator.

A(r1, t) = 04

V2

j(r2)ejtjkr12

r12dV2 (37)

where we have used Eq. (22).

We also have,

V(r1

, t) = 1

40V2

(r2)ejtjkr12

r12dV2

(38)

3.2 Current element radiator

The current element radiator is a prototype, infinitesimally small wire antenna.

By definition, the current has a uniform amplitude, Iejt and no phase variation along the lengthdl. Charges are created at the ends of the dipole since the current stops abruptly at the dipole ends.

Direct application of Eq. (37) gives,

AP = Az ez

Az = 0

4

Idlejkr

r (39)

where we henceforth omit the ejt term from all the equations, and

k=

c =

00=

2

0(40)

where, by definition of infinitesimal dipole, we require that 0 dl.

10

8/9/2019 AP Handout 2013

11/59

Now, since H = 10 A, and since Ais z-directed only, we have,

H= 1

0

Azy

ex Azx

ey

(41)

Azy

= 0Idl

4

y

ejkr

r

= 0Idl

4

r

y

r

ejkr

r

= 0Idl

4

r

y

1

rjk

ejkr

r (42)

Similarly, we find that,

Azx

= 0Idl4

rx

1

rjk

ejkr

r (43)

In order to find H we need to compute ry ex rx ey. This can be achieved using the standardspherical polar to Cartesian coordinate conversions,

r =

x2 + y2 + z2

x = r sin cos

y = r sin sin

Now, ry =12 2y

x2+y2+z2 = yr = sin sin , and similarly,

rx = sin cos . These results can be directly

substituted into ry ex rx ey to yield,

sin sin ex sin cos ey = sin e (44)

Combining Eqs. (41), (42), (43) and (44) thus yields,

H(r,,) =

Idl

4 sin 1

r +jkejkr

r e (45)

The radiated electric field can be found using the fact that at Pwe are situated in free-space andAmperes law from Eq. (6) becomes, H= Dt =j0E, which then gives,

E= 1

j0 H (46)

Using the curl operator in spherical polar coordinates together with the fact that only H is non-zero,

11

8/9/2019 AP Handout 2013

12/59

Er = 1

j0

1

r

H

+H

r cot

Er = Idl

20cos

j

r2

+k

re

jkr

r

(47)

E = 1

j0

H

r H

r

E = Idl

40sin

j

r2+

k

r +jk2

ejkr

r (48)

E= 0 (49)

Note that a Hertzian dipole is different from a current element in that the current varies in magnitudealong its length: It has a maximum in the middle and falls linearly to zero at its two ends.

3.2.1 Power considerations for a current element

Work done by the current Iagainst the generated electric field

The work done by the current flowing in a current element against the generated electric field isequal and opposite in sign to the radiated power. Therefore,

Prad = 1

2

E.j dV

Prad = 12

dl/2dl/2

EzI dl

Prad = 12{Ez(r 0) I dl}

Note thatEz =Ercos Esin . Since we need to evaluate the above expression on the z-axis asr 0, = 0, giving,

Ez = Idl

20

j

r3+

k

r2

1 jkr kr

2

2 +j

k3r3

6 + . . .

Ez = Idl

20

j

r3 j k

2

2r k

3

3 + . . .

The first two divergent terms in r do not contribute to the radiated power since they are in phasequadrature with I. The only non-vanishing in-phase contribution is,

{Ez(r 0)} = I dl k3

60

12

8/9/2019 AP Handout 2013

13/59

This is in antiphase with I, i.e., it opposes the current in the antenna and thus the energy is beinglost from the antenna in other words it is radiated.

Prad =

1

2

I dl k3

60I

dl

Prad = k2I2dl2

12(0/k)

However,0/k =

00

= 1Z0 = 1120 , giving,

Prad = 10k2I2dl2 (50)

Power transported into the far-field

When r

4

, i.e., k

r 1

r2 , Eqs (45), (47) and (48) simplify to,

H(r,,) = jkIdl

4 sin

ejkr

r

E(r,,) = jk2Idl

40 sin

ejkr

r

or,

E

(r,,) = jZ0kIdl

4 sin

ejkr

r

The time-averaged Poynting vector, given by, S = 12E H, is then,

S=1

2

Z0k2I2dl2

422r2 sin2 er

UsingZ0 = 120 simplifies this to,

S=15k2I2dl2

4r2 sin2 er

Integrating this over a sphere of radius r gives the total radiated power,

Prad =

S.da

=

0

20

15k2I2dl2 sin2

4r2 r2 sin dd

13

8/9/2019 AP Handout 2013

14/59

Prad = 15k2I2dl2

4

20

d

0

sin3 d

= 15k2I2dl2

4 2 4

3Prad = 10k2I2dl2 (51)

Which is in agreement with Eq. (50) derived earlier from different considerations. This proves thatthe work done by the current in the source against the generated electric field is al lconverted intoradiated energy in the far-field of the antenna.

3.2.2 Radiation resistance

In order to include antennas in circuit calculations, we account for the radiated power with anequivalent lumped radiation resistance, defined by,

Prad = 1

2I2Rr (52)

Using Eqs. (51) and (52) for the infinitesimally short current element,

Rr = 20k2dl2 (53)

3.2.3 Radiation pattern; antenna directivity and beamwidth

An antenna radiation pattern is a polar plot of the distribution of the radiated power in space. Forthe z-directed infinitesimally short current element under study, it is shaped as shown in the figurebelow. The radial distance of the polar plot from the origin is proportional to the magnitude of thePoynting vector in the same direction.

The 2D polar plot in a vertical plane that contains thez-axis is called theE-plane radiation pattern.The corresponding polar plot in the xy-plane is called the H-plane radiation pattern. The anglesubtended by the radial lines in the E-plane for which|S| = 12 |S|max and includes the direction ofmaximum radiation intensity is known as the 3 dB beamwidth. For the antenna discussed here,

|Smax(= /2)| =15k2I2dl24r2

|S(3dB)| = 15k2I2dl2

8r2

Therefore, sin2 3dB = 12 , giving

1,23dB =

4 ,34 . The 3 dB beamwidth, also known as the half power

beamwidth, is given by, HPBW =23dB 13dB= /2The directivity, D, is defined as,

14

8/9/2019 AP Handout 2013

15/59

Figure 2: Current element 3D radiation pattern [2].

D=Maximum power radiated per unit solid angle in the far-field

Average power radiated per unit solid angle (54)

D = |S|max(Prad/4r2)

= (15k2I2dl2)/(4r2)

(10k2I2dl2)/(4r2)

D=3

2 (55)

The above result is not only true for a current element antenna, but also holds for all electricallysmall antennas.

3.2.4 Antenna gain (transmit case); antenna efficiency

The gain of a transmitting antenna is defined as,

G= Maximum radiation intensity in the far-field

Radiation intensity of isotropic antenna with same input power (56)

If we assume no lossesPin = Prad and the gain of a lossless current element is given by, G = 32 . In

the presence of losses, Pin Prad, we define the antenna efficiency, to be,

= D

G (57)

with 0< 1.

15

8/9/2019 AP Handout 2013

16/59

3.2.5 Further antenna definitions

The gain of a receiving antenna is defined as,

G= Power delivered to a matched load by an antenna

Power delivered to a matched load by an isotropic antenna (58)

To determine the relationship between the transmit and receive gains for an antenna, we shallconsider the reciprocity theorem applied to antennas, as shown in the following figure.

Figure 3: Reciprocity applied to antennas.

Maxwells equations are linear and reciprocal otherwise reciprocity would not apply to linearcircuit theory which is just the special case of Maxwells electrodynamics when .The power received by antenna 2 when antenna 1 is transmitting is, Pr2 =

Pt1Gt1Ar24R2 , whereAr2 is

the effective receiving area for antenna 2 and Gt1 is the transmitting power gain of antenna 1 andPt1 is the corresponding transmitted power.

When the operation of the two antennas is reversed, we similarly have, Pr1 = Pt2Gt2Ar14R2 .

The four power quantities in the above equations can be related to the network description of thetwo antennas through the real part of the antenna impedances, (Z1) = R1 and(Z2) = R2,respectively, through the following equations:

16

8/9/2019 AP Handout 2013

17/59

Pt1 =|V1|2

8R1and Pr2 = |i2|2R2

Pt2 =|V2|2

8R2and Pr1 = |i1|2R1

Pr2Pt1

=8R1R2|i2|2

|V1|2 and Pr1

Pt2=

8R1R2|i1|2|V2|2

However, by the reciprocity theorem we have,

i2V1

= i1V2

(59)

Taking the modulus square of the above expression and using the equations on the last two slidesthus gives,

Pr2Pt1

=Pr1Pt2

= Gt1Ar2

4R2 =

Gt2Ar14R2

(60)

This simplifies to,

Gt1Ar1

= Gt2Ar2

= universal constant for al lantennas (61)

Now consider a receiving current element antenna situated on the xy-plane and being z-directed,as is a transmitting current element antenna a distance R apart along the x-axis.

Voc = |Eincidentz | dl (62)

Since the receiving current element antenna is at = 2 w.r.t. the transmitting one, Ez =E(R,

2 , 0), where E is the expression derived earlier for the current element, Eq. (48), evalu-

ated in the far-field, giving,

Voc = Z0 k I dl

4R sin

2dl

Voc = Z0 k I dl24R

(63)

The power that can be delivered into a matched load by a lossless receiving current element is,

Pr2 = 1

2

Voc2Rr

2Rr =

V2oc8Rr

(64)

Substituting for Pt1 from Eq. (51), for Pr2 from Eqs. (64) and (63) and Rr from Eq. (53) intoEq. (60), then gives,

17

8/9/2019 AP Handout 2013

18/59

Z0kIdl2

4R

2

8(20k2dl2)

10k2I2dl2 =

32Ar24R2

This simplifies to,

Ar2 = 3

2

2

4 (65)

Substituting Eq. (65) into Eq. (61) and using the fact that for our particular lossless current elementG= 32 , finally gives,

GtAr

=4

2 (66)

Note that an isotropic antenna has unity gain, by definition, thus,

Aisotropic antenna= 2

4 (67)

Using the definition of the receiving gain for an antenna, we have,

GRx= V2oc/8Rr

E2z2Z0

Area of isotropic antenna(68)

SubstitutingRr from Eq. (53) and Voc from Eq. (63) into Eq. (68) and using Eq. (67) finally gives,

GRx=3

2 (69)

We can thus see that for a current element,

GTx= GRx= G (70)

This is a general result that is valid for all passive, reciprocal antennas (i.e., ones without ferriteor active components). From it we can also see that the effective aperture area of any antenna isrelated to its gain by,

Ae= G2

4 (71)

Finally, we need to remark that in all above equations the antenna gain is expressed in linear,dimensionless units, whereas in engineering calculations it is usually expressed in logarithmic units,GdB= 10log10 G dBi, where dBi refers to decibels relative to an isotropic antenna.

18

8/9/2019 AP Handout 2013

19/59

3.2.6 Input impedance of an antenna

A complete electromagnetic analysis of a radiating antenna of finite conductivity and non-zerodiameter and length yields the result that the input impedance of an antenna is in general of theform,

Zin= Rin+jXin (72)

whereXinmay be either capacitive or inductive depending on the frequency and the specific antennageometry.

Furthermore,

Rin = Rloss+ Ri (73)

where Ri is determined by the radiation resistance through,

1

2I2inRi=

1

2I2maxRr (74)

In practice we adjustRin to be either 75 or, more usually, 50 .

The exact determination ofZin() is often a very difficult problem that is solved through electro-magnetic simulation and which we will not address analytically in this course. It involves knowledgeof the current distribution on an antenna, which is not known a priori(only the terminal voltage isknown exactly), but can be usually approximated by some function, which is typically trigonometric.

The calculation is summarised as follows:

Vterminal, assume &approximate

(z),j(z) V(z), A(z) E(z)

E(z),j(z) Zin=

Ez(z) dz

Iin

The detailed analysis for a short dipole of length and radius a, made out of perfectly conductingmaterial can be found in a number of antenna books (c.f. [1]). For such a dipole,

Zin 5.5k22 j 60k

ln

2a

3.3 Lossless, filamentary linear dipoles

We consider the dipole shown in the figure below only in the case when a H, , i.e. a 0.The boundary conditions we need to enforce on the surface of the perfectly conducting cylindricaldipole are:

(a) I(z = +H) =I(z = H) = 0

19

8/9/2019 AP Handout 2013

20/59

Figure 4: Linear dipole.

(b) Ez(z, r= a) = 0

CombiningE = V At together with boundary condition (b) yields,

Ez = 0 = Vz

jAz (75)

The Lorenz gauge condition,.A= 1c2 Vt , simplifies to,

Azz

= jc2

V (76)

Differentiating Eq. (76) w.r.t. z and substituting for Vz from Eq. (75), yields,

2Azz2

+2

c2Az = 0

2Azz2

+ k2Az = 0 (77)

The general solution of which is,

Az(z) =B cos(kz) + Csin(kz) (78)

Now, from Eq. (37), we have,

Az(z, r= a) =

04

+HH

I(z)ejk

a2+(zz)2a2 + (z z)2 dz (79)

where we have assumed (not strictly correctly) that for a filamentary dipole all the current is situatedon the z-axis. Asa0 the main contribution to the integral in Eq. (79) will arise from the poleatz = z . For very small, but non-zero sized a, it also follows that the main contribution to Az(z)

comes from the currentI(z) in the immediate vicinity ofz = z and hence,

20

8/9/2019 AP Handout 2013

21/59

Az(z, r= a) I(z)

04

+HH

ejk

a2+(zz)2a2 + (z z)2 dz

(80)

where the term in the curly brackets is approximately constant for H < z < +H, this term beingan average inductance per unit length, Lav, on dimensional grounds. Thus,

Az(z) I(z)Lav (81)

Eqs. (77) and (81) imply that for such a dipole,

2I

z2 + k2I= 0 (82)

I(z) =B cos(kz) + C sin(kz) (83)

Applying the boundary condition (a) for the current at z =Hand using the fact that Az andhence Imust satisfy the continuity condition at z= 0, we have,

I(z) I0sin [k(H z)] z >0, z /2, we have fromEq. (45),

Hr = H= 0

H jkIdl

4 sin ejkr

r (85)

Using the principle of superposition, the radiated magnetic field of the dipole of length 2 H is thengiven by,

H(r,,) = jk

4

+HH

I(z)sin (z)ejkR(z)

R(z) dz (86)

For a distant point of observation P, r H, z and also r . However, z is of the order of .Thus, the following approximations can be made to evaluate this integral.

21

8/9/2019 AP Handout 2013

22/59

Figure 5: Dipole radiation.

sin (z) sin 1R(z)

1r

Care needs to be exercised in simplifying kR(z) in the exponent.

kR(z) k

(r sin cos )2 + (r sin sin )2 + (r cos z)2 k r2 2rz cos + z2 12 kr

1 2z

r cos +

z2

r2

12

kr

1 12

2z cos

r +

1

2

z2

r2 1

2

z2 cos2

r2 + . . .

kr kz cos + kz2

2r sin2

ApproximatingkR bykr for electrically large antennas will cause significant errors in the integralof Eq. (86). However, all higher order terms in the expansion can be neglected if,

kz2

2r sin2

worst case

/8

2H2

2rminsin2

2 /8

rmin 8H2

(87)

22

8/9/2019 AP Handout 2013

23/59

The condition r 2L2/, defines thefar-field region(or Fraunhofer region) of any antenna, where

L= 2His its maximum linear dimension (for aperture antennas this is taken to be the diagonal).

It can be shown (see [2]) that the reactive near-field region for any antenna is defined by r 0.62

L3/. The intermediate region 0.62

L3/ r 2L2/is known as the radiating near-field,

or Fresnel, region.

Hence, in the far-field Eq. (86) becomes,

H(r,,) =jk sin

4

ejkr

r

+HH

I(z)ejkz cos dz (88)

Substituting for I(z) from Eq. (84) into Eq. (88), then gives,

H(r,,) = jkI0sin

4

ejkr

r +H

0sin[k(H z)] ejkz cos dz

+

0H

sin[k(H+ z)] ejkz cos dz

Both integrals are of the form,

eax sin(bx+ c) dx=

eax

a2 + b2[a sin(bx+ c) b cos(bx+ c)]

which gives the final result for H as,

H=jI0

2

ejkr

r

cos(kHcos ) cos(kH)

sin

(89)

Also, in the far-field we can use the result E=Z0H.

3.3.2 Radiation resistance of a centre-fed resonant half-wave dipole

In this case, H=/4 gives from Eq. (89),

E=j60I0 ejkr

r cos2cos

sin (90)

The power radiated through a large sphere of radius r is,

Prad =

0

20

|E|22Z0

r2 sin dd

= 602I20

2120

20

d

0

cos22cos

sin2

1

r2r2 sin d

= 30I

2

0 0

cos2 2cos sin d

23

8/9/2019 AP Handout 2013

24/59

Prad = 60I20

2

0

cos22cos

sin

d

60I20 0.609

36.5I2

0

Since Prad = 12I20Rr and I0= Imax(z),

Rr 73 for a 2 dipole (91)

3.3.3 Power gain of a 2 dipole

We assume that the dipole is lossless and thus, G = D. Thus, the power gain of the dipole can beexpressed symbolically as, G= max

avg.

max = |E|2max

2Z0=

I20Z20

4r22Z0= 4.775

I20r2

avg =12I20Rr

4r2 =

73I208r2

= 2.905I20r2

Therefore, G = 4.7752.905 , giving,

G= 1.64 or 2.16 dBi (92)

The effective aperture area of this resonant dipole is given by, Aeff=G2

4 1.642

4 = 0.132.

The polar plot of the E-plane radiation pattern for a 2 -dipole, i.e.,

cos(2 cos )

sin

2, is shown in the

figure below both in linear and decibel units.

Figure 6: Dipole radiation pattern: (a) linear (b) decibel units.

The 3 dB beamwidth of this dipole antenna is approximately 78.

24

8/9/2019 AP Handout 2013

25/59

3.3.4 Self (terminal) impedance of a thin linear half-wave dipole

For an approximate analytical treatment see [3]. There are two important results to highlight:

For a thin 2 -dipole, Zin (73 + 42.5j) The imaginary part of the terminal impedance of resonant, centre-fed antennas is inductive

3.4 Other wire antennas

3.4.1 Yagi-Uda array

The Yagi-Uda array is shown below. It is inexpensive to build, easy to feed, has a moderately highgain and is difficult to design.

Figure 7: Yagi-Uda array antenna.

SR 0.250SD 0.10 0.30hF =0/2

hR = 0/2 + h, where h < 0.10hD = 0/2 h, where h < 0.10L= 1.20 6.00G= 10 dBi 16 dBiNo of director elements 1 25Bandwidth 5% off0= c/0(compared to 15% for a/2 dipole)

Weillustratethe operation of a Yagi-Uda array by considering just two dipoles, each of approximatelength 0/2, spaced 0.10 apart. The driven element is chosen to be exactly0/2 long, while thesecond element is short-circuited (i.e. it is parasitic) and its length is 0/2 + h.

25

8/9/2019 AP Handout 2013

26/59

Figure 8: Two-element prototype Yagi-Uda array.

V1 = I1Z11+ I2Z12 (93)

0 = I2Z22+ I1Z21 (94)

See [4] for the calculation of the four impedances, Z11, Z12, Z21, Z22.Reciprocity requires thatZ12= Z21.

Zin=V1I1

(95)

From Eq. (94), we have,

I2I1

= Z21Z22

(96)

Combining Eqs. (95), (93) and (96) then gives,

Zin= Z11 Z212

Z22(97)

The two-element array radiation pattern depends on the element radiation patterns, spacing andI1 and I2 only and, through the latter, on the self and mutual impedances.

Again from [4], we quote the following three results (we assume an antenna spacing of0/10):

Z12

(67.5 +j10) for l

0

(98)

26

8/9/2019 AP Handout 2013

27/59

Z11 (73 +j42.5) at resonance (0/2) (99)

Z22 (73 + 350l20

) +j1500l20 for 0 = l 0 (100)

Since the azimuth (H-plane) radiation pattern of each dipole is omnidirectional, the overall radiationpattern in the xy-plane is given by the array factor of two point sources at (0, 0) and (0.10, 0)whose phasor amplitudes are I1 and I2 respectively.

The radiated field at the point of observation P is given by the principle of superposition usingEq. (81). When P lies in the far-field, and using the cosine rule with dr , gives a phasedifference for the two contributions ofk(r r) kd cos :

Figure 9: Two-element array radiation pattern.

|Ez(P)| I2+ I1ejkd cos

I2I1

+ ejkd cos

Using Eq. (96) gives,

|Ez(P)| Z12Z22 + ejkd cos

(101)Two cases are of particular interest in our discussion of the principles behind the operation of theYagi-Uda array.

(a)d = 0.10,l20

= 0.02: Eqs. (97)-(100) then yield, Zin (30 +j0) and the radiation patternis given by combining Eqs. (98), (100) and (101) to give,

27

8/9/2019 AP Handout 2013

28/59

0.5

1

1.5

30

210

60

240

90

270

120

300

150

330

180 0

(a) Radiation pattern with para-sitic element acting as a directord= 0.10,

l20

= 0.02

0.5

1

1.5

30

210

60

240

90

270

120

300

150

330

180 0

(b) Radiation pattern with para-sitic element acting as a reflectord= 0.10,

l20

= +0.03

Figure 10: Effect of parasitic element parameters on two-element array radiation pattern.

28

8/9/2019 AP Handout 2013

29/59

|Ez()| 0.93ej2.5744c + ej20.1cos (102)

(b) Similarly, d = 0.20,l20

= +0.03 gives, Zin (70 +j74) and

|Ez()| 0.59e+j2.2515c + ej20.2cos (103)The polar plots of these two H-plane radiation patterns are thus:

We can now see that in case (a) the parasitic element is acting as a director, by pullingthe boreside of the radiation pattern in its direction.

In contrast, in case (b)the parasitic element is acting as a reflector, by pushing the boresideof the radiation pattern away from itself.

3.4.2 Practical considerations for wire antennas

Balanced mode operation; BALUNs Tolerance to dimensions; rigidity, mechanical strength Presence of other conductors in the reactive near field zone affecting the antenna in situ

operation

Use of corner reflectors

4 Aperture antennas

At frequencies in the SHF band (3-30 GHz) and higher, the skin effect and the associated ohmiclosses dominate over radiation processes in all wire antennas. As a result G D and the antennaefficiency 0.The use of these high frequencies in long-range communications, e.g. satellite communications, im-plies that very high gains are required in order to compensate the huge propagation losses associatedwith free-space propagation over thousands of kilometres.

The only antennas that can be used for such applications are aperture and/or reflector antennas.

4.1 Physical optics

Consider a linearly polarised uniform plane wave incident normally on a perfectly conducting infiniteplane coincident with the xy-plane.

The incident fields are, E= Eiejkz ex andH= Hiejkzey, where the field amplitudes are relatedby, Ei =HiZ0, and Z0 120 is the impedance of free space.The total electric field must take the form Etotal = Eincident + Ereflected, and must satisfy theboundary conditionEtotal(z = 0) = 0.

Thus,E = Eiejkz ex

Eie+jkz ex.

29

8/9/2019 AP Handout 2013

30/59

Figure 11: Uniform plane wave normal incidence on a perfect conductor.

Using Faradays law, Eq. (4),

E =

Bt =

j0H and substituting for E gives, H =

Hiejkzey+ Hie+jkzey.

Using Amperes law, Eq. (6), H= j + Dt at z = 0, it can be easily shown that the surfacecurrent density J is given by J = n H, where n is the outward unit normal of the conductorsurface.

Thus,J = ez 2Hiey = 2Hiex, or

J= 2 (n H) (104)

For conductors of finite extend and curvature the above expression is only approximately true. How-

ever, if the conductor dimensions and radii of curvature are much bigger than 0, the approximationis very accurate and is referred to as the physical optics approximation.

4.2 The duality theorem

Maxwells equations, Eqs. (3-6), in a source-free region become,

H= Dt

E= Bt

.D= 0

.B= 0

Maxwells equations exhibit a symmetry, which can be expressed as follows: The two equationsto the left can be used to arrive at the two equations to the right and vice-versa, if the followingsystematic exchange of symbols is made:

E H, H E, ,

The symmetry described above breaks down when charges and currents are introduced.

In order to retain this symmetry in the presence of sources, we must introduce the concepts of(non-physical) magnetic charges and currents and the latter must obey the following exchange

rules:

30

8/9/2019 AP Handout 2013

31/59

j jm, jm j, m, m

Thus, Maxwells equations now become,

.D = (Coulombs/m3) (105).B = m (Webers/m3) (106)

E = jm Bt

(107)

H = j + Dt

(108)

In the case of time-harmonic fields the above equations can be solved in terms of potentials in theusual way:

Scalar potentials:

= 1

4

V

ejkr

r dV (109)

u= 1

4

V

mejkr

r dV (110)

Vector potentials:

A= 4

Vje

jkr

r dV (111)

F=

4

V

jmejkr

r dV (112)

The solutions for the fields in terms of potentials now become:

E= At

1

( F) (113)

B= u Ft

+1

( A) (114)

D= F (115)

B= A (116)

The duality theorem states that there is complete symmetry between the sets of solutions ofMaxwells equations for j= 0,jm = 0 and j= 0,jm= 0 under the appropriate exchange rules fordual quantities.

31

8/9/2019 AP Handout 2013

32/59

Figure 12: Illustration of the surface equivalence theorem.

4.3 Surface equivalence theorem

The surface equivalence theorem states that the actual sources bounded by a closed surface S= Vcan be replaced by equivalent sources on S, in such a way that the fields outside the volume Vdefined by S remain unchanged.

Actual problem: Sources J1, 1 excite fields E1, H1 over all space (V1+ V2).

Equivalent problem: Equivalent surface currentsJ, Jm excite identical fields E1, H1 everywherein volume V2. Volume V1 is now source-free and the fields in V1 can be set to any arbitrary,convenient value. If this value is 0, the electrical properties of V1 can be altered (, can takearbitrary values).

We know from Maxwells equations that a tangential magnetic field discontinuity can be accountedfor by an electric surface current density by:

J= n (H1 H)

Using duality,

Jm= n (E1 E)

Since what happens within V1 does not affect the fields in V2, we may set, arbitrarily, E = H = 0:

J= n H1 (117)

Jm= n E1 (118)

The above results are exact.

In the event that E = H = 0 in V1, changing the electrical properties of the medium enclosed by Swill not alter E1 and H1 in V2. Thus, if we set = in V1, we have J= 0 andJm= 2n E1.

32

8/9/2019 AP Handout 2013

33/59

Figure 13: Equivalent formulations of paraboloid antenna radiation.

Note that in the above case the currents are also exact, but the expressions for the vector potentialsare not. This is because the surface currents cannot radiate omnidirectionally when = andtherefore expressions such as,

F=

4

V

Jmejkr

r dV

are not strictly correct, since they make use of the free-space Greens function ejkr

4r . They are,though, a very good approximation if most ofSin the actual problem ismetallic.

4.3.1 Applications of the equivalence theorem

(a) Paraboloid antenna: See Fig. 13.

(i) On S: J= n H & Jm= 0(ii) On S1: J= n H& Jm= n E

while on S2: J= n H& Jm= 0(iii) OnS1: J= n H& Jm= n E

(b) Open-ended circular waveguide radiator: See Fig. 14.

(i) If the tangential magnetic fields over the outside and inside cylindrical surfaces,Ht and Hu,respectively are known, the radiation pattern can be calculated using J= n Ht +n HuandJm= 0.

(ii) On the cylindrical surface, J = n Ht and Jm = 0. On the aperture, J = n Ht andJm= n Et .

(iii) On the infinite plane containing the waveguide aperture,J = n Ht and Jm= n Et.

Note that all of the above are exactly equivalent formulations of the same physical problem described

by (i).

33

8/9/2019 AP Handout 2013

34/59

Figure 14: Equivalent formulations of open-ended circular waveguide radiation.

4.4 Radiation from an aperture

Consider an aperture which lies in thexy-plane of the chosen coordinate system as shown in Fig. 15,and over which the electric and magnetic fields are completely known.

4.4.1 Radiation due to tangential magnetic field

The equivalent magnetic current flowing over the aperture is,

Jm = n E= ez (Exex+ Eyey+ Ez ez)Jm = Exey+ Eyex

The electric vector potential set up at point Pdue to the magnetic current density flowing over an

infinitesimally small area around point S, is,

dF=

4(Eyex Exey)e

jkr0

r0dx dy

The electric field at point P is then given by,

dE= 1

( dF)

It turns out that it is easier to work in spherical polar coordinates to evaluate the previous equation,

which necessitates using the following coordinate transformations:

34

8/9/2019 AP Handout 2013

35/59

Figure 15: Radiation from an aperture.

dFr = dFxsin cos + dFysin sin + dFzcos

dF = dFxcos cos + dFycos sin dFzsin dF = dFxsin + dFycos

dEr = 1

1

r

(dF)

+

dFr

cot 1r sin

(dF)

dE = 1

1

r sin

(dFr)

(dF)

r dF

r

dE = 1

(dF)

r +

dFr

1r

(dFr)

Neglecting terms O(r2) and taking r0r 1 in the far-field, gives,

dEr 0dE 1

(dF)

r =

1

4(Excos + Eysin )jk

ejkr0

r0dx dy

dE 1

(dF)

r =

cos

4 (Exsin + Eycos )jk e

jkr0

r0dx dy

Usingk= 2 jk4 = j2 and applying the far-field approximation to the phase and 1r0 dependencies,

35

8/9/2019 AP Handout 2013

36/59

r 1

r0 1

r

r0 = r

r0=

(r sin cos cos )2 + (r sin sin sin )2 + (r cos )2

r0 =

r2 + 2 2r sin cos( )

r0 r sin cos( ) + O

2

2r

Using the above expressions and summing up the contributions over all the aperture area, gives:

E1(r,,) j

2

ejkr

r

aperture

Ex(

, )cos

+ Ey(, )sin

ejk

sin cos() d d

E1(r,,) j

2

ejkr

r cos aperture

Ex(, )sin

Ey(, )cos

ejk sin cos() d d

(119)

4.4.2 Radiation due to tangential magnetic field

The equivalent electric current density flowing on the aperture is,

J = n H=ez (Hxex+ Hyey+ Hzez)J = Hyex+ Hxey

The magnetic vector potential excited at Pdue to the current flowing over an infinitesimally smallarea on the aperture centred at S is,

dA=

4(Hyex+ Hxey)e

jkr0

r0dx dy

We may now use the same coordinate transformations employed in the previous section for dF to

determine the spherical polar components ofdA.

36

8/9/2019 AP Handout 2013

37/59

The corresponding magnetic field at point P is given by,

dH= 1

( dA)

Using the same far-field approximations as before, we have,

dHr 0dH 1

(dA)

r =

jk

4(Hxcos + Hysin )

ejkr0

r0 d d

dH 1

(dA)

r =

jk cos

4 (Hx sin + Hycos )e

jkr0

r0 d d

Therefore, the total magnetic fields due to the electric currents in the aperture are,

H2(r,,) j

2

ejkr

r

aperture

Hx(

, )cos

+ Hy(, )sin

ejk

sin cos() d d

H2(r,,) j

2

ejkr

r cos

aperture

Hx(, )sin + Hy(

, )cos ejk sin cos() d d

In the far-field though, we know that E = Z0H and E =Z0H. Hence, the far-field electricfields due to the tangential magnetic field on the aperture are:

E2(r,,) j

2

ejkr

r cos

aperture

Z0Hx(, )sin

+ Hy(, )cos

ejk

sin cos() d d

E2(r,,) j

2

ejkr

r

aperture

Z0

Hx(, )cos

+ Hy(, )sin

ejk

sin cos() d d

(120)

4.4.3 Radiation due to total fields on an aperture

The total radiated electric field must be calculated by taking into account both the electric andmagnetic equivalent current sources. Using superposition, Eqs. (119) and (120) can be used to

calculate:

37

8/9/2019 AP Handout 2013

38/59

Figure 16: Radiating slot geometry [2].

E(r,,) = [E1(r,,) + E2(r,,)] e

+ [E1(r,,) + E2(r,,)] e

H(r,,) = er E(r,,)/Z0

4.4.4 Radiation from a uniformly excited rectangular aperture

Assume that a rectangular aperture of width a and height b centred at the origin is cut out of an

infinite perfectly conducting ground plane situated at z = 0 and is illuminated with a uniform planewave polarised along the y -direction fromz

8/9/2019 AP Handout 2013

39/59

Figure 17: Linearly polarised rectangular slot radiation pattern [2].

The total radiation radiation pattern for an aperture with a= 3 and b = 2is shown below:

4.4.5 Power gain

G=max

av

The maximum radiation intensity is at = 0, = 0, so we need to use the limit limx0sin(x)

x = 1.

max = |E|2max

2Z0=

1

2Z0

|E|2max+ |E|2max

= 1

2Z0

a2b2E2y2r2

The average radiation intensity is given by,

avg

= (Power density in aperture)

(aperture area)

4r2

=

E2y/2Z0

(ab)

4r2

Therefore,

G=4

2ab

Note that in this instance, Ae = Aphysical =ab and, therefore, the aperture efficiencya= AeAphysical

=

1.

39

8/9/2019 AP Handout 2013

40/59

Figure 18: Microstrip patch antenna geometry [2].

4.5 Microstrip antennas

A microstrip patch antenna is a metallic strip or patch mounted on a dielectric layer (substrate)which is supported by a ground plane.

Advantages DisadvantagesLow profile Low power handling

Conformable to non-planar surfaces Narrow bandwidth (Q 100)

Simple and inexpensive Large EM signature at certain frequencies

outside operating band

Mechanically robust In large arrays there is a tradeoff between

bandwidth and scan volume

Can be integrated with MMICs Tradeoff substrate dielectric constant

Versatile in terms of resonant frequency, po-larisation, patterns and impedance

Large physical size at VHF and lower UHFbands

4.5.1 Basic characteristics

Very thin metallic patches, t 0 on top of a thin dielectric layer, h 0, usually 0.0030 h 0.0950, of dielectric constant 2.2 r 12. For a rectangular patch, the length L, is typically

0/3< L < 0/2. The width W is chosen to be W = c2f

2r+1

, to yield good radiation efficiency.

For good antenna performance, i.e. better efficiency and larger bandwidth, the fringing fields needto be loosely bound, requiring low dielectric constant, thick substrates. However, thick substratesgive rise to surface waves and low substrate dielectric constant gives rise to large elements.

4.5.2 Feeding methods

Microstrip line feed (see Fig. 18) Easy to fabricate

Simple to match by varying inset feed position

Low spurious radiation ( 20dB) Narrow bandwidth (2 5%) Suited to thin substrates, otherwise radiates

Gives rise to cross-polarisation

40

8/9/2019 AP Handout 2013

41/59

Coaxial probe feed Similar to microstrip feed lines

Aperture coupling Independent optimisation of feed and radiating element

Difficult to fabricate

Proximity coupling Largest bandwidth

Difficult to fabricate

4.5.3 Methods of analysis

1. Transmission line model

(a) Good prediction of terminal characteristics

2. Cavity model

(a) Good prediction of radiation characteristics

3. Full wave models (numerical)

(a) Integral equation (method of moments, MoM)

(b) Modal analysis

(c) Finite difference time domain (FDTD)

(d) Finite element analysis

The first two methods are approximate but give good physical insight, whereas the full wave methodsare accurate but complex.

4.5.4 Transmission line model for rectangular patch: Effective length, resonant fre-quency and effective width

For W/h >1,

reff r+ 12

+r 1

2

1 + 12

h

W

1/2(123)

The width of each radiating slot (the effective fringing field width), L is given approximately by,

L

h 0.412(reff+ 0.3)

Wh + 0.264

(reff 0.258)

Wh + 0.8

(124)For the dominant TM010mode, the resonant frequency of the rectangular microstrip patch antenna

is determined by Leff=L + 2L= 0/2,

41

8/9/2019 AP Handout 2013

42/59

Figure 19: Microstrip effective dielectric constant vs. frequency [2].

Figure 20: Effective microstrip patch antenna length [2].

42

8/9/2019 AP Handout 2013

43/59

Figure 21: Transmission line model used to calculate patch antenna input impedance [2].

f010= c

2(L+ 2L)

reff(125)

Equations (123)-(125), together with the efficient radiation condition W = c2f

2r+1

can be used to

determine the patch dimensions L and Wfrom its resonant frequency f010, the substrate thicknessh and the substrate dielectric constant r.

The terminal impedance is calculated using transmission line theory, to translate the two radiatingslot impedances to an impedance at the terminal plane of the microstrip feed line. The details ofthe calculation can be found in Balanis C.A., Antenna Theory, 3rd Ed., Wiley-Interscience, 2005

14.2.1.

4.5.5 Cavity model: Radiated fields

The microstrip patch antenna is modelled as rectangular cavity resonator with perfect magneticwalls (Hx = 0) and the surface equivalence theorem is used to model its far-field radiated fieldsfrom its two radiating slots (see Balanis C.A., Antenna Theory, 3rd Ed., Wiley-Interscience, 200514.2.2). The detailed analysis is beyond the scope of this module.

Figure 22: Cavity model (left) and equivalent radiating currents (right) for a rectangular microstripantenna [2].

43

8/9/2019 AP Handout 2013

44/59

Figure 23: Wave propagation over a homogeneous flat earth.

5 Radiowave propagation

Radiowave propagation is a subject treated mostly empirically or heuristically as the number ofrigorous boundary value problems that are tractable is very limited. In the remaining lectures ofthis module we shall concentrate on two topics that lend themselves to tractable electromagneticanalysis: the problem of medium frequency propagation over a homogeneous, lossy, flat ground andthe problem of high frequency wave propagation through the ionosphere.

5.1 Surface wave propagation over flat earth

We define the complex relative dielectric constant to be,

= r j 0

(126)

The magnetic vector potential which we will denote by has only a z-directed component andsatisfies the equations,

2 + k20 = 0J= 0(x)(y)(z h) z 0

2

3+ k203 = 0 z

8/9/2019 AP Handout 2013

45/59

2

z2+ k20 2

(x, y, z) = 0

42(z h) z 0

2

z2 + k2 23(x, y, z) = 0 z

8/9/2019 AP Handout 2013

46/59

C=B (1 ) (140)

Substituting the expressions for , 0 and can verify that is the familiar Fresnel reflectioncoefficient for a vertically polarised plane wave.

Inverting the Fourier transform gives,

1= 082

1

0

1 e20h 0

+ 0

ejxxjyy0(zh)dxdy

Changing to cylindrical polar coordinates (,,z) and defining x = cos and y = sin ,

yields,

1 = 0

82 2

0

0

1

01

e20h 0+

0

ej cos()0(zh)dd

Using the standard integral20 e

ju cos() d = 2J0(u), we obtain Sommerfelds 1909 result,

1(,,z) = 04

0

1

0

1 e20h

e0(zh)J0() d (141)

For = 0, the integral can be solved in closed form to obtain,

1= 04

ejk2+(zh)22 + (z h)2

that is, the free-space solution. Moreover, when = 1 (i.e. we have a perfectly conducting ground)we arrive at a superposition of two such free space solutions, the second one arising from a secondpoint source situated at z =h, an image source accounting for a specularly reflected sphericalwave.

The solution of Eq. (141) for ground based antennas, (z = h = 0), is more counter-intuitive as ityields a non-zero results: This results is a surface waveguidedby the air-earth interface.

Numerical evaluation of Eq. (141) shows that this surface wave amplitude dies exponentially withdistance away from the interface (z), on both sides of the interface, while for large the fieldmagnitude dependence is 2. The surface wave turns out to be very important for LF, MF andsometimes HF propagation (i.e. below 10 MHz).

For z = h = 0 and , asymptotic evaluation of the Sommerfeld integral in Eq. (141) and useofEz jAz (remember that Az ), yields,

Ez jk0Z04

ejk0

2( 1)

As (142)

where As is the surface wave attenuation factor, given by,

46

8/9/2019 AP Handout 2013

47/59

Figure 24: Surface wave attenuation factor|As| vs. numerical distance p [5].

As = 1 jseserfc (js) (143)

and where erfc() is the complementary error function and s is the numerical distancedefined as,

s= jk0

22 ( 1) pejb

(144)

The surface wave attenuation factor|As| is shown in the figure above plotted as a function of themagnitude of the numerical distance p for a number of ground phase angles b.

5.2 Ionospheric propagation

5.2.1 The ionosphere

The air in the upper parts of the earths atmosphere (at heights between 90 km and 1,000 km) is

ionised typically due to interactions with solar radiation.

The free electrons in the ionosphere interact with radiowaves and cause radiowaves of frequencies upto about 40 MHz to be reflected back to earth. In this way, the ionosphere can be used to establishlong-distance communication links over distances of the order of 1,000 km along the surface of theearth. At higher frequencies radiowaves penetrate the ionosphere, but its presence can degrade theperformance of geo-location satellite links.

The ionosphere acts as a lossy, depolarising mirror.

47

8/9/2019 AP Handout 2013

48/59

Figure 25: Ionospheric free electron density variation with height above mean sea level [5].

Figure 26: Long-range ionospheric communication [5].

48

8/9/2019 AP Handout 2013

49/59

5.2.2 The dielectric constant of an electron gas

We neglect the interaction of the radiowaves with the ionised gas molecules in the ionosphere com-pared to the corresponding interaction with the free electrons, as the molecules are typically 2,000times more massive than the electrons and thus their accelerations and velocities are correspondinglysmaller.

Newtons second law for one free electron is,

F = medv

dt =FLorenz+ Fdamping

jmev = e(E + v B) mev

If we neglect the strong magnetic field of the earth, B0, for the moment, at typical non-relativisticelectron velocities, we have|v B| |E|, giving,

(jme+ me) v eE

For Nelectrons per unit volume, the current density is related to the drift velocity of the chargesthrough,

j= eNv

These two equations can be combined to give,

j= j e2

Nme( j) E

Substituting this into Amperes law, Eq. (6) and using D= 0E in free-space and ()t = j () for

time-harmonic fields, gives,

H = j e2N

me( j) E +j0E

= j0 1 N e2

me

0

(

j)EThe square brackets can be identified as being the effective relative dielectric constant of the electrongas,

= 1 N e2

me0( j)

where is the collision frequency and Ne2

me0has dimensions of s2.

We define the plasma frequency, p, by,

49

8/9/2019 AP Handout 2013

50/59

p =

N e2

me0(145)

which gives,

= 1 2p

( j) (146)

A plane wave will have a propagation constant given by,

k=

k0 (147)

Even if= 0, for < p, k is purely imaginary and thus the waves are evanescent. Note that for > p and = 0, < 1 and tends to 1 as

p: Thephase velocityis greater than c.

vp k

=

k0=

c

> c (148)

This does not contradict the special theory of relativity, as information can only be conveyed bythe transport of wave energy, and the speed with which this travels is the group velocity.

vg k

(149)

However, k =

c and thus

1

vg =

1

c

d()

d . Substituting for from Eq. (146) and assuming forsimplicity that = 0, gives,

1

vg=

1

c

d

d

1 2p

2

1

vg=

1

c

d

d

(2 2p)

12

=

1

c

(2 2p)1

2

Therefore,

vg =c

1

2p

2

12

=c

< c (150)

as expected. Note that, as is the case with waveguides too,

vpvg = c

2

50

8/9/2019 AP Handout 2013

51/59

Figure 27: Radio ray trajectories in a planar slab ionosphere [5].

5.2.3 Planar slab ionosphere over a planar earth

(i) At p, 1 and the ionosphere behaves as if it were not present at all. Typically thisoccurs at

>

N e2

me0=

(5 1011 m3)(1.6 1019 C)2

(9.1 1031 kg)(8.854 1012 Fm1)

or, 2 >6.34MHz.

(ii) Normal incidence at < p: This is depicted in subfigure (a) above. As the electron densityvaries continuously with height z, there is a heightz0, at which(z0) = 0, and beyond this (z > z0),

< 0 making the wavenumber purely imaginary. In this case, the wave is evanescent and carriesno energy as E and H are /2 out of phase. Conservation of energy requires that at z = z0 thewave must undergo total internal reflection.

(iii) Oblique incidence at < p: Given that the refractive index n =

and using Snells law ofrefraction while treating the ionosphere as a homogenous stratified medium as shown in subfigure(c), we have,

(z1)sin 1=

(z2)sin 2 =

(z3)sin 3= . . ..

In the limit of continuously changing electron density/refactive index with height, the ray depictingthe radiowave will follow a smoothly curved path defined by,

(z)sin (z) = sin i, as shown in

subfigure (b). Total internal reflection will occur when(z) =/2, thus giving an implicit equationfor the true height of the ionosphere,

51

8/9/2019 AP Handout 2013

52/59

Figure 28: Radio path virtual height [5].

(zmax) = sin i (151)For a given value of the angle of incidence, i, a higher electron density (i.e. p) is needed as thefrequency increases in order to guarantee the existence of a point of total internal reflection. This,in turn, implies that for a given Nmaxthere exists a maximum usable frequency(MUF), definedby (again assuming = 0),

1

2c

2max= sin i

where c= Nmaxe2me0 is the critical frequency. Rearranging gives,

1 f2c

f2max= sin2 i

1 sin2 i = cos2 i = f2c

f2maxMUF = fmax= fcsec i (152)

(iv) Path heights: See Fig. 28

The true height of an ionospheric path is z0.

The virtual or effective height h is shown in the figure.

For normal incidenceonly,

the equivalent height, he is, he =z00

((,z)) dz

the phase height, Hp is, hp=z00 n(, z) dz

5.2.4 Planar ionosphere over a spherical earth

Fig. 31 depicts the geometry of half a skip distance, d2 , around a spherical earth, where ae =43 6, 400 km is the effective earth radius due to the standard atmosphere refractivity andh is the

52

8/9/2019 AP Handout 2013

53/59

Figure 29: Geometry of half a skip distance around a spherical earth [5].

effective height of the ionosphere.

Using the sine rule for the triangle shown, sinae+h = sini

ae.

Using the rule that the sum of the angles of a triangle is , we have = i. Thus,

sin( i)sin i

= ae+ h

aesin(+ i)

sin i= 1 +

h

aesin i cos + cos isin

sin i = 1 + h

ae1 +

h

ae cos

csc = cot i

But by definition of the angular unit of radian, = d/2ae = d2ae

, finally giving,

1 +

h

ae cos

d

2ae

csc

d

2ae

= cot i (153)

Eqs. (152) and (153) are used to establish the parameters for long-distance communication links.

Note that to calculate the skip distance d, we normally start from the latitude and longitude of thetwo endpoints, A and B of the link on the surface of the earth and compute using the fact thatin spherical polar coordinates, cos = rA.rB.

5.2.5 Example

A short-wave broadcasting station in Ohio, USA wants to establish a broadcasting service to centralEurope. Determine the radiation angle and the frequency for such a service.

Solution: For a single skip operation, d = 6, 760km.

Assuming = /2 we have = 6,720 km

2(436,400km) 0.396.

53

8/9/2019 AP Handout 2013

54/59

But (h+ ae)cos = ae giving h = aecos ae.h = (43 6, 400 km)

1cos(0.396)

1

= 715 km.

The reflection point lies above the ionosphere (even the night-time height of the peak of the F-layer)and, therefore, single-skip operation is not possible.

For double-skip operation, d = 3, 380 km. Following the same calculation with this value of deventually gives h 170 km. The F1 and F2 layers have effective (virtual) heights ranging from200 km to 400 km, which can be used provided that > /2 (i.e. the launching antenna is pointingabove the horizon).

Assuming thath 300 km and since d = 3, 380 km, Eq. (153) yields,

cot i =

1 +

300km43 6, 400km

cos

3, 400km

2 43 6, 400km

csc

3, 400km

2 43 6, 400km

Therefore, the angle of incidence on the ionosphere is i = 1.3c or 74.49.

The launch angle is then, l /2 = i /2.

l =

2

3, 400km

2 43 6, 400km

1.3c

= 0.072c

Therefore, the launch angle is 4.1 above the horizon.

The MUF is given by,

fmax = 1

2

(5 1011 m3)(1.6 1019 C)2

(9.1 1031 kg)(8.854 1012 Fm1) sec(1.3c)

= 6.34 MHz sec(1.3c)

= 23.7MHz

To account for electron density fluctuations, we usually take the MUF to be at least 15% less thanthe calculated value, i.e. in this case we use f

8/9/2019 AP Handout 2013

55/59

where B0 is the constant magnetic field of the earth, which we take arbitrarily to be z-directed,B0 = B0ez. Using j = eNv, we have,

(j + )j + cj ez =2p0E

where p is the plasma frequency given by Eq. (145) and c is the cyclotron frequency definedby,

c=eB0

me(154)

The equation of motion for the electrons can be written in an expanded, matrix form as,

j + c 0

c j + 00 0 j +

j= 0

2pE

wherej and E are (3 1) column vectors. Inverting the above matrix equation then gives, j = E,where is an effective conductivity tensor of rank 2, expressed in matrix form as,

=0

2p

2c 2 + 2 + 2j

j + c 0c j + 0

0 0 j +

Amperes law, then becomes,

H = E +j0E H = j0rE

where the tensor relative dielectric constant for the ionospheric plasma is,

r

=

1 j2 0j2 1 0

0 0 1

and

1= 1 2p(1 j /)

2 2c 2 2j & 2=

2p(c/)

2 2c 2 2j

When the collision frequency is zero, both 1 and2 are real. If2 =2c+

2, both 1 and 2 havelarge imaginary components leading to high attenuation at approximately 1.4 MHz.

We now consider the solution to Maxwells equations in such a plasma.

55

8/9/2019 AP Handout 2013

56/59

E = j0H E = j0 H E = j0 j0rE

2

E + (.E) = 2

00rE

But .E= /0 = 0 since in the ionosphere the electron charge density and the ionised gas moleculecharge density add up to exactly zero.

2E + k20rE= 0

We shall now consider the solution to the above equation in the special case of a uniform planewave solution propagating along the direction of the earths magnetic field, i.e. along thez-axis. Inthis case,

E=

ExEy

0

, E

x = 0,

E

y = 0

Thus,

2Exz2

+ k20(1Ex j2Ey) = 02Eyz2

+ k20(1Ey+ j2Ex) = 0 (155)

We seek trial solutions consistent with our assumptions:

Ex= Ae+jtjz & Ey =Be

+jtjz (156)

where a and B are amplitudes to be determined. Substituting Eq. (156) into (155) then yields,

2A+ k201A jk202B = 02B+ k201B+jk202A= 0(

2 + k201)A= (jk

202)B

(2 + k201)B = (jk202)AA

B = B

A A2 = B2

A= jB

When A = +jB , we then have,

(2A+ k201)A (jk202)(jA) = 02 =k20(1 2)= + = k01 2

56

8/9/2019 AP Handout 2013

57/59

Figure 30: Circularly polarised waves in the ionosphere [5].

Hence,

E= A (ex jey) ejtj+z (157)

Further assuming that = 0 gives 1, 2 and + real, thus,

(E) =A (cos(t +z)ex+ sin(t +z)ey) (158)

This is a circularly polarised wave of positive helicity (or right-hand circularly polarised (RCP)wave).

When A = jB , working in a similar way gives,

= = k0

1+ 2

E= A (ex+jey) ejtj+z (159)

(E) =A (cos(t +z)ex sin(t +z)ey) (160)

This is a circularly polarised wave of negative helicity (or left-hand circularly polarised (LCP) wave).In the special case of= 0,

+= k0

1

2p

( c) = k0

1 2p

(+ c)

For frequencies next the cyclotron frequency (true even if= 0)+ has a large imaginary part andthe positive helicity wave is heavily attenuated.

If we were to consider uniform plane wave propagation in a direction perpendicular to B0, say along

the x-axis, we also have two cases to consider:

57

8/9/2019 AP Handout 2013

58/59

Figure 31: The ordinary (a) and extraordinary (b) waves in the ionosphere[5].

The first case is when the electric field vector and the earths magnetic field vector are parallel: Inthis case the electron velocity is parallel to B0 so that the force term evB0 = 0 and the solutionis a (Ez) linearly polarised wave with x =

k0. This is called the ordinary wave.

The second case is when the incident E is perpendicular to B0, i.e. only Ey is non-zero. We omitthe analysis as it is similar to the z-directed propagation and state the final result, which gives only

one non-zero amplitude solution, namely, EyEx =j 12 with x =

21221

k0. This is an elliptically

polarised wave in the plane perpendicular to B0 (i.e the xy-plane) and is called the extraordinarywave.

5.2.7 Faraday rotation

Consider an ionospheric layer m thick and an incident plane wave,

E = 2E0exejk0z

= E0(ex+jey)ejk0z + E0(ex jey)ejk0z

The natural propagation modes into the ionosphere are circularly polarised, so on exiting the iono-sphere (neglecting any attenuation or reflections at the ionosphere-clear air interfaces), we have,

E = E0(ex+jey)ej + E0(ex jey)ej+

= 2E0ej (++)2

cos

( +)

2

ex+ sin

( +)

2

ey

This is a linearly polarised wave whose electric field vector has rotated by an angle = arctan EyEx ,i.e.,

58

8/9/2019 AP Handout 2013

59/59

= ( +) 2

References

[1] T.S.M. Maclean, Principles of antennas: wire and aperture, Cambridge (1986)2.22;[2] C.A. Balanis,Antenna Theory, 3rd Ed., Wiley-Interscience (2005)4.4.2;[3] C.A. Balanis,Antenna Theory, 3rd Ed., Wiley-Interscience (2005)8.5.2;[4] J.D. Kraus and R.J. Marhefka, Antennas, 3rd Ed. McGraw-Hill (2002) Chapters 13-14;

[5] R.E. Collin,Antennas and Radiowave Propagation, McGraw-Hill (1985);