Electromagnetic Waves

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11Electromagnetic Waves

Education should be exercise;it has become massage.

−− Martin H. Fischer

CHAPTER OBJECTIVES

To enable the students to understand the following:

• Uniform plane wave• Electromagnetic spectrum• Wave polarisation—linear, circular, and

elliptic polarisations

• Wave propagation in conducting mediaand lossy dielectrics

• Reflection and refraction of waves• Poynting Vector and the flow of power

11.1 Introduction

When we set up a communication link between a transmitter and receiver withoutthe help of cables we usually set up two antennas, a transmitting antenna and a re-ceiving antenna. The transmitting antenna transmits electromagnetic waves whichare essentially waves whose nature is that of light waves but at very lower frequency.These waves are generally sinusoidal, described by frequency and the velocity ofpropagation, which in turn depend on the medium in which the waves propagate.Thus if the waves propagate in a dielectric, then the medium is described by itspermittivity, ε = εrε0. If the medium is a conductor, then the conductivity σ comesinto play, and so on.

In this chapter we concentrate on the fundamental properties of electromagneticwaves as they propagate in various types of media, and their interaction with matter.

11.2 Uniform Plane Wave

In the very beginning we consider the simplest of waves, namely, the uniformplane wave. Let a sinusoidally time-varying wave travel in the +z direction butwith no variation in either x or y directions. Figure 11.1 depicts how the wave

COPYRIGHTED MATERIAL

OXFORD UNIVERSITY PRESS

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342 Fundamentals of Engineering Electromagnetics

Direction of propagation

H

E

x

y

z

Plane perpendicular to propagation

Fig. 11.1 The geometry of the uniform plane wave

travels. The shaded region shows a single phase front, perpendicular to the directionof propagation.

The procedure outlined below is standard for the analysis of most wavephenomena.

• Since we are considering a wave, the electric and magnetic fields must satisfythe wave equation.

• Since the nature of the wave is oscillatory, having a exp( jωt) time dependence, thefields must satisfy Helmholtz’s equation (see Section 10.9 for a better understand-ing of this point). The Helmholtz’s equation for the electric field is Eqn (11.72)

∇2E + k2E = 0

where k = ω√

με

• If we recall the results of the last chapter, a travelling wave must have a functionaldependence,

E(R, t) = �{E0ej(ωt−βz)} (11.1)

where β is the propagation constant of the wave and ω is the frequency of oscil-lation. Here β may or may not be equal to k . In the case of a uniform plane wave,E0 is a constant but real phasor,

E0 = axEx0 + ayEy0 + azEz0 (Ex0, Ey0, Ez0 are real and constant) (11.2)

Let us first satisfy the Helmholtz equation

∇2E = −k2E

E0∇2ej(ωt−βz) ?= −k2E0ej(ωt−βz)

E0ejωt ∂2

∂z2e−jβz ?= −k2E0ej(ωt−βz)

E0ejωt(−β2 e−jβz)?= −k2E0ej(ωt−βz)

−β2 E = −k2E

which is satisfied if β = k . The term exp{j(ω t − kz) } ensures that the wave is atravelling wave, travelling in the +z direction. The phase of the wave at time t0 isgiven by

ω t0 − kz = constant = K



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Electromagnetic Waves 343

which is the equation of a plane in three dimensions and which is parallel to thez = 0 plane.

z = (ω t0 − K)/k = another constant = K1

Now working only with the phasor

E = E0e−jkz (11.3)

The sinusoidal fields in time are

�(Eejωt) = �(E0e−jkzejωt)

= E0 cos(ω t − kz) (11.4)

The term E0 can be pulled out of the bracket since it is real and constant. If thiselectric field is to represent an actual wave, it must satisfy Maxwell’s equations. Thefirst Maxwell equation which has to be satisfied is

∇ · E = ρv

ε

But we know that the wave is travelling in a region free of charges. So ρv mustbe zero.

∇ · E = 0

∂

∂x(Ex0e−jkz) + ∂

∂y(Ey0e−jkz) + ∂

∂z(Ez0e−jkz) = 0

0 + 0 − jkEz0 = 0

Ez0 = 0 (11.5)

In other words there is no electric field in the direction of propagation; there areonly transverse components.

Hence E = axEx0e−jkz + ayEy0e−jkz (11.6)

The second Maxwell equation which the electric and magnetic fields have tosatisfy is

∇ × E = −jω μH (11.7)

which is Eqn (11.65). Writing out the various components starting with thex-component

(∇ × E)x = −jω μ Hx

∂Ez

∂y− ∂Ey

∂z= −jω μ Hx

0 + jkEy0e−jkz = −jω μ Hx

Similarly, we can get the other components

(∇ × E)y = ∂Ex

∂z− ∂Ez

∂x= −jkEx0e−jβz = −jω μ Hy

(∇ × E)z = ∂Ey

∂x− ∂Ex

∂y= 0 = −jω μ Hz

From the last equation, we get Hz = 0 or the magnetic field is also transverse tothe direction of propagation. Figure 11.1 clearly shows how both the electric and



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magnetic fields lie on z = constant planes. From the above equations, the magneticfield components are

−jω μ Hx = jβ Ey0e−jkz

Hx = jω√

με

−jω μEy0

= −√

ε

μEy0 (11.8)

and − jω μ Hy = −jβ Ex0e−jkz

Hy = −jω√

με

−jω μEx0

=√

ε

μEx0 (11.9)

Concentrating on the term√

ε /μ, we calculate its units

Units of

{√ε

μ

}=

√F/m

H/m

=√

F

H

=√

� · s

� ·s=

√�2

= � (11.10)

Since√

ε /μ has the units of �, we represent√

μ /ε by Z , a resistance

Hx = − 1

ZEy (11.11)

Hy = 1

ZEx (11.12)

Z =√

μ

ε(11.13)

Z is called the characteristic or intrinsic impedance of the medium. For air orvacuum, a quick calculation of the characteristic impedance, Z ≡ Z0 gives

Z0 =√

μ0

ε0≈ 377 � (11.14)

Since μ0 = 4π×10−7 and ε0 = (1/36π) ×10−9;therefore

√μ0 /ε0 = √

4 × 36π2 × 100 = 120π= 377 �. The magnitude of themagnetic field is

H =√

H 2x + H 2

y = 1

Z

√E2

y + E2x

= E

Z(11.15)

Taking a different tack, we calculate E × H for the uniform plane wave.

E × H = ax(EyHz − EzHy) + ay(EzHx − HzEx) + az(ExHy − EyHx)



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= az(ExHy − EyHx) (Ez and Hz are both zero)

= az

[Ex

(Ex

Z

)− Ey

(−Ey

Z

)]( From the previous set of equations)

= az1

Z

(E2

x + E2y

)V2/( � m2) ≡ W/m2

|E × H| = 1

Z

(E2

x + E2y

)= 1

Z|E|2 (11.16)

It is important to note that the direction of E × H is the direction of propagation.To compute the angle θ , between E and H we must compute the dot product

between the two. Since

E · H = EH cos θ

Using Eqns (11.11) and (11.12)

E · H = (axEx0e−jkz + ayEy0e−jkz) · (axHx0e−jkz + ayHy0e−jkz)

= e−j2kz

Z(axEx0 + ayEy0) · (−axEy0 + ayEx0)

= e−j2kz

Z(−Ex0Ey0 + Ey0Ex0)

= 0 (11.17)

Now E �= 0 and H �= 0, therefore cos θ must be zero. In other words, the angle fromE to H is π/2. It is an important conclusion:

See MagPoyntVect.m in Chapter 11

Not only are the electric and magnetic field perpendicular to the direction ofpropagation, but they are perpendicular to each other; the vector E × H isdirected towards the propagation direction.

Figure 11.2 depicts the advance of the plane wave where the electric field is drawnas the solid line while the magnetic field as the dotted line. Notice that the magneticfield is always in phase with the electric field; that is when the electric field increases,the magnetic field increases, and when the electric field decreases the magnetic field

0

−1

−10

Direction of propagationE fieldH field

1EmaxHmax1

Fig. 11.2 Electric and magnetic fields of a uniform plane wave



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does the same, keeping the ratio of the electric to the magnetic field constant (= Z ,the characteristic impedance).

Example 11.1 For a plane wave travelling in air, if ω = 2π×109 rad/s, find thepropagation constant k .Solution The propagation constant k is given by

k = 2π

λ

where λ is the wavelength of the wave. Also

f λ = c

where f = ω /(2π) is the frequency of the wave in Hz and c is the velocity of thewave in air (= 3 × 108 m/s). So

k = 2π f

c

= ω

c

= 2π×109

3 × 108

= 20π

3m−1

Example 11.2 If the electric field E = (10az + 20ay) cos(2π×107t − kx), showthat this is the electric field of a wave.Solution If we compare the functional dependance of the electric field withEqn (11.51), we can identify that

ω = 2π×107 r/s

and k as the propagation constant. Therefore, this is the electric field of a plane wave.

Example 11.3 Express E = (10az + 20ay) cos(2π×107t − kx) as a phasor.ω = 2π×107.Solution E = �{(10az + 20ay) ej(2π·107t− kx)}

= (10az + 20ay) �{ej(2π·107t− kx)}so the phasor is

E = (10az + 20ay) e−jkx

Example 11.4 Find the H field of an plane wave whose E vector is given by

E = (10az + 20ay) cos(2π ·107t − kx)

in air. Find the value of k and the wavelength λ.Solution From inspection of the E field it is clear that ω = 2π×107t( = 2π f )

where f is the frequency of the wave. So f = 107 Hz. Since the velocityc =3 × 108 m/s the wavelength is given by

λ = c/f = 3 × 108/107 = 30 m

from λ, we can calculate k by

k = 2π/λ = 0.20944 rad/m



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Again from inspection it is clear that the wave is travelling in the +x direction. So

H = 1

Z0ax × E = 1

377(−10ay + 20az) cos(2π×107t − kx) A/m

See PhaseConstant.m in Chapter 11

Example 11.5 A plane wave caries a power density of 1 MW/m2. Find themagnitude of the electric and magnetic field vectors.Solution The power density S is given by

S = E2/Z0 = 106 W/m2

so E =√

106 × 377 = 1.9416 × 104 V/m

and H = E/Z0 = 51.501 A/m

11.3 Electromagnetic Spectrum

Electromagnetic waves in actual use for different purposes are characterised by theirfrequency f and wavelength, λ. These bands of frequencies have been given namesby engineers, and there is an agreement all over the world about their use. Figure 11.3shows the frequency bands, their names and their uses.

In the figure, the frequency spectrum is depicted on a logarithmic scale from3 Hz to 3 PHz. (Read as petahertz). At the top the approximate analytical techniqueis outlined: for example the lumped element approach is used from DC to a few

Visible and UV

3 H

z

30 H

z

300

Hz

3 K

Hz

30 K

Hz

300

KH

z

3 M

Hz

30 M

Hz

300

MH

z

3 G

Hz

30 G

Hz

300

GH

z

3 T

Hz

30 T

Hz

300

TH

z

3 PH

z

I.R.Micro WavesRadio Waves

Lumped Distributed Optical

Voi

ce f

requ

ency

AM

Rad

io

Shor

twar

e R

adio

FM R

adio

Tele

visi

onM

obile

Sate

llite

3–30

Hz

– E

LF

30–3

00 H

z –

SLF

0.3–

3 K

Hz

– U

LF

0.3–

3 M

Hz

– M

F

3–30

KH

z –

VL

F

3–30

MH

z –

MF

30–3

00 K

Hz

– L

F

0.3–

3 G

Hz

– U

HF

> 3

GH

z –

EH

F

30–3

00 M

Hz

– V

HF

Fig. 11.3 The electromagnetic frequency spectrum



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Table 11.1 IEEE microwave band designations

Band Frequency

L 1–2 GHzS 2–4 GHzC 4–8 GHzX 8–12 GHzKu 12–18 GHzK 18–27 GHzKa 27–40 GHzV 40–75 GHzW 75–110 GHz

mm 110–300 GHz

hundred megahertz; the distributed element approach is used from a few hundredmegahertz to about a few hundred gigahertz, and so on. Below the logarithmic spec-trum is the rough demarcation of the frequency bands: radio waves, microwaves,infrared, visible, and beyond. At the bottom of the figure, the designated frequencybands are named: VLF, very low frequency; LF, low frequency; MF, medium fre-quency; HF, high frequency; VHF, very high frequency; UHF, ultra high frequencyand EHF extremely high frequency and so on.

Apart from the radio frequency bands there is yet another set of designationswhich is primarily due to the US military, but now adopted worldwide. These bandsare shown in Table 11.1.

11.4 Wave Polarisation

The electric field in the case just discussed always moves on a plane as the field ad-vances in the direction of propagation. Examination of Fig. 11.2 illustrates this point.The figure shows the electric field in the x-direction while the magnetic field movesin phase in the y direction, and the direction of propagation is in the az direction:

axEx × ayHy = az|ExHy| (11.18)

11.4.1 Circular PolarisationLet us examine the solution to the wave equation, Eqn (11.6) under different casesand conditions. So far we have stipulated that the amplitudes are real and we foundthat the electric field moved along a plane as the wave advanced.

Let us now consider the case when one amplitude of the set [Ex, Ey] is complex,and the other one real. Taking a specific case

Ex0 = A0ej0 (11.19)

Ey0 = A0ej(π/2) (11.20)

E0 = axEx0 + ayEy0 (11.21)

where A0 is of course real. The sinusoidal fields in time are

E(R, t) = �{E0ej(ωt−kz)}= �{(axEx0 + ayEy0) ej(ωt−kz)}



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= �{[axA0 + ayA0ej(π/2)]ej(ωt−kz)}= A0�{axej(ωt−kz) + ayej(π/2)ej(ωt−kz)}= A0�{axej(ωt−kz) + ayej(ωt−kz+π/2)}= A0[ax cos(ω t − kz) −ay sin(ω t − kz) ] (11.22)

The magnitude of the field in time is

|E(R, t) | =√

A20[cos2(ω t − kz) + sin2(ω t − kz) ]

= A0 (11.23)

which is constant. How do we interpret this result? Let us analyse this situation ingreater detail.

The original electric field in time is

E = axEx + ayEy

= A0[ax cos(ω t − kz) −ay sin(ω t − kz) ] (11.24)

We get rid of the z coordinate by setting it to zero. By setting z to zero we areanalysing the behaviour of the electric field on the plane z = 0.

E( x, y, z = 0, t) = A0[ax cos(ω t) −ay sin(ω t) ] (11.25)

This seems simpler. Let us now proceed to calculate the electric field E at differenttimes: at t = 0, t = T /4, t = T /2, and t = 3T /4, where T is the time period ofone cycle. Since ω = 2π /T , ω t = 0, π /2, π , and 3π /4 at these times.

E = A0ax (At t = 0)

E = −A0ay (At t = T /4)

E = −A0ax (At t = T /2)

E = A0ay (At t = 3T /4)

t = 0 w t = 0

t = T/4w t = p /2

t = T/2w t = p

t = 3T/4 w t = 3p /2

t = T w t = 2pEx

E(t = 3T/4)

E(t = T/4)E(t = T/2) E(t = 0)

Ey

Fig. 11.4 Figure illustrating left circularpolarisation

We notice that the electric field vector ro-tates in a direction of the fingers of theleft hand when made into a fist and thethumb points in the direction of propaga-tion in accordance left hand thumb rule.Figure 11.4 illustrates this point.

The next Fig. 11.5 shows how the waveadvances helically in a polarised wave.The figure shows the wave at time t = T .The tip of the electric field vector at t = 0is at point a; at t = T /4 it is at b. In this wayit traverses a full circle in time T , goingthrough points c and d. In time T differ-ent parts of the wave advances by varyingamounts. The electric field at point a advances to a′ which is one wavelengthaway. The electric field at point b advances to b′ three quarters of a wavelengthaway, and so on. The electric field vectors all lie parallel to the x-y plane but alltheir tips lie on a helix. On the x-y plane, the electric field traces a clockwise cir-cle. This type of polarisation is called left circular polarisation (LCP) because as



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x−1

y d

c

c�

d�

d

b

b�

Direction of propagationE field on the xy plane

E field helix

a

l

a�

z

A0

A0o

Fig. 11.5 The advance of the wave in a left circular polarisation (LCP) plane wave

the left hand is held in the form of a fist, with the thumb extended, the fingers curl inthe direction of motion of the electric field while the thumb points in the directionof the propagation of the wave.

Consider the case of a right hand circularly polarised wave. For such a wave, ad-vancing in the +z direction, the electric field rotates in the anti-clockwise directionon any plane parallel to the x-y plane and the wave travels in the direction of thethumb. The electric field then can be written as

Ex0 = A0ej0

Ey0 = A0e−j(π/2)

E0 = axEx0 + ayEy0

E = E0e−jβz (11.26)

The sinusoidal time-dependent field is

E = �{Eejωt}= �{A0[axej0 + aye−j(π/2)]ej(ωt−kz)}= A0

[ax cos(ω t − kz) + ay cos

(ω t − kz − π

2

)]= A0[ax cos(ω t − kz) + ay sin(ω t − kz) ] (11.27)

choosing z = 0 as the plane on which the electric field is evaluated

E(z = 0, t) = A0[ax cos(ω t) + ay sin(ω t) ] (11.28)

This is a vector which is rotating in the counter-clockwise direction and obeys theright-hand thumb rule.

Example 11.6 If Ex = 5∠45◦ and Ey = 5∠45◦, find the type of polarisation.Solution If Ex = 5∠45◦ and Ey = 5∠45◦ then

Ex = 5 cos(ω t + 45◦)Ey = 5 cos(ω t + 45◦)

and Ey = 1 × Ex

which is the equation of a straight line. So the wave is linearly polarised.



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Example 11.7 If Ex = 5∠45◦ and Ey = 15∠45◦ find the type of polarisation.

Solution If Ex = 5∠45◦ and Ey = 15∠45◦ then

Ex = 5 cos(ω t + 45◦)Ey = 15 cos(ω t + 45◦)

and Ey = 3 × Ex

which is the equation of a straight line. So the wave is linearly polarised.

Example 11.8 If Ex = 5∠0◦ and Ey = 5∠90◦, find the type of polarisation.

Solution If Ex = 5∠0◦ and Ey = 5∠90◦ then

Ex = 5 cos(ω t)Ey = 5 cos(ω t + 90◦)

= −5 sin ω tand E2

y + E2x = 50

which is the equation of a circle. So the wave is circularly polarised.

11.4.2 Elliptical PolarisationWe have studied two types of polarisations—linear and circular polarisation. Let ustake a look at the most general type of polarisation, namely elliptical polarisation.Let

E = axEx + ayEy

= axEx0 cos(ω t − kz) + ayEy0 cos(ω t − kz+θ ) (11.29)

where Ex0 and Ey0 are the wave amplitudes in the x and y directions and θ is the phaseangle with which Ey leads Ex. Choosing the plane z = 0 to evaluate the polarisation

E( z = 0) = axEx0 cos(ω t) + ayEy0 cos(ω t+ θ )

= axEx0 cos ω t + ayEy0(cos ω t cos θ − sin ω t sin θ ) (11.30)

cos ω t = Ex

Ex0(11.31)

sin ω t =√√√√1 −

(Ex

Ex0

)2

(11.32)

Ey

Ey0= cos ω t cos θ − sin ω t sin θ

= Ex

Ex0cos θ −

√√√√1 −(

Ex

Ex0

)2

sin θ (11.33)

or

(Ey

Ey0

)2

+(

Ex

Ex0cos θ

)2

− 2

(Ey

Ey0

)(Ex

Ex0cos θ

)=

⎡⎣1 −

(Ex

Ex0

)2⎤⎦ sin2θ

(11.34)



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352 Fundamentals of Engineering Electromagnetics(Ey

Ey0 sin θ

)2

+(

Ex

Ex0 sin θ

)2

− 2

(Ey

Ey0

)(Ex

Ex0

)cos θ

sin2θ= 1 (11.35)

Let us discuss these equations.Case 1. θ = 0 or π ; Ex0 and Ey0 can be anything. This case gives us linear polari-sation.

To illustrate this statement we use Eqn (11.34) since Eqn (11.35) is unsuitable.Letting sin θ = 0, we have(

Ey

Ey0

)2

+(

Ex

Ex0cos θ

)2

− 2

(Ey

Ey0

)(Ex

Ex0cos θ

)= 0

(Ey

Ey0− Ex

Ex0cos θ

)2

= 0

Ey

Ey0= Ex

Ex0cos θ

Ey =

⎧⎪⎪⎨⎪⎪⎩

Ey0

Ex0Ex for θ = 0

−Ey0

Ex0Ex for θ = π

(11.36)

which is a straight line between the variables Ey and Ex. This is shown in Figs 11.6(a) and (b).Case 2. θ = ±π /2, Ex0 and Ey0 can be anything. This gives elliptical polarisationwith the major and minor axes of the ellipse along the coordinate axes. We use

(a) x=cos(t), y=2∗cos(t) (b) x=cos(t), y=2∗cos(t + π)

(c) x=cos(t), y=2∗cos(t + π/2) (d) x=2∗cos(t), y=cos(t + π/2)

0−1

−1

0

1

2

−2−2

1 2

0−1

−1

0

1

2

−2−2

1 2

0−1

−1

0

1

2

−2−2

1 2

0−1

−1

0

1

2

−2−2

1 2

Fig. 11.6 Linear and elliptical polarisations



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Eqn (11.35) with cos θ = 0 and sin θ = ±1(Ey

Ey0

)2

+(

Ex

Ex0

)2

= 1 (11.37)

0Ex

Ey

−0.5−1−1.5−2−2

−1.5

−1

−0.5

0

0.5

1

1.5

21.5∗cos(t), 2∗cos(t + p /6)

0.5 1 1.5 2

Fig. 11.7 The case for θ = π /6,

Ey0 = 2 and Ex0 = 1.5

which is the equation of an ellipse with majorand minor axis along the x and y directions,respectively. This is shown in Figs 11.6(c)and (d). If Ey0 = Ex0 then we obviously getcircular polarisation.Case 3. θ , Ey0, Ex0 all of which can be any-thing. In this general case, we have ellipticalpolarisation, with the major axis of the el-lipse tilted in any general direction. The casefor θ = π/6, Ey0 = 2 and Ex0 = 1.5 is shownin Fig. 11.7.Example 11.9 If Ex = 5∠15◦ and Ey =7∠90◦, find the type of polarisation.

Solution We first remove the excess phaseby 15◦. Ex = 5∠0◦ and Ey = 7∠75◦. Then

Ex = 5 cos ω t

Ex

5= cos ω t

Ey = 7 cos(ω t + 75◦)Ey

7= cos ω t cos 75◦ − sin ω t sin 75◦

= 0.2588 cos ω t − 6.7615 sin ω t

= 0.2588

(Ex

5

)− 0.9659

√√√√1 −(

Ex

5

)2

from the previous theory we realise that this is elliptical polarisation.

11.5 Wave Propagation in Conducting Media

In conducting media, as the wave propagates, currents are set up. These currentscirculate in the medium and produce heat. Thus, energy carried by the wave mustnecessarily get diminished. We already know that where an electric field is presentin a medium with conductivity σ , there the current density generated at a point inthe medium is related to the electric field through the equation

J = σ E

Using Maxwell’s Eqn (11.67), in phasor form

∇ × H = jω ε E + J

= jω ε E + σ E (11.38)



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�


= jω

(ε − jσ

ω

)E (11.39)

= jω ε

(1 − jσ

ω ε

)E (11.40)

= jω εC E (11.41)

where εC is the complex dielectric constant which is so called because of the aboveset of equations, where εC is introduced to simplify things. The conducting mediumcan be modelled as a dielectric but one with a complex permittivity.

εC = ε′ −jε′′ (11.42)

= ε −jσ

ω(11.43)

The ratio σ /(ω ε ) is dimensionless (the same as εr). This ratio is called the dissi-pation factor, D

D = σ

ωε

{1, for a good dielectric

�1, for a good conductor(11.44)

Generally, the dissipation factor for lossy dielectrics is also called its loss tangent.

D = σ

ωε= ε′′

ε′ (11.45)

D = tan q

s /w or e ″

q

e or e ′

Fig. 11.8 Loss tangent fora dielectric

Why it is so called may be seen from examiningFig. 11.8.

Using the notation ofβC as the complex propagationconstant

βC = βR + jβI = β + jα

= ω√

μεC

To compute βC

β2C = (β + jα )2

= β2 − α2 + j2β α

ω2 μεC = β2 − α2 + j2β α

ω2 μ(ε −j

σ

ω

)= β2 − α2 + j2β α

Equating the real and imaginary parts, and solving the resulting equations, we get

β = βR =ω

√με

√√1 + (

σωε

)2 + 1√

2(11.46)

α = βI =ω

√με

√√1 + (

σωε

)2 − 1√

2(11.47)

What is the meaning of a complex propagation constant βC? Let us examine a wavetravelling in the z-direction, with the electric field in the x-direction. The wave incomplex notation is then

Ex = Ex0e−jβC z



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“Chapter11” — 2012/5/5 — 17:35 — pg. no. 355 — #15�

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= Ex0e−j(β−jα)z

= Ex0e− αze−jβz (11.48)

Using sines and cosines, the wave in the real world is

Ex = �[Ex0e−αze−jβze jωt]= Ex0�[e−αzej(ωt−βz)]= Ex0e−αz cos(ω t−β z) (11.49)

Observing the previous equation we can see that as the wave progresses in thez direction, the electric field is still oscillating (the cosine factor), but its amplitudedecays (the exponential factor). This is shown in Fig. 11.9. The figure shows a snap-shot at some time instant of a wave incident from air into a conductive medium. Asthe wave progresses into the medium, the following happens

1. The wavelength of the wave decreases. This is so because

βCond Med = 2π

λC=

ω√

με

√√1 + (

σωε

)2 + 1√

2> ω

√με

(= 2π

λ

)where λC is the wavelength in the conducting medium.

2. The wave amplitude decays as exp(−α z). The term exp(−α z) forms an enve-lope of the cosine term.

See AttenuationConstant.m and DissipationFactor.min Chapter 11

Concentrating on the decaying term, the amplitude of the wave falls by Ex0 exp(−1)

in a distance given by

zskin depth = δ = 1/α

Ex0e−δα = Ex0e−1

= 0.3679Ex0

δ is called the skin depth and depending on the value of α, the skin depth varies. Letus obtain the value of the magnetic field. Using the notation ∂x ≡ ∂ /∂x etc.,

H = − 1

jω μ∇ × E

= − 1

jωμ[az(∂x Ey−∂y Ex) + ax(∂y Ez−∂z Ey) + ay(∂z Ex−∂x Ez) ]

Conducting mediumAir

Envelope

Skin depth

Fig. 11.9 Profile of a wave propagating from air to a conducting medium



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“Chapter11” — 2012/5/5 — 17:35 — pg. no. 356 — #16�

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= − 1

jω μ

⎡⎢⎣az (∂x Ey−∂y Ex)︸︷︷︸

∂x=0, ∂y=0

+ ax (∂y Ez−∂z Ey)︸︷︷︸∂y=0, Ey=0

+ ay (∂z Ex−∂x Ez)︸︷︷︸∂z=−jβC , Ez=0

⎤⎥⎦

In the first factor the E-field is not a function of either x or y. The same argumentapplies to the next factor taking into account the additional fact that Ey = 0. In thelast bracket, differentiation with respect to z is the same as multiplication by −jβC .Therefore,

H = − 1

jω μ[ay(−jβC Ex) ]

= ayβC

ω μEx

= ay

√εC

μEx (11.50)

which means that the characteristic impedance is complex

ZC =√

μ

εC=

√μ

ε′ −jε′′ =√

μ

ε −j(σ /ω )

The meaning of a complex characteristic impedance is that the electric and magneticfields are not in phase with each other. Thus, in a propagating wave, propagating inthe n direction if the electric field is E then the magnetic field is (1/Z) n × E. Letus apply what we have learnt to different materials.

Example 11.10 A wave of frequency f = 1 MHz travels in a material withσ = 2 × 10−5 �/m and εr = 15. Compute the complex propagation constant.Solution For the material

ε = ε0 εr

= [15/(36π)] × 10−12

= [1/(24π)] × 10−11

and εC = ε

(1 − jσ

ω ε

)

=(

1 − 24j

24π

)× 10−11

The formula for β is

βC = ω√

μ0 εC

=√

1 − 24 j π

500√

6with βR = 0.009073

βI = −0.008703

11.5.1 Low Conductivity MaterialsLow conductivity materials are generally lossy dielectrics with a small dissipationfactor. Consider such a dielectric. In the following equations, we have used the



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“Chapter11” — 2012/5/5 — 17:35 — pg. no. 357 — #17�

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Taylor series expansions, where x 1√

1 + x � 1 + x

2(11.51)

1

1 + x� 1 − x (11.52)

Reproducing Eqns (11.46) and (11.47) with

σ

ωε= ε′

ε′′ 1

β =ω

√με

√√1 + (

σωε

)2 + 1√

2

�

ω√

με

√[1 + 1

2

(σωε

)2]

+ 1√

2(Taylor’s expansion of sq. root)

=ω√

με

√1 + 1

4

( σ

ω ε

)2

�ω√

με

[1 + 1

8

( σ

ω ε

)2]

(Taylor’s expansion of sq. root) (11.53)

α =ω

√με

√√1 + (

σωε

)2 − 1√

2

�

ω√

με

√[1 + 1

2

(σωε

)2]

− 1√

2

=(

ω√

με

2

)( σ

ω ε

)(11.54)

The characteristic impedance

Z =√

μ

εC

=√

μ

ε× 1√

1 − j(σ /ω ε )

=√

μ

ε

(1 + j

σ

2ω ε

)(applying the Taylor series expansion)

Example 11.11 Apply these results to a case of a wave travelling in a dielectricwith a loss tangent of 0.05 and a dielectric constant of 2. Let the frequency of thewave be 10 GHz. The loss tangent is 0.05 1.

Solution The permeability μ ≡ μ0; the permittivity ε = εr ε0 = 2ε0.

ω = 2π f = 2π×106 rad/s

also (see Fig. 11.8)σ

ω ε= 0.05 1



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“Chapter11” — 2012/5/5 — 17:35 — pg. no. 358 — #18�

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β0 =ω√

μ0 ε = 298.3412π

β = β0

[1 + 1

8

( σ

ω ε

)2]

= 298.4345π

α �β0

2

( σ

ω ε

)= 7.4585π

δ = 1/α = 0.04268 m ( Skin depth)√μ0

ε= 266.6 �

The characteristic impedance

Z =√

μ0

εC

=√

μ0

ε× 1√

1 − j(σ /ω ε )

�

√μ0

ε

(1 + j

σ

2ω ε

)= 266.6(1 + j0.0025)

= 266.6∠0.14◦

The magnetic field may be obtained from

H = 1

Z(n × E) =

√εC

μ0(n × E)

11.5.2 High Conductivity MaterialsFor the case of high conductivity materials like metals,

σ

ω ε� 1

and β =ω

√με

√√1 + (

σωε

)2 + 1√

2

�

ω√

με

√(σωε

) + 1√

2

(since

σ

ω ε� 1

)

�

ω√

με

√(σωε

)√

2

(since

σ

ω ε� 1

)�

√ω μ σ

2(11.55)

and α =ω

√με

√√1 + (

σωε

)2 − 1√

2

�

ω√

με

√(σωε

) − 1√

2



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“Chapter11” — 2012/5/5 — 17:35 — pg. no. 359 — #19�

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�

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�

ω√

με

√(σωε

)√

2

=√

ω μ σ

2(11.56)

Therefore, for high conductivity materials

α = β =√

ωμσ

2(11.57)

The characteristic impedance is given by

Z =√

μ0

εC

=√

μ0

ε0 − j(σ /ω )

=√

μ0

ε0

√√√√ 1

1 − j(

σωε0

)� Z0

√j

√ω ε0

σ

= Z0

(1 + j√

2

)√ω ε0

σ

The characteristic impedance of a high conductivity material is therefore

Z � Z0

(1 + j√

2

)√ω ε0

σ�

where Z0 is the characteristic impedance of free space (� 377 �).

Example 11.12 Taking the example of copper with a conductivity σ = 5.814 ×107 �/m; ε ≡ ε0 at a frequency of f = 1 MHz (ω = 6.28 × 106 rad/s), find β, thepropagation constant; δ, the skin depth and λ, the wavelength.Solution

σ /(ω ε0 ) = 1.045 × 1012 � 1

and so β = α �

√ωμσ

2= 15150

The skin depth

δ = 1/α = 6.6 × 10−5 m (11.58)

The wave decays to e−1 = 0.369 in a distance of 0.066 mm. The wavelength in air is

λair = c

f= 3 × 108

1 × 106= 300 m

the wavelength in copper at the same frequency is

λcopper = 2π

β

= 2π

15150= 4.15 × 10−4 m



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“Chapter11” — 2012/5/5 — 17:35 — pg. no. 360 — #20�

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Skin depth for copper as a function of frequency0.1

0.01

0.001

0.0001

1 10 100 1000 10000Frequency (Hz)

Skin

dep

th (

m)

100000 1e+06 1e+071e–05

Fig. 11.10 Skin depth for copper as a function of frequency

Example 11.13 Find the skin depth as a function of frequency for copper.

Solution The skin depth by Eqn (11.58) is given by

δ = 1/α =√

2

ω μ0 σ

A typical example of the skin depth as a function of frequency for copper is shownin Fig. 11.10. As we can see from the figure, the skin depth rapidly decreases withincreasing frequency, and at dc (zero frequency) the skin depth becomes infinity. Onthe other hand as the frequency is increased into the GHz range, the skin depth goesinto the μm range. It is for this reason that in many microwave components, theconducting surface is coated with a very thin layer of gold (in which the wavepenetrates) to minimise the energy dissipated. In high conductivity materials, thewavelength and skin depth are related by

λmaterial = 2πδmaterial (11.59)

The characteristic impedance for copper is given by

Z � Z0

(1 + j√

2

)√ω ε0

σ

= 377 × 1∠45◦ × 9.57 × 10−13

= 3.61 × 10−10∠45◦

11.6 Boundary Conditions

When we consider electromagnetic problems we need to look at the conditions ofthe electromagnetic fields at the boundary between two media. Principally whatwe are interested in is: what are the relations between the electromagnetic fieldcomponents in the two media which are (a) tangential to the surface separating thetwo media? and (b) normal to this surface?



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“Chapter11” — 2012/5/5 — 17:35 — pg. no. 361 — #21�

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nt1

(a) (b) (c)

e2, m2, s2

e1, m1, s1

R

E and H fields

e2, m2, s2e1, m1, s1

MediumBoundary

xy

ze2, m2, s2

e1, m1, s1

t2n

ad

cb

MediumBoundary

x

yz

Fig. 11.11 The behaviour of electromagnetic fields near a boundary consistingof a change of medium

There is one special case which we always have to keep in mind: the case of aperfect electric conductor, the conductivity, σ → ∞. In this case, no time vary-ing fields can exist inside a conductor: E and H will both be zero. In this casethere will be both a surface charge, ρs, as well as a surface current, Js, at theboundary of the two media.

To investigate these problems we take a look at the case where electromagnetic fieldsare present inside and outside a body with permittivity and permeability (ε2 , μ2 , σ2 ).This we call region 2. The body is immersed in a region with permittivity and per-meability (ε1 , μ1 , σ1 ), called region 1. This configuration is shown in Fig. 11.11. Inthe (a) part of the figure the macro level diagram is shown where a small regionlabelled R is shown. This region is shown blown up in parts (b) and (c) of the figure.

The fields in regions 1 and 2 must satisfy Maxwell’s equations.

∇ · D =ρv

∇ × E = −∂B

∂t∇ · B = 0

∇ × H = ∂D

∂t+ J (11.60)

These are the differential form of the equations. The integral forms are∫∫∫∇ · DdV = ©

∫∫D · dS =

∫∫∫ρv dV∫∫

∇ × E · dS =∮

ˆ E · dl = −∫∫

∂B

∂t· dS∫∫∫

∇ · BdV = ©∫∫

B · dS = 0∫∫∇ × H · dS =

∮ˆ H · dl =

∫∫∂D

∂t· dS +

∫∫J · dS (11.61)

We first take a look at the first of the above equations (Maxwell’s equations in integralform). The equation is applied to the pillbox shown in part (c) of the figure. The



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“Chapter11” — 2012/5/5 — 17:35 — pg. no. 362 — #22�

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mathematical description of the pillbox is

Height of pillbox � 0

Area of the top and bottom �� A

Applying Gauss’s law to this pill-box with the knowledge that there may or may notbe accumulated charge on the interface. There is no surface charge (ρs = 0) if σ2

is finite. If we have a perfect conductor then σ2 → ∞ then there will be a surfacecharge, ρs. Writing Gauss’s law in the most general case (i.e., if a surface chargeexists)

D1 · az� A + D2 · (−az) �A = ρs �A (11.62)

Dz1 − Dz2 = ρs (11.63)

in other words, the normal component of the flux density is discontinuous across adielectric–dielectric boundary by the amount of the surface charge density. Writingthe above equation in a more general form

Dn1 − Dn2 = ρs

n · (D1 − D2) = ρs (11.64)

If there is no surface charge (ρs = 0), then the right-hand side is replaced by zeroand in that case, the normal component of the D field is continuous.

Let us take the case where the second medium is a perfect electric conductor.Then there will be no fields there and Dn2 = 0 but there will be a surface charge,ρs. So

Dn1 = ρs (11.65)

Now considering the the second of the Maxwell’s equations in integral form (thesecond of the Eqn set (11.61)) to part (b) of Fig. 11.11. In this case

Length (b − c) = Length (d − a) � 0

Length (a − b) = Length (c − d) � |�x|Applying the line integral to the loop a-b-c-d in the anti-clockwise direction,

E1 · ax|�x| + E2 · (−ax) |�x| = 0

where E1 and E2 are the electric fields in regions 1 and 2, respectively. The right-hand side of the integral∫∫

∂B

∂t· dS �

∫∫ ∣∣∣∣∂B

∂t

∣∣∣∣ × |dS| = 0

since one side enclosing the area is vanishingly small, the area is vanishingly small,i.e.,

|dS| ≈ �x × Length (b − c) = 0 (11.66)

is zero. The condition on the electric field may then be summarised as

Et1 − Et2 = 0 (11.67)

The subscript t is used to signify the tangential component. Et1,2 are the tangen-tial components of the electric field next to the boundary but in media 1 and 2respectively. Writing this in a more compact form

n × (E1 − E2) = 0 (11.68)

Notice that the normal goes from medium 2 to medium 1.



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“Chapter11” — 2012/5/5 — 17:35 — pg. no. 363 — #23�

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Let us take the case where the second medium is a perfect electric conduc-tor. Then there will be no fields there and Et2 = 0. So

Et1 = 0 (11.69)

Taking a look at the next Maxwell equation∫∫∫∇ · BdV = ©

∫∫B · dS = 0

we apply the same arguments as for the D vector and get

Bn1 − Bn2 = 0 (11.70)

or n · (B1 − B2) = 0 (11.71)

Let us take the case where the second medium is a perfect electric conduc-tor. Then there will be no fields there and Bn2 = 0. So

Bn1 = 0 (11.72)

Finally we treat the last Maxwell equation∫∫∇ × H · dS =

∮ˆ H · dl =

∫∫∂D

∂t· dS +

∫∫J · dS

Integrating over the small loop abcd in the anticlockwise sense,

Hx1|�x| − Hx2|�x| = ∂Dy

∂t�A + Jy�A + Jsy|�x|

Here |�x| is the longer side of the loop and �A is the area of the loop. Jy is they-directed volume current while Jsy is the surface current just at the boundary. Theline integrals of H over the shorter sides of the loop are equated to zero since therelengths are virtually zero. Using the same argument, �A is also considered to bezero. Therefore

Hx1|�x| − Hx2|�x| = Jsy|�x|or Ht11 − Ht12 = Jst2 (11.73)

t1, t2, and n form a right-handed orthogonal coordinate system. t1, t2, and n areakin to ax, ay, and az. Scrutinising the figure, Jst2 is the y- (or t2-) directed surfacecurrent. Writing equation in vector notation

az × (H1 − H2) = Js (11.74)

or n × (H1 − H2) = Js (11.75)

Let us take the case where the second medium is a perfect electric conduc-tor. Then there will be no fields there and H2 = 0. But in this case there willbe the surface current Js. So

n × H1 = Js (11.76)

note that n will be from the pec (perfect electric conductor) into medium 1.



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“Chapter11” — 2012/5/5 — 17:35 — pg. no. 364 — #24�

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11.7 Reflection and Refraction of Waves

11.7.1 Reflection from a Metal SurfaceElectromagnetic waves have the same nature as light and therefore reflect frommetallic objects and suffer refraction in the presence of dielectrics. The simplest caseis one where a plane wave falls on a metal or dielectric plane surface. Let us considerfirst the case of a uniform plane wave obliquely incident on a air metal boundary.For the sake of convenience the metal may be considered to be a perfect one withσ → ∞. The wave may approach the boundary in either of two configurations:where the electric field is perpendicular to the plane of incidence or when it isparallel to the plane of incidence. Both configurations are shown in Fig. 11.12.

We need to look at a plane wave travelling in a general direction whose unitvector is n. The equation of a plane is

f ( x, y, z) = ax + by + cz = constant = p (11.77)

then the normal to this plane is

∇f = aax + bay + caz (11.78)

(Recall that the normal to any surface f ( x, y, z) is ∇f ) and the unit normal to theplane is

n = aax + bay + caz√a2 + b2 + c2

(11.79)

note that a/√

a2 + b2 + c2 is the direction cosine of the vector n in the x-direction,and the other direction cosines are b/

√a2 + b2 + c2 and c/

√a2 + b2 + c2. The

equation of the plane, Eqn (11.77), may now be written as

n · r = p√a2 + b2 + c2

= p′ (11.80)

where p′ is the shortest distance of the plane from the origin. With this introduction,the equation of a uniform plane wave travelling in the n direction is

E = E0e−jkn · r (11.81)

where E0 is perpendicular to n and

n · r = axx + ayy + azz

Medium 1

Medium 2

Parallelpolarisation

s → ∞

e1

Erz

EiHi

ni

y y

nrqr = qi

qi

z

EiHi

niqi

Hr

Er

nrqr = qi

Hr

(b) Perpendicularpolarisation

(a)

Fig. 11.12 A wave obliquely incident from air on a metal (σ → ∞)



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“Chapter11” — 2012/5/5 — 17:35 — pg. no. 365 — #25�

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describes different planes all parallel to to each other. We can see for instance if thewave travels in the ay direction then ay · r = y, then

E = E0e−jky

and so on.

Example 11.14 Find the equation of the electric field with amplitude E0 when thewave travels in the (a) x direction, (b) z direction and (c) in the direction ax+ ay+ az.Solution(a) When the wave travels in the x-direction, n = ax, n · r = x, and E = E0n⊥

exp{−jkx}, where n⊥ is a unit vector perpendicular to ax.(b) When the wave travels in the z-direction, n = az, n · r = z, and E = E0n⊥

exp{−jkz}, where n⊥ is a unit vector perpendicular to az.(c) When the wave travels in the ax+ay+az direction, n = (ax+ay+az) /

√3, n · r =

(x + y + z) /√

3 and E = E0n⊥ exp{−jk(x + y + z) /√

3}, where n⊥ is a unitvector perpendicular to n.

Referring now to Fig. 11.12(a),

ni = −az cos θi + ay sin θi (11.82)

nr = az cos θi + ay sin θi (11.83)

Ei = axEx0e−jk(y sin θi− z cos θi) (11.84)

Let the reflected E-field be

Er = axREx0e−jk(y sin θi+ z cos θi) (11.85)

where R is the reflection coefficient which is to be determined.At the boundary of the metal, the tangential electric field must vanish as per the

previous section. The coordinate system is: the y and z coordinates as shown in thefigure and the x coordinate comes out of the plane of the paper. The metal surfaceis defined by the equation z = 0. Therefore,

(Ei + Er)z=0 = 0 or (11.86)[axEx0e−jk(y sin θi−z cos θi) + axREx0e−jk(y sin θi+z cos θi)

]z=0

0 or (11.87)

R = −1 (11.88)

the reflected electric field is therefore for perpendicular incidence,

Er = −axEx0e−jk(y sin θi+z cos θi) (11.89)

We now analyse the case of parallel incidence where the magnetic field is perpen-dicular to the plane of incidence as shown in Fig. 11.12(b). The electric field isgiven by

E0i = Z0H0i × ni

= Z0Hx0i[ax × (−az cos θi + ay sin θi ) ]= Z0Hx0i(−ay cos θi + az sin θi ) (11.90)

where Z0 = √μ/ε1, ( 377�), the intrinsic impedance of free space. We now proceed

to apply the boundary condition at the metal boundary: the tangential electric fieldis zero at the boundary.

Ey0r = −Z0Hx0i cos θi (11.91)



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“Chapter11” — 2012/5/5 — 17:35 — pg. no. 366 — #26�

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so E0r = Z0Hx0i(−ay cos θi + az sin θi ) (11.92)

and the reflected magnetic field is nr × E0r/Z0

H0r = (az cos θi + ay sin θi ) ×Hx0i( −ay cos θi + az sin θi )

= Hx0i(cos2θi + sin2θi ) ax

= Hx0iax (11.93)

We notice that the reflected magnetic field remains as it is at z = 0! The incidentand reflected magnetic fields are

Hi = axHx0e−jk(y sin θyi− z cos θi) (11.94)

Hr = axHx0e−jk(y sin θi+ z cos θi) (11.95)

11.7.1.1 Normal IncidenceFrom the above discussion it is clear that for normal incidence, the two polarisationsbecome one and the same. Referring to Eqns (11.84) and (11.85), we set θi = θr = 0,

Ei = axEx0e+ jkz

Er = axREx0e− jkz

and set R = −1

Ei = axEx0e+ jkz

Er = −axEx0e− jkz

Adding these two waves,

ET = Ei + Er

= axEx0e+ jkz − axEx0e− jkz

= 2jaxEx0 sin(kz)

which is a standing wave.The electric field is zero on the surface of the metal and goes through maximas

and minimas as we move vertically perpendicular to the metal surface. That is

sin(kz) =

⎧⎪⎨⎪⎩

1, kzmin = π /2, 5π /2, . . .

−1, kzmax = 3π /2, 7π /2, . . .

0, kzzero = 0, π , 2π , . . .

Therefore, the minimas occur at

zmin = λ /4, 5λ /4, . . .

the maximas occur at

zmax = 3λ /4, 7λ /4, . . .

and the zeros at

zzero = 0, λ /2, λ , . . .

since the electric field in real time is �[ET exp{jω t}] therefore

ET (z, t) = 2ax sin(kz) cos(ω t+π /2)

which means that every half cycle the maxima becomes a minima and the minimabecomes a maxima. The zeros remain where they are.



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Example 11.15 For a wave of 1 GHz normally incident on a metal surface, findthe maximas and minimas and zeros of the standing wave for t = T /4.Solution For 1 GHz, the wavelength is

λ = c/f = 3 × 108/1 × 109 = 0.3 m

and the time period T is

T = 1/f = 1 ns

The equation of the electric field at t = T /4 is

E = 2 sin(kz) cos(ω T /4+π /2)

= 2 sin

(2π

λz

)cos

[2π

T(T /4) +π /2

]

= 2 sin

(2π

0.3z

)cos[π ] = −2 sin

(2π

0.3z

)Now the maximas are defined by

2π

0.3zmax = π /2, 5π /2, 9π /2 . . .

zmax 1 = 0.075 m

zmax 2 = 0.375 m

zmax 3 = 0.675 m

...

Similarly, the minimas are defined by2π

0.3zmin = 3π /2, 7π /2, 11π /2 . . .

zmin 1 = 0.225 m

zmin 2 = 0.525 m

zmin 3 = 0.825 m

...

and the zeros are defined by2π

0.3zzero = 0, π , 2π , . . .

zzero1 = 0 m

zzero2 = 0.15 m

zzero3 = 0.3 m

...

11.7.2 Refraction from a Dielectric Surface11.7.2.1 Perpendicular PolarisationWe now consider a wave obliquely incident on a dielectric surface as shown inFig. 11.13. First let us consider the case where the electric field is perpendicular



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ni

Medium 1

Medium 2y y

e1

e2

zEi

Hiqi

qt

Er

qr = qi

Hr

EtHt

nt

qt

EtHt

nt

z

EiHi

ninr qi

y

Er

nrqr = qi

Hr

y

Parallelpolarisation

(b) Perpendicularpolarisation

(a)

Fig. 11.13 A wave obliquely incident from air (ε1) on a dielectric (ε2)

to the plane of incidence as shown in part (a) of the figure. The incident, reflected,and transmitted electric fields are given by

Ei = axEx0ie−jk(y sin θi− z cos θi) (11.96)

Er = axEx0re−jk(y sin θi+ z cos θi) (11.97)

Et = axEx0te−jk(y sin θt− z cos θt) (11.98)

At the dielectric interface (z = 0) the tangential fields must be continuous, so weequate the sum of the fields in Region 1 to the fields in Region 2.

Ex0i + Ex0r = Ex0t (11.99)

similarly, the incident, reflected, and transmitted magnetic fields are given by

Hm = 1

Zjnm × Em (11.100)

where Zj = √μ0 /εj characterises the medium (1 or 2) and m characterises the wave:

incident, transmitted, or reflected waves. So

Hi = Ex0i

Z1(ay cos θi + az sin θi ) e−jk(y sin θi− z cos θi) (11.101)

Hr = Ex0r

Z1(−ay cos θi + az sin θi ) e−jk(y sin θi+ z cos θi) (11.102)

Ht = Ex0t

Z2(ay cos θt + az sin θt ) e−jk(y sin θt− z cos θt) (11.103)

At the dielectric interface, the tangential magnetic fields must also be continuous(Ex0i − Ex0r) cos θi

Z1= Ex0t cos θt

Z2(11.104)

using Eqn (11.99),(Ex0i − Ex0r) cos θi

Z1= (Ex0i + Ex0r) cos θt

Z2(11.105)

or Ex0r(cos θi /Z1 + cos θt /Z2) = Ex0i(cos θi /Z1 − cos θt /Z2) (11.106)

Ex0r

Ex0i= cos θi /Z1 − cos θt /Z2

cos θi /Z1 + cos θt /Z2(11.107)



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R⊥ = cos θi − Z1 cos θt /Z2

cos θi + Z1 cos θt /Z2(11.108)

= cos θi −√(ε2 /ε1 ) cos θt

cos θi +√(ε2 /ε1 ) cos θt

(11.109)

In these equations, R⊥ is the reflection coefficient. Also from Snell’s law

sin θi

sin θt=

√ε2

ε1(11.110)

which makes cos θt =√

1 − sin2θt =√

1−(ε1 /ε2) sin2θi (11.111)

which gives R⊥ = Er

Ei= cos θi −

√(ε2 /ε1) − sin2θi

cos θi +√

(ε2 /ε1 ) − sin2θi

(11.112)

Example 11.16 Prove that R⊥ of Eqn (11.112) will never be zero for any angle0< θi < π /2 and ε2 > ε1.Proof We know that √

1 − sin2θi = cos θi

therefore√

(ε2 /ε1) − sin2θi > cos θi , for ε2 > ε1

therefore cos θi −√

(ε2 /ε1) − sin2θi < 0

11.7.2.2 Parallel PolarisationProceeding to analyse the case of parallel polarisation, the magnetic fields for theincident, reflected, and transmitted waves are

Hi = axHx0ie−jk(y sin θi− z cos θi) (11.113)

Hr = axHx0re−jk(y sin θi+ z cos θi) (11.114)

Ht = axHx0te−jk(y sin θt− z cos θt) (11.115)

equating the tangential magnetic fields at the boundary

Hx0i + Hx0r = Hx0t (11.116)

we now find the corresponding electric fields

Ei = −Z1Hx0i(ay cos θi + az sin θi ) e−jk(y sin θi− z cos θi) (11.117)

Er = −Z1Hx0r(−ay cos θi + az sin θi ) e−jk(y sin θi+ z cos θi) (11.118)

Et = −Z2Hx0t(ay cos θt + az sin θt ) e−jk(y sin θt− z cos θt) (11.119)

equating the tangential electric fields at the boundary

Z1(Hx0i cos θi −Hx0r cos θi ) = Z2Hx0t cos θt (11.120)

substituting Eqn (11.116) in this equation

Z1 cos θi (Hx0i − Hx0r) = Z2 cos θt (Hx0i + Hx0r) (11.121)

Hx0r(Z1 cos θi + Z2 cos θt ) = Hx0i(Z1 cos θi − Z2 cos θt ) (11.122)

Hx0r

Hx0i= (Z1 cos θi − Z2 cos θt )

( Z1 cos θi + Z2 cos θt )(11.123)



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from where we get using Snell’s law

R|| = Er

Ei= (ε2 /ε1) cos θi −

√(ε2 /ε1) − sin2θi

(ε2 /ε1) cos θi +√

(ε2 /ε1) − sin2θi

(11.124)

Example 11.17 Prove that R|| of Eqn (11.124) may be zero for some angle0 < θi < π /2 and ε2 > ε1.Proof We know that √

1 − sin2θi = cos θi

therefore√

(ε2 /ε1) − sin2θi > cos θi , for ε2 > ε1

or cos θi −√

(ε2 /ε1) − sin2θi < 0

and (ε2 /ε1) cos θi −√


may be zero. (See next section)

Brewster Angle We now consider a case when R|| of Eqn (11.124) is zero at someangle when θi = θb which is the Brewster angle and which implies no reflection. ForR|| to be zero, the numerator must be zero,

(ε2 /ε1) cos θb −√

(ε2 /ε1) − sin2θb = 0

(ε2 /ε1) cos θb =√

(ε2 /ε1) − sin2θb

(ε2 /ε1)2 cos2θb = (ε2 /ε1) − sin2θb

(ε2 /ε1)2 (1 − sin2θb) = (ε2 /ε1) − sin2θb

sin2θb

[(ε2

ε1

)2

− 1

]= ε2

ε1

(ε2

ε1− 1

)

sin2θb = ε2 /ε1(ε2ε1

)+ 1

we now compute the value of sin θb and cos θb

sin2θb = ε2

ε1 + ε2

cos2θb = 1 − sin2θb = ε1

ε1 + ε2

and from which we get

tan θb =√

ε2

ε1(11.125)

where θb is the Brewster angle, when no reflection takes place.

Example 11.18 Find the Brewster angle for fresh water with εr = 80.Solution The Brewster angle is

tan θb =√

ε2

ε1



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=√

ε0 εr

ε0

θb = tan−1√

80

= 83.62◦

11.8 Poynting Vector and the Flow of Power

If we stand out in the sun we know that our skin is warmed by the rays from the sun.How did the energy from the sun reach us? It is obvious that the light which falls onour skin has warmed our skin. And light is a form of electromagnetic radiation—thesame waves which we have studied. But our studies have not given us a clue as tohow waves carry energy and power. All we know—as a hint—is that the product

|E| · |H|has the units of power density:

(V/m) × (A/m) = W/m2

From the knowledge obtained from this chapter, we can speculate that the dot prod-uct will not do, since the electric and magnetic fields are perpendicular to each other

E · H = 0 (11.126)

Though the dot product is zero, the wave still caries power. We can consider thecross product, which denoted by P is

P = E × H (11.127)

The units are right. That is, though P has the units of power density, but it is a vector?To understand about the concept of “Power which is a vector” we need to study thebasic equations in more detail. In many books S = E × H is used instead of P.

The basic work on P, called the Poynting vector, was carried out by HenryPoynting (1852–1914) in 1884. He was an English physicist, and a professor ofphysics at Mason Science College (now the University of Birmingham) from1880 until his death. The Poynting vector describes the direction and magnitudeof electromagnetic energy flow and is used in the Poynting theorem, a statementabout energy conservation for electric and magnetic fields.

11.8.1 Poynting’s TheoremLet us concentrate on the vector identity

∇ · (E × H) = H · ∇ × E − E · ∇ × H (11.128)

The units of this equation is W/m3 throughout. Now

∇ × E = −μ∂H

∂t(11.129)

∇ × H = ε∂E

∂t+ J (11.130)



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which are the standard Maxwell’s equations. Substituting these equations in the pre-vious identity

∇ · (E × H) = H ·(

−μ∂H

∂t

)− E ·

(ε

∂E

∂t+ J

)(11.131)

Considering each term

H ·(

μ∂ H

∂t

)is the power density (W/m3), of the magnetic field at any point in space. If we in-tegrate over any given volume we will get the total power of the magnetic field inthat region. Obviously, the term

μ H · H

is proportional to the energy density stored in the magnetic field (J/m3) at any instantof time, and at a given point in space. Similarly,

E ·(

ε∂E

∂t

)is the power density of the electric field, and

ε E · E

is proportional to the energy density stored in the electric field. The last term ofEqn (11.131),

E · J

is the ohmic power density. Integrating Eqn (11.131) over a region in space inaccordance with Fig. 11.14∫∫∫

V∇ · (E × H) dV =

∫∫∫V

[H ·

(−μ

∂ H

∂t

)− E ·

(ε

∂ E

∂t+ J

)]dV

− ©∫∫S

(E × H) · dS︸︷︷︸Poynting Vector

=∫∫∫

V

⎡⎢⎢⎣ μ

2

∂ |H|2∂t︸︷︷︸

Magnetic Field

+ ε

2

∂ |E|2∂t︸︷︷︸

Electric Field

+ E · J︸︷︷︸Heat

⎤⎥⎥⎦ dV

(11.132)

where we have used

A · ∂A

∂t= 1

2

∂ |A|2∂t

To prove this, we expand the left- and right-hand sides∑i=x,y,z

[Ai

∂Ai

∂t

]= 1

2

∑i=x,y,z

[∂A2

i

∂t

]

= 1

2

∑i=x,y,z

[2Ai

∂Ai

∂t

]

= LHS



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SV

+=− × +

V

V EH dSS E

− × H dSS E

J

E

m

dV

dV2H 2îî t

v

v

e dV2E 2îî t

m H2

2îî t

e2

E 2îî t

JdV

Fig. 11.14 Figure Illustrating Poynting’s theorem

To correctly interpret Eqn (11.132) (refer to Fig. 11.14) the term on the left isthe total power entering the surface S which encloses the volume V . The surfaceintegral denotes the total flux leaving the surface. The negative sign implies thatthe power is entering. The terms on the right are (in that order): the total powergained—(a) by the magnetic field, and (b) by the electric field. The last term is thepower dissipated in ohmic losses. Hence we can write this equation as

Power entering S = Power gained by{E in V + H in V}+ Heat dissipated in V

The two termsμ

2H · H,

ε

2E · E

are the energy densities, J/m3, stored in the electric and magnetic fields respectively.

11.8.2 Poynting VectorThe Poynting vector, P, is associated with the flow of power. Does the vectorhave actual physical significance? Famous scientists have expressed doubt aboutits reality.

Sir James Jeans in his book The Mathematical Theory of Electricity andMagnetism said that

“The integral of the Poynting Flux over a closed surface gives the total flowof energy into or out of a surface, but it has not been proved, and we are notentitled to assume, that there is an actual flow of energy at every point equalto the Poynting Flux.”

“For instance, if an electrified sphere is placed near to a bar magnet, thislatter assumption would require a perpetual flow of energy at every point inthe field except the special points at which the electric and magnetic lines offorce are tangential to one another. It is difficult to believe that this predictedcirculation of energy can have any physical reality. . . ”

The italicised part states that Poynting’s theorem has meaning, but not the Poyntingvector. That is the Poynting vector has no real physical significance whatsoever.



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H P

y

z

xI

E

Fig. 11.15 Poynting theorem applied to the case of a wire carrying a steady current

However, in the engineering world, and in most areas of electromagnetics—especiallyin antennas—the Poynting vector has been used with great success. We look at thecase of the energy flow of the case of wire carrying a steady current.

Consider the case of a current carrying conductor of conductivity σ which carriesa current I as shown in Fig. 11.15. The conductor has a steady electric field Ez. Themagnetic field in the cylindrical coordinate system, at a point (ρ , φ , z) outside (andjust inside the conductor) is given by

Hφ = I

2πρ

Now I = Jz×πa2 = πa2σ Ez ( Jz = σ Ez)

where the radius of the conductor is a, and Jz is the (constant) current density insidethe conductor. Therefore

Ez = I

πa2σ

Jz = I

πa2

H at the surface of the conductor is

Hφ = I

2πaWe apply Poynting’s theorem, Eqn (11.132), to the Gaussian surface shown as acylinder S in the figure. The Poynting vector is

P = E × H

=∣∣∣∣∣∣aρ aφ az

0 0 Ez

0 Hφ 0

∣∣∣∣∣∣= aρ(−EzHφ)

It is interesting to note that the power flow is inward from the surface of the cylin-der. This is due to the negative sign with the unit vector aρ . The only region wherethe electric field and magnetic field are present together is inside the conductor. TheGaussian surface (shown as S in the figure) is made to coincide with the surfaceof the conductor, but infinitesimally smaller and its length is d. Integrations on thetwo flat surfaces contribute nothing, because the direction of the Poynting vectoris parallel to those surfaces. The Pointing vector contribution is from the curvedsurface, an element of area which is

dS = aρadφdz



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On the surface of the conductor, the aρ part of the Poynting vector is

Pρ = −EzHφ = −(

I

πa2σ

)×

(I

2πa

)= − I 2

2π2a3σ

where we have used the earlier equations of Ez and Hφ . Now

− ©∫∫S

P · dS =φ=2π,z=d∫∫φ=0,z=0

I 2adφdz

2π2a3σ

= I 2( 2π) d

2π2a2σ= I 2d

πa2σ

= Total power flow into the conductor

If we calculate the total E · J term∫∫∫V

E · J = EzJzπa2d

=(

I

πa2σ

)(I

πa2

)×πa2d = I 2d

πa2σ

= Total power dissipated inside the conductor

We notice that the two results are equal.Let us look at another example, that of a uniform plane wave travelling in space

in the az. The electric and magnetic fields in phasor notation are given by

Ex = E0e−jkz

Hy = E0

Z0e−jkz (11.133)

where k is the notation used for the propagation constant β, for free space and Z0 isthe characteristic impedance of free space. We cannot work with phasor quantitiessince the Poynting theorem has been defined for real time variables. These fields inthe real time are

Ex = E0 cos(ω t − kz)

Hy = E0

Z0cos(ω t − kz) (11.134)

Then P = E × H =∣∣∣∣∣∣ax ay az

Ex 0 00 Hy 0

∣∣∣∣∣∣= azExHy

= azE2

0

Z0cos2(ω t − kz)

= azE2

0

Z0

[1

2− cos 2(ω t − kz)

2

](11.135)

We notice that (a) The Poynting vector travels in the direction of wave propaga-tion, which is intuitively satisfying; (b) the Poynting vector is always positive butpulsating, increasing from 0 to E2

0/Z0 and then back to 0, and so on, sinusoidally;



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and (c) the average value of the P is E20/(2Z0), since the average value is given by,

(ω T = 2π)

Pav = 1

T

∫ T

0

E20

Z0

[1

2− cos 2(ω t − kz)

2

]dt

= 1

T× E2

0

Z0× T

2(integral of the cosine term is zero)

= E20

2Z0

These ideas expressed above are reminiscent of circuits where the average powersupplied, for example, to a load is given by

P = VI cosφ

where V and I are the rms phasor voltage across and current through the load and φ

is the angle between them. Similarly, working with electromagnetic phasors (butnot rms values)

P = 1

2(E × H∗) (11.136)

Because these are not rms values, the factor 1/2 has to be included. P is called thePoynting vector, and E and H are the phasor electric and magnetic fields. On theother hand the time averaged Poynting vector is given by

Pav = 1

2�{(E × H∗) } (11.137)

Let us find out whether we are right. Using Eqn set (11.133) and applying thedefinition given above, we get

Pav = 1

2�{(E × H∗) }

= 1

2�⎧⎨⎩∣∣∣∣∣∣ax ay az

Ex 0 00 H∗

y 0

∣∣∣∣∣∣⎫⎬⎭

= 1

2�{(ExH∗

y ) }

= 1

2× �{E0e−jkz × H∗

0 e+jkz} = |E0|22Z0

POINTS TO REMEMBER

• The Helmholtz equation for the electric field for a travelling wave in air is

∇2E + k2E = 0

where k =ω√

με

• A travelling wave must have a functional dependence,

E(R, t) = �{E0ej(ωt−βz)} (11.138)

where β is the propagation constant of the wave and ω is the frequency of oscillation. Hereβ may or may not be equal to k .



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• For the uniform plane wave, there is no electric field in the direction of propagation; thereare only transverse components.

E = axEx0e−jkz + ayEy0e−jkz

• The magnetic field is given by

Hx = − 1

ZEy

Hy = 1

ZEx

and Z =√

μ

ε

• The magnitude of the magnetic field is

H =√

H 2x + H 2

y

= 1

Z

√E2

y + E2x

= E

Z• The magnitude of the Poyting vector E × H is given by

|E × H| = 1

Z

(E2

x + E2y

)W/m2

• Z is called the characteristic or intrinsic impedance of the medium. For air or vacuum,

Z0 =√

μ0

ε0≈ 377 �

• Wave polarisation is defined as

– linear when the tip of the electric field vector moves on a line in a plane perpendicularto the direction of propagation

– circular when the tip of the electric field vector moves on a circle in a plane perpen-dicular to the direction of propagation

– elliptical when the tip of the electric field vector moves on an ellipse in a plane per-pendicular to the direction of propagation

• The complex permittivity is given by

εC = ε′ −jε′′ = ε −jσ

ω

• The dissipation factor, D

D = σ

ω ε

{1, for a good dielectric

�1, for a good conductor

for lossy dielectrics it is also called the loss tangent.• Using the notation of βC as the complex propagation constant

βC = βR + jβI = β + jα

=ω√

μεC

• For βC = βR −jβI , the complex propagation constant (also called γ )

β = βR =ω

√με

√√1 + (

σωε

)2 + 1√

2



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α = βI =ω

√με

√√1 + (

σωε

)2 − 1√

2

• The wave in complex notation is then E = E0e−αze−jβz

• The skin depth is given by δ = 1/α• For low conductivity materials,

β �ω√

με

[1 + 1

8

( σ

ω ε

)2]

α �

(ω

√με

2

)( σ

ω ε

)Z �

√μ

ε

(1 + j

σ

2ω ε

)• For high conductivity materials,

α = β =√

ω μ σ

2The characteristic impedance is given by

Z � Z0

(1 + j√

2

)√ω ε0

σ

• The boundary conditions for time-varying fields between two media are

– For E Et1 − Et2 = 0

n×(E1 − E2) = 0

– For D Dn1 − Dn2 = ρs

n · (D1 − D2) = ρs

– For B Bn1 − Bn2 = 0

n · (B1 − B2) = 0

and– For H n×(H1 − H2) = Js

in all these cases the unit normal n extends from medium 2 to medium 1. Also ifmedium 2 is a perfect electric conductor, all the fields are zero.

• For perpedicular polarisation (electric field perpendicular to the plane of polarisation) thereflection coefficient is

R⊥ = Er

Ei= cos θi −

√(ε2 /ε1) − sin2θi

cos θi +√


for a wave incident from medium 1 (ε1) to medium 2 (ε2).• For parallel polarisation (electric field parallel to the plane of polarisation) the reflection

coefficient is

R|| = Er

Ei= (ε2 /ε1) cos θi −

√(ε2 /ε1) − sin2θi

(ε2 /ε1) cos θi +√


for a wave incident from medium 1 (ε1) to medium 2 (ε2).• The Brewster angle, θb, the angle at which R|| = 0 is given by the relation

tan θb =√

ε2

ε1

for a wave incident from medium 1 (ε1) to medium 2 (ε2).



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• Poynting’s theorem is

− ©∫∫S

(E × H) · dS =∫∫∫

V

[μ

2

∂ |H|2∂t

+ ε

2

∂ |E|2∂t

+ E · J

]dV

where S is the surface enclosing a volume V . The term on the left is the power enteringthe surface S. The first term on the right is the power stored in the magnetic field; thesecond term on the right is the power stored in the electric field; and the third term on theright is the power dissipated as heat.

• The average power propagated in a plane wave is given by

P = |E|22Z0

where E is the magnitude of the electric field and Z0 is the intrinsic impedence of themedium.

SELF ASSESSMENT

Objective Type Questions

1. A wave is called a plane wave because(a) it travels in planes (b) it appears like a plane(c) its phase is constant over a plane (d) none of the above

2. For a wave travelling in free space at a frequency of 1 GHz and in a material withσ = 103 and ε = ε0 compare the velocity (v) wavelength (λ) and radian frequency (ω)in the two media.(a) all three are equal (b) two out of three are equal(c) one out of three are equal (d) none of the three are equal

3. The magnitude of the magnetic field of a circularly polarised wave travelling in spaceat a point z = constant is(a) sinusoidally varying (b) circularly varying(c) is a constant (d) none of the above

4. The tip of the electric field vector of a linearly polarised wave travelling in space at apoint z = constant is(a) sinusoidally varying (b) circularly varying(c) is a constant (d) lies on a straight line

5. The power density in W/m2 of a plane wave(a) propagates with the wave (b) is equal to ( 1/2) |E × H ∗|(c) is equal to E2/( 2Z0) (d) none of the above

6. Distributed components are used in circuits in the frequency range(a) 3 Hz to 3 kHz (b) 3 kHz to 300 kHz(c) 3 GHz to 30 GHz (d) none of the above

7. The complex permittivity is(a) ε0 εr where εr is a real number (b) ε0 εr (1 − jσ)

(c) ε0 εr [1 + j( σ /ω)] (d) ε0 εr {1 + j[σ /(ω ε0 εr)]}8. The unit of σ /ε0 is

(a) � (b) � (c) dimensionless (d) s−1

9. The skin depth δ is(a) 1/βC (b) 1/α (c) 1/(ω ε ) (d) none of the above



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10. In a metal of high conductivity(a) α = β (b) Z0 is complex (c) λmetal λ (d) none of the above

11. The boundary condition of sinusoidally varying electric field phasor at an air–metalsurface is(a) Etan = 0 (b) En �= 0 (c) Dn = ρs (d) none of the above

Short-Answer Questions

1. Many “Direct to Home” channels, like “Tata Sky” and “Airtel” operate in a range offrequencies. Find the downlink satellite frequency band in terms of name and numbers.

2. If the electric field of a uniform plane wave is z directed, and the wave is also travellingin the z direction, is this possible?

3. Find the electric field strength

E = 10 cos(ω t−β z)

for

(a) For t = 0 and for all values of z(b) For z = 0 and for all values of t(c) For ω t = π /4 and for all values of z(d) For β z = π /2 and for all values of t

4. Find the magnetic field associated with

E = E0axe−αze−j(ωt−βz)

5. Find the propagation constant γ , for free space at f = 1MHz, 10 MHz, and 1 GHz.

Review Questions

1. Why is the “plane wave” so called?2. Write the Helmholtz equation and outline its solution in rectangular coordinates.3. Write the the functional dependance of the electric field which is sinusoidal in nature

and travelling in the z-direction.4. Write a short note on the complex propagation constant.5. How would you identify RCP and LCP waves?6. Explain why a circularly polarised wave can still be a plane wave.7. Explain how standing waves form for normal incidence on a metal surface.

Numerical Problems

If the medium is not specified, please assume ε = ε0 and μ = μ0.

1. Write the Helmholtz equation of the electric field which is z directed. Write its solutionfor a wave travelling in the y direction with a velocity c and frequency ω.

2. In what range of frequencies is the wavelength in the range 1–10 cm? Which are thefrequency bands involved?

3. E = E0 cos(2.1 × 108t−β z) is the equation of the electric field of a wave travelling inair. What is the velocity of the wave? Find the value of (a) β, (b) λ, (c) ω

4. If the electric field expression for a plane wave is given as

Ex = E0 cos( 1.5 × 1010t + 60z)

then determine (a) its wavelength, (b) phase velocity, and (c) direction of travel. Findthe magnetic field H associated with this electric field.



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5. A plane wave is written as

E( z, t) = E0ax cos(ω t − kz) +2E0ay cos(ω t − kz)

Express this wave in phasor form and find its polarisation.6. A plane wave is written as

E( z, t) = E0ax cos(ω t − kz) −2E0ay cos(ω t − kz)

Find its polarisation.7. Find the magnetic field associated with

E = E0aze−αxe+j(ωt−βx)

8. A plane wave is travelling from free space to a medium with εr = 3 and μr = 3. If theexpression in free-space is

E = E0axej(ω0t−β0z)

write the general expression of the electric field in the medium of the same wave.9. Show that E( x, t) = f1( x − ct) +f2(x + ct)

is a solution of the wave equation

(∇2+ω2 με ) E = 0

with k =ω√

με and c =ω /k .10. Find the intrinsic impedence of a medium at a frequency of 1 GHz with σ = 5.8 ×

107�/m, εr and μr = 4. What happens to the impedence for the two cases of (a) σ = 0

and (b) σ = ∞ if the other parameters are constant.11. A plane wave propagating through a medium with εr = 6 and μr = 4 has a magnetic

field given by

H = 10aze−z/5 sin(2π×108t+ β x)

find the expression for the electric field, the wave velocity, and the wave impedance.12. An EM plane wave travels with

E = 20ayej(√

32 y+ 1

2 z)ej2π×108t

determine (a) β, (b) the direction in which the wave travels, and (c) ω.13. Given that E = ayE0 sin(10πx) cos(6π×109t− βz)

Find H and β.14. Specify the polarisation of the wave if its electric field is defined as

E = E0xax cos(ω t− βz + θx ) + E0yay cos(ω t− βz + θy )

for the following cases:

(a) E0x = E0y, and θx = θy + 2nπ, and where n = 0, 1, . . .(b) E0x = E0y, and θx = θy + (2n + 1) π, and where n = 0, 1, . . .

15. Specify the polarisation of the wave if its electric field is defined as

E = E0xax cos(ω t−β z+ θx ) +E0yay cos(ω t−β z+ θy )

for the following cases:

(a) E0x = E0y and θx = θy + (2n + 1) (π /2), and where n = 0, 1, . . .(b) E0x = E0y and θy = θx + (2n + 1) (π /2), and where n = 0, 1, . . .



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Path 1

Path 3

Path 2

RHCPWave Perfect

reflector

Fig. 11.16

16. An RHCP wave is incident on a perfect reflector as shown in Fig. 11.16.Specify the polarisations on Paths 2 and 3. Write equations for all cases

17. Show that any linearly polarised wave can be written as a sum of two RHCP and LHCPwaves.

18. For a material having μr = 2, εr = 4, and σ = 100 S/m, calculate the attenuationconstant, phase constant, and intrinsic impedence at 10 GHz.

19. Find the skin depth δ for silver and gold at frequencies 1 MHz and 1 GHz. Also findthe propagation constant and wave velocity for both cases.

20. If a material has σ = 341 nS/m, what are the frequency ranges in which the materialcan be considered to be (a) a low-loss dielectric and (b) a good conductor.

21. A 500 MHz plane propagates normal to a material of σ = 105 S/m. Calculate the skindepth and the distance where the wave amplitude decreases to 20% of its initial value.

22. Plot the dissipation diagram (ω −β) for silver.23. A medium has a propagation constant γ = 200 + j500/m. (Note that the dependence

is exp{−γ z}). The intrinsic impedence η = Z = 75∠45◦ at a frequency f = 500MHz. The electric field is given as

Ex = 100e−αz cos(2π×5 × 108t − βz)

Calculate the magnetic field and average power/m2 at z = 1 mm.24. In free space

E( x, t) = 100ay sin(ω t − β x)

Find the average Poynting vector and power crossing a square area of 1 m2 in the planex = 3.

25. A plane wave is incident from air on a dielectric with εr = 10. If the angle of incidenceis 75◦, find the reflection coefficient for both polarisations. Find also the power reflectedand transmitted in both cases.

26. Find the Brewster angle for dielectric with εr = 10.

Answers

Objective Type Questions

1. (c) 2. (c) 3. (b) and (c) 4. (a) and (d) 5. (a), (b) and (c)6. (c) 7. (d) 8. (d) 9. (b) 10. (d) 11. (a), (b) and (c)

Short-Answer Questions

1. C-band (4–8 GHz) or Ku-band (12–18 GHz)



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2. No3. (a) E = 10 cos( β z), (b) E = 10 cos( ω t), (c) E = 10 cos( β z−π /4),

(d) E = 10 cos( ω t−π /2)

4. H =( E0/Z) aye−αze−j(ωt−βz) where Z is the intrinsic impedence of the medium.5. γ = jβ; At 1 MHz β = 0.0209; at 10 MHz β = 0.209; At 1 GHz β = 20.9r/m.

Numerical Problems

1. ∇2Ez + k2Ez = 0; E = E0zaze−j(w/c)y

2. λ = 1 cm corresponds to f = 30 GHz (Ka band); λ = 10 cm corresponds to f = 3 GHz(C band); bands involved are C bank, X band, Ku band, and Ka band.

3. c = 3 × 108m/s ( a) β = 0.7 r/m, (b) λ= 8.98 m, ( c) ω = 2.1 × 108r/s

4. H = ayE0 × 1.2

120πcos(1.5 × 1010t + 60z)

(a) λ = 10.47 cm (b) Vp = 2.5 × 108 (c) travelling in the z-direction.5. E = E0ax∠0 + 2E0ay∠0; linear polarisation.6. Linear polarisation.

7. H = E0

Zaye−αxe−j(ωt−βx)

8. E = E0axej(w0t−3β0z)

10. Z = 0.0165 + j0.0165; (a) Z = 376.73 (b) Z = 011. E = −Z10aye−z/5 sin( 2π ×108t+β x) where Z = 307.8 + j6; β = 10.26 r/m.

V = 6.12 × 107.12. (a) β = 1 (b) n = −

√3

2 ay − 12 az (c) ω = 2π ×108

13. H = −ax( E0/377) sin( 10πx) cos( 6π×109t−β z) where β = 62.83 r/m14. (a) Linear polarisation

(b) Linear polarisation15. (a) RHCP

(b) LHCP16. Path 2: LHCP; Path 3: RHCP18. α = 2778 Np/m, β = 2840 r/m and Z = 28.4 + j27.8 �

19. For silver: at 1 GHz δ = 2.02 μ m, α =β = 4.935 × 105/m; at 1MHz δ = 64.07 μ m,α =β = 1.56×104/m. For gold: at 1 GHz δ = 2.485 μ m, α =β = 4.02×105/m; at 1 MHzδ = 78.6 μ m, α =β = 1.27× 104/m. Wave velocity for silver = 1.27×104m/s at GHz,= 402 m/s at 1 MHz. Wave velocity for gold = 1.56 × 104 m/s at 1 GHz, = 493 m/sat 1MHz.

20. The material may be considered a low loss dielectric for f > 0.48 MHz Hz and a goodconductor for f < 4.8 kHz.

21. δ = 7.117 × 10−5m. Distance at which the wave decays to 0.2 of its initial value is1.15 × 10−4 m.

23. Hy = 4e−αz cos( βz−ω t)

3(

i√2

+ 1√2

) average power = 31.6W/m2.

24. P = 13.26axW/m2; average power crossing 1 m2 at x = 3 is 13.26 W.25. R⊥ = −0.8417, R|| = 0.0755; Power reflected. R2

⊥Pinc = 0.7085Pinc; R2||Pinc =0.0057Pinc; Transmitted power. P⊥ t = 0.2915Pinc; P||t = 0.9943Pinc

26. 72.5◦



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