Chapter
4
Guided Waves
Contents
Waves between parallel
ties of propagation,
planes. TE and TM waves and their characteristics. TEM waves,
veloc
Attenuation in parallel plane
guides, wave impedance.
POINTS TOREMEMBER
1.
Guided
2.
The field
waves use some
components
guiding surtaces
of wave between
are
Ex
x
Ey
joxHz
for the actual
propagation of wavees.
parallel planes propagating in positive z-direction
H
and
Ez
Hy = joe
h2 ax
electric wave
Transverse
4
(H waves) having Ez
=
0.
Field components of TE waves are:
Ey
C
sinTX-iB
a
H
and
5 Tra
dnsverse
C, sin mae-iB
a
Ccos
H2jo ua
a
-iD
magnetic waves are such wave having Hz =0
77
LORD ElectromagneticField Theory
78
6.
Field components of TM waves are
Hy
Ca cos
=
a
E, Cjot cos
E
Ez
and
7.
TEM
8.
8.
When the
waves
i
a
- jo ea C4 sin TXeiP2
=
a
entirely
waves are
iB
strikes
transverse in nature.
on
the walls of
waveguide
certain losses
are
taken
place.
Power lost per unit length
Thus attenuation takes place and attenuation factor is a
9.
Wave
impedance is the
ratio of electric
field strength
2 (power transmitted)
to the magnetic field strength.
QUESTION-ANSWERS
Q 1. What
are
the characteristics of TM waves?
Ans. 1. Propagation constant, =
2. Cut off wavelength, A
2a
m
3. Wavelength, =
4.Velocity;V=m/s
When frequency becomes high enough
for air as medium
C 3x 108 m/s.
E-
(PTU, Dec. 2006; May 2006))
Auided Waves
79
a 2. What is wave
Ans. The Wave
feld strength i.e.
impedance?
(PTU, Dec. 2006 ; May 2006)
impedance is the ratio of transverse
components of electric to magneu
H
Q 3. Define velocity of wave
propagation.
Ans. The velocity of propagation is times the frequency i.e.
(PTU, May 2007)
p Af
and for free space the velocity of propagation is equal to velocity of light i.e.
Vp
3
x
108 m/s.
Q 4. Why em waves are also called as guided waves?
(PTU, May 2007)
Ans. As electromagnetic waves are guided along the conducting or dielectric surfacesS
so it is called a guided waves.
Q5. What do you mean by TEM waves?
(PTU, Dec. 2007)
Ans. TEM wave is transverse electromagnetic wave, it having both axial components of
electric and magnetic field as zero.
Q6. Why em waves are also called as guided
waves?
(PTU, Dec. 2007)
Ans. As em waves propagate in guided media such as parallel conducting planes. Also,
the em waves propagate by guiding media along the conducting or dielectric surfaces. So it is
called as guided waves.
the dissipation factor of the dielectric?
Ans. D> Dissipation factor
Q7. What is
and
(PTU, May 2008)
D
cconduction current density J is displacement current densityand for dielectric D
ISvery less than 1.
8/Differentiate between phase velocity and group velocity. (PTU, Dec. 2008)
Ans.
v=VgVp
Where Vggroupvelocity
pphasevelocity
Vvelocity of light
velocity offree space whereas phase velocitv
ways greater than velocity of free space. When frequency is cut off frequency group
than
ane group velocity is always less
always
the
LORD Electromagnetic Field Theory
80
velocity becomes zero and phase velocity infinite so no propagation of energy along the
waveguide occurs.
(PTU, May 2009)
Q9. What are uniform plane waves?
Ans. Uniform plane wave have components of E and H at every point in free space
are perpendicular to each other and perpendicular to the direction of propagation i.e. these
waves are transverse in nature and do not have components in the direction of propagation
(i.e. E
= H, = 0).
Q 10. Define clearly dominant and degenerate modes with examples.
(PTU, May 2009, 2005)
Ans. Dominant modes are the modes which having highest cut off wavelengths for
example:TE10 mode of transverse electric field. Degenerate modes are the modes for which
the propagation constant y is same. In these modes the small losses result in strong coupling
between modes.
a 11. Discuss the
properties of
uniform
plane
wave.
(PTU, Dec. 2008, 2006)
Ans. The properties of uniform plane
to
() The electric E and magnetic field H, at every point in free space are perpendicular
each other and also perpendicular to the direction of propagation.
(i) The velocity of propagation of plane wave is
C=
()
The EM
wave
1
travels with
3x 108 m/s
a
velocity of light in the free space for any value of
frequency
(iv) E and H are in phase and amplitude is 120t.
Q 12. A uniform plane wave at 300 MHz is passed through a medium of 4, = 1 and
E,
78. Find the wave parameters in that medium assuming a = 0.
Ans.
AS
27tf
= 2x x x 300x 10
as
andwave
H 1,E 78
velocity C
=
(PTU, Dec. 2006)
81
A d e dW a v e s
3x
10
78
Y
Y
and
3x10
30 x10
a 13. Distilled water has the constants o = 0, e, = 81, 4, =1. Find refraction index,
shase
velocity
and
group
(PTU, May 2007)
velocity.
Ans. Refractive index
120T
as =
1
120T
81
120
40 x22
21
9
42
Phase velocity, V,B
VHo So Ho
3x10 m/s
3x 10
x
9
81
do
groupvelocity Vp dB
Q 14. Derive
an
m/s
expression for the
parallel conducting planes.
Ans.
3
E and H components for TM
attenuation factor for the
waves are
E, bAs
E
Ee sin m
TM10
Wave
between
(PTU, May 2007)
82
LORDS Electromagnetic Field Theory
Hy
=
Ex
iP
A4 cos e
A4cos
xei
Jo E
and for TM1o m = 1, n = 0
The current density is equal to the tangential component of Ez at x = 0 and x > a
Js IEz
nA4
Ea
The loss is each plane is power loss per unit length =
2 0
Power transmitted
=
Ea/
(Ex Hy)
BA cosx
20
a
E
Power between x = 0 and X = a space is
BA2
cos
20 E
X=0
Power transmitted = PA4a
40
E
Attenuation factor
Power lost per unit length
2xpower transmitted
2
2 (o ea
BAAa
40 E
V20
1
Js
Rs
83
a d e d Waves
22
2
m
Bo ea
The attenuation factor of TM10 waves reaches a minimum at a frequency which is 3 fc
then
increases with
frequency.
a 15. Discuss the propagation characteristics of TE and TM waves.
(PTU, Dec. 2007)
Ans. Characteristics of TE and TM waves
The characteristics are
1. Propagation constant
2. Wavelength
3. Velocity of propagation
as
h2
and
= h2-oue
oue +y2
Y a
mT
a 16. A 10 GHz plane wave travelling in free space has an amplitude Ey = 1 V/m.
Find the
phase velocity, wavelength
Ans. f
and the
propagation
(PTU, Dec. 2008)
10 GHz, Ex = 1 V/m
C f
3x10 0.03 m
10x 109
and
v
Wavelength, a
3
x
0.03
m
C
=
=
=
108 m/s for
Propagation constant
Y jo (o+jo e
and
o = 0 for free space
?=-oHe
and,
Y = jB
constant.
free space
LORDS Electromagnetic Field Theory
84
ory
Also,
B yoH0
Eo
(4,=¬, =1)
2x3.14x 10x10
=
o
x
VHo
Eo
209.3
3x10
Q 17. A long straight tubular conductor of circular cross-section with outside.
diameter of 5 cm and wall thickness of 0.5 cm carries a direct current of 100 amps.
Find H:
() just outside the wall of tube.
(i) at a point in the tube wall half-way between the inner and outer surface.
(PTU, May 2005)
Ans. I=100 amp, R2 =
H
()
5
cm
2TtR2
Ra cm
100x2 636.619 A/m
H
2T x5x10-2
(i)
H-
2
Ra
R
=
R
R-R
2.5x 10-2
m
2.5-0.5 2
cm
=
0.02
m
+R1 _0025+ 0.02
and
2
0.0255 m
H
100
2x0.0225
(O.0225)-(o.02)|
(0.025-(0.02) J
H 334A/m
Q 18. Discuss the propagation characteristics of TE modes.
Ans. Take the expression of h2 as
h2
Y+ 0 ue
(PTU, May 2007)
85
Guided Waves
h2-
ue
and
MT
h
and also,
a
2
7
-oe
At very high frequencies
mr
2
a
and
becomes pure imaginary.
i.e.
a
and
Thus for
perfectly conducting planes,
direction.
Also, HE-
(2n1
0
ue=m
and
m
c2a E
also
.Phase or wave velocity
attenuation
is
zero
and
wave
propagates along
z
86
LORD Electromagnetic Field Theor,
( = f and =27
A s frequency increases above critical frequency the v tends to speed of light.
Q 19. Derive the expression for the attenuation in the parallel plate guide for TE
(PTU, May 2008)
mode.
Ans. The attenuation in parallel plate guide
The E and H components for TE mode between conducting plates as;
Ey A sin mXi2
a
=
H, = A ,sin |
HThe
e
jmnA1
Acos ea
@pa
magnetic induction of linear current density
Js= tangential component of H at x = 0 and x = a
i.e
(Js) = (H2)
mrA
ua
The current does not flow in the direction of
in the each
propagation
plane of conducting parallel plane isJs2 Rs
The power per unit
length
-
maA
m
oua) 20m
Where
Hm
Rs 26om
The power transmitted
per unit
(EH)
-2Ey H
area
is
for these modes and the loss
Guided Waves
87
BAL
sin?(mXx
20 Sin
a
Now power transmitted for
a
conductor between
BAT sin?
20
a
x
=0 and
x
=
a
is
dx
BAa
4
Power lost per unit
Now,
length
2x power transmitted
mA
Pm
20m
20a2
2x
Aa
40
2m2 oHm
V20m
2
3
oua"oyE-mn
d
of uniform plane wave.
Q 20. List the properties
of an electromagnetic
The electric field intensity
Ey E
=
0, E
Determine
=
Eo cos w (t- zN)
as
Ey Ez
=
in free space is
given by
H.
the components of magnetic intensity
for
expressions
the
(PTU, Dec. 2008)
Ex Eo cos w (t- zV)
The components of H:
Ans.
wave
=
0,
=
xE aa
OH
Ho(4,= 1)
88
LORDS Electromagnetic Field Theory
ax ay
Ox
oz
Eo
Ho
H+aHy+ az H2)
and
1)
and
E, Ho
2)
Hy
PO
Q21.
Derive the relation between E
and H in uniform
Define intrinsic
plane wave
impedance and
its
Ans. The first
give
maxwell equation
is
physical significance.
VxE - OB
ot
i.e.
for
VxE =
Ex Ey Ez
uniform plane
wave
travelling
in
-(a, B,+a,B +a
ot
z-direction.
= 0
i.e.
no x
and
and y
components
Ez = 0
ax ay az
0
Ex Ey
o2 - la, By +a, By +0]
propagation.
(PTU, May 2009)
B)
89
Guided Waves
Byexpanding
[a, Ey-ay Exl -ax OBx ay By
ot
=
0Z
Ey
OBx
B HH
We have,
Ey-and Ex
- oHy
From 2nd Maxwell's equation is
VxH
We
have,
ax
ay
Ox
ay
OD
ot
lay
Dx+ ay Dy +az D,]
ot
Hx Hy Hz
=
Now,
O and D = 0
OX
a
So,
ay
az
=lD+a, D, +0
0
H Hy
la
ODy
Hy ay Hd= ax Dx +ay
Hence,
oty eOEx
ot
and
oHx
as,
0Z
eOHy
ot
D =¬E
90
LORDS
oHy
ot
and
0Z
We have,
Ey f (Z Ct)
C
where
also,
=f(Z-c)(Z-Ct)
ot
=
f
(Z- ct) -C)
Ey
=-Cf
if we take fh = f (Z- C)
from,
by putting
EyCii
ot
dx =eC
z
By integration both sides
H=eCdz
H
=e Cf'+K
K is constant of integration, C =
VHE
H,=e
Now,
'Ey-f(Z-c)
Electromagnetic Field Theory
91
Guided Waves
H
and
E
EHx
ly
Since,
E -E+E2
and
H H+H2
Thus,
E-H
and,
Hence,
n is intrinsic impedance or characteristic impedance of non-conducting medium.
For free space
n
4Tx107
36T x 109
Where
Thus,
H 4Ttx
n=
n
10-7,
1
e
=
V4Tx 36 x 100
120 7
36T x 10
LORDS Electromagnetic Field Theory
92
For,
22.
T = 3.14
n
120 x 3.14
n
377 Q
Derive the expression for attenuation factor for TEM
conducting planes.
Ans. Consider a transmission line terminated
in its
waves between
characteristics
parallel
(PTU, May 2009)
impedance.
The voltage and current expressed as,
V Voea7 e-iBz
.(1)
loea7 elf
2)
I
and
=
The average power transmitted is
Pav -
Re
given by
.(3)
IVI
By putting (1) and (2) in (3)
We have
Re [Vo 1o]e-2z
Pav
Now,
the rate of decrease of power is obtain
by differentiating with respect to
oPay
0Z
and
GFav
OZ
-
-2az
-2a)Ro
and
OPav 20
and
oFav 20 Pav
power lost per unit length = 2a Pav
Power lost per unit length
and Power transmitted (average
2a.
power
power
Powerlost per unit length
2 (power transmitted
This is the attenuation factopr.
-
2a
z.
Guided Waves
93
a 23. Discuss the attenuation of
wave
in
parallel planes.
Ans. e-Yz =eAd**c)z-jßz
Where
hg is the attenuation constant for lossy dielectric
A is the power loss in the conducting planes
As, time average power transmitted is
WT
e-2a'z
The rate of decrease of transmitted power is
dW
dz
Also,
2a' WT
Time rate of transmitted power
Time averagetransmitted power
1
'
2a'WT
2a WT
WT
Time rate of transmitted power
2 Time averagetransmitted power
Q 24. Three regions are shown in figure are lossless and non magnetic
medium1>er^ =1
medium 2 er2 =4
medium 3
erg = 9
d=
4
frequency = 2GHz.
Find
medium 3.
(a) Wave impedance in
ratio (VSWR)
(b) Voltage standing wave
medium 3
Ans. (a) Wave impedance in
HO
= 125.6 2
(b)
where
and
vsWR 1-|Pl
VSWR+|p
I= n-20
pZin +Zo
ZinZa
at interface of medium 1 and 2.
4Tx 10
V9x 8.85 x 10-12
94
LORDS Electromagnetic Field Theory
4Tx 10
Where
4
Z
and
188.4 Q
8.85x1 0 - 1 2 =
5.42826 Q
125.6
282.6-377
p
0.141
282.6+377
1+0.14
vsWR 1-0.14
=
=0.14
1.3
Q 25. Show that a uniform plane wave passing in any direction does not have
any component at that direction.
(PTU, Dec. 2009)
Ans. In uniform plane wave only one component is present among x, y and z directions
and it is transverse in nature i.e. uniform plane wave is perpendicular to the direction of
propagation. Hence it does not have component among the direction of propagation.
a 26. Differentiate between phase velocity and group velocity. (PTU, May 2009)
Ans. Phase velocity is the velocity of propagation with which the
In a direction parallel to the conducting surface.
i.e.
wave
changes phase.
2tf.a
2T
a s B-
N
Whereas, group velocity is defined as the velocity with which the group of the wave as
whole propagation.
do
i.e.
Vp
Group velocity
is
also,
Where vo is the
SS
dß
always
v
less than
Vg Vp
velocity
of
velocity of light.
frequency phase velocity becomes infinite
propagation of energy along the waveguide.
At cut off
i.e.
no
light
and group
Q27.
Define the term 'Brewster
Angle'.
Ans. Brewster angle is the angle at which there is
represented as
tane 2
O00
velocity
becomes
zero.
(PTU, May 2005)
no
reflection of
wave
and it is