Lecture 5

EE-344 Wave Propagation and Antennas

Dr. Muhammad Anis Chaudhary

Waveguides and Transmission Lines

Waveguides and Transmission Lines

Waveguides and Transmission Lines

• Interchange of electric and magnetic energy results in the propagation of EM waves in

space (or any unbounded medium)

• This interchange of energy is also possible along conducting or dielectric boundaries and
can result in waves that are guided by such boundaries

• These guiding structures (waveguides and transmission lines) are used to guide EM wave
from one point to another e.g. from a transmitter to antenna

Transmission Lines

• A guiding structure that supports TEM mode

Waveguides

• A guiding structure that supports a variety of different higher order modes

Waveguides and Transmission Lines

Mode

• A mode is a field configuration

• its a solution of Maxwell’s equations (or wave equations) that satisfy the given boundary
conditions imposed by the guiding structure

• All these different field configurations (or solutions) are called modes

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 5 Page 1 of 17
Waveguides and Transmission Lines
Transmission Lines

• can only support TEM (Transverse Electromagnetic) mode

– Figure shows some examples of guiding stuctures that can support TEM mode

• Note that all these guiding structures can also support higher order modes as well

Waveguides and Transmission Lines

Waveguides

• can support many possible field configurations (modes)

Conductor Waveguides (Hollow pipe waveguides)

• Rectangular Waveguides

• Circular Waveguides

• Eliptical Waveguides

• Ridged Waveguides

Dielectric Waveguides

• Dielectric Slab

• Optical Fibre

Conductor Waveguides

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 5 Page 2 of 17
Dielectric Waveguides

Waveguides and Transmission Lines

• Transmission Lines =⇒ TEM Mode

• Waveguides =⇒ Non-TEM Modes

Waveguides and Transmission Lines

Classification of Modes

• the wave solutions or modes may be classified as

– TEM Modes
– TM Modes
– TE Modes
– Hybrid Modes

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 5 Page 3 of 17
Waveguides and Transmission Lines
TEM Modes

• neither Ē nor H̄ field is in the direction of wave propagation

• Both Ē and H̄ lie in a plane transverse to the direction of wave propagation

• neitherĒ nor H̄ has a component in the direction of wave propagation

TM Modes

• H̄ lies entirely in transverse plane

• H̄ has no component in the direction of wave propagation

TE Modes

• Ē is transverse to the direction of wave propagation

• Ē has no component in the direction of wave propagation

Waveguides and Transmission Lines

Hybrid Waves or Modes

• is a combination of TE and TM modes

• both Ē andH̄ have components in the direction of wave propagation

Waveguides and Transmission Lines

Conductor Waveguides

• Rectangular Waveguides

– used at microwave frequencies (1 till 100GHz)

– used for applications requiring
* low attenuation
* high power

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 5 Page 4 of 17
Waveguides and Transmission Lines
General Formulation for Guided Waves

• Consider a cylindrical co-ordinate system with axis taken along z-axis

• Assume no free charges, ρv = 0

• Any conduction currents can be included by replacing ε by εc thus for now J=0

• Assume lossless dielectric (σ ≈ 0) and perfect conducting walls (σc ≈ ∞)

• The wave equations in phasor form are

¯ 2 Ēs + k2 Ēs = 0
– ∇
¯ 2 H̄s + k2 H̄s = 0
– ∇
√
* where k = ω µε
¯ 2 may be broken in to 2 parts
• The 3-dimensional ∇
¯ 2 Ēs + ∂ 2 Ē2s
¯ 2 Ēs = ∇
– ∇ t ∂z
¯ 2 Ēs is the 2-dimensional Laplacian in the transverse plane
– where ∇ t

Waveguides and Transmission Lines

General Formulation for Guided Waves;Continued

• Now if wave propagation is in +z direction, all field components will have the z depen-
dance of the form e−γz
∂ 2 e−γz
– ∂ z2
= (−γ)(−γ)e−γz = γ 2 e−γz
∂ 2 Ēs
– or ∂ z2
= γ 2 Ēs
∂2
– or ∂ z2
∼ γ 2 and ∂
∂z ∼ −γ

• above wave equation can be re-written as

¯ 2 Ēs + ∂ 2 Ē2s = ∇
– ∇ ¯ 2 Ēs = −k2 Ēs
t ∂z
¯ 2 Ēs + γ 2 Ēs = −k2 Ēs
– or ∇ t

• =⇒
¯ 2 Ēs = −(γ 2 + k2 )Ēs = −h2 Ēs
– ∇ t
¯
– ∇2 H̄s = −(γ 2 + k2 )H̄s = −h2 H̄s
t

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 5 Page 5 of 17
Waveguides and Transmission Lines
General Formulation for Guided Waves;Continued

• To find the solution, the usual procedure is

– Determine the two components of Ē and H̄ (usually z-components) that satisfy the
wave equations and the boundary conditions
– Determine other components of Ē and H̄ by using Maxwell’s Equations
– To determine these other components it would be a good idea to express all other
components in terms of Ēzs and H̄zs

Waveguides and Transmission Lines

General Formulation for Guided Waves;Continued

• to express all other components in terms of Ēzs and H̄zs ,

• The Curl Equations with e jωt time dependence and ∂

∂z ∼ −γ
¯ × Ēs = − jω µ H̄s and ∇
– ∇ ¯ × H̄s = jωε Ēs
âx ây âz âx ây âz
∂ ∂ ∂ ∂
– ∂x ∂y −γ = − jω µ H̄s and ∂x ∂y −γ = jωε Ēs
Exs Eys Ezs Hxs Hys Hzs
• first equation implies

– âx [ ∂∂y Ezs − (−γ)Eys ] = − jω µHxs âx or ∂

∂ y Ezs + γEys = − jω µHxs
– −ây [ ∂∂x Ezs − (−γ)Exs ] = − jω µHys ây or - ∂∂x Ezs − γExs = − jω µHys
– âz [ ∂∂x Eys − ∂∂y Exs ] = − jω µHzs âz or ∂ ∂
∂ x Eys − ∂ y Exs = − jω µHzs

• similarly 2nd equation implies ∂∂y Hzs +γHys = jωε Ēxs , - ∂∂x Hzs −γHxs = jωε Ēys , ∂
∂ x Hys −
∂
∂ y Hxs = jωε Ēzs

Waveguides and Transmission Lines

General Formulation for Guided Waves;Continued

¯ × Ēs = − jω µ H̄s
• ∇
∂
1. ∂ y Ezs + γEys = − jω µHxs
2. − ∂∂x Ezs − γExs = − jω µHys
∂ ∂
3. ∂ x Eys − ∂ y Exs = − jω µHzs

¯ × H̄s = jωε Ēs

• ∇
∂
4. ∂ y Hzs + γHys = jωε Ēxs

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 5 Page 6 of 17
 5. − ∂∂x Hzs − γHxs = jωε Ēys
∂ ∂
6. ∂ x Hys − ∂ y Hxs = jωε Ēzs

• 1st and 5th equations give

∂
– ∂ y Ezs + γEys = − jω µHxs and − ∂∂x Hzs − γHxs = jωε Ēys
– or Eys = − 1γ jω µHxs − 1γ ∂∂y Ezs and − jωε
1 ∂
∂ x Hzs −
γ
jωε Hxs = Ēys
– or − 1γ jω µHxs − 1γ ∂∂y Ezs = − jωε
1 ∂
∂ x Hzs −
γ
jωε Hxs
– rearranging
     
jω µ j2 ω 2 µε−γ 2 −k2 −γ 2
– − 1γ ∂∂y Ezs + 1 ∂
jωε ∂ x Hzs = γ − γ
jωε Hxs = jγωε Hxs = jγωε Hxs

– − jωε ∂∂y Ezs + γ ∂∂x Hzs = − γ 2 + k2 Hxs

Waveguides and Transmission Lines

General Formulation for Guided Waves;Continued

• Similarly other components can also be expressed in terms of Ēzs and H̄zs and can be
given as
 
∂ Ezs ∂ Hzs
– Exs = γ 2−1
+k2
γ ∂x + jω µ ∂y
 
1 ∂ Ezs ∂ Hzs
– Eys = γ 2 +k 2 −γ ∂y + jω µ ∂x
 
– Hxs = 1
γ 2 +k2
jωε ∂∂Eyzs − γ ∂∂Hxzs
 
– Hys = −1
γ 2 +k2
jωε ∂∂Exzs + γ ∂∂Hyzs

• and

¯ 2 Ezs = −h2 Ezs

– ∇ t
¯
– ∇2 Hzs = −h2 Hzs
t
– where h2 = γ 2 + k2 and k2 = ω 2 µε

Waveguides and Transmission Lines

TEM Modes

• Ē and H̄ are transverse to the direction of wave propagations

• Ēzs = 0 and Hzs = 0

• substituting this in above equations

– either Exs = Eys = Hxs = Hys = 0 or γ 2 + k2 = 0

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 5 Page 7 of 17
 – 1st case is of no importance, as all components equal zero
* 2nd implies
* γT EM = ± jk
• thus for TEM
¯ 2 Ēs = −h2 Ēs = 0 where h2 = γ 2 + k2
– ∇ t
¯ 2 H̄s = −h2 H̄s = 0
– ∇ t

¯ 2 Ēs = 0 and ∇
• thus∇ ¯ 2 H̄s = 0
t t

– i.e. both Ē and H̄ fields satisfy Laplace Equations so that both have the spatial
distribution of 2-dimensional static fields.
– Thus the fields satisfy static laws.

Waveguides and Transmission Lines

TEM Modes;Continued

• As a consequence, transmission lines (TEM lines) can be analyzed in terms of voltages

and currents using the simplified circuit theory.

Waveguides
Rectangular Waveguides

Rectangular Waveguides
Rectangular Waveguides

• Metal Waveguide

• Hollow metal pipe of rectangular cross-section

• cross-section stays the same along the length of the waveguide

• Rectangular waveguides (and other metal waveguides) can not support TEM mode

– As TEM waves have transverse variations like static fields

– and Static fields cannot exist inside a single conductor (metal waveguide)

• TM and TE modes are possible in metal waveguides

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 5 Page 8 of 17
Rectangular Waveguides

Rectangular Waveguides
Summary of General Formulation for Guided Waves

• All other components can be expressed in terms of Ēzs and H̄zs and can be given as
 
∂ Ezs ∂ Hzs
– Exs = γ 2−1
+k2
γ ∂x + jω µ ∂y
 
1 ∂ Ezs ∂ Hzs
– Eys = γ 2 +k 2 −γ ∂y + jω µ ∂x
 
– Hxs = 1
γ 2 +k2
jωε ∂∂Eyzs − γ ∂∂Hxzs
 
– Hys = −1
γ 2 +k2
jωε ∂∂Exzs + γ ∂∂Hyzs

• and
¯ 2 Ezs = −h2 Ezs
– ∇ t
¯
– ∇2 Hzs = −h2 Hzs
t
– where h2 = γ 2 + k2 and k2 = ω 2 µε

Rectangular Waveguides

• we start with
¯ 2 Ezs = −h2 Ezs
– ∇ t
¯
– ∇2 Hzs = −h2 Hzs
t

• 1st eq implies
∂2 2
– E + ∂∂y2 Ezs
∂ x2 zs
= −h2 Ezs
– now using the method of separation of variables
−γz
* Ezs (x, y, z) = X(x)Y (y)Z(z) = X(x)Y (y)e

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 5 Page 9 of 17
 ∂2 ∂2
– ∂x 2 E zs + E
∂ y2 zs
= −h2 Ezs implies
∂2 2
– ∂ x2
(XYe−γz ) + ∂∂y2 (XYe−γz ) = −h2 XYe−γz
– X”Y + XY ” = −h2 XY

• dividing both sides by XY

– X”
X + YY” = −h2 = −(γ 2 + k2 )
– X”
X + YY” + γ 2 = −k2

Rectangular Waveguides

• X”
X + YY” + γ 2 = −k2 or X”
X + YY” + γ 2 = −kx2 − ky2 − kz2

• As R.H.S is a constant each term on L.H.S must be a constant

X” Y”
• or X = −kx2 , Y = −ky2 , γ 2 = −kz2

• or X” + kx2 X = 0 and Y ” + ky2Y = 0

– so m2 + kx2 = 0 =⇒ m2 = −kx2 =⇒ m = ± jkx

• X = Ae jkx x + Be− jkx x = A [coskx x + jsinkx x] + B [coskx x − jsinkx x]

• = (A + B)coskx x + j(A − B)sinkx x

• X = C1 coskx x +C2 sinkx x where C1 = A + B , C2 = j(A − B)

• X = C1 coskx x +C2 sinkx x, this solution is selected because the wave is not travelling wave
in x-direction

• Similarly

• Y = C3 cosky y +C4 sinky y , this solution is selected because the wave is not travelling wave
in y-direction

Rectangular Waveguides

• As X = C1 coskx x +C2 sinkx x ,Y = C3 cosky y +C4 sinky y and Ezs (x, y, z) = XYe−γz

• implies

– Ezs (x, y, z) = [C1 coskx x +C2 sinkx x] [C3 cosky y +C4 sinky y] e−γz

• Similarly

– Hzs (x, y, z) = [B1 coskx x + B2 sinkx x] [B3 cosky y + B4 sinky y] e−γz

 
−1 ∂ Ezs ∂ Hzs
• Exs = h2
γ ∂ x + jω µ ∂ y

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 5 Page 10 of 17
  
1 ∂ Ezs ∂ Hzs
• Eys = h2
−γ ∂ y + jω µ ∂ x
 
• Hxs = h12 jωε ∂∂Eyzs − γ ∂∂Hxzs
 
−1 ∂ Ezs ∂ Hzs
• Hys = h2 jωε ∂ x + γ ∂ y

• where h2 = γ 2 + k2 = kx2 + ky2

Rectangular Waveguides
TM Modes

Rectangular Waveguides
TM Modes
• H̄ is transverse to the direction of wave propagation
• Hz = 0

• Start by determining Ez
• Once Ez is determined use the above equations to find other components

Boundary Conditions
• At the walls (perfect conductor) of the waveguide, tangential components of E field must
be continues (as the tangential E diminishes in a perfect conductor
– Etan |y=0,y=b,x=0,x=a = 0

Rectangular Waveguides
TM Modes; Continued
• Etan |y=0,y=b,x=0,x=a = 0
• As Ez is tangential to all four waveguide walls
– y=0 bottom wall, Ezs |y=0 = 0
– y=b top wall, Ezs |y=b = 0
– x=0 right wall, Ezs |x=0 = 0
– x=a left wall, Ezs |x=a = 0

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 5 Page 11 of 17
Rectangular Waveguides
TM Modes; Continued

• As Ezs (x, y, z) = [C1 cos (kx x) +C2 sin (kx x)] [C3 cos (ky y) +C4 sin (ky y)] e−γz

– Ezs |y=0 = 0 =⇒ C3 cos (ky y) +C4 sin (ky y) y=0

= 0 =⇒ C3 .1 +C4 .0 = 0 =⇒ C3 =
0
– Ezs |y=b = 0 =⇒ sin (ky y) y=b
= sin (ky b) = 0 =⇒ ky b = nπ , n = 1, 2, 3, 4, .....
– Ezs |x=0 = 0 =⇒ C1 cos (kx x) +C2 sin (kx x)|x=0 = 0 =⇒ C1 .1 +C2 .0 = 0 =⇒ C1 =
0
– Ezs |x=a = 0 =⇒ sin (kx x)|x=a = sin (kx a) = 0 =⇒ kx a = mπ , m = 1, 2, 3, 4, .....

• thus

• =⇒ Ezs (x, y, z) = C2C4 sin (kx x) sin (ky y) e−γz , ∵ C1 = C3 = 0

−γz
• or =⇒ Ezs (x, y, z) = E0 sin mπ nπ
, ∵ kx = mπ nπ


a x sin b y e a , ky = b , E0 = C2C4

Rectangular Waveguides
TM Modes; Continued
mπ nπ
e−γz



• Hzs = 0 and Ezs = E0 sin a x sin b y

• other components can easily be determined using the previously derived equations, namely

 
• Exs = −1
h2
γ ∂∂Exzs + jω µ ∂∂Hyzs
 
• Eys = 1
h2
−γ ∂∂Eyzs + jω µ ∂∂Hxzs
 
• Hxs = 1
h2
jωε ∂∂Eyzs − γ ∂∂Hxzs
 
• Hys = −1
h2
jωε ∂∂Exzs + γ ∂∂Hyzs

 
• Exs = −1
h2
γ ∂∂Exzs
 
• Eys = 1
h2
−γ ∂∂Eyzs
 
• Hxs = 1
h2
jωε ∂∂Eyzs
 
• Hys = −1
h2
jωε ∂∂Exzs

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 5 Page 12 of 17
Rectangular Waveguides
TM Modes; Continued
mπ nπ
e−γz



• Hzs = 0 and Ezs = E0 sin a x sin b y
 
∂ Ezs
• Exs = −1
h2
γ ∂x
 
• Eys = 1
h2
−γ ∂∂Eyzs
 
• Hxs = 1
h2
jωε ∂∂Eyzs
 
• Hys = −1
h2
jωε ∂∂Exzs

 
∂ Ezs
• Exs = −1 = −1 mπ nπ
e−γz = −1 mπ mπ nπ
e−γz







h2
γ ∂x γ ∂ E0 sin
h2 ∂ x a x sin b y h2
γE0 a cos a x sin b y

• Similarly
−γ nπ
mπ

nπ

−γz
– Eys = h2 b E0 sin a x cos b y e
jωε nπ
mπ

nπ

−γz
– Hxs = h2 b E0 sin a x cos b y e
− jωε mπ
mπ

nπ

−γz
– Hys = h2 a E0 cos a x sin b y e

Rectangular Waveguides
TM Modes; Continued
• Thus for TM, field components are given as

−γz
– Hzs = 0 , Ezs = E0 sin mπ nπ


a x sin b y e

• and
−γ mπ
mπ

nπ

−γz
– Exs = h 2 a E 0 cos a x sin b y e
−γ nπ
mπ

nπ

−γz
– Eys = h2 b E0 sin a x cos b y e
jωε nπ
mπ

nπ

−γz
– Hxs = h 2 b E 0 sin a x cos b y e
− jωε mπ
mπ

nπ

−γz
– Hys = h2 a E0 cos a x sin b y e

• where
mπ 2

2
– h2 = γ 2 + k2 = kx2 + ky2 = + nπ


a b

• Each set of integers m and n lead to a different field configuration or mode, written as
T Mmn mode.
• Thus an infinite number of different modes are possible in a rectangular waveguide

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 5 Page 13 of 17
Rectangular Waveguides
TM Modes; Continued

• also (m, n) ̸= (0, 0), (m, n) ̸= (1, 0), (m, n) ̸= (0, 1)

– if either m = 0 or n = 0, Ez = 0 and Hz = 0
– all field components are zero as rectangular waveguide can not support TEM mode

Propagaton Constant
mπ 2

2 mπ 2

2
• γ 2 = h2 − k2 = kx2 + ky2 − k2 = + nπ − k2 + nπ − ω 2 µε



a b = a b
q
mπ 2 nπ 2



• γ= a + b − ω 2 µε

• Depending upon the operating frequency ω or f , we can have 3 cases

– Case 1: Cutoff γ = 0
– Case 2: Evanescent γ = α
– Case 3: Propagation γ = jβ

Rectangular Waveguides TM Modes

Propagation Constant TM; Continued

• Case 1: Cutoff (γ = 0) occurs when (ω = ωc )

– here γ = 0 i.e . no propagation takes place at this frequency

– γ|ω=ωc = 0 =⇒ α + jβ = 0
– ωc is called cutoff angular frequency
q
mπ 2 nπ 2



– As γ = a + b − ω 2 µε
q
mπ 2

2
+ nπ


– γ|ω=ωc = 0 = a b − ωc2 µε

2 nπ 2
– or mπ − ωc2 µε = 0


a + b
h
i
2 1 mπ 2 nπ 2


– =⇒ ωc = µε a + b
q
mπ 2

2
– =⇒ ωc = √1µε + nπ


a b

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 5 Page 14 of 17
Rectangular Waveguides TM Modes
Propagation Constant TM; Continued

• Case 2: Evanescent γ = α occurs when (ω < ωc )

q
mπ 2 nπ 2



– As γ = a + b − ω 2 µε
– in this case for given ω
mπ 2 nπ 2 2



* a + b − ω µε > 0
mπ 2 nπ 2 2 2



* a + b h> ω µε = k i h
2= 1 mπ 2 nπ 2 mπ 2

i
nπ 2
= µεωc2




* using ω c µε a + b =⇒ a + b

2 µε − ω 2 µε =⇒ α = ω √ µε √
q 2 q
p ωc µε ωc2
* =⇒ α = ω c ω 2 µε − 1 = ω µε ω2
−1 =
√
q 2
f
ω µε fc2 − 1
* where f is the frequency of operation and fc is the cutoff frequency
q
mπ 2 nπ 2



– in this case a + b − ω 2 µε = real =⇒ β = 0
– No wave propagation, only attenuation. this case occurs when (ω < ωc )
– Such nonpropagating modes are called evanescent modes

Rectangular Waveguides TM Modes

Propagation Constant TM; Continued

• Case 3: Propagation γ = jβ occurs when (ω > ωc )

– here γ = imaginary = jβ , α = 0
q
mπ 2

2
+ nπ


– As γ = a b − ω 2 µε = jβ

2 nπ 2
– i.e mπ − ω 2 µε < 0


a + b

2
2
– mπ a + nπb < ω 2 µε = k2
r h n
2
oi
nπ 2
– − ω 2 µε − mπ a + b = jβ
r n
2
o
nπ 2
– β = ω 2 µε − mπ a + b
h
2
i
nπ 2
– but mπ a + b = ωc2 µε
√ √
q q
p ω 2 µε fc2
– =⇒ β = ω 2 µε − ωc2 µε = ω µε 1 − ωc2 µε = ω µε 1 − f2
– where f is the frequency of operation and fc is the cutoff frequency

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 5 Page 15 of 17
Rectangular Waveguides TM Modes
Cutoff Frequency

• is the frequency fc above which propagation takes place

• below fc attenuation occurs

q
1 mπ 2 nπ 2
ωc



• fc = 2π = 2π µε
√
a + b

• for each mode there is a corresponding cutoff frequency

• Among all the TM modes, it can be seen from the above equation that T M11 has the
lowest cutoff frequency

Waveguide is a High pass filter

• The rectangular waveguide is a high pass filter because

– below cutoff frequency there is only attenuation and no wave propagation

– above cutoff frequency wave propagation occurs

Rectangular Waveguides TM Modes

• cutoff wavelength=λc = u′ / fc where u′ = √1

µε

• Phase velocity =u p = ω/β

2π
• Wavelength in the guide = guide wavelength=λ = β

Intrinsic Impedance

• the intrinsic impedance of the mode (for γ = jβ )

Ex E β
– ηT M = = − Hyx = ωε
Hy
√
q q q
f2 fc2
– ηT M = µε 1 − fc2 ∵ β = ω µε 1 − f2

Rectangular Waveguides TM Modes

• The pattern for the T M21 mode is that of the two T M11 modes side by side and of opposite
sense.

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 5 Page 16 of 17
 •

Rectangular Waveguides
Transverse Electric (TE) Modes

Rectangular Waveguides
Transverse Electric (TE) Modes

• Electric field is transverse to the direction of wave propagation

• No component of electric field can lie in the direction of wave propagation

– i.e. Ez = 0

• Figures in this lecture are from

– Sadiku MN. Elements of electromagnetics

– Demarest KR. Engineering electromagnetics
– Ramo S, Whinnery JR, Van Duzer T. Fields and waves in communication electronics

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 5 Page 17 of 17