Analysis of

Transmission Lines

Dr. H. Rahaman, Associate Professor

Department of Electronics and Communication, BKBIET Pilani
 Objective

 Analysis of Transmission line model.

 Circular waveguide.
 Find different mode – TE and TM.
 Different parameters related to each mode
 Recap

 We discussed on rectangular waveguides.

 Concept of modes- TE and TM modes.
 Design of waveguides for dominant modes.
 Higher modes are also the solution of the rectangular
waveguide.
 Waveguide components
Waveguide – Basically a hollow metal pipe in circular form. Based
on dimension, it allows rf frequencies in a certain range.

Straight sections waveguide Coupler waveguide

Mode converter Terminator Horn antenna

 Circular Waveguide
A hollow, round metal pipe is called
a circular waveguide. It supports TE
and TM modes.

The waveguide is filled with a material of permittivity 𝜖 and

permeability 𝜇 .
Operators in different coordinate system

Cylindrical coordinate system is used in circular waveguide system.

𝑬 = 𝒇𝒏(𝒓, ∅, 𝒛) 𝑯 = 𝒇𝒏(𝒓, ∅, 𝒛)
 TE mode
TE mode is characterized by Ez = 0, Hz  0. The wave equation is
𝜵𝟐 𝑯𝒛 + 𝒌𝟐 𝑯𝒛 = 0 where 𝑯𝒛 = 𝒉𝒛 𝝆, ∅ 𝒆−𝒋𝜷𝒛

Recalling two dimensional 𝜕2 𝜕2

+ +
2
𝑘𝑐 ℎ𝑧 = 0 𝒌𝒄 = 𝒌 𝟐 − 𝜷𝟐
wave equation. 𝜕𝑥 2 𝜕𝑦 2
Cutoff wave number
for TE waves .
The wave equation in 1 𝜕 𝜕 1 𝜕2 2
cylindrical coordinates. 𝜌 + + 𝑘 ℎ𝑧 𝜌, ∅ = 0
𝜌 𝜕𝜌 𝜕𝜌 2
𝜌 𝜕∅ 2 𝑐

Using method of separation of variables.

𝒉𝒛 𝜌, ∅ = 𝑹 𝝆 𝑷(∅)
2𝑃 2
1 𝑑 𝑑𝑅 1
2
𝑑
𝜌 + + 𝑘𝑐 = 0
𝜌𝑅 𝑑𝜌 𝑑𝜌 𝜌 𝑃 𝑑∅ 2

Each of the term must be equal to constant. 𝑑𝜌

The equation becomes

𝑑 𝑑𝑅
𝜌 𝜌 𝑐 ∅
𝑑2 𝑃
= −𝑘 2 𝑃 𝑑∅2 ∅

+ 𝜌2 𝑘 2 − 𝑘 2 𝑅 = 0
 TE mode
𝑑2𝑃 2𝑃 𝑑 𝑑𝑅
= −𝑘 ∅ 𝜌 𝜌 + 𝜌2 𝑘𝑐2 − 𝑘∅2 𝑅 = 0
𝑑∅2 𝑑𝜌 𝑑𝜌
The solution is

The solution is periodic in hz, so kΦ ℎ𝑧 𝜌, ∅ = ℎ𝑧 𝜌, ∅ ± 2𝑛𝜋

in must be an integer n.

𝑑 𝑑𝑅
𝜌 𝜌 + 𝜌2 𝑘𝑐2 − 𝑛2 𝑅 = 0
𝑑𝜌 𝑑𝜌
This is Bessel’s differential equation of order n. The solution is

Where 𝐽𝑛 and 𝑌𝑛 are nth order Bessel’s function of 1st and 2nd kinds. The
transverse solution is
ℎ𝑧 𝜌, ∅ = 𝑅 𝜌 𝑃(∅)
or
ℎ𝑧 𝜌, ∅ = 𝐴𝑆𝑖𝑛 𝑛∅ + 𝐵𝐶𝑜𝑠 𝑛∅ (𝐶𝐽𝑛 𝑘𝐶 𝜌 + 𝐷𝑌𝑛 𝑘𝐶 𝜌 )

A, B, C, D are constants and to be decided by boundary conditions,

𝐽𝑛 𝑘𝐶 𝜌 Bessel’s Function
Bessel’s function of first
kind.

𝑘𝐶 𝜌

𝑌𝑛 𝑘𝐶 𝜌

Bessel’s function of second

kind.

𝑘𝐶 𝜌
 TE mode
ℎ𝑧 𝜌, ∅ = 𝐴𝑆𝑖𝑛 𝑛∅ + 𝐵𝐶𝑜𝑠 𝑛∅ (𝐶𝐽𝑛 𝑘𝐶 𝜌 + 𝐷𝑌𝑛 𝑘𝐶 𝜌 )
From boundary condition 𝑌𝑛 𝑘𝐶 𝜌 becomes infinite at 𝜌 = 0. This term is
physically not acceptable. So, D=0

ℎ𝑧 𝜌, ∅ = 𝐴𝑆𝑖𝑛 𝑛∅ + 𝐵𝐶𝑜𝑠 𝑛∅ 𝐶𝐽𝑛 𝑘𝐶 𝜌

Recalling Maxwell’s 𝛻 × 𝐸 = −𝑗𝜔𝜇𝐻

equation. 𝛻 × 𝐻 = 𝑗𝜔𝜖𝐸
Rectangular Cylindrical
coordinates Coordinates
 TE mode
Enforcing boundary condition

Ez = 0, Hz  0 on the waveguide wall.

𝐸∅ 𝜌, ∅ = 0 𝑎𝑡 𝜌=0

𝐻𝑧 = ℎ𝑧 𝜌, ∅ 𝑒 −𝑗𝛽𝑧

Where 𝐽𝑛′ 𝑘𝐶 𝜌 refers to the derivative of 𝐽𝑛 𝑘𝐶 𝜌 w.r.t. arguments.

For 𝐸∅ 𝜌, ∅ = 0 𝑎𝑡 𝜌 = 0, we must have 𝐽𝑛′ 𝑘𝐶 𝜌 = 0

Roots of 𝐽𝑛′ 𝑥 are defined as 𝜌′𝑛𝑚 where 𝜌′𝑛𝑚 is the mth root
of 𝐽𝑛′ . Then 𝑘𝐶 will have value
′
𝝆𝒏𝒎
𝑲𝒄𝒎𝒏 =
𝒂
n=0,1,2,3,… and m=1,2,3,…
 TE mode ′
𝝆𝒏𝒎
𝑲𝒄𝒎𝒏 =
Zeros of 𝐽𝑛′ 𝑘𝐶 𝜌 for 𝑇𝐸𝑚𝑛 modes in circular 𝒂
waveguide.

′
The mode 𝜌𝑛𝑚 with the lowest value is the dominant mode and is called TE11 mode.
All the transverse field components in circular waveguide are given by

Ez = 0, Hz  0 𝐻𝑧 = ℎ𝑧 𝜌, ∅ 𝑒 −𝑗𝛽𝑧
ℎ𝑧 𝜌, ∅ = 𝐴𝑆𝑖𝑛 𝑛∅ + 𝐵𝐶𝑜𝑠 𝑛∅ 𝐶𝐽𝑛 𝑘𝐶 𝜌
 TE mode
The mode propagation constant is 𝜌𝑛
′
𝑚
given by 𝛽𝑔 = 𝜔 2 𝜇𝜖 − 𝑘𝑐2 = 𝜔 2 𝜇𝜖 − 2
𝑎
𝑘
The cut off condition is given by
𝑐 𝑐
′
𝜌𝑛𝑚 𝑣𝑝 𝜌′𝑛𝑚
𝑓 = or 𝑓𝑐 𝑛𝑚 = =
2𝜋 𝜇𝜖 2𝜋𝑎 𝜇𝜖 2𝜋𝑎

Where 𝑣𝑝 = 1/ 𝜇𝜖 is the phase velocity in unbounded dielectric.

𝜔 𝑣𝑝
The phase velocity in the waveguide 𝑣𝑔 = =
𝛽
is 𝑔 2
𝑓𝑐
1− 𝑓

The guided wavelength is The wave impedance of the waveguide is

2𝜋 𝜆 𝜂
𝜆𝑔 = 𝛽 = 𝑍𝑔 =
2 2 𝑣
𝑔
𝑓𝑐 𝑓 𝜔𝜇 𝜇
1− 1 − 𝑓𝑐 =
𝑝
𝜂= =
𝑓 𝜆 𝑓 𝑘 𝜖

For unbounded medium

 TE Mode- Problem
The radius of the air filled circular waveguide is 5cm.
Find cut off frequency and other parameters at 3 GHz.
𝒌𝑪 𝝆 = 𝟏. 𝟖𝟒𝟏 1.841 1.841
𝐾𝑐 𝑚𝑛 = = = 36.82
𝜌 5𝑥10− 2
Cut off frequency
𝑣𝑝 𝐾𝐶𝑚𝑛 𝐶𝐾𝐶𝑚𝑛 3𝑥108 𝑥 36.82
𝑓𝑐 𝑛𝑚 = = = = 1.758𝑥109 𝐻𝑧
1
=
1
= 1.234
2𝜋 2𝜋 2𝜋
𝑓𝑐 2
1.758
3
2
Phase velocity
1−
𝑓
1− 𝑣𝑝
𝑣𝑔 = = 3𝑥108 𝑥1.23 = 3.70𝑥108 𝑚/𝑠
2
𝑓𝑐
1−
Guide wavelength 𝑓
𝜆 𝜆 1.234𝑥3𝑥1010 Phase constant
𝑔 = = 3𝑥109 = 12.34𝑐𝑚
𝑐 2
2𝜋
𝑓 𝛽𝑔 = = 0.509𝑟𝑎𝑑/𝑐𝑚
1− 𝑓 𝜆𝑔
𝜂
𝑍𝑔 = = 120𝜋𝑥1.23 = 464Ω
Waveguide impedance 2
𝑓𝑐
1−
𝑓
 TM mode
TE mode is characterized by Ez  0, Hz = 0. The wave equation is
𝜵𝟐 𝑬 𝒛 + 𝒌 𝟐 𝑬𝒛 = 0 where 𝑬𝒛 = 𝒆𝒛 𝝆, ∅ 𝒆−𝒋𝜷𝒛

Recalling two dimensional 𝜕2 𝜕2

+ +
2
𝑘𝑐 𝑒𝑧 = 0 𝒌𝒄 = 𝒌 𝟐 − 𝜷𝟐
wave equation. 𝜕𝑥 2 𝜕𝑦 2
Cutoff wave number
for TM waves .
The wave equation in 1 𝜕 𝜕 1 𝜕2 2
cylindrical coordinates. 𝜌 ∅ 𝑒𝑧 𝜌, ∅ = 0
𝜌 𝜕𝜌 𝜕𝜌 + + 𝑘𝑐
2
𝜌 𝜕 2

Using method of separation of variables.

𝒆𝒛 𝜌, ∅ = 𝑹2 𝝆 𝑷(∅)
2 𝑑2𝑃
1 𝑑 𝑑𝑅 1
𝜌 + + 𝑘𝑐 = 0
𝜌𝑅 𝑑𝜌 𝑑𝜌 𝜌 𝑃 𝑑∅2
Each of the term must be equal to constant. 𝑑𝜌

The equation becomes

𝑑 𝑑𝑅
𝑑2 𝑃
= −𝑘 2 𝑃 𝑑∅2 ∅

+ 𝜌2 𝑘 2 − 𝑘 2 𝑅 = 0
 TM mode
𝑑2𝑃 2𝑃 𝑑 𝑑𝑅
= −𝑘 ∅ 𝜌 𝜌 + 𝜌2 𝑘𝑐2 − 𝑘∅2 𝑅 = 0
𝑑∅2 𝑑𝜌 𝑑𝜌
The solution is

The solution is periodic in ez, so kΦ 𝑒𝑧 𝜌, ∅ = ℎ𝑧 𝜌, ∅ ± 2𝑛𝜋

in must be an integer n.

𝑑 𝑑𝑅
𝜌 𝜌 + 𝜌2 𝑘𝑐2 − 𝑛2 𝑅 = 0
𝑑𝜌 𝑑𝜌
This is Bessel’s differential equation of order n. The solution is

Where 𝐽𝑛 and 𝑌𝑛 are nth order Bessel’s function of 1st and 2nd kinds. The
transverse solution is
𝑒𝑧 𝜌, ∅ = 𝑅 𝜌 𝑃(∅)
or
𝑒𝑧 𝜌, ∅ = 𝐴𝑆𝑖𝑛 𝑛∅ + 𝐵𝐶𝑜𝑠 𝑛∅ (𝐶𝐽𝑛 𝑘𝐶 𝜌 + 𝐷𝑌𝑛 𝑘𝐶 𝜌 )

A, B, C, D are constants and to be decided by boundary conditions,

 TM mode
ℎ𝑧 𝜌, ∅ = 𝐴𝑆𝑖𝑛 𝑛∅ + 𝐵𝐶𝑜𝑠 𝑛∅ (𝐶𝐽𝑛 𝑘𝐶 𝜌 + 𝐷𝑌𝑛 𝑘𝐶 𝜌 )
From boundary condition 𝑌𝑛 𝑘𝐶 𝜌 becomes infinite at 𝜌 = 0. This term is
physically not acceptable. So, D=0

𝑒𝑧 𝜌, ∅ = 𝐴𝑆𝑖𝑛 𝑛∅ + 𝐵𝐶𝑜𝑠 𝑛∅ 𝐶𝐽𝑛 𝑘𝐶 𝜌

Recalling Maxwell’s 𝛻 × 𝐸 = −𝑗𝜔𝜇𝐻

equation. 𝛻 × 𝐻 = 𝑗𝜔𝜖𝐸
Rectangular Cylindrical
coordinates Coordinates
 TM mode
Enforcing boundary condition

Hz = 0, Ez  0 on the waveguide wall.

𝐸∅ 𝜌, ∅ = 0 𝑎𝑡 𝜌=0

𝐸𝑧 = 𝑒𝑧 𝜌, ∅ 𝑒 −𝑗𝛽𝑧

Consider 𝐽𝑛 𝑘𝐶 𝜌 w.r.t. arguments.

For 𝐸∅ 𝜌, ∅ = 0 𝑎𝑡 𝜌 = 0, we must have 𝐽𝑛 𝑘𝐶 𝜌 = 0

Roots of 𝐽𝑛 𝑥 are defined as 𝜌𝑛𝑚 where 𝜌𝑛𝑚 is the mth root of

𝐽𝑛 . Then 𝑘𝐶 will have value

𝝆𝒏𝒎
𝑲𝒄𝒎𝒏 =
𝒂
n=0,1,2,3,… and m=1,2,3,…
 TM mode 𝝆𝒏𝒎
𝑲𝒄𝒎𝒏 =
Zeros of 𝐽𝑛 𝑘𝐶 𝜌 for 𝑇𝑀𝑚𝑛 modes in circular 𝒂
waveguide.

′
The mode 𝜌𝑛𝑚 with the lowest value is the dominant mode and is called TE11 mode.
All the transverse field components in circular waveguide are given by

Hz = 0, Ez  0 𝐸𝑧 = 𝑒𝑧 𝜌, ∅ 𝑒 −𝑗𝛽𝑧
𝑒𝑧 𝜌, ∅ = 𝐴𝑆𝑖𝑛 𝑛∅ + 𝐵𝐶𝑜𝑠 𝑛∅ 𝐶𝐽𝑛 𝑘𝐶 𝜌
 TM mode
The mode propagation constant is 𝜌𝑛𝑚 2
given by 𝛽𝑔 = 𝜔 2 𝜇𝜖 − 𝑘𝑐2 = 𝜔 2 𝜇𝜖 −
𝑎
′
The cut off condition is given by 𝑘𝑐 𝜌𝑛𝑚 𝑣 𝜌
𝑓𝑐 = or 𝑓𝑐 𝑛𝑚 = = 𝑝 𝑛𝑚
2𝜋 𝜇𝜖 2𝜋𝑎 𝜇𝜖 2𝜋𝑎

Where 𝑣𝑝 = 1/ 𝜇𝜖 is the phase velocity in unbounded dielectric.

𝜔 𝑣𝑝
The phase velocity in the waveguide 𝑣𝑔 = =
𝛽
is 𝑔 2
𝑓𝑐
1− 𝑓

The guided wavelength is The wave impedance of the waveguide is

2𝜋 𝜆
𝜆𝑔 = = 𝛽 𝑓 2
𝑣𝑝 𝜇
𝛽 𝑔 2 𝑔 𝑐 𝜔𝜇
𝑓𝑐 𝑍𝑔 1− 𝜆= 𝜂= =
1− 𝑓 𝑓 𝑓 𝑘 𝜖
= =𝜂
𝜔𝜖 For unbounded medium
The mode with lowest cut off frequency is called the dominant mode. TE11 mode
is the dominant mode since it has smallest value of the product, 𝑘𝐶 𝜌 = 1.841
TM mode- Field Pattern
 Circular Waveguide- problem
1. Find the cutoff frequencies of different modes for an air filled
circular waveguide with radius, a=2 cm that carry energy at a frequency of
10 GHz. Find different parameters value with the interior of the waveguide
coated with gold of 30 cm in length.

2. Find the cutoff frequencies of different modes for a teflon filled

circular waveguide with radius, a=0.5 cm that carry energy at a frequency of
14 GHz. Find the overall loss in dB for 30 cm length of the waveguide.

Solution
 Summary

 Design of circular waveguide.

 Different modes: TE and TM modes of propagation.
 Dominant mode, higher order modes.
 Different parameters calculation like phase constant,
phase velocity, impedance & wavelength correspond
to each mode.
 TE & TM modes pattern in circular waveguides.
Thank you !