19-203-0602
MICROWAVE TECHNIQUES AND DEVICES
Module II
Waveguides - Rectangular Waveguide: TE waves, TM waves,
Velocities of propagation; dominant and degenerate modes,
Impossibility of TEM waves in wave guides; Power Transmission and
Power Losses in Rectangular Waveguides, Excitation of modes in
Rectangular Waveguides
Rectangular Cavity Resonator: Resonant frequency and Q factor,
Cavity excitation and tuning.
Microwave Hybrid Circuits: E plane Tee, H plane Tee, Hybrid Tee,
Hybrid Ring, Two hole directional coupler, Isolator, Circulator, Phase
shifter, Attenuator.
Scattering matrix representation: Properties of S matrix, S matrix
formulation of E plane Tee, H plane Tee, Magic Tee, Directional
coupler, Circulator
Course Outcomes:
On successful completion of teaching-learning and valuation activities, a student would be able
1. To understand the working of transmission lines and waveguides
2. To characterize Microwave networks and devices using S parameters
3. To acquire knowledge on Microwave solid state devices, tubes and Microwave Measurements
RECTANGULAR WAVEGUIDE
• Rectangular waveguides are the most commonly used waveguides. They consist of a hollow metallic structure with a
rectangular cross-section. A rectangular waveguide is usually constructed with a length of a > b, where b is the breadth of the
rectangle. A common trend for the dimension of a rectangular waveguide is a=2b
• Rectangular waveguides are one of the earliest type of the transmission lines. They are used in many applications. A lot of
components such as isolators, detectors, attenuators, couplers and slotted lines are available for various standard waveguide
bands between 1 GHz to above 220 GHz.
• A rectangular waveguide supports TM and TE modes but not TEM waves because we cannot define a unique voltage since
there is only one conductor in a rectangular waveguide.
In a rectangular waveguide, electromagnetic waves are reflected
from the walls. Since there is only one conductor present in a
rectangular waveguide, it does not support the transverse
electromagnetic (TEM) mode of propagation. Only TE and TM
modes are supported by rectangular waveguides.
RECTANGULAR WAVE GUIDE
• A waveguide is an electromagnetic feed line used in microwave
communications, broadcasting, and radar installations. A waveguide consists y
of a rectangular or cylindrical metal tube or pipe. The electromagnetic field
propagates lengthwise.
• Here we assume that inner surface is perfectly conducting and the region
inside the guide is lossless dielectric, from Maxwell’s equation b
𝝏𝑩 ---------- (1) a x
𝜵×𝑬 = −
𝝏𝒕
𝜕𝑫 z
𝜵×𝑯 = 𝑱+ ---------- (2)
𝜕𝑡
The harmonic equation for the same (J = 0)
𝜵 × 𝑬 = −𝒋𝝎𝝁𝑯 ---------- (3)
𝜵 × 𝑯 = 𝒋𝝎ε𝑬 ---------- (4)
𝜵 × 𝑬 = −𝒋𝝎𝝁𝑯 --------- (3) 𝜵 × 𝑯 = 𝒋𝝎ε𝑬 ---------- (4)
𝑎𝑥 𝑎𝑦 𝑎𝑧 𝑎𝑥 𝑎𝑦 𝑎𝑧
𝜕 𝜕 𝜕 𝜕 𝜕 𝜕
𝛻×𝐸 = = −𝑗𝜔𝜇(𝐻𝑥 + 𝐻𝑦 + 𝐻𝑧) 𝛻×𝐻 = = 𝑗𝜔ε(𝐸𝑥 + 𝐸𝑦 + 𝐸𝑧)
𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧
𝐸𝑥 𝐸𝑦 𝐸𝑧 𝐻𝑥 𝐻𝑦 𝐻𝑧
𝜕𝐸𝑧 𝜕𝐸𝑦 𝜕𝐸𝑥 𝜕𝐸𝑧 𝜕𝐸𝑦 𝜕𝐸𝑥 𝜕𝐻𝑧 𝜕𝐻𝑦 𝜕𝐻𝑥 𝜕𝐻𝑧 𝜕𝐻𝑦 𝜕𝐻𝑥
− 𝑎𝑥 + − 𝑎𝑦 + − 𝑎 = −𝑗𝜔𝜇(𝐻𝑥 + 𝐻𝑦 + 𝐻𝑧) − 𝑎𝑥 + − 𝑎𝑦 + − 𝑎𝑧 = 𝑗𝜔ε(𝐸𝑥 + 𝐸𝑦 + 𝐸𝑧)
𝜕𝑦 𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝑧 𝜕𝑦 𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑦
Equating x, y, z terms
𝜕𝐸𝑧 𝜕𝐸𝑦 𝜕𝐻𝑧 𝜕𝐻𝑦
− = −𝑗𝜔𝜇𝐻𝑥 ---------- (5) − = 𝑗𝜔ε𝐸𝑥 ---------- (8)
𝜕𝑦 𝜕𝑧 𝜕𝑦 𝜕𝑧
𝜕𝐸𝑥 𝜕𝐸𝑧 𝜕𝐻𝑥 𝜕𝐻𝑧
− = −𝑗𝜔𝜇𝐻𝑦 ---------- (6) − = 𝑗𝜔ε𝐸𝑦 ---------- (9)
𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑥
𝜕𝐸𝑦 𝜕𝐸𝑥 𝜕𝐻𝑦 𝜕𝐻𝑥
− = −𝑗𝜔𝜇𝐻𝑧 ---------- (7) − = 𝑗𝜔ε𝐸𝑧 ---------- (10)
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦
Let a s a s s u m e t h e f i e l d varying along
z-axis as 𝑒−𝛾𝑧
𝐻𝑥 = 𝐻𝑥0𝑒−𝛾𝑧
𝐸𝑥 = 𝐸𝑥0𝑒 −𝛾𝑧
𝐻𝑦 = 𝐻𝑦0𝑒 −𝛾𝑧
𝐸𝑦 = 𝐸𝑦0𝑒 −𝛾𝑧
𝜕𝐻𝑥
= −𝛾𝐻𝑥
𝜕𝑧
𝜕𝐸𝑥
= −𝛾𝐸𝑥 𝜕𝐻𝑦
𝜕𝑧
= −𝛾𝐻𝑦
𝜕𝑧
𝜕𝐸𝑦
= −𝛾𝐸𝑦
𝜕𝑧
𝜕𝐸𝑧 𝜕𝐻𝑧
(5) + 𝛾𝐸𝑦 = −𝑗𝜔𝜇𝐻𝑥 ---------- (11) (8) + 𝛾𝐻𝑦 = 𝑗𝜔ε𝐸𝑥 ---------- (14)
𝜕𝑦 𝜕𝑦
𝜕𝐸𝑧 𝜕𝐻𝑧
(6) 𝛾𝐸𝑥 + = 𝑗𝜔𝜇𝐻𝑦 (9) −𝛾𝐻𝑥 − = 𝑗𝜔ε𝐸𝑦 ---------- (15)
𝜕𝑥 ---------- (12) 𝜕𝑥
𝜕𝐸𝑦 𝜕𝐸𝑥 𝜕𝐻𝑦 𝜕𝐻𝑥
(7) − = −𝑗𝜔𝜇𝐻𝑧 ---------- (13) (10) − = 𝑗𝜔ε𝐸𝑧 ---------- (16)
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦
𝑗𝜔ε𝐸𝑥 1 𝜕𝐻𝑧
𝑗𝜔𝜇𝐻𝑦 1 𝜕𝐸𝑧 𝐻𝑦 = − ---------- (18)
(12) 𝐸𝑥 = − ---------- (17) 𝛾 𝛾 𝜕𝑦
𝛾 𝛾 𝜕𝑥
Substitute (18) in (17) 𝑗𝜔𝜇𝐻𝑦 1 𝜕𝐸𝑧
𝐸𝑥 = − ---------- (17)
𝑗𝜔𝜇 𝑗𝜔ε𝐸𝑥 1 𝜕𝐻𝑧 1 𝜕𝐸𝑧 𝛾 𝛾 𝜕𝑥
𝐸𝑥 = − −
𝛾 𝛾 𝛾 𝜕𝑦 𝛾 𝜕𝑥 𝑗𝜔ε𝐸𝑥 1 𝜕𝐻𝑧
𝐻𝑦 = − ---------- (18)
𝛾 𝛾 𝜕𝑦
𝜔2𝜇ε𝐸𝑥 − 𝑗𝜔𝜇 𝜕𝐻𝑧 − 1 𝜕𝐸𝑧
=
𝛾2 𝛾2 𝜕𝑦 𝛾 𝜕𝑥
𝜔2𝜇ε = − 𝑗𝜔𝜇 𝜕𝐻𝑧 − 1 𝜕𝐸𝑧
𝐸𝑥 1 + 2
𝛾 𝛾2 𝜕𝑦 𝛾 𝜕𝑥
𝛾2 𝑗𝜔𝜇 𝜕𝐻𝑧 1 𝜕𝐸𝑧
𝐸𝑥 = 2 − 2 −
𝛾 + 𝜔2𝜇ε 𝛾 𝜕𝑦 𝛾 𝜕𝑥
1 𝜕𝐻𝑧 𝜕𝐸𝑧
𝐸𝑥 = 2 −𝑗𝜔𝜇 − 𝛾
𝛾 + 𝜔2𝜇ε 𝜕𝑦 𝜕𝑥
Substitute ℎ2 = 𝛾 2 + 𝜔2𝜇ε
−𝑗𝜔𝜇 𝜕𝐻𝑧 𝛾 𝜕𝐸𝑧 𝑗𝜔ε 𝜕𝐸𝑧 𝛾 𝜕𝐻𝑧
𝐸𝑥 = − 𝐻𝑥 = −
ℎ2 𝜕𝑦 ℎ2 𝜕𝑥 ℎ2 𝜕𝑦 ℎ2 𝜕𝑥
𝑗𝜔𝜇 𝜕𝐻𝑧 𝛾 𝜕𝐸𝑧 −𝑗𝜔ε 𝜕𝐸𝑧 𝛾 𝜕𝐻𝑧
𝐸𝑦 = − 𝐻𝑦 = −
ℎ2 𝜕𝑥 ℎ2 𝜕𝑦 ℎ2 𝜕𝑥 ℎ2 𝜕𝑦
• These four field components are in terms of Ez
& Hz
−𝑗𝜔𝜇 𝜕𝐻𝑧 − 𝛾 𝜕𝐸𝑧 𝑗𝜔ε 𝜕𝐸𝑧 − 𝛾 𝜕𝐻𝑧 • IF we assume that both Ez and Hz components
𝐸𝑥 = 𝐻𝑥 =
ℎ2 𝜕𝑦 ℎ2 𝜕𝑥 ℎ2 𝜕𝑦 ℎ2 𝜕𝑥 are zero then all filed vanish or in other words
transverse electromagnetic (TEM) wave
𝑗𝜔𝜇 𝜕𝐻𝑧 𝛾 𝜕𝐸𝑧 −𝑗𝜔ε 𝜕𝐸𝑧 𝛾 𝜕𝐻𝑧 cannot exist in a wave guide
𝐸𝑦 = − 𝐻𝑦 = − • A wave pattern is possible if either 𝐸𝑧 ≠ 0 or
ℎ2 𝜕𝑥 ℎ2 𝜕𝑦 ℎ2 𝜕𝑥 ℎ2 𝜕𝑦
𝐻𝑧 ≠ 0.
• If 𝐸𝑧 = 0 then the electric field is transverse,
These equations gives the relationships among the but there is a non zero Hz, this is called
fields within the guide Transverse Electric (TE) wave
• Similarly for Transverse Magnetic (TM) wave
, 𝐻𝑧 = 0 and 𝐸𝑧 ≠ 0
• Thus a rectangular wave guide can support TE
or TM mode wave
Transverse Magnetic (TM) Waves in Rectangular wave guide
In general wave equation is
∇2𝐸 = 𝛾2𝐸 ---------- (1)
∇2𝐻 = 𝛾2𝐻 ---------- (2)
(1) 𝜕2𝐸 𝜕2𝐸 𝜕2𝐸
2
+ 2 + 2 = −𝝎𝟐𝝁ε𝑬 ---------- (4)
𝜕𝑥 𝜕𝑦 𝜕𝑧
𝛾= 𝒋𝝎𝝁 𝝈 + 𝒋𝝎ε ---------- (3) 𝜕2𝐻 𝜕2𝐻 𝜕2𝐻
(2) + 2 + 2 = −𝝎𝟐𝝁ε𝑯
---------- (5)
𝜕𝑥2 𝜕𝑦 𝜕𝑧
For dielectric σ=0
We have to consider only z – component since the wave is
𝛾= −𝝎𝟐𝝁ε travelling in z – direction
𝛾 2 = −𝝎𝟐𝝁ε 𝜕2𝐸 𝑧 𝜕2𝐸 𝑧 𝜕2𝐸 𝑧
2
+ 2 + 2 = −𝝎𝟐𝝁ε𝑬 𝒛 ---------- (6)
𝜕𝑥 𝜕𝑦 𝜕𝑧
𝜕2𝐻𝑧 𝜕2𝐻𝑧 𝜕2𝐻𝑧
2
+ 2
+ 2
= −𝝎𝟐𝝁ε𝑯𝒛 ---------- (7)
𝜕𝑥 𝜕𝑦 𝜕𝑧
ℎ2 = 𝛾 2 + 𝜔2𝜇𝜀
1 𝑑2𝑋 1 𝑑2𝑌
Let us assume 2
+ 2
+ ℎ2 = 0
𝑋 𝑑𝑥 𝑌 𝑑𝑦
𝐸𝑧 𝑥, 𝑦 = 𝐸𝑧0𝑒 −𝛾𝑧 ---------- (8)
1 𝑑2𝑌
1 𝑑2 𝑋
Where + ℎ2 = 𝐴2 − 2 = 𝐴2
𝑋 𝑑𝑥 2 𝑌 𝑑𝑦
𝐸𝑧0 = 𝑋𝑌
1 𝑑2𝑋
Substitute (8) in (6) 2
+ 𝐵2 = 0 𝐵2 = ℎ2 − 𝐴2
𝑋 𝑑𝑥
𝜕2 −𝛾𝑧
𝜕2 −𝛾𝑧
𝜕2
2
𝑋𝑌𝑒 + 2 𝑋𝑌𝑒 + 2 𝑋𝑌𝑒−𝛾𝑧 = −𝝎𝟐𝝁ε 𝑋𝑌𝑒−𝛾𝑧 Multiplying both sides by 𝑋
𝜕𝑥 𝜕𝑦 𝜕𝑧
𝑑2𝑋 𝑑2𝑌 𝑑2𝑋
𝑌𝑒 −𝛾𝑧 2
+ 𝑋𝑒 −𝛾𝑧
2
+ 𝛾 2𝑋𝑌𝑒 −𝛾𝑧 = −𝜔2𝜇𝜀𝑋𝑌𝑒−𝛾𝑧 + 𝐵2𝑋 = 0 ---------- (9)
𝑑𝑥 𝑑𝑦 𝑑𝑥 2
𝑑2𝑋 𝑑2𝑌 and
𝑌 2 + 𝑋 2 + [𝛾2 + 𝜔2𝜇𝜀]𝑋𝑌 = 0
𝑑𝑥 𝑑𝑦 𝑑2𝑌
2
+ 𝐴2𝑌 = 0 ---------- (10)
1
𝑑𝑦
Dividing both sides by
𝑋𝑌
The above ordinary differential equation can be expressed as
1 𝑑2𝑋 1 𝑑2𝑌 𝑋 = 𝐶1 cos 𝐵𝑥 + 𝐶2 sin 𝐵𝑥 ---------- (11)
2
+ 2
+ [𝛾2 + 𝜔2𝜇𝜀] = 0
𝑋 𝑑𝑥 𝑌 𝑑𝑦
𝑌 = 𝐶3 cos 𝐴𝑦 + 𝐶4 sin 𝐴𝑦 ---------- (12)
( For Reference )
𝑋 = 𝐶1 cos 𝐵𝑥 + 𝐶2 sin 𝐵𝑥
---------- (11)
𝑌 = 𝐶3 cos 𝐴𝑦 + 𝐶4 sin 𝐴𝑦 ---------- (12)
From (8)
𝐸𝑧0 = 𝑋𝑌
= (𝐶1 cos 𝐵𝑥 + 𝐶2 sin 𝐵𝑥)(𝐶3 cos 𝐴𝑦 + 𝐶4 sin 𝐴𝑦)
𝐸0 = 𝐶 𝐶 cos 𝐵𝑥 cos 𝐴𝑦 + 𝐶 𝐶 cos 𝐵𝑥 sin 𝐴𝑦 + 𝐶 𝐶 sin 𝐵𝑥 cos 𝐴𝑦 + 𝐶 𝐶 sin 𝐵𝑥 sin 𝐴𝑦 ---------- (13)
𝑧 1 3 1 4 2 3 2 4
𝑚𝜋 𝑛𝜋
The constants C1, C2, C3, C4, A & B can be calculated form the 𝐸𝑧0 = 𝐶 sin 𝑥 sin 𝑦
𝑎 𝑏
boundary conditions
x=0 y=0 x=a y=b
𝑚𝜋
(13) 𝐶1𝐶3 cos 𝐴𝑦 + 𝐶1𝐶4 sin 𝐴𝑦 = 0 (14) 𝐶2𝐶3 sin 𝐵𝑥 = 0 (15) 𝐶2𝐶4 sin 𝐵𝑥 sin 𝐴𝑦 = 0 𝐶 sin 𝑥 sin 𝐴𝑏 = 0
𝑎
𝐶1(𝐶3 cos 𝐴𝑦 + 𝐶4 sin 𝐴𝑦) = 0 𝐶 2𝐶 4 = 𝐶
𝐶 sin 𝐵𝑎 sin 𝐴𝑦 = 0 Which leads to sin 𝐴𝑏 = 0
Which leads to 𝐶1 = 0 (in eq 13) Which leads to 𝐶3 = 0 (in eq 13)
Which leads to sin 𝐵𝑎 = 0
𝐸0 = 𝐶 𝐶 sin 𝐵𝑥 cos 𝐴𝑦 + 𝐶 𝐶 sin 𝐵𝑥 sin 𝐴𝑦 𝐸0 = 𝐶 𝐶 sin 𝐵𝑥 sin 𝐴𝑦 𝑛𝜋
𝑧 2 3 2 4 𝑧 2 4
𝑚𝜋 𝐴= When n=1,2,3,..
𝑏
------- (14) ------- (15) 𝐵= Where m=1,2,3,..
𝑎
In TM mode H= 0
𝛾 𝜕𝐸𝑧 𝑗𝜔ε 𝜕𝐸𝑧
𝐸𝑥 = − 𝐻𝑥 =
ℎ2 𝜕𝑥 ℎ2 𝜕𝑦
𝛾 𝜕𝐸𝑧 −𝑗𝜔ε 𝜕𝐸𝑧
𝐸𝑦 = − 𝐻𝑦 =
ℎ2 𝜕𝑦 ℎ2 𝜕𝑥
Substitute the values of Ez we get
𝛾𝐶 𝜕 𝑚𝜋 𝑛𝜋 𝑗𝜔ε𝐶 𝜕 𝑚𝜋 𝑛𝜋
𝐸𝑥 = − sin 𝑥 sin 𝑦 𝐻𝑥 = sin 𝑥 sin 𝑦
ℎ2 𝜕𝑥 𝑎 𝑏 ℎ2 𝜕𝑦 𝑎 𝑏
𝐸𝑧 𝑥, 𝑦 = 𝐸𝑧0𝑒 −𝛾𝑧 ---------- (8)
𝛾𝐶 𝜕 𝑚𝜋 𝑛𝜋 −𝑗𝜔ε𝐶 𝜕 sin 𝑚𝜋 𝑥 sin 𝑛𝜋 𝑦
𝐸𝑦 = − 2 sin 𝑥 sin 𝑦 𝐻𝑦 =
ℎ 𝜕𝑦 𝑎 𝑏 ℎ2 𝜕𝑥 𝑎 𝑏
Assuming perfect conducting condition σ=0 and substitute
the value of E0 in (8)
Top view
𝑗𝛽 𝑚𝜋 𝑚𝜋𝑥 𝑛𝜋𝑦 −𝛾𝑧
𝐸𝑥(𝑥, 𝑦, 𝑧) = − 𝐶 cos sin 𝑒
ℎ2 𝑎 𝑎 𝑏
𝑗𝛽 𝑛𝜋 𝑚𝜋𝑥 𝑛𝜋𝑦 −𝛾𝑧
𝐸𝑦(𝑥, 𝑦, 𝑧) = − 𝐶 sin cos 𝑒
ℎ2 𝑏 𝑎 𝑏
Side view
𝑗𝜔ε 𝑛𝜋 𝑚𝜋𝑥 𝑛𝜋𝑦 −𝛾𝑧
𝐻𝑥(𝑥, 𝑦, 𝑧) = 𝐶 sin cos 𝑒
ℎ2 𝑎 𝑎 𝑏
−𝑗𝜔ε 𝑚𝜋 𝑚𝜋𝑥 𝑛𝜋𝑦 −𝛾𝑧
𝐻𝑦(𝑥, 𝑦, 𝑧) = 𝐶 cos sin 𝑒
ℎ2 𝑎 𝑎 𝑏
Front view
To find propagation constant
velocity
Consider
𝜔
ℎ2 = 𝛾 2 + 𝜔2𝜇𝜀 2 2 𝑣=
𝑚𝜋 𝑛𝜋 𝛽
𝛾2 = ℎ2 − 𝜔2𝜇𝜀 𝜔𝑐2 𝜇𝜀 = +
𝑎 𝑏
𝛾= ℎ2 − 𝜔2𝜇𝜀 𝜔
𝑣=
1 𝑚𝜋 2 𝑛𝜋 2 2 2
𝐵2 = ℎ2 − 𝐴2 𝑚𝜋 𝑛𝜋
𝜔𝑐 = + 𝜔2𝜇𝜀 − + 𝑏
𝜇𝜀 𝑎 𝑏 𝑎
ℎ2 = 𝐴2 + 𝐵2
𝛾= 𝐴2 + 𝐵2 − 𝜔2𝜇𝜀 The lower limit of angular frequency (ωc) is
called cut-off frequency, below which wave Wavelength
𝑛𝜋 𝑚𝜋 propagation is absent
𝐴= 𝐵= 𝑣
𝑏 𝑎
𝜆=
1 𝑚𝜋 2 𝑛𝜋 2 𝑓
𝑓𝑐 = +
𝑛𝜋 2 𝑚𝜋 2 2𝜋 𝜇𝜀 𝑎 𝑏
𝛾= + − 𝜔2𝜇𝜀
𝑏 𝑎 2𝜋
Wavelength corresponding to the cut off
𝜆=
2 2
frequency 𝑚𝜋 𝑛𝜋
We know propagation constant is a 𝜔2𝜇𝜀 − 𝑎 + 𝑏
complex value
When σ = 0 𝛾 = 𝑗𝛽
2
𝜆𝑐 =
𝑚 2 𝑛 2
𝑛𝜋 2 𝑚𝜋 2 𝑎 + 𝑏
𝛽= 𝜔2𝜇𝜀 − +
𝑏 𝑎
Transverse Electric (TE) Waves in Rectangular wave guide
For a TE mode the component of electric field strength along the
direction of propagation is zero (Ez = 0)
Let us assume
𝐻𝑧 𝑥, 𝑦 = 𝐻𝑧0𝑒−𝛾𝑧
Where
𝐻𝑧0 = 𝑋𝑌
𝐻0 = 𝐶 𝐶 cos 𝐵𝑥 cos 𝐴𝑦 + 𝐶 𝐶 cos 𝐵𝑥 sin 𝐴𝑦 + 𝐶 𝐶 sin 𝐵𝑥 cos 𝐴𝑦 + 𝐶 𝐶 sin 𝐵𝑥 sin 𝐴𝑦
𝑧 1 3 1 4 2 3 2 4
−𝑗𝜔𝜇 𝜕𝐻𝑧 𝛾 𝜕𝐻𝑧
𝐸𝑥 = 𝐻𝑥 = −
ℎ2 𝜕𝑦 ℎ2 𝜕𝑥
𝑗𝜔𝜇 𝜕𝐻𝑧 𝛾 𝜕𝐻𝑧
𝐸𝑦 = 2 𝐻𝑦 = −
ℎ 𝜕𝑥 ℎ2 𝜕𝑦
𝐻𝑧0 = 𝐶1𝐶3 cos 𝐵𝑥 cos 𝐴𝑦 + 𝐶1𝐶4 cos 𝐵𝑥 sin 𝐴𝑦 + 𝐶 2𝐶 3 sin 𝐵𝑥 cos 𝐴𝑦 + 𝐶 2𝐶 4 sin 𝐵𝑥 sin 𝐴𝑦 ---------- (1)
Differentiating eq (1) with respect
to x
𝜕𝐻𝑧
= −𝐵𝐶1𝐶3 sin 𝐵𝑥 cos 𝐴𝑦 − 𝐵𝐶1𝐶4 sin 𝐵𝑥 sin 𝐴𝑦 + 𝐵𝐶2𝐶3 cos 𝐵𝑥 cos 𝐴𝑦 + 𝐵𝐶2𝐶4 cos 𝐵𝑥 sin 𝐴𝑦 ---------- (1.1)
𝜕𝑥
Differentiating eq (2) with respect
to y
𝜕𝐻𝑧
= −𝐴𝐶1𝐶3 cos 𝐵𝑥 sin 𝐴𝑦 + 𝐴𝐶1𝐶4 cos 𝐵𝑥 cos 𝐴𝑦 ------- (3)
𝜕𝑦
The constants C1, C2, C3, C4, A & B can be calculated form the
boundary conditions
x=0 y=0
(4) 𝐶 1𝐶 3 = 𝐶
(1.1) 𝐵𝐶2𝐶3 cos 𝐴𝑦 + 𝐵𝐶2𝐶4 sin 𝐴𝑦 = 0
(3) 𝐴𝐶1𝐶4 cos 𝐵𝑥 = 0
𝐶2(𝐵𝐶3 cos 𝐴𝑦 + 𝐵𝐶4 sin 𝐴𝑦) = 0
Which leads to𝐶4 = 0 (in eq 2) 𝐻𝑧0 = 𝐶 cos 𝐵𝑥 cos 𝐴𝑦 ------- (5)
Which leads to 𝐶2 = 0 (in eq 1)
𝐻0 = 𝐶 𝐶 cos 𝐵𝑥 cos 𝐴𝑦 + 𝐶 𝐶 cos 𝐵𝑥 sin 𝐴𝑦 𝐻0 = 𝐶 𝐶 cos 𝐵𝑥 cos 𝐴𝑦 Where
𝑧 1 3 1 4 𝑧 1 3
𝑛𝜋 𝑚𝜋
------- (2) ------- (4) 𝐴= 𝐵=
𝑏 𝑎
𝐻𝑧0 = 𝐶 cos 𝐵𝑥 cos 𝐴𝑦 ------- (5)
−𝑗𝜔𝜇 𝜕 𝛾 𝜕
𝐸𝑥 = 𝐶 cos 𝐵𝑥 cos 𝐴𝑦 𝐻𝑥 = − (𝐶 cos 𝐵𝑥 cos 𝐴𝑦)
ℎ2 𝜕𝑦 ℎ2 𝜕𝑥
𝑗𝜔𝜇 𝜕 𝛾 𝜕
𝐸𝑦 = (𝐶 cos 𝐵𝑥 cos 𝐴𝑦) 𝐻𝑦 = − (𝐶 cos 𝐵𝑥 cos 𝐴𝑦)
ℎ2 𝜕𝑥 ℎ2 𝜕𝑦 𝐻𝑧 𝑥, 𝑦 = 𝐻𝑧0𝑒 −𝛾𝑧 ---------- (8)
For TE wave the expression for β, fc, λc, vp are same as those of TM wave . However there is a difference
for TE wave, it is possible to make either m or n but not both zero.
The lower order TE is possible than TM ( TE10 mode ,m=1,n=0)
𝑗𝜔𝜇
𝐸𝑥 = 𝐶𝐴 cos 𝐵𝑥 sin 𝐴𝑦 𝑒−𝛾𝑧
ℎ2
Top view
𝑗𝜔𝜇
𝐸𝑦 = − 𝐶𝐵 sin 𝐵𝑥 cos 𝐴𝑦 𝑒−𝛾𝑧
ℎ2
𝛾
𝐻𝑥 = 𝐶𝐵 sin 𝐵𝑥 cos 𝐴𝑦 𝑒−𝛾𝑧
ℎ2
𝛾
𝐻𝑦 = 𝐶𝐴 cos 𝐵𝑥 sin 𝐴𝑦 𝑒−𝛾𝑧
Side view
ℎ2
𝑛𝜋 𝑚𝜋
𝐴= 𝐵=
𝑏 𝑎
Front view
For TE10 mode
1 𝑚𝜋 2 𝑛𝜋 2
𝑓𝑐 = +
2𝜋 𝜇𝜀 𝑎 𝑏
Substituting m=1, n=0 • The mode with the lowest cut off frequency is called the
dominant mode
1 𝜋 • In rectangular wave guide the dominant mode is TE10
𝑓𝑐 =
2𝜋 𝜇𝜀 𝑎 • For a given wave guide it is possible to operate only in the
dominant mode over certain range of frequencies
𝑣0 The cut off frequency of TE10 • Most wave guide component are designed for operating in
𝑓𝑐 = mode is independent of the this mode
2𝑎
tube dimension b
𝑣0
𝜆𝑐 = = 2𝑎
𝑓𝑐
𝜋 2
𝛽= 𝜔2𝜇𝜀 −
𝑎
Top view Side view Top view Side view
Front view
Front view
Top view Side view Top view Side view
Front view
Front view
Dominant and Degenerate modes
Modes:
When a wave is being propagated in a rectangular waveguide it experience as a different types of patterns
is called modes.
Dominant mode :
Among these modes(no:of modes) which mode is having the least cut-off frequency or maximum cut-off
wavelength is known as dominant mode.
