Download Wave Guides and Resonating Cavities

Wave Guides and Resonating Cavities
Lecture38: Electromagnetic Theory
Professor D. K. Ghosh, Physics Department, I.I.T., Bombay
Resonating Cavity (contd.)
We have shown in the last lecture that the electric field components for 𝑇𝐸𝑙,𝑚,𝑛 are given by
𝐸𝑥 = 𝐸𝑥0 cos(𝑘𝑥 𝑥) sin(𝑘𝑦 𝑦) sin(𝑘𝑧 𝑧)
𝐸𝑦 = 𝐸𝑦0 sin(𝑘𝑥 𝑥) cos(𝑘𝑦 𝑦) sin(𝑘𝑧 𝑧)
𝜕𝐸𝑦
1
𝑘𝑧
𝐻𝑥 =
𝐸 sin(𝑘𝑥 𝑥) cos(𝑘𝑦 𝑦) cos(𝑘𝑧 𝑧)
(−
)=
−𝑖𝜔𝜇
𝜕𝑧
𝑖𝜔𝜇 𝑦0
1
𝜕𝐸𝑥
𝑘𝑧
𝐻𝑦 =
(
)=−
𝐸 cos(𝑘𝑥 𝑥) sin(𝑘𝑦 𝑦) cos(𝑘𝑧 𝑧)
−𝑖𝜔𝜇 𝜕𝑧
𝑖𝜔𝜇 𝑥0
𝜕𝐸𝑦 𝜕𝐸𝑥
1
1
𝐻𝑧 =
−
(𝐸 𝑘 − 𝐸𝑥0 𝑘𝑦 ) cos(𝑘𝑥 𝑥) cos (𝑘𝑦 𝑦) sin(𝑘𝑧 𝑧)
(
)=−
−𝑖𝜔𝜇 𝜕𝑥
𝜕𝑦
𝑖𝜔𝜇 𝑦0 𝑥
Let us consider 𝑇𝐸1,0,1 i.e., 𝑙 = 𝑛 = 1, 𝑚 = 0. This implies that 𝑘𝑦 = 0, leading to 𝐸𝑥 = 0 and 𝐻𝑦 = 0
.Thus the only non-zero components are
𝐸𝑦 = 𝐸𝑦0 sin(𝑘𝑥 𝑥) sin(𝑘𝑧 𝑧)
𝑘𝑧
𝐻𝑥 =
𝐸 sin(𝑘𝑥 𝑥) cos(𝑘𝑧 𝑧)
𝑖𝜔𝜇 𝑦0
1
𝐻𝑧 = −
𝐸 𝑘 cos(𝑘𝑥 𝑥) sin(𝑘𝑧 𝑧) ≡ 𝐻𝑧0 cos(𝑘𝑥 𝑥) sin(𝑘𝑧 𝑧)
𝑖𝜔𝜇 𝑦0 𝑥
where
𝐻𝑧0 =
𝑖𝑘𝑥
𝑖𝜋
𝐸𝑦0 =
𝐸
𝜔𝜇
𝜔𝜇𝑎 𝑦0
In terms of 𝐻𝑧0 , the fields are rewritten as,
𝐻𝑧 = 𝐻𝑧0 cos(𝑘𝑥 𝑥) sin(𝑘𝑧 𝑧)
𝑘𝑧
𝑎
𝐻𝑥 = − 𝐻𝑧0 sin(𝑘𝑥 𝑥) cos(𝑘𝑧 𝑧) = − 𝐻𝑧0 sin(𝑘𝑥 𝑥) cos(𝑘𝑧 𝑧)
𝑘𝑥
𝑑
𝜔𝜇𝑎
𝐸𝑦 = 𝐸𝑦0 sin(𝑘𝑥 𝑥) sin(𝑘𝑧 𝑧) =
𝐻 sin(𝑘𝑥 𝑥) sin(𝑘𝑧 𝑧)
𝑖𝜋 𝑧0
1
Q- Factor of a Resonator
The Q-factor, a short form for Quality factor of a resonator is defined as the ratio of the amount
of energy stored in the cavity and the amount of energy lost per cycle through the walls of the
cavity. In the following we will calculate the Q factor for the 𝑇𝐸1,0,1 mode of the cavity.
The deterioration of the cavity is because of two factors, viz. the finite conductivity of the walls
of the cavity and an imperfect dielectric in the space between. We will consider the space
between the walls to be vacuum so that it is only the finite conductivity of the walls that we
need to worry about. The conductivity, though not infinite, is nevertheless large so that the skin
depth is small. The tangential component of the magnetic field will be assumed to be confined
to a depth equal to the skin depth which leads to a surface current 𝐽𝑠 .
Let us first calculate the stored energy. The average energy stored is given by
𝜖
𝜖
2
〈𝑊〉 = ∫ |𝐸|2 𝑑𝑉 = ∫ |𝐸𝑦 | 𝑑𝑉
2
2
since the only non-zero component of the electric field is along the y direction. Substituting the
expression for 𝐸𝑦 , we get
𝑎
𝑑
𝑏
𝜖 𝜔𝜇𝑎 2
〈𝑊〉 = (
) |𝐻𝑧0 |2 ∫ 𝑑𝑥 sin2 (𝑘𝑥 𝑥) ∫ 𝑑𝑧 sin2 (𝑘𝑧 𝑧) ∫ 𝑑𝑦
2 𝜋
0
0
0
𝑎
2
𝑑
2
The first two integrals respectively give and while the last one gives 𝑏. We thus have,
𝜖 𝜔𝜇𝑎 2
𝑎𝑏𝑑 𝜖𝜇2 𝜔2 𝑎3 𝑏𝑑
〈𝑊〉 = (
|𝐻𝑧0 |2
) |𝐻𝑧0 |2
=
2 𝜋
4
8𝜋 2
We will now calculate the loss through the walls of the wave guide.
(1)
We know that the surface current causes a discontinuity in the magnetic field. As the magnetic
field decrease fast within the skin depth region, we will assume the field to be zero in this
region and calculate the surface current from the relation
⃗⃗𝐽𝑠 = 𝑛̂ × 𝐻
⃗𝑡
where 𝑛̂ is the unit normal pointing into the resonator.
For the rectangular parallelepiped we have three pairs of symmetric surfaces, contribution from
each pair being the same. For instance we have a front face at 𝑥 = 𝑎 and a back face at 𝑥 = 0.
z
Jy
Hz
O
d
y
Jy
Hx
a
x
b
2
For both the front face the normal direction is along −𝑖̂ and for the back face it is along +𝑖̂ since it is
inward into the cavity. The surface current density is thus given by
𝐽𝑠 = (∓𝑖̂) × (𝐻𝑦 𝑗̂ + 𝐻𝑧 𝑘̂ ) = ±𝐻𝑧 𝑗̂
since 𝐻𝑦 = 0 in this mode. We are, however interested in |𝑗𝑠 |2 so that contribution from the two walls
add up and we get the loss from these two sides as
1
Loss1 = 2 × 𝑅𝑠 ∫ |𝐽𝑠 |2 𝑑𝑦 𝑑𝑧
2
𝑏
𝑑
= 𝑅𝑠 |𝐻𝑧0 |2 ∫ 𝑑𝑦 ∫ sin2 (𝑘𝑧 𝑧)𝑑𝑧
0
0
1
= 𝑅𝑠 |𝐻𝑧0 |2 𝑏𝑑
2
we can in a similar way calculate the contribution from the left (y=0) and the right (y=b) faces for which
𝑛̂ = ±𝑗̂,
𝐽𝑠 = (±𝑗̂) × (𝐻𝑥 𝑖̂ + 𝐻𝑧 𝑘̂ ) = ∓𝐻𝑥 𝑘̂ ± 𝐻𝑧 𝑖̂
so that
|𝐽𝑠 |2 = |𝐻𝑥2 | + |𝐻𝑧 |2
𝑎 2
= |𝐻𝑧0 |2 [( ) sin2(𝑘𝑥 𝑥) cos2(𝑘𝑧 𝑧) + cos2(𝑘𝑥 𝑥) sin2(𝑘𝑧 𝑧)]
𝑑
Substituting this and integrating, the loss from the two side walls becomes
𝑎 2 𝑎𝑑 𝑎𝑑
Loss2 = 𝑅𝑠 |𝐻𝑧0 |2 [( )
+
]
𝑑 4
4
In a similar way the loss from the top and the bottom faces can be found to be
𝑎 2 𝑎𝑏
Loss3 = 𝑅𝑠 |𝐻𝑧0 |2 [( )
]
𝑑 4
Adding, all the losses, the total loss becomes
Loss =
𝑅𝑠 |𝐻𝑧0 |2 3
[𝑑 (2𝑏 + 𝑎) + 𝑎3 (2𝑏 + 𝑑) ] (2)
4𝑑2
The Q factor is given by 𝜔 times the quotient of [1) with (2)
𝑄=𝜔
= 𝜔3
𝜖𝜇2 𝜔2 𝑎 3 𝑏𝑑
8𝜋2
+ 𝑎) + 𝑎3 (2𝑏
𝑅𝑠
[𝑑3 (2𝑏
4𝑑2
2 3
3
+ 𝑑)]
𝜖𝜇 𝑎 𝑏𝑑 𝜎𝛿
+ 𝑎) + 𝑎3 (2𝑏 + 𝑑)]
2𝜋 2 [𝑑3 (2𝑏
3
where we have substituted 𝑅𝑠 =
1
.
𝜎𝛿
thus the quality factor can be determined from a knowledge of the
dimension of the cavity, the operating frequency and the nature of the material of the walls.
Circular Wave Guides
z
x
a
z
y
The waveguide that we consider are those with circular cross section. The direction of propagation is still
the z axis so that the geometry is cylindrical. This geometry is of great practical value as optical fibers
used in communication have this geometry. However, optical fibres are dielectric wave guides whereas
what we are going to discuss are essentially hollow metal tube of circular cross section.
We take the cylindrical coordinates to be (𝜌, 𝜙, 𝑧) and the radius of cross section to be a. We will first
express the two curl equations in cylindrical.
As there are no real current, we have,
𝜕𝐻𝜙
1 𝜕
𝐻𝑧 −
= 𝑖𝜔𝜖𝐸𝜌
𝜌 𝜕𝜙
𝜕𝑧
𝜕𝐻
𝜕𝐻
⃗ ) = 𝜌 − 𝑧 = 𝑖𝜔𝜖𝐸𝜙
(𝛻 × 𝐻
𝜙
𝜕𝑧
𝜕𝜌
⃗) =
(𝛻 × 𝐻
𝜌
4
⃗) =
(𝛻 × 𝐻
𝑧
1 𝜕
1 𝜕
𝐻 = 𝑖𝜔𝜖𝐸𝑧
(𝜌𝐻𝜙 ) −
𝜌 𝜕𝜌
𝜌 𝜕𝜙 𝑧
𝜕
As before, we will replace 𝜕𝑧 → −𝛾 so that these equations become
1 𝜕
𝐻 + 𝛾𝐻𝜙 = 𝑖𝜔𝜖𝐸𝜌
𝜌 𝜕𝜙 𝑧
𝜕𝐻𝑧
−𝛾𝐻𝜌 −
= 𝑖𝜔𝜖𝐸𝜙
𝜕𝜌
1 𝜕
1 𝜕
𝐻 = 𝑖𝜔𝜖𝐸𝑧
(𝜌𝐻𝜙 ) −
𝜌 𝜕𝜌
𝜌 𝜕𝜙 𝑧
The other set of the curl equations are obtained from Faraday’s law and can be easily written down from
the above set by replacing 𝜖 with – 𝜇 and interchanging E and H.
We get,
1 𝜕
𝐸 + 𝛾𝐸𝜙 = −𝑖𝜔𝜇𝐻𝜌
𝜌 𝜕𝜙 𝑧
𝜕𝐸𝑧
−𝛾𝐻𝜌 −
= −𝑖𝜔𝜇𝐻𝜙
𝜕𝜌
1 𝜕
1 𝜕
𝐸 = −𝑖𝜔𝜇𝐻𝑧
(𝜌𝐸𝜙 ) −
𝜌 𝜕𝜌
𝜌 𝜕𝜙 𝑧
As in the case of rectangular wave guides, we can still classify the modes as TE or TM. Consider one pair
of the above equations,
1 𝜕
𝐻 + 𝛾𝐻𝜙
𝜌 𝜕𝜙 𝑧
𝜕𝐸𝑧
𝑖𝜔𝜇𝐻𝜙 = 𝛾𝐻𝜌 +
𝜕𝜌
𝑖𝜔𝜖𝐸𝜌 =
we can eliminate 𝐻𝜙 from these two equations, and express 𝐸𝜌 as
(𝛾 2 + 𝜖𝜇𝜔2 )𝐸𝜌 = −
𝑖𝜔𝜇 𝜕
𝜕𝐸𝑧
𝐻𝑧 − 𝛾
𝜌 𝜕𝜙
𝜕𝜌
Define, 𝛾 2 + 𝜖𝜇𝜔2 = 𝑘 2 to rewrite this equation as
𝑘 2 𝐸𝜌 = −
𝑖𝜔𝜇 𝜕
𝜕𝐸𝑧
𝐻𝑧 − 𝛾
𝜌 𝜕𝜙
𝜕𝜌
Thus we have succeeded in expressing 𝐸𝜌 in terms of derivatives of the z component of E and H. In a
similar way we can express all the four components.
5
𝜕𝐸𝑧 𝑖𝜔𝜇 𝜕
–
𝐻
𝜕𝜌
𝜌 𝜕𝜙 𝑧
𝛾 𝜕
𝜕𝐻𝑧
𝑘 2 𝐸𝜙 = −
𝐸𝑧 + 𝑖𝜇𝜔
𝜌 𝜕𝜙
𝜕𝜌
𝑖𝜔𝜖
𝜕
𝜕𝐻
𝑧
𝑘 2 𝐻𝜌 =
𝐸 −𝛾
𝜌 𝜕𝜙 𝑧
𝜕𝜌
𝜕𝐸𝑧 𝛾 𝜕
𝑘 2 𝐻𝜙 = −𝑖𝜇𝜖
−
𝐻
𝜕𝜌 𝜌 𝜕𝜙 𝑧
𝑘 2 𝐸𝜌 = −𝛾
(1)
(2)
(3)
(4)
𝑘 2 = 𝛾 2 + 𝜖𝜇𝜔2
We will now obtain solutions for 𝐸𝑧 and 𝐻𝑧 using Helmholtz equation. In the following we will discuss
the TE modes for which 𝐸𝑧 = 0. We need to then find 𝐻𝑧 using
𝛻 2 𝐻𝑧 = −𝜔2 𝜇𝜖𝐻𝑧
Writing the Laplacian in cylindrical coordinates,
1 𝜕
𝜕
1 𝜕2
𝜕2
(𝜌 𝐻 ) + 2
𝐻 +
𝐻 = −𝜔2 𝜇𝜖𝐻𝑧
𝜌 𝜕𝜌 𝜕𝜌 𝑧
𝜌 𝜕𝜙 2 𝑧 𝜕𝑧 2 𝑧
We use technique of separation of variables by defining,
𝐻𝑧 (𝜌, 𝜙, 𝑧) = 𝑅(𝜌)𝐹(𝜙)𝑍(𝑧)
Substituting this into the Helmholtz equation and dividing throughout by 𝑅(𝜌)𝐹(𝜙)𝑍(𝑧), we get,
11 𝜕
𝜕
1 1 𝜕2
1 𝜕2
2
(𝜌 𝑅) +
𝐹
+
𝜔
𝜇𝜖
=
−
𝑍
𝑅 𝜌 𝜕𝜌 𝜕𝜌
𝐹 𝜌2 𝜕𝜙 2
𝑍 𝜕𝑧 2
The left hand side of this equation is a function of (𝜌, 𝜙) while the right hand side is a function of z
alone. Thus we can equate each side to a constant. Anticipating propagation along the z direction, we
equate the right hand side to −𝛾 2 so that, we have,
𝜕2
𝑍 = −𝛾 2 𝑍
𝜕𝑧 2
11 𝜕
𝜕
1 1 𝜕2
(𝜌 𝑅) +
𝐹 + (𝜔2 𝜇𝜖 + 𝛾 2 ) = 0
𝑅 𝜌 𝜕𝜌 𝜕𝜌
𝐹 𝜌2 𝜕𝜙 2
The last equation needs to be separated into a term involving 𝜌 alone and another depending on 𝜙
alone. This is done by multiplying the equation by 𝜌2 ,
6
1 𝜕
𝜕
1 𝜕2
𝜌 (𝜌 𝑅) + (𝜔2 𝜇𝜖 + 𝛾 2 )𝜌2 = −
𝐹
𝑅 𝜕𝜌 𝜕𝜌
𝐹 𝜕𝜙 2
Once again, we equate each of the terms to a constant 𝑛2 . We thus have the following pair of equations
𝜕2
𝐹 + 𝑛2 𝐹 = 0
𝜕𝜙 2
𝜕
𝜕
𝜌 (𝜌 𝑅) + 𝑘𝜌2 𝜌2 𝑅 − 𝑛2 𝑅 = 0
𝜕𝜌 𝜕𝜌
where 𝑘𝜌2 = 𝜔2 𝜇𝜖 + 𝛾 2 . The former equation has the solution
𝐹 = 𝐴 cos(𝑛𝜙) + 𝐵 sin(𝑛𝜙)
Singlevaluedness of F requires that if 𝜙 changes by 2𝜋, the solution must remain the same. This, in turn,
requires that n is an integer. We are now left with only the last equation which can be written in an
expanded form,
𝜕2
𝜕
𝜌2 2 𝑅 + 𝜌 𝑅 + 𝑘𝜌2 𝜌2 𝑅 − 𝑛2 𝑅 = 0
𝜕𝜌
𝜕𝜌
Defining a new variable 𝑥 = 𝑘𝜌 𝜌 (not to be confused with the Cartesian variable x), the equation can be
written as
𝑑2
1 𝑑𝑦
𝑛2
𝑦
+
+
−
(1
)=0
𝑑𝑥 2
𝑥 𝑑𝑥
𝑥2
where, we have written y in place of R. This equation is the well known Bessel equation, the solutions of
which are linear combinations of Bessel functions of first kind 𝐽𝑛 (𝑥) and that of the second kind 𝑁𝑛 (𝑥).
The Bessel functions of second ind are also known as Neumann function. The following graph shows
the variation of these functions with distance x.
7
𝑱𝒏 (𝒙)
It is seen that the Neumann function diverges at the origin and hence is not n acceptable solution. The
asymptotic behavior of these functions are
2
𝜋
𝐽𝑛 (𝑥) → √ cos (𝑥 − )
𝜋𝑥
4
2
𝜋
𝑁𝑛 (𝑥) → √ sin (𝑥 − )
𝜋𝑥
4
8
𝑵𝒏 (𝒙)
The complete solution is thus given by
𝐸𝑧 (𝜌, 𝜙, 𝑧) = 𝐽𝑛 (𝑘𝜌 𝜌)(𝐴𝑛 cos(𝑛𝜙) + 𝐵𝑛 sin(𝑛𝜙))𝑒 −𝛾𝑧
𝐻𝑧 (𝜌, 𝜙, 𝑧) = 𝐽𝑛 (𝑘𝜌 𝜌)(𝐶𝑛 cos(𝑛𝜙) + 𝐷𝑛 sin(𝑛𝜙))𝑒 −𝛾𝑧
For TE modes,
𝐸𝑧 = 0 ,
𝜕𝐻𝑧
=0
|
𝜕𝑛 𝑠𝑢𝑟𝑓𝑎𝑐𝑒
For TM modes,
𝐻𝑧 = 0, 𝐸𝑧 |𝑠𝑢𝑟𝑓𝑎𝑐𝑒 = 0
Let us look at the TE mode in a little more detail. The tangential component of the electric field must be
zero at the metallic boundary. Thus 𝐸𝜙 = 0 at 𝜌 = 𝑎. Substituting the expression for 𝐸𝜙 (with 𝐸𝑧 = 0)
we require, from Eqn. (2)
𝐸𝜙 =
𝑖𝜇𝜔 𝜕𝐻𝑧 𝑖𝜇𝜔
= 2 𝐽′𝑛 (𝑘𝜌 𝜌)(𝐴𝑛 cos(𝑛𝜙) + 𝐵𝑛 sin(𝑛𝜙))𝑒 −𝛾𝑧
𝑘𝜌2 𝜕𝜌
𝑘𝜌
where 𝐽′𝑛 is the derivative of Bessel function with respect to 𝜌. This expression vanishes at 𝜌 = 𝑎
provided
𝐽′𝑛 (𝑘𝜌 𝑎) = 0
9
which provides a restriction on propagation of TE mode. The zeros of the Bessel function 𝐽𝑛 are listed
in the following table, (m-th zero of 𝐽𝑛 is 𝑝𝑛𝑚 )
n
0
1
2
m=1
2.4048
3.8317
5.1356
m=2
5.5201
7.0156
8.4172
m=3
8.6537
10.1735
11.6198
The zeros of the derivative of the Bessel function are listed in the following table (m-th zero of 𝐽𝑛 is
𝑝𝑛𝑚 ′),
n
0
1
2
m=1
3.8317
1.8412
3.0542
m=2
7.0156
5.3314
6.7061
m=3
10.1735
8.5363
9.9695
For propagation to take place, 𝛾 = √𝑘𝜌2 − 𝜔 2 𝜇𝜖 must be imaginary. This is possible if 𝜔 > 𝜔𝑐 , where
𝜔𝑐 =
1
1 𝑝𝑛𝑚 ′
𝑘𝜌 =
√𝜇𝜖
√𝜇𝜖 𝑎
where 𝑝𝑛𝑚 ′ gives the m-th zero of the derivative of Bessel function. The corresponding mode is
classified as 𝑇𝐸𝑛𝑚 . Similarly, one can show that the cutoff frequency for the TM modes are given by
𝜔𝑐 =
1 𝑝𝑛𝑚
√𝜇𝜖 𝑎
From the tables given above, it is clear that the lowest mode is TE11 followed by TM01.
The field pattern of the waveguide are shown in the following (source – Web)
10
Wave Guides and Resonating Cavities
Lecture38: Electromagnetic Theory
Professor D. K. Ghosh, Physics Department, I.I.T., Bombay
Tutorial Assignment
1. TE10 mode propagates in an air filled rectangular waveguide of dimension 1cm x 0.5cm. The
frequency of the propagating wave is 3x 1012 Hz. It is desired to construct a cavity out of this
waveguide by closing the third dimension so that the cavity resonates in TE102 mode. (a) What
should be the length of the cavity? (b) Write down the electric field for the resonating mode.
2. For a cubical cavity, what is the degeneracy of the lowest resonant frequency, i.e. how many
different modes correspond to the lowest possible frequency?
3. A circular (cylindrical) waveguide of radius 4 cm is filled with a material of dielectric constant
2.25. The guide is operated at a frequency of 2 GHz. For the dominant TE mode, determine the
cutoff frequency, the guide wavelength and the bandwidth for a single mode operation
(assuming only TE modes).
4. If the cylindrical cavity of circular cross section is closed at both ends by perfectly conducting
discs, it becomes a cavity resonator. Find the resonant frequencies of a cavity resonator of
length d and radius a.
Solutions to Tutorial Assignments
1. For 𝑙 = 1, 𝑚 = 0, 𝑛 = 2, the resonant frequency is given by
1/2
𝜋 2
2𝜋 2
𝜔 = 2𝜋𝜈 = 𝑐 [( ) + ( ) ]
𝑎
𝑑
Substituting 𝑎 = 0.01, and the value of the frequency, we get, 𝑑 = 1.15 cm. With these values,
𝜋
we have 𝑘𝑥 = 𝑎 = 100𝜋, 𝑘𝑦 = 0, 𝑘𝑧 = 100√3𝜋 (m-1). The non-vanishing field components are
𝐸𝑦 = 𝐸𝑦0 sin(𝑘𝑥 𝑥) sin(𝑘𝑧 𝑧) cos(𝜔𝑡)
The other two non-vanishing field components can be calculated by using Maxwell’s equations,
11
𝜕𝐸𝑦
𝜕𝐻𝑥 𝜕𝐸𝑦
𝜕𝐻𝑧
=𝜇
,
= −𝜇
𝜕𝑧
𝜕𝑡
𝜕𝑥
𝜕𝑡
𝑘𝑧
𝐻𝑥 = ( ) 𝐸𝑦0 sin(𝑘𝑥 𝑥) cos(𝑘𝑧 𝑧) sin(𝜔𝑡)
𝜔𝜇
𝑘𝑥
𝐻𝑧 = ( ) 𝐸𝑦0 cos(𝑘𝑥 𝑥) sin(𝑘𝑧 𝑧) sin(𝜔𝑡)
𝜔𝜇
2. Since all the three dimensions are the same, we would have the same frequency for (100), (010)
and (001) modes. Counting TE and TM, there are 6 modes corresponding to the lowest
frequency.
3. For TEmn mode the cutoff frequency is given by
1 𝑝𝑛𝑚 ′
𝜔𝑐 =
√𝜇𝜖 𝑎
′
For the dominant mode m=n=1, 𝑝𝑛𝑚
= 1.8412. Substituting
1
√𝜇𝜖
=
𝑐
√2.25
= 2 × 108 m/s, and
a=0.04 m, we get 𝜔𝑐 = 9.206 × 109 rad/s corresponding to 1.47 GHz. To find the guide
wavelength, note that the propagation vector 𝛽 =
2𝜋
√4 ×
2×108
2𝜋
𝜆
= 2𝜋√𝜇𝜖 √𝜈 2 − 𝜈𝑐2 =
1018 − (1.47 × 109 )2. Solving, we get 𝜆 = 0.14 m. To find the bandwidth of single
mode TE operation note that the since the cutoff frequency for the next higher mode TE21 is
3.0542/1.8412=1.66 times the cutoff frequency for TE11 mode, the bandwidth is fixed.
4. If the cavity resonator is closed at both ends, there would be standing waves formed along the z
direction. Since the tangential component of the electric field is continuous (and therefore
vanishes near the surface of the perfect conductor,) we must have
𝜕𝐸𝑧
𝜕𝑧
= 0 at z=0 and z=d.
Taking the origin at the centre of one of the disks, for TM mode, Ez is given by
𝐸𝑧 = 𝐸0 𝐽𝑛 (𝑘𝜌 𝜌)(𝐴 sin 𝑛𝜑 + 𝐵 cos 𝑛𝜑) cos
𝑙𝜋𝑧
𝑑
Since 𝑘𝜌2 + 𝑘𝑧2 = 𝜔2 𝜇𝜖, the resonant frequencies for the TM modes are
𝜔=
1
1.8412 2
𝜋 2
√(
) + ( )
𝑎
𝑑
√𝜇𝜖
Note that unlike the dominant frequency of the TM mode which is fixed by the dimension a, the
TE111 frequency can be tuned by adjusting the value of the length of the resonator. If d is made
sufficiently large, the frequency can be made to be lower than that of the TM mode.
12
Wave Guides and Resonating Cavities
Lecture38: Electromagnetic Theory
Professor D. K. Ghosh, Physics Department, I.I.T., Bombay
Self Assessment Questions
1. Consider TE011 mode in a cubical cavity of side a. Write the expressions for the electric and
magnetic fields and show that they are orthogonal.
2. For the above problem determine the total energy stored in the electric and magnetic fields.
3. A cubical cavity of side 3 cm has copper walls (conductivity 𝜎 = 6 × 107 S/m). Calculate the
Q factor of the cavity for TE101 mode.
4. Consider a cylindrical cavity resonator. Show that the dominant frequency is TE mode of the
resonator.
Solutions to Self Assessment Questions
1. By definition of TE mode, Ez=0. Further, for 011 modes, 𝑘𝑥 = 0, 𝑘𝑦 = 𝑘𝑧 =
2𝜋
.
𝑎
This gives 𝐸𝑦 =
0 . We have,
𝐸𝑥 = 𝑅𝑙 𝐸0 sin(𝑘𝑦 𝑦) sin(𝑘𝑧 𝑧)𝑒 𝑖𝜔𝑡 = 𝐸0 sin(𝑘𝑦 𝑦) sin(𝑘𝑧 𝑧) cos(𝜔𝑡)
1 𝜕𝐸𝑥
𝐸0 𝑘𝑧
𝐻𝑦 = 𝑅𝑙 (
)=
sin(𝑘𝑦 𝑦) cos (𝑘𝑧 𝑧) sin(𝜔𝑡)
−𝑖𝜔𝜇 𝜕𝑧
𝜔𝜇
𝐻𝑥 = 0
𝐸0 𝑘𝑦
1 𝜕𝐸𝑥
𝐻𝑧 = 𝑅𝑙 (
)=
cos (𝑘𝑦 𝑦) sin(𝑘𝑧 𝑧) sin(𝜔𝑡)
𝑖𝜔𝜇 𝜕𝑦
𝜔𝜇
2. Total energy in the electric field is obtained by
13
𝑎
𝑎
𝑎
1 2
𝜖𝐸0 cos2 (𝜔𝑡) ∫ 𝑑𝑥 ∫ sin2(𝑘𝑦 𝑦)𝑑𝑦 ∫ sin2 (𝑘𝑧 𝑧)𝑑𝑧
2
0
0
0
1
2 3
2
= 𝜖0 𝐸0 𝑎 cos (𝜔𝑡)
8
𝜇
One can similarly find the magnetic field energy from 2 ∫ 𝐻 2 𝑑𝑉 . This gives,
1
(𝑘𝑦2
8
+ 𝑘𝑧2 )
𝜇𝜔 2
𝐸02 𝑎3 sin2 (𝜔𝑡)
1
8
Using the relation 𝑘𝑦2 + 𝑘𝑧2 = 𝜔2 /𝑐 2, this expression can be written as 𝜖0 𝐸02 𝑎3 sin2(𝜔𝑡). So
1
8
that the total energy in the field is 𝜖0 𝐸02 𝑎3 .
3. The resonant frequency of the cavity for TE101 mode is given by 𝜔 = 𝜋𝑐 √2/𝑎2 = 4.44 × 1010
rad/s. Using the expression given in the text, the Q factor can be written as follows for the case
of 𝑎 = 𝑏 = 𝑐.
𝜔3 𝜖𝜇2 𝜎𝛿𝑎3
𝑄=
24𝜋 2
1
Equivalent expressions for Q factor can be written using 𝑅𝑠 = 𝜎𝛿 and the preceding expression
for 𝜔 as
𝑄=
𝜔𝜇𝑎
6𝑅𝑠
𝜔𝜇
Where 𝑅𝑠 = √ 𝜎 = 3 × 10−2 . Substituting these values we get 𝑄 ≈ 9300.
4. For the TE mode of the cavity, 𝐵𝑧 must vanish at the surface of the end faces. The z –
component of the magnetic field is, therefore, given by
𝑙𝜋𝑧
𝐵𝑧 = 𝐵0 𝐽𝑛 (𝑘𝜌 𝜌)(𝐴 sin 𝑛𝜑 + 𝐵 cos 𝑛𝜑) sin
𝑑
The resonant frequencies are given by
2
𝜔=
1
𝑝′
𝑙𝜋 2
√( 𝑚𝑛 ) + ( )
𝑎
𝑑
√𝜇𝜖
where 𝑝′𝑚𝑛 are the zeros of the derivatives of Bessel functions. Note that unlike in the case of
TM modes, in this case 𝑙 ≠ 0 because that would make the fields vanish. For TM modes, the
𝑐
𝑐
dominant frequency is TM010, for which the frequency is 𝜔 = 𝑎 𝑝01 = 2.4048 𝑎. The dominant
mode for TE is TE111 for which the frequency is given by
2
𝑝′𝑚𝑛
𝑙𝜋 2
𝜔 = 𝑐√(
) + ( )
𝑎
𝑑
14
15