Wave Guides

 

Rectangular Wave Guide

Advantages of Waveguides over Transmission Line

- Waveguide is a structure used to transmit electromagnetic wave from one point (source) to another point (load) by the process of reflection of wave.

The advantages of waveguides are enlisted below:
1. Simple to construct
2. Supports Transverse electric, Transverse magnetic and Transverse electro-magnetic ware propagation
3. Low power loss due to propagation via reflection
4. Applicable even for very high frequency wave
5. High power handling capacity
6. Multiple signals at same time

The disadvantages of waveguides are enlisted below:
1. Unable to propagate lower frequency wave
2. Difficult to install
3. Expensive due to silver or gold coating to reduce skin effect loss

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Rectangular waveguide

- Rectangular waveguide is a hollow metallic conductor with rectangular cross section.
- It supports Transverse Electric and Transverse Magnetic wave propagation.
- It is filled with source less (p = 0 and J = 0) and distortionless dielectric medium (σ = 0).
- It consists of walls which is perfectly conduction (σ = ∞)

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Circular waveguide

- It consists of circular cross section.
- It supports Transverse Electric and Transverse Magnetic wave propagation.
- It is easy to construct and install.
- It has limited dominant mode bandwidth.

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Modes of Propagation

- Mode represents the configuration of electric or magnetic field within a waveguide.
- Dominant mode is defined as the mode which provides low loss and minimum distortion wave propagation. It is represented as TE(10) for rectangular waveguide and TE(11) for circular waveguide.

- Cutoff frequency is the minimum frequency of wave propagation below which attenuation occurs.
f(c) = v / 2 * √ [ (m/a)^2 + (n/b)^2 ]

- TE mode is the mode in which the electric field is transverse to the direction of propagation and magnetic field is absent. E(z) = 0
- It is represented as TE(mn)

- TM mode is the mode in which magnetic field is transverse to the direction of propagation and electric field is absent. H(z) = 0
- It is represented as TM(mn)

- m denotes number of half sine wave variation of electric or magnetic field across wider dimension a.
- n denotes number of half sine wave variation of electric or magnetic field across narrower dimension b.

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Numerical Problems of Waveguide

1. For rectangular waveguide, a = 3cm and b = 2cm. If the medium within guide σ = 0, μ(r) = E(r) = 1, will the waveguide support mode with frequency 3.2 GHz.

 

Consider transverse electric (TE) mode of wave propagation.

 

Given: a = 3 * 10 ^ -2 m
b = 2 * 10 ^ -2 m

For a medium within the guide,
σ = 0
μ(r) = E(r) = 1

So, v = 1/ √(μE) = c = 3 * 10 ^ 8 m/s

Now,
For TE(10):
f(c) = v / 2 * √ [ (m/a)^2 + (n/b)^2 ]
= ((3 * 10 ^ 8) / 2) * (1 / (3 * 10 ^ -2))
= 5 GHz

For TE(11):
f(c) = v / 2 * √ [ (m/a)^2 + (n/b)^2 ]
= 9.01 GHz

Since, the lowest cut off frequency is greater than the operating frequency i.e 3.2 GHz, the waveguide does not support it.

 

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2. A rectangular waveguide with a = 2.5cm and b = 1cm is to operate below 15.1 GHz. How many TE and TM modes can it transmit if filled with medium σ = 0, E = 4 E(0), μ(r) = 1. Calculate cut off frequencies. Also write the expression for instantaneous field TE(01) mode.

 

Given :
v = 1/ √(μE) = 1/ √(μ(0) * 4 E(0)) = 1/2 * c

 

The cutoff frequency is given by:
f(c) = v / 2 * √ [ (m/a)^2 + (n/b)^2 ]
= (3 * 10 ^ 8) / 4 * √ [ (m/(2.5 * 10 ^ -2))^2 + (n/(1 * 10 ^ -2))^2 ]
= 3 √ (m ^ 2 + 6.25 n ^ 2) GHz

By question, waveguide operates below 15.1 GHz, so f(c) < 15.1;

Now:
TE(10), f(c) = 3 GHz
TE(01), f(c) = 7.5 GHz
TE(20), f(c) = 6 GHz
TE(02), f(c) = 15 GHz
TE(30), f(c) = 9 GHz
TE(03), f(c) = 22.5 GHz (Not acceptable)
TE(40), f(c) = 12 GHz
TE(50), f(c) = 15 GHz
TE(60), f(c) = 18 GHz (Not acceptable)

This shows that, maximum value of m is 5 and n is 2.

Then:
TM(11) and TE(11), f(c) = 8.078 GHz
TM(21) and TE(21), f(c) = 9.6 GHz
TM(31) and TE(31), f(c) = 11.72 GHz
TM(41) and TE(41), f(c) = 14.14 GHz
TM(12) and TE(12), f(c) = 15.3 GHz (Not acceptable)

Hence, there are 11 TE and 4 TM modes it can transmit.

Again,
E(x) = -w b E(0) sin (π/b*y) sin (wt - βz)
E(y) = 0
E(z) = 0

 

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3. In rectangular waveguide, a = 15cm, b = 8cm with σ = 0, μ = μ(0), E = 4E(0) and H(x) = 2*sin (πx/a) * cos (3πy/b) * sin (π*10^11t - βz) A/m.
Find mode of operation, f(c), β, γ and η.

 

From the question:
       H(x) = 2*sin (πx/a) * cos (3πy/b) * sin (π*10^11t - βz) A/m

 

Comparing with H(x) = 2*sin (mπx/a) * cos (nπy/b) * sin (wt - βz) A/m
We get:
m = 1
n = 3
w = π*10^11
f = w / 2π = 50 GHz (operating Frequency)

So, the mode of operation is TM(13).

f(c) = v / 2 * √ [ (m/a)^2 + (n/b)^2 ]
= 28.57 GHz

β = w √(μE) √[1-(f(c)/f)^2]
= 1718.21 rad/m

γ = jβ = j1718.21 m

η = 377 / √E(o) * √[1-(f(c)/f)^2]
= 154.752

 

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-by SURAJ AWAL