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Waveguide

By S. NaimanTE (Assistant Lecturer)

Lecture 7

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Transmission line

Waveguide

Fibre optic etc

Waveguide

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Waveguides• The prediction and realization of EM waves in

the late 19th century, but limited to applications e.g. radio and radar transmissions.

• The idea of guided wave was proposed first by J.J. Thomson in 1893, and experimentally verified by O.J. Lodge in 1894.

• The mathematical analysis of the propagatingmodes within a hollow metal cylinder was first

performed by Lord Rayleigh in 1897.

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Waveguide• At microwave frequencies (3-300 GHz) transmission

lines become inefficient due to- skin effects- dielectric losses

• Therefore use waveguides, lower losses and increased bandwidth

• Waveguides don’t operate at lower frequencies; act as a high pass filter with a cutoff frequency.

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Any system of conductors and insulators for carrying electromagnetic waves could be called a waveguide But reserve for hollow metallic pipes

They are used at microwave frequencies as TL were used at lower frequencies (less than 20GHz). They have much less lossy at the highest frequencies as compared to TL

Waveguide

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A pipe with any sort of cross section could be used but simplest cross section are preferred. Hence constant rectangular , elliptical or circular cross sections are normally employed.Setup i.e. antenna at one end of a waveguide and load at the other end.

Circular Wave Guide

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NOTE: Conduction of energy takes place through the dielectric(air) filling the waveguide and not through the walls

The behavior and properties of waveguides is described by electric and magnetic fields as in wave propagation instead of V and I as in TL)

The wall of waveguide is perfect conductor , reflect electromagnetic energy from the surface.

If wall is a perfect conductor and very thin , little current flow in the interior walls and hence very little power is dissipated.

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• Some selected waveguide size together with their frequency of operation are listed in the table (Kennedy pg290).

• However waveguide dimensions decrease as the frequency is increased (lower the wavelength).

• Waveguide are generally restricted to frequencies above 1GHz.

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Basic Behavior in WaveguideAn electromagnetic plane wave in space is transverse-electromagnetic (TEM) which means the electric field, the magnetic field, and the direction of propagation are mutually perpendicular.

What happen if we send the waves straight down a waveguide? It will not propagate in it. The electric field will be short-circuited by the walls (perfect conductor).

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Trick is to send the waves down the waveguide in zigzag fashion, bouncing it off the walls setting up a field that is maximum at or near the centre of the guide and zero at the walls.

Major consequences of zigzag propagation are :

•The velocity of propagation in a waveguide must be less than in free space•Wave can no longer be TEM

Maxwell’s equation show that TEM wave cannot have a tangential component of the electric field at the wall of the waveguide

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ViVn VR

Vg

Ø Ø

Reflection from a conducting plane

American System labels MODES according to the field component that behaves as in free space.

Mode - no component of electric field in the direction of propagation are called transverse-Electric (TE) MODEMode with no such component of magnetic field are called transverse-magnetic (TM)

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British and European systems label according to the components that has behavior diff from that of free space the modes are called H instead of TE and E for TM

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X and Z dir

Y-dir

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ModesCASES depend on components of z-direction1. TEM: Transverse ElectroMagnetic

Ez = Hz =0 also Ex=Ey=Hx=Hy=0• All components vanish rectangular waveguide cannot support

TEM modes2. TE: Transverse Electric Hz≠0, Ez=0, E┴ to propagation3. TM: transverse Magnetic Hz=0, Ez ≠ 0, H ┴ to

propagation4. HE: hybrid mode, Ez ≠ 0 and Hz ≠ 0 neither ┴ to

propagation

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cos

sin

cn

cg

VV

VV

cos

sin

n

p

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Phase velocity (vp)

sinsin

cvfp pfv

Phase velocity (vp) :the velocity with which the wave changes phase in a direction parallel to the conducting surface is given by:

gv Velocity in the dir parallel to conducting surface ( velocity at which a wave propagate)

Phase velocity is greater than Vg and Vc but it is just apparent velocity.

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Addition of the second wall must be placed carefully so that does not disturb the existing wave pattern

The second wall must be placed at a position where the E-intensity due to first wall is zero.

2nma

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cos22mm na

am2cos

For a given wall separation , the angle of incidence is determined by free space wavelength of the signal ,the integer m and distance between the walls.

gamp

2

22 )(1cos1sin

g guide wavelength

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The smallest free –space wavelength that is just unable to propagate in the waveguide under given conditions is called cutoff wavelength

22 )(1 am o

ma

o2

When m=1 the signal propagate in its resonant mode (dominant mode)

--this is the longest cutoff wavelength of the guide.

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2122

])([1)(1 oamp

2)(1 op

Group and phase velocity in the waveguide

sinsin cvcvpvgv

2cvpvgv

2)(1 ocg vv

pp fv

The guide wavelength can be related with cutoff by..

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Example1A wave is propagated in a parallel- planewaveguide , under conditions as above discussion. The freq is 6GHz, and the plane separation is 3 cm.Calculate• The cutoff wavelength for dominant mode• The wavelength in a waveguide, also for

dominant mode• The corresponding group and phase velocities.

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Example2It is necessary to propagate a 10 GHz signal in a waveguide whose wall separation is 6 cm. What is

the greatest number of half-waves of electric intensity which it will be possible to establish between the two walls( i.e what is the largest value of m)?

Calculate the guide wavelength for this mode of propagation.

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Assume perfectly conducting walls and perfect dielectric filling the wave guide.Convention ‘a’ is always the wider side of the waveguide.

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The rectangular wave guide has the same TE modes corresponding to the two parallel plate wave guides obtained by considering opposite metal walls

In 1955 (IRE) label modes in rectangular waveguides as TEm,n for transverse –electric and TM m,n for transverse-magnetic

m and n- number of half wavelengths of intensity (electric for TE modes and magnetic for TM mode) between each pair of walls. m is measured along x axis of waveguide, n in other direction

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TE Modes• Ez = 0• Electric fields are zero at metal interfaces• Slope of magnetic fields are zero at metal

interfaces

Use this to find Ex, Ey, Hx, Hy

Modes of waveguide are written as TMmn and TEmn m is a half cycle variation in the x directionn is a half cycle in variation in the y directionThe lowest TE mode is TE10 or TE01 depending on the geometry of a and b

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2)(1120

ooZ

The characteristic impedance of waveguide

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Example

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TMm,n Mode

• Magnetic force are closed loops. If magnetic exist and it is changing in the x-direction, it must exist and be changing in y-direction then TMm,0 cannot exist.

2)(1120ooZ

The characteristic impedance of waveguide

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•End of lecture 7

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TEmn &TMmn Results

• The TE dominant mode (ie. The TE mode with the lowest cut-off frequency) is TE10

• And for TM is TM11

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Difference btn TEm,n & TMm,n Mode

• Line of magnetic force are closed loops.If magnetic field exist and it is changing in X-

direction, it must also exist and be changing in the Y-direction

With (m,n) = (0,0), (0,n), or (m,0) all fields vanish the lowest TM mode is TM11

2)(1120

ooZ

2)(1120ooZ

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TE Modes• Ez = 0• Electric fields are zero at metal interfaces• Slope of magnetic fields are zero at metal

interfaces

Use this to find Ex, Ey, Hx, Hy

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TM Modes

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3 cases depends on m, n and k(w), Radical equal, less than or greater than zero.

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Case III-Propagating

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Example A rectangular waveguide measures 3X4.5cm internally and has a 9GHz signal propagated in it. Calculate the cutoff wavelength, the guide wavelength, the group and phase velocities and the characteristic wave impedance for (a) the TE1,0 mode and (b) the TM1,1 mode.

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Why more TEx0 than TEy0? a>b, x>y

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The rectangular waveguide has a high-pass behavior, since signals

can propagate only if they have frequency higher than the cut-off

for the TE10 mode.

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Circular waveguide• In general terms the behavior is the same as in RG.• However different geometry means diff application

hence a separate investigation

• From the analysis of behavior :• The law governing the propagation of waves in

waveguides are independent of the cross sectional shape and dimensions of the guide.

• All the parameters and definitions evolved for RG apply to circular with minor modification

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Modes are labeled somewhat differently

• The cutoff wavelength must be different due to different geometry

• Where r= internal radius of waveguide• kr= solution of a bessel function equation

krr

o 2

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TE TM

Mode (kr) Mode (kr) Mode (kr) Mode (kr)

TE0,1 3.83 TE0,2 7.02 TM0,1 2.40 TM0,2 5.52

TE1,1 1.84 TE1,2 5.33 TM1,1 3.83 TM1,2 7.02

TE2,1 3.05 TE2,2 6.71 TM2,1 5.14 TM2,2 8.42

Values of (kr) for principal modes in circular waveguides

Example:Calculate the cutoff wavelength, the guide wavelength and characteristic wave impedance of a circular waveguide whose internal diameter is 4cm, for a 10-GHz signal propagated in it in the TE1,1 mode

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Mode labeling

• The interger m now denote the number of full wave intensity variations around circumference and

• n represent half wave intensity changes radially out from the center to the wall

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Attenuation in waveguide

Waveguide below the cutoff have the following attenuation

• Reflection from obstacles, discontinuites, misaligned waveguide section

• Losses due to currents flowing in the waveguide walls

• Losses in the dielectric filling the waveguide• 2&3 depends on wall material ,its roughness

and frequency used

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eAwhere

0

2

Length of waveguide

dBeeAdB 00

5.5440 loglog20

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Calculate the voltage attenuation provided by a 25-cm length of waveguide having a= 1cm and b= 0.5cm, in which a i-GHz signal is propagated in the dominant mode

Example