waveguides rectangular waveguides –tem, te and tm waves –cutoff frequency –wave propagation...

Waveguides

• Rectangular Waveguides– TEM, TE and TM waves– Cutoff Frequency– Wave Propagation– Wave Velocity,

Waveguides

• In the previous chapters, a pair of conductors was used to guide electromagnetic wave propagation. This propagation was via the transverse electromagnetic (TEM) mode, meaning both the electric and magnetic field components were transverse, or perpendicular, to the direction of propagation.

• In this chapter we investigate wave-guiding structures that support propagation in non-TEM modes, namely in the transverse electric (TE) and transverse magnetic (TM) modes.

• In general, the term waveguide refers to constructs that only support non-TEM mode propagation. Such constructs share an important trait: they are unable to support wave propagation below a certain frequency, termed the cutoff frequency.

Rectangular waveguide

Circular waveguide

Optical FiberDielectric Waveguide

Rectangular Waveguide

Location of modes

• Let us consider a rectangular waveguide with interior dimensions are a x b,

• Waveguide can support TE and TM modes. – In TE modes, the electric field is transverse

to the direction of propagation. – In TM modes, the magnetic field that is

transverse and an electric field component is in the propagation direction.

• The order of the mode refers to the field configuration in the guide, and is given by m and n integer subscripts, TEmn and TMmn.

– The m subscript corresponds to the number of half-wave variations of the field in the x direction, and

– The n subscript is the number of half-wave variations in the y direction.

• A particular mode is only supported above its cutoff frequency. The cutoff frequency is given by

2 2 2 21

2 2mn

r r

cm n c m n

fa b a b


1 1 1 1

o r o r o o r r r r

cu

8where 3 10 m/sc

Table 7.1: Some Standard Rectangular Waveguide

WaveguideDesignation

a(in)

b(in)

t(in)

fc10

(GHz)

freq range(GHz)

WR975 9.750 4.875 .125 .605 .75 – 1.12

WR650 6.500 3.250 .080 .908 1.12 – 1.70

WR430 4.300 2.150 .080 1.375 1.70 – 2.60

WR284 2.84 1.34 .080 2.08 2.60 – 3.95

WR187 1.872 .872 .064 3.16 3.95 – 5.85

WR137 1.372 .622 .064 4.29 5.85 – 8.20

WR90 .900 .450 .050 6.56 8.2 – 12.4

WR62 .622 .311 .040 9.49 12.4 - 18

Location of modes


Rectangular WaveguideThe cutoff frequency is given by

2 2

2mn

r r

cc m n

fa b

2 2

2mncc m n

fa b

8where 3 10 m/sc

r

r

For air 1

and 1

To understand the concept of cutoff frequency, you can use the analogy of a road system with lanes having different speed limits.

Rectangular WaveguideRectangular Waveguide• Let us take a look at the field pattern for two

modes, TE10 and TE20

– In both cases, E only varies in the x direction; since n = 0, it is constant in the y direction.

– For TE10, the electric field has a half sine wave pattern, while for TE20 a full sine wave pattern is observed.

Rectangular WaveguideExampleLet us calculate the cutoff frequency for the first four modes of WR284 waveguide. From Table 7.1 the guide dimensions are a = 2.840 mils and b = 1.340 mils. Converting to metric units we have a = 7.214 cm and b = 3.404 cm.

2 2

2mncc m n

fa b

8

10

3 10 1002.08 GHz

2 2 7.214 1c

mxc cmsfa cm m

8

01

3 10 1004.41 GHz

2 2 3.404 1c

mxc cmsfb cm m

20 4.16 GHzc

cf

a

8 2 2

11

3 10 1 1 1004.87 GHz

2 7.214 3.404 1c

mx cmsfcm cm m

TE10:

TE01:

8where 3 10 m/sc

TE20:

TE11:

TE10 TE01TE20 TE11

2.08 GHz 4.16 GHz 4.41 GHz 4.87 GHz

TM11

Rectangular WaveguideExample

8For air 3 10 m/sc

Rectangular Waveguide - Wave PropagationWe can achieve a qualitative understanding of wave propagation in waveguide by considering the wave to be a superposition of a pair of TEM waves. Let us consider a TEM wave propagating in the z direction. Figure shows the wave fronts; bold lines indicating constant phase at the maximum value of the field (+Eo), and lighter lines indicating constant phase at the minimum value (-Eo).

The waves propagate at a velocity uu, where the u subscript indicates media unbounded by guide walls. In air, uu = c.

Now consider a pair of identical TEM waves, labeled as u+ and u- in Figure (a). The u+ wave is propagating at an angle + to the z axis, while the u- wave propagates at an angle –.

These waves are combined in Figure (b). Notice that horizontal lines can be drawn on the superposed waves that correspond to zero field. Along these lines the u+ wave is always 180 out of phase with the u- wave.

Rectangular Waveguide - Wave Propagation


Since we know E = 0 on a perfect conductor, we can replace the horizontal lines of zero field with perfect conducting walls. Now, u+ and u- are reflected off the walls as they propagate along the guide.

The distance separating adjacent zero-field lines in Figure (b), or separating the conducting walls in Figure (a), is given as the dimension a in Figure (b).

The distance a is determined by the angle and by the distance between wavefront peaks, or the wavelength . For a given wave velocity uu, the frequency is f = uu/.

If we fix the wall separation at a, and change the frequency, we must then also change the angle if we are to maintain a propagating wave. Figure (b) shows wave fronts for the u+ wave.

The edge of a +Eo wave front (point A) will line up with the edge of a –Eo front (point B), and the two fronts must be /2 apart for the m = 1 mode.

(a)

(b)

a


The waveguide can support propagation as long as the wavelength is smaller than a critical value, c, that occurs at = 90, or

2 uc

c

ua

m f

Where fc is the cutoff frequency for the propagating mode.

sin c

c

f

f

We can relate the angle to the operating frequency and the cutoff frequency by

2sin

m

a

For any value of m, we can write by simple trigonometry

2sin uua

m f


A constant phase point moves along the wall from A to D. Calling this phase velocity up, and given the distance lAD is

2

cosAD

ml

Then the time tAD to travel from A to D is

2

cos AD

p p

ADl mt

u u

Since the times tAD and tAC must be equal, we have

cosu

p

uu

The time tAC it takes for the wavefront to move from A to C (a distance lAC) is

2

Wavefront Velocity

Distance from A to CAC

u u

ACl mt

u u

2

1

up

c

uu

ff


22 2cos cos 1 sin 1cf f

cosu

p

uu

The Phase velocity is given by

using

cosG uu u

2

1 cG u

fu u

f

The Group velocity is given by

The Wave velocity is given by

1 1 1 1u

o r o r o o r r r r

cu

8where 3 10 m/sc

Wave velocity

Phase velocity

pu

Group velocity

Beach

Ocean

Phase velocitypu

uuWave velocity

uu

Gu Group velocity

Analogy!

Point of contact


The ratio of the transverse electric field to the transverse magnetic field for a propagating mode at a particular frequency is the waveguide impedance.

2,

1

TE umn

c

Zf

f

For a TE mode, the wave impedance is For a TM mode, the wave impedance is

2

.1TM cmn u

fZ

f

2

1 cu

ff

The phase constant is given by

2

1

u

cff

The guide wavelength is given by

Example


Example


Let’s determine the TE mode impedance looking into a 20 cm long section of shorted WR90 waveguide operating at 10 GHz.

From the Waveguide Table 7.1, a = 0.9 inch (or) 2.286 cm and b = 0.450 inch (or) 1.143 cm.

2 2

2mncc m n

fa b

TE10 6.56 GHz

Mode Cutoff Frequency

TE01 13.12 GHz

TE11 14.67 GHz

TE20 13.13 GHz

TE02 26.25 GHz

At 10 GHz, only the TE10 mode is supported!

TE10 6.56 GHz

Mode Cutoff Frequency

TE01 13.12 GHz

TE11 14.67 GHz

TE20 13.13 GHz

TE02 26.25 GHz

TE10 TE20TE01 TE11

TM11

6.56 GHz 13.12 GHz

TE02

26.25 GHz14.67 GHz

Rearrange

13.13 GHz

Example


10 2

120 500 .

6.56GHz1-

10GHz

TEZ

10 tanIN

TEZ jZ l

The impedance looking into a short circuit is given by

The TE10 mode impedance

2 2

9 2

8

21 1

2 10 10 6.561 158

103 10

c cu

f ff

f c f

x Hz GHz rad

m GHz mxs

The TE10 mode propagation constant is given by

500 tan 31.6 100INZ j j

500 tan 158 0.2IN

radZ j m

m

waveguides rectangular waveguides –tem, te and tm waves –cutoff frequency –wave propagation...

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