waveguides

1. 1
RectangularWaveguide
In the lastlecture,we have coonsidered the case of guidedwave between apairof infiniteconducting
planes.Inthislecture,we consdierarectangularwave guide whichconsistsof ahollow pipe of infinite
extentbutof rectangular cross sectionof dimension 𝑎 × 𝑏.The longdirectionwill be takentobe the z
direction.
Unlike inthe previouscase 𝜕 𝜕𝑦⁄ isnotzero inthiscase. However,asthe propagationdirection
isalongthe z direction,we have
𝜕
𝜕𝑧
→ −𝛾 and
𝜕
𝜕𝑡
→ 𝑖𝜔. We can write the Maxwell’scurl
equationsas
𝜕𝐻𝑧
𝜕𝑦
+ 𝛾𝐻 𝑦 = 𝑖𝜔𝜖𝐸 𝑥
−𝛾𝐻 𝑥 −
𝜕𝐻𝑧
𝜕𝑥
= 𝑖𝜔𝜖𝐸 𝑦
𝜕𝐻 𝑦
𝜕𝑥
−
𝜕𝐻 𝑥
𝜕𝑦
= 𝑖𝜔𝜖𝐸 𝑧
The secondsetof equationsare obtainedfromthe Faraday’slaw,(thesecanbe writtendown
fromabove by 𝐸 ↔ 𝐻, 𝜖 → −𝜇
𝜕𝐸 𝑧
𝜕𝑦
+ 𝛾𝐸 𝑦 = −𝑖𝜔𝜇𝐻 𝑥
−𝛾𝐸 𝑥 −
𝜕𝐸 𝑧
𝜕𝑥
= −𝑖𝜔𝜇𝐻 𝑦
𝜕𝐸 𝑦
𝜕𝑥
−
𝜕𝐸 𝑥
𝜕𝑦
= −𝑖𝜔𝜇𝐻𝑧
Wave Guides
Lecture37: Electromagnetic Theory
Professor D. K. Ghosh, Physics Department, I.I.T., Bombay
z
xy
b
a

2. 2
As inthe case of parallel plate waveguides,we canexpressall the fieldquantitiesintermsof
the derivatives of 𝐸 𝑧and𝐻𝑧. Forinstance,we have,
𝑖𝜔𝜖𝐸 𝑥 = 𝛾𝐻 𝑦 +
𝜕𝐻𝑧
𝜕𝑦
=
𝛾
𝑖𝜔𝜇
( 𝛾𝐸 𝑥 +
𝜕𝐸 𝑧
𝜕𝑥
)+
𝜕𝐻𝑧
𝜕𝑦
whichgives,
( 𝑖𝜔𝜖 −
𝛾2
𝑖𝜔𝜇
) 𝐸 𝑥 =
𝛾
𝑖𝜔𝜇
𝜕𝐸 𝑧
𝜕𝑥
+
𝜕𝐻𝑧
𝜕𝑦
so that
𝐸 𝑥 = −
𝛾
𝑘2
𝜕𝐸 𝑧
𝜕𝑥
−
𝑖𝜔𝜇
𝑘2
𝜕𝐻𝑧
𝜕𝑦
(1)
where
𝑘2 = 𝛾2 + 𝜔2 𝜇𝜖
The other componentscanbe similarly writtendown,
𝐸 𝑦 = −
𝛾
𝑘2
𝜕𝐸 𝑧
𝜕𝑦
−
𝑖𝜔𝜇
𝑘2
𝜕𝐻𝑧
𝜕𝑥
(2)
𝐻 𝑥 =
𝑖𝜔𝜖
𝑘2
𝜕𝐸 𝑧
𝜕𝑦
−
𝛾
𝑘2
𝜕𝐻𝑧
𝜕𝑥
(3)
𝐻 𝑦 = −
𝑖𝜔𝜖
𝑘2
𝜕𝐸 𝑧
𝜕𝑥
−
𝛾
𝑘2
𝜕𝐻𝑧
𝜕𝑦
(4)
As before,we willlookintothe TEmode indetail.Since 𝐸 𝑧 = 0,we need tosolve for 𝐻𝑧 from
the Helmholtzequation,
(
𝜕2
𝜕𝑥2 +
𝜕2
𝜕𝑦2 + 𝑘2) 𝐻𝑧( 𝑥, 𝑦) = 0 (5)
Rememberthatthe complete solutionisobtainedbymultiplyingwith 𝑒−𝛾𝑧+𝑖𝜔𝑡. We solve
equation(5) usingthe technique of separationof variableswhichwe came acrossearlier.Let
𝐻𝑧( 𝑥, 𝑦) = 𝑋( 𝑥) 𝑌(𝑦)
Substitutingthisin(5) anddividingby 𝑋𝑌 throughout, we get,
1
𝑋
𝑑2 𝑋
𝑑𝑥2 + 𝑘2 = −
1
𝑌
𝑑2 𝑌
𝑑𝑦2 ≡ 𝑘 𝑦
2
where 𝑘 𝑦 isa constant. We furtherdefine 𝑘 𝑥
2 = 𝑘2 − 𝑘 𝑦
2

3. 3
We nowhave twosecondorderequations,
𝑑2 𝑋
𝑑𝑥2 + 𝑘 𝑥
2 𝑋 = 0
𝑑2 𝑌
𝑑𝑦2 + 𝑘 𝑦
2 𝑌 = 0
The solutionsof these equationsare well known
𝑋( 𝑥) = 𝐶1 cos 𝑘 𝑥 𝑥 + 𝐶2 sin 𝑘 𝑥 𝑥
𝑌( 𝑦) = 𝐶3 cos 𝑘 𝑦 𝑦 + 𝐶4 sin 𝑘 𝑦 𝑦
Thisgives,
𝐻𝑧( 𝑥, 𝑦) = 𝐶1 𝐶3cos 𝑘 𝑥 𝑥cos 𝑘 𝑦 𝑦+ 𝐶1 𝐶4 cos 𝑘 𝑥 𝑥sin 𝑘 𝑦 𝑦
+𝐶2 𝐶3sin 𝑘 𝑥 𝑥cos 𝑘 𝑦 𝑦 + 𝐶2 𝐶4 𝑠𝑖𝑛 𝑘 𝑥 𝑥sin 𝑘 𝑦 𝑦
The boundaryconditionsthatmust be satisfiedtodeterminethe constants isthe vanishingof
the tangential componentof the electricfieldonthe plates.Inthiscase,we have twopairsof
plates.The tangential directiononthe platesat 𝑥 = 0and 𝑥 = 𝑎 isthe y direction,sothatthe y
componentof the electricfield
𝐸 𝑦 = 0 at 𝑥 = 0, 𝑎
. Likewise, onthe platesat 𝑦 = 0 and 𝑦 = 𝑏,
𝐸 𝑥 = 0 at 𝑦 = 0, 𝑏
We needfirsttoevaluate 𝐸 𝑥 and 𝐸 𝑦 usingequations(1) and(2) and thensubstitute the
boundaryconditions. Since 𝐸 𝑧 = 0,we canwrite eqn.(1) and (2) as
𝐸 𝑥 = −
𝑖𝜔𝜇
𝑘2
𝜕𝐻𝑧
𝜕𝑦
𝐸 𝑦 = −
𝑖𝜔𝜇
𝑘2
𝜕𝐻𝑧
𝜕𝑥
𝐸 𝑥 = −
𝑖𝜔𝜇
𝑘2 [−𝐶1 𝐶3 ky cos 𝑘 𝑥 𝑥sin 𝑘 𝑦 𝑦 + 𝐶1 𝐶4 𝑘 𝑦 cos 𝑘 𝑥 𝑥 cos 𝑘 𝑦 𝑦
−𝐶2 𝐶3kysin 𝑘 𝑥 𝑥sin 𝑘 𝑦 𝑦 + 𝐶2 𝐶4 𝑘 𝑦 sin 𝑘 𝑥 𝑥cos 𝑘 𝑦 𝑦
𝐸 𝑦 = −
𝑖𝜔𝜇
𝑘2 [−𝐶1 𝐶3kx sin 𝑘 𝑥 𝑥cos 𝑘 𝑦 𝑦 − 𝐶1 𝐶4 𝑘 𝑥sin 𝑘 𝑥 𝑥sin 𝑘 𝑦 𝑦
+𝐶2 𝐶3kxcos 𝑘 𝑥 𝑥cos 𝑘 𝑦 𝑦 + 𝐶2 𝐶4 𝑘 𝑥 𝑐𝑜𝑠 𝑘 𝑥 𝑥sin 𝑘 𝑦 𝑦]
Since 𝐸 𝑦 = 0 at 𝑥 = 0, we musthave 𝐶2 = 0 and thenwe get,
𝐸 𝑦 = −
𝑖𝜔𝜇
𝑘2 [−𝐶1 𝐶3kx sin 𝑘 𝑥 𝑥cos 𝑘 𝑦 𝑦− 𝐶1 𝐶4 𝑘 𝑥 sin 𝑘 𝑥 𝑥sin 𝑘 𝑦 𝑦]
Further,since 𝐸 𝑥 = 0 at y=0, we have 𝐶4 = 0. Combiningthese,we get,ondefiningaconstant
𝐶 = 𝐶1 𝐶3

4. 4
𝐸 𝑥 =
𝑖𝜔𝜇
𝑘2 𝐶𝑘 𝑦 cos 𝑘 𝑥 𝑥sin 𝑘 𝑦 𝑦
𝐸 𝑦 =
𝑖𝜔𝜇
𝑘2 𝐶𝑘 𝑥 sin 𝑘 𝑥 𝑥cos 𝑘 𝑦 𝑦
and
𝐻𝑧 = 𝐶 cos 𝑘 𝑥 𝑥cos 𝑘 𝑦 𝑦
We still have the boundaryconditions, 𝐸 𝑥 = 0 at 𝑦 = 𝑏 and 𝐸 𝑦 = 0 at 𝑥 = 𝑎tobe satisfied. The
formergives 𝑘 𝑦 =
𝑛𝜋
𝑏
while the lattergives 𝑘 𝑥 =
𝑚𝜋
𝑎
,where mandn are integers.Thuswe have,
𝐻𝑧 = 𝐶 cos(
𝑚𝜋
𝑎
𝑥) cos(
𝑛𝜋𝑦
𝑏
)
and
𝑘2 = 𝑘 𝑥
2 + 𝑘 𝑦
2 = (
𝑚𝜋
𝑎
)
2
+ (
𝑛𝜋
𝑏
)
2
However,𝑘2 = 𝜔2 𝜇𝜖 + 𝛾2,sothat,
𝛾 = √(
𝑚𝜋
𝑎
)
2
+ (
𝑛𝜋
𝑏
)
2
− 𝜔2 𝜇𝜖
For propagationtotake place, 𝛾 must be imaginary,sothat the cutoff frequencybelow which
propagationdoesnottake place isgivenby
𝜔 𝑐 =
1
√ 𝜇𝜖
√(
𝑚𝜋
𝑎
)
2
+ (
𝑛𝜋
𝑏
)
2
The minimumcutoff isforTE1,0 (or TE0,1 ) mode whichare knownas dominantmode. For
these modes 𝐸 𝑥 (or 𝐸 𝑦) iszero.
TM Modes
We will notbe derivingthe equationsforthe TMmodesforwhich 𝐻𝑧 = 0 . Inthiscase, the
solutionfor 𝐸 𝑧,becomes,
𝐸 𝑧 = 𝐸 𝑧0 sin (
𝑚𝜋
𝑎
)sin (
𝑛𝜋
𝑏
)
As the solutionisintermsof productof sine functions,neithermnorn can be zeroin thiscase.
Thisis whythe lowestTE mode isthe dominantmode.
For propagatingsolutions,we have,
𝛽 = √ 𝜔2 𝜇𝜖 − (
𝑚𝜋
𝑎
)
2
− (
𝑛𝜋
𝑏
)
2
= √ 𝜇𝜖√ 𝜔2 − 𝜔 𝑐
2
where,
𝜔 𝑐 =
1
√ 𝜇𝜖
√(
𝑚𝜋
𝑎
)
2
+ (
𝑛𝜋
𝑏
)
2

5. 5
We have, 𝜔2 =
𝛽2
𝜇𝜖
+ 𝜔 𝑐
2.Differentiatingbothsides,we have,
𝜔
𝑑𝜔
𝑑𝛽
=
1
𝜇𝜖
𝛽
The group velocityof the wave isgivenby
𝑣𝑔 =
𝑑𝜔
𝑑𝛽
=
𝛽
𝜔𝜇𝜖
=
√ 𝜇𝜖√𝜔2 − 𝜔 𝑐
2
𝜔𝜇𝜖
=
1
√ 𝜇𝜖
√1 −
𝜔 𝑐
2
𝜔2
whichisless thanthe speedof light.The phase velocity,however,isgivenby
𝑣 𝜙 =
𝜔
𝛽
=
𝜔
√ 𝜇𝜖√𝜔2 − 𝜔 𝑐
2
=
1
√ 𝜇𝜖
1
√1 −
𝜔 𝑐
2
𝜔2
It may be notedthat 𝑣 𝜙 𝑣𝑔 =
1
𝜇𝜖
, whichinvacuumequal the square velocityof light.
For propagatingTE mode,we have,from(1) and (4) using 𝐸 𝑧 = 0
𝐸 𝑥
𝐻 𝑦
=
𝑖𝜔𝜇
𝛾
=
𝜔𝜇
𝛽
= √
𝜇
𝜖
1
√1 −
𝜔 𝑐
2
𝜔2
≡ 𝜂 𝑇𝐸
where 𝜂 𝑇𝐸 isthe characteristicimpedance forthe TE mode.Itisseenthat the characteristic
impedance isresistive. Likewise,
𝐸 𝑦
𝐻 𝑥
= −𝜂 𝑇𝐸
Power Transmission
We have seenthatinthe propagatingmode,the intrinsicimpedance isresistive,indicatingthat
there will be average flowof power. The PoyntingvectorforTE mode isgivenby
〈 𝑆〉 =
1
2
Re〈 𝐸⃗ × 𝐻⃗⃗ ∗〉
=
1
2
Re( 𝐸 𝑥 𝐻𝑦
∗ − 𝐸 𝑦 𝐻𝑥
∗)
=
1
2𝜂 𝑇𝐸
(| 𝐸 𝑥
2| + | 𝐸 𝑦
2|)
Substitutingthe expressionsforthe field components,the average powerflow throughthe x y
plane isgivenby
〈 𝑃〉 = ∫ 〈 𝑆〉 𝑑𝑥 𝑑𝑦 =
1
2𝜂 𝑇𝐸
∫ ∫ (| 𝐸𝑥
2| + | 𝐸 𝑦
2|) 𝑑𝑥 𝑑𝑦
𝑏
0
𝑎
0
=
1
2𝜂 𝑇𝐸
𝐶2 𝜔2 𝜇2
𝑘4
[(
𝑛𝜋
𝑏
)
2
∫ ∫ cos2 (
𝑚𝜋
𝑎
𝑥) sin2 (
𝑛𝜋
𝑏
𝑦) 𝑑𝑥 𝑑𝑦
𝑏
0
𝑎
0
+ (
𝑚𝜋
𝑎
)
2
∫ ∫ sin2 (
𝑚𝜋
𝑎
𝑥) cos2 (
𝑛𝜋
𝑏
𝑦) 𝑑𝑥 𝑑𝑦
𝑏
0
𝑎
0
]

6. 6
=
1
2𝜂 𝑇𝐸
𝐶2 𝜔2 𝜇2
𝑘4
[(
𝑚𝜋
𝑎
)
2
+ (
𝑛𝜋
𝑏
)
2
]
𝑎𝑏
4
ImpossibilityofTEM mode in Rectangular waveguides
We have seenthat in a parallel plate waveguide,aTEMmode for whichboththe electricand
magneticfieldsare perpendiculartothe directionof propagation,exists.This,howeverisnot
true of rectangularwave guide,orforthat matter forany hollow conductorwave guide without
an innerconductor.
We knowthatlinesof H are closedloops.Since there isnozcomponentof the magneticfield,
such loopsmustlie inthe x-yplane.However,aloopinthe x-yplane,accordingtoAmpere’s
law,impliesanaxial current.If there isnoinnerconductor,there cannotbe a real current.The
onlyotherpossibilitythenisadisplacementcurrent. However,anaxial displacementcurrent
requiresanaxial componentof the electricfield,whichiszeroforthe TEM mode.ThusTEM
mode cannot existinahollowconductor.(forthe parallel plate waveguides,thisrestriction
doesnotapplyas the fieldlinesclose atinfinity.)
ResonatingCavity
In a rectangularwaveguide we hadahollow tube withfoursidesclosedand apropagation
directionwhichwasinfinitelylong.We wil now close the thirdside andconsider
electromagneticwave trappedinsidearectangularparallopipedof dimension 𝑎 × 𝑏 × 𝑑 with
wallsbeingmade of perfectorconductor.(The thirddimensionistakenasd so as notto confuse
withthe speedof lightc).
Resonantingcavitiesare useful forstoringelectromagneticenergyjustasLCcircuit doesbutthe
formerhas an advantage inbeinglesslossyandhavingafrequencyrange muchhigher,above
100 MHz.
The Helmholtzequationforanyof the components 𝐸 𝛼 of the electricfieldcanbe writtenas
∇2 𝐸 𝛼 = −𝜔2 𝜇𝜖𝐸 𝛼
We write thisinCartesiananduse the technique of separationof variables,
𝐸 𝛼( 𝑥, 𝑦, 𝑧) = 𝑋 𝛼( 𝑥) 𝑌𝛼( 𝑦) 𝑍 𝛼(𝑧)
Introducingthisintothe equationanddividingby 𝑋 𝛼( 𝑥) 𝑌𝛼( 𝑦) 𝑍 𝛼(𝑧),we get,
1
𝑋 𝛼
𝜕2 𝑋 𝛼
𝜕𝑥2 +
1
𝑌𝛼
𝜕2 𝑌𝛼
𝜕𝑦2 +
1
𝑍 𝛼
𝜕2 𝑍 𝛼
𝜕𝑧2 = −𝜔2 𝜇𝜖
Since eachof the three termson the rightis a functionof an independentvariable,whilethe
righthand side isa constant,we musthave each of the three termsequalingaconstantsuch
that the three constantsadd up to the constanton the right. Let,
𝜕2 𝑋 𝛼
𝜕𝑥2 = −𝑘 𝑥
2 𝑋 𝛼

7. 7
𝜕2 𝑌𝛼
𝜕𝑦2 = −𝑘 𝑦
2 𝑌𝛼
𝜕2 𝑍 𝛼
𝜕𝑧2 = −𝑘 𝑧
2 𝑍 𝛼
such that 𝑘 𝑥
2 + 𝑘 𝑦
2 + 𝑘 𝑧
2 = 𝜔2 𝜇𝜖.
As eachof the equationshasa solutionintermsof linearcombinationof sine andcosine
functions,we write the solutionforthe electricfieldas
𝐸 𝛼 = (𝐴 𝛼 cos 𝑘 𝑥 𝑥 + 𝐵 𝛼 sin 𝑘 𝑥 𝑥)(𝐶 𝛼 cos 𝑘 𝑦 𝑦 + 𝐷 𝛼 sin 𝑘 𝑦 𝑦)(𝐹cos 𝑘 𝑧 𝑧 + 𝐺 𝛼 sin 𝑘 𝑧 𝑧)
Tha tangential componentof above mustvanishatthe metal boundary. Thisimplies,
1. At 𝑥 = 0 and at 𝑥 = 𝑎 , 𝐸 𝑦 and 𝐸 𝑧 = 0 for all valuesof 𝑦, 𝑧,
2. At 𝑦 = 0 and 𝑦 = 𝑏, 𝐸 𝑥 and 𝐸 𝑧 = 0 forall valuesof 𝑥, 𝑧,
3. At 𝑧 = 0 and z=d, 𝐸 𝑥 and 𝐸 𝑦 = 0 for all valuesof 𝑥, 𝑦
Let usconsider 𝐸 𝑥 whichmustbe zero at 𝑦 = 0, 𝑦 = 𝑏, 𝑧 = 0, 𝑧 = 𝑑. Thisisposible if
𝐸 𝑥 = 𝐸 𝑥0(𝐴 𝑥 cos 𝑘 𝑥 𝑥 + 𝐵 𝑥 sin 𝑘 𝑥 𝑥)sin 𝑘 𝑦 𝑦sin 𝑘 𝑧 𝑧
with 𝑘 𝑦 =
𝑚𝜋
𝑏
and 𝑘 𝑧 =
𝑛𝜋
𝑑
. Here m and n are non-zerointegers.
In a similarway,we have,
𝐸 𝑦 = 𝐸 𝑦0sin 𝑘 𝑥 𝑥 (𝐶 𝑦 cos 𝑘 𝑦 𝑦 + 𝐷 𝑦 sin 𝑘 𝑦 𝑦)sin 𝑘 𝑧 𝑧
𝐸 𝑧 = 𝐸 𝑧0sin 𝑘 𝑥 𝑥sin 𝑘 𝑦 𝑦 (𝐹𝑧 cos 𝑘 𝑧 𝑧 + 𝐺𝑧 sin 𝑘 𝑧 𝑧)
with 𝑘 𝑥 =
𝑙𝜋
𝑎
, with 𝑙 beingnon-zerointeger.
We nowuse
∇ ⋅ 𝐸⃗ =
𝜕𝐸 𝑥
𝜕𝑥
+
𝜕𝐸 𝑦
𝜕𝑦
+
𝜕𝐸 𝑧
𝜕𝑧
= 0
Thisrelationmustbe satisfiedfor all valuesof 𝒙, 𝒚, 𝒛 withinthe cavity. Thisrequires,
𝐸 𝑥0(−𝐴 𝑥 𝑘 𝑥 sin 𝑘 𝑥 𝑥
+ 𝐵 𝑥 𝑘 𝑥cos 𝑘 𝑥 𝑥)sin 𝑘 𝑦 𝑦sin 𝑘 𝑧 𝑧
+ 𝐸 𝑦0sin 𝑘 𝑥 𝑥(−𝐶 𝑦 𝑘 𝑦 sin 𝑘 𝑦 𝑦
+ 𝐷 𝑦 𝑘 𝑦 cos 𝑘 𝑦 𝑦)sin 𝑘 𝑧 𝑧 + 𝐸 𝑧0 sin 𝑘 𝑥 𝑥𝑠𝑖𝑛 𝑘 𝑦 𝑦(−𝐹𝑧 𝑘 𝑧 sin 𝑘 𝑧 𝑧
+ 𝐺𝑧 𝑘 𝑧 cos 𝑘 𝑧 𝑧) = 0
Let uschoose some special pointsand try to satisfythisequation.Let 𝑥 = 0, 𝑦, 𝑧 arbitrary,
Thisrequires 𝐵 𝑥 = 0.Likewise,taking 𝑦 = 0,x,z arbitrary,we require 𝐷 𝑦 = 0 andfinally, 𝑧 = 0,
𝑥, 𝑦 arbitrary gives 𝐺𝑧 = 0.
Withthese our solutionsforthe componentsof the electricfieldbecomes,
𝐸 𝑥 = 𝐸 𝑥0cos 𝑘 𝑥 𝑥sin 𝑘 𝑦 𝑦sin 𝑘 𝑧 𝑧
𝐸 𝑦 = 𝐸 𝑦0sin 𝑘 𝑥 𝑥cos 𝑘 𝑦 𝑦sin 𝑘 𝑧 𝑧
𝐸 𝑧 = 𝐸 𝑧0 sin 𝑘 𝑥 𝑥sin 𝑘 𝑦 𝑦cos 𝑘 𝑧 𝑧
where we have redefinedourconstants 𝐸 𝑥0,𝐸 𝑦0 and 𝐸 𝑧0.

8. 8
We alsohave,
𝑘 𝑥 =
𝑙𝜋
𝑎
, 𝑘 𝑦 =
𝑚𝜋
𝑏
, 𝑘 𝑧 =
𝑛𝜋
𝑑
the integers 𝑙, 𝑚, 𝑛 cannotbe simultaneouslyzeroforthenthe fieldwill identicallyvanish.
Note furtherthatsince ∇ ⋅ 𝐸⃗ = 0 must be satisfiedeverywhere withinthe cavity,we musthave,
calculatingthe divergence explicitlyfromthe above expressions,
𝑘 𝑥 𝐸 𝑥0 + 𝑘 𝑦 𝐸𝑦0 + 𝑘 𝑧 𝐸𝑧0 = 0
Thisrequiresthat 𝑘⃗ is perpendicularto 𝐸⃗ .
One can see that the modeswithinacavitycan existonlywithprescribed “resonant”frequency
correspondingtothe setof integers 𝑙, 𝑚, 𝑛
𝜔 =
1
√ 𝜇𝜖
[(
𝑙𝜋
𝑎
)
2
+ (
𝑚𝜋
𝑏
)
2
+ (
𝑛𝜋
𝑑
)
2
]
1
2
Thoughthere isnothinglike apropagationdirectionhere,one cantake the longestdimension
(sayd) to be the propagationdirection.
We can have modeslike 𝑇𝐸𝑙,𝑚,𝑛 correspondingtothe set 𝑙, 𝑚, 𝑛 for which 𝐸 𝑧 = 0.For thisset
we get,
𝐸 𝑥 = 𝐸 𝑥0cos 𝑘 𝑥 𝑥sin 𝑘 𝑦 𝑦sin 𝑘 𝑧 𝑧
𝐸 𝑦 = 𝐸 𝑦0sin 𝑘 𝑥 𝑥cos 𝑘 𝑦 𝑦sin 𝑘 𝑧 𝑧
with 𝑘 𝑥 𝐸𝑥0 + 𝑘 𝑦 𝐸𝑦0 = 0. The magneticfieldcomponentscanbe obtainedfromthe Faraday’s
law,
𝐻 𝑥 =
1
−𝑖𝜔𝜇
(−
𝜕𝐸 𝑦
𝜕𝑧
) =
𝑘 𝑧
𝑖𝜔𝜇
𝐸 𝑦0sin 𝑘 𝑥 𝑥cos 𝑘 𝑦 𝑦 cos 𝑘 𝑧 𝑧
𝐻 𝑦 =
1
−𝑖𝜔𝜇
(
𝜕𝐸 𝑥
𝜕𝑧
) = −
𝑘 𝑧
𝑖𝜔𝜇
𝐸 𝑥0cos 𝑘 𝑥 𝑥sin 𝑘 𝑦 𝑦cos 𝑘 𝑧 𝑧
𝐻𝑧 =
1
−𝑖𝜔𝜇
(
𝜕𝐸 𝑦
𝜕𝑥
−
𝜕𝐸 𝑥
𝜕𝑦
) = −
1
𝑖𝜔𝜇
(𝐸 𝑦0 𝑘 𝑥 − 𝐸 𝑥0 𝑘 𝑦)cos 𝑘 𝑥 𝑥cos 𝑘 𝑦 𝑦sin 𝑘 𝑧 𝑧
,
Wave Guides
Lecture37: Electromagnetic Theory
Professor D. K. Ghosh, Physics Department, I.I.T., Bombay

9. 9
Tutorial Assignment
1. A rectangularwaveguidehasdimensions8cm x 4 cm. Determine all the modesthatcan
propagate whenthe operatingfrequencyis (a) 1 GHz, (b) 3 GHz and (c) 8 GHz.
2. Two signals,one of frequency10 GHz and the otherof 12 GHz are simultaneouslylaunchedin
an air filledrectangularwaveguideof dimension2cm x 1 cm. Findthe time interval betweenthe
arrival of the signalsata distance of 10 m fromthe commonplace of theirlaunch.
3. What shouldbe the thirddimensionof acavityhavinga lengthof 1 cm x 1 cm whichcan
operate ina TE103 mode at 24 GHZ?
Solutionsto Tutorial Assignments
1. The cutoff frequencyfor(m,n) mode is 𝜈𝑐 =
𝑐
2𝜋
√(
𝑚𝜋
𝑎
)
2
+ (
𝑛𝜋
𝑏
)
2
= 1.5 × 1010√
𝑚2
64
+
𝑛2
16
.Some
of the lowestcutoffsare asunder(inGHz) :
m n Cutoff frequency(GHz)
0 1 3.75
0 2 7.5
0 3 11.25
1 0 1.875
1 1 4.192
1 2 7.731
1 3 11.405
2 0 3.75
2 1 5.30
2 2 8.39
3 0 5.625
3 1 6.76
3 2 9.375
4 0 7.5
4 1 8.38
5 0 9.375

10. 10
At 1 GHz, nopropagationtakesplace. At3 GHz onlyTE10 mode propagates(recall noTM mode is
possible wheneitherof the indicesiszero.) At8GHz, we have TE01, TE02, TE10, TE11, TE12, TE20,
TE21, TE30, TE31, TE40, TM11, TM12, TM21 and TM31, i.e.a total of 14 modespropagating.
2. The cutoff frequencyisgivenby
𝑓𝑐 =
𝑐
2
√(
𝑚
𝑎
)
2
+ (
𝑛
𝑏
)
2
= 7.5 𝐺𝐻𝑧
The propagationof the wave isgivenbythe groupvelocity 𝑣𝑔 = 𝑐√1 − (
𝑓𝑐
𝑓
)
2
,whichis 1.98x 108
and 2.07x108
m/s respectivelyforf=10 GHz and12 GHz respectively.Thusthe difference in
speedis9x 106
m/s,resultinginatime difference of approximately10-5
sin travelling10m.
3. The operatingfrequency is givenby
𝜔 =
1
√ 𝜇𝜖
[(
𝑙𝜋
𝑎
)
2
+ (
𝑚𝜋
𝑏
)
2
+ (
𝑛𝜋
𝑑
)
2
]
1
2
Substituting 𝑙 = 1, 𝑚 = 0, 𝑛 = 3,we get 𝑑 = 2.4cm.
Wave guides
Lecture37: Electromagnetic Theory
Professor D. K. Ghosh, Physics Department, I.I.T., Bombay

11. 11
SelfAssessmentQuestions
1. A rectangular,airfilled waveguide hasdimension2cm x 1 cm. For whatrange of
frequencies,there isa“single mode”operationinthe guide?
2. The cutoff frequencyfora TE10 mode inan airfilledwaveguide is1.875 GHz. What would
be the cutoff frequencyof thismode if the guide were tobe filledwithadielectricof
permittivity 9𝜖0?
3. In an air filledwaveguideof dimension2cm x 2 cm, the x componentof the electricfieldis
givenby
𝐸 𝑥 = −8sin (
2𝜋𝑦
𝑎
)sin(𝜔𝑡 − 100𝑧)
Identifythe propagatingmode,determine the frequencyof operation and find 𝐻𝑧 and 𝐸 𝑧.
Solutionsto SelfAssessmentQuestions
1. The cutoff frequencyfor(m,n) mode is 𝜈𝑐 =
𝑐
2𝜋
√(
𝑚𝜋
𝑎
)
2
+ (
𝑛𝜋
𝑏
)
2
= 1.5 × 1010√
𝑚2
4
+ 𝑛2.
Substitutingvalues,the cutoff for (1,0) mode is7.5 GHz andfor (2,0) is15 GHz. The frequency
for (0,1) is also15 GHz. All othermodeshave highercutoff.Thusin orderthat onlyone mode
propagates,the operatingfrequencyshouldbe inthe range 7.5 < 𝜈 < 15 GHz. In thisrange,
onlyTE10 mode operates.Recall thatthere isnoTM mode witheitherof the indicesbeingzero.
2. The cutoff frequencyisproportional to
1
√ 𝜇𝜖
.Thusthe frequencywouldbe reducedbyafactor of
3 makingit625 MHz.
3. Since one of the standingwave factoris missing,one of the mode indicesiszero.Thusitisa TE
mode.The electricfieldof TEmn mode isgivenby (witha=b)
𝐸 𝑥 =
𝑖𝜔𝜇
𝑘2 𝐶𝑘 𝑦 cos(
𝑚𝜋𝑥
𝑎
)sin (
𝑛𝜋𝑦
𝑎
)sin(𝛽𝑧 − 𝜔𝑡)
Thus the mode isTE02 mode. From the relation 𝑘 𝑥
2 + 𝑘 𝑦
2 + 𝛽2 =
𝜔2
𝑐2
.Given 𝑘 𝑥 = 0, 𝑘 𝑦 =
2𝜋
0.02
=
100𝜋, 𝛽 = 100,
𝜔2
𝑐2 = (100𝜋)2 + (100)2

12. 12
the operatingfrequencyis15.74 GHz.
Since 𝑘 𝑥 = 0, 𝐸 𝑦 = 0 . Further,bydefinitionforTE mode 𝐸 𝑧 = 0.
Here we have 𝑘2 = 𝑘 𝑥
2 + 𝑘 𝑦
2 = 𝑘 𝑦
2. The ratio
|
𝐻𝑧
𝐸 𝑥
| =
𝑘2
𝜔𝜇𝑘 𝑦
=
𝑘 𝑦
𝜔𝜇
= 0.0025
so that
𝐻𝑧 = 0.020cos(
2𝜋
𝑎
𝑦) cos (𝜔𝑡 − 100𝑧) A/m