Lecture 5c -- Rectangular waveguideemlab.utep.edu/ee4347appliedem/Lecture 5c -- Rectangular...

10/30/2017

1

Lecture 5c Slide 1

EE 4347

Applied Electromagnetics

Topic 5c

Rectangular Waveguide

These notes may contain copyrighted material obtained under fair use rules. Distribution of these materials is strictly prohibited

Lecture Outline

Lecture 5c Slide 2

• What is a rectangular waveguide?

• TEM Analysis

• TM Analysis

• TE Analysis

• Visualization of Modes

• Conclusions

10/30/2017

2

Lecture 5c Slide 3

What is a Rectangular Waveguide?

Lecture 5c Slide 4

Geometry of Rectangular Waveguide

Standard convention: a b

10/30/2017

3

Lecture 5c Slide 5

Analysis of Rectangular Waveguide

Rectangular waveguides are analyzed along each axis like a parallel plate waveguide.

Notes on the Rectangular Waveguide

• Most classic waveguide example

• Some of the first waveguides used for microwaves

• Not a transmission because only one conductor

• Does not support a TEM mode

• Exhibits a low‐frequency cutoff below which no waves will propagate

Lecture 5c Slide 6

10/30/2017

4

Lecture 5c Slide 7

TE Analysis

Lecture 5c Slide 8

Recall TE Analysis

The governing equation for TE analysis is

After a solution is obtained, the remaining field components are calculated according to

2 20, 0, 2

0,2 20z z

c z

H Hk H

x y

2 2 2ck k

0,0, 2

0,0, 2

zx

c

zy

c

HjH

k x

HjH

k y

0,0, 2

0,0, 2

0, 0

zx

c

zy

c

z

HjE

k y

HjE

k x

E

10/30/2017

5

Lecture 5c Slide 9

General Form of the Solution

From the geometry of the waveguide, we can immediate write the form of the solution as

Viewing the rectangular waveguide as the combination of two parallel plate waveguides, we can apply separation of variables to write H0,z(x,y) as the product of two functions.

0,, , , j zz zH x y z H x y e

0, ,zH x y X x Y y

Lecture 5c Slide 10

Separation of Variables (1 of 3)

We write our solution as the product of two 1D functions and substitute that back into our differential equation.

2 20, 0, 2

0,2 20z z

c z

H Hk H

x y

0, ,zH x y X x Y y

2 22

2 2

2 22

2 2

2 22

2 2

0

0

1 10

c

c

c

XY XYk XY

x y

X YY X k XY

x y

d X d Yk

X dx Y dy

We have ordinary derivatives because X(x) and Y(y)have only one independent variable each.

10/30/2017

6

Lecture 5c Slide 11


First, we focus our attention on the x‐dependence in our differential equation.

2 22

2 2

1 10c

d X d Yk

X dx Y dy

2xk

This definition will let us write the differential equation as a wave equation.

22

20x

d Xk X

dx

Second, we focus our attention on the y‐dependence in our differential equation.

2 22

2 2

1 10c

d X d Yk

X dx Y dy

2yk


22

20y

d Yk Y

dy

Lecture 5c Slide 12


We would like to be able to add our two new differential equations together to get the original differential equation.

22

2

22

2

10

10

x

y

d Xk

X dx

d Yk

Y dy

+

2 22 2

2 2

1 10x y

d X d Yk k

X dx Y dy

22

2

22

2

0

0

x

y

d Xk X

dx

d Yk Y

dy

We get our original differential equation back if

2 2 2c x yk k k

10/30/2017

7

Lecture 5c Slide 13

General Solution

We now have two differential equations to solve.2

22

0x

d Xk X

dx

22

20y

d Yk Y

dy

These are essentially the same differential equation so their solution has the same general form.

2

22

0 cos sinx x x

d Xk X X x A k x B k x

dx

2

22

0 cos siny y y

d Yk Y Y y C k y D k y

dy

Slab waveguide along x

Slab waveguide along y

The overall solution is the product of X(x) and Y(y).

0, , cos sin cos sinz x x y yH x y X x Y y A k x B k x C k y D k y

Lecture 5c Slide 14

Electromagnetic Boundary Conditions

0, , 0 0xE x

0, , 0xE x b

0, 0, 0yE y 0, , 0yE a y

Boundary conditions required that the tangential component of the electric field be zero at the boundary with a perfect conductor.

10/30/2017

8

Lecture 5c Slide 15

E0,x and E0,yIn order to apply the boundary conditions, we must derive the electric field components E0,x and E0,y from our expression for H0,z.

0,0, 2

2

2

,

cos sin cos sin

cos sin sin cos

zx

c

x x y yc

y x x y yc

HjE x y

k y

jA k x B k x C k y D k y

k y

jk A k x B k x C k y D k y

k

0,0, 2

2

2

,

cos sin cos sin

sin cos cos sin

zy

c

x x y yc

x x x y yc

HjE x y

k x

jA k x B k x C k y D k y

k x

jk A k x B k x C k y D k y

k

Lecture 5c Slide 16

Apply Boundary Conditions (1 of 2)

At the x = 0 boundary, we have

0,

2

0 0,

sin 0

y

xc

E y

jk A

k

2

cos 0 cos sin

cos sin

y y

x y yc

B C k y D k y

jk B C k y D k y

k

0B

At the x = a boundary, we have

0,

2

0 ,

sin cos sin

y

x x y yc

E a y

jk A k a C k y D k y

k

A = 0 leads to a trivial solution. It must be the sin(kxa) term that enforces the BC.

0 sin 0,1,2,... xx k aa mk m

10/30/2017

9

Lecture 5c Slide 17


At the y = 0 boundary, we have

0,

2

0 ,0

cos sin sin 0

x

y x xc

E x

jk A k x B k x C

k

2

cos 0

cos siny x xc

D

jk A k x B k x D

k

0D

At the y = b boundary, we have

0,

2

0 ,

cos sin sin

x

y x x yc

E x b

jk A k x B k x C k b

k

C = 0 leads to a trivial solution. It must be the sin(kyb) term that enforces the BC.

0 sin 0,1,2,... yy k bb nk n

Lecture 5c Slide 18

Revised Solution for H0,z

0, , cos cosz x yH x y AC k x k y

We have determined that B = D = 0 so our expression for H0,z becomes

We write the product AC as a single constant Amn.

0, , cos cosz mn x yH x y A k x k y

Also, recall our conditions for kx and ky.

x x

mk a m k

a

y y

nk b n k

b

0, , cos cosz mn

m x n yH x y A

a b

10/30/2017

10

Lecture 5c Slide 19

Entire Solution (1 of 2)

The final expression for H0,z is

From this, the other field components are

0, , cos cosz mn

m x n yH x y A

a b

0, 2

0, 2

0, 2

0, 2

, cos sin

, sin cos

, sin cos

, cos sin

x mnc

y mnc

x mnc

y mnc

j n m x n yE x y A

k b a b

j m m x n yE x y A

k a a b

j m m x n yH x y A

k a a b

j n m x n yH x y A

k b a b

0, , 0zE x y

Lecture 5c Slide 20


The overall electric and magnetic fields at any position are

2

2

2

2

, , cos sin

, , sin cos

, , 0

, , sin cos

, , cos sin

j zx mn

c

j zy mn

c

z

j zx mn

c

y mnc

j n m x n yE x y z A e

k b a b

j m m x n yE x y z A e

k a a b

E x y z

j m m x n yH x y z A e

k a a b

j n m x n yH x y z A

k b a

, , cos cos

j z

j zz mn

eb

m x n yH x y z A e

a b

10/30/2017

11

Lecture 5c Slide 21

Phase Constant, Recall the cutoff wave number

2 2 2c x yk k k

After analyzing the boundary conditions, this expression can be written as

2 22c

m nk

a b

The phase constant is therefore

2 2 2

2 2 2

2 22 2 2

c

c

mn c

k k

k k

m nk k k

a b

Lecture 5c Slide 22

Cutoff Frequency, fcRecall our expression for the phase constant

2

c

c

c c

k k

k

f k

2 2ck k

The phase constant must be a real number for a guided mode. This requires

ck k

Any time k < kc, our mode is cutoff and not supported by the waveguide. From this, we can derive the cutoff frequency fc.

2 2

,

1

2 2c

c mn

k m nf

a b

10/30/2017

12

Lecture 5c Slide 23

Characteristic Impedance, ZTEThe characteristic impedance is

2

TE

2

cos sin

cos sin

j zmn

x c

j zymn

c

j n m x n yA e

E k b a b kZ

j n m x n yH A ek b a b

Lecture 5c Slide 24

Cutoff for First‐Order TE Mode (1 of 2)

The cutoff frequency for the TEmnmode was found to be

2 2

c,

1

2mn

m nf

a b

What about the TE00 mode?

00TE 0m n

2 2

c,00

1 0 00

2f

a b

The TE00 mode does not exist.

10/30/2017

13

Lecture 5c Slide 25

Cutoff for First‐Order TE Mode (2 of 2)


01TE 0, 1m n

2 2

c,01

1 0 1 1

2 2f

a b b


10TE 1, 0m n

2 2

c,10

1 1 0 1

2 2f

a b a

CAUTION: We cannot yet say that the TE10 is the fundamental mode because we have not checked the cutoff frequency of the TM modes.

Since a > b, we conclude that the first‐order mode is TE10 because it has the lowest cutoff frequency.

Lecture 5c Slide 26

Single Mode Operation (1 of 2)

We wish to determine over what range of frequencies the waveguide supports only a single TE mode.

c1 c2f f f

Low‐Frequency Cutoff

We just found the lower frequency cutoff.c1

1

2f

a

High‐Frequency Cutoff

The high‐frequency cutoff is the frequency where the second‐order TE mode us supported. This could be the TE01, TE11 or TE20 mode. We must consider all.

,012

,20

2

2c

cc

f a bf

f a b

2 2

01 c,01

2 2 2

11 c,11

2 2

20 c,20

1 0 1 1TE :

2 2

1 1TE : 1

2 2

1 2 0 1 2TE :

2 2

fa b b

bf

a b ab

bf

a b ab

TE11 will always have a higher cutoff frequency than TE01.

The second‐order mode depends on our choice of a and b.

10/30/2017

14

Lecture 5c Slide 27

Single Mode Operation (2 of 2)

Typical rectangular waveguides will have a > 2b, so

Bandwidth

c1

1

2f

a 2

1cf

a

c2 c1

1 1 1

2 2f f f

a a a

Fractional Bandwidth

c2 c1 c2 c1

c c2 c1 c2 c1

1 1

2 2FBW 2 2 66.7%

1 12 32

a af f f ff

f f f f fa a

Continuing the assumption that we have a > 2b, the fractional bandwidth can be calculated from fc1 and fc2 above as follows

Lecture 5c Slide 28

Example #1 – TE Mode Analysis (1 of 4)

Suppose we have an air‐filled rectangular waveguide with a = 3 cm and b = 2 cm.

What is the cutoff frequency of the waveguide?

0

1

1 299792458 m s5.0 GHz

2 2 2 0.03 m 1.0 1.0c

r r

cf

a a

Over what range of frequencies is the waveguide single mode?

We see that a < 2b, so the second‐order mode is TE01.

0

2

1 299792458 m s7.5 GHz

2 2 2 0.02 m 1.0 1.0c

r r

cf

b b

5.0 GHz 7.5 GHzf

10/30/2017

15

Lecture 5c Slide 29


What is the fractional bandwidth of the waveguide?

2 1

2 1

7.5 5.0FBW 100% 200% 200% 40%

7.5 5.0

f ff

f f f

Plot the phase constant and effective refractive index for the first‐order and second‐order modes from DC up to 15 GHz.

2 2

12 202

2 2

20

2

2mn

f

c am nk

a bf

c b

The phase constant is calculated as:

00 eff eff

0

2 2

ck n n

f fc

The effective refractive index is calculated as:

Lecture 5c Slide 30


Plot the phase constant and effective refractive index for the first‐order and second‐order modes from DC up to 15 GHz.

2 2

10

2 f

c a

0eff,1 1 2

cn

f

2 2

20

2 f

c b

0eff,2 2 2

cn

f

10/30/2017

16

Lecture 5c Slide 31


Plot the velocity of the modes as a function of frequency.

01

eff,1

cv

n 0

2eff,2

cv

n

Are our modes travelling faster than the speed of light?

Lecture 5c Slide 32

Summary of TE Analysis

Field Solution

2

2

2

2

, , cos sin

, , sin cos

, , sin cos

, , cos si

, , 0

n

j zx mn

c

j zy mn

c

j zx mn

mnc

z

c

y

j n m x n yE x y z A e

k b a b

j m m x n yE x y z A e

k a a b

j m m x n yH x y z A e

k a a b

j n m x n yH x y z A

k b

E x y z

a

, , cos cos

j z

j zz mn

m x n yH x y z A e

a b

eb

Phase Constant2 2

2mn

m nk

a b

Cutoff Frequency2 2

c,

1

2mn

m nf

a b

Characteristic Impedance

TE,mnmn

kZ

• TE00 mode does not exist• TE10 is the lowest order TE mode

, a b

Same equation as for TM Same equation as for TM

10/30/2017

17

Lecture 5c Slide 33

TM Analysis

Lecture 5c Slide 34

Recall TM Analysis

The governing equation for TM analysis is

After a solution is obtained, the remaining field components are calculated according to

2 20, 0, 2

0,2 20z z

c z

E Ek E

x y

2 2 2ck k

0,0, 2

0,0, 2

zx

c

zy

c

EjH

k y

EjH

k x

0,0, 2

0,0, 2

zx

c

zy

c

EjE

k x

EjE

k y

0, 0zH

10/30/2017

18

Lecture 5c Slide 35

General Form of the Solution

From the geometry of the waveguide, we can immediate write the form of the solution as

Viewing the rectangular waveguide as the combination of two parallel plate waveguides, we can apply separation of variables to write E0,z(x,y) as the product of two functions.

0,, , , j zz zE x y z E x y e

0, ,zE x y X x Y y

Lecture 5c Slide 36


We write our solution as the product of two 1D functions and substitute that back into our differential equation.

2 20, 0, 2

0,2 20z z

c z

E Ek E

x y

0, ,zE x y X x Y y

2 22

2 2

2 22

2 2

2 22

2 2

0

0

1 10

c

c

c

XY XYk XY

x y

X YY X k XY

x y

d X d Yk

X dx Y dy

We have ordinary derivatives because X(x) and Y(y)have only one independent variable each.

10/30/2017

19

Lecture 5c Slide 37


First, we focus our attention on the x‐dependence in our differential equation.

2 22

2 2

1 10c

d X d Yk

X dx Y dy

2xk


22

20x

d Xk X

dx

Second, we focus our attention on the y‐dependence in our differential equation.

2 22

2 2

1 10c

d X d Yk

X dx Y dy

2yk


22

20y

d Yk Y

dy

Lecture 5c Slide 38


We would like to be able to add our two new differential equations together to get the original differential equation.

22

2

22

2

10

10

x

y

d Xk

X dx

d Yk

Y dy

+

2 22 2

2 2

1 10x y

d X d Yk k

X dx Y dy

22

2

22

2

0

0

x

y

d Xk X

dx

d Yk Y

dy

We get our original differential equation back if

2 2 2c x yk k k

10/30/2017

20

Lecture 5c Slide 39

General Solution

We now have two differential equations to solve.2

22

0x

d Xk X

dx

22

20y

d Yk Y

dy

These are essentially the same differential equation so their solution has the same general form.

2

22

0 cos sinx x x

d Xk X X x A k x B k x

dx

2

22

0 cos siny y y

d Yk Y Y y C k y D k y

dy

Slab waveguide along x

Slab waveguide along y

The overall solution is the product of X(x) and Y(y).

0, , cos sin cos sinz x x y yE x y X x Y y A k x B k x C k y D k y

Lecture 5c Slide 40

Electromagnetic Boundary Conditions

0, ,0 0zE x

0, , 0zE x b

0, 0, 0zE y 0, , 0zE a y

Boundary conditions required that the tangential component of the electric field be zero at the boundary with a perfect conductor. E0,z is tangential to all interfaces so we just use it. There is no need to calculate E0,x or E0,y.

10/30/2017

21

Lecture 5c Slide 41


At the x = 0 boundary, we have

0,0 0,

cos 0 sin 0

zE y

A B

cos sin

cos sin

y y

y y

C k y D k y

A C k y D k y

0A

At the x = a boundary, we have

0,0 ,

sin cos sin

z

x y y

E a y

B k a C k y D k y

B = 0 leads to a trivial solution. It must be the sin(kxa) term that enforces the BC.

0 sin 1, 2,...xx k m ma ak m = 0 leads to a trivial solution.

Lecture 5c Slide 42


At the y = 0 boundary, we have

0,0 ,0

cos sin cos 0 sin 0

x

x x

E x

A k x B k x C D

cos sinx xA k x B k x C

0C

At the y = b boundary, we have

0,0 ,

cos sin sin

x

x x y

E x b

A k x B k x D k b

D = 0 leads to a trivial solution. It must be the sin(kyb) term that enforces the BC.

0 sin 1, 2,...yy k n nb bk n = 0 leads to a trivial solution.

10/30/2017

22

Lecture 5c Slide 43

Revised Solution for H0,z

0, , sin sinz x yE x y BD k x k y

We have determined that A = C = 0 so our expression for E0,z becomes

We write the product BD as a single constant Bmn.

0, , sin sinz mn x yE x y B k x k y

Also, recall our conditions for kx and ky.

x x

mk a m k

a

y y

nk b n k

b

0, , sin sinz mn

m x n yE x y B

a b

Lecture 5c Slide 44


The final expression for E0,z is

From this, the other field components are

0, , sin sinz mn

m x n yE x y B

a b

0, 2

0, 2

0, 2

0, 2

, cos sin

, sin cos

, sin cos

, cos sin

x mnc

y mnc

x mnc

y mnc

j m m x n yE x y B

k a a b

j n m x n yE x y B

k b a b

j n m x n yH x y B

k b a b

j m m x n yH x y B

k a a b

0, , 0zH x y

10/30/2017

23

Lecture 5c Slide 45


The overall electric and magnetic fields at any position are

2

2

2

, , cos sin

, , sin cos

, , sin sin

, , sin cos

,

j zx mn

c

j zy mn

c

j zz mn

j zx mn

c

y

j m m x n yE x y z B e

k a a b

j n m x n yE x y z B e

k b a b

m x n yE x y z B e

a b

j n m x n yH x y z B e

k b a b

H x y

2, cos sin

, , 0

j zmn

c

z

j m m x n yz B e

k a a b

H x y z

Lecture 5c Slide 46

Phase Constant, Recall the cutoff wave number

2 2 2c x yk k k

After analyzing the boundary conditions, this expression can be written as

2 22c

m nk

a b

The phase constant is therefore

2 2 2

2 2 2

2 22 2 2

c

c

mn c

k k

k k

m nk k k

a b

10/30/2017

24

Lecture 5c Slide 47

Cutoff Frequency, fcRecall our expression for the phase constant

2

c

c

c c

k k

k

f k

2 2ck k

The phase constant must be a real number for a guided mode. This requires

ck k

Any time k < kc, our mode is cutoff and not supported by the waveguide. From this, we can derive the cutoff frequency fc.

2 2

c,

1

2 2c

mn

k m nf

a b

This is the same equation as for the TE modes.

Lecture 5c Slide 48

Characteristic Impedance, ZTMThe characteristic impedance is

2

TM

2

cos sin

cos sin

j zmn

x c

j zymn

c

j m m x n yB e

E k a a bZ

j m m x n yH kB ek a a b

10/30/2017

25

Lecture 5c Slide 49

Cutoff for First‐Order TM Mode (1 of 2)

The cutoff frequency for the TMmnmode was found to be

2 2

c,

1

2mn

m nf

a b

Note, we cannot have n = 0 or m = 0 for the TM mode. So…

The TM00 mode does not exist.The TM01 mode does not exist.The TM10 mode does not exist.The TM02 mode does not exist.The TM20 mode does not exist.The TM03 mode does not exist.The TM30 mode does not exist.

etc.

Lecture 5c Slide 50

Cutoff for First‐Order TM Mode (2 of 2)

What combination of m and nminimizes fc?

1, 1m n

Since a > b, the TM11 mode will have the lowest cutoff frequency.

2 2 2 2

c1

1 1 1 1 1 1

2 2f

a b a b

CAUTION: We cannot yet say that the TM11 is the fundamental mode because we have not checked this against the TE modes.

2 2

c,

1

2mn

m nf

a b

10/30/2017

26

Lecture 5c Slide 51

Example #2 – TM Mode Analysis (1 of 3)


What is the cutoff frequency of the waveguide?

2 2

1

2 2

0 0

2 2

0 0

2 2

0

2 2

1

2

1

2

1

2

1 1

2

299792458 m s 1 19.0 GHz

0.03 m 0.02 m2 1.0 1.0

c

r r

r r

r r

fa b

a b

a b

c

a b

0

0 0

1c

Recall that

Lecture 5c Slide 52


Plot the phase constant and effective refractive index for the first‐order mode from DC up to 15 GHz.

22 2 22

0

2

m n fk

a b c a

The phase constant is calculated as:

00 eff eff

0

2 2

ck n n

f fc

The effective refractive index is calculated as:

10/30/2017

27

Lecture 5c Slide 53


Plot the velocity of the modes as a function of frequency.

0

eff

cvn

Are our modes travelling faster than the speed of light?

Lecture 5c Slide 54

Summary of TM Analysis

Field Solution

2

2

2

, , cos sin

, , sin cos

, , si

, , sin si

n co

n

s

,

j zz m

j zx mn

c

j zy mn

c

j zx mn

c

y

n

j m m x n yE x y z B e

k a a b

j n m x n yE x y z B e

k b a b

j n m

m x n yE x y z B e

x n yH x y z B e

k b a b

H x y

a b

2

, , 0

, cos sin j zmn

c

z

j m m x n yz B e

k a a b

H x y z

Phase Constant2 2

2mn

m nk

a b

Cutoff Frequency2 2

c,

1

2mn

m nf

a b

Characteristic Impedance

TM,mn

mnZk

• m ≠ 0 and n ≠ 0, so TM00, TM01, TM02, TM10, TM20, etc. are not supported modes.

• TM11 is the lowest order TM mode

, a b

Same equation as for TE Same equation as for TE

10/30/2017

28

Lecture 5c Slide 55

Visualization of Modes

Lecture 5c Slide 56

Animation of TE10

Notice one bright spot along x and zero along y. (m = 1, n = 0)

10/30/2017

29

Lecture 5c Slide 57

Animation of TE20

Notice two bright spots along x and zero along y. (m = 2, n = 0)

Lecture 5c Slide 58

Animation of TE01

Notice zero bright spots along x and one along y. (m = 0, n = 1)

10/30/2017

30

Lecture 5c Slide 59

Animation of TE11

Notice one bright spot along x and one along y. (m = 1, n = 1)

Lecture 5c Slide 60

Animation of TE21

Notice two bright spots along x and one along y. (m = 2, n = 1)

10/30/2017

31

Lecture 5c Slide 61

Animation of TM11

Lecture 5c Slide 62

Conclusions

10/30/2017

32

Lecture 5c Slide 63

The Fundamental Mode

The fundamental mode is the mode which has the lowest cutoff frequency. This is either the TE01 or the TM11 mode.

2 2

c,TM

1 1 1

2f

a b

2

c,TE

1 1

2f

a We can see that the TE01 mode will have the

lowest cutoff frequency.

We conclude that the TE01 mode is the fundamental mode of the waveguide.

This is also called the dominant mode. When multiple modes are excited, usually most of the power ends up in the fundamental mode.

Lecture 5c Slide 64

Example #3 – Mode Analysis (1 of 3)


Over what range of frequencies is this waveguide single mode?

An easy way to do this is to calculate a table using a desktop computer.

m n TE Cutoff TM Cutoff--- --- --------- ---------0 0 1 0 3.74 GHz 2 0 7.48 GHz 3 0 11.22 GHz 4 0 14.96 GHz 0 1 7.48 GHz 1 1 8.37 GHz 8.37 GHz2 1 10.58 GHz 10.58 GHz3 1 13.49 GHz 13.49 GHz4 1 16.73 GHz 16.73 GHz0 2 14.96 GHz 1 2 15.43 GHz 15.43 GHz2 2 16.73 GHz 16.73 GHz3 2 18.71 GHz 18.71 GHz4 2 21.16 GHz 21.16 GHz0 3 22.45 GHz 1 3 22.76 GHz 22.76 GHz2 3 23.66 GHz 23.66 GHz3 3 25.10 GHz 25.10 GHz4 3 26.98 GHz 26.98 GHz0 4 29.93 GHz 1 4 30.16 GHz 30.16 GHz2 4 30.85 GHz 30.85 GHz3 4 31.96 GHz 31.96 GHz4 4 33.46 GHz 33.46 GHz

10/30/2017

33

Lecture 5c Slide 65





Then sort the table in order of increasing cutoff frequency.

Mode Cutoff------ ---------TE10 3.74 GHzTE20 7.48 GHzTE01 7.48 GHzTE11 8.37 GHzTM11 8.37 GHzTE21 10.58 GHzTM21 10.58 GHzTE30 11.22 GHzTE31 13.49 GHzTM31 13.49 GHzTE40 14.96 GHzTE02 14.96 GHzTE12 15.43 GHzTM12 15.43 GHzTE41 16.73 GHz

Lecture 5c Slide 66





Then sort the table in order of increasing cutoff frequency.

Mode Cutoff------ ---------TE10 3.74 GHzTE20 7.48 GHzTE01 7.48 GHzTE11 8.37 GHzTM11 8.37 GHzTE21 10.58 GHzTM21 10.58 GHzTE30 11.22 GHzTE31 13.49 GHzTM31 13.49 GHzTE40 14.96 GHzTE02 14.96 GHzTE12 15.43 GHzTM12 15.43 GHzTE41 16.73 GHz

We immediately see that the TE10 mode is the fundamental mode with the lowest cutoff frequency of 3.74 GHz.

The second‐order mode is taken from the table to be either the TE20 or TE01 mode because both of these have the same cutoff frequency of 7.48 GHz.

The overall range of frequencies for single‐mode operation is therefore 3.74 GHz 7.48 GHzf

10/30/2017

34

Key Points

• The rectangular waveguide is not a transmission line because it has less than two conductors.

• When filled with a homogeneous dielectric, the rectangular waveguide supports TE and TM modes, but not TEM.

• The cutoff frequencies for TE and TM modes are the same.

• The TE00 mode does not exist.• For TM modes, m ≠ 0 and n ≠ 0 .• The TE10 is the dominant mode because the TM10mode does not exist.

• It is only the phase velocity that exceeds the vacuum speed of light.

Lecture 5c Slide 67

Lecture 5c Slide 68

Summary from Text Book

Lecture 5c -- Rectangular waveguideemlab.utep.edu/ee4347appliedem/Lecture 5c -- Rectangular...

Documents

Transcript of Lecture 5c -- Rectangular waveguideemlab.utep.edu/ee4347appliedem/Lecture 5c -- Rectangular...