Under guidance of INTRODUCTION OF WAVEGUIDES Joydeep …ece.vnit.ac.in/people/jsengupta/wp-content/uploads/sites/... · 2017-02-09 · INTRODUCTION OF WAVEGUIDES Under guidance of

VNIT BT14ECE031 CHARAN SAI KATAKAM 1

INTRODUCTION OF WAVEGUIDES

Under guidance of

Joydeep Sengupta sir

INTRODUCTION TO WAVEGUIDES

In a waveguide energy is transmitted in the form of electromagnetic waves whereas in transmission line, it is transmitted in the form of voltage and current.

Conduction of energy takes place not through the walls whose function is only to confine this energy but through the dielectric filling the waveguide which is usually air.

ADVANTAGES OF WAVEGUIDE OVER A TRANSMISSION LINE :

In a transmission line high frequency signals cannot be transmitted because

Large mutual induction.

Length is very large.

Multiplexing the signals is not possible in transmission lines.

Flashover is less in case of waveguide.

Minimum loss of power in case of waveguide.

Power handling capability is 10 times than coaxial cable.


INTRODUCTION TO WAVEGUIDES

Mechanical simplicity.

Higher maximum operating frequency.(3GHZ to 100GHZ).

Normally waveguides are of two types.

Rectangular waveguide.

Circular waveguide.

REFLECTION OF WAVEFORM IN CONDUCTING PLANE :


a

b

Rectangular waveguide Circular waveguide

d

Vg

VR

Vn

VI

Vg is velocity guided by wall

(group velocity)

Vn is the normal velocity to

wall

WAVEGUIDES

Vn and Vg are components of VR.

VI – incident velocity

VR – reflected velocity

TEM – Transverse Electro Magnetic

In TEM, electric ,magnetic and propagation of wave should be perpendicular to each other.

In rectangular waveguide, if TEM strikes the wall, as it is a conductor it get absorbed by wall whereas in rectangular waveguide, reflection takes place when TEM strikes wall. Therefore TEM cannot be propagated through rectangular waveguide but in circular waveguide.

Only TE and TM are possible in rectangular waveguide.

TE- Transverse Electric: Electric field is perpendicular to propagation of wave.

TM-Transverse Magnetic: Magnetic field is perpendicular to propagation of wave. VNIT BT14ECE031 CHARAN SAI KATAKAM 4

DOMINANT MODE OF OPERATION

DOMINANT MODE : The natural mode of oper-

-ation for a waveguide is called dominant mode. This mode is

the lowest possible frequency that can be propagated in a

waveguide.


a

L

a

L

L

a

ELECTRIC

FIELD

>

>

>

MAGNETIC

FIELD m=no. of half

wavelength across

waveguide.

n=no. of half

wavelength along

the waveguide

height.

DOMINANT MODE OF OPERATION cont …

n=1 => 1 circle

n=2 => 2 circles

m- electric field

n – magnetic field

The signal of maximum wavelength that can pass through a

waveguide is

λ0=2a/m

An electromagnetic plane wave in space is transverse electrom-

agnetic or TEM. The electric field, the magnetic field and

direction of propagation are mutually perpendicular. If such a

wave were sent straight down a waveguide, it would not

propagate in it. This is because the electric field would be

short-circuited by the walls since the walls are assumed to be


Magnetic

field

m=1 m=2

BASIC BEHAVIOUR

BASIC BEHAVIOUR cont….

Perfect conductors and a potential cannot exist

across them.

What must be found in some method of propagation which

does t e ui e a ele t i field to e ist ea a all a d simultaneously the parallel to it. This is achieved by sending

the wave down the waveguide in a zigzag fashion bouncing it

off the walls and setting of a field i.e., maxima at the centre of

waveguide and zero at the walls.

In this case, the walls are nothing to be short circuit and they

do t i te fe e ith the a e patte setup et ee the . To measure consequences of zigzag propagation are apparent.

The first is that velocity of propagation in a waveguide must be

less than that of free space.


PLANE WAVE OF A CONDUCTING SURFACE

If actual velocity of wave is Vc , then the simple

trigonometry shows that the velocity of the wave in the

direction parallel to conducting is Vg and velocity normal to

the wall is Vn.

PARALLEL AND NORMAL WAVE LENGTH: Distance between 2

successive identical points i.e., successive crests or successive

troughs. In the figure, it is seen that the wavelength in

direction of propagation of wave is λ being the distance

between 2 successive crests in this direction.


Vg

Vn VC

λP

λ λn

θ

INCIDENT

WAVE

REFLECTED

WAVE

PARALLEL AND NORMAL WAVELENGTH cont….

So the distance between 2 consecutive crests in

the direction parallel to conducting plane is λp and wavelength

perpendicular to surface is λn.

Vp – phase velocity – velocity with which wave changes its

phase. Vg – group velocity – velocity parallel to wall.

In order to wave to propagate in a waveguide there should be

no voltage at the walls because walls are purely conductive if

there exists some voltage at walls the wave get shorted and

there will be no propagation of wave.


λp = λ

sinθ λn = λ

cosθ VC = f λ

Vg = VC sinθ

Vp= fλp= f λ sinθ

= VC

sinθ

PARALLEL PLANE WAVEGUIDE

Eq 1

PARALLEL PLANE WAVEGUIDE cont…..

Voltage variation in waveguide is almost similar

to the transmission lines.

Waveform of voltage in transmission lines is as follows

So dimension of waveguide can have following values such

that voltage at walls will be zero.


λ/2

2λ/2

3λ/2

a= λn / 2

a=2 λn / 2 a=3 λn / 2

DERIVATION OF CUT OFF WAVELENGTH

a is the dista e et ee the alls λn is the

wavelength in the direction normal to both walls.

is the o .of half a ele gth of ele t i i te sit to e established between walls which is nothing but integer.

From equation , it is easy to say that as the free space wave

length is increased, there comes a point beyond which, the

a e a o li ge p opagate i a a eguide ith fi ed a & . The f ee spa e a ele gth at hi h this takes pla e is

called cut off wavelength and it defines as the smallest free


a = mλn 2

= mλ

2cosθ From Eq 1

cosθ = mλ 2a

λP = λ

sinθ =

λ

√ 1- cos2 θ

λ

√ 1- (mλ)2 (2a)2

DERIVATION OF CUT OFF WAVELENGTH

Space wavelength that is just unable to propagate

in waveguide under such condition.

1-

So a waveguide allows a signal having a frequency more than

cut off frequency so a waveguide acts as high pass filter.

The lowest cut off frequency can be calculated through


mλo

2a ) ( 2

= 0

λo= 2a m

λo= cut off wavelength for dominant mode , m=1 , λo=2a

λp = λ

√ 1 - λ ( m 2a )

2

λp = λ

√ 1 - ( λ λo )

2

λp = guided wavelength

λ = f ee spa e a ele gth

fc =1.5*108 √ m a ( ) 2

+ ( n 2 b )

fc - lower cut off frequency

a,b – waveguide measurements

m,n – integers indicating the modes

PROBLEM

Calculate the lowest frequency and determine

the mode closest to the dominant mode of rectangular

waveguide 5.1cm*2.4cm.Calculate the cut off frequency of

dominant mode?

Ans) From the formula in previous slide for dominant mode is

TE1,0 so m=1,n=0,a=5.1*10-2 ,b=2.4*10-2 we get

fc = 2.94GHz

Mode closest to dominant mode can be determined by

substituting m , n values.

for m=0, n=1, we get fc = 6.25GHz



As fc = 5.8GHz in TE2,0 is close to fc= 2.94GHz in TE2,0 , therefore

mode closest to dominant mode is TE2,0 .


GROUP AND PHASE VELOCITY IN WAVEGUIDE

Vg = Vc sinθ Vp =

Vp Vg = (Vc )2 ------Eq 2

Vp=f λp

Vp=f

Substituting above equation in Eq 2 we get

Vg = Vc

The above equation represents that velocity of propagation or

group velocity in a waveguide is lower than in free space.

Group velocity decreases as the free space wavelength

approaches the cut off wavelength and is zero when the two

wavelength are equal. VNIT BT14ECE031 CHARAN SAI KATAKAM 14

sinθ Vc

θ

Vg

Vc

Vn

θ

λp

λ

λn λ

√ 1 - ( λ λo

) 2

Vp= Vc

√ 1 - λ ( ) λo 2

( √ 1- λ ( λo ) 2 )

PROBLEM

It is necessary to propagate a 10GHz signal in a

waveguide whose wall separation is 6cm.What is the greatest

no.of halfwaves of electric intensity which it will possible to

establish between 2 walls. Calculate the guide wavelength for

this mode of operation?

Solution:

a = 6 cm , f = 10GHz => λ = f/c = 3cm

λ0 = (2*a)/m

for m=1 => λ0 =12/1 =12 cm

for m=2 => λ0 =12/2 =6 cm

for m=3 => λ0 =12/3 =4 cm

for m=4 => λ0 =12/4 =3 cm

So maximum value of m is 4.


RECTANGULAR WAVEGUIDE

All equations are same for parallel plane waveguide and rect-

-angular waveguide are same except characteristic impedance.

TEm,0 MODE

Zo = where Z0 = characteristic impedance

Z = 120π = 377ohms (characteristic

impedance of free space).

If λ= λo in particular mode then Zo = ∞. i.e., a e p opagatio will have infinite resistance i.e., Vg = 0.


a

b

x

y

z

√ 1 - ( ) λ

λo

2

TEm,n MODES

The obvious difference between TMm,n mode

mode and those described thus foe is that magnetic field here

is transverse only and the electric field has a component in

the direction of propagation. Although most of the behaviour

of thus modes is same for TE modes, a no.of differences do

exist.

The first such difference is due to the fact that lines of magne-

tic field lines are closed loops consequently if magnetic field

exists and is changing in x- direction it must also exist and be

changing in Y- direction. Hence TMm,0 modes cannot exist in

rectangular wave guide.

If we want to propagate TM mode, if a wave is propagating in

z-direction it may propagate in x-direction and y-direction.

In TEm,0 mode if λ = λ0 then Z0 = 0, if potential difference

between two sides is zero there will be no variation of field. VNIT BT14ECE031 CHARAN SAI KATAKAM 17

CIRCULAR WAVEGUIDE

Let e i te al adius of i ula a eguide then cut of wavelength of waveguide is

λ0 = kr –solution of a vessel function equation.

The are values of Kr for different modes of transverse electric

and transverse magnetic modes of circular wave guide.


2πr kr

TE MODE TM MODE

MODE Kr MODE Kr

TEO,1 3.83 TMO,1 2.40

TE1,1 1.84 TM1,1 3.83

TE2,1 3.05 TM2,1 5.14

TE0,2 7.02 TM0,2 0.52

TE1,2 5.33 TM1,2 7.02

TE2,2 6.71 TM2,2 8.42

PROBLEM

Calculate the ratio of cross section of a circular

waveguide to that of rectangular waveguide if each is to have

same cut off wavelength for its dominant mode.

Solution: for circular for

λ0 = 2πr/(kr) = 2πr/1.84 = 3.14r

TE1,1 is dominant mode

r = (λ0 * 1.84)/ 2π = λ0/3.41

Ac = π*r2 = π*(λ0/3.41)2

for rectangular waveguide λ0 = 2a/m

TE1,0 is dominant mode => a = (λ0*1)/2 (since m=1)

Ratio of a : b = 2 :1,area = a *b = a*(a/2) = a2/2=(λ0/2)2/2

Ratio of cross sections is π*(λ0/3.41)2 : (λ0/2)2/2

2.17 : 1


CIRCULAR WAVEGUIDE

Smallest value of Kr, the TE1,1 mode is dominant in circular waveguide. The cut off wavelength of this mode is

λ0 = (2πr)/(1.84)

λ0 =1.7d d – inner diameter of circular waveguide.

Another difference lies in the different method of mode levelling, which must be used because of the circular cross se tio . The i tege de otes the no.of full wave intensity a iatio alo g the i u fe e e a d ep ese ts the no.of

half wave intensity changes radially out from the centre to the wall. It is seen that cylindrical co-ordinates are used here.

DISADVANTAGES OF CIRCULAR WAVEGUIDE OVER RECTANGULAR

WAVEGUIDE:

The first drawback associated with the circular waveguide is that its cross section will be much bigger in area than that of a corresponding rectangular waveguide used to carry the same signal.


CIRCULAR WAVEGUIDE

Another problem with circular waveguide is that

it is possible for the plane of polarisation to rotate during the

waves travel through waveguide. This may happen because of

roughness and discontinuities in the wall at departure from

circular cross section.

ADVANTAGES:

It is easier to manufacture than rectangular waveguide.

They are also easier to join together.


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Under guidance of INTRODUCTION OF WAVEGUIDES Joydeep …ece.vnit.ac.in/people/jsengupta/wp-content/uploads/sites/... · 2017-02-09 · INTRODUCTION OF WAVEGUIDES Under guidance of