Microwave PPT.ppt

WHAT IS MICROWAVE ????
Microwave is a descriptive term used to identify EM waves in
the frequency spectrum ranging approximately from 1 GHz to
30 GHz.
This corresponds to wavelength from 30 cm to 1 cm (λ =
c/f).Sometimes higher frequency ranging upto 600 GHz are
also called Microwaves.

APPLICATION
Communication
• UHF TV
• Microwave Relay
• Satt. Comm
• Tropo Sctter Comm
• Mobile Radio
• Telemetry
RADAR
• Search
• ATC
• Navigation
• Radar Altimeter
• Velocity Measurements
EWS
ECM
ESM
ECCM
Microwave Heating
• Industrial Heating
• Microwave oven
Industrial,Scientific,Medical
• Linear Acclerators
• Plasma Containment
• Radio Astronomy
• Whole body cancer theraphy

MAXWELL EQUATION
• Ampere’s Law
∫H.dl= I
I is the surface Integral of Conduction current density JT over the bounded area by
the path of Integration
The Total current Density
JT = J + ∂[D]/∂t
∫ H. dl = ∫s (J + ∂/∂t[D]) . ds
By applying Strokes Theorem
∫ H. dl = ∫s (Δ X H).ds
Hence Δ X H = J + ∂/∂t[D]

MAXWELL EQUATION
Faraday’ Law
• V= - = ∫E.dl Φ=
• ∫E.dl = - ∂/∂t ∫s B.ds
By applying Strokes Theorem
∫ E. dl = - ∂/∂t ∫s (Δ X E).ds
(Δ X E) = - ∂/∂t [B]
∫s B.ds

Gauss’s Law
• In Its Basic form
• ∫D.ds = ∫v ρdv
• Lt ∫ (D.ds)/dv = Δ.D
dv 0
• Δ.D= ρ
• In case of magnetic field the surface integral over of B over a
closed surface is zero. This is because a magnetic field has
neither a source nor a sink.
• Therefore Δ.B = 0

Wave Equation
We assume here the field vector vary with respect to time t in a sinusoidal manner
Then we have E = E0 exp (jωt)
Therefore we have
∂E/ ∂t = jωE
Therefore we have ∂/∂t = jω Similar way ∂2/∂t2 = -ω2
Now consider a medium which does not contain any free charges and is also non
conducting, for example free space or air.Then we have ρ = 0 , σ=0
Therefore maxwell ‘s 1st & 2nd Eqns give the following form
Hence Δ X H = J + ∂/∂t[D] = σ E + ∂/∂t (εE) = jωεE
Δ X E = - ∂/∂t [B] = - jωμH
Taking the curl of 2nd Eqn We have
Δ X Δ X E = ω2 μεE From Maxwell 3 rd Eqn We have Δ.D= ρ = 0
We know from vector Analysis Δ X Δ X E = ∆ . (∆.E) - ∆2E
Therefore ∆ . (∆.E) - ∆2E = ω2μεE Therefore ∆.E = 0
So ∆2E = -ω2μεE Similar way ∆2H = -
ω2 μεH

Wave Equation
Therefore maxwell ′s 1st & 2nd Eqns give the following
form
Hence Δ X H = J + ∂/∂t[D] = σ E + ∂/∂t (εE) = jωεE ..(1)
Δ X E = - ∂/∂t [B] = - jωμH…….(2)
Taking the curl of 2nd Eqn We have
Δ X Δ X E = ω2 μεE
From Maxwell 3 rd Eqn We have Δ.D= ρ = 0
We know from vector Analysis Δ X Δ X E = ∆ . (∆.E) - ∆2E
Therefore ∆ . (∆.E) - ∆2E = ω2μεE Therefore ∆.E = 0
So ∆2E = -ω2μεE
Similar way ∆2H = - ω2 μεH

Wave Equation in other form
• ∆2E = 1/c2 ∂2E/∂t2
• ∆2H = 1/c2 ∂2H/∂t2
• Where c is velocity of light in vaccum.
• These two Eqns represents wave propagating through free
space with a velocity equal to that of light and are also
known as HELMHOLTZ Equation.

TEM/ TE/ TM/ HE Modes
EZ,HZ
E
y
,
H
y
When Ez = 0 =Hz, mode is called TEM, When
only Ez = 0 it is called TE, When only Hz=0 It is
called TM, When Ez≠0,Hz≠ 0 , It is called HE/EH
Mode.

Relationship
• In rectangular co- ordinate system
• ∆2E = ∂2E/∂x2 + ∂2E/∂y2 + ∂2E/∂z2
• In Cylindrical co ordinate system
∆2E = ∂2E/∂ρ2 +(1/ρ) ∂E/∂ρ +(1/ρ2) ∂2E/∂φ2 + ∂2E/∂z2
Δ X E can be written as Δ X E can be written as
In rectangular Co-ordinate In Cylindrical Co-ordinate
i j k
∂/∂x ∂/∂y ∂/∂z
Ex Ey Ez
(1/ρ) uρ uɸ (1/ρ) uz
∂/∂ρ ∂/∂ɸ ∂/∂z
Eρ Eɸ Ez

A conclusion
• If a wave travelling along positive z direction
then the wave Eqn can be represented as
• ∆2Ez = - ω2μεEz
∆2Hz = - ω2 μεHz
Then its components that is plane wave is represented
by Ez= Eoz exp (-γz) Where γ =α+jβ
Therefore ∂/∂z = - γ
∂2/∂z2 = γ2

Some basic definition of Tx Lines
• Characteristics Impedance = Zo =[(R+jωL)/ (G + jωC)]½
• The Reflection Coefficient ρ = (ZL-Zo)/(ZL+Zo)
• VSWR S = (1+ I ρI)/(1- IρI)
• Attn Loss = 10 log10 [(Ei-Er)/Ei]
• Reflection Loss = 10 log10 [Ei/(Ei-Er)]
• Transmission Loss = 10 log10(Ei/Er)
• Return Loss = 10 log10 Ei/Er= -20 log10 IρI
• Insertion Loss = 10 log10E1/E2

Waveguide
• A waveguide consists of hollow metallic tube of rectangular or circular
shape used to guide an EM wave.
• In waveguides the E- fields & H- fields are confined to a space within
the guide. Thus no power is lost through radiation and even the
Dielectric Loss is negligible, since the guide are normally air-filled.
• It is possible to propagate several modes within the waveguide.
• Dominant Mode :- The dominant mode in particular waveguide is the
mode having the lowest cutoff frequency.
• It is advisable to choose the dimension of a guide in such a way that,
for a given input signal only the energy of dominant mode can be
transmitted through the guide.

Soln to the waveguide Equation
The process of solving waveguide Eqn may involve three steps:
• The desired wave equation are written in the form of either
rectangular or cylindrical co-ordinate.
• The boundary condition are then applied to the wave
equation set up in the above.
• The resultant eqn usually are in the form of partial differential
eqn in either freq. or time domain.

Solution of wave Equation in Rectangular co-ordinates
The E Field and H field in frequency domain are given by
∆2Ez = - ω2μεEz = γ2 Ez ∆2Hz = - ω2 μεHz = γ2 Hz
Where γ2= [jωμ(σ+ jωε)]
The Rectangular components of E & H satisfy the complex
scalar wave Eqn or Helmhotz Equation. i.e.
∆2 ϕ = - ω2με ϕ= γ2 ϕ
This can be written as
∂2ɸ/∂x2 + ∂2ɸ/∂y2 + ∂2ɸ/∂z2 = γ2 ϕ.
This is PDE of ϕ in 3D
Soln By the use of Separation of variables

• ϕ = X(x) Y(y) Z(z)
• Then We have
• (1/X)d2X/dx2 + (1/Y)d2Y/dy2+ (1/Z)d2Z/dz2 = γ2
• Since the sum of three terms on LHS is a constant and each
term is independently variable, It follows that each term
must be equal to constant
• Let us take -κ2
x -κ2
y -κ2
z = γ2
• d2X/dx2 = -κ2
xX
• d2Y/dy2 = -κ2
y Y
• d2Z/dz2 = -κ2
z Z

• The solution of the above three Equations
• X = A sin(κxx ) + B cos(κxx )
• Y = C sin(κyy) + D cos(κyy)
• Z = E sin(κzz) + F cos(κzz)
• The total Solution of the Helmhotz Eqn in Rec. Co-
ordinate
Φ = [ A sin(κxx ) + B cos(κxx )].
[C sin(κyy) + D cos(κyy)][E sin(κzz) + F cos(κzz)]

Soln of wave Eqn
• The propagation of wave in the guide is conventionally assumed in the
+ve direction. It should be noted that the propagation constant γg in
the guide differs from the intrinsic prop. constant γ of the dielectric.
Let
γg
2 = γ2+ κ2
x+ κ2
y = κ2
c+ γ2 κc = (κ2
x+ κ2
y) ½
κc Called cut off wave number.
For a lossless dielectric
γ2 = - ω2 με Then γg
2 = κ2
c+ γ2 γg
2 = - ω2 με + κ2
c γg=√(ω2 με - κ2
c)
There are three cases for the propagation constant in the guide------
I : There will be no wave propagation in the guide if ωc
2 με = κ2
c,
This is the critical condition for cut off propagation. Therefore the cut-
off frequency is given by
fc = (1/2π √εμ) κc

Case II : The wave will be propagating in the guide if ωc
2 με ˃ κ2
c,
and γg = (+-) jβg = (+-) jω √εμ √[1-(fc/f)2]
This means that the operating freq.must be above the cut-
off freq. in order to propagate in the guide.
CaseIII. The wave will be atteneuted if ωc
2 με ˂ κ2
c
and γg = α g = (+-) ω √εμ √[(f/fc)2-1]
This means that if the operating is below the cut-off freq, the wave will
decay exponentially wr t a factor of -α gz and there will be no wave
propagation becoz the constant is a real quantity. Therefore the soln of
the Helmhotz Eqn in Rec. co-ordinate
Φ = [ A sin(κxx ) + B cos(κxx )].
[C sin(κyy) + D cos(κyy)].[E sin(κzz) + F cos(κzz)] exp (- jβgz)

RWG Transmission
• Typically the di-electric is air.
• We have
• ∂2E/∂x2 +∂2E/∂y2 + ∂2E/∂z2 = γ2 E.
• ∂2E/∂x2 +∂2E/∂y2 + ∂2E/∂z2 = - ω2 με E
• ∂2E/∂x2 +∂2E/∂y2 + ∂2E/∂z2 + ω2 με E=0
• For lossless propagation in the +ve z direction all the three components
must have an exp (- jβgz) functional dependence. Therefore
• ∂2E/∂x2 +∂2E/∂y2 +(ω2 με - βg
2 ) E=0
• ∂2E/∂x2 +∂2E/∂y2 + κ2
c E=0 as ω2 με - βg
2 = κ2
c
• Similarly
• ∂2H/∂x2 +∂2H/∂y2 + κ2
c H=0
• Hence We will have total six components in x, y, z direction.

TE Mode solution
• The above condition require that B= 0, κx= mπ/a, D=0
and κy=nπ/b Where m & n are the +ve integer.
Hence We have Hz = AC exp (- jβgz) cos (nπ/b)x cos (mπ/a)y
= H0 exp (- jβgz) cos (nπ/b)y cos (mπ/a)x
This eqn represents all possible solutions of Hz for TE modes
in RWG.
The other components can be obtained from the original
curl eqn. For TE waves (Ez=0) the phasor form of these
equation reduces to
(i) βgEy = ωμHx (ii) ∂Hz/∂y+jβHy= jωεEx
(iii) βgEx = ωμHy (iv) jβgHx + ∂Hz/∂x = - jωεEy
(v) ∂Ey/∂x - ∂Ex/∂y = - jωμHz (vi) ∂Hy/∂x- ∂Hx/∂y = 0

RWG
• With some algebraic manipulation, all field components can be
expressed in terms of Hz
• Ex = - (jωμ/ κ2
c) ∂Hz/∂y Ey = (jωμ/ κ2
c) ∂Hz/∂x Ez= 0
Hx = - (jβg/ κ2
c)∂Hz/∂x Hy= - (jβg/ κ2
c)∂Hz/∂y
Hz= = H0 exp (- jβgz) cos (nπ/b)y cos (mπ/a)x
Substituting the value of Hz in the above Equation, We have,
Ex = - (jωμ/ κ2
c) H0(nπ/b) exp (- jβgz) sin (nπ/b)y cos (mπ/a)x
= E0x sin (nπ/b)y cos (mπ/a)x exp (- jβgz)
Similar way
Ey =Eoy cos (nπ/b)y sin (mπ/a)x exp (- jβgz)
Ez=0
Hx = Hxo cos (nπ/b)y sin (mπ/a)x exp (- jβgz)
Hy = Hoy sin (nπ/b)y cos (mπ/a)x exp (- jβgz)
Hz= = H0 exp (- jβgz) cos (nπ/b)y cos (mπ/a)x

Cut off Frequency ( Wave as High pass Filter)
• We know that
• κ2
c = γg
2 + ω2 με = (κ2
x+ κ2
y) =( mπ/a)2 +(nπ/b)2
• γg
2 = ( mπ/a)2 +(nπ/b)2 - ω2 με
• γg = √([( mπ/a)2 +(nπ/b)2 - ω2 με]
• At lower freq. ( mπ/a)2 +(nπ/b)2 ˃ ω2 με
• γg becomes real and +ve and the wave is completely
attenuated and there is no phase change. Hence the wave
can not propagate.
• However at higher freq, ( mπ/a)2 +(nπ/b)2 ˂ ω2 με, γg
becomes imaginary and propagation constant β
changesand wave propagates.
• At the transition, γg becomes Zero and the propagation
just starts.
• The frequency at which the γg just becomes zero is
defined as Cut-off freq.( Threshold Freq.) fc.

RWG
• However at higher freq, ( mπ/a)2 +(nπ/b)2 ˂
ω2 με, γg becomes imaginary and
propagation constant β changes and wave
propagates.
• At the transition, γg becomes Zero and the
propagation just starts.
• The frequency at which the γg just
becomes zero is defined as Cut-off freq.
( Threshold Freq.) fc.

Cut off Frequency ( Wave Guide as High pass Filter)
• γg = √([( mπ/a)2 +(nπ/b)2 - ω2 με]
• 0 = √([( mπ/a)2 +(nπ/b)2 - ωc
2 με]
• From this it can be written as
• fc= (c/2) √([( m/a)2 +(n/b)2 ] as c= 1/√(εμ)
• Cut off wave length = λc = c/ fc= c/(c/2) √([( m/a)2 +(n/b)2 ]
• = 2/ √([( m/a)2 +(n/b)2 ]
• All wavelength greater than λc are atteneuated and less than
will be propagated

• Guided WavelengtSome important definition
• h: (λg) : It is defined as the distance travelled by the wave in
order to undergo a phase shift of 2π It is related by the
Radians.It is related to the phase constant by the relation
λg=2π/ βg
Phase Velocity: (vp): In a waveguide vp the velocity of propagation is the
product of λg and f. But the speed of light is equal to the product of λo
and f. This Vp is greater than than the speed of light since λg ˃ λo. This is
contradicting since no signal can travel faster than light. However the
wavelength in the guide is the length of the cycle and vp represents the
velocity of phase.
In fact it is defined as the rate at which the wave chages its phase in
terms of guided wavelength.
i.e. vp= λg . f = (2πf . λ g)/2π = (2πf .)/2π/ λ g = ω/ βg
Since no intelligence or modulation travel at this velocity ‘ this velocity is
known as Phase Velocity.

Some important definition
• Group Velocity (vg):
If there is modulation in the carrier, the modulated envelope actually travels at
velocity slower than that of carrier alone and of course slower than speed of
light. The velocity of modulation envelope is called the Gp. Velocity.
This happens when a modulated signal travels in a waveguide, the modulation
goes on slipping backward w r t the carrier.
It is defined as the rate at which the wave propagates
through the waveguide and is given as
vg= dω/dβ

Expression for the Phase Velocity
• As we know vp= ω/ βg
• We also know
κ2
c = γg
2 + ω2 με = (κ2
x+ κ2
y) =( mπ/a)2 +(nπ/b)2
γg = α g +jβg For the wave to propagate γg = (+-)jβg
γg
2 = [(+-)jβg]2 = -β2
g = ( mπ/a)2 +(nπ/b)2 - ω2 με
At f= fc, ω = ωc, γg = 0
There fore ωc
2 με = ( mπ/a)2 +(nπ/b)2
Therefore γg
2 = [(+-)jβg]2 = -β2
g = ωc
2 με - ω2 με
β2
g = ω2 με -ωc
2 με
βg =√( ωc
2 με - ω2 με) = √( με) √(ω2 -ωc
2 )
Therefore vp= ω/ βg = ω/ [√( με) √(ω2 -ωc
2)] = c/ (√[1-(fc/f)2])
= c/(√[1-(λ0 / λc)2]

Expression for the Gp Velocity
• vg= dω/dβg
• As we know βg = √( με) (ω2 -ωc
2 ) then differentiating
dβg/ dω = 1/[2(√( με) (ω2 -ωc
2 )] . 2 ωμε
= √( με) / √[1-(fc/f)2]
• Therefore vg= dω/dβ = √[1-(fc/f)2]/ √( με) = c √[1-(fc/f)2] = c
√[1-(λ0/ λc)2
• Therefore we can conclude that
vg vp = c2

Relation between λ0, λc, λg
• vp= λg. f = (λg / λ0) . c
• vp = c/(√[1-(λ0 / λc)2]
From this two equation we have
(λg / λ0) . c = c/(√[1-(λ0 / λc)2]
After some manipulation We can write
1/ λ2
0 = (1/λ2
c)+ (1/λ2
g)
From this We can write λg = (λ0)/(√[1-(λ0 / λc)2]
For the case of wave guide with constant εr,
λg = (λ0)/(√[εr -(λ0 / λc)2] and λg dielectric = (λg air)/√(εr)
Since εr is always greater than 1 and εr˃ εdielectric hence
frequency less than cut-off frequency can pass through the
wave guide

TM modes solution
• From the boundary condition require that B= 0, κx= mπ/a,
D=0 and κy=nπ/b Where m & n are the +ve integer.
Hence We have Ez = E0z sin (nπ/b)y sin (mπ/a)x exp (- jβgz)
This eqn represents all possible solutions of Hz for TE modes
in RWG.
The other components can be obtained from the original
curl eqn. For TM waves (Ez=0) the phasor form of these
equation reduces to
(i) ∂Ez/∂y+jβgEy=- jωμ Hx (ii) jβgEx + ∂Ez/∂x = jωμHy
(iii) ∂Ey/∂x - ∂Ex/∂y =0 (iv) βgHy = ωεEx
(v) -βgHx + ∂Hz/∂x = ωεEy
(vi) ∂Hy/∂x - ∂Hx/∂y = - jωεEz

TM modes solution
• These equation can be solved simultaneously for Ex, Ey, Hx,
Hy in Terms of Ez and Hz
• Ex = - (jβg/ κ2
c)∂Ez/∂x
• Ey = - (jβg/ κ2
c) ∂Ez/∂y
• Ez= = E0z exp (- jβgz) sin(nπ/b)y sin (mπ/a)x
•
• Hx = (jωε/ κ2
c) ∂Ez/∂y
Hy = - ((jωε/ κ2
c))∂Ez/∂x
• Hz= 0

Ex = E0x sin (nπ/b)y cos (mπ/a)x exp (- jβgz)
Ey =Eoy cos (nπ/b)y sin (mπ/a)x exp (- jβgz)
Ez= E0z sin (nπ/b)y sin (mπ/a)x exp (- jβgz)
Hx = Hox cos (nπ/b)y sin (mπ/a)x exp (- jβgz)
Hy = Hoy sin (nπ/b)y cos (mπ/a)x exp (- jβgz)
Hz= = 0

TM & TE modes in RWG
Various TMmn modes:
• (i) TM00 Modes : If m=0 n=0 , We see all the components are
vanishing , Hence We will have trival solution.
• (ii) TM01 Modes: Again all the field components vanish
• (iii) TM10 Modes: Even now all are vanishing
• (iv) TM11 Modes: Now all the components Exist and for all
other values of m,n , the components exists.
Various TEmn Modes:
Similarway (i) TE00 Modes: TE00 does not exist.
• (ii) TE01 Modes : Ex Hy Exist, Hx=Ey=0
• (iii) TE10 Modes: Ex = Hy =0, Hx&Ey Exist
• (iv) TE11 Modes: All the components Exist
•

Dominant & degenerate Modes
• Dominant mode is that mode for which the cut off
wavelength assumes a max value.
• Cut off wave length = λcm,n = c/ fc= c/(c/2) √([( m/a)2 +(n/b)2 ]
• = 2/ √([( m/a)2 +(n/b)2 ]
• For (i) TE01 Modes : λc0,1 = 2b
• (ii) TE10 Modes: λc1,0 = 2a
• (iii) TE11 Modes: λc1,1 = 2ab/√(a2 +b2)
• Of these λc1,0 has max. value since a is the largest. Hence TE10 Mode is
the dominant mode for RWG.
• Degenerate modes : Whenever two or more modes have the
same cut-off freq, they are said to be degenerate modes.
• In a particular waveguide the corresponding TEmn & TMmn modes are
always degenerate. In square guide the TEmn, TEnm, TMmn, TMnm modes
form a foursome of degeneracy.

Wave Impedance Zz in TM and TE Waves.
• Wave Impedance is defined as the ratio of the strength of
electric field in one transverse direction to the strength of the
magnetic field along the other transverse direction.
• Wave Impedance for a TM wave in RWG
For a TM wave Hz = 0 and γg =jβg
Zz ™ = Ex / Hy
= [- (jβg/ κ2
c)∂Ez/∂x]/[- ((jωε/ κ2
c))∂Ez/∂x] = βg/ ωε
= √(ω2 με -ωc
2 με)/ ωε = √(μ/ε) √ [1-(λ0 / λc)2]
For air, √(μ/ε) = 377 Ω = η
Therefore Zz ™ = η √ [1-(λ0 / λc)2]
Similarway Zz
TE = η /√ [1-(λ0 / λc)2], This shows that wave
impedance for TE mode is always greater than η

Poynting Theorem
• At what rate will electromagnetic energy be
transmitted through free space or any
medium, be stored in the E field or M field,
and be dissipated as heat?
• From the standpoint of complex power in
terms of complex field vectors, the time
average of any two complex vectors is equal to
the real part of the product of one complex
vector multiplied by the complex conjugate of
the other vectors. Hence
• < P> = <ExH> = (½) Re (ExH*)

Poynting Theorem
• It is neceessary to define a complex poynting vector
as P => = (½) Re (ExH*)
• Maxwell Equation in frequency domain
• ∆XE = -jωμH
• ∆XH = J+jωεE
• (∆XE).H* = -jωμH. H*
• (∆XH*).E = (J*-jωεE*).E
• Substracting

Poynting Theorem
• E.(∆XH*)- H*.(∆XE) =E.J* -jω[ε|E|2-.μ|H|2]
• -∆.(ExH*) = E.J* -jω[ε|E|2-.μ|H|2]
• (½)∆.(ExH*) = -(½)E.J* +(½) jω[ε|E|2-.μ|H|2]
• After putting J= (Jo*+σE) and P=(½) (ExH*) Manipulating We
have
-(½)E.Jo*= (½) σE. E*+(½) jω[(½)μH.H*-(½)εE.E*] + ∆.P
Integrating over a volume and application of Gauss’s theorem
We have
∫(½)E.Jo*= ∫(½) σE. E*+∫(½) jω[(½)μH.H*-(½)εE.E*] +∫ ∆.P

Poynting theorem
• ½ σ║E2║ = σ˂║E2║˃ is the time average dissipated power.
• ¼ μH.H* = ½ μ ˂║H2║˃ = wm is the time average magnetic
stored energy
• ¼ εE.E* = ½ ε ˂║E2║˃ = we is the time average electric
stored energy
• - ½ E.J0* = The complex power impressed by the source Jo
in the field
• Pin = - ʃv ½ (E.J0*)dv =The total complex power impressed by
the source J0 within the region.

Poynting theorem
• ˂Pd˃ = ʃv ½ σ║E2║dv be the time average dissipated as
heat inside the region.
• Wm-We = ʃv (wm -we ) dv be the difference between time av.
Mag. And elec field stored within the region.
• Ptr = ∫ P.ds be the transmitted power from the region.
• The complex poynting theorem can be written as
• Pin= ˂Pd˃ +j2ω ˂ Wm-We ˃ + Ptr

Power transmission through RWG
• The power transmitted a RWG (assuming the waveguide is infinitely long
and no reflection are coming from load end)
Ptr = ∫ P.ds = ∫ (E X H*).ds
For a lossless dielectric, the time av. Power flow through a RWG is given by
Ptr = 1/(2Zz) ʃa║E2║da = (Zz)/2 ʃa║H2║da
Where Zz= Ex / Hy = -Ey/ Hx
Where ║E║2 = ║Ex║2+ ║Ey║2
and ║H║2= ║Hx║2 + ║Hy║2
For the TEmn the av power transmitted through a RWG is
Ptr = [√ [1-(λ0 / λc)2] / (2η)] x [ʃ0
a ʃ0
b[║Ex║2+ ║Ey║2]dx dy]
For the TMmn the av power transmitted through a RWG is
Ptr = 1/ [√ [1-(λ0 / λc)2] / (2η)] x [[ʃ0
a ʃ0
b ║Ex║2+ ʃa║Ey║2]dx dy]

Power Losses in RWG
• Losses in Dielectric
• Losses in the guided wall
• Power loss in Dielectric
For low loss dielectric (σ ˂˂με), The propagation constant for a
plane wave travelling in unbounded lossy dielectric is given by
α = (σ/2) √(μ/ε) =ησ/2
The attn caused by the low loss dielectric in RWG is given by
αg =[ησ/2]/ [√ [1-(λ0 / λc)2] for TE modes
αg =[ησ/2]x [√ [1-(λ0 / λc)2] for TM modes

Power Loss es in the guide wall
• The E & H field propagate through a lossy waveguide, their
magnitudes
• ║E║ = ║Eoz║ exp ( - αgz)
• ║H║ = ║Hoz║ exp ( - αgz)
• It is interesting to note that, for a low loss guide, the time av.
Power flow decreases proportionally to exp(( - 2αgz)
• Hence Ptr = (Ptr + Ploss) exp(( - 2αgz)
• For Ptr ˂˂ Ploss and 2αgz ˂˂1,
•
• (Ploss/ Ptr) + 1 = 1+ 2αgz
• Finally αg= PL/ 2Ptr PL is the loss per unit length

Power Losses in the guide wall
• Since the E field and H field established at the surface of low
loss guide wall decay exponencially wrt the skin depth while
the waves progress in the wall, it is better to define a surface
resistance of the guide wall as
• Rs = ρ/δ = 1/(σδ) = (αg)/σ = √ (πfμ)/σ Ω/Square
• The power loss/ lentgh of guide
• PL = (Rs/2) ʃ ║Ht║2 ds Ht is the tangential components of the
H field
• αg= [ Rs ʃs ║Ht║2 ds]/[2Zz ʃa║H║2da]
Where ║H║2 = ║Hz║2+ ║Ey║2
and ║Ht║2= ║Htx║2 + ║Hty║2

TE10 mode in RWG
• An air field waveguide with
a cross section 2x1 cms
transport energy in the TE10
mode at the rate of 0.5 hp.
The impressed frequency is
30 GHz. What is the peak
value of the E- Field
occuring in the guide?
• SOLUTION
• (1)Find the βg
• (2) The power delivered
in the z direction by the
guide is
P = Ptr =Re [1/2 ʃ0
b ʃ0
a (E X
H*)]dx.dy uz
• Put the value of Ey & Hx
and get the answer.

Characterisrics of standard Waveguide
• The side dimension of RWG designated as EIA WR
(90) by Electronic Industry association
• For Example The X band frequencies the outside
dimension of a RWG is 2.54 cm wide and 1.27 cm
high and its inside dimension are 2.286 and 1.016
cms respectively

Circular Waveguide
Circular Waveguide

CWG
• In cylindrical co-ordinate system, the Helmhothz
Eqn can be written as
• (1/r) ∂/∂r(r ∂Ez/∂r) + (1/r2) ∂2Ez /∂Φ2 = -κ2
c Ez
• (1/r) ∂/∂r(r ∂Hz/∂r) + (1/r2) ∂2Hz /∂Φ2 = -κ2
c Hz

CWG
TE modes
In this case Ez = 0 and Hz not equal to 0.
For Propagation in + ve z direction We have Hz = Hz* exp (-jβz)
Hz* is the fn of r and Φ only.
There the Equation becomes
• (1/r) ∂/∂r(r ∂ Hz* /∂r) + (1/r2) ∂2 Hz* /∂Φ2 = -κ2
c Hz*
Again use the separation of variable……..
i.e. Hz* = R Ψ Then the Equation becomes
d2R/dr2 + (1/r) dR/dr +[κ2
c- (n2/r2)] R = 0
And d2 Ψ /dΦ2 + n2 Ψ =0 n2 is a constant

CWG
• The solution of the Equation
R= A Jn(κcr) + B Nn (κcr)
Ψ = C cos (n Φ) +D sin (n Φ)
Since the field must remain finite at r=0, the Nn term must be excluded
and therefore B=0.
Since the field, and Hence Ψ must be single valued, n is restricted to
integer value. Finally, one of the two Ψ terms can be eliminated as
both of this leads to same field pattern except that one is rotatiing
90o in Φ direction.
Setting D=0 yields
Hz* = Ho Jn(κcr) cos (n Φ) Where Ho = AC is an arbitary amplitude.

CWG
• At r= a, the boundary condition
requires that
∂Hz/∂r= 0, which means J'n(κca) =0
The m th root of this equation is
designated by q'nm and thus
q'nm = κca several roots value are
available
The right hand side table gives the
different value of q'mn in TEmn
modes.
m
n
1 2
0 3.832 7.016
1 1.841 5.331
2 3.054 6.706

CWG
• Now The Equation for Hz may be written as
Hz= Hz* exp (-jβz) = Ho Jn(κcr) cos (n Φ) exp (-jβz)
Where κc= q'mn/a
The remaining field equation can be obtained from original curl
Equation. For TE waves The phasor form of the equation
are
(1) βgEΦ= -ωμHr (2) (1/r) (∂Hz/∂Φ) + jβgHΦ=jωεEr
(3) βgEr = ωμ HΦ (4) jβgHr +∂Hz/∂r= - jωεEΦ
(5) (1/r) {∂ (rEΦ)/∂r - ∂(Er)/∂Φ}= -jωμHz
(6) (1/r) {∂ (rHΦ)/∂r - ∂(Hr)/∂Φ}= 0
These equation can be solved for EΦ ,Hr ,HΦ,Er in terms of Hz and
Ez.

CWG
(1) EΦ= =(-jωμ)/ -κ2
cr ∂(Hz)/∂r (2) HΦ= (-+) EΦ/ ZTE
(2) Er =(-jωμ)/ -κ2
cr ∂(Hz)/∂Φ (4) Hr = (+-) HΦ / ZTE
By substituting the Hz in the above Equation We have the different field
components.
Er =(-jωμ)/ -κ2
cr ∂(Hz)/∂Φ
= jHo (nωμ)/ (κ2
cr) Jn(κcr) sin (n Φ) X exp (-jβz)
EΦ=jHo (ωμ)/ (κc) J'n(κcr) cos (n Φ) X exp (-jβz)
Er = (+-) HΦ / ZTE
HΦ= (-+) EΦ/ ZTE Where κc= q'mn/ a
and Zz
TE = η /√ [1-(λ0 / λc)2

CWG
• As we know β2
g =(ω2 με - κ2
c)
• βg =√(ω2 με - κ2
c)
• As we know κc= q‘nm/ a
• Therefore , The cut-off wave number of a mode is that mode for which the
mode propagation constant vanishes.
• Hence
0=β2
g =(ω2 με - κ2
c)
κc = ωc √ με = q‘nm/ a
The cut off frequency for TE mode in CWG is
fc = (1/2π √εμ) κc = (q'mn /2π a √εμ)
Therefore λc = 2π a / q‘nm

CWG
• Λc will be maximum if q‘nm
• is minimum.
• Therefore the value of
q‘nm is found to be minimum i.e.
1.841 for n =1, m=1
Hence TE11 is called dominant mode
in CWG
i.e. λc = 2π a / q‘nm
= 2π a / 1.841
m
n
1 2
0 3.832 7.016
1 1.841 5.331
2 3.054 6.706

CWG TM modes
qnm for TM modes
m
n
1 2
0 2.405 5.520
1 3.832 7.016
2 5.135 8.417

• Q1 Define the Degenerate Modes in CWG `with
Example.
• Q2 An air field CWG has a radius of 2 cms and is
to carry energy at a frequency of 10 GHz. Find all
the TEmn and TMmn modes for which energy
transmission is possible.

Ad, Disad, of CWG
• It is easier to manufacture
• Easier to join
• TM01 & TE01 are symmetrical,Hence rotation of polarization
can be overcome
• TE01 mode in CWG has the lowest attn /unit length of
waveguide, Hence suitable for long distance tx.
• DISAD…………………………….
• Occupy more space
• Plane of polarization rotates
• Very difficult to separate dominant modes from infinite no. of
modes available.

Application
• As a rotating joints in Radars
• TE01 mode in CWG has the lowest attn /unit length of waveguide,
Hence suitable for long distance tx.
• Short and Medium distance broadband comm.( could replace
/share coaxial and microwave links)

Characteristics of Standard CWG
• The inner diameter of a CWG is regulated by the freq. of the
signal being transmitted.
• For example : at X band freq., the inner diameter of a CWG
designated as EIA WC (94) by the Electronic Industry
Association is 2.383 cm.

Microwaves Cavities
• Cavities, or resonators, are used for storing energy
• Used in klystron tubes, band-pass filters and frequency
meters
• It’s equivalent to a RLC circuit at high frequency
• Their shape is that of a cavity, either cylindrical or
cubical or Rectangular
They serve as key elements in osc,,tuned amplifiers,
frequency meters, phase equalizers as well as BPF

CR
• It is a metallic enclosure that confines the EM energy.
• The stored E & M energies inside the cavity determine its
equivalent inductance and capacitance.
• The energy dissipated by the finite conductivity of the cavity
walls determine its equivalent resistance.
• Example RCR, CCR, Re-Entrant Cavity
• A given resonator has a infinite no of modes and each modes
corresponds to a definite resonant frequency.
• When the frequency of an impressed waveform is equal to a
resonant freq, a max amplitude of the standing waves occurs, and
the peak energy stored in the E & M Fields are equal.
• The mode having the lowest resonant freq. is known as dominant
mode.

CR
Its Q is very very high.
Q= 2π (energy stored)/(energy Loss per cycle)
Unloaded Qu =ωr (energy stored in the resonant
circuit)/(Power Loss in the resonant circuit)| at ωr
External QE =ωr (energy stored the resonant circuit)/(power
Loss in the external circuit)|at ωr
Loaded QL =ωr (energy stored in the resonant circuit)/(Total
Power Loss)|at ωr

CR
• It can easily proved that
(1/ QL) = (1/ QE) + (1/Qu)
Therefore We can say QL is a function of the degree of coupling
between the resonant circuit and external circuit.

CR
• Hollow metallic enclosure exhibit resonance behavior when excited
by an EM field. These enclosures are known as CR and are widely
used above 3 GHz.
• Typical value of Q ranging from 5000 to 50000
• CR has a infinite no of modes and each modes corresponds to a
definite resonant frequency
• These can be determined by Maxwell Equation.

RCR
• The cavity is resonant at freq. where length I is multiple of
λg/2 , λg is guided wavelength along Z- axis.
•
• At certain freq, an E- Field null occur at z=0

The resonant condition is given by l = p λg/2
• For βl =pπ, the waves adds constructively thus producing the
resonant condition

The resonant condition is given by l = p λg/2
For βl =pπ, the waves adds constructively thus producing the resonant
condition

Field Analysis in TE modes:
∂2Hz/∂x2 +∂2Hz/∂y2 + κ2
c Hz=0
:
obtain
we
where
from
)
(
)
(
)
(
)
,
,
(
:
Variables
of
Separation
by
Solving
z
Z
y
Y
x
X
z
y
x
H z 
z
k
c
z
k
c
z
Z
y
k
c
y
k
c
Y(y)
x
k
c
x
k
c
X(x)
z
z
y
y
x
x
sin
cos
)
(
sin
cos
sin
cos
6
5
4
3
2
1






2
2
2
2
z
y
x k
k
k
k
where 


Cavity TE Mode to z

Rectangular CR
• TEmnp Boundary
Conditions
,b
y
E
,a
x
E
,c
z
H
x
y
z
0
at
,
0
0
at
0
0
at
0






From these, we conclude:
kx=mπ/a
ky=nπ/b
kz=pπ/c
where c is the dimension in z-axis



















c
y
p
b
y
n
a
x
m
H
H o
z



sin
cos
cos

RCR In TE modes
• Substituting the value of kx ky and kz in the first
equation we have
• [(mπ/a)2 + (nπ/b)2 +(pπ/c)2 - ω2 με] Hz=0
If Hz = 0, all the filed components will vanish
and the solution is trival. In order for the field
to be non- zero the following condition must
hold
(mπ/a)2 + (nπ/b)2 +(pπ/c)2 = ωr
2 με
fr = [1/(2 √ με)][(m/a)2 + (n/b)2 +(p/c)2]

Cavity TM Mode to z
:
obtain
we
where
from
)
(
)
(
)
(
)
,
,
(
:
Variables
of
Separation
by
Solving
z
Z
y
Y
x
X
z
y
x
Ez 
z
k
c
z
k
c
z
Z
y
k
c
y
k
c
Y(y)
x
k
c
x
k
c
X(x)
z
z
y
y
x
x
sin
cos
)
(
sin
cos
sin
cos
6
5
4
3
2
1






2
2
2
2
z
y
x k
k
k
k
where 



TMmnp Boundary Conditions
• From these, we conclude:
• kx=mπ/a
• ky=n π /b
• kz=p π /c
• where c is the dimension in z-axis
,c
z
E
E
,a
x
E
,b
y
E
x
y
z
z
0
at
,
0
0
at
0
0
at
0








Cavity TM Mode to z








2
2
2
2
2
cos
sin
sin









































c
p
b
n
a
m
k
where
c
z
p
b
y
n
a
x
m
E
E o
z

Resonant frequency
• The resonant frequency is the same for TM or TE
modes, except that the lowest-order TM is TM110 and
the lowest-order in TE is TE101.
2
2
2
2





















c
p
b
n
a
m
c
fr

CCR
TE modes
In this case Ez = 0 and Hz not equal to 0.
For Propagation in + ve z direction We have Hz = Hz* exp (-jβz)
Hz* is the fn of r and Φ only.
There the Equation becomes
• (1/r) ∂/∂r(r ∂ Hz /∂r) + (1/r2) ∂2 Hz /∂Φ2 + ∂2 Hz /∂z2 = -ω2 με Hz
Again using the separation of variable..
i.e. Hz = R Ψ Z
Then the Equation becomes
(1/rR) ∂/∂r(r ∂ R /∂r) + (1/r2 Ψ) ∂2 Ψ/∂Φ2 +(1/Z) ∂2 Z /∂z2 = -ω2 με

CCR
• Let
(1/rR) ∂/∂r(r ∂ R /∂r) + (1/r2 Ψ) ∂2 Ψ /∂Φ2 = - κ2
c
and
(1/Z) ∂2 Z /∂z2 = - κ2
z
Therefore
(r/R) ∂/∂r(r ∂ R /∂r) + κ2
c r2 = -(1/Ψ) ∂2 Ψ /∂Φ2 …………(1)
(1/Z) d2 Z /dz2 = - κ2
z …………………………………(2)
Where κ2
c + κ2
z = ω2 με

CCR
• Let RHS of Equation (1) = n2
Therefore
-(1/Ψ) ∂2 Ψ /∂Φ2 = n2
d2 Ψ/dΦ2 + n2 Ψ = 0…………………(3)
Therefore Eqn (1) reduces to
(r/R) d2/dr2+(1/r) dR/dr) + (κ2
c - n2 /r2)R =0………………(4)

CCR
The soln of these Eqn (2)(3)(4)
Z = A sin(κzz) + B cos(κzz)
Ψ = C cos (n Φ) +D sin (n Φ)
R= E Jn(κcr) + F Nn (κcr)
Since the field must remain finite at r=0, the Nn term
must be excluded and therefore F=0.
Since the field, and Hence Ψ must be single valued, n is
restricted to integer value.
Finally, one of the two Ψ terms can be eliminated as
both of this leads to same field pattern except that
one is rotatiing 90o in Φ direction.
Setting D=0 yields

CCR
• Then the total solution
• Hz = (CE) Jn(κcr) cos (n Φ)[A sin(κzz) + B cos(κzz) ]
• ………….(5)
• Using boundary condition i.e. at z=0 and z=d require that
Hz =0.
Therefore B= 0 and kz=pπ/d
Now Eqn (5) reduces to
• Hz = Ho Jn(κcr) cos (n Φ)[ sin(κzz) ] ………….(6)

CCR
• Again boundary condition
at r=a, require that
dHz/dr = 0
Which means J'n(κcr) =0 .
The m th root of this Equation is
designated by q'mn
and thus q'mn = κca Several roots are
given below in the Table.
m
n
1 2
0 3.832 7.016
1 1.841 5.331
2 3.054 6.706

CCR
• The resonance condition is obtained by substituting Eqn (6) in
the equation below
• Then we have
κ2
c + κ2
z = ωr
2 με = (q'mn /a)2 +(pπ/d)2
fr = (c/2π) [(q'mn /a)2 +(pπ/d)2]1/2
(1/r) ∂/∂r(r ∂Hz /∂r) + (1/r2) ∂2 Hz /∂Φ2 + ∂2 Hz /∂z2 = -ω2 με Hz

CCR
• So We have the field components as below
Er =(-jωμ)/ -κ2
cr ∂(Hz)/∂Φ
= jHo (nωμ)/ (κ2
cr) Jn(κcr) sin (n Φ) X exp (-
jβz)
EΦ=jHo (ωμ)/ (κc) J'n(κcr) cos (n Φ) X exp (-jβz)
Hr = (+-) HΦ / ZTE
HΦ= (-+) EΦ/ ZTE Where κc= q'mn/ a

CCR
Hence the field components can be derived using the above equation.
Er =(-jωrμ Ho)/ (-κ2
cr) (n/r)/ Jn(κcr) sin (n Φ) X sin(κzz)
EΦ=(-jωrμ Ho)/ (-κ2
cr) J'n(κcr) cos (n Φ)
X cos(κzz)
Hr = ( κz/κc) HoJ'n(κcr) cos (n Φ)
HΦ= (- κz/ κ2
c) (n/r) Ho Jn(κcr) sin (n Φ)Xcos (κzz)
Where κc= q'mn/ a

Microwave Hybrid Junction
• A microwave ckt ordinarily consists of several
microwave devices connected in some way to
achieve the desired transmission of a
microwave signal.
• The interconnection of two or more
microwave devices may be regarded as
microwave junction.

S parameter
• A two port device can be described by a no. of parameter like
H Parameter
Where
V1 = h11 I1 + h12 V2
I2 = h21 I1 + h22 V2

S parameter
• Y parameter
Where
I1 = y11 V1 + y12 V2
I2 = y21 V1 + y22 V2

S parameter
Z Parameter:
Where
V1 = z11 I1 + z12 I2
V2 = y21 I1 + y22 I2

S parameter
All these N/W parameters relate total voltage and total current at
each of the two port.
For instant h11= V1 / I1 At V2= 0 i.e . O/p is short
h12= V1 / V2 At I1= 0 i.e . i/p is open

S parameter
Now the Question is , If the freq. is microwave range, Are these
Parameters useful ?
Answer is Y/N
The reason behind this is
(1)Eqpt is not ready available to measure the total voltage and current
at the poet of N/W
(2) Sh and Open are dificult to obtain over a broadband of freq.
(3) Active devices such as power transistor, tunnel diodes, etc will not
have stability for sh. and open ckt.

S Parameter
A “black box” or network may have any
number of ports.
This diagram shows a simple
network with just 2 ports.
S parameter

S Parameter
• S Parameter
S-parameters are measured by sending a single frequency
signal into the network or “black box” and detecting what
waves exit from each port.
Power, voltage and current
can be considered to be in
the form of waves travelling
in both directions.

S Parameter
Power, voltage and current
can be considered to be in
the form of waves travelling
in both directions.
For a wave incident on Port 1,
some part of this signal
reflects back out of that port
and some portion of the signal
exits other ports.
S-parameters are measured by sending a single frequency signal
into the network or “black box” and detecting what waves exit from
each port.

S Parameter
• First lets look at S11.
• S11 refers to the signal
reflected at Port 1 for the
signal incident at Port
Scattering parameter
S11 is the ratio of the
two waves b1/a1.

S Parameter
• Now lets look at S21.
• S21 refers to the signal
exiting at Port 2 for the
signal incident at Port 1.
Scattering parameter S21
is the ratio of the two
waves b2/a1.

S Parameter
• S Parameter
Now lets look at S21.
S21 refers to the signal
exiting at Port 2 for the
signal incident at Port 1.
Scattering parameter S21
is the ratio of the two
waves b2/a1.
S21? Surely that should be S12??

S Parameter
• A linear network can be characterised by a set of simultaneous
equations describing the exiting waves from each port in terms of
incident waves.
•
•
• S11 = b1 / a1
•
• S12 = b1 / a2
•
• S21 = b2 / a1
•
• S22 = b2 / a2
•
•
• Note again how the subscript follows the parameters in the ratio
(S11=b1/a1, etc...)

S Parameter
• S-parameters are complex (i.e. they have magnitude
and angle) because both the magnitude and phase
of the input signal are changed by the network.
(This is why they are sometimes referred to as
complex scattering parameters).

S Parameter
• What do S-parameters depend on?
S-parameters depend upon the network
and the characteristic impedances of the
source and load used to measure it, and
the frequency measured at.
i.e.
if the network is changed, the S-parameters change.
if the frequency is changed, the S-parameters change.
if the load impedance is changed, the S-parameters change.
if the source impedance is changed, the S-parameters change.

S Parameter
• A little math
This is the matrix algebraic representation
of 2 port S-parameters:
S matrices are symmetrical.

• A little math…………….
A little math…
This is the matrix algebraic representation
of 2 port S-parameters:
Some matrices are symmetrical.
In the case of a 2-port network, that means that S21 = S12 and interchanging
the input and output ports does not change the transmission properties.
A transmission line is an example of a symmetrical 2-port network.

• A little math…
Parameters along the leading diagonal,
S11 & S22, of the S-matrix are referred to as
reflection coefficients because they refer to
the reflection occurring at one port only.

Larger networks:
A Network may have any number of ports.
The S-matrix for an n-port network contains n2 coefficients (S-parameters),
each one representing a possible input-output path.
The number of rows and columns in an S-parameters matrix is equal to the
number of ports.
For the S-parameter subscripts “ij”, “j” is the port that is excited (the input port)
and “i” is the output port.

S Parameter
• This S parameter can be represented by
b1 = S11a1 + S12 a2
b2 = S21 a1 + S22 a2
In General We Have [b] = [S] [a]

Properties of the S Matrix
1. [S] is always a square Matrix of order of (nxn)
1. [S] is a symmetric Matrix i.e. Sij = Sji
1. [S] is a unitary Matrix i.e. [S][S*] = [I]
1. The sum of the products of Each term of any row ( or
column) multiplied by the complex conjugate of the
corresponding terms of any other row (column) is ZERO
i.e. Σ Sik S*ij = 0
1. In any of the terminal or reference plane ( say Kth port)
are moved away from the junction by an Electric distance βk
lk, Each of the co-efficient Sij involving k will be multiplied by
factor exp (-j βk lk)

WG TEEs
• TEEs
In Microwave ckt a waveguide or co-ax line junction
with three independent port is commonly referred
as TEE Junction.
So it should have the matrix of 9 Elements (3x3).

E Plane TEEs
• In Microwave ckt a waveguide or co-ax line junction with three
independent port is commonly referred as TEE Junction.
• So it should have the matrix of 9 Elements (3x3).
Main Guide

S Parameter for E plane TEE
• S=
• S parameter S11, S22, S33 will be Zero as Plane is
perfectly matched.
• S13 = - S23 (?)

• From the sym. Property of S Matrix We can have
• S12 = S21, S13 = S31, S23 = S32………(1)
From the Zero Property of S Matrix
S11 S12* + S21 S22* + S31 S32* = 0 …………(2)
Hence S13 S23* = 0
Hence Either S13 = 0 or S23* =0
From the Unity Property of S Matrix,
We have S21 S21* + S31 S31* = 1…….(3)
S12 S12* + S32 S32* = 1……..(4)
S13 S13* + S23 S23*= 1……….(5)

By some manipulation
we have |S12|2 = 1- |S13|2 = 1- |S23|2
This inconsistency proves that The TEE junction cannot be matched to
three arms.
In other words the diagonal elements of S matrix of a TEE junction are
not all ZEROS.

• Hence the S matrix of E plane TEE is
• S =

So the diagonal elements of S matrix of a TEE
junction are not all ZEROS.
But If one port for e.g. is zero i.e S33 = 0
Then what happen??????????????
• S=

• The S matrix will be
• S=

• From unitary property
[S][S*] = [I]
Therefore

R1 C1 |S11|2 + |S12|2 + |S13|2 = 1
R2 C2 |S12|2 + |S22|2 + |S13|2 = 1
R3 C2 |S13|2 + |S13|2 = 1
R3 C1 S13 S11*- S13 S12* = 0
AS S11 = S22
From Above Equation We have
S13 = 1/(√2)
S13( S11*-S12*) = 0
Therefore
S11 = S12 = S22

• Using these values, We have
S11 = ½
Substituting the all value of S Matrix Elements
We have
S =

As We know
[b] = [S] [a]
Hence We have

b1 = ½ a1 + ½ a2 + 1/(√2) a3
b2 = ½ a1 + ½ a2 - 1/(√2) a3
b3 = 1/(√2) a1 - 1/(√2) a2
Case I : a1= a2 = 0 but a3 ≠ 0
Therefore b1 = 1/(√2) a3
b2 = -1/(√2) a3
b3 = 0

• At input at port 3 equally divides between port 1
and 2 but introduces a phase shift of 1800
.
HenceE plane TEE is also working as a 3 dB splitter.
Case II : a1= a2 = a but a3 = 0
Therefore b1 = a, b2 = a b3 = 0
i.e. Equal i/p at the port

Q factor of a cavity Resonator
• It is a measure of the freq. selectivity of a resonant
and anti resonant circuit
• Q= 2π x( max. Energy stored)/ Energy dissipated per
cycle
• Q = ω W / P ……………………….(1)
• We = ʃv ½ ε║E2║dv = wm = ʃv½ μ ║H2║ = W…..(2)

Q factor of a cavity Resonator
• P = ½ Rs ʃ |Ht | 2da ( It is over the surface of the resonator)
…………..(3)
Putting the value of P and W of Equation 2 and 3 in Eqn (1)
We have
Q = [ω μ ʃv½ ║H2║dv] / [Rs ʃ |Ht | 2da ]
Since the the peak value of magnetic intensity is related to its
tangential and normal components by
║H║2= ║Ht║2 + ║Hn║2
The value of ║Ht║2 at the resonator wall is approx. twice the
value of ║H║2 averaged over the volume.
Hence Q = ω μ(volume)/ 2 Rs

DC
The Spacing b/w the centre of two holes must be
L = (2n+1) λg/4

DC
• S Matrix
• In DC, All ports are perfectly matched.
Hence S11 = S22= S33 =S44 = 0
There is no coupling in P1 & P3 and P2 & P4
Thus
S13 = S31 = 0
S24 = S42 = 0

DC
• With the help of Zero Property
S12 S14* + S32 S34* = 0 ………….(1)
ІS12 І ІS14 І = ІS32 І ІS34 І
S21 S23* + S41 S43* = 0
ІS21І ІS23І = ІS41І ІS43 І
From Unitary property
S12 S12* + S14 S14* = 1 ………………….(2)
As We have symmetric property
S12= S21, S14= S41, S23=S32, S34=S43 then We
Have

DC
• ІS12 І =ІS34 І
• ІS14 І = ІS32 І = ІS23 І
Let ІS12 І =ІS34 І = p p is a positive and real
Then From (1)
p(S23* + S41) = 0
Let S23 = S41=S14= jq q is +ve and Real
Then from (2)
p2+ q2 = 1………………………….
THEN…………………….

Ferrites
• Ferrite are non-metallic substance with
resistivity nearly equal to times
greater than Metals & μr of the order of 1000
and Dielectric constant 10- 15 times.
They have magnetic properties similar to that of
ferrous metal
They are oxide based compound……..a mixture
of metallic oxide and ferric oxide having
general formula Meo.Fe2o3

Ferrites
• Because of their spinning Properties, It finds application to
reduce the reflected power, modulation purposes, and
switching circuits.
• It can be used up to 100 GHz.
• It is having the non- reciprocal properties.
• Another Property: When two CP waves, one rotating
clockwise and another rotating anti-clockwise are made to
propagate through a ferrite, the material react differently to
the two rotating fields, thereby presenting different charac.
(like ε μ ρ ) to both the waves.

Ferrites
• FARADAY ROTATION IN FERRITES
Consider an infinite lossless medium . A satic B0 is applied along the
z direction. A plane TEM waves that is linearly polarised along the X
axis is made to propagate through the ferrite in the z direction.
THE PLANE OF POLARISATION OF THIS WAVE WILL ROTATE WITH
DISTANCE

Ferrites
• E = axEo = (ax+jay) Eo/2 + (ax-jay) Eo/2 at z=0
At z= l
E = (ax+jay) Eo/2 exp(jβ-l) + (ax-jay) Eo/2 exp(jβ+l)
= (ax) Eo/2 [exp(jβ-l) + exp(jβ+l)]+( jay) Eo/2 [exp(jβ-l) -
+ exp(jβ+l )]
= Eo/2 [exp(-j(β-+β+)l/2 [ax exp(-j(β--β+)l/2 +exp (j(β--
β+)l/2 )(jβ+l)] +
(jay) ( exp(-j(β--β+)l/2 - exp (-j(β-β+)l/2 )]
= Eo/2 [exp(-j(β-+β+)l/2[ax cos (β+-β-)l/2 –
aysin (β+-β-)l/2 ]

Design of Circulator using two Magic TEE

Microwave Tubes
1 MW
1 KW
1W
1mW
0.3 1 3 10 30 100
Frequency (GHz)
Average
power
Microwave
tubes
Microwave
semiconductor
devices
• Lower weight
• Smaller size
• Longer life time

Two possible methods of achieving high output
power in microwave system
Low power
semiconductor
oscillator
High power tube
amplifier
High power
tube
oscillator

Important Parameters
• Peak power • Average power
• Efficiency • Gain
• Bandwidth • Frequency
• Harmonic and spurious power • Intermodulation products
• Manufacturability at low cost
Relative
operating
voltage
Gain
(dB)
 (%)
Relative
BW (%)
Type
Low
6-15
20-50
1-10
Gridded tube
High
40-60
30-70
1-5
Klystron
High
30-50
20-40
30-120
Helix tube
High
30-50
20-40
5-40
Coupled
cavity tube

10 MW
1 MW
100 KW
10 KW
1 KW
100 W
0.3 1 3 10 30 100
300 Frequency (GHz)
Average power
Klystron
Coupled
cavity
TWT
Helix
TWT
Gridde
d tube
1000 MW
100 MW
10 MW
1 MW
100 KW
10 K W
0.3 1 3 10 30 100
300
Frequency (GHz)
Peak power
Coupled
cavity
TWT
Klystron
Gridde
d tube
Helix
TWT

Conventional Vaccum Tubes
The Limitation of conventional Low frequency
Vacuum Tubes are:
• Lead Inductance and inter- electrode Capacitance:
• Transit Time Effects
• Gain BW products:
•
• Effects due to RF Losses: Dielectric Loss & Skin Effect Loss
• Radiation Losses:

Conventional Vaccum Tubes
• The i/p impedance of the tube is given by
Zin = 1/( ω2Lk Cgk gm) - j [1 /( ω3L2
k Cgk g2
m) ]

Lead Inductance Effect
• As the freq. increases, the reactance
increases and hence the voltage s appearing
at the active Electrode are less than the
voltage set the base pins.
• This results in reduced the gain for the tube
amplifier
• This effect can be minimized by decreasing L

The Transit Time Effect
• Transit Time is the time taken for the
electron to travel from cathode to Anode

• StaticEnergy of the Electron = eV
• Kinetic Energy of the electron = ½ m V2
o
• Under Equilibrium Condition
• e V = ½ m V2
o
From Which Velocity Vo = (2 e V/ m)1/2
Then we have Transit Time
τ = d/ [(2 e V/ m)1/2]

Remedy
d to be minimized
V to be increased
Trade off B/W IEC and τ

Gain BW product
• Maximum gain is achieved when the tuned ckt is
at Resonance.
For the above ckt BW can be calculated by
BW = G/C
The max gain at resonance is A max = gm /G
The gain BW product = (gm / C)
Thus it is independent of frequency

Gain BW product
• As gm and C are fixed for a particular
Tube / ckt , Higher gain can be achieved
at the cost of BW.
• In mW ckt , this restriction can be
overcome by use of (i) Re- entrant
cavities (ii) Slow wave structure for a
greater gain over a larger BW

Skin Effect Losses and Dielectric Loss
• Skin Effect Losses and Dielectric Loss
will be discussed at home by YOU
( Repeat) by YOU Only.

O type Tubes
• Accelerating E field is in the same direction as that of
Static Magnetic Field used to focus the Electron
beams.
• The Magnetic field whose axis coincides with that of
the electron beam is used to hold the beam together
as it travels the length of the tube.

O type Tubes
• The Electron receives the potential energy
from the DC beam voltage before they arrive
the microwave interaction Region, and this
energy is converted in Kinetic Energy .
• In the Microwave interaction region , the
electron are either accelerated or
decelerated by the microwave field and
then bunched as they drift down the tube.

O type Tubes
• The bunched Electrons , in turn, induce
current in the o/p structure.
• Then the electron give up their kinetic
energy to the microwave field and are
collected by collector.
• Example : Klystron, TWT

Reflex Klystron : Mechanism of OSC

Reflex Klystron : Power and Efficiency
• Transit Time : The round up transit time in the
repeller space is given by
tr = 2 x [Velocity (u)/ Accln]
A we know
accln a = (e E/m) = (e/m) (V0 +VR + V1 sinωt) /L
= (e/m) (V0 +VR )/L

• Since the reference does not undergo any
velocity modulation, its transit time in repeller
space
to = 2 uo /a = 2 uo mL /e (V0 +VR )
= NT = 2 πN/ω ……..(2)
Combining 1 & 2, We have

Density Modulation:
The time of arrival of electron to the cavity gap
tb = tr + tg
= tg + to [ {1+(β1 V1)/(2Vo)} sin (ωtg –θg/2) ]
= tg + 2 πN/ω + (πN/ω)(β1 V1)/(Vo) sin (ωtg –θg/2) ]
= tg + 2 πN/ω + X/ω sin (ωtg –θg/2)…………..(3)
Where X= Bunching parameter= (πNβ1 V1)/Vo

• From the Eqn (3) We have
dtb/dtg = 1+ X cos (ωtg –θg/2)…………(4)
The bunched electron on return constitute
the bunched beam current ib such that the
conservation of charges gives
Io |dtg | = ib | dtb |……………(5)
Io is the dc beam current

From the Equation (4) and (5), we have
ib = Io /(dtb/dtg ) = Io [ 1+ X cos (ωtg –θg/2)]-1
Since V1<<< Vo, X<<1 and tb = tg + 2 πN/ω so that
ib = Io [ 1+ X cos {(ω tb –(2 πN/ω )- θg/2)}]-1
……………………(6)

After manipulating by using fourier Expansion
We have the fundamental component of
the induced RF current in the cavity i. e.
iRF (tb) = β1 2 Io J1 (X) cos (ωtb –2 πN)
The magnitude of fundamental RF current is:
iRF (tb) = β1 2 Io J1 (X)

Power o/p :
The rms RF power delivered to the
cavity
PRF = (V1 x iRF) /2
= V1 β1 2 Io J1 (X)
= [Vo Io X J1 (X)]/(πN)

From
to = 2 uo /a = 2 uo mL /e (V0 +VR )
= NT = 2 πN/ω
And uo = (2eVo/m )1/2.
We Have πN = (2eVo/m )1/2 ω L/ (V0 +VR )

• Therefore
PRF = [Vo Io X J1 (X)]/(πN)
= [Vo Io X J1 (X) (V0 +VR ) (2eVo/m )1/2]/2 π f L
…………………….(7)

• The Power supplied by beam voltage Vo
Pdc = VoIo
Therefore the Electronic efficiency of reflex
Klystron oscillator is
η = PRF/ Pdc
= X J1 (X) /(πN)
Where X= (πNβ1 V1)/Vo
N= n+ ¾ the mode number( n= 0 1 2 3 ……..)

• It can be shown from the Bessels function table , X J1
(X) will be max value of 1.252 at X= 2.408
Thus the optimum Value is
PRF = [Vo Io X J1 (X)]/(πN)
=[ 0.3986 Vo Io]/ N …………………….(8)
Therefore
η = 0.3986 /N

• Although these Equation shows that the
lowest value is N, the highest are PRF and η.
It has been observed that it is not possible to
get ¾ mode in Reflex Klystron, so that
N = 1 ¾ mode leads to max RF power. And
Efficiency.
Therfore PRF Max = 0.227 Vo Io
and η = 22.7 %

• From Equation (7) & (8)
PRF = [0.3986 Vo Io (V0 +VR ) (e/2mVo )1/2]/2 f L
Operation Frequency in MHz
f = [(V0 +VR ) N]/ [Lcm (Vo)1/2 x 6.74x 10-2]
Repeller Voltage = |VR|
= [ 6.74375 x 10 -6 x f x (Lm/N) (Vo)1/2 ] - Vo

VARACTOR DIODE : A Freq. Multiplier

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