this document talks about the design of a circular waveguide, rectangular waveguide, cubical cavity, it even describes the cutoff frequency, the wavenumber at the operating frequency, it shows the calculation of the attenuation constant in the waveguide, finally it shows derivation of TM mode propagation
this document talks about the design of a circular waveguide, rectangular waveguide, cubical cavity, it even describes the cutoff frequency, the wavenumber at the operating frequency, it shows the calculation of the attenuation constant in the waveguide, finally it shows derivation of TM mode propagation
this document talks about the design of a circular waveguide, rectangular waveguide, cubical cavity, it even describes the cutoff frequency, the wavenumber at the operating frequency, it shows the calculation of the attenuation constant in the waveguide, finally it shows derivation of TM mode propagation
this document talks about the design of a circular waveguide, rectangular waveguide, cubical cavity, it even describes the cutoff frequency, the wavenumber at the operating frequency, it shows the calculation of the attenuation constant in the waveguide, finally it shows derivation of TM mode propagation
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PAN AFRICAN UNIVERSITY
INSTITUTE FOR BASIC SCIENCES,
TECHNOLOGY AND INNOVATION
M.Sc. Electrical Engineering
Option: Telecommunication
ASSIGMENT #2
Course: Advanced Microwave Engineering Systems
Lecturer: Dr. Kibet Langat
KWIZERA Eva (EE300-0008/15)
Date, 08 October, 2015
Academic year: 2015 2016
ASSIGNMENT #2 1. Design an air filled circular waveguide Frequency separation = 1GHZ TE f c11
TM f c01
= =
TM TE f c01 f c11
Pmn
2a Pmn 2
1.841 2a o o
2.4049 2a o o
2.4049 2a o o
1.841 2a o o
0.5639 2a o o
0.56393108 =109 2a a=
0.56393108 =0.0269 m=2.69 cm 210 9
The waveguide radiusis 2.69 cm
The corresponding cutoff frequencies for this waveguide
Pmn 1.841 1.8413108 TE = =3.269GHZ f c11 = 2a = 20.0269 o o 20.0269 f
2.
TM c01
P mn 2a
2.4049 2.40493108 = =4.271 GHZ 20.0269 o o 20.0269
f =75 GHZ10 GHZ
i) ii) iii)
Design a rectangular waveguide such that there is only one mode of propagation The lowest usable frequency is 10% above cutoff frequency The highest usable frequency is 5% below the cutoff frequency where the next mode can propagate m a
1 f c=
TE for TE 10 , f c10 =
a=
C C C 1.1310 + 0.1 =1.1 = =7.5109 2a 2a 2a 2a
1.1310 8 =0.022m=2.2cm 27.510 9 TE for TE 01 , f c01 =0.95
b=
C 0.953108 = =10109 2b 2b
0.953108 =0.01425 m=1.425 cm 210109
3. Cubical cavity fr = 3GHZ Determine a (the smallest possible size of this cavity) =? Which mode does it resonate at this frequency? m a
p c
f r= 2 For a cubical cavity ,a=b=c
Therefore,
m a
p a
f r= 2
1 a
At
1 2+ a
()
TM 110 , f r= a=
3108 2 =0.071m=7.1 cm 2310 9 The mode of propagation 1 a
2 1 2+ a
1 TM 110 , f r=
()
1 a
1 2+ a
()
TE 101 , f r=
1 a
2+
1 a
()
TE 011 , f r =
4. An air-filled rectangular waveguide
a = 7.2cm b = 3.4cm = 40dB f = 3GHZ Determine: . The wavenumber at the operating frequency (K) . The cutoff frequency for the waveguide (fc) . The cutoff frequency for the attenuating section of the waveguide (fc) . The attenuation constant . The length of the attenuating section
2f 2310 9 1 = =62.8 m 8 C 310
i)
K= =
ii)
m a
n b
f c= 2 m a
2 At TE10,
1 f TE
c, 10 = 2
iii)
m 0.6 a
1 TE f c, 10=
iv)
0.6a
2k 2
0.60.072
2(62.8)2
v)
40 dB=20lo g10 ( ez ) 2=log 10 ( ez ) 102=ez z=
4.6 =0.1258 m=12.58 cm 36.596
5. Derive the TM modes in a lossy circular waveguide
Assuming the axis of the structure along z-axis, the general expressions for cylindrical Coordinate fields may be written as; z E z ( , , z )=[e ( , ) +e z ( , ) a z ]e H z ( , , z )=[h ( , ) +h z ( , ) az ]ez Where the vectors
e ( , ) and
the wave while the vectors
h ( , ) represent the transverse field components of
ez ( , ) az
and
components of the wave.
1 d Hz 1 d Ez j E = + H ( 1 ) j H = + E (1 a) d d
h z ( , ) a z are the longitudinal
j E = H
dHz d Ez (2) j H = E (2 a) d d
H ( ) d H z d
d d
E ( ) d E z d
d d
1 j E z=
To solve the longitudinal field components in terms of transverse field component
dH z dH z 1 dE 1 j dE z E = 2 z + j ( 4 ) H = 2
(4 a) d d d kc k c d
E =
dH z dE dH z 1 dE z 1 j (5)H = 2 j z (5 a) 2 d d d k c d kc
2 2 2 Where k c =k + and = + j
The value of
is contains purely imaginary when there is a totally lossless system.
As in reality some loss always occurs, the propagation constant,
will contain both
real and imaginary parts, and respectively.
TM Modes The transverse fields of TM modes are found by simplifying the general guided wave equations with Hz=0. The resulting transverse fields for TM modes are; dE z j dE z E = 2 ( 6 ) H = 2 (6 a) k c d k c d
( )
( )
E =
dEz j dE z ( 7 ) H = 2 ( 7 a) 2 d k c d kc
( )
( )
The longitudinal electric field of the TM modes within the cylindrical waveguide must satisfy: 2 2 E z +k E z=0 Where
E z ( , , z )=e z ( , ) ez
Inserting the expression for Ez into the differential equation yields:
2 d e z ( , ) 1 d e z ( , ) 1 e z ( , ) 2 + + 2 +k c ( , ) =0 2 2
d d d The electric field function may be determined using the separation of variables technique by assuming a solution of the form: z e z ( , ) e =R ( ) P( ) Inserting the assumed solution into the governing partial differential equation yields d 2 R ( ) P ( ) dR ( ) R ( ) d 2 P ( ) 2 P() + + 2 +k c R ( ) P ()=0
d d 2
d 2 Dividing by
R ( ) P() gives
2 d2 P () 2 1 d R ( ) 1 dR ( ) 1 + + + k c =0 R ( ) d 2 R() d 2 P( ) d 2
We multiply the expression above by
in order to make the third term dependent
on only. The result is: 2 2 2 d R ( ) dR ( ) 1 d P ( ) 2 2 + + + k c =0(1) P() d 2 R ( ) d 2 R ( ) d According to the separation of variables technique, we may set the
2 equal to a constant (- k c ). The resulting differential equation defining
d2 P () 2 + k c P ( )=0 d 2
dependent term P ( ) is
Which has the general solution of
The function
P ()
P ( )= Asin k + Bcos k
must be periodic in so that
must be an integer (n).
P ( )= Asinn + Bcosn 2 2 d R ( ) dR ( ) + + ( 2 k 2c + n2 ) =0(2) 2 R ( ) d R ( ) d
Equation (2) is known as Bessels equation which has solutions known as Bessel functions. We may write the general solution to Bessels equation as: R ( )=C J n ( k c ) + D Y n ( k c ) The Bessel function of the second kind approaches
as its argument approaches zero.
Since the circular waveguide fields must be bounded at the origin ( = 0), then the constant D must be zero. e z ( , ) ez =R ( ) P ( )=J n ( k c ) [ Asinn + Bcosn] The longitudinal electric field function for the cylindrical waveguide TM modes is: E z ( , )=J n ( k c ) [ Asinn + Bcosn]ez The TM boundary conditions for the cylindrical waveguide are E ( , , z )=0 E z ( , , z )=0 Application of the boundary condition on Ez yields: E z ( , )=J n ( k c ) [ Asinn + Bcosn ] ez =0 Thus, the TM modes of the cylindrical waveguide are defined by
J n ( k c )=0
If we define the mth zero of the nth order Bessel function as pnm, then the TM nm P nm k = c , nm mode cutoff wavenumber is found by
The resulting transverse fields of the TMnm modes are:
dE z E ( , , z )= 2 = J ' n ( k c ) [ Asinn + Bcosn ] ez kc k c d
( )
E ( , , z )=
dEz
= 2 J n ( k c ) [ Acosn Bsinn ] ez 2 d kc kc
( )
H ( , , z )=
j dE z j n = 2 J n ( k c ) [ Acosn Bsinn ] ez 2 k c d kc
H ( , , z )=
j dE z j n = J ' n ( k c ) [ Asinn + Bcosn ] ez 2 d k
kc c
( )
( )
The cutoff frequency of the TMnm mode is given by
P k c , nm= nm
f c, nm=
Pnm 2
Pnm
f c ,nm f
2 k nm= k 2k 2c ,nm= f c ,nm f
2 1 k
V p= = nm f c ,nm f
2 1
d 1 1 1 V g [ ] = d
6. Specify the cutoff frequencies for the first 4TE and first 4TM modes of an air- filled circular waveguides with radius 0.8cm. Draw a graph of frequency vs. propagation showing the frequencies at which each of these modes will propagate. Is there a frequency range in which only one mode will propagate? If so, what is the range and what mode is propagating ANSWER: p' k c = mn a f c, mn=
