11 engineering electromagnetics essentials chapter 10 waveguide resonator: analytic appreciation by...
TRANSCRIPT
11
Engineering Electromagnetics
EssentialsChapter 10
Waveguide resonator: analytic appreciation
by equivalent transmission-line approach
22
Objective
Topics dealt with
To analyse and characterise waveguide resonators
Transmission line theory as an approach to study waveguide resonators
Distributed transmission line parameters
Telegrapher’s equations
Distortionless transmission line
Input impedance of a transmission line terminated in a load impedance
Transmission line theory as applied to a duplexer of a radar system
Transmission line theory as applied to a radome of an antenna
Characteristic impedance of a transmission line
Reflection coefficient and voltage standing wave ratio (VSWR) of a transmission line
Finding the load impedance of a transmission line from the shift of its standing-wave pattern when a short replaces the load impedance
333
Theory of Smith chart and its applications to transmission line problems
Closed-ended resonator analysed by transmission line theory
Cylindrical waveguide
Inability of a hollow-pipe waveguide to support a TEM mode
Power flow and power loss in a waveguide
Power loss per unit area, power loss per unit length and attenuation constant therefrom
Background
Maxwell’s equations (Chapter 5), electromagnetic boundary conditions at conductor-dielectric interface (Chapter 7), basic concepts of propagation of electromagnetic waves through a waveguide (Chapter 9) and those of circuit theory
3
444
Difficulties alleviated by storing energy in a microwave resonator formed out of a designed length of a hollow metal-pipe waveguide:
• Closed-ended waveguide resonator
• Open-ended waveguide resonator
Difficulties of storing microwave energy at microwave frequencies in conventional tank circuit consisting of an inductor in parallel with a capacitor:
• Radiative loss due to part sizes becoming comparable to the operating wavelength • Increased resistive loss due to skin effect• Difficulty of fabricating inductors and capacitors due to their tiny sizes
How to find the length of the waveguideclosed-ended or open-endedthat would make it a resonator?
We have found this length by treating a waveguide resonator as equivalent to a transmission line an alternative to the field theory approach that uses Maxwell’s equations, wave equation, electromagnetic boundary conditions, Poynting theorem, etc.
Therefore, in what follows, we present the fundamentals of transmission line theory (that is in vogue for the analysis of a transmission line like a two-wire line or a coaxial cable).
55
Distributed transmission line parameters L, C, R and G to represent an infinitesimal length of a transmission line:
Series inductance per unit length, L ,of the line accounting for the energy stored in electric field
Shunt capacitance per unit length, C, of the line accounting for the energy stored in electric field
Series resistance per unit length, R, of the line accounting for the power losses in the conductors
Shunt conductance per unit length, G, of the line accounting for the power losses in the dielectric, if present in the region between the conductors
Let us present here the fundamentals of transmission line theory to be used for treating a waveguide resonator. It becomes convenient to treat a transmission line with the help of distributed transmission line parameters to represent an infinitesimal length of it.
Transmission line theory
By choosing the infinitesimal length of the line z0, we can treat the individual elements of line section of infinitesimal length z as lumped circuit elements and apply the laws of the circuit theory such as Kirchoff’s laws for the analysis of the line.
(See the Note to follow)
6
Note:
In the definition of the distributed line parameters representing a transmission line, the following is implied:
(i) Effect of the size of a circuit element relative to the operating wavelength is not ignored as could be done at low frequencies;
(ii) Effect of the finite time of travel of signal along the electric circuit is taken into account in the analysis for frequencies greater than a kilohertz while assigning electrical quantities such as the voltage, current, resistance, capacitance, etc;
(iii) Due to the transit-time effect, there will be a non-zero reactive drop for a lossless transmission line even though its resistive voltage drop could be zero;
(iv) Higher the operating frequencies, the manifestation of the transit-time effect in the reactance effect becomes more pronounced;
(v) Current I and voltage V continuously vary from point to point along the length of the line, warranting the use of a cascade of individual ‘discrete’ elemental line sections, each of infinitesimal length z, chosen small compared to the operating wavelength ;
(vi) By choosing z0, the individual elements of line section of infinitesimal length z of the distributed transmission line model can be treated as the lumped circuit elements, thereby allowing the application of the laws of the circuit theory such as Kirchoff’s laws for the analysis of the line.
77
Rdz Ldz
Cdz Gdz dzz
VV
V
Idzz
II
Ldz
Rdz
Cdz
Gdz
Series resistance due to the finite conductivity of the conductor used in making the transmission line, accounting for ohmic power loss in the conductor
Series inductance associated with the magnetic flux due to the current through the transmission line that links with the conductors making up the line, which takes into account the energy stored in the magnetic field
Shunt capacitance between the conductors of the transmission line because of the existence of an electric field between the conductors insulated by a dielectric medium
Shunt conductance between the conductors of the transmission line, often referred to as leakage conductance or ‘leakance’, which manifests itself because of leakage current flowing between the conductors through the insulating dielectric if it is not perfect, thus accounting for the dielectric power loss in the insulator
Representation of an infinitesimal length of an element of a transmission line by distributed line parameters
88
VCjGz
I)(
ILjRz
V)(
t
ILdzRdzIdz
z
VVV
)()(
GdzVt
VCdzdz
z
III
dzzVV )/(
t
VCGV
z
I
dzzII )/(
t
ILIR
z
V
Current at the input end of the transmission line of infinitesimal length
Voltage at the output end of the transmission line of infinitesimal length
VVoltage at the input end of the transmission line of infinitesimal length
I
Current at the output end of the transmission line of infinitesimal length
Rdz Ldz
Cdz Gdz dzz
VV
V
Idzz
II
jt
In view of time dependence:)exp( tj
Telegrapher’s equations
99
VCjGLjRz
ILjR
z
V))(()(
2
2
))((2 CjGLjR
VCjGLjRz
V))((
2
2
)exp(]exp)exp([ tjzBzAV
ILjRz
V)(
VCjGz
I)(
(rewritten) (rewritten)
Taking partial derivative
Vz
V 22
2
)exp()exp( zBzAV Solving
)exp( tjInvoking time dependence
1010
)(exp)exp(
)(exp)exp(
ztjzB
ztjzAV
G
CR
L
12
11
tan
tan
j
2/1)])([( CjGLjR
2/1)])([( CjGLjRj
))((2 CjGLjR )exp()]exp()exp([ tjzBzAV (rewritten) (rewritten)
2exp)])([(
)]exp())(exp()[(
214/1222222
2/12
2/12221
2/1222
jCGLR
jCGjLRj
2
sin2
cos)])([( 21214/1222222 jCGLRj
(expression for the voltage on the transmission line)
1111
2
cos)])([( 214/1222222 CGLR
2
sin)])([( 214/1222222 CGLR
ILjRz
V)(
2
sin2
cos)])([( 21214/1222222 jCGLRj (rewritten)
Equating real part Equating imaginary part
(expression for the voltage on the transmission line) (recalled)
Let us next find an expression for the current on the transmission line.
(recalled)
LjR
ztjBztjA
LjRz
V
I
)exp()exp()exp()exp(
)exp()exp()exp()exp( ztjBztjAz
V
)exp(]exp)exp([ tjzBzAV
1212
)exp()]exp()exp([ tjzBzALjR
I
)exp()]exp()exp([ tjzBzALjR
I
2/1
)exp()]exp()exp([
CjGLjR
tjzBzAI
)exp()exp()exp(
0
tjZ
zBzAI
2/1
0
CjG
LjRZ
LjR
ztjBztjA
LjRzV
I
)exp()exp()exp()exp(
(rewritten)
2/1)])([( CjGLjR
(Characteristic impedance of the line)
(expression for the current on the transmission line)
1313
2/tan 121 0GR
Distortionless transmission line
The information or signal that we transmit through a transmission line is composed, in general, of a number of frequency components. For distortionless transmission, two conditions need to be satisfied: (i) all the frequency components constituting the signal should simultaneously reach the same receiving point on the line, which also implies that the phase velocity of the wave, vph, supported by the line should be constant with frequency, or in other words, the line should be ‘dispersion-free’ and (ii) the attenuation constant of the line due to line losses, if any, should be the same for all the frequency components of the signal.
Does a lossless transmission line (R = G = 0) satisfy the distortionless condition?
G
CR
L
12
11
tan
tan
0tan
0tan
12
11
C
L
2
cos)])([( 214/1222222 CGLR
2
sin)])([( 214/1222222 CGLR
(recalled)
(recalled)
0)2/cos()])([ 4/12222 CL
0)2/cos(
2/1
4/12222
)(
)2/sin()])([
LC
CL
1)2/sin(
Lossless line
(recalled)
1414
LCv
1ph
2cos)]1)(1[( 214/1
2
22
2
22
G
C
R
LRG
For a lossless line (R = G = 0) we obtained earlier
0
2/1)(LC
The attenuation constant and phase velocity vph of the line are both constant with frequency, suggesting that the lossless line (R = G = 0) is a distortionless line.
Can you show that if the distributed line parameters of a lossy transmission line satisfy the relation RC = LG, then the line becomes a distortionless line?
2
cos)])([( 214/1222222 CGLR
2
sin)])([( 214/1222222 CGLR
2sin)]1)(1[ 214/1
22
2
22
2
C
G
L
RLC
15
2cos)]1)(1[( 214/1
2
22
2
22
G
C
R
LRG
2sin)]1)(1[ 214/1
22
2
22
2
C
G
L
RLC
(recalled) (recalled)
condition)(
LGRC G
C
R
L
C
G
L
R
2
cos)1( 212/12
22 R
LRG
2
sin)1( 212/122
2
L
RLC
16
21
condition)(
LGRC G
C
R
L
2221
2221
cos
sin
LR
R
LR
L
G
CR
L
12
11
tan
tan
(recalled)
2
cos)1( 212/12
22 R
LRG
2
sin)1( 212/122
2
L
RLC
12/1
2
22
cos)1( R
LRG 1
2/122
2
sin)1(
L
RLC
(recalled)(recalled)
1717
2221
2221
cos
sin
LR
R
LR
L
12/1
2
22
cos)1( R
LRG 1
2/122
2
sin)1(
L
RLC
RGLR
R
R
LRG
222
2/12
22
)1(
LCLR
L
L
RLC
222
2/122
2
)1(
LCv
1ph
The attenuation constant and phase velocity vph of the line are both constant with frequency, thereby suggesting that that the lossy transmission line can be made distortionless if the distributed line parameters satisfies the relation RC = LG.
(recalled) (recalled)
18
0
0
0
ZBA
BA
ZBABA
I
VZ
zl
0
0
ZZ
ZZ
A
B
l
ll
Reflection coefficient of a transmission lineZ l 0Z
ZlZin
BA
BA
Z
Z l
0
Let a transmission line of length l be terminated in load impedance Zl. We consider the source end to be located at z = = l and the load end at z = 0.
)exp()exp()exp(
0
tjZ
zBzAI )exp()]exp()exp([ tjzBzAV
)exp()(0
tjBAVz
)exp(
00
tjZ
BAIz
0z
(reflection coefficient)
(load impedance)
(recalled)(recalled)
19
)exp()exp(
)exp()exp(0 lBlA
lBlAZ
I
VZ
lzin
)exp()exp(
)exp()exp(
0
lAB
l
lAB
lZZ in
Input impedance of a transmission line
)exp()exp()exp(
0
tjZ
lBlAI
lz
)exp(]exp)exp([ tjzBzAV
)exp()]exp()exp([ tjlBlAVlz
lz
)exp()exp()exp(
0
tjZ
zBzAI
(input impedance)
20
)exp()exp(
)exp()exp(
0
lAB
l
lAB
lZZ in
)exp()()exp()(
)exp()()exp()(
)exp()exp(
)exp()exp(
00
000
0
0
0
0
0 lZZlZZ
lZZlZZZ
lZZZZ
l
lZZ
ZZl
ZZll
ll
l
l
l
l
in
)]exp()[exp()]exp()[exp(
)]exp()[exp()]exp()[exp(
0
00 llZllZ
llZllZZZ
l
lin
)cosh()sinh(
)sinh()cosh(
0
00 lZlZ
lZlZZZ
l
lin
2
)exp()exp()sinh(
2
)exp()exp()cosh(
lll
lll
(rewritten)0
0
ZZ
ZZ
A
B
l
l
(recalled)
Rearranging terms
(input impedance)
2121
0
00 )tanh(
)tanh(
ZlZ
lZZZZ
l
lin
)cosh( l
)tanh(
)tanh(
0
00 lZZ
lZZZZ
l
lin
)cosh()sinh(
)sinh()cosh(
0
00 lZlZ
lZlZZZ
l
lin
(input impedance) (rewritten)
Dividing by
(input impedance)
Rearranging terms
(input impedance)
(general expression for the input impedance of a transmission line which)
Let us next read this general expression for the input impedance the line for a lossless line.
2222
ljlj tantanh
)tanh(
)tanh(
0
00 ljZZ
ljZZZZ
l
lin
Let us take a lossless line (R = G = 0) for which we obtained earlier:
0
2/1)(LC
j
2/1)])([( CjGLjRj
(recalled) (lossless line)
j (lossless line)
(lossless line)
)tan(
)tan(
0
00 ljZZ
ljZZZZ
l
lin
(lossless line)
Input impedance of a lossless transmission line:
23
ljZZin tan0
)tan(
)tan(
0
00 ljZZ
ljZZZZ
l
lin
(lossless line) (recalled)
0lZ
Input impedance of a lossless transmission line that is short-circuited at the load end:
(short-circuited lossless transmission line at the load end)
Input impedance of a lossless transmission line that is open-ended at the load end:
)tan(
)tan(1
0
0
0
ljZ
Z
lZ
Zj
ZZ
l
lin
Dividing the numerator and denominator each by lZ
Putting for an open-ended line lZ
)cot()tan(
1
)tan(
1000 ljZ
ljZ
ljZZ in
(lossless transmission line open-ended at the load end)
24
Quarter-wave transformer and its use in a radar duplexer:
ljZZ in tan0 (short-circuited lossless transmission line at the load end)
4
2
l
))(()]2
tan()]4
)(2
tan[(tan 0000 jZjZjZljZZ in
(quarter-wave transmission line: l = /4)
(short-circuited lossless quarter-wave transmission line of length l = /4 at the load end)
Thus the input impedance of a short-circuited lossless quarter-wave transmission is found to be infinity.
2525
)cot()tan(
1
)tan(
1000 ljZ
ljZ
ljZZ in
(lossless transmission line open-ended at the load end)
4
2
l
(quarter-wave transmission line: l = /4)
0)0)(()]2
cot()]4
)(2
cot[(cot 0000 jZjZjZljZZ in
(lossless quarter-wave transmission line open-ended at the load end)
Thus the input impedance of a lossless quarter-wave transmission line open-ended at the load end is found to be zero.
26
In a radar system, the signal power is sent to a target in pulses and the echo signal is received between pulses.
The duplexer in the radar permits the use of a single antenna in both transmitting and receiving modes of the radar. In the transmitting mode the duplexer allows the transmitter to send signal power to the antenna for radiation while protecting the receiver from the transmitted power and, in the receiving mode, allows the signal power to be received by the receiver.
In the transmitting mode the duplexer allows the transmitter to send signal power to the antenna for radiation while protecting the receiver from the transmitted power
In the receiving mode the duplexer allows the signal power to be received by the receiver.
Transmission line theory as applied to a duplexer of a radar system
27
Transmit-receive (TR) and anti-transmit-receive (ATR) switches (both, typically, gas-discharge type) are turned on (by gas ionisation discharge) when the radar sends signal power pulses and turned off (by gas deionisation) between pulses. TR and the ATR switches primarily disconnect the receiver in the transmitting mode and disconnect the transmitter in the receiving mode.
Branch-type radar duplexer
Branch-type radar duplexer:
TR switch is located toward the antenna end in the branch line at a quarter wavelength (/4) distance from the main line the antenna end. It is terminated in the receiver.
ATR switch is located toward the transmitter end in the branch line at a quarter wavelength (/4) distance from the main line. It is separated from TR switch by a distance of /4 measured on the main line.
28
Input impedance of a short-circuited lossless quarter-wave (/4) transmission is infinity. (recalled)Input impedance of a lossless quarter-wave (/4) transmission line open-ended at the load end is zero. (recalled)
Transmitting mode:
(i) TR switch turns on and short circuits the transmitted thereby preventing it from entering the receiver. (ii) Branch line containing TR switch with its terminating end shorted presents infinite impedance at the main line /4 away from it and does not impede power flow in the main line to the antenna. (Iii) Similarly, ATR switch also turns on and presents infinite impedance at the main line and does not impede power flow in the main line to the antenna (as in (ii) above).
Receiving mode:
(i) ATR switch turns off and makes the branch line containing it open ended thereby presenting zero impedance at a point /4 away on the main line. (ii) TR switch turns off and similarly presents zero impedance at a point on the main line. (iii) Received power sees infinite impedance toward the transmitter end at a point on the main line where it is connected to the branch line containing TR switch since this point is /4 distance away from the zero impedance point on the main line connected to ATR switch via the branch line (see (i) above). (iv) Since received power sees zero impedance point toward receiver (as in (i) above) and infinite impedance toward transmitter (see (iii) above), and it goes to the receiver instead of transmitter taking the lower impedance path.
29
Let us exemplify the transmission line theory to determine the thickness of a radome that is used to protect an antenna while for this purpose by first appreciating that the input impedance of a half-wave lossless transmission line is the same as the terminating load impedance of the line and hence finding the thickness of a radome, treated as a lossless transmission line, made of a fibre glass of dielectric constant 4.9 and taking the operating frequency as 3 GHz.
Transmission line theory as applied to a radome of an antenna
ljZZ
ljZZZZ
l
lin
tan
tan
0
00
(recalled) (lossless line)
0tan)tan(
)2/)(/2()/2(
2/
l
ll
l
ll
in ZZ
ZZZ
00
(input impedance of a half-wave line treated here as a lossless line becoming equal to the load impedance)
From the physical point of view, taking the thickness of the radome as /2 will ensure a path difference of the wave reflected from the radome as , thereby making it in anti-phase with the incident wave and hence causing the cancellation of the incident and reflected waves resulting in the matching of the radome to the incident wave.
Here, l corresponds to radome thickness
30
rr
rr
f
c
f
f
111
2
2222
00
00000
Hz 103GHz 3
9.49f
r
mm 17.45m 04517.09.4
1
103
10319
8
rf
c
ll
in ZZ
ZZZ
00
We have found the input impedance of a half-wave line as equal to the load impedance. This concept may be used to find the thickness of the radome. In the present context, if we take the thickness of the radome treated as a transmission line of length l = /2, then since the load impedance is here the free-space intrinsic impedance 0, we obtain
0 lin ZZ
(rewritten)
In the present example
(given)
(input and load ends of the radome each being both a free space medium)
mm 59.222/17.452/
(required thickness of radome)
3131
01.2)3)(375.9/2()/2(
cm375.9)102.3/()103(/ 910
ll
fc
ljZZ
ljZZZZ
l
lin
tan
tan
0
00
tan2.01))(1525(50
tan2.01))(50(152550
jj
jjZ in
Let us further numerically appreciate the problem of finding the input impedance of a 3-cm long lossless transmission line of characteristic resistance 50 , which operates at 3.2 GHz and is terminated in load impedance Zl = 25-j15 .
Generator of
frequency 3.2GHz
50Ω transmission line
l=3cm
ZL=(25-j15)Ω
Zin
6.135.109 jZin
1525Z
Hz 102.3GHz 2.3
cm 3
50
9
0
j
f
l
Z
l
(given)
Separating the real and imaginary parts
(recalled)
3232
Characteristic impedance of a transmission line
The characteristic impedance of a transmission line may be thus identified as the input impedance of an infinitely long line ( l = ) .
)exp()exp(
)exp()exp(0 lBlA
lBlAZZin
(input impedance) (recalled)
l
(infinitely long line)
00 )exp(
)exp(Z
lA
lAZ
I
VZ
lzin
(the second term of each of the numerator and the denominator becoming significantly less than the first term)
(infinitely long line)
Next let us find the value of the input impedance of a transmission line of finite length that is terminated in the characteristic impedance.
3333
Thus the characteristic impedance of a transmission line may be identified as the terminating load impedance of the line that makes the input impedance of the line equal to the load impedance.
0
00 )tanh(
)tanh(
ZlZ
lZZZZ
l
lin
(input impedance) (recalled)
000
000 )tanh(
)tanh(Z
ZlZ
lZZZZ in
0ZZ l 2/1
0
CjG
LjRZ
(terminating load impedance being taken as the characteristic impedance Z0)
(characteristic impedance defined earlier in terms of the line parameters)
34
Voltage standing wave ratio (VSWR) of a transmission line
)exp()]exp()exp([ tjzjBzjAV
)exp()]exp()exp([ tjzjBzjAV
The superposition of the forward and backward waves on a transmission line will give rise to standing waves. Consequent to such superposition, the magnitude of the line voltage becomes alternately maximum and minimum down the length of the line. We can then define a quantity called the voltage standing-wave ratio (VSWR) as the ratio of the magnitude of the line voltage maximum to the magnitude of the line voltage minimum as .
)exp(]exp)exp([ tjzBzAV (recalled)
0 j (recalled) (lossless line)
j (lossless line)
minmax /VSWR VV
3535
)(exp)]2exp(1[ dtjdjA
Vl
)exp(by Dividing zjA
)(exp)]2exp(1[ dtjdjA
B
A
V
)exp()]exp()exp([ tjzjBzjAV (rewritten)
)exp(])exp(
)exp(1[
)exp(tj
zjA
zjB
zjA
V
)(exp)]2exp(1[ ztjzj
A
B
A
V
Rearranging terms
Defining the load end of the line as the reference point z = 0
Normalised potential at a distance z = d from the load end:
A
Bl (recalled)
(reflection coefficient)
3636
)exp( jll
)(exp)]2(exp1[ dtjdjA
Vl
)(exp)]2exp(1[ dtjdjA
Vl (rewritten)
(reflection coefficient expressed in terms of its magnitude and phase )
)2(exp1 djA
Vl
(magnitude of normalised voltage amplitude)
In general a complex quantity assuming real values for specific values of
l
d 2
3737
l
AP
)2(exp1 djA
Vl (magnitude of normalised voltage amplitude) (rewritten)
Can be represented by in the following vector diagram:
OPAOAP
AO has magnitude unity and is directed along the reference = 0.
OP has magnitude and its direction depends on the location of the point P on a circle of radius which also depends on the distance d of P from the load.
lIn order to reach the point PL we can rotate the vector through an angle from its reference = 0 anticlockwise with its base fixed at O around the circle of radius .In order to reach the point P we can then rotate the same vector around the same circle now clockwise from its position PL through an angle 2d.
OP
l
38
l1 ,....4 ,2 ,02 dAPThe length of the vector representing the magnitude of normalised
voltage amplitude takes on maximum value AM = for
AP,....5 ,3 ,2 d
l1The length of the vector representing the magnitude of normalised voltage amplitude takes on minimum value AM / = for
AV /
d
AV /
l
l
V
V
1
1
MA
AMVSWR
min
max
22 d
2
d representing the distance between consecutive maxima = distance between consecutive minima
2
d
39
1l
In a simple illustrative example let us find the reflection coefficient and VSWR of a transmission line for three situations of line termination: (i) line terminated in a short (ii) line open-ended at the load end and (iii) line terminated in characteristic impedance Z0.
(i) Line terminated in a short:
10
0
0
0
0
0
Z
Z
ZZ
ZZ
l
ll
0
2
11
11
1
1VSWR
l
l
(ii) Line open-ended at the load end:
101
01
1
1
1
1
0
0
0
0
0
0
Z
Z
Z
ZZ
Z
ZZ
ZZ
l
l
l
ll 1l
0
2
11
11
1
1VSWR
l
l
(iii) Line terminated in characteristic impedance:
0l02
0
000
00
0
0
ZZZ
ZZ
ZZ
ZZ
l
ll 1
01
01
1
1VSWR
l
l
0lZ
lZ
0ZZ l
40
In another example let us calculate the reflection coefficient and the VSWR a transmission line of characteristic resistance 50 if it is terminated in complex impedance of 25 + j100 .
10075
10025
5010025
5010025
j
j
j
jl
50
10025
0Z
jZl
36.08.0)100()75(
10010075100100257525
10075
10075
10075
10025
22j
jj
j
j
j
j
l
l
877.0)36.0()8.0( 22 l 3.15877.01
877.01
1
1VSWR
l
l
0
0
ZZ
ZZ
l
ll
(given)
414141
lA
V1max
lA
V1min
)2(exp1 djA
Vl
(magnitude of normalised voltage amplitude)
,....)4 ,2 ,02 toding(correspon d ,....)5 ,3 ,2 toingcorrespond( d
In yet another illustrative example let us show that, for a transmission line supporting a standing wave, the impedance at the voltage maximum is Zmax = (Z0)(VSWR) and the impedance at the voltage minimum is Zmin = Zo/ VSWR.
(recalled)
42
l
l
ZAIA
V
1
1
0
max
min
min
maxmax I
VZ
l
l
ZI
VZ
1
11
0max
minmin
l
l
ZAIA
V
1
1
0
min
max
(recalled)
)VSWR)((1
100
min
maxmax ZZ
I
VZ
l
l
l
l
1
1VSWR
VSWR
1
11 0
0max
minmin
ZZ
I
VZ
l
l
l
l
1
1VSWR
(recalled)
43
)exp( jll
0
0
ZZ
ZZ
l
ll
,....5,3,2 d
10
0
0
0
Z
Zl
1)exp( j
Finding the load impedance of a transmission line from the study of its standing-wave patterns
,....5,3,2 d
ddd 2)(2 ddd
(condition for voltage minimum at a point on the line that is distant d from the terminating load end)
(reflection coefficient in terms of load impedance recalled)
0lZ (for a short at the load end)
(reflection coefficient in terms of its amplitude and phase )
(for a short at the load end)
(condition for voltage minimum at a point on the line that is distant d/ from the terminating load replaced by a short)
In this method, the distance of the first voltage minimum from the load end or alternatively the spatial shift of the voltage minimum of the standing-wave pattern on the length of a transmission line when a short replaces the complex impedance at the terminating load end is interpreted to find the load impedance.
( = for a short at the load end)
(taking the difference between the above two conditions)
;1l
is replaced by a short)
44
ddd 2)(2
ddd
d is interpreted as positive when the voltage standing-wave pattern shifts towards the load end when a short replaces the complex load at the terminating end (d/<d). This also implies that the first voltage extremum from the load end is the voltage minimum at a distance d for a complex load, the first voltage minimum however shifting to the location of the short when the latter replaces the load (Fig. (a)).
d is interpreted as negative when the voltage standing-wave pattern shifts towards the source end when a short replaces the complex load at the terminating end (d/>d).This also implies that the first voltage extremum from the load end is the maximum at a distance d for a complex load which would shift to the location of the source such that a minimum coincides with the short when the latter replaces the load (Fig. (b)).
(rewritten)
(a)
(b)
45
sincos)exp( jj
))2sin()2(cos(1VSWR
1VSWRdjdl
)sin(cos1VSWR
1VSWR jl
)2sin2(cos1VSWR
1VSWRdjdl
)exp(1VSWR
1VSWR jl
l
l
1
1
1
VSWRl
l
ll
ll
2
2
)1()1(
)1()1(
1VSWR
1VSWR
l
l
1
1VSWR
(recalled)
)exp( jll
d 2 (recalled)
46
0000
00
2
2
)()(
)()(
1
1
Z
Z
Z
Z
ZZZZ
ZZZZ ll
ll
ll
l
l
)2sin2(cos1VSWR
1VSWR
1
10
djd
ZZ
l
l
ll
0
0
1 ZZ
ZZ
l
ll
0
0
ZZ
ZZ
l
ll
(recalled)
)2sin2(cos1VSWR
1VSWRdjdl
(recalled)
Put together
We can use the above expression for the load impedance Zl in conjunction with the above expression for l to find the load impedance.
The approach provides the theory of measurement of the load impedance terminating a transmission line. The method involves finding experimentally (i) VSWR of the line and (ii) the shift d in the minima of the standing-wave pattern which is also equal to the distance of the first minimum from the load end. However, it is easier to find the shift in the minima of the standing-wave pattern than the distance of the first minimum from the load end.
47
4818.02.151sin84.0sin2sin
8763.02.151cos84.0cos2cos0
0
d
d
84.0)21.0)(/2)(2(2 d
21.0d
As an illustration, let us find the complex impedance terminating a transmission line of characteristic impedance 50 that exhibits VSWR = 1.5 and gives the first extremum of the voltage standing-wave pattern as a minimum located at a distance of 0.21 from the load end.
0.21λ
Unknown load ZlGenerator
Standing wave
50Ω transmission line
(given)
5.1VSWR
)4818.08763.0(15.1
15.1
(recalled) )2sin2(cos1VSWR
1VSWR
j
djdl
48
01
1ZZ
l
ll
096.0175.0 jl
9.136.69506898.0
1920.09602.050
)096.0()825.0(
)096.0825.0)(096.0175.1(
50096.0825.0
096.0825.0
096.0825.0
096.0175.150
096.0825.0
096.0175.1
50096.0825.0
096.0175.150
)096.0175.0(1
)096.0175.0(1
22j
jjj
j
j
j
j
j
j
j
j
j
jZl
)4818.08763.0(15.1
15.1jl
(rewritten)
Simplifies to
(recalled)
49
Smith chart: theory andapplication to transmission line problems
Smith chart due to Phillip H. Smith:
• an elegant graphical tool for high frequency circuit applications, is popularly used to solve transmission line problems
• similar to a slide rule used for carrying out calculations involving addition, subtraction, multiplication, division, logarithmic and trigonometric functions, etc.
• can be applied to transmission line problems, for instance, to state the value of VSWR from the value of the complex load impedance terminating a lossless line of known characteristic resistance and find the input impedance of the line if its length is also given
In what follows, let us outline the theory of Smith chart describing the basic features of the chart and some of its applications.
Readers can find more applications of the chart in Section 10.1.8 of Chapter 10 of the book.
50
BA
BA
R
Zz ll
0
0GR
Constant- or VSWR circle of Smith chartl
BA
BA
Z
Z l
0
(recalled)2/1
0
CjG
LjRZ
(Characteristic impedance of a line)
(for a lossless line)2/1
0
CjG
LjRZ
say ,0
2/1
0 RC
LZ
(Characteristic resistance of a lossless line)
(load impedance normalised with respect to load resistance)
ABAB
zl
1
1
Dividing the numerator and denominator of the right hand side by the same quantity
51
ABAB
zl
1
1j
A
Bll exp
jvujxr
jxrjll
1
1exp
1)exp(1
)exp(1 jxr
j
j
l
l
(rewritten)
1
1exp
jxr
jxrjl
1)(
1)(
)]exp(1[)]exp(1[
)]exp(1[)]exp(1[
jxr
jxr
jj
jj
ll
ll
say ,)exp(1
)exp(1jxr
j
jz
l
ll
(reflection coefficient in terms of its amplitude and phase)
(reflection coefficient in terms of its real part u and imaginary part v)
(recalled)
52
jvujxr
jxrjll
1
1exp (reflection coefficient in terms of its real
part u and imaginary part v) (rewritten)
(complex reflection coefficient l plane of Smith chart in the domain of )
A point P on the reflection coefficient l plane of the Smith chart indicating its magnitude represented by the length OP and the phase represented by the angle between OP and u axis, O being the origin of coordinates.
The projection of the length OP on the abscissa and ordinate represent the real part u and the imaginary part v of the complex reflection coefficient l respectively.
POOP l
Can be read as the intercept OP/ with the u axis of a circle of radius called the ‘constant- circle’ drawn by the user such that it passes through the point P. The length can be read as the magnitude of the reflection coefficient of the line on a horizontal scale provided on a commercial Smith chart. The phase of the reflection coefficient can be read on a scale provided on the periphery of the chart.
Thus Smith chart is generated on the reflection coefficient l plane which can also be referred to as the VSWR plane since these two quantities are related as: . Therefore, the constant- circle can also be referred to as the ‘constant-VSWR circle’. .
1l
l
OPllPO
)1/()1(VSWR ll
l
l
53
jvujxr
jxr
jxr
jxr
1
1
1
1
Constant-r and constant-x circles of Smith chart
ljvujxr
jxr
1
1(normalised load impedance in terms of its real part r and imaginary part x)
jvuxr
xj
xr
xr
xr
xjxr
xr
xjxjxrjxrjxrrr
2222
22
22
22
22
22
)1(
2
)1(
1
)1(
21
)1(
1
jxrR
Zz ll
0
(real part r and imaginary part x of normalised load impedance related to real part u and imaginary part v of reflection coefficient)
Multiplying the left hand side by a quantity that is equal to unity having the same numerator and denominator
After a little algebra
(defined earlier)
(recalled)
54
22
22
)1(
1
xr
xru
22)1(
2
xr
xv
jvuxr
xj
xr
xr
2222
22
)1(
2
)1(
1(rewritten)
Equating the real and imaginary parts
(real part)
(imaginary part)
Thus, we have expressed the real (u) and imaginary (v) parts of reflection coefficient l in terms of the real (r) and imaginary (x) parts of the normalised load impedance zl .
Similarly, let us express in what follows next the real (r) and imaginary (x) parts of the normalised load impedance zl in terms of the real (u) and
imaginary (v) parts of reflection coefficient l .
55
11
1 jvu
jxr
jxr
)(1
)(1
)1()1(
)1()1(
jvu
jvu
jxrjxr
jxrjxr
Let us now proceed to express the real (r) and imaginary (x) parts of the normalised load
impedance zl in terms of the real (u) and imaginary (v) parts of reflection coefficient l .
jvujxr
jxr
1
1(recalled)
jvu
jvujxr
1
1
jvu
jvu
jvu
jvujxr
1
1
1
1
By algebraic manipulationBy simplification
Multiplying the right hand side by a quantity that is equal to unity having the same numerator and denominator
22
22
)1(
21
vu
jvvujxr
Simplifies to
Equating the real and imaginary parts
22
22
)1(
1
vu
vur
22)1(
2
vu
vx
Thus we have expressed the real (r) and imaginary (x) parts of the normalised load impedance zl in terms of the real (u) and imaginary (v)
parts of reflection coefficient l .
56
222
)1(
1)
1(
rv
r
ru
22
22
)1(
1
vu
vur
22)1(
2
vu
vx
With a little algebraic manipulation
(equation representing a constant-r circle on the reflection coefficient plane of Smith chart)
(equation representing a constant-x circle on the reflection coefficient plane of Smith chart)
(values of u and v varying from point to point for a constant value r)
(values of u and v varying from point to point for a constant value x)
222 1
)1
()1(xx
vu
However, before proceeding further with the above two equations representing constant-r and constant-x circles respectively, for the benefit to readers, let us provide in what follows next the necessary algebraic steps for the deduction of these equations.
(rewritten) (rewritten)
57
rrvu
rrurru
1
1)1()
1
11(2 222
2222 1])1[( vuvur
rrrvururru
1
1
1
1)1(12 222 0)1(12 222 rvururru
22
22
)1(
1
vu
vur
(recalled)
rrv
r
rurru
1
1)1(
12)1( 2
22
rrvu
r
rurru
1
1)1(
12 22
22
Necessary algebraic steps for deduction of the two equations representing constant-r and constant-x circles respectively
By cross multiplication
22
2
22
1
1
)1(12
r
vr
r
r
ruu
rv
r
r
r
ruur
1
1
)1(12)1( 2
2
22
Adding the quantity 1/(1+r) to both sides
Dividing by (1 + r)
22
2
1
1
1
rv
r
ru
Combining the first three terms
(equation representing the constant-r circle)
58
(recalled) 22)1(
2
vu
vx
02])1[( 22 vvux
02])1[( 222 vxvux
By cross multiplication and rearrangement of terms
112])1[( 222 vxvux
112)1( 2222 vxvxux
Multiplying both sides by x Adding the quantity 1 to both sides
By rearrangement of terms
1)1()1( 222 xvux
222 1
)1
()1(xx
vu
Combining the first three terms
Dividing by x2
(equation representing the constant-x circle)
Let us next discuss in what follows the features of constant-r and constant-x circles of Smith chart with the help of the equations representing them derived here.
59
00
1
1,
1
1;
1
1,1
1
1radius
interceptintercept
uv
r
r
r
rv
r
ru
r
Features of constant-r and constant-x circles of Smith chart with the help of the equations representing them
222
)1(
1)
1(
rv
r
ru
(equation representing a constant-r circle on the reflection coefficient plane of Smith chart)
(recalled)
Features of constant-r circles of Smith chart:
(constant-r circle)
Examining the equation
60
r
r
r
rv
r
ru
r
1
1,
1
1;
1
1,1
1
1radius
interceptintercept
(constant-r circle)(rewritten)
(complex reflection coefficient l plane of Smith chart in the domain of )1l
(1) The constant- circles are concentric all having the common centre at u = 0, v = 0.
(1) In general, there are two intercepts of a constant-r circle with u axis and two intercepts with v axis.
(2) The value of r referring to a constant-r circle is near the left intercept on the chart.
(3) The constant r = 0 circle is the outermost circle of radius = 1. It is also referred to as the constant circle. All the points of relevance in the chart are located within this circle. The two intercepts of this circle each on u and v axes are .
(4) All the constant-r circles meet u axis at a common point at the extreme right point of the axis corresponding to .
(5) With the increase of the value of r, the radius of the constant-r circle decreases and in the limit r = the value of the radius of the circle becomes zero, which means that the constant r = circle shrinks to a point on u axis .
1l
1;1 interceptintercept vu
)1( intercept u
1intercept u
l
61
(6) The constant r = 0 and the constant r = circles pass through the extreme left and right points of the real u axis respectively. .
(7) The VSWR of the transmission line is the value of r of the constant-r circle (read on the real u axis) passing through the positive intercept of the constant- circle with the real u axis of the chart.
Explanation:
POOP l1
1PO ntercept
r
rui
1
1
r
rl 1
11
r
r
l
rr
rr
rr
rr
rr
l
l
2
2
)1()1(
)1()1(
)1()1(
)1()1(
1
1VSWR
1
1
l
l (recalled)
rVSWR
l
62
00
11
1,
11
1;1
1radius
22interceptintercept
uv
xxxxvu
x
(8) The values of the impedance maxima and minima are the values of r of the constant-r circles (read on the real u axis) intersecting with the extreme right and left intercepts of the constant- circle with the real u axis.
(9) The extreme left and right points of the circle (constant r = 0 circle) represent the short circuit and open circuits at the load end respectively.
(10) The VSWR of the points on the chart representing the short and open circuits at the load end is infinity, noting that holds good for each of these terminating loads.
(11) The constant-r circles with negative r values are of no relevance to the Smith chart since they correspond to inadmissible values of both and , each falling outside the domain of the reflection coefficient.
l
1l
1l
interceptu nterceptiv
Features of constant-x circles of Smith chart:
222 1
)1
()1(xx
vu
(equation representing the constant-x circle)
(recalled)
Examining the equation
(constant-x circle)
63
(complex reflection coefficient l plane of Smith chart in the domain of )
xxxxvu
x
11
1,
11
1;1
1radius
22interceptintercept
1l
(constant-r circle)(rewritten)
(1) The values of the constant-x circles are displayed near these circles at the periphery of the outermost (r = 0) constant-r circle.
(2) The radius of the constant-x circle increases with the decrease of the value of x and the radius becomes infinite for the constant x = 0 circle, which in fact becomes a straight line coinciding with u axis.
(3) All the constant-x circles meet the real u axis of the chart at a common point which is the extreme right point of u axis. .
(4) The radius of the constant-x circle decreases with the increase of the value of x, shrinking to zero for the constant-x = circle at the extreme right point on the u axis of the chart. .
(5) The constant-x circles of relevance to the Smith chart correspond to the admissible values of and falling within the domain of . This, in fact, allows only the family of ‘arcs’ rather than the full circles to be displayed on the chart.
(6) The constant-x circles for the positive and negative x values lie respectively on the positive and negative imaginary v regions of the chart which appear as the mirror images of each other across u axis for the same though opposite values of x.
)1( intercept u
)1( intercept u
nterceptiu
nterceptiv 1l
64
Source: Emeloid Company, Hillside, New Jersey
65
50
10025
0Z
jZl
0.2
5.0
x
r
In an earlier example we had calculated the magnitude of reflection coefficient of a transmission line of characteristic resistance 50 as 0.877 and its VSWR as 15.3 when it is terminated in complex impedance of 25 + j100 using the expressions and . Let us now use Smith chart to find the same quantities of the same line.
)1/()1(VSWR ll )/()( 00 ZZZZ lll
0.25.050/)10025()/( 0 jjZZz ll (given)
We locate the point P as the point of intersection between the r = 0.5 and x = 0 circles on Smith chart. Next, we draw a circle taking the origin (u = 0, v =0) as its centre such that it passes through the point P intersecting u axis at the point P/.
OP)(PO We can then read the length by superposing it on the horizontal scale provided at the bottom of commercially available Smith chart. This gives the value of the magnitude of reflection coefficient as , which agrees with the value calculated earlier from the formula of the reflection coefficient in terms of the load impedance and characteristic impedance. The value of VSWR can be read as (the value of r displayed on the scale of the chart), being the value of r referring to the constant-r circle passing through P/ (as explained earlier while describing the features of constant-r circle). This value of VSWR also agrees to that calculated earlier using the formula of the VSWR in terms of the magnitude of the reflection coefficient.
87.0PO l
2.15VSWR
66
Location of the load impedance of a transmission line on Smith chart
jxrR
Zz ll
0
(load impedance normalised with respect to characteristic resistance expressed in terms of its real part r and imaginary part x)
Locate the point of intersection between constant-r and constant-x circles now that the values of r and x have been identified.
We can read the values of r on u axis and those of x on the periphery of the outermost constant-r circle.
In the accompanying figure, the point P being the intersection between constant r = 0.5 and constant x = 2 circles, can be made to represent the normalised load impedance zl = Zl/R0 = 0.5 + j2. Therefore, P can be made to represent the load impedance of a line of characteristic resistance 50 (typical):
Zl = R0(0.5 + j2) = 50 (0.5 + j2) = 25 + j100 .
67
Representation of purely resistive, purely inductive and purely capacitive, short-circuit, open-circuit, and matched loads:
Load at terminating end Representing point on chart
Purely resistive A (typical)
Purely inductive B (typical)
Purely capacitive C (typical)
Short circuit PSC
Open circuit POC
matched O
Explanation:
(i)For purely resistive load, x = 0 and the point A lies on u axis which coincides with x = 0 line (see constant-x circle features discussed earlier).
(ii)For purely inductive load, r = 0 and the point B lies on constant r = 0 circle (outermost constant-r circle) such that constant-x circle referring to a positive value of x passes through B.
68
(iii) For purely capacitive load too, r = 0 and the point C lies on constant r = 0 circle (outermost constant-r circle) such that constant-x circle referring to a negative value of x passes through C.
(iv) For the line terminated in a short circuit, r = 0 and x = 0 and the point PSC lies on the extreme left end point on u axis.
(v) For the line terminated in an open circuit, r = and x = and the point POC lies on the extreme right end point on u axis. (We recall the constant-r and constant-x circle features that the constant r = 0 and the constant r = circles pass through the extreme left and right points of the real u axis respectively; and also that in the limit x = 0 circle becomes a straight line coinciding with u axis and that constant-x = circle shrinks to the the extreme right point on u axis).
(vi) For matched load and the point O is the origin u = 0, v = 0, which is also the centre of constant circles.
0ll
69
)exp()exp(
)exp()exp(
0 ljBljA
ljBljA
Z
Zz inin
)2exp(1
)2exp(1
)2exp(1
)2exp(1
lj
lj
ljAB
ljAB
zl
lin
)2(exp1
)2(exp1
)2exp()exp(1
)2exp()exp(1
lj
lj
ljj
ljjz
l
l
l
lin
Location of the input impedance of a transmission line on Smith chart
)exp()exp(
)exp()exp(0 lBlA
lBlAZZin
(input impedance)
(recalled)
(input impedance normalised with respect to characteristic impedance)
Putting = j for a lossless line
Dividing the numerator and denominator of the right hand side by A exp(jl) and remembering B/A = l
)exp( jll (recalled)
70
)2(exp1
)2(exp1
lj
ljz
l
lin
(rewritten))exp(1
)exp(1
j
jz
l
ll
(recalled)
(load impedance)(input impedance)
Impedance at the load endImpedance at a distance l from the load end)
Two expressions are identical but for the phase factor exp (jl).
We had earlier showed how to locate the point P, say, to represent the normalised load impedance zl.
Let be the angle of P measured from u axis ( = 0) moving ‘anticlockwise’ around the constant- circle.
Now, we have to move from P through angle 2l ‘clockwise’ to reach a new point representing the impedance of the line at a distance l from the load end, which is the same as the input impedance of a line of length l.
The values of r and x referring to the constant-r and constant-x circles passing through this new point give the real and imaginary parts of the input impedance zin normalised with respect to characteristic resistance R0.
We can then obtain the input impedance Zin by multiplying zin by R0.
l
The above method of finding the input impedance of a transmission line using Smith chart has been further illustrated in an example to follow.
71
Note on wavelength scale displayed around the periphery of the outermost constant-r circle
ll )/4(2
pl
/2 pll 4)/4(2
242 pl
2/1p2/ pl
• One rotation through an angle 2 round the periphery of the outermost constant-r circle (constant r = 0 circle) of the chart corresponds to the length of the transmission line of half wavelength (/2).
• The distance in wavelengths is indicated on the periphery of the outermost constant-r circle.
• The extreme left and right points of the circle on the real u axis of the chart respectively refer to 0 and 0.25 times the wavelength on the distance scale of the length of the transmission line.
72
Let us retake, however now using Smith chart, the problem of finding the input impedance of a 3-cm long lossless transmission line of characteristic resistance 50 , which operates at 3.2 GHz and is terminated in load impedance Zl = 25-j15 .
(given)
cm375.9)102.3/()103(/ 910 fc
3.05.050/)1525(// 00 jjRZZZz lll
0.32
λ
B
AzL
O
• Locate the point A of intersection between the constant r = 0.5 and constant x = 0.3 circles.
• Move 0.32 (wavelengths) clockwise towards generator around the constant- circle through the point A to reach the point B. Identify the constant-r and constant-x circles passing through the point B and note that these circles are assigned the values: r = 2.2 and x = - 0.3 respectively.
Length of the line in number of wavelengths 32.0375.9/3/ l
l
3.02.2)/( 0 jZZz inin 15110)50)(3.02.2())(3.02.2( 0 jjZjZin
The value so obtained by Smith chart is very close to that calculated earlier using formulae.
1525Z
Hz 102.3GHz 2.3
cm 3
50
9
0
j
f
l
Z
l
73
Closed-ended resonator
a
b
Z
Y
X
l
A closed-ended rectangular waveguide resonator (abl), which is a rectangular waveguide of broad dimension a and narrow dimension b and whose length is l and which has six conducting walls located at x = 0 and x = a; y = 0 and y = b; and z = 0 and z =l.
74
22
b
n
a
mcc
0tan l ...),3,2,1( ppl
02222 cc
Let us use the theory of transmission line to the rectangular waveguide with both of its ends closed by conducting walls to form a closed-ended waveguide resonator.
ljZZin tan0(input impedance of the line short-circuited at the load end)
0inZ(input impedance of the line if its input end is short-circuited)
(recalled)
cl
p
b
n
a
mr
2/1222
0222
22
2
c
b
n
a
mc
l
pr
(Waveguide cutoff frequency) (See Chapter 9)
The length l of the waveguide resonates at angular frequency r.
(angular resonant frequency of closed-ended rectangular waveguide resonator)
(closed-ended resonator)
Dispersion relation for both TE and TM modes
p is the mode number in addition to m and n
75
lcot
ljZZin cot0
We can similarly use the theory of transmission line to the rectangular waveguide with both of its ends open to form an open-ended waveguide resonator.
(input impedance of the line open-ended at the load end)
(recalled) inZ
0tan l ...),3,2,1( ppl
cl
p
b
n
a
mr
2/1222
(open-ended resonator)Condition identical with that obtained for closed-ended resonator
(angular resonant frequency of open-ended rectangular waveguide resonator)
Following the same steps as taken for closed-ended rectangular waveguide resonator
Expression identical to that obtained for closed-ended resonator
76
In an illustrative example let us find the resonant frequency of a rectangular cavity which has the dimensions 3 cm (broad), 2 cm (narrow) and 5 cm (linear) and which is excited in the mode TE101.
GHz83.5Hz1083.5Hz103.58 98 rf
cla
f rr
2/1
22
11
2
1
2
Hz103)5
100()
3
100(
2
1 82/1
22
rf
m = 1, n = 0, p = 1 (given)
a = 3 cm, b = 2 cm, l = 5 cm (given)
cl
p
b
n
a
mr
2/1222
77
)exp( tj
Fields in a waveguide resonator
xa
zjSH z
cos)exp(
)(expcos0 ztjxa
HH zz
SH z 0
Axial magnetic field in a rectangular waveguide excited in TE10 mode (recalled from Chapter 9) (representing only the forward wave propagating along z)
(recalled)
(symbol for field amplitude Hz0 changed so for the sake of convenience)
xa
zjSzjSH z
cos)]exp())exp([
Keeping understood the time dependence
(S standing for the field amplitude of the forward wave)
(S/ standing for the field amplitude of the backward wave)
A standing wave is formed in a waveguide resonator due to the combination of forward and backward waves caused by reflection from the conducting walls perpendicular to axial z direction.
(forward wave)
Similarly for a backward wave:
xa
zjSH z
cos)exp(
Combining the contribution of the forward and backward waves:
78
yy
zn
aEE
aa
2
xa
zjSzjSH z
cos)]exp()exp([ (rewritten)
02 Ean
We can establish a relation between S and S/ with the help of electromagnetic boundary condition at the conducting wall at z = 0
(electromagnetic boundary condition recalled (see Chapters 7 and 9) at the interface between conducting and dielectric/free-space media)
na
2E
Unit vector at the conducting surface directed from the conducting wall to dielectric/free-space medium
Electric field vector in dielectric/free-space medium
)0(0 zaEa yyz
xyz aaa
At the waveguide wall z = 0
)0(0 zEa yx
00
zyE
x
H
k
jE zy
22
0
)0(0
zx
H z
(relation recalled from Chapter 9)
(for all values of x)(for all values of x)
T20: Thank you!
79
)]exp()exp([sin zjSzjSxaax
H z
zxa
HH z sincos0
jSH 20
xa
zjSzjSH z
cos)]exp()exp([ (rewritten)
)0(0
zx
H z
xa
zjzjSH z
cos)]exp()[exp(
(recalled)
0 SS
0z
(for all values of x)
SS
sincos)exp( jj
xa
zjSzjSH z
cos)]exp()exp([
(trigonometrical relation)
In view of
80
zxa
Hk
j
aEy
sinsin0220
(rewritten)
x
H
k
jE zy
22
0
zxaa
Hx
H z sinsin0
zx
aHH z
sincos0
)exp(sinsin00 tjzx
aH
ajEy
222 )/( ak
0,1 nm
0sin l
0lzyE
222ckk
2/122
b
n
a
mkc
(TE10 mode)
akc
(for all values of x)
Boundary condition at the conducting wall at z = l
...),3,2,1( ppl
81
....),3,2,1( pp
l
g /2
0sin l ...),3,2,1( ppl (rewritten)
The condition exactly agrees to what has been derived earlier using transmission line theory
....),3,2,1(2
ppl g
(phase propagation in terms of guide wavelength g)
(length of the resonator excited in TE10p mode)
....),3,2,1( pl
p
zxa
HH z sincos0
zl
px
aHH z )sin(cos0
(TE10p mode) (TE10p mode)
zxa
Haj
Ey
sinsin00
zl
px
aH
ajEy )sin(sin0
0
82
xy Hjz
E0
z
l
px
aH
l
apH x
cossin0
zl
px
aH
l
paj
z
Ey
cossin00
)exp(cossin
)exp(sinsin
)exp(sincos
0
00
0
tjzl
xa
Hl
aH
tjzl
xa
Haj
E
tjzl
xa
HH
x
y
z
(TE10p mode)zl
px
aH
ajEy )sin(sin0
0
(rewritten)
Substituting in the following Maxwell’s equation recalled from Chapter 9
(TE10p mode)
(TE101-mode field expressions of a rectangular waveguide resonator)
Putting, in the field expressions deduced, p = 1 and invoking time dependence: )exp( tj
we get
83
Quality factor of a waveguide resonator
Energy stored (time-averaged) in electric field:
dEEWE
.
2
10 (static electric field energy)
(recalled from Chapter 8)
daEaEdEEW yyyyE average-time0average-time0 ).(2
1).(
2
1
dEy average-time2
0 )(2
1
Time averaging for time-varying fields
)exp(cossin
)exp(sinsin
)exp(sincos
0
00
0
tjzl
xa
Hl
aH
tjzl
xa
Haj
E
tjzl
xa
HH
x
y
z
(TE101 mode)
84
dxdydzd
dEW yE average-time2
0 )(2
1
2sin and,
2sin
0
2
00
2 ldxx
abdy
adxx
a
lba
)exp(sinsin00 tjz
lx
aH
ajEy
dzzl
dydxxa
Ha
Wb la
E
0 0
2
0
2202
220
20 sinsin
4
(rewritten)
2
2cos
2
1sin2
20
230
204
1HfblaWE
An extra factor of 1/2 introduced for time averaging
Making use of relation:
Evaluating the integral
Making use of relations:
f 2
(TE101 mode)
(average energy stored in electric field)
(TE101 mode)
a
b
Z
Y
X
l
85
dHHWB
.
2
10 (static magnetic field
energy) (recalled from Chapter 8)
Time averaging for time-varying fields
Energy stored (time-averaged) in magnetic field:
)exp(cossin
)exp(sinsin
)exp(sincos
0
00
0
tjzl
xa
Hl
aH
tjzl
xa
Haj
E
tjzl
xa
HH
x
y
z
(TE101 mode)average-time0 ))()(2
1(
daHaHaHaHW yyxxyyxxB
lbalba
B dzzl
dydxxa
dzzl
dydxxal
aHW
0
2
00
2
0
2
00
22
22
00 sincoscossin4
1
dHH
daHaHaHaHW
zx
yyxxyyxxB
average-time22
0
average-time0
)(2
1
))()(2
1(
dxdydzd
An extra factor of 1/2 introduced for time averaging
86
202
20 1
16H
l
aablWW EB
lbalba
B dzzl
dydxxa
dzzl
dydxxal
aHW
0
2
00
2
0
2
00
22
22
00 sincoscossin4
1
202
20 1
16H
l
aablWB
Making use of relations:
2
2cos
2
1cos2 Making use of relation:
(TE101 mode)
2/1
222/100
2/1
22
11
)(2
111
2
1
lac
lafr
20
230
204
1HfblaWE rff
20
230
204
1HfblaW rE
(resonant frequency of TE101 mode) (recalled)
202
20 1
16H
l
aablWE
(recalled)
2sin ,
2cos,,
2sin
0
2
0
2
00
2 ldxx
a
ldxx
abdy
adxx
a
llba
(rewritten)
,2
2cos
2
1sin2
(average energies stored in electric and magnetic fields are equal)
87
202
20 1
16H
l
aablWW BE
(rewritten)
202
20 1
822 H
l
aablWWWWW BEBE
*LA 2
1ssS JJRP sn JHa
2
wall)back(wall)front(
wall)top(
wall)bottom(wall)left(wall)right(
0
0
0
zlzn
zzn
ybyn
yyn
xaxn
xxn
aaaa
aa
aaaaaa
(total time-averaged energy stored in resonant cavity)
Total energy (time-averaged) stored in a resonant cavity
Power dissipated in the conducting walls of a resonant cavity
In finding the power dissipated in the conducting walls of a rectangular waveguide resonator, we can use the same approach as followed in finding the power dissipated in the conducting walls of a rectangular waveguide in Chapter 9.
(power loss in a conducting wall in terms of the surface current density developed at surface of the conducting wall and surface resistance)
(boundary condition at a conducting wall)
a
b
Z
Y
X
l
2H
na
magnetic field inside waveguide
unit vector at the conducting wall
88
zzxx aHaHH
2
wall)back()(
wall)front()(
wall)top()(
wall)bottom()(
wall)left()(
wall)right()(
0
0
0
yxzzxxzlzs
yxzzxxzzs
xzzxzzxxybys
xzzxzzxxyys
yzzzxxxaxs
yzzzxxxxs
aHaHaHaJ
aHaHaHaJ
aHaHaHaHaJ
aHaHaHaHaJ
aHaHaHaJ
aHaHaHaJ
)exp(cossin
)exp(sinsin
)exp(sincos
0
00
0
tjzl
xa
Hl
aH
tjzl
xa
Haj
E
tjzl
xa
HH
x
y
z
(TE101-mode field expressions of a rectangular waveguide resonator)
(recalled)
(wall surface current densities in terms of magnetic field components)
sn JHa 2
(boundary condition)
Magnetic field has two components
89
);wallback()exp(sin
0
)0;wallfront()exp(sin
0
);walltop(
)exp(cossin
)exp(sincos
)0;wallbottom(
)exp(cossin
)exp(sincos
);wallleft(
0
)exp(sin
)0;wallright(
0
)exp(sin
0
0
0
0
0
0
0
0
lztjx
aH
l
aH
H
ztjx
aH
l
aH
H
by
tjzl
xa
Hl
aH
tjzl
xa
HH
y
tjzl
xa
Hl
aH
tjzl
xa
HH
ax
H
tjzl
HH
x
H
tjzl
HH
x
z
x
z
x
z
x
z
x
z
x
z
(magnetic field components to be put in the expressions for surface current densities at resonator walls)
90
wall)back()exp(sin
wall)front()exp(sin
wall)top()exp(sincos
)exp(cossin
wall)bottom()exp(sincos
)exp(cossin
wall)left()exp(sin
wall)right()exp(sin
0
00
0
0
0
00
0
00
ylzs
yzs
x
zbys
x
zys
yaxs
yxs
atjxa
Hl
aJ
atjxa
Hl
aJ
atjzl
xa
H
atjzl
xa
Hl
aJ
atjzl
xa
H
atjzl
xa
Hl
aJ
atjzl
HJ
atjzl
HJ
We have obtained immediately above (i)the expressions for the magnetic field components and (ii)the expressions for the surface current densities in terms of magnetic field components both at each of the waveguide resonator walls.
We can substitute (i) in to (ii).
(wall surface current densities to be used later for deriving the expressions for power lost per unit area at resonator walls)
Let us next proceed to derive the expressions for power loss per unit area at resonator walls.
91
*LA 2
1ssS JJRP
In order to derive the expressions for power loss or dissipated per unit area at resonator walls, let us recall the expression for power loss per unit area at a conducting surface PLA in terms of the surface resistance and surface current density deduced in Chapter 8 and already used in Chapter 9 with respect to waveguide walls.
(recalled from Chapter 9)
xa
Hl
a
JJJJ
zl
xa
Hzl
xa
Hl
a
JJJJ
zl
H
JJJJ
lzsszss
byssyss
axssxss
220
2
*
0
*
2220
2220
2
*
0
*
220
*
0
*
sin
wall)back ()(wall)front ()(
sincoscossin
wall)top()(wall)bottom()(
sin
wall)left()(wall)right()(
*sJ
sJ
The expression for can be obtained from that of already obtained earlier simply by replacing the factor exp(jt) with exp(jt). Hence we obtain
(expressions to be substituted in the expressions for power loss per unit area at the surfaces of the waveguide walls)
92
xa
Hl
aR
PP
zl
xa
Hzl
xa
Hl
aR
PP
zl
HR
PP
S
S
S
220
2
LALA
2220
2220
2
LALA
220
LALA
sin2
1
wall)back(wall)front (
sincoscossin2
1
wall)top(wall)bottom(
sin2
1
wall)left(wall)right(
*ss JJSubstitute the expression for already derived in the expression for power loss per unit area
*LA 2
1ssS JJRP
(power loss per unit area at resonator walls)
(recalled)
Next, we are going to use these expressions for power loss per unit area at the surfaces of the resonator walls to find the power loss over the entire wall surfaces.
93
wallright
LAL wall)right(wall)right( dydzPP
zl
HRP S
220LA sin
2
1wall)right(
Let us now use these expressions for power loss per unit area PLA at the surfaces of the resonator walls to find the power loss over the entire wall surfaces.
For this purpose, with reference to a resonator wall, with the help the expression for PLA , let us find the power loss over an infinitesimal area of the wall and then integrate it to find the power loss PL over the entire area of the wall.
wallright
LAL wall)left(wall)left( dydzPP
zl
HRP S
220LA sin
2
1wall)left(
lb
S dzzl
dyHRP0
2
0
20L sin
2
1wall)right(
2)(
2
1wall)left( 2
0L
lbHRP S
2)(
2
1wall)right( 2
0L
lbHRP S
lb
S dzzl
dyHRP0
2
0
20L sin
2
1wall)left(
2sin,
0
2
0
ldxx
abdy
lb
Making use of the relation:
2sin,
0
2
0
ldxx
abdy
lb
Making use of the relation:
94
zl
xa
Hl
aR
P
S
2220
2
LA
cossin2
1
wall)top(
z
lx
aH
2220 sincos
la
S dzzl
dxxal
aHR
P
0
2
0
22
20
L
cossin2
1
wall)bottom(
zl
xa
Hl
aR
P
S
2220
2
LA
cossin2
1
wall)bottom(
z
lx
aH
2220 sincos
wallbottom
LAL wall)bottom(wall)bottom( dxdzPPwallbottom
LAL wall)top(wall)top( dxdzPP
la
dzzl
dxxa 0
2
0
2 sincos
la
S dzzl
dxxal
aHR
P
0
2
0
22
20
L
cossin2
1
wall)top(
la
dzzl
dxxa 0
2
0
2 sincos
1
22
1wall)top(
22
0L l
al
l
aHRP S
1
22
1wall)bottom(
22
0L l
al
l
aHRP S
Using integral values as done earlier Using integral values as done earlier
95
x
aH
l
aRP S
220
2
LA sin2
1wall)front (
x
aH
l
aRP S
220
2
LA sin2
1wall)back(
wallfront
LAL wall)front(wall)front( dxdyPP wallfront
LAL wall)back(wall)back( dxdyPP
)(2
1wall)front(
22
0L bl
a
l
aHRP S
2
1
2
1
2
22
0L a
bl
l
b
l
aaHRP S
a b
S dydxxal
aHRP
0 0
22
20L sin
2
1wall)front(
a b
S dydxxal
aHRP
0 0
22
20L sin
2
1wall)back(
Using integral values as done earlier Using integral values as done earlier
)(2
1wall)back(
22
0L bl
a
l
aHRP S
wall)back(wall)front(wall)bottom(wall)top(wall)left(wall)right( LLLLLLL PPPPPPP
(power losses in all the six walls of a rectangular waveguide resonator summed up)
(TE101 mode)
96LL
P
W
P
WQ
2
2
2/2 LLTL PTPW
2/
2
TLW
WQ
2/TLW
2
1
2
1
2
22
0L a
bl
l
b
l
aaHRP S
202
20 1
822 H
l
aablWWWWW BEBE
(total time-averaged energy stored in resonant cavity)
(power losses in all the six walls of a rectangular waveguide resonator summed up)
Let us see how these two expressions derived for a rectangular waveguide resonator excited in TE101 mode can be used to obtain an expression for the quality factor Q of the resonator defined as follows.
energy lost per cycle, that is, in the wave time period T = 2/
W = time-averaged energy stored
2
T
97
21
21
2
182
2
20
ab
llb
laa
R
la
abl
P
WQ
SL
LL
P
W
P
WQ
2
2
202
20 1
8H
l
aablW
2
1
2
1
2
22
0L a
bl
l
b
l
aaHRP S
(TE101 mode)
(recalled) (recalled)
98
2/100 )/(1 c
21
21
2
1811
)( 2
2
20
2/1
222/100
ab
llb
laa
R
la
abl
laQ
S
rr f 2
2/1
222/100
11
)(2
lafrr
21
21
2
182
2
20
ab
llb
laa
R
la
abl
P
WQ
SL
(TE101 mode) (re-written)
cla
f rr
2/1
22
11
2
1
2
(TE101 mode) (recalled)
21
21
2
182
2
20
ab
llb
laa
R
la
abl
Q
S
r
Interpreted at resonant frequency
)(2)(
)(2
4 3322
2/3220
lablaal
lab
RQ
S
2/1000 )/(
(TE101 mode)
(quality factor of a rectangular waveguide resonator)
99ext ,ohmic ,loaded, LLL PPP
2/1
01
r
S
fR
)(2)(
)(2
4 3322
2/3220
lablaal
lab
RQ
S
(TE101 mode)
cla
f rr
2/1
22
11
2
1
2
(both recalled from Chapter 6 for a good conductor at relatively high frequencies: f = fr )
(rewritten) (TE101 mode) (recalled)
The quality factor of a waveguide resonator depends on the dimensions of the resonator and on the conductivity of the material used in making it.
0
1
f
Loaded quality factor
In an actual application, some part of energy stored in a cavity is coupled out from the cavity to an external load and the cavity thereby becomes loaded. For such a loaded cavity, the power lost from the loaded cavity PL,loaded consists of
(i)the ohmic power loss PL,ohmic due the finite conductivity of the material making the cavity and
(ii)the power PL,loaded that couples out from the cavity to the load.
100
loadedloaded,
extext ,
unloadedohmic ,
Q
WP
Q
WP
Q
WP
L
L
L
extunloadedloaded Q
W
Q
W
Q
W
ext ,ohmic ,loaded, LLL PPP (rewritten)
LP
WQ (recalled)
extunloadedloaded
111
QQQ
(quality factors: unloaded Qunloaded, external Qext and loaded Qloaded related to ohmic power loss PL,ohmic, power coupled out to the external load PL,ext and total power loss PL,loaded of the loaded cavity, respectively
(relation between the loaded, unloaded and external quality factors)
Frequency response of equivalent impedance of the resonator The waveguide resonator may be represented by a resonant circuit comprising an inductor of inductance L having reactance jL, a capacitor of capacitance C having reactance 1/(jC) , and a resistor of resistance R, all in parallel.
In what follows next, let us find the frequency response of the impedance Zeq of the L, C, R parallel resonant circuit equivalent to the waveguide resonator and hence relate its quality factor to the resonant frequency and the bandwidth of the resonator.
2
2
1CVW (total energy W stored in the circuit, taken here as the energy in the capacitor which is
transferred back and forth between the inductor and the capacitor during each cycle)
101
CRQ r 2/1)(
1
LCr
RCj
LjZ
11111
eq
)1(1 2eq
CLjQ
RZ
r
2
1
rLC
L
RQ
r
2
2
1CVW
R
VPL 2
2
Lr P
WQ
(power loss in the resistor of the parallel equivalent circuit) CRPW L
(recalled)
(recalled)
(resonant angular frequency of the circuit)
An expression that is going to be put in the expression for the impedance Zeq of the equivalent L, C, R parallel resonant circuit
After a little algebra
rQL
R LC
R
102
(rewritten)
2
2eq
11r
rjQ
RZ
r
rrQjR
Z
))((
1
1eq
11 2
2eq
r
rQj
RZ
2/1)(
1
LCr )1(1 2
eq
CLjQ
RZ
r
At resonance = r, Zeq = R and we can normalize Zeq with respect to R that is its value at resonance.
103
2
2
eq
2/12
eq
))((1
1
))((1
1
r
rr
r
rr
QR
Z
QR
Z
r
rrQjR
Z
))((
1
1eq
(rewritten)Normalized impedance magnitude and its square
Impedance magnitude and voltage across the circuit
• maximum at resonance
• decreases both below and above resonance.
Circuit power is proportional to the square of voltage across the circuit
Voltage across the circuit is proportional to impedance
Circuit power is proportional to the square of impedance
2
1
))((1
12
2
eq
r
rrQR
Z
Corresponds to half the maximum power at resonant frequency r
104
2
1
))((1
12
r
rrQ
(corresponding to half the maximum power at resonant frequency r)
(recalled)
We expect two solutions 1 and 2 of around the resonant r.
r1
r2 r1
r2
)( (first solution)
r
rr
r
1
1
1
2
1)(4 2
2
2
r
Q
r
rr
r
2
2
2
2
(second solution)
1)(4 2
2
2
r
Q(relation identical to that obtained taking from first solution
(from first solution)
(from second solution)
105
2
rQbandwidthfrequencyangular
rQ
1)(4 2
2
2
r
Q(rewritten)
bandwidthfrequency rfQ
Interpreting 2 as the angular frequency bandwidth
In terms of circular frequency or simply frequency
Let us take an illustrative example of finding the dimension of a cubical cavity made of copper ( = 5.8 108 mho/m) excited in the TE101 mode so that it resonates at 15 GHz. Calculate also its quality factor and bandwidth.
m = 1, n = 0, p = 1 and a = b = l (given)
cl
p
b
n
a
mr
2/1222
ca
fr
21
m/s103 8c
cf
ar
1
2
1
Hz1015GHz15 9rf m10414.11015
103
2
11
2
1)( 2
9
8
cf
lbar(given)
(given)
106
Hz1015GHz15 9 rff
MHz713.1Hz10713.18756
1015bandwidthfrequency 6
9
Q
fr
ohm1019.3108.5
1041015143.3 27
790
f
Rs
87561019.3414.13
377143.3
23)(2)(
)(2
4 20
3322
2/3220
SS Raaaaaaa
aaa
RQ
)(2)(
)(2
4 3322
2/3220
aaaaaaa
aaa
RQ
S
)(2)(
)(2
4 3322
2/3220
lablaal
lab
RQ
S
(TE101 mode)
a = b = l (given)
Simplifies to
(see Chapter 6 for the expression for Rs used)
bandwidthfrequency rfQ
(recalled)
Hz1015GHz15 9 rff (given)
(given)
107107
Summarising Notes Waveguide resonator at a defined resonant frequency can be made out of an appropriately chosen length of a waveguide with its both ends either closed by a conductor or kept open.
Transmission line theory is an easy approach of treating a waveguide resonator.
Basic concepts of transmission line theory have been developed such as
distributed transmission line parameters;
telegrapher’s equation;
condition for distortionless transmission;
input impedance of the line terminated in a load impedance;
characteristic impedance of the line;
voltage standing-wave ratio (VSWR) of the line;
Impedance matching such as in Radome for the protection of an antenna
and branch-type radar duplexer of a radar system; and
Smith chart: theory and application to transmission line problems to make
them simpler.
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Resonator length has been found by transmission line theory as an integral multiple of half the guide wavelength for both closed-ended and open-ended resonators.
Resonant frequency of the waveguide resonator can be found with the help of transmission line theory using the dispersion relation of the waveguide.
Field theory can be applied to treat a waveguide resonator as an alternative to transmission line theory.
Field solutions and electromagnetic boundary conditions typically for a rectangular waveguide closed-ended resonator has yielded
(i) the same resonator length as predicted by the transmission line theory and
(ii) an additional mode number p of the waveguide resonator, to be read with reference to TE10-mode excitation of the waveguide as TE10p mode of the resonator (which may be generalised as TEmnp mode of the resonator with reference to TEmn-mode excitation of the waveguide).
Field solution and relevant electromagnetic boundary conditions can be used to obtain
(i) the expression for the time-averaged energy stored in electric and magnetic fields and
(ii) the expression for the power loss in resonator walls.
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Expression for the quality factor of a resonator in the TE101 mode at the resonant frequency, in terms of the resonator dimensions and the surface resistance of the conducting material making the resonator, has been derived with the help of the expressions for the time-averaged energy stored and power loss in the resonator.
Relation between the unloaded quality factor, external quality factor and loaded quality factor of a cavity has been obtained keeping in view some part of energy stored in a cavity being coupled out from it to an external load in practice.
Quality factor of a cavity resonator may also be expressed in terms of the resonant frequency and bandwidth of the frequency response of half the value of square of the ratio of the magnitude of the equivalent impedance of the resonant circuit comprising an inductor, capacitor and resistor in parallel.
Readers are encouraged to go through Chapter 10 of the book for more topics and more worked-out examples and review questions.