chapter 2 transmission line theory. transmission-line (tl) theory at dc or very low frequencies, the...
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Chapter 2
Transmission Line Theory
Transmission-Line (TL) Theory
At DC or very low frequencies, the equivalent circuit can be simplified as
At medium and high frequencies, the equivalent circuit becomes
Rs R
ZL
Rs R L
G C ZL
Lumped-element equivalent circuit
ZL
Rs l
c
TL theory bridges the gap between field analysis and basic circuit theory.
Distributed equivalent circuitAt RF and microwave frequencies, a general two-conductor uniform line divided into many sections can be used to describe the transmission-line behavior.
Rs L Z
R Z
G Z C Z
R Z
G Z C Z
R Z
G Z C Z ZL
L Z
L Z
ZL
Rs l = NZ
Z
N sections
L,C,R,G are called distributed parameters.
whereR: Conductor resistance (Series resistance) per unitlength.I2R/2: Time-average power dissipated due to conductor loss per
unitlength.L: Self inductance (Series inductance) per unitlength.I2L/4: Time-average magnetic energy stored in a unitlength
transmission line.C: Self capacitance (Shunt capacitance) per unitlength.V2C/4: Time-average electric energy stored in a unitlength
transmission line.G: Dielectric Conductance (Leakage conductance, Shunt
conductance) per unitlength.V2G/2: Time-average power dissipated due to dielectric loss in a
unitlength transmission line.At very low frequencies:
0)()(
0
0
0
0
0
G
CjY
LjZ
C
L
Thus, L,C,G can be ignored at very low frequencies. But at high frequencies, effects due to L,C,G have to be considered.
( represents dielectric conductivity)
Solutions of L,C,G parameters
),(),(0),( 0002 00
vuEvuVvuV tVE
tttttt
0),(
),(
220
0110
vuV
VvuV
t
t
),(0 vuHt
00 ˆ tzc
t EaH
PDE: (Laplace’s Equation)
BCs:
,,
),(0 vuVt0V
Z=0
Z=l
jc
SC
(Watt/m) 2
1
2
1
(Joule/m) 4
1
4
1
(Joule/m) 4
1
4
1
(A)
*002
0
*0020
*0020
00
dSEEGV
dSEECV
dSHHLI
ldHI
S
tt
t
S
t
t
S
t
C
t
(S/m)
(F/m)
(H/m)
*00
20
*002
0
*0020
dSEEV
G
dSEEV
C
dSHHI
L
S
tt
t
S
t
t
S
t
L,C,G Distributed parameters can be found as
For distributed parameters of TEM transmission lines
// , GCLC
Example2.1: Find the TL parameters of coaxial Line?
PDE: 0),( 1
)(1 0
2
2
2
rV
rrr
rr t
BCs:
0),(
),(0
00
brV
VarV
t
tab
,,
),(0 rVt
Due to symmetry,
0 ),(),( 00
rVrV tt
PDE becomes ODE:
0)( ) d
d(
d
d 0 rVr
rr t
BCs become
0)( ,)( 00
0 brVVarV tt
Solution( another solution can refer to p.54 of the text book)
General solutions for electric potential at z=0
210 )ln()( CrCrVt
Substitute BCs into general solutions to find the coefficients C1 and C2
)ln()/ln(
,)/ln(
02
01 b
ab
VC
ab
VC
Final solution
(V) )/ln()/ln(
)( 00 brab
VrVt
Electric and magnetic fields at z= 0
(A/m) ˆ
)/ln()(ˆ)(
(V/m) ˆ
)/ln()( )ˆˆ()()(
000
0000
r
a
ab
VrEarH
r
a
ab
VrV
ra
rarVrE
ctz
ct
rtrttt
0tE
0tH
Current along the inner conductor at z=0
(A) )/ln(
2ˆˆ
)/ln(002
00
0 ab
Vrda
r
a
ab
VldHI cc
ct
Find distributed parameters L,C,G
(S/m) )/ln(
2dd
1
)/ (ln
(F/m) )/ln(
2dd
1
)/ (ln
(H/m) )/ln(2
dd 1
)2(
2
0 22
*002
0
2
0 22
*002
0
2
0 22
*0020
abrr
rabdSEE
VG
abrr
rabdSEE
VC
abrrr
dSHHI
L
b
aS
tt
b
atS
t
b
atS
t
Check the following relations between LC and C/G
// , GCLC
Loss tangent of dielectric
0)(tan
c
Material =r0tanc
FR4 r= 4.5 0.014
Ceramic r= 9.9 0.0001
Teflon r= 2.2 0.0003
GaAs r= 12.9 0.002
Silcon r= 11.9 0.015
Conductor resistance per unitlength
/m)( )11
(
1
) 2(
1
) 2(
1
21
CCt
btatR
c
cc
t
t
cc ,
C1
C2
Skin effect: At high frequencies, currents tend to concentrate on surface of the conductor within a skin depth or penetration depth
cc ,
)( ) (1resistance surface
(m) 1depthskin where
/m)( )11
()11
(
1
) 2(
1
) 2(
1
, If
2121
cccs
cc
sc
cc
fR
f
CCR
CC
baR
t
f
t
(f)
tec
fec
ec
ecec fff
fftt
),(
,
Effective conductor thickness
(Defined as amplitude of fields decay to 1/e)
/m)( ,) (
,) () ( )
2
1
2
1(
1
(S/m) )2(cosh
)tan(
)2(cosh
(F/m) )2(cosh
(H/m) )2
(cosh
1
11
11
1
1
ecs
ecc
ecc
c
ffaR
ffat
aatR
aDaDG
aDC
a
DL
Example2.2: Two-wire line
Example2.3: Parallel-plate line
/m)( ),/2(
),/2( ) ( )
11(
1
(S/m) )tan(
(F/m)
(H/m)
1
ecs
ecc
ecc
c
ffwR
ffwt
wwtR
d
w
d
wG
d
wC
w
dL
D
a a ,,
cc
d
w
,,
cc ,
LC
//
GC
LC
//
GC
General solutions (traveling-wave solutions) of transmission-line equations yield:
zz
zz
eIeIzI
eVeVzV
0
0
0
0
)(
)(
0 00
0 0
(Characteristic Impedance, Unit: )
V V R j LZ
G j CI I
where
zz eIeV 0
0 ,
zz eIeV 0
0 ,
0 , Z
z
constant Phase: constant, nAttenuatio :
constant, npropagatioComplex :
)(m ) )( ( -1
CjGLjRj
ps. The parameters and Z0 are called transmission line parameters.
0)()(
0)()(
22
2
22
2
zIdz
zId
zVdz
zVd
Wave Equations of Transmission Line
)()()(
)()()(
zVCjGdz
zdI
zILjRdz
zdV
Telegrapher equations
Low-Loss Transmission Lines
LCjC
LG
L
CR
Cj
G
Lj
RLCj
Cj
GCj
Lj
RLjj
)22
() 2
1)( 2
1(
)
1( )
1(
(Low-loss conditions: R<<L, G<< C,)
LCGZ
Z
Rdc ,
220
0
Therefore,
where c is attenuation due to conductor loss d is attenuation due to dielectric loss
)](2
11[
] 1[
)] (1[
0 C
L
C
G
L
Rj
C
L
CjGCj
LjRLj
CjG
LjRZ
Lossless Transmission Lines
LC
0,0
2 1 =
LC LCp
LZ v
C
(Lossless conditions: R= G=0)
ab ,,
D
a a ,,
cc
d
w
,,
cc ,
For low-loss coaxial lines,
)/ln(2
1
)/ln(/) 2(
)/ln( )2/(0 ab
ab
ab
C
LZ
For low-loss parallel-plate lines,
)2/(cosh1
)2/(cosh/) (
)2/(cosh)/(
1
1
1
0
aD
aD
aD
C
LZ
For low-loss two-wire lines,
w
d
dw
wd
C
LZ
/
/ 0
Example2.4:
Distortionless Lines
Lossy line has a linear phase factor as a function of frequency.
C
G
L
RRelation :
LCjL
CRj Verification :
Advantage : Distortionless line transmitted signal without dispersion.
0 C
LZ
Dispersion : If the phase velocity is different for different frequencies,then the individual frequency components will not maintain their original phase relationships as they propagate down the TL, and signal distortion will occur.
Why 50 characteristic impedance for coaxial lines?
ab ,,/m)( )
11(
2
1 (S/m)
)/ln(
2
(F/m) )/ln(
2 (H/m) )/ln(
2
baR
abG
abCabL
c
)/ln(2
10 ab
C
LZ
)()
22( dcC
LG
L
CRj
2
1
ln
)11()(
2
1 1
d
cc
ab
ba
From distributed parameters
From transmission-line parameters
Attenuation constant due to conductor loss
Attenuation constant due to dielectric loss
ab
ab
bcln
)1(1 Assuming that the outer dimension b is fixed, c
has a minimum when b/a=5.591. The value comes from
0da
d c
filledTeflon ; 52591.3ln2.22
1
filledAir ; 77591.3ln2
1
0
00
0
00
Z
Z
Considering the breakdown voltage Vb
a
baEV
r
a
ab
VrE BB
rt ln
ˆ
)/ln()( 00
Then the maximum power capacity Pmax
a
baP ln2
max Assuming that the outer dimension b is fixed, Pmax has a minimum when b/a=1.649. The value comes from
filledTeflon ; 20649.1ln2.22
1
filledAir ; 30649.1ln2
1
0
00
0
00
Z
Z
Therefore, use 50 to compromise between 77 and 30.(Also reference textbook p.130 “point of interest” and problem 2.28 and 3.28.)
0max da
dP
1.52.02.53.0
3.54.04.55.05.5
6.06.57.07.58.08.5
9.09.5
1.0
10.0
0.02
0.04
0.06
0.08
0.10
0.12
0.14
0.16
0.18
0.00
0.20
10
20
30
40
50
60
70
80
90
100
110
120
130
0
140
x
z1
1.6440.184
m7
z2
m7indep(m7)=z1=0.184
1.649
Eqny1=x**-2*ln(x)
Eqnx=[1::0.001::10]
Eqny2=ln(x)*377/(2*pi)
Eqnz1=plot_vs(y1,x)
Eqnz2=plot_vs(y2,x)
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
1.0
10.0
3
4
5
2
6
10
20
30
40
50
60
70
80
90
100
110
120
130
0
140
x
z1
3.4443.594
m7
z2
m7indep(m7)=z1=3.591
3.591Eqny1=(x+1)/ln(x)
Eqnx=[1::0.001::10]
Eqny2=ln(x)*377/(2*pi)
Eqnz1=plot_vs(y1,x)
Eqnz2=plot_vs(y2,x)
The Terminated Lossless TL
aveincident w is
] [ )(
][ )(
0
0
0
j 0
zj
zjzj
zzj
eV
eeZ
VzI
eeVzV
1
1WR
dB log20
min
max
0
0
0
0
V
VS
RL
ZZ
ZZ
V
V
L
L
ljZZ
ljZZZ
e
eZZ
eeV
eVl
L
Llj
lj
in
ljlj
lj
tan
tan
1
1
)0()(
0
002
2
0
2
0
0
Transmission line impedance equation
Reflection coefficient 0 1
Return loss (RL,dB) 0
Standing wave ratio 1
Match Mismatch (Total Reflection)
Match conditions
Zin
Terminated in short circuit Terminated in open circuit
ljZZin tan0 ljZZin cot0
2|in L l
Z Z
Lin Z
ZZ
21
Quarter-Wave TransformerA useful and practical circuit for impedance matching.
Defined as TL with length equals to ℓ=/4(+ n/2).
Perfect matching occurs at one frequency (odd multiple) but mismatch will occur at other frequencies.
Impedance matching is limited to real load impedances (complex load impedance can be transferred to real one, by transformation through an appropriate length of line.)
Substituting ℓ=(2/)(/4)= /2
into equation Zin can find
LRZZ 01
In order for =0, one must have Zin = Z0, then Insertion Loss (IL)
1T
(dB) ;log20 TIL
Formulations for TL
Generator and Load Mismatch
l
l
g
gg
l
ll
S
ZZ
ZZ
ZZ
ZZ
1
1WR
0
0
0
0
Load matched to line Generator matched to line
Conjugate matching
Zl=Z0 ; l=0; Zin=Z0 ; SWR=1
Zin=Zg ; l 0 ; SWR>1
220
02
)(2
1
ggg
XRZ
ZVP
2
2
)(42
1
gg
gg
XR
RVP
Zin=Zg* ; g 0 ; l 0 ; 0 ; SWR>1
gg R
VP4
1
2
1 2
0
gin
gin
ZZ
ZZ
Multiple reflections may add in phase to deliver more power P to the load.
The Smith ChartThe Smith chart is a plot of
1800 , where
)1
1( ;ΓΓ
1
1 2
jxrZ
ZeeZ
Z ljj
constant-r circles
constant-x circles
1
1 is radius ,0),
1(at iscenter circules where
;)1
1(Γ)
1( 222
rr
rrr
rir
xx
xxir
1 is radius ),
1,1(at iscenter circules where
;)1
()1
()1( 222
xr ; 0 impedance passiveFor
Compressed Smith Chart
It is applied for both active and passive networks. While a standard Smith chart is used in passive networks where Re(Z)0.
Example2.5: A load impedance of 50+j100 terminates a lossless /4 line (Z0=50). Find the input impedance, the load reflection coefficient, and VSWR?Solve 1Solution from equations
904
2 l
201010050
2500
)10050(
5050
90tan)10050(50
90tan50)10050(50
tan
tan
0
00
jjjj
j
jj
jj
ljZZ
ljZZZZ
L
Lin
2010)4.02.0(50
;4.02.0
;4571.0
;8.5
jjZ
jZ
VSWR
in
in
L
8283.5707.01
707.01
1
1
45707.0452100
90100
100100
100
5010050
5010050
0
0
VSWR
j
j
j
j
ZZ
ZZ
L
L
Solve 2
Solution from Smith chart
scale. generator" toward
wavelength" theon 0.438at point a to
)2- (i.e.-by moves load theFrom
2150
10050
0
l
jj
Z
ZZ L
L
(a) 0.62 30
(b) 4.25
(c) 0.375 , 2 4.7124 270
, clockwise from the load gets
75(0.3 0.54) 22.5 40.5
(d) (0.25 0.208) 0.042
Normalized impedance 4.25(shortest)
(
L
in
VSWR
d - βl rad
Z j j
d
d
0.5 0.208) 0.292
Normalized impedance 0.23(next one)
(e)The normalized impedance and VSWR are the same at this point, therefore
75 4.25 318.75R
Example2.6: A lossless 75 line is terminated by an impedance of 150+j150. Find (a) L , (b) VSWR, (c) Zin at a distance of 0.375 from the load, (d) the shortest length of the line for which impedance is purely resistive, and (e0 the value of this resistance?Solve
2275
150150
0
jj
Z
ZZ L
L
Transverse Electromagnetic (TEM) Wave (Chapter 3)
. thealong npropagatio waveAssumed ;0 axiszHE zz
22
2
22
2222 ;0),( ,0),(
yxyxhyxe ttt
TEM wave have a uniquely defined voltage, current, and characteristic impedance.
TEM wave exists in TL consisted of two or more conductors.
impedance) istic(character impedance) (wave 0 I
VZ
H
EZ
y
xTEM
),(ˆ1
),( yxezZ
yxhTEM
two-conductor TLsingle-conductorclosed waveguide
Characterized as
TE Wave : 0,0 zz HE
222222
2
22
2
;0)(
kkhkyx
czc
Transverse Electric (TE) and Transverse Magnetic (TM)
Wave
22; cy
xTE kk
k
H
EZ
wave)for ; : Note( TEMZk
0,0 zz HE
222222
2
22
2
;0)(
kkekyx
czc
22; cy
xTM kk
kH
EZ
TM Wave :
TE and TM waves have not a uniquely defined voltage, current, and characteristic impedance.
TE and TM waves often exist in single-conductor structure.
Equation must be solved subject to the boundary conditions of specific geometry
• kcis cutoff wavenumber.•Wave propagation needs: is real k>kc
f>fc (cutoff frequency)
•Cutoff or evanescent mode f<fc is imaginary
2c
ck
f
Parallel Plate Waveguide
Parallel Plate Waveguide
