Transmission Line Theory – Frequency Domain
i(z,t)
V(z,t)
(a)
i(z+
i(z,t)
V(z,t)
R
L
c
G
z,t)
V(z+ z,t)
(b)
For an incremental length of transmission line , (a) voltage and
current definitions , (b) lumped – element equivalent circuit .
R : series resistance per unit length , for both conductors , in
L : series inductance per unit length , for both conductors , in
G : shunt conductance per unit length , in
C : shunt capacitance per unit length , in
S /m
F /m
/ m
H /m
Kichhoff’s voltage law :
v ( z, t )  Rzi ( z, t )  Lz
i ( z, t )
 v ( z  z, t )  0
t
Kichhoff’s current law :
i ( z, t )  Gzv ( z  z, t )  Cz
v ( z  z, t )
 i ( z  z, t )  0
t
dividing by z and taking the limit as z  0
v ( z, t )
i ( z, t )
  Ri ( z, t )  L
z
t
i ( z, t )
v ( z, t )
 Gv( z, t )  C
z
t
time – domain form of the
transmission line equation
(telegrapher equation)
For the sinusoidal steady – state condition , the phasor equation :
dV ( z )
 ( R  jwL) I ( z )
dz
dI ( z )
 (G  jwC)V ( z )
dz
Solving simultaneously ,
d 2V ( z )
  2V ( z )  0
2
dz
d 2 I ( z)
  2 I ( z)  0
2
dz
Where     j 
( R  jwL)(G  jwC)
is the complex propagation constant , which is a function of frequency.
Traveling wave solutions :
V ( z )  V  ( z )  V  ( z )  V0 e z  V0 e z
I ( z )  I  ( z )  I  ( z )  I 0 e z  I 0 e z
Since I ( z ) 
1
dV ( z )


[V0 e z  V0 e z ]
 ( R  jwL) dz
R  jwL
 characteristic
Z0 
R  jwL

therefore

impedance is defined as
R  jwL


G  jwC G  jwC
V0 z V0 z
I ( z) 
e 
e
Z0
Z0
V0
V0
Z0     
I0
I0
The Terminated Lossless Transmission Line
Figure A transmission line with characteristic impedance
terminated with a
load impedance
ZL
V ( z )  V  ( z )  V  ( z )  V0 e  jz  V0 e  jz
I ( z )  I  ( z )  I  ( z )  I 0 e  jz  I 0 e  jz 
Where Z 0 
Let
1
(V0 e  jz  V0 e  jz )
Z0
V0
V0


I 0
I 0
V  ( z)
( z )  
V ( z)
, then
V0 e  jz  V0 e  jz
V ( z)
1  ( z )
Z ( z) 
 Z 0   jz
 Z0
  jz
I ( z)
V0 e
 V0 e
1  ( z )
Solving for (z ) , then
( z ) 
Note:
Z ( z)  Z 0
Z ( z)  Z 0
 (z ) :voltage
T (z ) :voltage
reflection coefficient
transmission coefficient
Z0
is
Since
Z L at
z=0 , then
VL  V (0)  V  (0)  V  (0)  V0  V0
I L  I (0)  I  (0)  I  (0)  I 0  I 0 
1
(V0  V0 )
Z0
V  (0) V0
L  (0)  

V (0) V0
Z L  Z (0) 
therefore
VL
1  L
1   ( 0)
 Z0
 Z0
IL
1   ( 0)
1  L
, L  Z L  Z 0
Z L  Z0
Note : Power delivered to the load
V L  V0  V0  V0 (1  L )
IL 
V0  V0 V0

(1  L )
Z0
Z0
 2
V0
1
1
Pdel  Re(VL I L *)  Re * (1  L )(1  L *)
2
2
Z0

V0
2
 2
1
1 V0
2
2
Re[ * (1  L  L  L *)] 
Re(1  L )
2
2
2 Z0
Z0
 Pinc (1  L )
2
 2
(V  ) * 1 V0
1
1
 Pinc  Re(V0 I 0 *)  Re V0 0

R0
2
2
Z0 *
2 Z* 2
0
therefore , although T=1+ 
T  (1  )(1  *)  1  
2
2
*Input impedance at z=- l looking toward load
V0 e  jl  V0 e  jl
e  jl  L e  jl
V ( l )
Z in ( l ) 
 Z 0   jl
 Z 0  jl
I ( l )
V0 e
 V0 e  jl
e
 L e  jl
Z L  Z 0  jl
e
Z L  Z0
Z ( e  jl  e  jl )  Z 0 ( e  jl  e  jl )
 Z 0 L  jl
Z  Z 0  jl
Z 0 (e
 e  jl )  Z L ( e  jl  e  jl )
 L
e
Z L  Z0
e  jl 
 Z0
e  jl
 Z0
Similarly ,
Z L cos l  jZ 0 sin l
Z  jZ 0 tan l
 Z0 L
Z 0 cos l  jZ L sin l
Z 0  jZ L tan l
V0 e  jl  V0 e  jl
e  jl  L e  jl
I ( l )
Yin ( l ) 
 Y0   jl
 Y0  jl
V ( l )
V0 e
 V0 e  jl
e
 L e  jl
Y0  YL  jl
e
Y0  YL
YL ( e  jl  e  jl )  Y0 ( e  jl  e  jl )
 Y0
Y0  YL  jl
Y0 ( e  jl  e  jl )  YL ( e  jl  e  jl )

e
Y0  YL
e  jl 
 Y0
 Y0
or simply
Yin ( l ) 
e  jl
YL cos l  jY0 sin l
Y0 cos l  jYL sin l
1 Z 0 cos l  jZ L sin l
1

Z in ( l ) Z 0 Z L cos l  jZ 0 sin l
 Y0
YL cos l  jY0 sin l
Y0 cos l  jYL sin l
*Special Cases
1. open – circuit termination
Z in  jX in   jZ0 cot l
( ZL   )
( Z 0  R0 , actually)
Figure: Input reactance of open – circuited transmission line.
2. short – circuit termination
Z in  jX in   jZ0 tan l
( ZL  0 )
( Z 0  R0 , actually)
Figure: Input reactance of short – circuited transmission line.
3. quarter – wave section

tan l  tan[( 2n  1) ]  
2
2
Z
Z in  0
ZL
(l 

4
, l 

2
) (more general ,
l  ( 2n  1)

4
)
Voltage Standing Wave Ratio
V ( z )  V  ( z )  V  ( z )  V  ( z )(1  ( z ))
 V ( z )  V  ( z ) (1  ( z ))
1
V  ( z)


I ( z)  I ( z)  I ( z) 
[V ( z )  V ( z )] 
[1  ( Z )]
ZO
ZO

 I ( z) 

V  ( z)
1  ( Z )
ZO
Since ( z )  e j , the maximum voltage is
(   00 )
 V ( z ) max  V  ( z ) 1  ( z ) max  V  ( z ) (1   )
this maximum occurs when   0 0 ,i.e., (z )   , at the same time ,
the current is a minimum ,
I ( z ) min 
V  ( z)
Z0
1  ( z ) min 
V  ( z)
Z0
(1   )
similarly , the minimum voltage is
V ( z ) min  V  ( z ) 1  ( z ) min  V  ( z ) (1   )
this minimum occurs when
  180 0
(   180 0 )
,i.e.,
current
is a maximum.
I ( z ) max 
V  ( z)
Z0
1  ( z ) max 
V  ( z)
Z0
(1   )
(z )   
, meanwhile , the
The ratio of the maximum to minimum voltages along a terminated
transmission line is called voltage standing wave ratio (VSWR).
VSWR 
V ( z ) max
V ( z ) min

V  ( z ) (1   )
V  ( z ) (1   )

(1   )
(1   )
(   (z) )
* Maximum Impedance
Z ( z ) max 
V ( z ) max
I ( z ) min
 Z0
1 
1 
 R0
1 
1 
(for lossless transmission line)
normalized maximum impedance :
Z max 
Z
max
R0

(1   )
 VSWR
(1   )
* Minimum Impedance
Z ( z ) min 
V ( z ) min
I ( z ) max
 Z0
1 
1 
 R0
1 
1 
normalized maximum impedance :
Z min 
Z min
R0

(1   )
1

(1   ) VSWR
(for lossless transmission line)
V ( z )  V  ( z )  V  ( z )  V0 e  jz  V0 e  jz  V0 e  jz 1  (0)e  j 2 z
 V0 1  e j (  2 z )
V0
j
let (0)    e
V0
 V0 1   cos(  2 z )  j sin(   2 z )
 V 1    2  cos(  2 z )

0
1
2
2
 V0 (1   ) 2  2  [1  cos(  2 z )]2
1
1
 2

 V0 (1   ) 2  4 sin 2 ( z  )]
2 


where z   n ,
2
z 

2
 n 

V ( z ) max  V0 (1   )

2
,

V ( z ) min  V0 (1   )
* therefore , distance between two successive maximum or minimum is
d  
or d 
 

 2
distance between nearby maximum or minimum is d 
or
d

2

4
* Standing wave pattern along a terminated transmission line
plot
V (z )
along a transmission line , for an arbitrary unknown load
(but is not short or open).
The Smith Chart
* Impedance chart
For a lossless transmission line of characteristic impedance Z 0 , the
voltage
reflection coefficient of a load impedance Z L measured at the load can
L 
be written as
where Z 0  R0 
Z L  Z0
  e j
Z L  Z0
L
C
normalized impedance : z L 
reflection coefficient :
Z L RL
X

 j L  r  jx
R0 R0
R0
  r  ji 
Z L  R0 z L  1

Z L  R0 z L  1
1   1  r  ji (1  r  i )  j(2i )
 zL 


2
2
1   1  r  ji
(1  r )  i
2
1  r  i
2
r 
2
(1  r )  i
2
2
r(1  2r  r  i )  1  r  i
2
 ( r  1)r
2
2
2
2
2
 (r  1)i  2rr  1  r
2
 r 
2rr
1 r
2
 i 
r 1
r 1
 ( r 
r 2
1 r
r 2
1 2
2
)  i 
(
) (
)
r 1
r 1 r 1
r 1
2
(dimensionless)
Figure : Constant resistance r circles.
x
2i
2
2
(1  r )  i
x(1  r )  i  2i
2
2
 xr  2 xr  xi  2i   x
2
2
2
2
2
 r  2r  i  i  1
x
1
1 1
 ( r  1)  ( i  ) 2  1  1  2   
x
x
 x
2
2
*Observations about r and x
1. All r-circles are centered at (
r
,0 )
r 1
with radius
1
r 1
2. The r=0 circle having a unity radius and centered at the
origin is the largest circle.
3. For each r-circle , the radius decreases as r increases form
0 to

and the center moves from (0,0) to (1,0) .
4. All r-circles pass through the (1,0) point .
5. The center of all x-circles lie on the r  1 line , centered
at (1,
1
1
) with radius
x
x
for x>0 (inductive reactance) above the r -axis
for x<0 (capacitive reactance) below the r -axis
6. The x=0 circle becomes the r -axis
7. For each x-circle , the radius decreases as x increases from
0 to  and the center moves form(1   ) to (1,0) .
8. All x-circles pass through the (1,0) point .
9. The open circuit z=r+jx =  , r =  or x=  is the
(1,0) point . (It is more clear from   1 )
10. The short circuit z=r+jx =0 , r =0 and x=0 is the
(-1,0) point .
11. The constant r and the constant x form two families of
orthogonal circles in the chart.
*Observation about
 (reflection
coefficient) (    )
1. All  circles are centered at the origin , and their radii vary
uniformly from 0 to 1 .
2. The line drawn form the origin to the point representing Z L ,
its radius is    , the angle with positive real axis ( r -axis)
is  .
3. The value of the r-circle passing through the intersection of the
 - circle and positive r axis equals the standing wave ratio S .
j
1  1  e
zL 

1   1   e j

=0
 zL 
    zL 
1 
1 
1 
1 
rS
r
1
S
4. ( l )  L e  j 2 l , from load toward generator ,

rotates
clockwise , phase angle changes 2 z .
'
z' 

2
, 2 z = 2
'
2 
 2  one cycle in Smith Chart
 2
If from generator to load , the rotation is counterclockwise.
5. Voltage maximum occurs at the intersection of the  circle and
the positive r axis .
Voltage minimum occurs at the intersection of the  circle and
the negative r axis .
V  V   V   V  1   e j
 =0 , maximum value
V  V  (1   )
   , minimum value
V  V  (1   )
Note:
V  ( z ) V0 e  jz
( z )  
   jz  (0)e  j 2 z  L e  j 2 z  e j e  j 2 z
V ( z ) V0 e
(l )  e j e j 2 l
Figure: Smith Chart
*Admittance Chart
Since normalized impedance
zL 
Z L G0
1


R0 YL y L
Therefore , the normalized admittance
yL 
YL
1   1  r  ji
 g  jb 

G0
1   1  r  ji
(1  r  i )  j(2i )

2
2
(1  r )  i
2
2
1  r  i
g 
2
2
(1  r )  i
2
2
g (1  2r  r  i )  1  r  i
2
2
2
2
( g  1)r  ( g  1)i  2 gr  1  g
2
2
 r 
2 gr
1 g
2
 i 
( g  1)
( g  1)
 ( r 
g 2
1 g
g 2
1 2
2
)  i 
(
) (
)
g 1
g 1 g 1
g 1
2
b
 2i
2
2
(1  r )  i
b(1  2r  r  i )  2i
2
2
 br  2br  bi  2i  b
2
2
2
2
2
 r  2r  i  i  1
b
1
1
1
 ( r  1) 2  ( i  ) 2  1  1  2  ( ) 2
b
b
b
* Observations about g and b
1. All g-circles are centered at (-
g
,0 )
g 1
with radius
1
g 1
Figure: Constant conductance g circles
2. The g=0 circle having a unity radius and centered at the origin .
3. For each g-circles , the radius decreases as g increases from
0 to  and the center moves form (0,0) to (-1,0)
4. All g-circles pass through the (-1,0) point .
5. The center of all b-circles lie on the r  1 line , centered
1
at (-1,  ) with radius
b
1
b
for b<0 (inductive reactance) above the r -axis
for b>0 (capacitive reactance) below the r -axis
6. The b=0 circle becomes the r -axis
7. For each b-circles , the radius decreases as b increases from
0 to  and the center moves form(-1   ) to (-1,0) .
8. All b-circles pass through the (-1,0) point .
9. The open circuit y=g+jb =0 , g =0 and b=0 r =  or x=  is the
(1,0) point .
10. The short circuit z=r+jx =0 , g =  or b=  is the
(-1,0) point . (It is more clear from   1 )
11. The constant g and the constant b form two families of
orthogonal circles in the chart.
Figure: Constant susceptance b circles
*Constant Q Contours on Smith Chart
Figure: Constant Q contours on Smith Chart
If Q 
x
R
1
Q
2
2
Then u  (v  )  1 
1
Q2
L 
Z L  R0 Z L / R0  1 z L  1


 r  ji   e j
Z L  R0 Z L / R0  1 z L  1
zL 
1 
 r  jx
1 
( r 
r 2
1 2
2
)  i  (
)
r 1
r 1
1
1
( r  1)  ( i  ) 2   
x
 x
2
zL 
ZL
R0
solid line circles , r - circles
2
dash line circles , x – circles
Toward generator
ZL
Z’
=0
Z’
THE SMITH CHART
Smith Chart with polar coordinates.
V ( z )
1  e  j 2 z
Z i ( z ) 
 R0 [
]
I ( z )
1  e  j 2 z
j
Zi 1  e j 2 z  1   e
zi 


 j 2 z 
R0 1  e
1   e j
     2z
S=
V ( z ) max
V ( z ) min

1 
1 
Normalized
Admittances on Smith Chart
yL 
1
1

 R0YL
z L Z L / R0
yL 
YL
Y0
2
R
Zi  0
ZL
P: z L  1.7  j0.6
P : y L 
1
 0.52  j 0.18
zL
Zi
1

R0 Z L
R0
zi 
1
zL
Find the input impedance of a 50-  T.L that is 0.1  long
And is terminated in a shorted circuit
50 ohm
Zin
zL 
L=0.1 lamda
0
0
50
P1 : zi  j0.725
Z i  R0 zi  j36.3
Design a circuit so that an incident wave or power
will totally deliver to the load Z L
Impedance Matching holds if
Yi 
1
 Y0  Y B  Y S
R0
R0Y0  R0YB  R0YS (Normalization)
1  yB  yS
y S is purely imaginary
1  y B  jbS
or y B  1  jbS
y s   jbS
We adjust the lengths d and l so that the above equation
is satisfied.
A 100-  T.L having length 0.434  is terminated with a
load 260+j180 (  ) Find
(a)voltage reflection coefficient
(b) SWR
(c)input impedance
(d) the location of a voltage maximum closest to the load.
P2 : z L 
260  j180
 2.6  j1.8
100
 L  0.6e j 21.6
0
VSWR: PM  S  4
0.434
Z in : PM  P3
zin  0.7  j1.2
z max : P2 
 PM
z max  0.25  0.22  0.03
Z in  70  j1 2 0
Example: Z 0  50  is connected to a load impedance
Z L  35  j 47.5()
(a) Find the position and length of a
short circuited stub required to match the line.
50ohm
zL 
35-j47.5 ohm
350  j 47.5
 0.7  j 0.95  P1
50
0.25
P1 
 P2
P3 : y B  1  j1.2
P4 : y B  1  j1.2
y L  0.5  j0.7
(change from impedance to admittance)
P2 to P3
d1  0.168  0.109  0.059
P2 to P4
d 2  0.332  0.109  0.223
Psc to P3
l1  0.301  0.25  0.111
Psc to P4
l2  0.25  0.139  0.389
Example: Illustrate the effect of adding a series inductor L( z L  j0.8 )
to an impedance z  0.3  j0.3 in the immittance Smith Chart
ZL=j0.8
Z=0.3-j0.3
Zin=0.3+j0.5
Example: Illustrate the effect of adding a series capacitor C( zL   j0.8 )
to an impedance z  0.3  j 0.3 in the immittance Chart
ZC=-j0.8
Z=0.3-j0.3
Zin =0.3-j1.1
Example: Illustrate the effect of adding a shunt inductor L( z L   j2.4 )
to an admittance y  1.6  j1.6 in the immittance Chart
yL=-j2.4
yin=1.6-j0.8
y=1.6+j1.6
Example: Illustrate the effect of adding a series capacitor C( yc  j3.4 )
to an admittance y  1.6  j1.6 in the immittance Chart
yc=j3.4
yin=1.6+j5
y=1.6+j1.6
Example: A load Z L  10  j10() is to be matched to a 50  line. Design
two matching networks and specify the values of L and C at
500MHz.
Solution : Choose the series L-shunt C network ,
ZL
at A(0.2+j0.2) + series L( Z  j 0.2 )  zB (0.2  j0.4;10  j20)
1
or y B (1  j 2;20  j 40m ) + shunt C(y=j2)  yin  1 or
zin  1
10
 3.18nH
2  500  10 6 Hz
1
C
 12.74 pF
25  2  500  10 6 Hz
L
(a)
(b)
Example : Design the circuit using series C – shunt L network
Solution : Z L at A(0.2+j0.2) + series C( Z  0.6 )
 z B (0.2  j0.4;10  j 20)
or y B (1  j 2;20  j 40m 1 ) + shunt L(y=-j2)  yin  1 or zin  1
L
1
 7.95nH
40  10 3  w
1
 0.6  50  30
WC
C 
1
 10.6 pF
30  w
(a)
(c)
Find the input admittance (treat the Smith Chart as a normalized
impedance)
Z
yL=2.6+j1.8
0.072入
yin

or1800
0.072

or1800
4
y L  
 z L  zin 

 yin
4
yin =2-j1.9
Treat the Smith Chart as a normalized admittance
Z
yL=2.6+j1.8
0.072入
yin
For the configuration , please find the input admittance
1. Enter the chart with the complex value y L , exactly as if it were a
normalized impedance.
2. Rotate the reflection coefficient in the regular direction , according
only to a distance z .
3. Read y in from the chart , exactly as if it were a normalized
impedance.