Chapter 5 Transmission Lines
5-1 Characteristics of Transmission Lines
Transmission line: It has two conductors carrying current to support an EM wave,


which is TEM or quasi-TEM mode. For the TEM mode, E   Z TEM aˆ n  H ,

H
1
Z TEM

aˆ n  E , and Z TEM   

.

The current and the EM wave have different characteristics. An EM wave
propagates into different dielectric media, the partial reflection and the partial
transmission will occur. And it obeys the following rules.
Snell’s law:
sin  t 1 n1 v p 2  2
1
 r1
and θ i =θ r






sin  i  2 n2 v p1 1
2
 r2
The reflection coefficient: Γ=
Er0
E
and the transmission coefficient: τ= t 0
Ei 0
Ei 0
 2 / cos  t   1 / cos  i n1 cos  i  n 2 cos  t sin( t   i )

   / cos    / cos   n cos   n cos   sin(   )

2
t
1
i
1
i
2
t
t
i

2

/
cos

2
n
cos

2
cos

sin
t
2
t
1
i
i
 




 2 / cos  t   1 / cos  i n1 cos  i  n 2 cos  t
sin( t   i )
for perpendicular polarization (TE)
 2 cos  t   1 cos  i n1 / cos  i  n 2 / cos  t tan( t   i )

||   cos    cos   n / cos   n / cos   tan(   )

2
t
1
i
1
i
2
t
t
i



n

 i sin  t
2
cos
2
/
cos
2
cos
2
1
i
t
 


||

 2 cos  t   1 cos  i n1 / cos  i  n 2 / cos  t sin( i   t ) cos( i   t )
for parallel polarization (TM)
 2  1

  //    
1
2
2
1
In case of normal incidence, 
, where η 1 =
and η 2 =
.


2

1
2
2
   
//
 
 2  1
Equivalent-circuit model of transmission line section:
R ( / m) , L ( H / m) , G ( S / m ) , C ( F / m)
Transmission line equations: In higher-frequency range, the transmission line model
is utilized to analyze EM power flow.
i(z, t)
i
 v(z  z, t)  v(z, t)
 v
Ri
L
 Ri(z, t)  L




  z
z
t
t


 i(z  z, t)  i(z, t)  Gv(z, t)  C v(z, t)
   i  Gv  C  v
  z

t
z
t
Set v(z,t)=Re[V(z)ejωt], i(z,t)=Re[I(z)ejωt]

 



 d 2V ( z )
dV
2
 ( R  j L ) I ( z )
 dz 2  ( R  jL)(G  jC )V ( z )   V ( z )
dz
 2
dI
 d I ( z )  ( R  jL)(G  jC ) I ( z )   2 I ( z )
 (G  j C )V ( z )
 dz 2
dz




where γ=α+jβ= ( R  jL)(G  jC ) V (z)  V0 ez V0 ez , I (z)  I 0 ez  I 0 ez

Characteristic impedance: Z 0 =

V0
V
R  jL
R  jL

  0 




G  jC
G  jC
I0
I0
Note:
1. International Standard Impedance of a Transmission Line is Z 0 =50Ω.
2. In transmission-line equivalent-circuit model, G≠1/R.
Eg. The following characteristics have been measured on a lossy transmission
line at 100 MHz: Z 0 =50Ω, α=0.01dB/m=1.15×10-3Np/m, β=0.8π(rad/m). Determine
R, L, G, and C for the line.
(Sol.) 50 
R  j2108 L
, 1.15×10-3+j0.8π= (R  jL)(G  jC)  50 (G  j2108 C)
8
G  j 210 C
0.8
1.15
 80( pF / m) , G 
 10 3  2.3  10 5 ( S / m) ,
8
50
2  10  50
R  2500G  0.0575( / m) , L  2500C  0.2( F / m)
C 
Eg. A d-c generator of voltage and internal resistance is connected to a lossy
transmission line characterized by a resistance per unit length R and a
conductance per unit length G. (a) Write the governing voltage and current
transmission-line equations. (b) Find the general solutions for V(z) and I(z).
(Sol.) (a)   0    ( R  jL)(G  jC )  RG
d 2V ( z )
d 2 I ( z)
 RGV ( z ),
 RGI ( z )
dz 2
dz 2
(b) V ( z )  V0 e 
RG z
 V0 e
RG z
, I ( z )  I 0 e 
RG z
 I 0 e
RG z
Lossless line (R=G=0):
    j  j LC   0,    LC, v p 
L
L

1

, Z0 
 R0  jX0  R0  , X 0  0
C
C

LC
Low-loss line (R<<ωL, G<<ωC):
    j  j LC (1 
Z0 
L
1
C
1 R G
G
),    LC , v p 
(  )]    ( R
C
2
L
2 j L C
L
1 R G
(  )]
[1 
C
2 j L C
Distortionless line (R/L=G/C):
    j 
C
C
( R  j L )    R
,    LC , v p 
L
L
1
LC
, Z0 
L
C
1
LC
Large-loss line (ωL<<R, ωC<< G):
  ( R  jL)(G  jC )    j  RG  (1  j
RG [1 
R
1
) 2 (1 
jC 12
) 
G
j L C
(  )]
2 R G

   RG ,  
Z0 
L
2
(L 
G
R
1
G
R
C
) , v p  (L 
C
)
R
G
2
R
G
R  j L
R
jL 12
jC  1 2
R
j L C

 (1 
)  (1 
) 
 [1 
(  )]
G  jC
G
R
G
G
2 R G
Eg. A generator with an open-circuit voltage v g (t)=10sin(8000πt) and internal
impedance Z g =40+j30(Ω) is connected to a 50Ω distortionless line. The line has a
resistance of 0.5Ω/m, and its lossy dielectric medium has a loss tangent of 0.18%.
The line is 50m long and is terminated in a matched load. Find the instantaneous
expressions for the voltage and current at an arbitrary location on the line.
(Sol.) 0.18%=
G


 C  2.21  10 2 G , V g =10j
 C
∵ Distortionless, ∴
   LC  L
V0 
Z 0V g
Z0  Z g

L C
C
R

  L  1.11  10  2 H / m ,   R
=0.01Np/m,
R G
L Z0
C L

 5.58rad / m , γ=α+jβ=0.01+j5.58
L Z0
5
5
 j 5 , V0  0 , V ( z )  V0 e z  (  j 5)e ( 0.01 j 5.58) z
3
3
V ( z , t )  Re[V ( z )e j 8000t ] 
I ( z, t ) 
5 10 0.01z
e
 cos(8000t  5.58 z  71.6)
3
1
V ( z, t )

e 0.01z cos(8000t  5.58  71.6)
Z0
2 10
Relationship between transmission-line parameters:
 12
G 12
  ( R  jL)(G  jC )  j LC (1 
)  j  (1 
)  G/C=σ/ε
jC
j
and LC=με
Two-wire line: I  2aJ s , P 
R
1 f c
1 2 Rs
)  R  2( s ) 
I (
2
2a
a  c
2a
Coaxial-cable line: I  2aJ si  2bJ so , Pi 
R
Rs 1 1
1
(  )
2
2 a b
R
1
1 2 Rs
) , Po  I 2 ( s )
I (
2
2a
2
2b
f c 1 1
(  )
c a b
Eg. It is desired to construct uniform transmission lines using polyethylene
(ε r =2.25) as the dielectric medium. Assume negligible losses. (a) Find the distance
of separation for a 300Ω two-wire line, where the radius of the conducting wires
is 0.6mm; and (b) find the inner radius of the outer conductor for a 75Ω coaxial
line, where the radius of the center conductor is 0.6mm.


D
(Sol.) Two-wire line: C 
, L  cosh 1 ( ) , a=0.6mm, ε=2.25ε 0
1
2a

cosh ( D / 2a)
L
Z 0  300 

C
Coaxial line: C 
cosh 1 (

D
)
2a 
4  10 7
 D  25.5mm
1
9
2.25 
 10
36
2
b

, L
ln( )
ln(b / a )
2
a
b
ln( )
L
a 
a=0.6mm, Z 0  75 

C
2
4  10 7
 b=3.91mm
1
9
2.25 
 10
36
Parallel–plate transmission line:

 E  yˆ E 0 e z  yˆ E y

,   j  j  ,  
E 0 z

 H   xˆ  e  xˆH x
0



At y=0 and y=d, E x =E y =0, H y =0

 yˆ  D   sl   sl  E y  E 0 e  jz

At y=0, aˆ n  yˆ , 
 

E
ˆ
ˆH z  zˆ 0 e  jz



y
H
J
J
sl
sl   z




 yˆ  D   su   su  E y  E 0 e  jz

At y=d, aˆ n   yˆ , 
 

E
ˆ
ˆH x   zˆ 0 e  jz




y
H
J
J
su
su  z





dE y
d
d d
   E   jH , 
 jH x 
E y dy  j  H x dy

0
dz
dz 0
dV ( z )
d
d

 jJ su ( z )d  j (  )[ J su ( z ) w]  jLI ( z )  L  
( H / m)
dz

w


w
dH x
d w
   H  jE , 
 jE y 
H x dx  j  E y dx

0
dz
dz 0
dI ( z )
w
w

  jE y ( z ) w  j ( )[ E y ( z )d ]  jCV ( z )  C  
( F / m)
dz
d
d
1
1

 d 2V ( Z )
 dV

   LC    , v p  
  2 LCV ( z )



j
LI

2
 dz


LC
dz

 2

 dI  jCV
 d I ( z )   2 LCI ( z )
V ( z)
L d  d
Z0 


 
 dz
 dz 2
I ( z)
C w 
w
Lossy parallel–plate transmission line: G 
Surface impedance: Z s 
 P 
R  2(

w
C 
d

f c
Et
E
 z   c  Rs  jX s  (1  j )
c
Js Hx
R
1
1
1
1
2
2
Re( J su Z s )  J su Rs  I 2 ( s )  I 2 R
2
2
2
2
w
Rs
2 f c
)
( / m)
w
w c
Eg. Consider a transmission line made of two parallel brass strips σ c =1.6×107S/m
of width 20mm and separated by a lossy dielectric slab μ=μ 0 , ε r =3, σ=10-3S/m of
thickness 2.5mm. The operating frequency is 500MHz. (a) Calculate the R, L, G,
and C per unit length. (b) Find γ and Z 0 .
(Sol.) (a) R 
L  0
w
2 f  0
 1.11( / m) , G    8  10 3 ( S / m)
w c
d
d
w
 1.57  10 7 ( H / m) , C    2.12  10 10 ( F / m)
w
d
(b) γ= (R  jL)(G jC) =18.13∠-0.41°, ω=2π×500×106, Z 0 =
R jL
=27.21∠0.3°
G jC
Eg. Consider lossless stripline design for a given characteristic impedance. (a)
How should the dielectric thickness d be changed for a given plate width w if the
dielectric constant ε r is doubled? (b) How should w be changed for a given d if ε r
is doubled? (c) How should w be changed for a given ε r if d is doubled?
(Sol.) Z 0 
L d

C w


(a)   2  d  2d , (b)   2  w 
w
2
(c) d  2d  w  2 w
Attenuation constant of transmission line: α =
PL ( z )
, where P L (z) is the
2 P( z )
time-average power loss in an infinitesimal distance.
    j    Re( )  Re[ ( R  jL)(G  jC ) ]
Suppose no reflection, V ( z )  V0 e  (  j ) z , I ( z ) 
2
 P( z ) 

V
1
Re[V ( z ) I * ( z )]  0
2
2 Z0
2
V 0  (  j  ) z
e
Z0
 R0 e  2z  e 2z
P ( z)
P( z )
 PL ( z )  2P( z )    L
z
2 P( z )
Microstrip lines: are usually used in the mm wave range.
p 
c
 ff
,  
p

1
L

, 
 
 pC
C
f
 ff
Assuming the quasi-TEM mode:
Case 1: t/h＜0.005, t is negligible.
Given h, W, and ε r , obtain Z 0 as follows:
W
 h
ln 8  0.25  ,
h
 ff  W
60
For W/h  1:  
where  ff 
 r  1  r 1 
h

1  12 
2
2 
W
1 / 2
120
For W/h  1:  
where  ff 
 ff

W  1.393  0.667 ln W  1.444
h
h
 r 1  r 1
2
2
 W 
 0.041   
 h  

h
1  12 
W
2 
,
1 / 2
Given Z 0 , h, and ε r , obtain W as follows:
For W/h  2: W 
8he A

, where   0
2A
e 2
60
For W/h>2: W 
r 1
2

 r 1
0.11 
 0.23 

r 1
 r 
 r 1 
2h 
0.61 
 B  1  ln2 B  1 
 , where
lnB  1  0.39 
2 r 
 
 r  
377
2 0  r
Case 2: t/h＞0.005. In this case, we obtain W eff firstly.
Weff W
2h 
t 
  1  ln 
For W/h  1 :
2
h
h h 
t 

4W 
W
t 
 1  ln

2
h
h h 
t 
And then we substitute W eff into W in the expressions in Case 1.
For W/h  1
:
Weff

Assuming not the quasi-TEM mode:
Weff 0  W

377 h
, f p   (h in cm)
, where Weff  f   W 
2
8h
Weff  f   ff


1  f 
 fp 
 r   ff
377h
, G=0.6+0.009Z 0,  ff  f    r 
(f in GHz)
and Weff 0  
2
  0  ff 0
 f 
1  G

 fp 
The frequency below which dispersion may be neglected is given by
  f  
f 0 G z   0.3

h  r 1
, where h must be expressed in cm.
Attenuation constant: α=α d +α c
For a dielectric with low losses:  d  27.3
 r  ff  1 tan 
 ff  r  1 
 ff  1
 

For a dielectric with high losses:  d  4.34
 ff  r  1   
For W/h → ∞:  c 
For W/h  1
2
: c 
8.68Rs P 
h
h  4W t 

 
 ln
1 
t
W 
2  h  Weff Weff 
Weff 
8.68Rs

＜W/h  2:  c 
PQ , where P 1 
2
2  h
 4h 
and Q  1 
h
h

Weff Weff
 2h t 
 
 ln
 t h
dB
)
cm
1/ 2

 

f 
8.68
Rs , where Rs 

 W
2
For 1
(
(
dB
)
cm
For W/h  2:


Weff
2






W
W
W
 eff
 
8.68Rs Q  eff 2
eff
h

c 
 ln 2e
 0.94   


W
h
   2h
  h  h





 
 eff 2h   0.94 





Eg. A high-frequency test circuit with microstrip lines.
5-2 Wave Characteristics of Finite Transmission Line
Eg. Show that the input impedance is Z i = ( Z ) z 0  Z 0
z ' 
Z L  Z 0 tanh 
.
Z 0  Z L tanh 


V ( z )  V0 e z  V0 e z ...(1)
V0
V0
(Proof) 
,



Z0


I0
I0
 I ( z )  I 0  e z  I 0  e z .....(2)


Let z=l, V(l)=V L , I(l)=I L
  1

VL  V0  e   V0  e 
V0  2 (VL  I L Z 0 )e




V0  V0   


I
e
e
 L
V   1 (V  I Z )e 
L
L 0
Z
Z
 0
0
0

2
IL

 (  z )
 ( Z L  Z 0 )e  (   z ) ]
V ( z )  2 [( Z L  Z 0 )e

 I ( z )  I L [( Z L  Z 0 )e  (   z )  ( Z L  Z 0 )e  (   z ) ]

2Z 0
IL

z '
z '
'
V ( z ' )  I L ( Z L cosh z ' Z 0 sinh z ' )
V ( z )  2 [( Z L  Z 0 )e  ( Z L  Z 0 )e ]



IL
 I ( z ' )  I L [( Z L  Z 0 )e z '  ( Z L  Z 0 )e z ' ]  I ( z ' )  Z ( Z L sinh z ' Z 0 cosh z ' )
0


2Z 0
Z  Z 0 tanh z '
Z  Z 0 tanh 
 Z ( z' )  Z 0 L
, Z i = ( Z ) z 0  Z 0 L
Z 0  Z L tanh z '
Z 0  Z L tanh 
z ' 
Lossless case (α=0, γ=jβ, Z 0 =R 0 , tanh(γl)=jtanβl): Z i = R0 
Z L  jR0 tan  l
R0  jZ L tan  l
Note: In the high-frequency circuit, the input current I i =
Vg
Z g  Zi

Vg
Zg  ZL
: the
value in the low-frequency case. And the high-frequency I i is dependent on the length
l, the characteristic impedance Z 0 , the propagation constant γ of the transmission line,
and the load impedance Z L . But the low-frequency I i is only dependent on Z 0 and Z L .
Eg. A 2m lossless air-spaced transmission line having a characteristic impedance
50Ω is terminated with an impedance 40+j30(Ω) at an operating frequency of
200MHz. Find the input impedance.
 4
(Sol.)  
  , R0  50 , Z L  40  j 30 ,   2m
vp 3
4
 2)
3
 26.3  j 9.87
Z i  50
4
50  j (40  j 30)  tan(  2)
3
(40  j 30)  j 50  tan(
Eg. A transmission line of characteristic impedance 50Ω is to be matched to a
load Z L =40+j10(Ω) through a length l’ of another transmission line of
characteristic impedance R 0 ’. Find the required l’ and R 0 ’ for matching.
40  j10  jR0'  tan '
 R0 '  1500  38.7(), '  0.105
(Sol.) 50  R  '
R0  j (40  j10)  tan '
'
0
Eg. Prove that a maximum power is transferred from a voltage source with an
internal impedance Z g to a load impedance Z L over a lossless transmission line
when Z i =Z g *, where Z i is the impedance looking into the loaded line. What is the
maximum power transfer efficiency?
Zi
V
(Proof) I i 
, Vi 
V
Zi  Z g
Zi  Z g
2
( Power ) out
Ri V
1
 Re[Vi I i *] 
2
2[( Ri  R g ) 2  ( X i  X g ) 2 ]
When Ri  R g and X i   X g , ( Power ) out  Max , ∴ Z i  Z g *
2
In this case, ( Power ) out
2
V
V
( Power ) out 1
1


, Ps  Re[VI i *] 
, e
4 Rg
2
2 Rg
Ps
2
Transmission lines as circuit elements:
Z  Z 0 tanh 
Consider a general case: Z i = Z 0 L
Z 0  Z L tanh 
1. Open-circuit termination (Z L →∞): Z i =Z io =Z 0 coth(γl)
2. Short-circuit termination (Z L =0): Z i =Z is =Z 0 tanh(γl)
Z is
1
∴ Z 0 = Z i 0  Z is , γ= tanh 1

Zi0
2
3. Quarter-wave section in a lossless case (l=λ/4, βl=π/2): Z i 
R0
ZL
4. Half-wave section in a lossless case (l=λ/2, βl=π): Z i  Z L
Eg. The open-circuit and short-circuit impedances measured at the input
terminals of an air-spaced transmission line 4m long are 250∠-50°Ω and
360∠20°Ω, respectively. (a) Determine Z 0 , α, and β of the line. (b) Determine R,
L, G, and C.
(Sol.) (a) Z 0  250e  j 50  360e j 20  289.8  j77.6 ,
36020
 0.139  j 0.235    j
250  50
57.3 57.3
(b) R  jL  Z 0    58.5  j 57.3 , L 

 0.812( H / m)

c
1
4
  tanh1
G  j C 

Z0
 24.5  10 5  j8.76  10  4 , C 
8.76  10 4
 12.4( pF / m)
c
Eg. Measurements on a 0.6m lossless coaxial cable at 100kHz show a capacitance
of 54pF when the cable is open-circuited and an inductance of 0.30μH when it is
short-circuited. Determine Z 0 and the dielectric constant of its insulating
medium.
(Sol.) (a) C 
54  10 12
0.3  10 6
 9  10 11 ( F / m) , L 
 5  10 7 ( H / m)
0.6
0.6
Lossless  Z 0  R0 
L
=74.5Ω,    0  r  0  r  LC   r  4.05
C
General expressions for V(z) and I(z) on the transmission lines:
Z  Z0
  e j  , z’=l-z
Let Γ= L
Z L  Z0
IL

z '
 2z '
V ( z ' )  2 (Z L  Z 0 )  e  [1  e ]

I ( z ' )  I L (Z L  Z 0 )  ez '  [1  e 2z ' ]

2Z 0
IL

j (   2z ')
z '
]
V ( z ' )  2 ( Z L  Z 0 )  e  [1   e

 I ( z ' )  I L ( Z L  Z 0 )  e z '  [1   e j (   2z ') ]

2Z 0
For a lossless line, V(z)=
Z 0Vg
Z0  Z g
e  jz [1  e  j 2  (  z ) ]
Eg. A 100MHz generator with V g =10∠0° (V) and internal resistance 50Ω is
connected to a lossless 50Ω air line that is 3.6m long and terminated in a
25+j25(Ω) load. Find (a) V(z) at a location z from the generator, (b) V i at the
input terminals and V L at the load, (c) the voltage standing-wave radio on the
line, and (d) the average power delivered to the load.
(Sol.)
V g  100(V )
,
Z g  50()
,
f  10 8 ( Hz )
,
Z 0  50()
,
Z L  25  j 25  35.3645() ,
2 10 8 2

(rad / m) ,   2.4 (rad / m)
c 3  10 8
3
Z  Z 0 (25  j 25)  50
 L

 0.4470.648 , g  0
Z L  Z 0 (25  j 25)  50
  3.6(m) ,  
(a) V ( z ) 


Z 0V g
Z0  Z g
e  jz [1  e  j 2  (   z ) ]  5[e  j 2z / 3  0.447e j ( 2 z / 30.152 ) ]
(b) Vi  V (0)  5(1  0.447e  j 0.152 )  7.06  8.43(V )
(c) V L  V (3.6)  5[e  j 0.4  0.447e j 0.248 ]  4.47  45.5(V )
1 
2
1 VL
1 4.47 2
1  0.447
RL  (
)  25  0.200(W )
(d) S 

 2.62 , Pav 
2 ZL
2 35.36
1   1  0.447
Eg. A sinusoidal voltage generator V g =110sin(ωt) and internal impedance
Z g =50Ω is connected to a quarter-wave lossless line having a characteristic
impedance Z 0 =50Ω that is terminated in a purely reactive load Z L =j50Ω. (a)
Obtain the voltage and current phasor expressions V(z’) and I(z’). (b) Write the
instantaneous voltage and current expressions V(z’,t) and I(z’,t).
(Sol.) (a) Vg  110 j,  
(c) V ( z ' ) 
I ( z' ) 
Z 0V g
Z0  Zg
Vg
Z0  Z g
Z g  Z0
Z L  Z 0 50 j  50


 j, g 
 0,  
Z L  Z 0 50 j  50
Z g  Z0
4
e  jz '  (1  e  2 jz ' )  j 55( e  jz '  je jz ' ) ,
e  jz '  (1  e 2 jz ' )   j1.1( e  jz '  je jz ' )
P  V ( z '  0, t ) I ( z '  0, t )  60.5 cos(2t ) , Pav 
1 T
Pdt  0
T 0
(b) V ( z ' , t )  Im[V ( z ' )e jt ]  55[sin(t  z ' )  cos(t  z ' )]
I ( z ' , t )  Im[I ( z ' )e jt ]  1.1[sin(t  z ' )  cos(t  z ' )]
Eg. A sinusoidal voltage generator with V g =0.1∠0° (V) and internal impedance
Z g =Z 0 is connected to a lossless transmission line having a characteristic
impedance Z 0 =50Ω. The line is l meters long and is terminated in a load
resistance Z L =25Ω. Find (a) V i , I i , V L 錯誤! 尚未定義書籤。, and I L ; (b) the
standing-wave radio on the line; and (c) the average power delivered to the load.
(Sol.) (a)  
Z g  Z0
ZL  Z0
1
1
0.1
  , g 
 0 , Vi  V ( z '  ) 
(1  e  2 j )
Z L  Z0
3
Z g  Z0
2
3
I i  I ( z '  ) 
0.1
1
 (1  e  2 j )
100
3
V L  V ( z '  0) 
I L  I ( z '  0) 
(b) S 
1 
1 
0.1  j
1
1
e (1  )  e  j
2
3
30
0.1  j
1
1  
e (1  ) 
e
100
3
750
 2 , (c) Pav 
1
Re[V L I L *]  2.22  10 5 W
2
Eg. Consider a lossless transmission line of characteristic impedance R 0 . A
time-harmonic voltage source of an amplitude V g and an internal impedance
R g =R 0 is connected to the input terminals of the line, which is terminated with a
load impedance Z L =R L +jX L . Let P inc be the average incident power associated
with the wave traveling in the +z direction. (a) Find the expression for P inc in
terms of V g and R 0 . (b) Find the expression for the average power P L delivered
to the load in terms of V g and the reflection coefficient Γ. (c) Express the ratio
P L /P inc in terms of the standing-wave ratio S.
Vg
1
(Sol.) V ( z )  V0 e  jz  V0 e jz , I ( z ) 
(V0 e  jz  V0 e jz ) , Vinc ( z  0)  V0 
,
2
R0
Vg
I inc ( z  0)  I 0 
2 R0
2
2
V0
Vg
1



(a) Pinc  Re[V0 ( I 0 )*] 
2
2 R0
8 R0
(b) PL  1 Re[V ( z ) I * ( z )]  1 Re[(V0 e  jz  V0 e jz )(V0* e jz  V0* e  jz )]  1 {V0 2  V0 2 }
2

V0
(c)
2 R0
2
2 R0
2
{1   } 
Vg
2
8 R0
2
{1   }
PL
S 1 2
4S
2
 1   1 (
) 
Pinc
S 1
( S  1) 2
2 R0
Case 1 For a pure resistive load: Z L =R L
V ( z ' )  VL  cos z ' jI L R0  sin z ' 
2
2
2

 V ( z ' )  VL  cos z '( R0 / RL ) sin z '


VL
 I ( z ' )  I L  cos z ' j R  sin z '
 I ( z ' )  I L  cos 2 z '( RL / R0 ) 2 sin 2  z '
0

S
1 
1 
,  
S 1
 1. Γ=0  S=1 when Z L = Z 0 (matched load)
S 1
2. Γ=-1  S=∞ when Z L =0 (short-circuit), 3. Γ=1  S=-∞ when Z L =∞ (open-circuit)
Vmax & I min occurs at    2 z 'max  2n
Vmin & I max occurs at    2 z 'min  (2n  1)
If R L  R0    0     0, z ' max 
n
, n  0,1,2,3....
2
If R L  R0    0      , z ' min 
If R L    z ' max 
n
2
n
2
Eg. The standing-wave radio S on a transmission line is an easily measurable
quality. Show how the value of a terminating resistance on a lossless line of
known characteristic impedance R 0 can be determined by measuring S.
(Sol.) If RL  R0 ,    0 , Vmax occurs at z '  0 and Vmin occurs at  z '   .
2
Vmax  VL , Vmin  VL
Vmax
I max
R0
R
R
, I min  I L , I max  I L L ,

S L
Vmin
I min
RL
R0
R0
or
R L  SR0 .
If R L  R0 ,     , Vmin occurs at z ' 0 , and Vmax occurs at  z '   .
2
Vmin  VL , Vmax  V L
RL 
R0
S
Vmax
I max
R0
R
R
, I max  I L , I min  I L L .

S 0
I min
RL
R0 Vmin
RL
or
Case 2 For a lossless transmission line, and arbitrary load:
R  jR0 tan  m
Z L = R0  m
, z m ’+l m =λ/2
R0  jRm tan  m
Find Z L =?
1.  
S 1
, 2. At θ Γ =2βz m ’-π, V(z’) is a minimum.
S 1
3. Z L =R L +jX L = R0 
1   e j 
1 e
j 
 R0 
1 
1 
Eg. Consider a lossless transmission line. (a) Determine the line’s characteristic
resistance so that it will have a minimum possible standing-wave ratio for a load
impedance 40+j30(Ω). (b) Find this minimum standing-wave radio and the
corresponding voltage reflection coefficient. (c) Find the location of the voltage
minimum nearest to the load.
(Sol.)
 
1   dS
Z L  R0
40  R0  j 30
(40  R0 ) 2  30 2 1 2
1

[
 0  R0  50   
] , S
,
2
2
Z L  R0
40  R0  j 30
1   dR0
3
(40  R0 )  30
 S  2,  
z min ' 
Z L  R0  10  j 30 1
1



   90    ,    
90  j 30
3
Z L  R0
3
2
2


  3
1
(  )  ,  m   
2
2
8
2 8
8
Eg. SWR on a lossless 50Ω terminated line terminated in an unknown load
impedance is 3. The distance between successive minimum is 20cm. And the first
minimum is located at 5cm from the load. Determine Γ, Z L , l m , and R m .
(Sol.)


2
 0.2    0.4m,  
2

 5

31
 0.5, z ' m  0.05   m   z m ' =0.15m
31
2
   2z m '  0.5 ,    e j  0.5e  j 0.5  

j
2
j
)
2  30  j 40  50  Rm  j 50 tan  m
R0  50, Z L  50 
j
50  jRm tan  m
1  ( )
2
1 (
 Rm 
50
 16.7()
3
Eg. A lossy transmission line with characteristic impedance Z 0 is terminated in
an arbitrary load impedance Z L . (a) Express the standing-wave radio S on the
line in terms of Z 0 and Z L . (b) Find the impedance looking toward the load at the
location of a voltage maximum. (c) Find the impedance looking toward the load
at a location of a voltage minimum.
(Sol.) (a)  
1 
Z L  Z 0  Z L  Z 0 e 2z '
Z L  Z 0  2z '
, S
e

ZL  Z0
1 
Z L  Z 0  Z L  Z 0 e  2z '
(b)    2z max '  2n  e j  e  2z

max
(c)    2 z min '  (2n  1)  e
Z ( z min ' )  Z 0 
1   e 2zmin '
1  e
 2z min '

'
e
j 
 2z max '
e
,  Z ( z max ' )  Z 0 
2z min '
 e
1   e 2z max '
1  e
 2z max '

Z0
S ( z max ' )
2z min '
Z0
S ( z min ' )
5-3 Introduction to Smith Chart

j
Z L  R0
Z / R 1 z 1
1  1  e 
  e j  L 0  L

 r  ji  z L 
 r  jx
1   1   e j 
Z L  R0
Z L / R0 1 zL 1
r
1  r2  i2
(1  r ) 2  i2
 (r 
, x
2 i
2
(1  r ) 2  i
r 2
1 2
1
1
)  i2  (
) : r-circle, ( r  1) 2  ( i  ) 2  ( ) 2 : x-circle
x
x
1 r
1 r
Several salient properties of the r-circles:
1. The centers of all r-circles lie on the Γ r -axis.
2. The r=0 circle, having a unity radius and centered at the origin, is the largest.
3. The r-circles become progressively smaller as r increases from 0 toward ∞,
ending at the (Γ r =1, Γ i =0) point for open-circuit.
4. All r-circles pass through the (Γ r =1, Γ i =0) point.
Salient properties of the x-circles:
1. The centers of all x-circles lie on the Γ r =1 line, those for x>0 (inductive reactance)
lie above the Γ r –axis, and those for x<0 (capacitive reactance) lie below the
Γ r –axis.
2. The x=0 circle becomes the Γ r –axis.
3. The x-circle becomes progressively smaller as |x| increases from 0 toward ∞,
ending at the (Γ r =1, Γ i =0) point for open-circuit.
4. All x-circles pass through the (Γ r =1, Γ i =0) point.
Summary
1. All |Γ|–circles are centered at the origin, and their radii vary uniformly from 0 to
1.
2. The angle, measured from the positive real axis, of the line drawn from the origin
through the point representing z L equals θ Γ .
3. The value of the r-circle passing through the intersection of the |Γ|–circle and the
positive-real axis equals the standing-wave radio S.
Application of Smith Chart in lossless transmission line:
Z i ( z' ) 
j
Z i ( z ) 1  e  j 2  z ' 1   e
V ( z' )
1  e  j 2  z '



z
(
z
'
)
 z0 [
]
,
when
i
I ( z' )
Z0
1   e  j 2  z ' 1   e j
1  e  j 2  z '
     2z '
keep |Γ| constant and subtract (rotate in the clockwise direction) an angle
4z '
 2z ' 
from θ Γ . This will locate the point for |Γ|ejφ, which determine Z i .

Increasing z’  wavelength toward generator in the clockwise direction

 A change of
A change of half a wavelength in the line length z ' 
2
2  (z ' )  2 in φ.
Eg. Use the Smith chart to find the input impedance of a section of a 50Ω lossless
transmission line that is 0.1 wavelength long and is terminated in a short-circuit.
(Sol.) Given z L  0 , R0  50() , z '  0.1
1. Enter the Smith chart at the intersection of r=0 and x=0 (point Psc on the
extreme left of chart; see Fig.)
2. Move along the perimeter of the chart (   1) by 0.1 “wavelengths toward
generator’’ in a clockwise direction to P 1 .
At P 1 , read r=0 and x  0.725 , or z i  j 0.725 , Z i  50( j 0.725)  j 36.3() .
Eg. A lossless transmission line of length 0.434λ and characteristic impedance
100Ω is terminated in an impedance 260+j180(Ω). Find (a) the voltage reflection
coefficient, (b) the standing-wave radio, (c) the input impedance, and (d) the
location of a voltage maximum on the line.
(Sol.) (a) Given l=0.434λ, R 0 =100Ω, Z L =260+j180
1. Enter the Smith chart at z L =Z L /R 0 =2.6+j1.8 (point P 2 in Fig.)
2. With the center at the origin, draw a circle of radius OP 2    0.60 .
( OP sc =1)
3. Draw the straight line OP2 and extend it to P 2 ’ on the periphery. Read
0.22 on “wavelengths toward generator” scale.   = 21 ,
   e j   0.6021 .
(b) The   0.60 circle intersects with the positive-real axis OPoc at r=S=4.
(c) To find the input impedance:
1. Move P 2 ’ at 0.220 by a total of 0.434 “wavelengths toward generator,” first to
0.500 and then further to 0.154 to P 3 ’.
2. Join O and P 3 ’ by a straight line which intersects the   0.60 circle at P 3 .
3. Read r=0.69 and x=1.2 at P 3 . Z i  R0 z i  100(0.69  j1.2)  69  j120() .
(d) In going from P 2 to P 3 , the   0.60 circle intersects the positive-real axis
OPoc at P M , where the voltage is a maximum. Thus a voltage maximum
appears at (0.250-0.220)  or 0.030  from the load.
Application of Smith Chart in lossy transmission line
zi 
2z '
 e j
1  e 2z '  e 2 jz ' 1   e

1  e  2z '  e  2 jz ' 1   e  2z '  e j
∴ We can not simply move close the |Γ|-circle; auxiliary calculation is necessary for
the e-2αz’ factor.
Eg. The input impedance of a short-circuited lossy transmission line of length 2m
and characteristic impedance 75Ω (approximately real) is 45+j225(Ω). (a) Find α
and β of the line. (b) Determine the input impedance if the short-circuit is
replaced by a load impedance Z L = 67.5-j45(Ω).
(Sol.) (a) Enter z i1  (45  j 225) / 75  0.60  j 3.0 in the chart as P 1 in Fig.
Draw a straight line from the origin O through P 1 to P 1 ’.
1
1
1
ln(
)  ln(1.124)  0.029( Np / m)
2 0.89
4
Record that the arc Psc P1 ' is 0.20 “wavelengths toward generator”.  /   0.20 ,
Measure OP1 / OP1 '  0.89  e 2 ,  
2   4 /   0.8 .  
0.8 0.8

 0.2 (rad / m) .
2
4
(b) To find the input impedance for:
1. Enter z L  Z L / Z 0  (67.5  j 45) / 75  0.9  j 0.6 on the Smith chart as P 2 .
2. Draw a straight line from O through P 2 to P 2 ’ where the “wavelengths toward
generator” reading is 0.364.
3. Draw a  –circle centered at O with radius OP2 .
4. Move P 2 ’ along the perimeter by 0.2 “wavelengths toward generator” to P 3 ’ at
0.364+0.20=0.564 or 0.064.
5. Joint P 3 ’ and O by a straight line, intersecting the  –circle at P 3 .
6. Mark on line OP3 a point P i such that OPi / OP3  e 2  0.89 .
7. At P i , read z i  0.64  j 0.27 . Z i  75(0.64  j 0.27)  48.0  j 20.3()
5-4 Transmission-line Impedance Matching
Impedance matching by λ/4-transformer: R 0 ’= R0 RL
Eg. A signal generator is to feed equal power through a lossless air transmission
line of characteristic impedance 50Ω to two separate resistive loads, 64Ω and
25Ω. Quarter-wave transformers are used to match the loads to the 50Ω line. (a)
Determine the required characteristic impedances of the quarter-wave lines. (b)
Find the standing-wave radios on the matching line sections.
(Sol.) (a) Ri1  Ri 2  2 R0  100() .
R01'  Ri1 RL1  100  64  80() , R02'  Ri 2 RL 2  100  25  50()
(b) Matching section No. 1:
1 
1  1 1  0.11
RL1  R01'
64  80


 0.11 , S1 
 1.25
'
1  1 1  0.11
RL1  R01 64  80
Matching section No. 2:
2 
1  2 1  0.33
RL 2  R02'
25  50


 0.33 , S 2 
 1.99
'
1  2 1  0.33
RL 2  R02 25  50
Application of Smith Chart in obtaining admittance:
YL  1 / Z L , z L 
ZL
1
1


, where y L  YL / Y0  Y0 / G0  R0YL  y  jb
R0 R0 YL y L
Eg. Find the input admittance of an open-circuited line of characteristic
impedance 300Ω and length 0.04λ.
(Sol.) 1. For an open-circuited line we start from the point P oc on the extreme right of
the impedance Smith chart, at 0.25 in Fig.
2. Move along the perimeter of the chart by 0.04 “wavelengths toward generator” to
P 3 (at 0.29).
3. Draw a straight line from P 3 through O, intersecting at P 3 ’ on the opposite side.
4. Read at P 3 ’: y i  0  j 0.26 , Yi 
1
(0  j 0.26)  j 0.87 mS .
300
Application of Smith Chart in single-stub matching:
Yi  YB  YS  Y0 
1
 1  y B  y S , where y B =R 0 Y B , y s =R 0 Y s
R0
∵ 1+jb s = y B , ∴ y s =-jb s and l B is required to cancel the imaginary part.
Using the Smith chart as an admittance chart, we proceed as y L follows for
single-stub matching:
1. Enter the point representing the normalized load admittance.
2. Draw the |Γ|-circle for y L , which will intersect the g=1 circle at two points. At
these points, y B1 =1+jb B1 and y B2 =1+jb B2 . Both are possible solutions.
3. Determine load-section lengths d 1 and d 2 from the angles between the point
representing y L and the points representing y B1 and y B2 .
Determine stub length l B1 and l B2 from the angles between the short-circuit point on
the extreme right of the chart to the points representing –jb B1 and –jb B2 , respectively.
Eg. Single-Stub Matching:
Eg. A 50Ω transmission line is connected to a load impedance Z L = 35-j47.5(Ω).
Find the position and length of a short-circuited stub required to match the line.
(Sol.) Given R0  50() , Z L  35  j 47.5() , z L  Z L / R0  0.70  j 0.95
1. Enter z L on the Smith chart as P1 . Draw a  –circle centered at O with radius
OP1 .
2. Draw a straight line from P1 through O to P ' 2 on the perimeter, intersecting the  –circle at
P2 , which represents y L . Note 0.109 at P' 2 on the “wavelengths toward generator” scale.
3. Two points of intersection of the  –circle with the g=1 circle.
At
P3 : y B1  1  j1.2  1  jbB1 . At P4 : y B 2  1  j1.2  1  jbB 2 ;
4. Solutions for the position of the stubs:
For
P3 (from P' 2 to P '3 ): d1  (0.168  0.109)  0.059
For P4 (from P ' 2 to P ' 4 ): d 2  (0.332  0.109)  0.223
For
P3 (from Psc to P ' '3 , which represents  jbB1   j1.2 ):
 B1  (0.361  0.250)  0.111
For P4 (from Psc to P"4 , which represents  jbB 2  j1.2 ):
 B 2  (0.139  0.250)  0.389
5-5 Introduction to S-parameters
S S 
S-parameters: S    11 12  for analyzing the high-frequency circuits.
 S 21 S 22 
Define a  x  
1
V x   Z 0 I x  , bx   1 V x   Z 0 I x 
2 Z0
2 Z0
b1 l1   S 11 a1 l1   S 12 a 2 l 2  , b2 l2   S 21a1 l1   S 22 a2 l2 
b l 
b l 
 b1 l1     S11 S12   a1 l1   , where S11  1 1 a2 l2 0 , S 21  2 2
a1 l1 
a1 l1 
b2 l2   S 21 S 22  a2 l2 
S 22 
b2 l 2 
a2 l 2 
a1 l1 0
, and S12 
b1 l1 
a2 l 2 
a1 l1 0
a2 l2 0
,
.
New S-parameters obtained by shifting reference planes:
b1 l1   b1 0e j1 , a1 l1   a1 0e  j1 , b2 l 2   b2 0e j 2 , a 2 l 2   a 2 0 e  j
2
 j 2
 j   
b1 0   S11e 1 S12 e 1 2  a1 0 


 , where

 j   
 j 2
b2 0  S 21e 1 2 S 22 e 2  a 2 0
 j 2
 j   
j 2
j   
 S '11 S '12   S11e 1 S12 e 1 2 
 S11 S12   S '11 e 1 S '12 e 1 2 
 and 

S ' S '  = 
 =
 j   
 j 2
j   
j 2
 21 22   S 21e 1 2 S 22 e 2 
 S 21 S 22   S '21 e 1 2 S '22 e 2 
 1
a1 l1  T11 T12  b2 l 2 
T11 T12   S 21
T-parameters: 
,
where


 
 

T T    S
b1 l1  T21 T22  a 2 l 2 
 21 22   11
S
 21
 T21

T
S S 
and  11 12    11
 S 21 S22   1
T
 11
T21T12 
T11 


T12


T11

T22 



S11 S 22 
S12 
S 21 

S 22
S 21