D-tutorial.pdf

Transmission Lines
© Amanogawa, 2006 – Digital Maestro Series 65
We have obtained the following solutions for the steady-state
voltage and current phasors in a transmission line:
Loss-less line Lossy line
( )
z z
z z
0
(z)
1
(z)
j j
j j
V V e V e
I V e V e
Z
β β
β β
+ − −
+ − −
= +
= − ( )
z z
z z
0
(z)
1
(z)
V V e V e
I V e V e
Z
γ γ
γ γ
+ − −
+ − −
= +
= −
Since V (z) and I (z) are the solutions of second order differential
(wave) equations, we must determine two unknowns, V +
and V−
,
which represent the amplitudes of steady-state voltage waves,
travelling in the positive and in the negative direction, respectively.
Therefore, we need two boundary conditions to determine these
unknowns, by considering the effect of the load and of the
generator connected to the transmission line.

Transmission Lines
Before we consider the boundary conditions, it is very convenient
to shift the reference of the space coordinate so that the zero
reference is at the location of the load instead of the generator.
Since the analysis of the transmission line normally starts from the
load itself, this will simplify considerably the problem later.
We will also change the positive direction of the space coordinate,
so that it increases when moving from load to generator along the
transmission line.
z
d
0
ZR
New Space Coordinate

Transmission Lines
We adopt a new coordinate d = − z, with zero reference at the load
location. The new equations for voltage and current along the lossy
transmission line are
( )
d d
d d
0
(d)
1
(d)
j j
j j
V V e V e
I V e V e
Z
β β
β β
+ − −
+ − −
= +
= − ( )
d d
d d
0
(d)
1
(d)
V V e V e
I V e V e
Z
γ γ
γ γ
+ − −
+ − −
= +
= −
At the load (d = 0) we have, for both cases,
( )
0
(0)
1
(0)
V V V
I V V
Z
+ −
+ −
= +
= −

Transmission Lines
For a given load impedance ZR , the load boundary condition is
(0) (0)
R
V Z I
=
Therefore, we have
( )
0
R
Z
V V V V
Z
+ − + −
+ = −
from which we obtain the voltage load reflection coefficient
0
0
R
R
R
Z Z
V
Z Z
V
Γ
−
+
−
= =
+

Transmission Lines
We can introduce this result into the transmission line equations as
( )
( )
d 2 d
d
2 d
0
(d) 1
(d) 1
j j
R
j
j
R
V V e e
V e
I e
Z
β β
β
β
Γ
Γ
+ −
+
−
= +
= −
( )
( )
d 2 d
d
2 d
0
(d) 1
(d) 1
R
R
V V e e
V e
I e
Z
γ γ
γ
γ
Γ
Γ
+ −
+
−
= +
= −
At each line location we define a Generalized Reflection Coefficient
2 d
(d) j
R e− β
Γ = Γ 2 d
(d) R e− γ
Γ = Γ
and the line equations become
( )
( )
d
d
0
(d) 1 (d)
(d) 1 (d)
j
j
V V e
V e
I
Z
β
β
Γ
Γ
+
+
= +
= −
( )
( )
d
d
0
(d) 1 (d)
(d) 1 (d)
V V e
V e
I
Z
γ
γ
Γ
Γ
+
+
= +
= −

Transmission Lines
We define the line impedance as
A simple circuit diagram can illustrate the significance of line
impedance and generalized reflection coefficient:
Zeq=Z(d)
ΓReq = Γ(d)
d 0
ZR
0
(d) 1 (d)
(d)
(d) 1 (d)
V
Z Z
I
+ Γ
= =
− Γ

Transmission Lines
If you imagine to cut the line at location d, the input impedance of
the portion of line terminated by the load is the same as the line
impedance at that location “before the cut”. The behavior of the line
on the left of location d is the same if an equivalent impedance with
value Z(d) replaces the cut out portion. The reflection coefficient of
the new load is equal to Γ(d)
If the total length of the line is L, the input impedance is obtained
from the formula for the line impedance as
The input impedance is the equivalent impedance representing the
entire line terminated by the load.
0
(L) 1 (L)
(L) 1 (L)
in
in
in
V V
Z Z
I I
+ Γ
= = =
− Γ
0
0
( )
Req
Req
Req
Z Z
d
Z Z
−
Γ = Γ =
+

Transmission Lines
An important practical case is the low-loss transmission line, where
the reactive elements still dominate but R and G cannot be
neglected as in a loss-less line. We have the following conditions:
L R C G
ω >> ω >>
so that
2
( )( )
1 1
1
j L R j C G
R G
j L j C
j L j C
R G RG
j LC
j L j C LC
γ ω ω
ω ω
ω ω
ω
ω ω ω
= + +
  
= + +
  
  
≈ + + −
The last term under the square root can be neglected, because it is
the product of two very small quantities.

Transmission Lines
What remains of the square root can be expanded into a truncated
Taylor series
1
1
2
1
2
R G
j LC
j L j C
C L
R G j LC
L C
γ ω
ω ω
ω
 
 
≈ + +
 
 
 
 
 
= + +
 
 
so that
1
2
C L
R G LC
L C
α β ω
 
= + =
 
 

Transmission Lines
The characteristic impedance of the low-loss line is a real quantity
for all practical purposes and it is approximately the same as in a
corresponding loss-less line
0
R j L L
Z
G j C C
+ ω
= ≈
+ ω
and the phase velocity associated to the wave propagation is
1
p
v
LC
ω
= ≈
β
BUT NOTE:
In the case of the low-loss line, the equations for voltage and
current retain the same form obtained for general lossy lines.

Transmission Lines
Again, we obtain the loss-less transmission line if we assume
0 0
R G
= =
This is often acceptable in relatively short transmission lines, where
the overall attenuation is small.
As shown earlier, the characteristic impedance in a loss-less line is
exactly real
0
L
Z
C
=
while the propagation constant has no attenuation term
( )( )
j L j C j LC j
γ = ω ω = ω = β
The loss-less line does not dissipate power, because α = 0.

Transmission Lines
For all cases, the line impedance was defined as
0
(d) 1 (d)
(d)
(d) 1 (d)
V
Z Z
I
+ Γ
= =
− Γ
By including the appropriate generalized reflection coefficient, we
can derive alternative expressions of the line impedance:
A) Loss-less line
2 d
0
0 0
2 d
0
tan( d)
1
(d)
tan( d)
1
j
R
R
j
R
R
Z jZ
e
Z Z Z
jZ Z
e
− β
− β
+ β
+ Γ
= =
β +
− Γ
B) Lossy line (including low-loss)
2 d
0
0 0
2 d
0
tanh( d)
1
(d)
tanh( d)
1
R
R
R
R
Z Z
e
Z Z Z
Z Z
e
− γ
− γ
+ γ
+ Γ
= =
γ +
− Γ

Transmission Lines
Let’s now consider power flow in a transmission line, limiting the
discussion to the time-average power, which accounts for the
active power dissipated by the resistive elements in the circuit.
The time-average power at any transmission line location is
{ }
*
1
(d, ) Re (d) (d)
2
P t V I
〈 〉 =
This quantity indicates the time-average power that flows through
the line cross-section at location d. In other words, this is the
power that, given a certain input, is able to reach location d and
then flows into the remaining portion of the line beyond this point.
It is a common mistake to think that the quantity above is the power
dissipated at location d !

Transmission Lines
The generator, the input impedance, the input voltage and the input
current determine the power injected at the transmission line input.
{ }
*
1
1
Re
2
in
in G
G in
in G
G in
in in in
Z
V V
Z Z
I V
Z Z
P V I
=
+
=
+
〈 〉 =
Vin
Iin
Generator Line
VG
ZG
Zin

Transmission Lines
The time-average power reaching the load of the transmission line
is given by the general expression
{ }
( ) ( )
( )
*
*
*
0
1
(d=0, ) Re (0) (0)
2
1 1
Re 1 1
2
R R
P t V I
V V
Z
Γ Γ
+ +
〈 〉 =
 
 
= + −
 
 
 
This represents the power dissipated by the load.
The time-average power absorbed by the line is simply the
difference between the input power and the power absorbed by the
load
(d 0, )
〈 〉 = 〈 〉 − 〈 = 〉
line in
P P P t
In a loss-less transmission line no power is absorbed by the line, so
the input time-average power is the same as the time-average
power absorbed by the load. Remember that the internal impedance
of the generator dissipates part of the total power generated.

Transmission Lines
It is instructive to develop further the general expression for the
time-average power at the load, using Z0=R0+jX0 for the
characteristic impedance, so that
0 0 0 0 0
* * 2 2 2
0 0 0 0 0
0
1 Z R jX R jX
Z Z Z R X
Z
+ +
= = =
+
Alternatively, one may simplify the analysis by introducing the line
characteristic admittance
0 0 0
0
1
Y G jB
Z
= = +
It may be more convenient to deal with the complex admittance at
the numerator of the power expression, rather than the complex
characteristic impedance at the denominator.

Transmission Lines
*
*
0
2
0 0
2
0
2
0 0
2
0
1 Re
1 1
(d=0, ) Re 1 1
2
Re 1 Re Im 1 Re Im
2
Re
2
R
R R
R R R R
P t V V
Z
V
R jX j j
Z
V
R jX
Z
 
 
 
   
 
 
 
   
 
   
 
 
 
 
 
   
         
 
   
 
         
     
       
 
 
   
 
   
   
  
 
+ +
+
+
− Γ
〈 〉 = +Γ −Γ =
= + + Γ + Γ − Γ Γ
= +
+
2 2
2
2
0 0
2
0
2
2
0 0 0
2
0
Im 2Im
Re 1 2Im
2
2 Im
2
R R
R R
R R
j
V
R jX j
Z
V
R R X
Z
 
 
 
 
 
 
   
 
   
 
   
 
     
 
   
    
 
 
   
 
 
 
 
 
 
 
 
 
 
 
   
 
 
 
   
   
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
+
+
+ Γ + Γ
= + − Γ + Γ =
= − Γ − Γ

Transmission Lines
Equivalently, using the complex characteristic admittance:
*
0
*
2
0 0
2
0 0
1 Re
1
(d=0, ) Re 1 1
2
Re 1 Re Im 1 Re Im
2
Re
2 R
R R
R R R R
P t V Y V
V
G jB j j
V
G jB
 
 
 
   
 
 
 
   
 
   
 
 
 
 
 
   
         
 
   
 
         
     
       
 
 
   
 
 
   

   
    
   
+ +
+
+
− Γ
〈 〉 = +Γ −Γ =
= − + Γ + Γ − Γ Γ
= −
+
2 2
2
2
0 0
2
2
0 0 0
Im 2Im
Re 1 2Im
2
2 Im
2
R R
R R
R R
j
V
G jB j
V
G G B
 
 
 
 
 
 
 
 
   
 
  
 
     
 
  
   
 
 
 
 
 
 
 
 
 
 
 
 
 
 
   
 
 
 
   
   
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
+
+
+ Γ + Γ
= − − Γ + Γ =
= − Γ + Γ

Transmission Lines
The time-average power, injected into the input of the transmission
line, is maximized when the input impedance of the transmission
line and the internal generator impedance are complex conjugate of
each other.
Z Z
G in
= *
Zin
ZR
ZG
Transmission line
Load
Generator
for maximum power transfer
VG

Transmission Lines
The characteristic impedance of the loss-less line is real and we
can express the power flow, anywhere on the line, as
( )
( )
*
d 2 d
*
* d 2 d
0
2 2 2
0 0
1
(d, ) Re{ (d) (d) }
2
1
Re 1
2
1
( ) 1
1 1
2 2
j j
R
j j
R
R
P t V I
V e e
V e e
Z
V V
Z Z
β β
β β
Γ
Γ
Γ
Incident Reflected w
wave ave

+ −
+ − −
+ +
〈 〉 =

= +



− 

= −
This result is valid for any location, including the input and the load,
since the transmission line does not absorb any power.

Transmission Lines
In the case of low-loss lines, the characteristic impedance is again
real, but the time-average power flow is position dependent
because the line absorbs power.
{ }
( )
{
( )
*
d d 2 d
*
* d d 2 d
0
2 2 2
2 d 2 d
0 0
1
(d, ) Re (d) (d)
2
1
Re 1
2
1
( ) 1
1 1
2 2
j
R
j
R
R
P t V I
V e e e
V e e e
Z
V e V e
Z Z
α β γ
α β γ
α α
Γ
Γ
Γ
Reflec
Inciden ted wave
t wave

+ −
+ − −
+ + −
〈 〉 =
= +

− 

= −

Transmission Lines
Note that in a lossy line the reference for the amplitude of the
incident voltage wave is at the load and that the amplitude grows
exponentially moving towards the input. The amplitude of the
incident wave behaves in the following way
input inside the line
L d
load
V e V e V
+ α + α +
⇔ ⇔

The reflected voltage wave has maximum amplitude at the load, and
it decays exponentially moving back towards the generator. The
amplitude of the reflected wave behaves in the following way
input inside the line lo
L d
ad
R R R
V e V e V
+ −α + −α +
Γ Γ
⇔ ⇔
Γ

Transmission Lines
For a general lossy line the power flow is again position dependent.
Since the characteristic impedance is complex, the result has an
additional term involving the imaginary part of the characteristic
admittance, B0, as
{ }
( )
{
( ) }
*
d d
*
* * d d
0
2 2 2
2 d 2 d
0 0
2 2 d
0
1
(d, ) Re (d) (d)
2
1
Re 1 (d)
2
( ) 1 (d)
2 2
Im( (d))
j
j
R
P t V I
V e e
Y V e e
G G
V e V e
B V e
α β
α β
α α
α
Γ
Γ
Γ
Γ
+
+ −
+ + −
+
〈 〉 =
= +
−
= −
+

Transmission Lines
For the general lossy line, keep in mind that
*
0 0 0 0 0
0 0 0
* 2 2 2
0 0 0 0 0
0
0 0
0 0
2 2 2 2
0 0 0 0
1 Z R jX R jX
Y G jB
Z Z Z R X
Z
R X
G B
R X R X
− −
= = = = = +
+
−
= =
+ +
Recall that for a low-loss transmission line the characteristic
impedance is approximately real, so that
0 0
B ≈ and 0 0 0
1
Z G R
≈ ≈ .
The previous result for the low-loss line can be readily recovered
from the time-average power for the general lossy line.

Transmission Lines
To completely specify the transmission line problem, we still have
to determine the value of V+ from the input boundary condition.
¾The load boundary condition imposes the shape of the
interference pattern of voltage and current along the line.
¾The input boundary condition, linked to the generator, imposes
the scaling for the interference patterns.
We have
0
1 (L)
(L)
1 (L)
in
in G in
G in
Z
V V V Z Z
Z Z
Γ
Γ
with
+
= = =
+ −
0
0
0
0
0
0
tan( L)
tan( L)
tanh( L)
tanh( L)
R
in
R
R
in
R
Z jZ
Z Z
jZ Z
Z Z
Z Z
Z Z
β
β
γ
γ
loss - less line
lossy
o
lin
r
e
+
=




+
+
=
+




Transmission Lines
For a loss-less transmission line:
[ ]
L L 2 L
(L) 1 (L) (1 )
j j j
R
V V e V e e
β β β
Γ Γ
+ + −
= + = +
L 2 L
1
(1 )
in
G j j
G in R
Z
V V
Z Z e e
β β
Γ
+
−
⇒ =
+ +
For a lossy transmission line:
[ ] ( )
L L 2 d
(L) 1 (L) 1 R
V V e V e e
γ γ γ
Γ Γ
+ + −
= + = +
L 2 L
1
(1 )
in
G
G in R
Z
V V
Z Z e e
γ γ
Γ
+
−
⇒ =
+ +

Transmission Lines
In order to have good control on the behavior of a high frequency
circuit, it is very important to realize transmission lines as uniform
as possible along their length, so that the impedance behavior of
the line does not vary and can be easily characterized.
A change in transmission line properties, wanted or unwanted,
entails a change in the characteristic impedance, which causes a
reflection. Example:
Γ1
ZR
Z01 Z02
Z01 Zin
Γ1
in 01
in 01
Z Z
Z Z
=
−
+

Transmission Lines
Special Cases
0
R
Z → (SHORT CIRCUIT)
The load boundary condition due to the short circuit is V (0) = 0
0 2 0
(d 0) (1 )
(1 ) 0
1
j j
R
R
R
V V e e
V
+ β − β
+
⇒ = = + Γ
= + Γ =
⇒ Γ = −
ZR = 0
Z0

Transmission Lines
Since
R
V
V
V V
−
+
− +
Γ =
⇒ = −
We can write the line voltage phasor as
d d
d d
d d
(d)
( )
2 sin( d)
j j
j j
j j
V V e V e
V e V e
V e e
jV
+ β − − β
+ β + − β
+ β − β
+
= +
= −
= −
= β

Transmission Lines
For the line current phasor we have
d d
0
d d
0
d d
0
0
1
(d) ( )
1
( )
( )
2
cos( d)
j j
j j
j j
I V e V e
Z
V e V e
Z
V
e e
Z
V
Z
+ β − − β
+ β + − β
+
β − β
+
= −
= +
= +
= β
The line impedance is given by
0
0
(d) 2 sin( d)
(d) tan( d)
(d) 2 cos( d)/
V jV
Z jZ
I V Z
+
+
β
= = = β
β

Transmission Lines
The time-dependent values of voltage and current are obtained as
( )
(d, ) Re[ (d) ] Re[2 | | sin( d) ]
2| |sin( d) Re[ ]
2| |sin( d) Re[ cos( ) sin( )]
2| |sin( d)sin( )
j t j j t
j t
V t V e j V e e
V j e
V j t t
V t
ω + θ ω
+ ω +θ
+
+
= = β
= β ⋅
= β ⋅ ω + θ − ω + θ
= − β ω + θ
0
( )
0
0
0
(d, ) Re[ (d) ] Re[2 | | cos( d) ]/
2| |cos( d) Re[ ]/
2| |cos( d) Re[(cos( ) sin( )]/
| |
2 cos( d)cos( )
j t j j t
j t
I t I e V e e Z
V e Z
V t j t Z
V
t
Z
ω + θ ω
+ ω +θ
+
+
= = β
= β ⋅
= β ⋅ ω + θ + ω + θ
= β ω + θ

Transmission Lines
The time-dependent power is given by
2
0
2
0
(d, ) (d, ) (d, )
| |
4 sin( d)cos( d)sin( )cos( )
| |
sin(2 d)sin (2 2 )
P t V t I t
V
t t
Z
V
t
Z
+
+
= ⋅
= − β β ω + θ ω + θ
= − β ω + θ
and the corresponding time-average power is
0
2
0
0
1
(d, ) (d, )
| | 1
sin(2 d) sin (2 2 ) 0
T
T
P t P t dt
T
V
t
Z T
+
=
= − β ω + θ =
∫
∫

Transmission Lines
R
Z → ∞ (OPEN CIRCUIT)
The load boundary condition due to the open circuit is I (0) = 0
0 2 0
0
0
(d 0) (1 )
(1 ) 0
1
j j
R
R
R
V
I e e
Z
V
Z
+
β − β
+
⇒ = = − Γ
= − Γ =
⇒ Γ =
ZR → ∞
Z0

Transmission Lines
Since
R
V
V
V V
−
+
− +
Γ =
⇒ =
We can write the line current phasor as
d d
0
d d
0
d d
0 0
1
(d) ( )
1
( )
2
( ) sin( d)
j j
j j
j j
I V e V e
Z
V e V e
Z
V jV
e e
Z Z
+ β − − β
+ β + − β
+ +
β − β
= −
= −
= − = β

Transmission Lines
For the line voltage phasor we have
d d
d d
d d
(d) ( )
( )
( )
2 cos( d)
j j
j j
j j
V V e V e
V e V e
V e e
V
+ β − − β
+ β + − β
+ β − β
+
= +
= +
= +
= β
The line impedance is given by
0
0
(d) 2 cos( d)
(d)
(d) tan( d)
2 sin( d)/
V V Z
Z j
I jV Z
+
+
β
= = = −
β
β

Transmission Lines
The time-dependent values of voltage and current are obtained as
( )
(d, ) Re[ (d) ] Re[2 | | cos( d) ]
2| |cos( d) Re[ ]
2| |cos( d) Re[(cos( ) sin( )]
2| |cos( d)cos( )
j t j j t
j t
V t V e V e e
V e
V t j t
V t
ω + θ ω
+ ω +θ
+
+
= = β
= β ⋅
= β ⋅ ω + θ + ω + θ
= β ω + θ
0
( )
0
0
0
(d, ) Re[ (d) ] Re[2 | | sin( d) ]/
2| |sin( d) Re[ ]/
2| |sin( d) Re[ cos( ) sin( )]/
| |
2 sin( d)sin( )
j t j j t
j t
I t I e j V e e Z
V j e Z
V j t t Z
V
t
Z
ω + θ ω
+ ω +θ
+
+
= = β
= β ⋅
= β ⋅ ω + θ − ω + θ
= − β ω + θ

Transmission Lines
The time-dependent power is given by
2
0
2
0
(d, ) (d, ) (d, )
| |
4 cos( d)sin( d)cos( )sin( )
| |
sin(2 d)sin (2 2 )
P t V t I t
V
t t
Z
V
t
Z
+
+
= ⋅ =
= − β β ω + θ ω + θ
= − β ω + θ
and the corresponding time-average power is
0
2
0
0
1
(d, ) (d, )
| | 1
sin(2 d) sin (2 2 ) 0
T
T
P t P t dt
T
V
t
Z T
+
=
= − β ω + θ =
∫
∫

Transmission Lines
0
R
Z Z
= (MATCHED LOAD)
The reflection coefficient for a matched load is
0 0 0
0 0 0
0
R
R
R
Z Z Z Z
Z Z Z Z
− −
Γ = = =
+ +
no reflection!
The line voltage and line current phasors are
d 2 d d
d 2 d d
0 0
(d) (1 )
( ) (1 )
j j j
R
j j j
R
V V e e V e
V V
I d e e e
Z Z
+ β − β + β
+ +
β − β β
= + Γ =
= − Γ =
ZR = Z0
Z0

Transmission Lines
The line impedance is independent of position and equal to the
characteristic impedance of the line
d
0
d
0
(d)
(d)
(d)
j
j
V V e
Z Z
I V
e
Z
+ β
+
β
= = =
The time-dependent voltage and current are
d
( d )
d
0
( d )
0 0
(d, ) Re[| | ]
| | Re[ ] | | cos( d )
(d, ) Re[| | ]/
| | | |
Re[ ] cos( d )
j j j t
j t
j j j t
j t
V t V e e e
V e V t
I t V e e e Z
V V
e t
Z Z
+ θ β ω
+ ω +β +θ +
+ θ β ω
+ +
ω +β +θ
=
= ⋅ = ω + β + θ
=
= ⋅ = ω + β + θ

Transmission Lines
The time-dependent power is
0
2
2
0
| |
(d, ) | | cos( d ) cos( d )
| |
cos ( d )
V
P t V t t
Z
V
t
Z
+
+
+
= ω + β + θ ω + β + θ
= ω + β + θ
and the time average power absorbed by the load is
2
2
0
0
2
0
1 | |
(d) cos ( d )
| |
2
t V
P t dt
T Z
V
Z
+
+
= ω + β + θ
=
∫

Transmission Lines
R
Z jX
= (PURE REACTANCE)
The reflection coefficient for a purely reactive load is
0 0
0 0
2 2
0 0 0 0
2 2 2 2
0 0 0 0
( )( )
2
( )( )
R
R
R
Z Z jX Z
Z Z jX Z
jX Z jX Z X Z XZ
j
jX Z jX Z Z X Z X
− −
Γ = = =
+ +
− − −
= = +
+ − + +
ZR = j X
Z0

Transmission Lines
In polar form
exp( )
R R j
Γ = Γ θ
where
( )
( ) ( )
( )
( )
2 2
2 2 2 2
2 2
0 0
0
2 2 2
2 2 2 2 2 2
0 0 0
1 0
2 2
0
4
1
2
tan
R
X Z Z X
X Z
Z X Z X Z X
XZ
X Z
−
− +
Γ = + = =
+ + +
 
θ =  
 
−
 
The reflection coefficient has unitary magnitude, as in the case of
short and open circuit load, with zero time average power absorbed
by the load. Both voltage and current are finite at the load, and the
time-dependent power oscillates between positive and negative
values. This means that the load periodically stores power and
then returns it to the line without dissipation.

Transmission Lines
© Amarcord, 2006 – Digital Maestro Series 107
Reactive impedances can be realized with transmission lines
terminated by a short or by an open circuit. The input impedance of
a loss-less transmission line of length L terminated by a short
circuit is purely imaginary
( )
0 0 0
2 2
tan L tan L tan L
in
p
f
Z j Z j Z j Z
v
 
π π
 
= β = =  
   
λ
   
For a specified frequency f, any reactance value (positive or
negative!) can be obtained by changing the length of the line from 0
to λ/2. An inductance is realized for L λ/4 (positive tangent)
while a capacitance is realized for λ/4 L λ/2 (negative tangent).
When L = 0 and L = λ/2 the tangent is zero, and the input
impedance corresponds to a short circuit. However, when L = λ/4
the tangent is infinite and the input impedance corresponds to an
open circuit.

Transmission Lines
Since the tangent function is periodic, the same impedance
behavior of the impedance will repeat identically for each additional
line increment of length λ/2. A similar periodic behavior is also
obtained when the length of the line is fixed and the frequency of
operation is changed.
At zero frequency (infinite wavelength), the short circuited line
behaves as a short circuit for any line length. When the frequency
is increased, the wavelength shortens and one obtains an
inductance for L λ/4 and a capacitance for λ/4 L λ/2, with
an open circuit at L = λ/4 and a short circuit again at L = λ/2.
Note that the frequency behavior of lumped elements is very
different. Consider an ideal inductor with inductance L assumed to
be constant with frequency, for simplicity. At zero frequency the
inductor also behaves as a short circuit, but the reactance varies
monotonically and linearly with frequency as
X L
= ω (always aninductance)

Transmission Lines
Short circuited transmission line – Fixed frequency
L
L
L
L
L
L
L
L
= =

= → ∞

= =

= → ∞

0 0
0
4
0
4
4 2
0
2
0
2
3
4
0
3
4
3
4
0
Z
Z
Z
Z
Z
Z
Z
Z
in
in
in
in
in
in
in
in
short circuit
inductance
open circuit
capacitance
short circuit
inductance
open circuit
capacitance
λ
λ
λ λ
λ
λ λ
λ
λ
λ
Im
Im
Im
Im
k p
k p
k p
k p
…
L

Transmission Lines
β Impedance of a short circuited transmission line
(fixed frequency, variable length)
Line Length L
L
θ [deg]
inductive
capacitive
inductive inductive
cap.
-40
-30
-20
-10
0
10
20
30
40
0 200 300 400 500
100
2π/β= λ 5π/(2β)= 5λ/4
3π/(2β)= 3λ/4
π/β =λ/2
π/(2β)= λ/4
Normalized
Input
Impedance
Z(L)/Zo
=
j
tan(
L)

Transmission Lines
β
θ [deg]
inductive
capacitive
inductive inductive
cap.
-40
-30
-20
-10
0
10
20
30
40
0 200 300 400 500
100
vp / (4L)
Impedance of a short circuited transmission line
(fixed length, variable frequency)
vp / (2L) 3vp / (4L) vp / L 5vp / (4L)
f
Frequency of operation
Normalized
Input
Impedance
Z(L)/Zo
=
j
tan(
L)

Transmission Lines
For a transmission line of length L terminated by an open circuit,
the input impedance is again purely imaginary
( )
0 0 0
2
tan L 2
tan L tan L
in
p
Z Z Z
Z j j j
f
v
= − = − = −
π
β    
π
   
λ  
 
 
We can also use the open circuited line to realize any reactance, but
starting from a capacitive value when the line length is very short.
Note once again that the frequency behavior of a corresponding
lumped element is different. Consider an ideal capacitor with
capacitance C assumed to be constant with frequency. At zero
frequency the capacitor behaves as an open circuit, but the
reactance varies monotonically and linearly with frequency as
1
X
C
=
ω
(always a capacitance)

Transmission Lines
Open circuit transmission line – Fixed frequency
L
L
L
L
L
L
L
L
= → ∞

= =

= → ∞

= =

0
0
4
0
4
0
4 2
0
2
2
3
4
0
3
4
0
3
4
0
Z
Z
Z
Z
Z
Z
Z
Z
in
in
in
in
in
in
in
in
open circuit
capacitance
short circuit
inductance
open circuit
capacitance
short circuit
inductance
λ
λ
λ λ
λ
λ λ
λ
λ
λ
Im
Im
Im
Im
k p
k p
k p
k p
…
L

Transmission Lines
capacitive capacitive
inductive inductive inductive
θ [deg]
β
capacitive
L
Line Length L
0 π/(2β) = λ/4 5π/(2β) = 5λ/4
2π/β = 2λ
3π/(2β) = 3λ/4
π/β = λ/2
-40
-30
-20
-10
0
10
20
30
40
0 200 400 500
300
100
Impedance of an open circuited transmission line
(fixed frequency, variable length)
Normalized
Input
Impedance
Z(L)/
Zo
=
-
j
cotan(
L)

Transmission Lines
capacitive capacitive
inductive inductive inductive
θ [deg]
β
0
capacitive
vp / (4L) vp / (2L) 3vp / (4L) vp / L 5vp / (4L)
f
Frequency of operation
-40
-30
-20
-10
0
10
20
30
40
0 200 400 500
300
100
Impedance of an open circuited transmission line
(fixed length, variable frequency)
Normalized
Input
Impedance
Z(L)/
Zo
=
-
j
cotan(
L)

Transmission Lines
It is possible to realize resonant circuits by using transmission
lines as reactive elements. For instance, consider the circuit below
realized with lines having the same characteristic impedance:
( ) ( )
1 0 1 2 0 2
tan L tan L
in in
Z j Z Z j Z
= β = β
I
V
L2
L1
short
circuit
short
circuit
Zin1 Zin2
Z0 Z0

Transmission Lines
The circuit is resonant if L1 and L2 are chosen such that an
inductance and a capacitance are realized.
A resonance condition is established when the total input
impedance of the parallel circuit is infinite or, equivalently, when
the input admittance of the parallel circuit is zero
( ) ( )
0 2
0 1
1 1
0
tan L
tan L r
r j Z
j Z
+ =
β
β
or
1 2
2
tan L tan L
r r r
r
p p r p
v v v
   
ω ω π ω
= − β = =
   
    λ
   
with
Since the tangent is a periodic function, there is a multiplicity of
possible resonant angular frequencies ωr that satisfy the condition
above. The values can be found by using a numerical procedure to
solve the trascendental equation above.

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