Time–Domain solutions of Transmission Line Equation
The partial–differential equations of a lossless line are
v ( z, t )
i ( z, t )
 L
z
t
i ( z, t )
v ( z, t )
 C
z
t
These gives as
 2 v ( z, t )
 2 v( z, t )

LC
z 2
 2t
 2i ( z, t )
 2i ( z, t )

LC
z 2
 2t
Solutions to the above equations are
z
z
v ( z , t )  f  (t  )  f  (t  )
v
v
z
z
i ( z, t )  YO [ f  (t  )  f  (t  )]
v
v
Where
v
1
LC
and
Z0 
1

Y0
L
C
z
f  (t  )
v
represents a wave traveling in the +z direction , where
z
f  (t  )
v
is a wave traveling in the –z direction.
Boundary conditions at the Source End
A semi-infinite line driven by a generator with internal resistance
The line has no voltage nor current anywhere on it (z>0) at t=0.
We are interested in determining the voltage and current waves along
the line for t  0 z  0. The voltage – current relation is simply
VS (t )  I1 (t ) RS  V1 (t )
for t  0 z  0.
Note that
z
z
V1 ( z, t )  f  (t  )  f  (t  ) |Z 0
v
v
z
z
I 1 ( z, t )  YO [ f  (t  )  f  (t  )] |Z 0
v
v
No wave travels in the –z direction and
f  (t )  0
. We then have
I1 (t )  YO f  (t )
Rs
V1 ( z )  f  (t )
This gives us
V1 ( z )  f  (t ) 
VS ( t )
1  Y0 RS
Vs
I1(t)
V1(t)
Z0
t0
Equivalent circuit for input condition at z=0
for t  0
The equivalent circuit is shown on the right side
Terminated transmission lint with transient input
Rs I1(t)
I(0,t)
+
V(0,t)
V1(t)
Vs
Ir(t)
V(l,t)
L,C
-
+
Vr(t)
-
Z=0
Z=l
We assume that the generator voltage is zero and that the line and load
are not energized at time 0<t< l we have
v
VS (t )  I1 (t ) RS  V (0, t )  f  (t )
I1 (t )  I (0, t )  Y0 f  (t )
This leads to the following result
f  (t ) 
VS ( t )
1  YO RS
At the load
for 0<t< l
v
z= l , we have
l
l
V (l , t )  f  (t  )  f  (t  )
v
v
l
l
I (l , t )  YO [ f  (t  )  f  (t  )]
v
v
The essential boundary conditions at the load are then simply
V (l , t )  Vr (t )
I (l , t )  I r (t )
By solving the above equation, we first obtain
l
l
f  (t  )  f  (t  )  Z 0 I (l , t )
v
v
We then get
l
V ( l , t )  Vr ( l , t )  2 f  ( t  )  Z 0 I r ( l , t )
v
The above equation is extremely important . It says that the
transmission line
l
2 f  (t  )
v
behaves with respect to any termination like a voltage source
in series with an impedance
Ir(t)= I(l,t)
Z0 I1(t)
2V+(l,t)=
2f+(t-l/v)
Z0
+
Vr(t)=V(l,t)
Zr
-
Equivalent circuit at a termination
Solving the above circuit , we may obtain the terminal voltage
Vr (t )
The reflected wave at the load is
l
l
f  (t  )  Vr (t )  f  (t  )
v
v
Note that
l
f  (t  )  0
v
if
Z0  Z r .
The reflected wave
l
f  (t  )
v
will
travel back to the source end and it reach the source end at time
t
2l
v
.In particular it carries information about the load.
At the source end , we have the following boundary conditions.
VS (t )  I1 (t ) RS  V1 (t )  V (0, t )  f  (t )  f  (t )
I1 (t )  I (0, t )  Y0 [ f  (t )  f  (t )]
The above two equations yield
VS (t )  I1 (t ) RS  2 f  (t )  Z 0 I1 (t )
It says that as for as the input generator circuit is concerned ,
the line behaves like a voltage generator
2 f  (t )
in series with a
resistance Z 0 . The equivalent circuit at the source end now is as
followings.
Rs
I1(t)
Z0
+
+
Vs(t)
2f-(t)
V1(t)
-
-
We get
I 1 (t ) 
VS ( t )
2 f  (t )

RS  Z 0 RS  Z 0
f  (t )  Z 0 I (l , t )  f  (t )
or
f  (t ) 
VS ( t )
R  Z0
( S
) f  (t )
1  RS Y0
RS  Z 0
The first term in
f  (t )
is the incident wave due to source
Second term is due to the reflection of f  (t )
VS (t )
, the
Example: An open transmission line
RS
VS
L,C (Z0)
Z=0
Z=l
A DC voltage source is connected to an open transmission line and the
circuit is under steady state condition . We have
z
z
v ( z , t )  f  (t  )  f  (t  )
v
v
z
z
i ( z, t )  YO [ f  (t  )  f  (t  )]
v
v
Since the transmission line is open circuited at z  l , we have
i ( z, t )  0
v( z, t )  VS
for 0  z  l
These conditions yield
z
z V
f  (t  )  f  (t  )  S
v
v
2
This indicates that two waves having amplitude
VS
2
travel
along the transmission line in the opposite directions.
Example:
D-C voltage applied to a shorted line
RS
L,C (Z0)
V0
Z=0
Z=l
At t=0 , the switch is closed , we obtain the following voltage on the
line.
V0
Z
Z=vt
Once the traveling wave reaches the load , we have
IR
Z0
2V0
As a result , the current I R at the load end is
IR 
2Vo
Z0
The overall current at the load end is
IR
4V0/Z0
2V0/Z0
t
3T
T
where
T
l
v
Example: Capacitive – terminated transmission line
Z0
U-1(t)
L,C (Z0)
Z=0
C
Z=l
The source is a unit step
For 0<t< l , we obtain
v
z
1
z
f  (t  )  u  (t  )
v
2
v
At
t
l
v
this wave reaches the load and the conditions are as
following
Z0
Ir(t)
2f+=
U-1(t-l/v)
C
+
Vr(t)
-
l
Vr ( t )  u  ( t  )  Z 0 I r ( t )
v
I r (t )  C
dVr (t )
dt
Solving the equation , we get
z
 ( t  ) / Z 0C
l
Vr (t )  u  (t  )(1  e v
)
v
The incident wave
z
1
z
f  (t  )  u  (t  ) will
v
2
v
generate a reflected wave
at z= l , which is
z
l
l 1  ( t  ) / Z 0C
f  (t  l )  Vr (t )  f  (t  )  u  (t  )(  e v
)
v
v 2
For values of z< l , f  is delayed from its form by the time
(l  z ) / v . Therefore it becomes
2l z
f  ( z, t )  u  (t 
 ( t   ) / Z 0C
2l z 1
 )(  e v v
)
v v 2
Line voltage versus distance and time is as follows.
IR
IR
1
1
1/2
1/2
z
0
z
0
l
l
0<t< l / v
l / v <t< 2l / v
IR
1
1/2
z
0
l
2l / v <t< 3l / v
Example:
An open circuited transmission line fed by a current source with an
internal conductance G .
I(t)
+
V(z,t)
-
G
lossless
Z0
Z=0
Z=l
If an input to the line is of the form
I (t )  u (t ) sin w0 t
w0 
2


4T 2T
T  l /v
The line is a quarter – wave length long . u (t ) is a unit step function
Then the voltage in the signal line is

z
z
V ( z, t )   (G  Y0 ) 1  n [u  (t  2nT  ) sin( w0 t  n  w0 )
v
v
n 0
z
z
 u  (t  ( 2n  2)T  ) sin( w0 t  ( n  1)  w0 )]
v
v
V(0,t)
-1
(G+Y0)
t
0
2T
4T
6T
8T
10T
V(0,t)
0
t
2T