W.C.Chew
ECE 350 Lecture Notes
6. Terminated Uniform Lossless Transmission Lines
Zo, v
lossless
Z L LOAD
z=0
z = –l
Consider a lossless transmission line terminated in a load of impedance
.
L A wave traveling to the right will be reected at the termination. In
general, there will be both positive going and negative going waves on the
line. Hence,
~ (z) = V0 e;jz + V1e+jz :
V
(1)
Here, = j , = 0, because of no loss. The corresponding current, as in
(5.32), is
~(z) = V0 e;jz ; V1 e+jz I
(2)
Z
Z
Z
qL
where Z0 =
At z = 0,
C
and = !
0
p
0
for a lossless line.
LC
~ (z = 0)
V0 + V1
=
ZL =
Z0 :
~(z = 0)
V0 ; V1
I
We can solve for V1 in terms of V0 , i.e.
V
V1
; Z0 V :
= ZZL +
0
Z
(4)
; Z0 = ZZL +
Z
(5)
L
If we dene
v
(3)
L
0
0
then V1 = v V0 , and Equation (1) becomes
~ (z) = V0e;jz + v V0e+jz :
V
(6)
In the above, v is the ratio of the negative going voltage amplitude to the
positive going voltage amplitude at z = 0, and it is known as the voltage
reection coecient.
1
The current reection coecient is dened as the ratio of the negative
going current to the positive going current at z = 0, and it is
i=
The current can be written as
~(z) =
I
I1
I0
V0
Z0
= ; VV1 = ;v :
(7)
0
;jz ; v
e
V0
Z0
jz :
(8)
e
The voltage and current in (6) and (8) are not constants of position. We can
dene a generalized impedance at position z to be
;jz + v e+jz
~ (z)
V
e
(9)
Z (z ) =
=
Z0
;jz ; v e+jz :
~(z)
e
I
At z = ;l, this becomes
jl + v e;jl
e
Z (;l ) = Z0
(10)
jl ; e;jl :
e
v
With v dened by (5), we can substitute it into (10) to give after some
simplications,
ZL + j Z0 tan l
:
(11)
Z (;l ) = Z0
Z + j Z tan l
0
L
Shorted Terminations
If ZL is a short, or ZL = 0, then,
Z (;l ) = j Z0 tan l = j X:
(12)
x
inductive
2π
π
2
π
3π
2
5π
2
β
capacitive
Open-Circuit Terminations
If ZL is an open circuit, ZL = 1, then
Z (;l ) = ;j Z0 cot l = j X:
2
(13)
x
inductive
π
2π
π
2
3π
2
5π
2
β
capacitive
Standing Waves on a Lossless Transmission Line
The positive going wave in Equation (6) is
( ) = V0 e;jz (14)
V+ z
and the negative going wave in Equation (6) is
; (z ) = v V0 e
V
+jz
(15)
:
We can dene a generalized reection coecient to be the ratio of V+(z)
to V;(z) at position z. Hence,
;(z) = VV;((zz)) = v e2jz :
(16)
(z) = V0 e;jz 1 + ;(z)]:
The magnitude of V (z) is then
(17)
+
Hence,
V
j ( )j = j 0 j j1 + ;( )j
V
z
V
z
(18)
:
A plot of jV (z)j is as shown.
Im Axis
–dmin
1
ρv
+z
z=0
–z
Re Axis
Γ(z)
–d1
1+Γ
(z)
3
|V(z)|
Vmax
Vmin
–d1 – λ
2
–dmin
–d1
z=0
z
We can use the triangular inequality and show that
jV0j (1 ; j;(z)j) jV (z)j jV0j (1 + j;(z)j):
(19)
From (16), j;(z)j = jv j, hence (19) becomes,
jV0j (1 ; jv j) jV (z)j jV0j (1 + jv j):
(20)
The voltage standing wave ratio is dened to be Vmax =Vmin, and from (20),
it is
+ jv j :
(21)
VSWR = 11 ;
jv j
If v = 0, then VSWR= 1, and we have no reected wave. We say that
the load is matched to the transmission line. Note that v = 0 when ZL = Z0 .
If jv j =1, then VSWR= 1, and we have a badly matched transmission
line. In a passive load,
0 jv j 1:
(22)
jv j =1 only when ZL = 0, or ZL = 1 according to Equation (5). Hence,
1 VSWR < 1:
(23)
VSWR is an indicator of how well a load is being matched to the transmission
line. We can solve (21) for jv j in terms of VSWR, i.e.
VSWR ; 1 :
jv j = VSWR
(24)
+1
Therefore, given the measurement of V S WR on a terminated transmission
line, we can deduce the magnitude of v . Furthermore, if we know the phase
of v , we would be able to derive ZL from (5), or
1 + v (25)
ZL = Z0
1 ; v
or
1 + jv j ejv ZL = Z0
(26)
1 ; j j ejv
v
4
where
v
Determining
v
= jv j ejv :
(27)
from j ( )j
V
z
can be determined from the voltage standing wave measured. The
voltage standing wave pattern is proportional to j1 + ;(z)j, but ;(z) is related
to v as
;(z) = v e2jz :
(28)
Writing the polar representation of v , we have,
v
;(z) = jv j ej(2z+v ):
(29)
However, we know that the rst minimum value of V (z) occurs when ;(z) is
purely negative, or the phase of ;(z) is ;. This occurs at z = ;dmin rst.
In other words,
;2dmin + v = ;:
(30)
Since dmin can be obtained from the voltage standing wave pattern measurement, and that = 2=, we deduce that
4 d :
= ; +
(31)
v
min
Transmission Coecients
It is sometimes useful to dene a transmission coecient on a transmission line. The transmission coecient may be dened as the ratio of the
voltage on the load to the amplitude of the incident voltage. Since
V
(z) = V0e;jz + v V0e+jz :
(32)
The voltage at the load is V (z = 0), and it is given by
V
(0) = V0(1 + v ):
Since the amplitude of the incident voltage is V0 , we have
2ZL :
V (0)
=
1
+
v =
v =
V0
ZL + Z0
5
(33)
(34)