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
ZL . A wa v etra v elingto the right will be reected at the termination. In
general, there will be both positive going and negative going wa ves on the
line. Hence,
V~ (z ) = V e;jz + V e jz :
(1)
Here, = j , = 0, because of no loss. The corresponding current, as in
(5.32), is
V
V
(2)
I~(z ) = e;jz ; e jz ;
Z
Z
0
qL
where Z =
At z = 0,
0
C
p
+
1
0
1
0
0
+
and = ! LC for a lossless line.
V~ (z = 0)
V0 + V1
=
Z
Z:
=
L
V0 ; V1 0
I~(z = 0)
We can solve for V1 in terms of V0 , i.e.
Z ; Z0
V1 = L
V:
ZL + Z0 0
If we dene
Z ; Z0
;
v = L
ZL + Z0
(4)
(5)
then V = v V , and Equation (1) becomes
V~ (z ) = V e;jz + v V e
1
(3)
0
0
0
jz :
+
(6)
In the above, v is the ratio of the negative going v oltage amplitude to the
positive going v oltage amplitude at z = 0, and it is known as the voltage
reection co ecient.
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 =
I1
V1
=
;
= ;v :
I0
V0
(7)
The current can be written as
V
V
I~(z ) = e;jz ; v ejz :
Z
Z
0
0
0
0
(8)
The voltage and current in (6) and (8) are not constants of position. We can
dene a generalized impedance at position z to be
V~ (z )
e;jz + e jz
(9)
Z (z ) = ~ = Z ;jz v jz :
e
; v e
I (z )
At z = ;l, this becomes
ejl + e;jl
Z (;l) = Z jl v ;jl :
(10)
e ; e
+
0
0
+
v
With v dened by (5), we can substitute it into (10) to give after some
simplications,
Z + jZ tan l
:
(11)
Z (;l) = Z L
Z + jZ tan l
0
0
0
L
Shorted Terminations
If ZL is a short, or ZL = 0, then,
Z (;l) = jZ tan l = jX:
(12)
0
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) = ;jZ cot l = jX:
0
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
V+ (z ) = V0 e;jz ;
(14)
and the negative going wave in Equation (6) is
V; (z ) = v V0 e+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 e jz :
(16)
V (z ) = V0 e;jz [1 + ;(z )]:
(17)
2
+
Hence,
The magnitude of V (z) is then
jV (z)j = jV j j1 + ;(z)j :
(18)
0
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
jV j (1 ; j;(z)j) jV (z)j jV j (1 + j;(z)j):
(19)
From (16), j;(z)j = jv j, hence (19) becomes,
jV j (1 ; jv j) jV (z)j jV j (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 = Z .
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 VSWR 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 = Z
1 ; v
or
1 + jv j ejv ;
ZL = Z
(26)
1 ; j j ejv
0
0
0
0
0
0
0
v
4
where
v = jv j ejv :
(27)
Determining v from jV (z)j
v 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,
;(z) = jv j ej
z+v ) :
(29)
(2
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 ) = V0 e;jz + v V0 e+jz :
(32)
The voltage at the load is V (z = 0), and it is given by
V (0) = V0 (1 + v ):
(33)
Since the amplitude of the incident voltage is V , we have
V (0)
2ZL :
v =
=
1
+
v =
V
ZL + Z
0
0
0
5
(34)