ESE 331 Notes on
Transmission Lines
Two-Conductor Transmission Lines
Consider an electrically-small section of such a guided
wave structure
i(z,t) RΔz LΔz
i(z+Δz,t)
+
v(z,t)
+
GΔz
v(z+Δz,t)
CΔz
-
Δz
Δz
Note: R, G, L, and C are “per unit length” values.
Using KVL:
v(z,t) − RΔzi(z,t) − LΔz
€
∂i(z,t)
− v(z + Δz,t) = 0
∂t
(1)
or
v(z + Δz,t) − v(z,t)
∂i(z,t)
−
= Ri(z,t) + L
Δz
∂t
(2)
as Δz → 0 :
∂v(z,t)
∂i(z,t)
−
= Ri(z,t) + L
∂z
∂t
(3)
Using KCL:
i(z,t) − GΔzv(z + Δz,t) − CΔz
∂v(z + Δz,t)
− i(z + Δz,t) = 0
∂t
as Δz → 0
∂i(z,t)
∂v(z,t)
−
= Gv(z,t) + C
∂z
∂t
€ Equations (3) and (5) are called the general
(4)
(5)
transmission line equations or telegrapher’s equations
Time-Harmonic Equations
For time-harmonic sources and signals
v(z,t) = Re[V (z)e
i(z,t) = Re[I(z)e
jωt
jωt
]
]
Substituting these forms into (3) and (5)
€
dV (z)
−
= (R + jωL)I(z)
dz
dI(z)
−
= (G + jωC)V (z)
dz
(6)
Waves on an Infinite Line
Solve (6) for I(z) in 1st equation and substitute in
the 2nd equation and vice versa
d 2V (z)
2
=
γ
V (z)
2
dz
d 2 I(z)
2
=
γ
I(z)
2
dz
where
(7)
(8)
γ = (R + jωL)(G + jωC) = α + jβ
(9)
Solutions to (7) and (8), respectively, are:
€
V = Vo+e−γz + Vo−e +γz
+ −γz
o
I=I e
− + γz
o
+I e
(10)
(11)
Using (6), (VO+, IO+) and (VO-, IO-) are related
according to
Vo+
Vo− R + jωL
Characteristic impedance = + = − − =
Io
Io
γ
= Zo =
€
γ
R + jωL
=
G + jωC
G + jωC
There are two important special cases of T-Lines
From (9):
a) Lossless line
b) Low loss line
(12)
Lossless line (R = G = 0)
γ = jω LC ⇒ α = 0;β = ω LC
ω
1
v p = phase velocity = =
= constan t
β
LC
L
Zo =
= real resistance only
C
€
Low loss line (R<<ωL; G<<ωC)
1 R G
γ ≈ jω LC[1+
( + )]
j2ω L C
1
C
L
α ≈ (R
+G
)
2
L
C
β ≈ ω LC
ω
1
vp = ≈
β
LC
Zo ≈
L
1 R G
[1+
( − )] = Ro + jX o
C
j2ω L C
Ro ≈
L
C
Xo ≈ −
L 1 R G
( − )
C 2ω L C
For both of these cases, the voltage and current
can be expressed as
+ − jβz
o
V (z) = V e
− + jβz
o
+V e
I(z) = Io+e− jβz + Io−e + jβz
(13)
If there is only a +z directed wave, then
€
+ − jβz
o
V =V e
and
I = Io+e− jβz
€
(14)
How do we know this describes a traveling wave and
at what velocity does it travel?
If
V = Vo+e− jβz
Then
V (z,t) = Re[Vo+e− jβze jωt ] = Re[Vo+e j(ωt− βz ) ]
V (z,t) = Vo+ cos(ωt − βz)
To track the movement, need to stay at
€
ωt − kz = constan t
(15)
Let us plot the behavior versus z as time unfolds
At t=0
V
+Vo+
€
+
t=0
z
−Vo+
V
+
t=t1>0
+Vo+
€
z
−Vo+
t=0
V
+
t=t1>0
+Vo+
€
z
−Vo+
t=0
t=t2>t1
Then, to determine the velocity at which the
wave travels
ωt constan t
z=
−
β
β
or
ω
ω
1
up = =
=
β ω LC
LC
For a pulse with all the Fourier terms, each
€ frequency component travels at up. Then, the
pulse will not distort as it travels.
For a coax line,
2πε
C=
ln( ba )
b
a
ε
and
µo
L=
ln( ba )
2π
Then
€
€
up =
1
1
1
c
3 ×10 8 m /sec
=
=
=
=
LC
µoε
µoεo εr
εr
εr
L
1
60 b
Z o = Ro =
=
= up L =
ln
C upC
εr a
We use
RG58/U
εr=2.3
Ro=50 Ω
In the lecture prior to Experiment 6, I will further
discuss the propagation of voltage and current
pulses on transmission lines.
Transmission line parameters
Need two paramters for most applications
L
Zo =
C
1
up =
LC
€
If you know these, you know L and C; or vice versa.
There are analytic expressions for parallel plate guide,
two-wire, and coax. For microstrip see R. Ludwig and
P. Bretchko, “RF Circuit Design, Theory and Applications,”
Prentice-Hall, 2000.
Wave Characteristics on a Finite
Transmission Line
Consider
Vg
+
IL
Ii
Zg
VL
γ, Zo
VI
z
z’=l-z
l
z=0
z=l
ZL
In general, at location z
+ − γz
o
V (z) = V e
+ − γz
o
I(z) = I e
− +γz
o
+V e
− +γz
o
+I e
(16)
Both forward and backward waves are needed for a
terminated line
€
Now, recall that
+
o
+
o
−
o
−
o
V
V
=−
= Zo
I
I
(17)
Consider the line to be terminated in a general load
Impedance.
VL = V (z = l) = Vo+e−γl + Vo−e +γl
+ − γl
o
IL = I(z = l) = I e
− +γl
o
+I e
Solving for Vo+ and Vo−
1
+
Vo = (VL + IL Z o )eγl
2
1
−
Vo = (VL − IL Z o )e−γl
2
Using these in (16)
(19)
IL
[(Z L + Z o )eγ (l−z ) + (Z L − Z o )e−γ (l−z) ]
2
I
I(z) = L [(Z L + Z o )eγ (l−z) + (Z L − Z o )e−γ (l−z ) ]
2Z o
V (z) =
€
(18)
(20)
IL
V (z') = [(Z L + Z o )eγz' + (Z L − Z o )e−γz' ]
2
I
I(z') = L [(Z L + Z o )eγz' + (Z L − Z o )e−γz' ]
2Z o
(21)
Using the relations for cosh and sinh functions
€
eγz' + e−γz' = 2cosh γz';eγz' − e−γz' = 2sinh γz'
(21) can be written as
€
V (z') = IL (Z L cosh γz'+Z o sinh γz')
I
I(z') = L (Z L sinh γz'+Z o cosh γz')
Zo
Dividing
V (z')
Z L cosh γz'+Z o sinh γz'
= Z(z') = Z o
I(z')
Z L sinh γz'+Z o cosh γz'
or
Z L + Z o tanh γz'
Z(z') = Z o
Z o + Z L tanh γz'
(22)
(23)
At the source end, z=0 and z’=l
€
Z L + Z o tanh γl
Z(z'= l) = Z o
Z o + Z L tanh γl
(24)
Note that in (23) when ZL=Zo, then Z(z’)=Zo
for any z’ on the line and the line seems to be extended
to infinity.
Lumped Elements are Hard to
Realize at RF and MW
Frequencies
f Hz
Absolute value of the impedance
magnitude versus frequency for a lumped
capacitor
f, Hz
Frequency response of the impedance of an RFC
Transmission Lines as Circuit Elements
Consider a lossless line
L
γ = jβ ,Z o = Ro =
C
At the input of such a line having length L and
€
terminated in load impedance ZL, (24) yields
Z L + jRo tan βL
Z in = Z i = Ro
Ro + jZ L tan βL
(25)
Consider the following cases:
1) Open-circuit termination ( Z L → ∞ }
jRo
(Z i ) oc = −
= − jRo cot βL
tan βL
(26)
€
-RocotβL
€
inductive
λ/2
λ/4
3λ/4
λ
5λ/4
L
capacitive
Then, for L=λ/4, 3λ/4, 5λ/4, etc., the open circuit appears
at the line input as a short circuit. For values slightly less
than these values the impedance at the input looks
capacitive and for lengths slightly greater than those values
the input impedance looks inductive. Why would λ/4 be
preferable to the longer lengths?
What’s the point of this?
2) Short-circuit termination (ZL=0)
(Z i ) sc = jRo tan βL
€
(27)
RotanβL
inductive
L
λ/4 λ/2 3λ/4 λ 5λ/4
capacitive
Then, for λ/4, 3λ/4, 5λ/4, etc. the short circuit
appears at the line input as an open circuit. For
values of L slightly less than these values the
impedance at the input looks inductive and for
lengths slightly greater than these values the input
impedance looks capacitive.
3) Quarter-Wave Section
When L=λ/4, 3λ/4,5 λ/4, etc., from (25)
2
o
R
Zi =
ZL
(28)
The line functions as an impedance transformer or inverter.
What is a transformer?
€
Why not use op amps?
Now, repeating eq. (20)
IL
[(Z L + Z o )eγ (l−z ) + (Z L − Z o )e−γ (l−z) ]
2
I
I(z) = L [(Z L + Z o )eγ (l−z) + (Z L − Z o )e−γ (l−z ) ]
2Z o
or
I
(Z − Z o ) −γ (l−z )
V (z) = L (Z L + Z o )[eγ (l−z ) + L
e
]
2
(Z + Z o )
I
(Z − Z o ) −γ (l−z)
I(z) = L (Z L + Z o )[eγ (l−z) − L
e
]
2Z o
(Z + Z o )
V
At,z = L, L = Z L
IL
 ZL − Zo 
1+ 

 ZL + Zo 
ZL = Zo
 ZL − Zo 
1− 

Z
+
Z
 L
o
V (z) =
€
(20)
(29)
(30)
Now, define a reflection coefficient as
Vo−
Vo+
z= L
z= L
ZL − Zo
=
= ρ z= L = ρ e jφ
ZL + Zo
(31)
This is the voltage reflection coefficient at the load.
€
A Smith chart is a graphical representation based upon
a two-dimensional space plotted in the plane of this
reflection coefficient
The Smith Chart
ZL − Zo
RL + jX L − Z o
ρ=
=
Z L + Z o RL + jX L + Z o
(RL − Z o ) 2 + X L2
=
(RL + Z o ) 2 + X L2
2X L Z o
φ = tan
X L2 − Z o2
−1
Because
ZL − Zo
ρ=
ZL + Zo
Then
ZL
−1
z −1
Zo
ρ=
= L
ZL
+ 1 zL + 1
Zo
Here zL = normalized load impedance
€
ZL
ρ
Termination
∞
1
Open circuit
0
-1
Short circuit
Zo
0
Matched load
jX
€
1ejφ Reactive load
Key locations on the Smith chart
Im ρ
-1
ZL=Zo
1
Re ρ
Reactive load
Im ρ
Curves of
constant jX
Re ρ
Im ρ
Curves of
constant
resistance
RL1
Zo
€
RL 2
Zo
€
Re ρ
€
Notice that although Zo and ZL may be fixed numbers,
Zin and zin are not according to
€
Z in (ω,z') Z L + jZ o tan βz'
=
Zo
Z o + jZ L tan βz'
where
2π 2π 2πf
β=
=
=
λ v/ f
v
This leads to frequency and wavelength dependence of Zin.
This is incorporated into the Smith chart by rotation around
the center of the chart. Consider the following equation,
obtained by dividing and rearranging the two equation in
(29).
e− jβz + ρ L e + jβz 
Z in (z,ω ) = Z o  − jβz
+ jβz 
e − ρL e 
By factoring out
e
− jβz
(32)
from both numerator and
denominator of
€
€
1+ ρ L e + j 2 βz 
Z in (z,ω ) = Z o 
+ j 2 βz 
1− ρ L e

(33)
Based upon (33), how much does z have to increase
in order for the value of Zin to repeat in value?
Im ρ
ρ
φ
€
€
2βz
Re ρ
€
You move around the periphery as f, λ, ω, or z change
and return to ρL (starting point) every λ/2.
λ/4-long T-lines terminated in impedance ZL
Zin
Z o2
Z in =
ZL
Zo
(34)
λ/4
If Z L = jωL
Z o2
1
Z in ==
=
=> capacitance
jωL jω ( L )
Z o2
€
ZL
If ZL = 1/jωC
Z in =
Z o2
1
= jω (CZ o2 ) => inductor
jωC
Rearranging (34)
€
Z in Z o
1
=
=
Zo ZL ZL
Zo
For a quater − wave line
1
zin = = y L
zL
yLnorm
zLnorm
Application
Zin
Zo
λ/4
RL
If we want Zin to equal a value, say, Rin, then
make
Z o = Rin RL
If
RL = 100Ω,Rin = 50Ω
Z o = (100)(50) = 5000 = 70.7Ω
€
Understanding Why Movement away from
ZLalong a T-line is Equivalent to a Clockwise
Rotation on the Smith Chart from ρL
The Smith is the reflection coefficient plane. For a location
z on a transmission line at some distance from the load,
represent the reflection coefficient at that location as
ρz = ρz e
jθ z
To interpret this in some sense as a rotation, consider the
reflection coefficient at the “z” location as
€
Understanding the rotation
(Z in ) z
−1
(Z in ) z − Z o
(zin ) z −1
Zo
ρz =
=
=
(Z
)
(Z in ) z + Z o
in z
+ 1 (zin ) z + 1
Zo
(307)
We have already shown that (Zin)z can be written as
€
1+ ρ L e + j 2 βz
(Z in ) z = Z o
1− ρ L e + j 2 βz
where
Z − Zo
ρL = L
ZL + Zo
(308)
Understanding rotation
(Z in ) z
1+ ρ L e + j 2 βz
−1
−1
+ j 2 βz
1+ ρ L e + j 2 βz −1+ ρ L e + j 2 βz
Zo
1− ρ L e
ρz =
=
=
+ j 2 βz
+ j 2 βz
+ j 2 βz
(Z in ) z
1+ ρ L e
1+
ρ
e
+
1−
ρ
e
L
L
+1
+
1
Zo
1− ρ L e + j 2 βz
2 ρ L e + j 2 βz
ρz =
= ρ L e + j 2 βz
2
€
Understanding rotation
We can think of the location of ρz on the Smith chart
as starting from ρL, retaining both the magnitude and
angle of ρL
Im ρ
ρL
Re ρ
Understanding rotation
As we move from the load (at z=L) to a location of
smaller z, then +2βz gets smaller and rotates clockwise
(smaller θ z ) around the chart
Im ρ
ρL
€
Re ρ
ρz
smaller z
increasing z’
Overview of the Smith Chart
• A Smith chart is the plane of reflection
coeffiecient, where each point corresponds to
both a reflection coefficient a normalized
impedance value at the same time
• We think of each point in terms of either a real
and imaginary part or a magnitude and phase
angle
Smith Chart (continued)
• Understand the following
– Key locations
• Short circuit impedance
• Open circuit impedance
• Match load
• Purely imaginary impedance
– Key movements
• Constant reactance movement
• Constant resistance movement
• Rotations around the center at fixed radius
Smith Chart (continued)
• Understanding special properties
– Impedance inversion =>admittance location for a
corresponding impedance
• halfway around at a constant radius
• transform a real imoedance to a different real impedance
– Rotation clockwise (counter clockwise)
corresponds to movement along a T-line from a
load (or source) or equivalently to a change in line
length connected to a load impedance
Smith Chart (continued)
• Understanding the use of techniques for
matching impedances
– Use of λ/4 T-line lengths to match one real
impedance to another by changing the
value of Zo
– Use of open and short circuited stubs (Tline pieces) for matching impedances
Smith Chart (continued)
• Understanding concepts
– VSWR
– Return loss
– Voltage and power reflection coefficients