Transmission Line
Theory
EE142
Dr. Ray Kwok
Transmission Line - Dr. Ray Kwok
RF Spectrum
km
mm
µm
Å
Advanced Light Source
Berkeley Lab
Transmission Line - Dr. Ray Kwok
RF / Microwave Circuit
1 – port
2 wires
network
GND
input
source
2 – port
network
output
load
Transmission Line - Dr. Ray Kwok
Series connection
low f
A
B
A
B
RF
Transmission Line - Dr. Ray Kwok
Parallel connection
low f
A
B
RF
A
B
Transmission Line - Dr. Ray Kwok
Common transmission lines
Zo
l
most correct schematic
twisted pair
VLF
lossy & noisy
paralllel wire
LF - HF
noisy & lossy
coaxial cable
no distortion
wide freq range
microstrip (line)
no distortion
wide freq range
lowest cost
co-planar waveguide
low cost
flip chip access
complex design
waveguide
lowest loss
freq bands
Transmission Line - Dr. Ray Kwok
Equivalent circuit
i(z,t)
V(z,t)
i(z+∆z,t)
L∆z
Ideal transmission line
V(z+∆z,t)
C∆z
Taylor
Kirchhoff”s law: V (z, t ) − (L∆z) ∂i(z, t ) = V (z + ∆z, t ) ≈ V (z, t ) + ∂V( z, t ) ∆z
∂t
∂z
∂i(z, t ) ∂V(z, t )
−L
=
∂t
∂z
Junction rule:
i(z + ∆z, t ) − i(z, t ) = −(C∆z)
−C
∂V (z, t ) ∂i(z, t )
≈
∆z
∂t
∂z
∂V(z, t ) ∂i(z, t )
=
∂t
∂z
Q=CV
dQ/dt=i=C dV/dt
Transmission Line - Dr. Ray Kwok
Coupled equations (V – i)
−L
∂i(z, t ) ∂V (z, t )
=
∂t
∂z
−C
∂V(z, t ) ∂i(z, t )
=
∂z
∂t
∂ 2i ∂ 2 V ∂  1 ∂i 
1 ∂ 2i
−L 2 =
= −
=−
∂t
∂t∂z ∂z  C ∂z 
C ∂z 2
∂ 2i
∂ 2i
= LC 2
current wave
2
∂z
∂t
similarly
∂ 2 V ∂ 2i
∂  1 ∂V 
1 ∂ 2V
−C 2 =
= −
=−
∂t
∂t∂z ∂z  L ∂z 
L ∂z 2
∂ 2V
∂ 2V
= LC 2
2
∂z
∂t
voltage wave
Transmission Line - Dr. Ray Kwok
Wave equation
f ( x ± vt ) ≡ f (u )
reverse / forward traveling wave
∂f
∂u
= f ' (u )
= f ' (u )
∂x
∂x
∂ 2f
∂u
= f " (u )
= f " (u )
2
∂x
∂x
∂f
∂u
= f ' (u )
= ± vf ' (u )
∂t
∂t
∂ 2f
∂u
2
=
±
vf
"
(
u
)
=
v
f " (u )
2
∂t
∂t
∂ 2f
1 ∂ 2f
= 2 2
wave equation
2
∂x
v ∂t
note:
x ± vt =
1  2π
2π 
x
±
vt 

λ 
k λ
1
(kx ± 2πft )
k
1
= ± (ωt ± kx )
k
f ( x ± vt ) = f (ωt ± kx )
r r
f ωt − k ⋅ r
(3D)
=
(
)
Transmission Line - Dr. Ray Kwok
Voltage & Current Waves
+
o
V ( z, t ) = V e
j( ωt −βz )
− j( ωt + βz )
o
+V e
i(z, t ) = I o+ e j( ωt −βz ) − I o− e j( ωt +βz )
why “-” ?
2π
λ
1
 λ  ω
v = fλ = (2πf )  = =
LC
 2π  β
∂i
= − jβI o+ e j( ωt −βz ) − jβI o− e j( ωt +βz )
∂z
∂i
∂V
= −C
= −C jωVo+ e j( ωt −βz ) + jωVo− e j( ωt +βz )
∂z
∂t
β I o+ = CωVo+
[
β=
where
]
β I = CωV
v=
1
LC
β ±
LC ±
L ±
V =
Io =
Io =
I o ≡ Zo I o±
Cω
C
C
Zo =
L
C
−
o
±
o
−
o
Transmission Line - Dr. Ray Kwok
Fields and circuits
r
r
2
 
∂ E
2 E
∇  r  = µε 2  r 
∂t  H 
H
r
r j( ωt − kr ⋅rr ) r j( ωt − kr ⋅rr )
i
r
E(z, t ) = E oi e
+ E or e
r
r j( ωt − kr ⋅rr ) r j( ωt − kr ⋅rr )
i
r
H(z, t ) = H oi e
− H or e
∂2  V 
∂2  V 
  = LC 2  
2 
∂z  i 
∂t  i 
V(z, t ) = Vo+ e j( ωt −βz ) + Vo− e j( ωt +βz )
i(z, t ) = I o+ e j( ωt −βz ) − I o− e j( ωt +βz )
1
v=
µε
µ
η=
ε
v=
1
LC
L
Zo =
C
Transmission Line - Dr. Ray Kwok
What is Zo?
•
•
•
•
•
•
•
Characteristic Impedance.
50 ohms for most communications system,
75 ohms for TV cable.
Measure 75 ohms with a ohmmeter?
Two 75Ω cables together (in series) makes a 150Ω cable?
75 + 75 = 75 !!!!
What does Zo represent?
Transmission Line - Dr. Ray Kwok
Reflection at Load
V( x ) = Vo+ e − jβx + Vo− e jβx
Zo
l
x = −l
ZL
+ − jβ x
o
i( x ) = I e
−
o
−I e
jβ x
V (0) ≡ VL = Vo+ + Vo−
x=0
Define normalized impedance
at the load
(
1
Vo+ − Vo−
i(0) ≡ I L = I − I =
Zo
+
o
−
o
 Vo+ + Vo− 
VL

≡ Z L = Zo  +
− 
IL
 Vo − Vo 
(
)
(
Z L Vo+ − Vo− = Zo Vo+ + Vo−
)
)
Z
Z≡
Zo
Vo+ (Z L − Zo ) = Vo− (Z L + Zo )
ZL − 1
ΓL =
ZL + 1
Vo−
Z L − Zo
≡
Γ
=
L
Vo+
ZL + Zo
ZL ≠ Zo
reflection
Transmission Line - Dr. Ray Kwok
Example
does it work?
75Ω
50Ω
75Ω
Transmission Line - Dr. Ray Kwok
Impedance at Input
Zin
x = −l
Zo
Vin V (−l)
Vo+ e jβl + Vo− e − jβl
=
=
Zin ≡
1
I in
i ( −l )
Vo+ e jβl − Vo− e − jβl
Zo
(
l
ZL
x=0
V( x ) = Vo+ e − jβx + Vo− e jβx
i( x ) = I o+ e − jβx − I o− e jβx
)
 e jβl + ΓL e − jβl 

Zin = Zo  jβl
− jβ l 
 e − ΓL e 
 1 + j tan βl   ZL − 1  − jβl
e
 + 
e − jβl 
 1 − j tan βl   ZL + 1 
Zin =
 1 + j tan βl   ZL − 1  − jβl
e
 − 
e − jβl 
 1 − j tan βl   ZL + 1 
(Z + 1)(1 + j tan βl ) + (Z
(Z + 1)(1 + j tan βl ) − (Z
2(Z + j tan βl )
=
2(1 + jZ tan βl )
Zin =
L
L
Zin
L
L
Zin =
ZL + j tan βl
1 + jZL tan βl
 Z + jZ o tan βl 

Zin = Zo  L
 Zo + jZ L tan βl 
− 1)(1 − j tan βl )
L − 1)(1 − j tan β l )
L
Transmission Line - Dr. Ray Kwok
Exercise
Zin
Zo
l
x = −l
ZL
x=0
ZL + j tan βl
Zin =
1 + jZL tan βl
Zo = 50 Ω
ZL = 100Ω
Zin = ?
For length = λ/8? λ/4? λ/2?
What if Zo = ZL = 50Ω?
Would the length make any difference?
 Z + jZo tan βl 

Zin = Zo  L
 Zo + jZ L tan βl 
50Ω (-37o)
25Ω
100Ω
Zin=Zo=ZL
Transmission Line - Dr. Ray Kwok
Transmission Line Impedance
Zin
Zo
l
x = −l
Zin =
ZL
x=0
ZL + j tan βl
1 + jZL tan βl
 Z + jZo tan βl 

Zin = Zo  L
 Zo + jZ L tan βl 
case 1: β l = 0, or l = 0
tan β = 0
Zin = ZL
case 2: β l = π, or l = λ/2
tan β l = 0
Zin = ZL
case 3: β l = π/2, or l = λ/4
tan β l → ∞
Zin = Zo2/ ZL
Quarter-wave transformer (impedance),
real-to-real, complex-to-complex.
note: at low freq, β → 0, Zin = ZL regardless of line length or line impedance.
Transmission Line - Dr. Ray Kwok
Reflection at Input
ZL + j tan βl
Zin =
1 + jZL tan βl
Z’o
Zin
Zo
l
ZL
Γin
x = −l
x=0
In general
 Z L + jZ o tan βl 

Zin = Zo 
 Zo + jZ L tan βl 
Zin − Z'o Zin' − 1
Γin =
= '
'
Zin + Zo Zin + 1
Zin − Zo Zin − 1
Γin =
=
Zin + Zo Zin + 1
just have to know what Z to use
Transmission Line - Dr. Ray Kwok
Exercise
Z’o
Zin
Zo
l
ZL
Γin
x = −l
Zin =
x=0
ZL + j tan β l
1 + jZL tan βl
 Z + jZo tan βl 

Zin = Zo  L
 Zo + jZ L tan βl 
Zin − Z'o Zin' − 1
Γin =
= '
'
Zin + Zo Zin + 1
Zo = 50 Ω
Z’o = 50 Ω
ZL = 100Ω
Length = λ/8
ΓL = ? Γin = ?
What if Z’o is 75 Ω?
1/3
o
1/3 (-90 )
only change phase !?!
0.388 (235o)
Transmission Line - Dr. Ray Kwok
Voltage wave in transmission line
V( x ) = Vo+ e − jβx + Vo− e jβx
Zo
Zin
l
ZL
(
V( x ) = Vo+ e − jβx 1 + ΓL e 2 jβx
)
V = Vo+ 1 + ΓL e 2 jβx
x = −l
ΓL ≡ ρe jθ
x=0
V = Vo+ 1 + ρe j( θ+ 2βx )
V = Vo+
V = Vo+ 1 + 2ρ cos(θ + 2β x ) + ρ 2
min when sine = 1
(1 + ρ)
+
o
2
Vmin = V
− 4ρ = V (1 − ρ)
+
o
max when sine = 0
+
o
Vmax = V
(1 + ρ cos(θ + 2βx ) )2 + ρ2 sin 2 (θ + 2βx )
(1 + ρ)
2
= V (1 + ρ)
+
o
V = Vo+
(1 + ρ)2 − 2ρ(1 − cos(θ + 2βx ) )
V = Vo+
(1 + ρ)2 − 4ρ sin 2  θ + 2βx 

2

Transmission Line - Dr. Ray Kwok
Voltage Standing Wave
V( x ) = Vo+ e − jβx + Vo− e jβx
Vmin
Vmax
ZL
standing wave
Vo+ = Vo− , ΓL ≡ ρ = ±1
If
perfect standing wave with nodes
x = −l
x=0
V = Vo+
min when
(x < 0)
θ + 2βx
(2n + 1)π
=±
2
2
λ
x = −[θ m (2n + 1)π]
4π
(1 + ρ)2 − 4ρ sin 2  θ + 2βx 
max when

2
θ + 2βx
= ± nπ
2
λ
x = −[θ m 2nπ]
4π

Transmission Line - Dr. Ray Kwok
VSWR (Voltage Standing Wave Ratio)
Vmin
Vmax
ZL
V = Vo+
(1 + ρ)2 − 4ρ sin 2  θ + 2βx 
Vmin = Vo+
x = −l
x=0
Vmax = Vo+

2

(1 + ρ)2 − 4ρ = Vo+ (1 − ρ)
(1 + ρ)2
= Vo+ (1 + ρ)
Vmax 1 + ρ 1 + Γ
VSWR ≡
=
=
Vmin 1 − ρ 1 − Γ
perfect match: ρ = 0, VSWR = 1.0
open / short: ρ = 1, VSWR → ∞
It is an indicator on how well the load matches the line.
VSWR is the standing wave pattern INSIDE the line.
Only Γ at the reflected junction that counts
Transmission Line - Dr. Ray Kwok
Exercise
Z’o
Zin
Zo
l
ZL
Γin
x = −l
x=0
Zo = 50 Ω
Z’o = 75 Ω
ZL = 100Ω
Length = λ/8
VSWR = ?
ΓL = 1/3
VSWR = 2
Transmission Line - Dr. Ray Kwok
Return Loss
RL ≡ − 20 log ρ
(dB)
perfect match: ρ → 0, VSWR → 1.0, RL → ∞
open / short: ρ = 1, VSWR → ∞, RL → 0 dB
1+ ρ 1+ Γ
VSWR ≡
=
1− ρ 1− Γ
Typical VSWR = 1.1 to 2
ρ = 0.048 to 0.33
RL = 26 dB to 9.5 dB
VSWR − 1
ρ= Γ =
VSWR + 1
Transmission Line - Dr. Ray Kwok
Stub
Transmission line connecting nowhere(?)
Open stub
Short stub
Series stub
Shunt stub
(short)
(short)
Transmission Line - Dr. Ray Kwok
Open Shunt Stub
L-Band
Transmission Line - Dr. Ray Kwok
Short Shunt Stub
20 GHz
Interdigital
Filter
Transmission Line - Dr. Ray Kwok
Radial
Stub
18 GHz
Rat Race
Transmission Line - Dr. Ray Kwok
Tuning stub (open)
Transmission Line - Dr. Ray Kwok
Short Stub
ZL + j tan β l
1 + jZL tan βl
Zin =
Zo
Zin
l
ZL
(ZL → 0) Zsh = j tan βl
Ysh ≡
Zin
1
= − j cot βl
Zsh
Zcoil = jωL
Zcap = -j/ωC
1 

Zsres = jωL1 − 2 
 ω LC 
π/2
π
3π/2
2π
period of π
βl
Z pres =
1
1 

jωC1 − 2 
 ω LC 
Transmission Line - Dr. Ray Kwok
Open Stub
Zo
Zin
l
Zin =
ZL
(ZL → ∞)
ZL + j tan βl
1 + jZL tan βl
Zop = − j cot βl
Yop = j tan βl
Zin
Zcap = -j/ωC
Zcoil = jωL
1 

Zsres = jωL1 − 2 
 ω LC 
π/2
π
3π/2
2π
period of π
βl
Z pres =
1
1 

jωC1 − 2 
 ω LC 
Transmission Line - Dr. Ray Kwok
Exercise
Find Zin & Γin.
75 Ω, λ/8
Zin
Zsh = j Zotan(βl)
o
= j 75 tan(45 ) = j 75 Ω
50 Ω
λ/2
100 Ω
0.1λ
Zop = -j Zocot(βl)
o
= -j 100 cot(36 ) = -j 138 Ω
Y =1/j75 = -j 0.0133
Zin
50 Ω
λ/2
Y = j/138 = j0.0073
Z = 1/Y = 1/(-j0.006) = j166 Ω
Zin
50 Ω
λ/2
Zin =
ZL + j tan βl
= ZL
1 + jZL tan βl
Zin = j 166 Ω
Zin − Zo j166 − 50
Γin =
=
= 1∠(107 o − 73o ) = 1∠34o
Zin + Zo j166 + 50
Transmission Line - Dr. Ray Kwok
Admittance (Y = 1/Z)
Z L + j tan β l
Zin =
1 + jZ L tan β l
Yin ≡
1
1 + jZ L tan β l
=
Zin
Z L + j tan β l
Yin =
1 + j(1 / YL ) tan β l
1 / YL + j tan β l
YL + j tan β l
Yin =
1 + jYL tan β l
 YL + jYo tan β l 

Yin = Yo 
 Yo + jYL tan β l 
Yin
Y1
Yo
l
YL
Γin
Zin − Z1
Γin =
Zin + Z1
1 / Yin − 1 / Y1
Γin =
1 / Yin + 1 / Y1
Y1 − Yin 1 − Yin
Γin =
=
Y1 + Yin 1 + Yin
useful for shunt circuits
Transmission Line - Dr. Ray Kwok
Earlier exercise
Find Zin & Γin.
75 Ω, λ/8
Zin
Ysh = - j Yocot(βl)
o
= - j (1/75) cot(45 ) = - j 0.0133
50 Ω
λ/2
100 Ω
0.1λ
Yop = j Yotan(βl)
o
= j 0.01 tan(36 ) = j 0.0073 Ω
Y = -j 0.0133
Zin
50 Ω
λ/2
Zin
Y = j0.0073
Y = - j 0.006
50 Ω
λ/2
Yin =
Γin =
YL + j tan βl
= YL
1 + jYL tan βl
YL = - j 0.006
Zin = j 166 Ω
Yo − Yin 1 / 50 + j0.006
=
= 1∠(17 o + 17 o ) = 1∠34 o
Yo + Yin 1 / 50 − j0.006
Transmission Line - Dr. Ray Kwok
Earlier Exercise – power consideration
Z’o
Zin
Zo
l
ZL
Γin
x = −l
Z + j tan βl
Zin = L
1 + jZL tan βl
 Z + jZo tan βl 

Zin = Zo  L
 Zo + jZ L tan βl 
Zin − Z'o Zin' − 1
Γin =
=
Zin + Z'o Zin' + 1
x=0
Zo = 50 Ω
Z’o = 50 Ω
ZL = 100Ω
Length = λ/8
o
ΓL = ?1/3 Γin = ?1/3 (-90 )
only change phase
Power reflected = ?
V−
V+
2
2
1
=Γ =
= 11%
3
2
2
Power delivered = ? 1 − Γ = 89%
Don’t double count reflection…. ΓL & Γin
Return Loss (RL) = - 20 log|ρ| = + 9.5 dB
Transmission Line - Dr. Ray Kwok
High f circuit elements
1 GHz lumped element
Band pass filter
CAP
CAP
CAP
CAP
IND
IND
CAP
ID=
C2ID=
ID=
C3C4
IND
ID=
C6
ID=
L3
CAP
ID=
ID=
C1
1L7pF
ID=
C=
1L4
pF
C=
1pF
pF
C=
1C=
ID=
C5
L=
1
nH
L=1 nH
C=
1 1pF
L=
nH
C=1 pF
…..
…..
12 GHz lumped element
Low pass filter
much smaller
…..
A small loop of thin wire is an inductor !!
CAP
IND
IND
CAP
IND
IND
CAP
ID=
C4
ID=
L1
ID=
L7
ID=
C6
ID=
L3
ID=
L4
ID=
C1
C=
1 pF
L=L=
1L=
nH
C=
nH
pF
1pF
111nH
nH
C=1L=
…..
Transmission Line - Dr. Ray Kwok
High-Z Line as inductor
Zin
Zo
Z1
l
ZL
Γin
 Z L + jZ1 tan β l 

Zin = Z1 
 Z1 + jZ L tan β l 
Z1 >> ZL
line length < λ/4 (π/2)
ZL ~ Zo (order of magnitude)
 a∠ + Ψ 
 = Zin ∠ + θ
Zin = Z1 
 b∠ + ϕ 
L∆z
C∆z
small C, large L, series inductor
Zin has a positive phase
→ inductor-like !!!
Transmission Line - Dr. Ray Kwok
Low-Z Line as capacitor
Zin
Zo
Z1
l
ZL
Γin
 Z L + jZ1 tan β l 

Zin = Z1 
 Z1 + jZ L tan β l 
L∆z
C∆z
small L, large C, shunt capacitor
Z1 << ZL
line length << λ/4 (π/2)
ZL ~ Zo
 a∠ + ϕ 
Zin = Z1 
 = Zin ∠ − θ
 b∠ + Ψ 
Yin = Yin ∠ + θ
Yin has a positive phase
→ capacitor-like !!!
Transmission Line - Dr. Ray Kwok
Low pass filter
5 GHz low pass filter
…..
CAP
IND
IND
CAP
IND
IND
CAP
ID=
C4
ID=
L1
ID=
L7
ID=
C6
ID=
L3
ID=
L4
ID=
C1
C=
1 pF
L=L=
1L=
nH
nH
C=
pF
1pF
111nH
nH
C=1L=
…..
14 GHz low pass filter
high-low impedance lines
waveguide
high power
low loss
Transmission Line - Dr. Ray Kwok
High-Low-Z lines
20 GHz band pass filter
high Z lines → inductors
Short shunt stubs λ/4 resonators
…..
…..
IND
IND
CAP
IND
CAP
CAP
IND
ID=
L2
IND
IND
IND
ID=
L4
ID=
C6
ID=
C1
L3
ID=
C4
ID=
L7
L5
ID=
L6
ID=
L1
L=
1L=
nH
L=
1nH
nH
C=
1
1pF
pF
C=
pF
L=
nH
L=
L=
1111
nH
nH
L=
nH
13 GHz coupler
Tuning with stubs (shunt open)
Think of them as shunt capacitors
→ low Z lines
Transmission Line - Dr. Ray Kwok
Homework
1. A 100 Ω tranmission line has an effective dielectric constant of 1.65. Find the
shortest open-circuited length of this line that appears at its input as a capacitor
of 5 pF at 2.5 GHz. Repeat for an inductance of 5 nH.
2. A radio transmitter is connected to an antenna having an impedance 80 + j40
Ω with a 50 Ω coaxial cable. If the 50 Ω transmitter can deliver 30 W when
connected to a 50 Ω load, how much power is delivered to the antenna?
3. A 75 Ω coaxial transmission line has a length of 2 cm and is terminated with a
load impedance of 37.5 + j75 Ω. If the dielectric constant of the line is 2.56 and
the frequency is 3 GHz, find the input impedance to the line, the reflection
coefficient a the load, the reflection coefficient at the input, and the SWR on the
line.
4. The VSWR on a lossless 300 Ω transmission line terminated in an unknown
load impedance is 2.0, and the nearest voltage minimum is at a distance 0.3λ
from the load. Determine (a) ΓL, (b) ZL.
Transmission Line - Dr. Ray Kwok
5. Calculate VSWR, ρ, and return loss
values to complete the entries in the
following table.
VSWR
ρ
RL (dB)
1.00
0.00
∞
1.01
0.01
1.05
32
30
1.10
1.20
0.10
1.50
10
6. Measurements on a 0.6 m lossless
coaxial cable at 100 kHz show a
capacitance of 54 pF when the cable is
open-circuited, and an inductance of
0.30 µH when it is short-circuited. (a)
Determine Zo and εr of the medium.
2.00
2.50