Leyda V. León Colón, Ph. D.
Electrical and Computer Engineering Department
Email: leyda.leon@upr.edu
1

I.
Tx Lines parameters
II.
Tx Lines Equations
III.
Input Impedance, SWR,
Power
I.
Smith Chart
2

!
Tx Lines have two conductors
( σ c
,µ c
, ε c
= ε o
) in parallel with a dielectric separating them ( σ ,µ, ε )
!
They transmit TEM waves inside the lines
3

Two-wire (ribbon)
Microstrip
4
Stripline (Triplate)
Coaxial

5

"
V proportional to E ,
"
I proportional to H
6

The parameters that characterize the TL are given in terms of per length.
7
•   R = ohms/m,
•   L = Henries/m
•   C = Farads/m
•   G = Siemens/m
Represent the losses in the conductors
Created by the magnetic field
Created by the electric field
Represent the losses in the dielectric between conductors

R, L, G , and C depend on the particular transmission line structure and the material properties. R, L, G, and C can be calculated using fundamental EMAG techniques.
8

9

•   Given V g
, Z g
, and Z
L
, find V
L
? Assuming λ = 20 m
Δ z ≈ λ /20 = 1m
# sections = l/ Δ z = 1000
V g
Z g
I in
+
V in
- v(z,t) , i(z,t)
+
V
L
-
Z
L
10 l=1000 m
•   We can't apply circuit analysis because the voltage propagates along the TL line at a certain speed
+
V
L
-
V
L
is affected by TL parameters
(losses and phase)
•   To find V
L
, we need to derivate equations for v(z,t) or i(z,t) using the TL model

Using KVL:
11

!
Taking the limit as
Δ z tends to 0 leads to
KVL
!
Similarly, applying Kirchoff’s current law (KCL)to the main node gives
KCL
12

!
Using phasors
V ( z , t ) = Re[ ˜ ( z ) e j ω t
]
I ( z , t ) = Re[˜ ( z ) e j ω t
]
!
The two expressions reduce to
€
−
∂
∂ z
€
= R
˜
+ j ω L
∂
2
∂ z
˜
2
− γ
2 ˜
= 0
−
∂
= G
˜
+ j ω C
∂ z
Wave Equation for voltage
€
13

!
Note that these are the wave eq. for voltage and current inside the lines. d
2 ˜
− γ dz
2
2 ˜
= 0 d
2 ˜
− γ
2 ˜
= 0 dz
2
!
The propagation constant is γ and the wavelength and velocity are
γ = α + β j = ( R + j ω L )( G + j ω C )
λ =
2 π
β u =
ω
β
= f λ
14
€

•   Defined as
15
Propagation constant
Where
▫   α = attenuation constant = Nepers/meters
▫   β = phase constant or wave numbers = radians/meters
•   Phase constant is related to
Wavelength wavelength
Wave velocity

16
!
The general solution is
˜
= V
+ e
− γ z
+ V
− e
γ z
!
In time domain is
V ( z , t ) = Re[ ˜ ( z ) e j ω t
]
= V
+ e
− α z cos( ω t − β z ) + V
− e
+ α z cos( ω t + β z ) z
!
Similarly for current, I
I ( z , t ) = I
+ e
− α z cos( ω t − β z ) + I
− e
α z cos( ω t + β z )
€

€
Z o
−
•   Is the ratio of positively traveling voltage wave to current wave at any point on the line d
˜
( z )
= ( R + j ω L )˜ ( z ) dz
˜
( z ) = V
+ e
− γ z
˜
( z ) = I
+ e
− γ z z
− ( − γ V
+ e
− γ z
) = ( R + j ω L ) I
+ e
− γ z
Z o
=
V
+
I
+
=
R + j ω L
γ
=
R + j ω L
G + j ω C
= R o
+ jX o
= −
V
I
−
−
Admitance:
Y o
=1/Z o
17

Has perfect conductors and perfect dielectric medium between them.
!
Propagation:
!
Velocity:
!
Impedance
18

Case #2: Distortionless Lines... R/L=G/C
Is one in which the attenuation is independent on frequency.
!
Propagation:
γ = α + j β
α = RG β = ω LC
!
Velocity: u =
ω
β
=
1
LC
= f λ
!
Impedance X o
= 0 Z o
= R o
=
L
C
=
R
G
€
19

General
Lossless
Distortionless
=
+ j
Z o
20

€
Define reflection coefficient at the load,
Γ
L
21
˜
( z ) = V
+ e
− γ z
+ V
− e
+ γ z
Γ
L
=
V
V
−
+

Terminated, Lossless Tx line
Then,
˜
( z ) = V
+ ( e
− γ z
+ Γ
L e
+ γ z )
Similarly,
€
˜
( z ) =
V
+
Z o
( e
− γ z
− Γ
L e
+ γ z )
The impedance anywhere along the line is given by
€
˜
( z ) =
˜
( z )
˜
( z )
= Z o
( e
− γ z
( e
− γ z
+ Γ
L e
+ γ z
− Γ
L e
+ γ z )
)
The impedance at the load end, Z
L
, is given by
€
22

Terminated, Lossless Tx Line
Then,
Conclusion: The reflection coefficient is a function of the load impedance and the characteristic impedance.
Recall for the lossless case,
Then
˜
( z ) = V
˜
( z ) =
V
+ ( e
+
− j β z
( e
− j β z
Z o
+ Γ
L e
+ j β z
− Γ
L e
+ j β z )
)
€
€
23

Terminated, Lossless Tx Line
It is customary to change to a new coordinate system, z = - l , at this point.
z z = - l
Rewriting the expressions for voltage and current, we have
24
Rearranging,
V ( − l ) = V
+ e
+ j β l (
1 + Γ
L e
− 2 j β l )
I ( − l ) =
V
+
Z o e
+ j β l (
1 − Γ
L e
− 2 j β l )
€

The impedance anywhere along the line is given by
The reflection coefficient can be modified as follows
Then, the impedance can be written as
After some algebra, an alternative expression for the impedance is given by
Conclusion: The load impedance is “transformed” as we move away from the load.
25

The impedance anywhere along the line is given by
The reflection coefficient can be modified as follows
Then, the impedance can be written as
After some algebra, an alternative expression for the impedance is given by
Conclusion: The load impedance is “transformed” as we move away from the load.
26

€
What if the length is
λ
/4?
(Quarter-wave)
Z in
=Z
0
Z o
’ chosen such that
Z
L l= λ /4
Z in tan (
= Z o
π /2 )
Z
L
Z o
=
+ jZ o
+ jZ
L tan
⎡
⎣
⎛ 2
⎜
⎝ λ
π tan
⎡
⎣
⎛ 2
⎜
⎝ λ
π
⎞
⎟
⎠
⎛
⎜
⎝
λ
4
⎞
⎟
⎠
⎤
⎦
⎞
⎟
⎠
⎛
⎜
⎝
λ
4
⎞
⎟
⎠
⎤
⎦
∞
IS frequency dependent!!!!
WHAT IF the length is λ /2?
Z in
=Z
L
Conclusion: **A piece of line of
λ
/4 can be used to change the impedance to a desired value (e.g. for impedance matching)
27

€
Whenever there is a reflected wave, a standing wave will form out of the combination of incident and reflected waves.
28
The (Voltage) Standing Wave Ratio - SWR (or VSWR) is defined as s = SWR =
V max
V min
=
I max
I min s =
1 + Γ
L
1 − Γ
L

•   The average input power at a distance l from the load is given by
•   which can be reduced to
•   The first term is the incident power and the second is the reflected power . Maximum power is delivered to load if
Γ =0.
29

•   Shorted Line ( Z
L
=0 )
Z ( l ) = Z in
= Z o
( Z
L
( Z o
+ jZ o
+ jZ
L tan β l ) tan β l )
•   Open-circuited Line ( Z
L
= ∞ )
€
Γ
L
=
Z
L
Z
L
− Z o
+ Z o
•   Matched Line ( Z
L
= Z o
) € s =
1 + Γ
L
1 − Γ
L
30
€

Shorted Line ( Z
L
=0 ), we had
•   So substituting in V(z)
Voltage maxima
|V(z)|
-z
λ
λ /2 λ /4
*Voltage minima occurs at same place that impedance has a minimum on the line 31

Open Line ( Z
L
= ∞ ) ,we had
•   So substituting in V(z)
Voltage minima |V(z)|
-z
λ
λ /2 λ /4
32

33
Matched Line ( Z
L
= Z o
), we had
•   So substituting in V(z)
-z
λ
λ /2 λ /4
|V(z)|

•   http://www.amanogawa.com/transmission.html
•   http://physics.usask.ca/~hirose/ep225/
•   http://www.educatorscorner.com/index.cgi?
CONTENT_ID=2483
34

35

•   Commonly used as graphical representation of a
Tx Lines.
•   Used in hi-tech equipment for design and testing of microwave circuits
•   One turn (360 o ) around the SC = to
λ
/2
36

37

•   Suppose you use as coordinates the reflection coefficient real and imaginary parts.
and define the normalized i
Z
L
:
38
| Γ |
Γ r

z=r+jx
•   After some algebra, we obtain two eqs.
Circles of r
Circles of x
•   Similar to general equation of a circle of radius a , center at (h,k)
39

40
Circles of r
Circles of x

41
Circles of r or
“Resistance Circles
Circles of x or
“Reactance Circles”

•   Numerically s=r on the +axis of Γ r
Proof:
in the SC
42

•   A lossless Tx Line is represented as a circle of constant radius, | Γ |, or constant s
To generator
•   Moving along the line from the load toward the generator, the phase decrease, therefore, in the SC equals to moves clockwisely.
43

•   One turn (360 o ) around the SC = to
λ
/2 because in the formula below, if you substitute length for half-wavelength, the phase changes by 2 π , which is one turn.
44

•   The Γ r
+axis, where r >0 corresponds to V max
•   The Γ r
-axis, where r <0 corresponds to V min
V min
V max (Maximum impedance)
45

•   Quarter Wave Transformer: Series
Connection
•   Single Stub Tuner: Parallel Connection
46

Quarter-wave Transformer:
What if the length is
λ
/4?
€
Z
' in
= Z o
Z o
’ chosen such that
Z
L l= λ /4
€
Z in tan (
= Z o
π /2 )
Z
L
Z o
=
+ jZ o
+ jZ
L tan
⎡
⎣
⎛ 2
⎜
⎝ λ
π tan
⎡
⎣
⎛ 2
⎜
⎝ λ
π
⎞
⎟
⎠
⎛
⎜
⎝
λ
4
⎞
⎟
⎠
⎤
⎦
⎞
⎟
⎠
⎛
⎜
⎝
λ
4
⎞
⎟
⎠
⎤
⎦
∞
Z
' in
=
Z
Z
IS frequency dependent!!!!
2 o
L
WHAT IF the length is λ /2?
Z in
=Z
L
€
Conclusion: **A piece of line of
λ
/4 can be used to change the impedance to a desired value (e.g. for impedance matching)
47

48
€
Z in
€
= Z o
Z o l
Z
' in
€
= Z o
Z
' o l= λ /4
€
Z in
€
= Z o
Z o
€ l
Z
' in
€
= Z o
Z
' o l= λ /4
€
Z=R
Z o d
€
Z
L
Z=R
Z
L
Z=R+jX

Y=1/Z
Y=G+jB !
y=g+jb
Z=R+jX !
z=r+jx

•   We work with Y, because in parallel connections they add.
•   Y
L
(=1/Z
L
) is to be matched to a TX line having characteristic admittance Y o
by means of a "stub" consisting of a shorted (or open) section of line having the same characteristic admittance Y o

51
Z in y in
’ =y o
= y d
+ y s
Z o
, γ l y in
’ y d y s l s d s
Z
L
SHORT or
OPEN

Y=G+jB !
y=g+jb
•   Enter the point representing the normalized load admittance y
L
.
•   Draw the constant | Γ | circle for y
L
, which will intersect g=1 circle at two points. At these points y d1
=1+jb d1
AND
y d2
=1-jb d2
. Both are possible solutions.
•   Determine position distance from the load d s
going from the load toward those two points. Choose shortest.
•   Determine the stub length l s
from the short or open circuit point (in Admittance) to the points representing the y jb d1
or y s2
= +jb d2 s1
= –
, depending which was chosen before.
•   NOTE! From the two possible solutions always choose the shortest! ☺

Solution 1
0.067
λ y oc y
L y s
@0.378
λ
-j0.96
y d1 y d2 z
L
0.160
λ
1) Normalize Z
L with Z o
=99.75-j80.25
=75 ΩΩ
53 z
L
= Z
L
/ Z o
= 1.33
− j 1.07
2) Transform to admittance y
€ sc
3) Move through the TL until it intercept the circle r=1. This point represents y’=1+jb y d 1
= 1 + j0.96
← y d 2
= 1 − j0.96
4) Measure length of TL connecting the load
€ d s d s
≈ 0.160
λ - 0.067
λ
= 0.093
λ
5) Represent stub susceptance
( y s
=-j0.96
) on the Smith C.
6) Measure length of the stub
For a short-circuited stub l sc
≈ 0.378
λ - 0.25
λ = 0.128
λ ←
For an open-circuited stub
€ with y stub
= -j0.96/75 ΩΩ = j .0128
mhos

Solution 2
0.067
λ y oc y
L
0.1215
λ
+j0.96
y d2 y d1 z
L
1) Normalize Z
L
54 z
L
= Z
L
/ Z o
= 1.33
− j 1.07
y
€
2) Transform to admittance y
L
= 0.46
+ j 0.367
€
3) Move through the TL until it intercept the circle r=1. This point represents y’=1+jb sc y d 1
= 1 − j0.96
← y d 2
= 1 + j0.96
4) Measure length of TL connecting the load
€ d s d s
≈ 0.34
λ - 0.067
λ
= 0.273
λ
5) Represent stub susceptance
( y s
=+j0.96
) on the Smith C.
6) Measure length of the stub
For a short-circuited stub
0.34
λ
€
For an open-circuited stub l oc
≈ 0.1215
λ - 0 λ = 0.1215
λ ← with y stub
= +j0.96/75 ΩΩ =+ j .0128
mhos

•   Used to measure frequency and load impedance l min-min
= λ /2
55
HP Network Analyzer in
Standing Wave Display
55

Given s , the distance between adjacent minima, and l min for an “air” 100
Ω transmission line, Find f and Z
L
•   s=2.4, l min
=1.5 cm, l min-min
=1.75 cm
#1
0.071
λ
Solution:
Wavelength
Frequency
Vmin z
L
Draw a circle on r =2.4, that’s your
T.L. move from V min
to z
L
0.429
λ
56
€ z
L
Z
L
= 0.5
+ j 0.38
= Z o z
L
= 50 + j 38 Ω
V min
= l min
= l load
0.5
λ − 0.429
λ = 0.071
λ
€

57
Microstrip lines are designed in PCB technology, it consists of a single ground plane and an open strip conductor separate by dielectric substrate

Matching Networks
58
High Frequency Applications
1:16 way power dividers
RF switches Antennas
Filters

EM Fields and Effective Relative Permittivity
•   Waves propagate using Transversal Electromagnetic Mode
(TEM) w h
59
ε r
Since the EM field is partially confined in the dielectric region and surrounding air, the TX line presents an effective relative permittivity
ε eff
=
( ε r
2
+ 1)
+
2 1
( ε
+ r
−
12
1) h / w
€

€
€
60
•   Characteristic impedance
Z o
=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
60
ε eff
1
ε eff
[ ln
⎛
⎜
8 h
⎝ w
+ w
4 h
⎞
⎟
⎠ w h w / h ≤ 1
120 π
+ 1.393
+ 0.667 ln( w / h + 1.444)
] w / h ≥ 1
•   TX parameters
β = w ε eff c u = c
ε eff
α ( dB / m ) ≅ α c
+ α d
8.686
=
€ σ c
δ wZ o
+ 27.3
( ε eff
− 1) ε
( ε r €
− 1) ε eff r tan θ
λ where and
λ is used to calculate the
TX length. Stub length is given as a function of λ

•   The w/h ratio necessary to achieve Z o
is given by where h is provided by substrate laminates
61

62