Transcript
CHAPTER 1
INTRODUCTION of TL
DEFINITION: Metallic conductor system to transfer
electrical energy (EM signal) from one point to another.
Includes: i. Telephone wires, ii. Coaxial cables carrying audio and video
information to TV or digital data to computer monitors.
TL is two-port Network
A transmission line is a two-port network connecting a generator circuit at the sending end to a load at the receiving end.
Guided vs Unguided
Can be classified into two:
i. Guided
ii. Unguided
EM Waves
Self-propagating, mutual oscillations of electric and magnetic field (TEM)
Propagation of EM waves often referred as radiation.
EM Spectrum
TEM
Properties of EM WavesTaken from: eesc.columbia.edu/.../radiation/em_energy.html
GUIDED TL
EM waves are guided in the conductor to prevent signals from radiating away
For example:
i. Parallel Conductor
ii. Coaxial cable
iii. Waveguides
iv. Fibre Optic Cable
PARALLEL CONDUCTOR (PC)
Two or more conductor separated by nonconductive dielectric (air, rubber, polyethylene, etc).
The most common are open-wire, twin lead and twisted pair (UTP and STP).
PC: OPEN WIRE
Two wire parallel conductor, closely space, separated by air.
Nonconductive spacers is placed at periodic intervals for support and keeping separation distance constant.
No shielding cause high radiation losses, easily picking up signals through mutual induction causing cross talk.
Open wire parallel transmission line
PC: OPEN WIRE
PC: TWIN LEAD
Similar to open wire except the spacers are replaced with continuous solid dielectric ensuring uniform spacing along the entire cable.
Example of twin lead is the black, flat cable used to connect rooftop antenna to television set.
Common dielectric materials are Teflon and polyethylene.
PC: TWIN LEAD
(a) (b)
(a) Twin lead (b) 300 twin lead used for television antenna
PC: TWISTED PAIR
Formed by twisting two insulated conductor around each other.
Often stranded by units, and the units are then cabled into cores containing up to 3000 pairs of wire.
Cores are covered by various types of sheaths forming cables.
Reduce crosstalk between cable pairs. Two type, unshielded twisted pair (UTP) and
shielded twisted pair (STP).
PC: TWISTED PAIR
(a) twisted pair (b) four insulated twisted pairs (UTP) (c) two insulated twisted pair with shielding (STP) (d) example of twisted pair
(a)
(b)
(c)
(d)
Various type of twisted pair
Twisted Pair Ethernet Cable
The 10BASE-T system operates over two pairs of wires, one pair used for receive data signals and the other pair used for transmit data signals. The two wires in each pair must be twisted together for the entire length of the segment.
COAXIAL CABLES
Useable up to 110 GHz. Center conductor surrounded by a dielectric
material (insulation), then a concentric (uniform distance from the center) shielding and rubber outer jacket.
Shielding refers to the woven or stranded mesh (or braid) that surrounds some type of coaxial cables.
Center conductor carries signal, outer shield is the signal return (ground). Braid conductors provides excellent shielding against external interference.
Two basic types; rigid air-filled or solid flexible with the latter is more expensive to manufacture.
Anatomy of Coaxial Cable
COAXIAL CABLEILLUSTRATIVE DIAGRAM
Taken from: www.smarthomepro.com/851082.html
RG refers to Radio Guide while U for Universal (multiple uses).
Coaxial cable (a) rigid air-filled (b) solid dielectric (c) example in the market.
Coaxial Cables in the market
Coaxial cable can have characteristic impedance from 50 to 200.
Connectors
The ends of coaxial cables are connected to connectors.
Typical RF/Microwave connectors are: BNC (Baby N Connector) for d.c. to 300MHz N Connector for d.c. to 6GHz 3.5mm or SMA connectors for d.c. to 20GHz 2.5mm connectors for d.c. to 40GHz
BNC Connector for RG58
N Connector
N Series coaxial connectors are medium-sized, threaded coupling connectors designed for use from DC to 12.4 GHz. Their consistently low broadband VSWR have made them popular over the years in many applications. The N series connector is impedance matched to 50 ohm cables.
Cable terminations are available in crimp, clamp and solder configurations. The threaded coupling ensures proper mating in applications where shock and extreme vibration are design considerations.
N connectors are used in aerospace, broadcast audio and video applications as well as many microwave components such as filter, couplers, dividers, amplifiers and attenuators.
WAVEGUIDES
DO YOU KNOW?
The stethoscope that a doctor uses to listen to sounds in your chest is a waveguide for sound waves.
Familiar Waveguides: Stethoscope
COMMON TYPES OF TL
NOtE ThAt:
ONLY TEM TL COVERED IN THIS CHAPTER 1
HIGHER-ORDER TL (WAVEGUIDES) WILL BE COVERED IN CHAPTER 2
Propagation Modes
TL can be classified into two basic types:
i. TEM TL
(E and H perpendicular to each other. Thus, no component in direction of propagation)
ii. Higher-order TL (Waveguides!)
(One significant field components in the direction of propagation)
TEM TL
In a coaxial line, the electric field lines are in the radial directionbetween the inner and outer conductors and the magnetic field forms circles around the inner conductor.
Factors to consider when choosing the suitable transmission line
Operating frequency and bandwidth Acceptable loss Input power Physical structure Impedance (Characteristic Impedance, Zo)
UNGUIDED TL
Unguided transmission media are wireless system
Medium: Air or Vacuum (Earth’s atmosphere) The direction of propagation of the unguided
signals depends on the direction at which the signals are emitted.
The signals are then vulnerable to any obstacles while propagating.
Receiver is device to accept the signal
UNGUIDED TL
Balanced vs Unbalanced TL
A balanced transmission line is one whose currents are symmetric with respect to ground so that all current flows through the transmission line and the load and none through ground.
Note that line balance depends on the current through the line, not the voltage across the line.
An example of a Balanced Line Here is an example of a balanced line. DC rather than AC is used to simplify the analysis:
Notice that the currents are equal and opposite and the that the total current flowing through ground = 25mA-25mA = 0A
I = 25 mA
V = +6 VDC
6 V
6 V
V = -6 VDCI = -25 mA
Another Balanced Line Example
Here is another example:
Note that the total current flowing through ground is again 0A Because the ground current is 0, the ground is not required.
I = 25 mA
V = +9 VDC
V = -6 VDCI = -25 mA
Another Balanced Line Example
Here is another example. Is the line balanced?
No – although the voltages are equal and opposite, the currents are not!
V = +6 VDC
V = -6 VDC
I = 20 mA
I = -25 mA
BALUNS
How they work? How they are made?
What is a balun?
A Balun is special type of transformer/TL that performs two functions: Impedance transformation Balanced to unbalanced transformation
The word balun is a contraction of “BALanced to UNbalanced transformer”
Why do we need a balun? Baluns are important because many types of antennas
(dipoles, yagis, loops) are balanced loads, which are fed with an unbalanced transmission line (coax).
Baluns are required for proper connection of parallel line with a 50 ohm (not fixed) unbalanced output to a transceiver.
The antenna’s radiation pattern changes if the currents in the driven element of a balanced antenna are not equal and opposite.
Baluns prevent unwanted RF currents from flowing in the “third” conductor of a coaxial cable.
Real Life Scenario
Most common baluns are the one used to connect television with the antenna, i.e. to convert from 300 transmission line (flat, paralled twin lead from a folded dipole) to 75 (coaxial cable RG6 or television set) coaxial cable of the television set.
Impedance and balanced to unbalanced conversion is essential to avoid signal loss and reflection in the transmission line.
Note that:
Coaxial cable is unbalanced TL! However, it can be balanced as well
provided that neither wire is connected at ground potential.
The shield is generally connected to ground to prevent static interference from penetrating the center conductor.
Other: Real Life Scenario
In television, amateur radio, and other antenna installations and connections, baluns convert between 300 ohm ribbon cable (balanced) and 75 Ω coaxial cable (unbalanced) or to directly connect a balanced antenna to (unbalanced) coax. To avoid EMC problems it is a good idea to connect a centre fed dipole antenna to coaxial cable via a balun. Match 300 Ω twin-lead cable to 75 Ω coaxial cable
In electronic communications, baluns convert Twinax cables to Category 5 cables, and back, or they convert between coaxial cable and ladder line. Baluns can be used to convert video coming from an S-video, RCA or VGA connector to run over Cat5 cables.
Baluns are present in radars, transmitters, satellites, in every telephone network, and probably in most wireless network modem/routers used in homes. It can be combined with transimpedance amplifiers to compose high voltage amplifiers out of low voltage components.
Taken from: http://en.wikipedia.org/wiki/Balun
300 to 75 TV balun
300Ω Twin-lead from Antenna
75Ω Coaxial Cable of TV Set
BALANCED TL
UNBALANCED TL
Voltage Baluns
A Voltage Balun is one whose output voltages are equal and opposite (balanced with respect to ground).
True balance occurs only if the balun’s load is symmetric with respect to ground.
Voltages baluns are easily constructed and commonly used in spite of their inability to provide true current balance.
4:1 Transformer Voltage Balun This is the simplest
voltage balun, consisting of two coils of wire connected as shown.
The coils may use an air core or a ferrite core.
Current flowing through the lower coil induces an equal and opposite voltage in the upper coil.
4 Z BALANCED
Z UNBALANCED
The primary circuit contains N turns and the secondary 2N, so the input impedance is ZL(N/2N)2= ¼ ZL
4:1 TL Voltage Balun
This voltage balun is constructed solely from transmission line and requires to cores.
Unlike the transformer-type baluns, this balun may be used only over a narrow range of frequencies.
The extra half wave section causes the voltage at its output to be equal and opposite to the voltage at the input
TL EQUATIONS
At microwave frequencies the line parameters are distributed along the whole length of the line in the z-direction.
A small length of the line can be represented by an equivalent symmetrical T-network with constant parameters R, L, G, and C per unit length
Lumped-Element Model
From previous figure:
(a) Regardless of its actual shape, a TEM TL is represented by the parallel-wire configuration.
(b) To analyze the voltage and current relations, the line is subdivided into small differential sections (SDS)
(c) Each SDS is then represented by an equivalent circuit
Distributed element in a TL model
The Line Parameters (Differential Quantities – unit per length)
R or R’ : the combined resistance of both conductors per unit length (Ω/m)
L or L’ : the combined inductance of both conductors per unit length (H/m)
G or G’ : conductance of the insulation medium per unit length (S/m)
C or C’ : capacitance of the two conductors per unit length (F/m)
For each line, the conductors are characterized by c, c, c = 0 and the homogeneous (the same at all points) dielectric separating the conductors is characterized by , , .
The value of L is the external inductance per unit length; that is, L = Lext. The effects of internal inductance Lin (R/) are negligible at high frequencies at which most communication systems operate.
For each line, and
LC G
C
Cross-section of Coaxial Line
Skin-depth, δ Concept
What is Skin-depth?
Skin depth concept rises from the skin effect – a phenomenon where at RF and microwave frequencies, current tends to flow only in the surface of a conductor
Neither the electric field (E) nor the magnetic field (H) penetrate far into a "good" conductor.
The point where these fields are reduced by a factor of 1/e ~ 1/2.71 is called the skin depth.
Figure shows a good conductor and how a pulse traveling along this conductor is attenuated going into the conductor.
Skin effect: skin depth decreases with increasing frequency!
Skin Effect on a Round Conductor
Skin effect: skin depth decreases with increasing frequency!
How to minimize the losses from skin effect?
The surface of the conductor must be smooth and the edges must be well-defined
TRANSMISSION CHARACTERISTICS
Also called the secondary constants Determined from the four primary
constants. R,L,G and C The secondary constants are:
Characteristic impedancePropagation constant (coefficient)
Questions!
What is TL? Guided vs Unguided? Examples of Guided/Unguided? EM waves? TEM? Properties of EM waves? Balanced/Unbalanced line? Balun? Distributed line parameters? Skin effect phenomena? Skin depth?
CHARACTERISTIC IMPEDANCE, Zo
It is defined as the impedance seen looking into an infinitely infinitely long line (1) or the impedance seen looking into a finite lengthfinite length of line that is terminated in a purely resistive load equal to the characteristic impedance of the line (2)
Transmission line stores energy in its distributed inductance and capacitance if the line is infinitely long, it can store energy
indefinitely, i.e. energy from the source is entering the line and none is returned
the line acts as a resistor that dissipates all the energy
Z seen into an infinitely long line (1)
Z1=110Ω Z2=62.38Ω Z3=48.32Ω Z4=42.62Ω
It can be seen that after each additional section, the total impedance looking into the line decreases until asymptotically toward 37Ω which is the Zo of the line.
Z seen into a finite line (2)
If the TL above is terminated in ZL= 37Ω, the impedance looking into any number of sections (single section) would be equal to 37Ω.
ZL= 37Ω
ASSUME THE LINE IS FINITE!
Zo
The ratio of V(z)/I(z) is called the characteristic impedance of the transmission line provided that the reflected wave in the line is zero or in other word, it can said as the ratio of positively travelingpositively traveling voltage wave to current wave at any point on the line.
0
V zZ
I z Forward Voltage
Forward Current=
Equivalent circuit for a single section of transmission line terminated in a load equal to Z0
Zo (in terms or R’,L’,G’ and C’
Zo (in terms or R’,L’,G’ and C’
From figure, a single section of a transmission line is terminated in ZL that equal to Z0. Impedance looking into a line of n such sections is determined from the following expression:
For an infinite number of sections ZL2 / n approaches
0 if:2lim 0L n
Z
220 1 2 where = number of sectionsLZ
Z Z Z nn
Thus:
Where:
Therefore:
0 1 2Z Z Z
1
22
2
1 1 1
1
1
S
Z R j L
YZ R j C
G j C
ZG j C
0 0 0 0
1 ( )( ) or
R j LZ R j L Z R jX
G j C G j C
Zo (in terms or R’,L’,G’ and C’
Zo
Where Ro and X0 are the real and imaginary part of Z0 .
But you have to remember, R0 should not be mistaken as R as their units are different.
The unit for R0 is ohms while R is in ohms per meter. The reciprocal of Z is the characteristic admittance Y
and that is
00
1 .Y Z
Characteristic impedance can be calculated either by using the primary constant (R’,L’,G’,C’)using the primary constant (R’,L’,G’,C’) of the line or by the geometry size and spacinggeometry size and spacing of conductors and by the dielectric constant of the insulatordielectric constant of the insulator separating them.
Zo
Geometery of (a) parallel wire (b) coaxial cable
For a parallel wire line
For a coaxial cable
0
2276log
DZ
d= W
0
138log
r
DZ
de=
where: D : distance from center of the two conductorsd : diameter of the conductor
where: r : dielectric constant of the dielectric materiald: diameter of the inner conductorD: diameter of the outer conductor
Zo
Zo Parallel Wire
Zo Coaxial Cable
Zo
Microstrip line Characteristic impedance for a microstrip line depend on its physical
structure. Impedance of 50 or 200 can be achieved with microstrip line by simply changing its dimensions.
However, every configuration of microstrip or a stripline has its own unique formula.
Characteristic impedance of an unbalanced microstrip TL:
where w = width of copper trace
t = thickness of copper trace
h = distance between copper trace and ground plane
0
87 5.98ln
0.81.41r
hZ
w t
Printed Circuit Board (PCB) TL
FAQ’S A question that is often asked is what the significance of a 50,
75 and 300 characteristic impedance. The best coaxial cable impedances to use in high-power,
high-voltage, and low-attenuation applications were experimentally determined in 1929 at Bell Laboratories to be 30, 60, and 77 respectively.
30 cable is exceedingly hard to make however, so a compromise between 30 and 60 was reached at 52 (which lead to 5050), which has persisted.
73 is an exact match for a centre fed dipole aerial/antenna so 7575 was adopted as a compromise between 73 and 77 ohms.
300300 is the input impedance for a folded dipole (television antenna).
Why 50Ω ?
PROPAGATION CONSTANT,
Also known as propagation coefficient It is used to express attenuation,, (signal loss) and
the phase shift, , per unit length of a transmission line
Used to determine the reduction in voltage or current with distance as a TEM wave propagates down a transmission line
The line voltage and current per unit length decrease as z increases and are expressed by:
and
where:
are called series impedance per unit length and shunt admittance per unit length of the line.
( ) (1)dV
R j L I ZIdz
( ) (2)dI
G j C V YIdz
Z R j L
Y G j C
If we eliminates I from (1) and V from (2), we will get:
In this case,
and is called the propagation constant.
2 2 2d V dz V
( ) [( )( )]ZY R j L G j C
j
2 2 2d I dz I
: attenuation constantattenuation constant which determines how the voltage or current decreases with distance along the line
Unit: Nepers per unit length
: phase shift constantphase shift constant (a function of time and distance) and determine the phase angle of the voltage or current variation with distance
Unit: radians per unit length
Because a phase shift of 2 radian occur over a distance of one wavelength:
2
Velocity Factor, vf
Electromagnetic waves traveling in metallic conductor or in dielectric materials suffer degradation in speed compared when travelling in free space or vacuum.
Velocity factor, vf or velocity constant is a ratio of the actual velocity of propagation, vph in the given medium to the velocity of propagation through vacuum (free space speed).
vf = vph /c Velocity factor also depends on the the dielectric constant of
the given material: 1f
r
v
Phase Velocity, vph
Phase velocity, vph, is the actual velocity of propagation in the medium, is given by:
Since wavelength is directly proportional to velocity and the phase velocity of a TEM wave varies with dielectric constant, the wavelength of a TEM waves also varies with dielectric constant.
Therefore, guided wavelength, g
ph fg
r
v cv c
f f f
1 ph f
r
cv v c
LC
r
o
Where λo is the wavelength in air
vf and εr (or k)
SPECIAL CASES OF TL
i. Lossless Line
ii. Distortionless Line
LOSSLESS LINE
A transmission line is said to be lossless if the conductors of the line are perfect and the dielectric medium separating them is lossless
For such a line, the line parameters are such that R = 0 = G. This is necessary for the line to be lossless.
Thus, it has to follow these equations on the right:
0
0
1
ph g
j j LC
v fLC
LZ
C
DISTORTIONLESS LINE The signal usually consists of a band of
frequencies. Thus, wave amplitudes of different frequency components will be attenuated differently in a lossy line as it is frequency independent.
This leads to the definition of distortionless line. A distortionless lines is one which the attenuation constant is frequency independent while the phase constant is linearly dependent on frequency.
A distortionless line will result in the line parameters which can be seen on the left:
0
R G
L Cj
RG j LC
RG
LC
LZ
C
SUMMARY
Note that: A lossless line is also a distortionless line but a
distortionless line is not necessarily lossless. Although lossless lines are desirable in power transmission,
we need a distortionless line in telephone line.
EXAMPLE 1
A piece of RG-59B/U coaxial cable has 75 characteristic impedance and a nominal capacitance of 69 pF/m. o What is it’s inductance per meter?o If the diameter of the inner conductor is
0.584 mm and the dielectric constant of the insulator is 2.23, what is the outer conductor diameter?
EXAMPLE 2
For a lossless transmission line with
g = 20.7 cm at 1GHz, find r of the insulating material.
Answer: 2.1
EXAMPLE 3
A lossless transmission line uses a dielectric insulating material with r = 4. If its line capacitance is C = 10 pF/m, find:The phase velocity, vph
The line inductance, LThe characteristic impedance, Z0
EXAMPLE 4
An electromagnetic waves is propagating in a coaxial cable with velocity factor of 0.8 at 100MHz. Find
Wavelength in free space Phase velocity Wavelength in the coaxial cable.
Answer: 3m; 240Mm/s; 2.4m
EXAMPLE 5
A distortionless line has Z0 = 60,
= 20 mNp/m, vph = 0.6c,
where c is the speed of light in a vacuum.
Find R, L, G, C and at 100MHz.
Answer: 1.2Ω/m; 333.33nH/m; 333.33μS/m; 92.59pF/m; 1.8m
EXAMPLE 6
An air line has characteristic impedance of 70 and phase constant of 3rad/m at 100MHz.
Calculate L and C of the line.
Answer: 334.2nH/m; 68.2pF/m
INPUT IMPEDANCE, Zin
The impedance of the line at z looking towards the load is called the input impedance of the line and it can be defined as:
For lossy medium,
For lossless medium, =j, tanh jl = jtan l
00
0
tanh
tanhL
inL
Z ZZ Z
Z Z
00
0
tan
tanL
inL
Z jZZ Z
Z jZ
zerVzesV
zerVzesVoZ
)z(I
)z(VinZ
At the generator end, the terminated transmission line can be replacedwith the input impedance of the line Zin
Zin
At voltage maximum or current minimum,
And at voltage minimum or current maximum,
V (1 )V imaxZ Z Sin oI I (1 )min i
SoZ
)1(iI
)1(iV
maxIminV
inZ
Where S=VSWR is the Voltage Standing Wave Ratio
SPECIAL CASE Zin
ltan0jZ0LZinZSCinZ
lcotojZltanj
0ZinZ
LZlimOC
inZ
SHORT CIRCUIT
OPEN CIRCUIT
Example 7
A 50Ω lossless transmission line uses an insulating material with εr=2.25. When terminated in an open circuit, how long should the line be for its input impedance to be equivalent to a 10pF capacitor at 50MHz?
Answer: 5.68m
Question…
What is the difference between Zo and Zin?
INCIDENT & REFLECTED WAVE
RECALL : Electrical energy travels in the transmission line in the form of voltage or current.
This energy whose travel from the source to the load is called the incident wave.
A line which is terminated by a load that is not equal to the characteristic impedance, some of the applied power will be absorbed by the termination and the remaining power will be reflected.
The reflected wave travels from the load back to the generator and will continue to oscillate in the transmission line until completely attenuated or terminated.
( )
( )
zs
zs
V z V e
I z I e
INCIDENT & REFLECTED WAVE
Thus, voltage and current at any point on the transmission line are given by:
The first term in the right represents the incident wave and the second term represents the reflected wave.
Vs = sending (incident) voltage amplitude
Is = sending (incident) current amplitude
Vr = reflected voltage amplitude
Ir = reflected current amplitude
zerVzesV)z(V zerIzesI)z(I
Open and Short Line
If, at the end of the line, the load is replace with a short circuit line (ZL = 0) or an open line (ZL = ), all of the energy that reaches the end of the line will be reflected back to the source (no power is transferred to the load).
STANDING WAVE (SW)
Standing wave is the resultant wave or the algebraic sum of the incident wave and reflected wave at any point along the line
i. It is a stationary waveii. Its pattern along the line varies in magnitude and
contains maximum and minimum at certain points along the line, where adjacent max and min are separated by /2.
SW FORMATION
INCIDENT WAVE
REFLECTED WAVE
STANDING WAVE !STANDING WAVE !
+
=
SW
Standing wave can be used as an indication of how much signal reflection occurs in the line, i.e. voltage standing wave ratio.
The formation of standing waves in the transmission line results in inefficient power transfer and possible equipment breakdown.
SW
There are three special cases for the standing wave which are:
i. Matched Line
ii. Open Circuit Line
iii. Short Line
OC
Open circuit (ZL = ) When incident wave reaches the open circuit at the end of
the line, the magnetic field collapses since the current is reduce to zero.
This induces a voltage on the line which adds to the existing voltage and equal in its magnitude Voltage Doubling Effect
The induced voltage on the line then travels back along the line and may be absorbed by the generator impedance if it equals to Z0.
If not, it travels to and along the line until it finally attenuated completely.
SC
Short circuit (ZL = 0) Since the voltage is zero at the end of the line,
there must be a phase reversal of the incident voltage at the end of the line so that the incident and the reflected waves cancel one another at the end of the line.
The reflected wave then travels back along the line and may be absorbed by the generator impedance or completely attenuated.
Matched Line
Matched Line (ZL = Z0) When this happen, all the energy is
absorbed or dispersed by the load. Thus no reflected wave. Ideal condition (always desirable)
VoltageVoltageStandingStanding
Wave Wave PatternPattern
REFLECTION COEFFICIENT, Γ Also known as reflection constant is a measure of how much reflection occur in
the transmission line. It is the ratio of the reflected wave over the incident wave at the end of the line.
Mathematically,
is a complex and may be written as:
where is the magnitude of
is the phase angle
sVrV
rje
r
Γ
In terms of impedances, reflection coefficient can be rewritten as:
Note : Reflection coefficient is a complex number, make sure you
know how to use your calculator to do complex calculation and convert from rectangular form to polar form.
Magnitude of the reflection coefficient is a real number.
oZLZoZLZ
Using the impedance definition of reflection coefficient, the following conclusion is true for the three special condition (Open, Short and Matched line).
Open-Circuit line ( )
Γ
1
LZoZ
1
LZoZ
1
LZ
TOTAL REFLECTION OCCURS!
Γ
Short-Circuit line ( )
Total reflection but opposite’s direction (180o out of phase)
Matched line ( )
no reflected wave.
1oZ0oZ0
0LZ
inZoZLZ
0oZLZoZLZ
Example 8
50 transmission line is terminated with a 100 load. Find the reflection coefficient.
The load is replace with another one with impedance of 50 + j50. Find the magnitude of the reflection coefficient and its argument.
Transmission Coefficient, T
The transmission coefficient, T can be defined as:
wherelerVlesVLV
V 2ZL LT 1l Z ZL 0V es
POWER TRANSMISSION, PL
Power transmitted to a mismatched load, PL is given by
PL = Pincident – Preflected The above equation can be rewritten in term of reflection
coefficient
If the transmission line is lossless, Pincident is the actual output power from the transmitter (Pincident = PL )
2
2
1 1
1
ref refL
in in in
P VP
P P V
Average Power, Pave
Average power delivered to the load is given by:
22 VoV2 2 2siP P (1 ) (1 ) (1 )ave av 2Z 2Z0 0
0Z2
2oV
iavP
0Z2
2oV
2ravP
Where: andand
Pave or Pav
1
2
1 2+Pave =
Example 9
For 50Ω lossless transmission line terminated in a load impedance ZL=100+j50Ω, determine the fraction of the average incident power reflected by the load.
Answer: 20.25%
Example 10
For the line on the previous example, what is the magnitude of the average
reflected power if = 1V?
Answer: 2mW
oV
STANDING WAVE RATIO (SWR)
RECALL : Standing wave is the resultant wave or the algebraic sum of the incident wave and reflected wave at any point along the line.
The standing wave ratio (SWR) is defined as: Ratio of the maximum voltage to the minimum voltage or Ratio of the maximum voltage to the minimum voltage or
the maximum current to the minimum current of a the maximum current to the minimum current of a standing wave on a transmission linestanding wave on a transmission line
Most of the time, SWR is called the voltage standing wave ratio (VSWR).
VSWR Basically, VSWR is a measure of the mismatch between the load
impedance and the characteristic impedance of the transmission line.
SWR can be written as:
and is unitless
Voltage maxima exists when the incident and the reflected waves are in phase (their maximum peaks pass the same point on the line with the same polarity).
Voltage minima occur when the incident and reflected waves are 180o out of phase.
rs VVV max
rs VVV min
max
min
VVSWR S
V
dmax versus dmin
The location or distance of the voltage maxima (dmax) and voltage minima (dmin) from from
the loadthe load can be calculated using the formula below:
max
min max max
min max max
1,2.........if 0
0,1,2......if 04 2
if 4 4
if 4 4
rr
r
nnd
n
d d d
d d d
Example 11
If Γ=0.5∟-60° and λ=24cm, find the locations of the voltage maximum and minimum nearest to the load.
Answer: 10cm; 4cm
VSWR VSWR can be rewritten as:
When the incident and reflected waves are equal in amplitude (a total mismatch), VSWR = .
This is the worst case condition. When there is no reflected wave, (Vr = 0)
VSWR = 1. This condition only occurs when ZL = Z0 and it is an ideal
condition.
rVsVrVsV
minVmaxV
VSWR
VSWR
In term of reflection coefficient:
1
1VSWR
Voltage and current standing waves on a transmission line that is terminated in an open circuit.
Voltage and current standing waves on a transmission line that is terminated in a short circuit.
It can be shown that
ZOCZSC = Z02
Thus
0 OC SCZ Z Z
SUMMARY
OPEN CIRCUIT
MATCHED LINE
SHORT CIRCUIT
Γ +1 0 -1
1 0 1
VSWR 1
Example 12
A 50 transmission line is connected to 75 load at 100MHz.
Given vf = 0.8, find:
Reflection coefficient SWR Distance of the first voltage maximum and
minimum .
Example 13
A 50 transmission line is connected to 150 + j50 load at 100MHz.
Given vf = 0.8, find
Reflection coefficient SWR Distance of the first voltage maximum and
minimum.
Example 14 An RG 6A has the following properties
Characterisitic Impedance = 75Attenuation for 100 ft @ 400MHz = 6.5dB
Max Voltagerms = 1900 V Find the following
Maximum power transmit allowed by the cable? Attenuation of the cable if the cable length is 350 ft. Reflected power if transmit power is 1000W and the load is a
dipole with impedance of 73 . SWR
IMPEDANCE MATCHING
Whenever the characteristic impedance of a transmission line and its load are not matched, standing waves are present on the line. Thus, maximum power is not transferred to the
load. Two common transmission line techniques used to
match the unmatched transmission lines are: Stub matching Quarter-wavelength transformer matching
Matched Line Condition
inZoZLZ
Matching Network
The function of a matching network is to transform the load impedance ZL
such that the input impedance Zin looking into the network is equalto Zo of the transmission line.
STUB MATCHING A load which is purely inductive or purely capacitive does not absorb
any energy. When the load is a complex impedance (which is usually the case),
it is necessary to remove the reactive component to match the transmission line to the load. Transmission line stubs are commonly used for this purpose.
A transmission line stub is simply a piece of additional transmission line that is placed across the primary line as close to the load as possible. The susceptance of the stub is used to tune out the susceptance of the load.
Stubs are shorted or open circuit lengths of transmission line intended to produce a pure reactance at the attachment point for the line frequency of interest. Any value of reactance can be made as the stub length is varied from zero to half a wavelength.
Stub Matching Networkl
x
xd
lx
xd
Stub Matching Idea
Consists of 2 basic steps: i. The distance l is selected so as to transform the
load admittance, YL = 1/ZL into an admittance of Yd=Yo+jB when looking towards the load at MM’
ii. The length d of the stub line is selected so that its input admittance Ys at MM’ is equal to –jB
The parallel sum of the two admittances at MM’ yields Yo=Yin = 1, which the characteristic admittance of the line.
If you are stuck, remember this figure!
2BdAd Note that:
In this case, we have two possible shunted stubs. To avoid confusion, we normally choose to match the shorter stub
and one at a position closer to the load lA and dA
yd.lB and dB is the
alternative stub distance and stub
length
SC stub or OC stub is better?
An open-circuited stub radiates some energy at high frequencies.
With this drawback, shunt short-circuited stubs are preferred.
QUARTER WAVELENGTH TRANSFORMER MATCHING
Quarter-wavelength transformers are used to match transmission lines to purely resistive loads whose resistance is not equal to the characteristic impedance of the line.
Keep in mind that a quarter-wavelength transformer is not actually a transformer but rather a quarter-wavelength section of transmission line that acts as if it were a transformer.
λ/4 Transformer Matching
A quarter- wavelength transformer is simply a length of transmission line one-quarter wavelength long.
The characteristic impedance of the quarter-wavelength section is determined mathematically from the formula:
where
Z0’ is the characteristic impedance of a quarter-wavelength transformer
Z0 is the characteristic impedance of the transmission line that is being matched
ZL is the load impedance
LZZZ 0'0
λ/4 Transformer Matching
Quarter wave section of 150 Ω transmission line matches 75 Ω source to 300 Ω load.
This technique of impedance matching is often used to match the differing impedance values of transmission line and antenna in radio transmitter systems, because the transmitter's frequency is generally well-known and unchanging. The use of an impedance “transformer” 1/4 wavelength in length provides impedance matching using the shortest conductor length possible.
Example 15
Find the characteristic impedance of a quarter wavelength transmission line that can matched a 50 line to a 75 antenna.
What is the length of the transformer?
SMITH SMITH
CHARTCHART
THE SMITH CHART The Smith Chart has been introduced to reduce the tedious
manipulations involved in calculating the characteristics of a transmission lines by means of graphical representation.
It is basically a graphical indication of the impedance of a basically a graphical indication of the impedance of a transmission line as one moves along the line. transmission line as one moves along the line.
In this case, we have to make an assumption that the transmission lines to which the Smith chart will be applied is lossless.
The Smith chart is constructed within a circle of radius (|| 1) as can be seen in the figure. The construction of the Smith chart is based on the relation of:
or
where r and i are the real and imaginary parts of the reflection coefficient .
r ij 0
0
L
L
Z Z
Z Z
Smith Chart Visualization
More on Smith Chart
The two outermost scales on the Smith chart indicate distance in wavelengths
Outside scale gives distance from the load toward the generator and increases in a clockwise direction
Second scale gives distance from the source toward the load and increases in counter clockwise direction
One complete revolution (360o) represent a distance of λ/2 (0.5λ)
Half revolution (180o) represent a distance of λ/4 (0.25λ)
Toward Generator
Away From
Generator
SWR Circle or Constant Reflection Coefficient
Circle
Scale in Wavelengths
Full Circle is One Half Wavelength Since Everything
Repeats
Why we need Smith Chart?
Thus, instead of having separate Smith charts for transmission lines with different characteristic impedance such as 60, 100, 300 etc it is preferable and more helpful to have just one that can be used for any line.
This is achieved by using a normalized chartnormalized chart in which all impedances are normalized with respect to the characteristic impedance Z0 of that particular line under consideration.
For the load impedance ZL, for instance, the normalized impedance zL is given by :
jxr0ZLZ
Lz
Parameters plotted on the Smith Chart include the following: Reflection coefficient magnitude, Reflection coefficient phase angle, θr
length of transmission line between any two points in wavelength
VSWR Input Impedance, Zin
The location of Vmax and Vmin. Ie: dmax and dmin
PARAMETERS PLOTTED ON SMITH CHART
The Smith Chart Derivation
Impedance of transmission line made up from both real and imaginary components of either sign (Z = R ± jX)
Figure shows impedance three typical circuit elements and shows their impedance graphed on a rectangular coordinate system
All values correspond to passive networks must be plotted on or to the right of the imaginary axis of the Z plane
Plot must extends to infinity in three directions (+R, +jX, -jX) to display the impedance of all possible passive networks on a rectangular plot.
(a) Typical circuit elements (b) Impedances graphed on rectangular coordinate plane
(Note: is the angular frequency at which Z is measured.)
Smith chart overcome this problems by plotting the complex reflection coefficient.
Where zL equals the impedance normalized to the characteristic impedance (i.e zL = ZL/Z0)
For all passive impedance values, zL, the magnitude of between 0 and 1
Since | | ≤ 1, the entire right side of the z plane can be mapped onto a circular area of the plane.
Resulting circle has radius r = 1 and a center at = 0, which corresponds to zL = 1, or ZL = Z0.
1Lz
1Lzijr
Resistive Circle , r-circle
Figure shows the rectangular plotrectangular plot of four lines of constant resistance Re(zL)=0, 0.5, 1 and 2.
jxr0ZLZ
Lz
r-circle
Resistive Circle , r-circle
r-circler-circle
Values on rectangular plotValues on rectangular plot
Values on Values on plane planeRecall:
Reactance Circle, x-circle
Impedances with a positive reactive component (xL) fall above real axis
Impedances with a negative reactive component (xC) fall below the real axis
With the circular plot, the lines no longer extends to infinity
The infinity points all meet on the plane at a distance of 1 to the right of the origin
This imply that for zL = , = 1.
(a) Rectangular plot (b) plane
Reactance Circle, x-circle Figure (a) below shows the rectangular plot of three lines of
constant inductive reactance (x = 0.5, 1, 2), three line of constant capacitive reactance (x= -0.5, -1, -2) and a line of zero reactance (x = 0)
Figure (b) shows the same values on plane
jxr0ZLZ
Lz
x- circle
Recall:
Impedance Inversion (Admittance)
Admittance (Y) is the mathematical inversion of Z:
Y = 1/Z In the Smith chart, admittance is deduced simply rotating
the zL value on the plane by 180o through = 0 (center of the Smith Chart).
By rotating every point on the chart by 180o, a second set of coordinates ( y coordinates) can be developed that is inverted mirror image of the original chart.
Figure shows how an admittance can be found on the Smith chart.
(a) Impedance inversion (b) Impedance-Admittance Chart
NormalizedNormalizedImpedance (Impedance (----))
And And Admittance (Admittance (----))
ChartChart
Complex Conjugate Can simply be found by reversing the sign of the
angle . On the Smith chart, is usually written in polar
form and angles become more negative (phase lagging) when rotated in a clockwise direction around the chart.
Hence, 0Hence, 0oo is on the right end of the real axis and ± is on the right end of the real axis and ± 180180oo is on the left end. is on the left end.
Figure shows how a complex conjugate can be found using the Smith chart.
Complex conjugate on Smith ChartComplex conjugate on Smith Chart
Plotting Impedance and AdmittancePlotting Impedance and Admittanceon the Smith Charton the Smith Chart
Any impedance ZL can be plotted on the Smith chart by simply normalizing the impedance value to the characteristic impedance (i.e. zL = ZL /Z0) and plotting the real and imaginary parts.
If zL is purely resistive (±jx = 0), then its plot must fall on the horizontal axis
Rotating the plot 180o around the chart gives a normalized admittance value (yL = YL/Y0).
For example, if zL = 0.5 (from ZL = 25Ω and Zo=50Ω), point A represent the normalized impedance value whereas point B represent the normalized admittance value.
HorizontalAxis
Plotting SWR circle on the Smith Chart
Any lossless line can be represented on the Smith chart by a circle having its origin at 1± j0 (center of the chart) and radius equal to the distance between the origin and the impedance plot.
Therefore, SWR corresponding to any particular SWR corresponding to any particular circle is equal to the value of Zcircle is equal to the value of ZLL /Z /Z00 at which the at which the
circle crosses the horizontal axis on the right circle crosses the horizontal axis on the right side of the chart.side of the chart.
If Z0 = 50 and ZL = 25 + j25.
Find yL, YL and SWR.
Example 16
Answer:
zL is plotted on the Smith chart by locating the point where R = 0.5 arc intersects the X = 0.5 arc on the top half of the chart.
ZL = 0.5 + j0.5 is plotted on the Figure 34 at point A and yL is plotted at point B(1- j1).
From the Smith Chart, SWR is approximately 2.6 (point C)
PROPOSED PROPOSED SOLUTION SOLUTION
FOR FOR Example 16Example 16
SC and OC points SC and OC points on the Smith charton the Smith chart
The location of SC and OC points are different depending on Impedance or Admittance chart.
Points of OC and SC on Points of OC and SC on Impedance ChartImpedance Chart
SC OC
OC Termination Transmission line terminated in open circuit has an impedance at the
open end that is purely resistive and equal to infinity, ZZLL = = ∞∞ On Smith chart, this point is plotted on the right end of X=0 line
(point A) Moving toward the generator, Zin can be found by rotating the Zin can be found by rotating the
chart in a clockwise directionchart in a clockwise direction, it can be seen that the ZinZin become capacitive
Rotating further, capacitance decreases to a normalized value of unity ZinZin = –j1 (1/8 , point C)
At distance of 1/4λ, ZinZin is purely resistive and equal 0 (point B) Moving past ¼ , impedance become inductive; then increases to
normalized value of unity (ZinZin = +j1) (point D) at point 3/8 from the load.
It then becomes purely resistive and equal to infinity (return to point A) at 1/2 .
NORMALIZED IMPEDANCE CHART
Figure 35 : Transmission-line input impedance for shorted and open line
A
C
B
DNORMALIZED IMPEDANCE CHART
Input Impedance Variation for OC Transmission Line
SC Termination
Similar analysis can be done for transmission line terminated in a short circuit, although opposite impedance variation are achieved as with the open load At the load, ZZLL are purely resistive and equal to 0 Ω
(point B) Point A represent distance of ¼ from the load Point D represent distance of 1/8 wavelength from the
load and point C represent distance 3/8 wavelength from the load
IMPEDANCE CHART
Input Impedance Variation for SC Transmission Line
Points of OC and SC on Points of OC and SC on Admittance ChartAdmittance Chart
OC SC
For transmission line terminated in a purely resistive load not equal to Z0 , Smith Chart analysis is very similar to the process described in the preceding section
For example, for load impedance ZL = 37.5, characteristic impedance Z0 = 75 , input impedance at various distance from the load can be determined as follows Normalized impedance, zL=0.5 zL = 0.5 is plotted on the Smith chart. A circle is drawn that
passes through point A with its centre located at the intersection of the R = 1 and X = 0 arc.
SWR is read at the intersection of the circle and the X = 0 line on the right side, SWR = 2
Plotting Zin on the Smith Chart
SWRPoint A
The input impedance, Zin at a distance of 0.125 from the load is determined by extending point A to a similar position on the outside scale (point B) and moving around the scale in clockwise direction a distance of 0.125=0.127x720°=90°
Rotate from point B a distance equal to the length of the transmission line ( 0.125λ at point C). Transfer this point to a similar position on the SWR=2 circle (point D)
Normalized input impedance is located at point D (0.8 +j0.6). Actual impedance is found by multiplying the normalized impedance by the characteristic impedance of the line
For distance of 0.3= 0.3x720°= 216° from the load, the normalized impedance is plotted at point E
Plotting Zin on the Smith Chart(cont.)
45j6075)6.0j8.0(inZ
Plotting ZinPlotting Zin
Example 17
Determine the input impedance and SWR for a transmission line 1.25 long with a characteristic impedance Zo = 50 and load impedance, ZL = 30 + j40
Answer: Zin = 31.5 – j38.5, SWR = 2.9
Answer: Zin = 30 – j40, SWR = 3.0
PROPOSED PROPOSED SOLUTION SOLUTION
FOR FOR Example 17Example 17
Example 18
A 30m long lossless transmission line with Zo = 50Ω operating at 2MHz is terminated by a load of ZL=60+j40Ω. If v=0.6c on the line, find:
Γ VSWR Zin
75.1j5.23inZ;1.2S;5635.0Answer:Answer:
Vmax versus
Vmin
Determining ZL Using Smith Chart (moving counterclockwise)
Given that S=3 on a 50-Ω line, that the first voltage minimum occurs at 5cm from the load, that the distance between successive minima is 20cm, find the load impedance, ZL.
Answer: zAnswer: zLL=0.6-j0.8; Z=0.6-j0.8; ZLL=30-j40=30-j40ΩΩ
λ/4 Transformer Matching
Steps:1) A load ZL=75+j50Ω can be matched to a 50 Ω source with
a λ/4 transformer. 2) zL=1.5+j1 is plotted on the Smith chart (Point A) and the
impedance/VSWR circle is drawn3) Extend point A to the outermost scale (Point B). The
characteristic impedance of an ideal transmission line is purely resistive. Thus, if a λ/4 transformer is located at a distance from the load where the input impedance is purely resistive, the transformer can match the TL to the load. There are 2 points where the input impedance is purely resistive (where the circle intersects the x=0) which are Point C and D.
4) The distance of the λ/4 transformer from the load is the distance from Point B to point C or D (whichever the shortest) Point C at zin=2.4. Distance B to C = 0.058λ
λ/4 Transformer Matching
Steps (cont.)5) zin can be read directly (Point C) where it
is also VSWR of the mismatched line.
6) The actual actual input impedance,
Zin = 50(2.4)=120 Ω. The characteristic impedance of the λ/4
transformer is:
5.7712050LZoZ'oZ
PROPOSED PROPOSED SOLUTION SOLUTION
FOR FOR λλ/4 Transformer /4 Transformer
MatchingMatching
λ/4 Transformer Matching
In conclusion: If a If a λλ/4 of a 77.5Ω TL is inserted at 0.058/4 of a 77.5Ω TL is inserted at 0.058λλ
from the load, the line is matched.from the load, the line is matched. It should be noted that a It should be noted that a λλ/4 transformer /4 transformer
does not totally eliminate standing wave on does not totally eliminate standing wave on the TL.the TL.
It simply eliminates the SW from the It simply eliminates the SW from the transformer back to the source.transformer back to the source.
Standing wave are still present on the line Standing wave are still present on the line between the transformer and the load.between the transformer and the load.
SW on λ/4 Transformer
Voltage standing wave pattern of mismatched load (a) Without a transformer (b) With a transformer
Example 19
Determine the SWR, characteristic impedance of a quarter wavelength transformer, and the distance the transformer must be placed from the load to match a 75 transmission line to a load ZL = 25 - j50
Answer: 4.6; 35.2Ω; 0.1
EXERCISE (Submit after 15 mins.)
An antenna, connected to a 150Ω lossless line, produces a standing wave ratio of 2.6. If measurements indicate that voltage maxima are 120cm apart and that the last maximum is 40cm from the antenna, calculate: The operating frequency The antenna impedance The reflection coefficient. Assume v=c.
STUB MATCHING
Matching a load ZL= 50-j100 to a 75Ω TL can be done by shorted stub using Smith chart.
Steps:1) Plot the normalized impedance, zL=0.67-j1.33 (Point A)
and draw VSWR circle. Stubs are shunted across the load, thus admittances are used rather than impedances to simplify calculations. The circles and arcs on the Smith chart are now used for conductance and susceptance.
2) Normalized admittance,yL is determined by rotating the impedance plot, zL 180o(Draw a line from point A through the center to point B)
3) Rotate admittance point clockwise to a point on the impedance circle where it intersects the r = 1 circle (point C). The real component of the input impedance at this point is equal to the characteristic impedance Z0, Zin = r ± jx, where R = Z0. At point C, admittance yd = 1 + j1.7.
STUB MATCHING Steps (cont.):
4) Distance from B to C is how far the stub must be placed, for this example 0.09. The stub must have an impedance with zero resistive component and susceptance that has the opposite polarity (i.e. ys = 0 – j1.7)
5) To find the length of the stub where ys = 0 – j1.7, move around the outside circle of the Smith chart (R = 0), having a wavelength identified at point D until an admittance ys = 1.7 is found (wavelength value identified at point E).
6) If open stubs are used, rotation would begin at opposite direction (point F)
7) The distance from point D to point E is the length of the stub which is 0.084
PROPOSED PROPOSED SOLUTION SOLUTION
FOR FOR Stub Stub
MatchingMatching
Example 20
For a transmission line with a characteristic impedance Z0 = 300 and a load with complex impedance ZL = 450 + j600, determine: SWR, the distance a shorted stub must be placed from
the load to match the load to the line the length of the stub.
2BdAd Note that:
In this case, we have two possible shunted stubs. To avoid confusion, we normally choose to match the shorter stub
and one at a position closer to the load lA and dA
yd.lB and dB is the
alternative stub distance and stub
length
RECALL BACK!
Stub Matching Networkl
x
xd
lx
xd
Please Study for your Quiz 1!
EXTRA INFORMATION!EXTRA INFORMATION!
BROWSE DURING FREE TIMEBROWSE DURING FREE TIME
the load reflection coefficient behavior is given as r and x vary, what exactly does L do?
let’s plot in the complex plane
for any non-negative value of Re(Z) (i.e., r ≥ 0) and any value of Im(Z), falls on or within the unit circle in the complex plane
1
1L o
Lo L
r jxZ Z
Z Z r jx
0 1L
x = 0
r = 10
-j
j
-1 Re()
Im()
x=-20
r = 2
r = 0 r = .5
r = 1x=20
x = +2
r = 0 r = 100
r = 0r = 0
1
-j
j r = 0
-1 Re()
Im()
r = 0r = 0
r = 0
x = -2
x = 0
x = .5
x = -.5
x = -1
x = 1
• pick a value of x, then vary r • pick a value of r, then vary x
Relationships between and Z• Note relation between and the “normalized impedance”
u j v
2222
22
1
2
1
1
vu
vj
vu
vuxjr
11
11
1 11
1
r j x
r j x
r j x
r j x
• consider in complex plane
• so the relationship between the normalized impedance and the real and imaginary parts of is
1
1r j x
1
1
xjr
xjr 11
11
xjrxjr
xjrxjr
2
22 xjr
1
1L
o
Zr j x
Z
vju
vju
1
1
vju
vju
vju
vju
1
1
1
1
22
22
1
21
vu
vjvu
Relationships between and Z
From the real part:
22
2
1
1
1 rv
r
ru
2222
22
1
2
1
1
vu
vj
vu
vuxjr
222
2
2
2
22
2
2
2
2
2
22
22
2222
2222
1
1
1
1
11
1
11
1
112
1
1
12
1121
121
11
r
rrv
r
ru
r
rrrv
r
ru
r
r
r
rv
r
r
r
ruu
r
rv
r
ruu
rrvurru
vuvuur
vuvur
u j v
2 2
2 2
1
1
u vr
u v
Plot of reflection coefficient in complex plane
-1
1
u, Re(reflection coef.)
v, I
m(r
efle
ctio
n c
oef
.)
1•j
-1•jcurves of constant r = Re(Z)
22
2
1
1
1 rv
r
ru
plot as a function of r these are circles!
22
2
2
1
2
1
vu
When: r = 1
2
22
1
10
vu
When: 0 < r < 1
When: r > 1
When r = 0
• radius
0,1 r
r
r1
1
• center
u j v
Plot of Reflection Coefficient
-1
1
u, Re(reflection coef.)
v, I
m(r
efle
ctio
n c
oef
.)
j
-j
plot as a function of x x =
When: x < 0
When: x > 0
When: x = 0
• from the Im part:
2
22
0
1
0
11
vu
111
22 vu
curves of constant x = Im(Z)
2222
22
1
2
1
1
vu
vj
vu
vuxjr
u = 1
– these are also circles!
2
22 11
1xx
vu
• radius
x
1,1
x
1• center
u j v
1.0 1.2
1.4
1.6
1.8
2.0
0.5
0.6
0.7
0.8
0.9
0.4
0.3
0.2
0.1
3.0
4.0
10
20
5.0
1.0
0.5
0.6
0.7
0.8
0.9
0.4
0.3
0.2
0.1
1.2
1.4
1.6
1.8
2.0
4.0
10
20
5.0
3.0
0.2
0.2
0.6
0.8
1.0
1.0
0.4
0.4
0.6
0.8
0.2
0.2
0.6
0.8
1.0
1.0
0.4
0.4
0.6
0.8
0 .1
0.2 0.3
0.4 0.5
0.6
0.7
0.8 0.9
1.0 1.2 1.4 1.6
1.8
2.0 4.0
3.0
5.0 10 20
0
negativereactive
com
ponent
posi
tive
reac
tive
com
pone
nt
Plot of Reflection Coefficient in Complex Plane
u, Re(reflection coef.)
v, I
m(r
efle
ctio
n c
oef
.)curves of constant r = Re(Z)
curves of constant x = Im(Z)
The Smith Chart is a plot of the reflection coefficient in the complex plane, with contours of constant load resistance and load reactance superimposed
1
1L o
Lo L
r jxZ Z
Z Z r jx
2
22 11
1xx
vu
22
2
1
1
1 rv
r
ru
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