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EKT 241/4: ELECTROMAGNETIC THEORY. UNIVERSITI MALAYSIA PERLIS. CHAPTER 5 – TRANSMISSION LINES. PREPARED BY: Saidatul Norlyana Azemi snorlyana@unimap.edu.my. Chapter Outline. General Considerations Lumped-Element Model Transmission-Line Equations Wave Propagation on a Transmission Line

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slide1

EKT 241/4:ELECTROMAGNETIC THEORY

UNIVERSITI MALAYSIA PERLIS

CHAPTER 5 – TRANSMISSION LINES

PREPARED BY: SaidatulNorlyanaAzemi

snorlyana@unimap.edu.my

chapter outline
Chapter Outline
  • General Considerations
  • Lumped-Element Model
  • Transmission-Line Equations
  • Wave Propagation on a Transmission Line
  • The Lossless Transmission Line
  • Input Impedance of the Lossless Line
  • Special Cases of the Lossless Line
  • Power Flow on a Lossless Transmission Line
  • The Smith Chart
  • Impedance Matching
  • Transients on Transmission Lines
general considerations
General Considerations
  • Transmission line – a two-port network connecting a generator circuit to a load.
so what is the use of transmission line
So…What is the use of transmission line??
  • A transmission line is used to transmit electrical energy/signals from one point to another
    • i.e. from one source to a load.
  • Types of transmission line include: wires, (telephone wire), coaxial cables, optical fibers n etc…
the role of wavelength
The role of wavelength

length of line, l

The impact of a transmission line on the current and voltage in the circuit depends on the:

  • At low frequency, the impact is negligible
  • At high frequency, the impact is very significant

frequency, fof the signal provided by generator.

propagation modes
Propagation modes

Electric field lines

Magnetic field lines

propagation modes1
Propagation modes

A few examples of transverse electromagnetic (TEM) and higher order transmission line

lumped element model
Lumped- element model
  • A transmission line is represented by a parallel-wire configuration regardless of the specific shape of the line, (in term of lumped element circuit model)
    • i.e coaxial line, two-wire line or any TEM line.
  • Lumped element circuit model consists of four basic elements called ‘the transmission line parameters’ : R’ , L’ , G’ , C’ .

Series element

Shunt element

lumped element model1
Lumped- element model
  • Lumped-element transmission line parameters:
    • R’ : combined resistance of both conductors per unit length, in Ω/m
    • L’ : the combined inductance of both conductors per unit length, in H/m
    • G’ : the conductance of the insulation medium per unit length, in S/m
    • C’ : the capacitance of the two conductors per unit length, in F/m
  • For example, a coil of wire has the property of inductance. When a certain amount of inductance is needed in a circuit, a coil of the proper dimension is inserted
lumped element model for 3 type of lines
Lumped- element model for 3 type of lines

Note: µ, σ, ε pertain to the insulating material between conductors

exercise 1
Exercise 1:
  • Use table 5.1 to compute the line parameter of a two wire air line whose wires are separated by distance of 2 cm, and, each is 1 mm in radius. The wires may be treated as perfect conductors with σc=.

R’ = ?, L’=?, G’=?, C’=?

solution exercise 1
Solution exercise 1:

σc=

σc=

exercise 2
Exercise 2:
  • Calculate the transmission line parameters at 1 MHz for a rigid coaxial air line with an inner conductor diameter of 0.6 cm and outer conductor diameter of 1.2 cm. The conductors are made of copper.

(μc=0.9991 ; σc=5.8x107)

f = 1MHz

r1 = 0.006m/2 = 0.003m

r2 = 0.012m/2 = 0.006m

slide18

BARE IN UR MIND

From calculator

slide19

BARE IN UR MIND

From calculator

slide20

Because, the material separating the inner and outer is perfect dielectric (air) with σ=0, thus G’ = 0

  • G’ : the conductance of the insulation medium per unit length, in S/m
transmission line equations
Transmission line equations

Is used to describes the voltage and the current across the transmission line in term of propagation constant and impedance

  • Complex propagation constant, γ
  • α – the real part of γ

- attenuation constant, unit: Np/m

  • β – the imaginary part of γ

- phase constant, unit: rad/m

transmission line equations1
Transmission line equations
  • The characteristic impedance of the line, Z0:
  • Phase velocity of propagating waves:

where f = frequency (Hz)

λ = wavelength (m)

β= phase constant

example 1
Example 1

An air lineis a transmission line for which air is the dielectric material present between the two conductors, which renders G’ = 0.

In addition, the conductors are made of a material with high conductivity so that R’ ≈0.

For an air line with characteristic impedance of 50Ωand phase constant of 20 rad/m at 700MHz, find the inductance per meter and the capacitance per meter of the line.

solution to example 1
Solution to Example 1
  • The following quantities are given:
  • WithR’ = G’ = 0,
solution to example 11
Solution to Example 1

2

  • The ratio is given by:
  • We get L’ from Z0
lossless transmission line
Lossless transmission line

Transmission line can be designed to minimize ohmic losses by selecting high conductivities and dielectric material, thus we assume :

  • Lossless transmission line - Very small values of R’ and G’.
  • We set R’=0 and G’=0, hence:
transmission line equations2
Transmission line equations
  • Complex propagation constant, γ
  • α – the real part of γ

- attenuation constant, unit: Np/m

  • β – the imaginary part of γ

- phase constant, unit: rad/m

0

0

lossless transmission line1
Lossless transmission line

Transmission line can be designed to minimize ohmic losses by selecting high conductivities and dielectric material, thus we assume :

  • Lossless transmission line - Very small values of R’ and G’.
  • We set R’=0 and G’=0, hence:
lossless transmission line2
Lossless transmission line
  • Using the relation properties between μ, σ, ε :
  • Wavelength, λ

Where εr = relative permittivity of the insulating material between conductors

exercise 3
Exercise 3:
  • For a losses transmission line, λ = 20.7 cm at 1GHz. Find εrof the insulating material.

λ=20.7cm 0.207m ; f=1 GHz

2

exercise 4
Exercise 4
  • A lossless transmission line of length 80 cm operates at a frequency of . The line parameters are & Find the characteristic impedance, the phase constant and the phase velocity.

The condition apply that the line is lossless, So: R= 0 & G=0

slide32
characteristic impedance :
  • phase constant:

With R n G = 0

= 18.85 rad/m

voltage reflection coefficient
Voltage Reflection Coefficient
  • Every transmission line has a resistance associated with it, and comes about because of its construction. This is called its characteristic impedance, Z0.
  • The standard characteristic impedance value is 50Ω. However when the transmission line is terminated with an arbitrary load ZL, in which is not equivalent to its characteristic impedance (ZL ≠ Z0), a reflected wave will occur.
voltage reflection coefficient1
Voltage reflection coefficient
  • Voltage reflection coefficient, Γ – the ratio of the amplitude of the reflected voltage wave, V0- to the amplitude of the incident voltage wave, V0+ at the load.
  • Hence,
voltage reflection coefficient2
Voltage reflection coefficient
  • The load impedance, ZL

Where;

= total voltage at the load

V0- = amplitude of reflected voltage wave

V0+ = amplitude of the incident voltage wave

= total current at the load

Z0 = characteristic impedance of the line

voltage reflection coefficient3
Voltage reflection coefficient
  • And in case of a RL and RC series, ZL:

ZL = R + jL ; ZL = R -1/jC

  • A load is matched to the line if ZL = Z0 because there will be no reflection by the load (Γ= 0and V0−= 0.
  • When the load is an open circuit, (ZL=∞), Γ = 1 and V0- = V0+.
  • When the load is a short circuit (ZL=0), Γ = -1 and V0- = V0+.
what is the difference between an open and closed circuit
What is the difference between an open and closed circuit?
  • closed allows electricity through, and open doesn't.
  • open circuit - Any circuit which is not complete is considered an open circuit. The open status of the circuit doesn't depend on how it became unclosed, so circuits which are manually disconnected and circuits which have blown fuses, faulty wiring or missing components are all considered open circuits.
  • close circuit: A circuit is considered to be closed when electricity flows from an energy source to the desired endpoint of the circuit. A complete circuit which is not performing any actual work can still be a closed circuit. For example, a circuit connected to a dead battery may not perform any work, but it is still a closed circuit.
example 2
Example 2
  • A 100-Ωtransmission line is connected to a load consisting of a 50-Ωresistor in series with a 10pF capacitor. Find the reflection coefficient at the load for a 100-MHz signal.
solution to example 2
Solution to Example 2
  • The following quantities are given
  • The load impedance is
  • Voltage reflection coefficient is
exercise 5
Exercise 5
  • A 150 Ωlossless line is terminated in a load impedance ZL= (30 –j200) Ω. Calculate the voltage reflection coefficient at the load.

Zo = 150 Ω

ZL= (30 –j200) Ω

standing waves
Standing Waves
  • Interference of the reflected wave and the incident wave along a transmission line creates a standing wave.
  • Constructive interference gives maximum value for standing wave pattern, while destructive interference gives minimum value.
  • The repetition period is λ for incident and reflected wave individually.
  • But, the repetition period for standing wave pattern is λ/2.
standing waves1
Standing Waves
  • For a matched line, ZL = Z0, Γ= 0and

= |V0+| for all values of z.

standing waves2
Standing Waves
  • For a short-circuited load, (ZL=0), Γ = -1.
standing waves3
Standing Waves
  • For an open-circuited load, (ZL=∞), Γ = 1.

The wave is shifted by λ/4 from short-circuit case.

standing waves4
Standing Waves
  • First voltage maximum occurs at:
  • If θr ≥ 0  n=0;
  • If θr ≤ 0  n=1
  • First voltage minimumoccurs at:

Where θr = phase angle of Γ

slide49
VSWR
  • Voltage Standing Wave Ratio (VSWR) is ratio between the maximum voltage an the minimum voltage along the transmission line.
  • VSWR provides a measure of mismatch between the load and the transmission line.
  • For a matched load with Γ= 0, VSWR = 1 and for a line with |Γ| - 1, VSWR = ∞.

The VSWR is given by:

example 3
Example 3

A 50-transmission line is terminated in a load with ZL = (100 + j50)Ω. Find the voltage reflection coefficient and the voltage standing-wave ratio (VSWR).

solution to example 3
Solution to Example 3
  • We have,
  • VSWR isgiven by:
exercise 6
Exercise 6:
  • A 140 Ω lossless line is terminated in a load impedance ZL= (280 +j182) Ω, if λ = 72cm, find

a) Reflection coefficient, Г

b) The VSWR,

c) The locations of voltage maxima and minima

input impedance of a lossless line
Input impedance of a lossless line
  • The input impedance, Zin is the ratio of the total voltage (incident and reflected voltages) to the total current at any point z on the line.
  • or
special cases of the lossless line
Special cases of the lossless line
  • For a line terminated in a short-circuit, ZL = 0:
  • For a line terminated in an open circuit, ZL = ∞:
application of short circuit and open circuit measurements
Application of short-circuit andopen-circuit measurements
  • The measurements of short-circuit input impedance, and open-circuit input impedance, can be used to measure the characteristic impedance of the line:
  • and
length of line
Length of line
  • If the transmission line has length , where n is an integer,
  • Hence, the input impedance becomes:
quarter wave transformer
Quarter wave transformer
  • If the transmission line is a quarter wavelength, with , where , we have , then the input impedance becomes:
example 4
Example 4

A 50-Ωlossless transmission line is to be matched to a resistive load impedance with ZL=100Ωvia a quarter-wave section as shown, thereby eliminating reflections along the feedline. Find the characteristic impedance of the quarter-wave transformer.

quarter wave transformer1
Quarter wave transformer
  • If the transmission line is a quarter wavelength, with , where , we have , then the input impedance becomes:
solution to example 4
Solution to Example 4
  • Zin = 50Ω; ZL=100Ω
  • Since the lines are lossless, all the incident power will end up getting transferred into the load ZL.
matched transmission line
Matched transmission line
  • For a matched lossless transmission line, ZL=Z0:

1) The input impedance Zin=Z0 for all locations z on the line,

2) Γ=0, and

3) all the incident power is delivered to the load, regardless of the length of the line, l.

slide66

When ZL=0(short circuit)

Ratio of the total voltage to total current on the line

Special case

When ZL=(open circuit)

Input Impedance, Zin

Application

Be used to measure the characteristic impedance of the line :

But, If the transmission line is

power flow on a lossless transmission line
Power flow on a losslesstransmission line
  • Two ways to determine the average power of an incident wave and the reflected wave;
    • Time-domain approach
    • Phasor domain approach
  • Average power for incident wave;
  • Average power for reflected wave:
  • The net average power delivered to the load:
power flow on a lossless transmission line1
Power flow on a losslesstransmission line
  • The time average power reflected by a load connected to a lossless transmission line is equal to the incident power multiplied by |Г|2
exercise 7
Exercise 7
  • For a 50Ω lossless transmission line terminated in a load impedance ZL = (100 + j50)Ω, determine the percentage of the average power reflected over average incident power by the load.

Z0=50Ω; ZL = (100 + j50)Ω

slide70

Reflection coefficient, Г

the percentage of the average incident power reflected by the load = 20%

exercise 8
Exercise 8
  • For the line of exercise previously (exercise 7), what is the average reflected power if |V0+|=1V
smith chart
Smith Chart
  • Smith chart is used to analyze & design transmission line circuits.
  • Reflection coefficient, Γ :

Гr = real part, Гi = imaginary part

  • Impedances on Smith chart are represented by normalized value, zL:
  • the normalized load impedance, zL is dimensionless.
smith chart1
Smith Chart
  • Reflection coefficient, ΓA :0.3 + j0.4

Reflection coefficient, ΓB :-0.5 - j0.2

In order to eliminate –ve part, thus

the complex plane
The complex Γplane.

ΓA :0.3 + j0.4

ΓB :-0.5 - j0.2

smith chart2
Smith Chart
  • Reflection coefficient, Γ :
  • Since , Γ becomes:
  • Re-arrange in terms of zL:

rL= Normalized load resistance

xL = Normalized load admittance

input impedance
Input impedance
  • The input impedance, Zin:
  • Γis the voltage reflection coefficient at the load.
  • We shift the phase angle of Γby 2βl, to get ΓL. This will zL to zin. The |Γ| is the same, but the phase is changed by 2βl.
  • On the Smith chart, this means rotating in a clockwise direction (WTG).
input impedance1
Input impedance
  • Since β = 2π/λ, shifting by 2 βl is equal to phase change of 2π.
  • Equating:
  • Hence, for one complete rotation corresponds to l = λ/2.
  • The objective of shifting Γ to ΓL is to find Zin at an any distance l on the transmission line.
example 5
Example 5
  • A 50-Ω transmission line is terminated with ZL=(100-j50)Ω. Find Zin at a distance l =0.1λ from the load.

Solution:

Normalized the load impedance

solution to example 5
Solution to Example 5

l =0.1λ

zin = 0.6 –j0.66

de normalize (multiplying by Zo) Zin= 30 –j33

vswr voltage maxima and voltage minima
VSWR, Voltage Maxima andVoltage Minima

zL=2+j1

VSWR = 2.6

(at Pmax).

lmax=(0.25-0.213)λ

=0.037λ.

lmin=(0.037+0.25)λ

=0.287λ

vswr voltage maxima and voltage minima1
VSWR, Voltage Maxima and Voltage Minima
  • Point A is the normalized load impedance with zL=2+j1.
  • VSWR = 2.6 (at Pmax).
  • The distance between the load and the first voltage maximum is lmax=(0.25-0.213)λ=0.037λ.
  • The distance between the load and the first voltage minimum is lmin=(0.037+0.25)λ =0.287λ.
impedance to admittance transformations
Impedance to admittancetransformations

zL=0.6 + j1.4

yL=0.25 - j0.6

example 6
Example 6
  • Given that the voltage standing-wave ratio, VSWR= 3. On a 50-Ωline, the first voltage minimum occurs at 5 cm from the load, and that the distance between successive minima is 20 cm, find the load impedance.
  • Solution:
  • The distance between successive minima is equal to λ/2.
  • the distance between successive minima is 20 cm, Hence, λ = 40 cm
solution to example 6
Solution to Example 6

Point A =VSWR= 3

de normalize (multiplying by Zo) Zin= 30 –j40

solution to example 61
Solution to Example 6
  • First voltage minimum (in wavelength unit) is at

on the WTL scale from point B.

  • Intersect the line with constant SWR circle = 3.
  • The normalized load impedance at point C is:
  • De-normalize (multiplying by Z0) to get ZL:
slide89

Solution:

Normalized the load impedance

a) reflection coefficient from smith Chart

slide90

lmax

lmin

slide91

Move a distance 0.301λ towards the generator (WTG) (refer to Smith chart)

  • → 0.301λ + 0.082λ=0.383λ
  • At 0.383λ, read the value of which at the point intersects with constant circle, we have = zin = 0.72- j0.62.
  • Denormalized it, hence = Zin= 72- j62
  • Distance from load to the first voltage maximum, (refer to Smith chart)

→ 0.25λ-0.082λ=0.168λ

impedance matching
Impedance Matching
  • Transmission line is matched to the load when Z0 = ZL.
  • This is usually not possible since ZL is used to serve other application.
  • Alternatively, we can place an impedance-matching networkbetween load and transmission line.
single stub matching
Single- stub matching
  • Matching network consists of two sections of transmission lines.
  • First section of length d, while the second section of length lin parallel with the first section, hence it is called stub.
  • The second section is terminated with either short-circuit or open circuit.
single stub matching1
Single- stub matching

YL=1/ZL

stub

l

d

feed line

Yd = Y0+jB

single stub matching2
Single- stub matching
  • The length l of the stub is chosen so that its input admittance, YS at MM’ is equal to –jB.
  • Hence, the parallel sum of the two admittances at MM’ yields Y0, which is the characteristic admittance of the line.

Yd = Y0+jB

single stub matching3
Single- stub matching
  • Thus, the main idea of shunt stub matching network is to:
  • (i) Find length d and l in order to get yd and yl .
  • (ii) Ensure total admittance yin= yd + ys= 1 for complete matching network.
example 7
Example 7

50-Ωtransmission line is connected to an antenna with load impedance ZL = (25 − j50)Ω. Find the position and length of the short-circuited stub required to match the line.

Solution:

The normalized load impedance is:

(located at A).

solution to example 71
Solution to Example 7
  • Value of yL at B is which locates at position 0.115λ on the WTG scale.
  • Draw constant SWR circle that goes through points A and B.
  • There are two possible matching points, C and D where the constant SWR circle intersects with circle rL=1 (now gL =1 circle).
slide100

B

C = 1+j1.58

D = 1+j1.58

A

solution to example 72
Solution to Example 7

First matching points, C.

  • At C, is at 0.178λ on WTG scale.
  • Distance B and C is
  • Normalized input admittance

at the juncture is:

E is the admittance of short-circuit stub, yL=-j∞.

Normalized admittance of −j 1.58 at Fand position 0.34λ on the WTG scale gives:

slide102

d1

= 0.063λ

B

C = 1+j1.58

E

l1

= 0.090λ

A

F = -j1.58

F

first matching points c
First matching points, C
  • Thus, the values are:
  • d1 = 0.063 λ
  • l1 = 0.09 λ
  • yd1 = 1 + j1.58 Ω
  • ys1= -j1.58 Ω
  • Where Yin = yd + ys = (1 + j1.58) + (-j1.58) = 1
solution to example 73
Solution to Example 7

Second matching point, D.

  • At point D,
  • Distance B and D is
  • Normalized input admittance at G.
  • Rotating from point E to point G, we get
slide105

B

G = +j1.58

G

d2

= 0.207λ

E

l2

= 0.41λ

D = 1-j1.58

A

first matching points d
First matching points, D
  • Thus, the values are:
  • d2 = 0.207 λ
  • l2 = 0.41 λ
  • yd2 = 1 - j1.58 Ω
  • ys2= +j1.58 Ω
  • Where Yin = yd + ys = (1 - j1.58) + (+j1.58) = 1
slide107

d1=0.063 λ

d2=0.207 λ

l1=0.09λ,

l2=0.41 λ