ANTENNAS and MICROWAVES ENGINEERING …...Voltage Standing Wave Ratio (VSWR): It is the ratio of the...

Philadelphia University

Faculty of Engineering Communication and Electronics Engineering

Part 2 Dr. Omar R Daoud 1

ANTENNAS and MICROWAVES

ENGINEERING

(650427)

Transmission Lines

General Considerations It is a two-port network connecting a generator circuit to

a load.

The impact of a transmission line on the current and

voltage in the circuit depends on the length of line, l and

the frequency, f of the signal provided by generator.

At low frequency, the impact is negligible

At high frequency, the impact is very significant

2/27/2018 Dr. Omar R Daoud 2

Transmission Lines

General Considerations It may be classified into two

types:

Transverse electromagnetic

(TEM) transmission lines:

waves propagating along these

lines having electric and

magnetic field that are entirely

transverse to the direction of

propagation

Higher order transmission

lines:

waves propagating along these

lines have at least one

significant field component in

the direction of propagation


Transmission Lines

General Considerations It is represented by a parallel-wire

configuration regardless of the specific

shape of the line, i.e coaxial line, two-

wire line or any TEM line.

It can be modelled as a Lumped

element:

It consists of four basic elements called

‘the 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


The following parameters µ, σ, ε

refer to the insulating material

between conductors

Transmission Lines

General Considerations In the TEM transmission lines:

The followings should be fulfilled

The propagation constant is

The line characteristics impedance is

The wave propagation phase velocity is




between conductors

''CL

'

'

C

G

j

CjG'LjR'

''

''

''0

CjG

LjRZ

fu p

Transmission Lines

Example For an air line with characteristic impedance of 50Ω and phase

constant of 20 rad/m at 700MHz, find the

Capacitance per meter and

the inductance per meter of the line.


An air line (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 ).

'

'

'

' and ''''Im 0

C

L

Cj

LjZCLCjLj

pF/m 9.90501072

20'

8

0

ZC

nH/m 227109.9050''' 122

0 LCLZ

Transmission Lines

Lossless Transmission Lines In the lossless transmission lines, R’

and G’ are set to zero (very small

values) : The propagation constant is

The line characteristics impedance is




between conductors

line) (lossless '

'

0,G' and 0R' since

''

''

0

0

C

LZ

CjG

LjRZ

line) (lossless ''

line) (lossless 0

CL

Transmission Lines

Lossless Transmission Lines In the lossless transmission lines, R’

and G’ are set to zero (very small

values) :

The wave propagation phase velocity is

The wavelength is




between conductors

(m/s) 1

(rad/m)

pu

rr

p

f

c

f

u

01

Transmission Lines

Reflection Coefficient

Voltage Reflection Coefficient (Γ) :

It is the ratio of the amplitude of the reflected voltage wave, V0- to the

amplitude of the incident voltage wave, V0+ at the load.

Z0 for lossless line is a real number while ZL in general is a complex

number. Hence,

A load is matched to the line if ZL = Z0 because there will be no

reflection by the load (Γ = 0 and 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

+.


less)(dimension 1

1

0

0

0

0

0

0

ZZ

ZZ

ZZ

ZZ

V

V

L

L

L

L

rje

Transmission Lines

Example A 100-Ω transmission line is connected

to a load consisting of a 50-Ω resistor in

series with a 10pF capacitor. Find the

reflection coefficient at the load for a

100-MHz signal.


Hz10MHz100 ,100 F,10 ,50 80

11LL fZCR

15950

10102

150

/

118

LLL

jj

CjRZ

7.6076.0

159.15.0

159.15.0

1/

1/

0L

0L

j

j

ZZ

ZZ

Transmission Lines Standing Waves

The Interference between the reflected

wave and the incident wave along a

transmission line creates a standing

wave.

Constructive interference gives maximum

value, and occurs at

Destructive interference gives minimum

value, and occurs at

The repetition period for standing wave

pattern is λ/2 (it is λ for incident and

reflected wave individually).


For a matched line, ZL = Z0, Γ = 0 and

= |V0+| for all values of z. zV

~

For a short-circuited load, (ZL=0), Γ = -1.

For an open-circuited load, (ZL=∞), Γ = 1.

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

0 if ...... 2, 1, 0, n

0 if ...... 3, 2, 1, n

0 where24

r

r

rmax

n

nl

4/ if 4/

4/ if 4/

maxmax

maxmax

min

ll

lll

Transmission Lines

Standing Waves Voltage Standing Wave Ratio

(VSWR):

It is the ratio of the maximum voltage

amplitude to the minimum voltage

amplitude and provides a measure of

the mismatching between the load and

the transmission line.

For a matched load, if

Γ = 0, VSWR = 1

|Γ| - 1, VSWR = ∞


For a matched line, ZL = Z0, Γ = 0 and

= |V0+| for all values of z. zV

~

For a short-circuited load, (ZL=0), Γ = -1.

For an open-circuited load, (ZL=∞), Γ = 1.

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

less)(dimension 1

1~

~

min

max

V

VVSWR

Transmission Lines

Example 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).


6.26

0L

0L 45.05050100

5050100

1/

1/ jej

j

ZZ

ZZ

6.245.01

45.01

1

1

VSWR

Transmission Lines

Lossless TL: input impedance (Zin) It is the ratio of the total voltage (incident and reflected voltages)

to the total current at any point z on the line.

For a line terminated in a short-circuit, ZL = 0:

For a line terminated in an open circuit, ZL = ∞:

Then:


ljZ

lI

lVZ tan~

~

0

sc

scsc

in

ljZlI

lVZ cot~ 0

oc

ococin

ocin

scin ZZZo oc

in

scintan

Z

Zl

0

0

0 ZeeV

eeV

lI

lVZ

ljlj

ljlj

in

02

2

1

1Z

e

elj

lj

ljZZ

ljZZZ

L

L

tan

tan

0

00

Transmission Lines

Example A source with 50 source

impedance drives a 50

transmission line that is 1/8 of

wavelength long, terminated in a

load ZL = 50 – j25 . Calculate:

the voltage reflection

coefficient

the voltage standing-wave

ratio (VSWR)

The input impedance seen

by the source.


Transmission Lines

Example


076

0

0

242.0502550

502550 j

L

LL

ej

j

ZZ

ZZ

64.11

1

L

LVSWR

48

2

14

tan

8.38.30

255050

50255050

tan

tan

0

00

j

j

jj

jZZ

jZZZZ

L

Lin

Transmission Lines

Lossless TL: length of line At the line length of where n is an integer, then

If the transmission line is a quarter wavelength, with

then


2/nl

0tan

2//2tantan

n

nl

2/for ZZ Lin nl

2/4/ nl

24

2

l

2/4/for Z

ZZ

L

2

0in nl

Transmission Lines Example

A 50-Ω lossless transmission line is to be

matched to a resistive load impedance

with ZL=100Ω via a quarter-wave section

as shown, thereby eliminating reflections

along the feedline. Find the characteristic

impedance of the quarter-wave

transformer.


To eliminate reflections at terminal AA’, the input impedance Zin looking into the quarter-wave line should be equal to Z01 (the characteristic impedance of the feedline). Thus, Zin = 50Ω .

Since the lines are lossless, all the incident power will end up getting transferred into the load ZL.

7.701005002

L

202

in

Z

Z

ZZ

Transmission Lines

Lossless TL: Power Flow Average power for incident wave

Average power for reflected wave

The net average power delivered to the

load:


(W) 2 0

2

0iav

Z

VP

iav

2

0

2

02rav

2P

Z

VP

(W) 12

2

0

2

0rav

iavav

Z

VPPP

Transmission Lines

Lossless TL: Return Loss (RL)

When the load is mismatched, not all the of the available

power from the generator is delivered to the load.

This “loss” is called return loss, and is defined (in dB) as:

At matching load case (|Γ|=0),

RL is dB (no reflected power).

At total reflection case (|Γ|=1),

RL is 0 dB (all incident power is reflected).

So return loss will have only values between 0 < RL <


log20RL

Transmission Lines Summary


Transmission Lines

Smith Chart

It is the TL calculator and

used to analyze and design

TL circuits.

Smith Chart parameters are:

The reflection coefficient:

Γ = |Γ|ejθ, (|Γ| ≤ 1), (-180°≤ θ ≤180°)

The normalized impedance:

z = Z / Z0

The normalized admittance: y = 1 / z

SWR and RL scale


Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth

Copyright © 2005 by John Wiley & Sons. All rights reserved.

1

1

L

L

z

z

LL jxrz

1

1L

1

11

L

Lz

y

Transmission Lines Smith Chart


The families of circle for rL and xL.

The complex Γ plane. Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth


For Z0 = 50Ω ,

a ZL = 0 (S.C.)

b ZL = ∞ (O.C.)

c ZL = 100 + j100 Ω

d ZL = 100 - j100 Ω

e ZL = 50 Ω

yL=0.25 - j0.6

zL=0.6 + j1.4

Transmission Lines

Smith Chart

The Smith Chart is a plot of

normalized impedance. For example, if a Z0 = 50 Ω transmission line is

terminated in a load ZL = 50 + j100 Ω. To

locate this point on Smith Chart, Normalize the load impedance, ZNL = ZL/ZN to obtain

ZNL = 1 + j2 Ω

The normalized load impedance is located at the

intersection of the r =1 circle and the x =+2 circle.




A T-line terminated in a load (a) shown with values normalized to Z0 in (b). (c) The

location of the normalized load impedance is found on the Smith Chart.

(c)





(c)





(c)

Transmission Lines

Smith Chart: Input Impedance The input impedance, Zin:

Γ is the voltage reflection coefficient at the

load. The phase angle of Γ is shifted by 2βl

to get ΓL ( from zin to zL; the same |Γ|, but the

phase is changed by 2βl = 2(2π/λ)l= 2π). On

the Smith chart, this means rotating in a

clockwise direction (WTG).

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.


lj

lj

ine

eZZ

2

2

01

1

A 50-Ω transmission line is

terminated with ZL=(100-j50)Ω.

Find Zin at a distance l =0.1λ from

the load.

Transmission Lines

Smith Chart: Reflection Coefficient

The reflection coefficient has a magnitude

and an angle : The magnitude can be measured using a scale for

magnitude of reflection coefficient provided below

the Smith Chart,

The angle is indicated on the angle of reflection coefficient scale shown outside the circle

on chart.


L

jL e

1L



Scale for magnitude of reflection coefficient

0457.0 j

jL

e

e

Transmission Lines

Smith Chart: VSWR 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.

The distance between the load and the first

voltage minimum is lmin=(0.037+0.25)λ.

Move along the constant circle is

akin to moving along the transmission line.


j

L e

Transmission Lines Example

Given that the voltage standing-wave

ratio, VSWR = 3. On a 50-Ω line, 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.

The distance between successive minima

is equal to λ/2. Hence, λ = 40 cm.

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:


125.040

5min l

8.06.0L jz

40308.06.050L jjZ

Transmission Lines Smith Chart

The Smith Chart is a plot of normalized

impedance. For example, if a Z0 = 50 Ω transmission line is terminated in

a load ZL = 50 + j100 Ω and has a line length of 0.3λ.

Normalize the load impedance, ZNL = ZL/ZN to obtain ZNL = 1 + j2 Ω

The normalized load impedance is located at the intersection of

the r =1 circle and the x =+2 circle.

Moving away from the load (towards generator) corresponds

to moving in the clockwise direction on the Smith Chart.

Moving towards the load corresponds to moving in the

anti-clockwise direction on the Smith Chart.

To find ZIN, move towards the generator by:

Drawing a line from the center of chart to outside

Wavelengths Toward Generator (WTG) scale, to get

starting point a at 0.188λ

Adding 0.3λ moves along the constant circle

to 0.488λ on the WTG scale.

Read the corresponding normalized input

impedance point c, ZNIN = 0.175 - j0.08Ω

Denormalizing, to find an input impedance,

VSWR is at point b;






(c)





(c)

j

L e

475.8

0

jZ

ZZZ

IN

NININ

9.5VSWR





(c)



Transmission Lines Smith Chart ZL= 50 - j25 and Z0=50 Ohm. Find Zin, VSWR and ΓL

using the Smith Chart.

Locate the normalized load, and label it as point a,

where it corresponds to

Draw constant circle.

It can be seen that

Move from point a (at 0.356λ) on the WTG scale,

clockwise toward generator a distance λ/8 or

0.125λ to point b, which is at 0.481λ.

We could find that at this point, it corresponds to


5.01 jZNL

j

L e

076245.0 j

L e

66.1VSWR

07.062.0 jZNIN

5.331 jZIN

Transmission Lines Smith Chart The input impedance for a 100 Ω lossless transmission

line of length 1.162 λ is measured as 12 + j42Ω.

Determine the load impedance.

Locate th Normalize the input impedance:

Locate the normalized input impedance and label it

as point a

Take note the value of wavelength for point a at

WTL scale.

At point a, WTL = 0.436λ Move a distance 1.162λ

towards the load to point b (WTL = 0.436λ +

1.162λ = 1.598λ ). But, to plot point b, 1.598λ –

1.500λ = 0.098λ.

Read the point b:

Denormalized it:


42.012.0100

4212

0

jj

Z

Zz in

in

7.015.0 jZNL

7015

0

j

ZZZ NLL

Transmission Lines Smith Chart On a 50 lossless transmission line, the VSWR is

measured as 3.4. A voltage maximum is located

0.079λ away from the load (towards generator).

Determine the load.

Use the given VSWR to draw a constant circle.

Then move from maximum voltage at WTG =

0.250λ (towards the load) to point a at WTG =

0.250λ - 0.079λ = 0.171λ.

At this point we have

ZNL = 1 + j1.3 Ω,

or ZL = 50 + j65 Ω.


j

L e

Transmission Lines

Smith Chart: Impedance Matching

Transmission line is matched to the load when Z0 =

ZL; which no reflection occurs.

This is usually not possible since ZL is used to serve

other application.

Alternatively, we can place an impedance-

matching network between load and transmission

line (The load impedance will be transformed to be

equal to Z0 of the transmission line.

The important of impedance matching or tuning:

Maximum power is delivered when the load is matched

to the line.

The power loss in the feed line is minimized. (increase

power handling capability by optimizing VSWR)

Improved the signal-to-noise ratio (SNR). (e.g with

controlled mismatch, an amplifier can operate with

minimum noise generation)

Reduced the amplitude and phase errors.


Matching

Network

Load

ZL

Z0



Adding an impedance-matching network ensures that all power will make it to the

load.Techniques of impedance matching :

• Quarter-wave transformer

• Single / double stub tuner

• Lumped element tuner

• Multi-section transformer

Transmission Lines


Quarter Wave Transformer It can only be constructed if the load impedance is all

real (no reactive component)

Designed to achieve matching at a single

frequency/narrow bandwidth

Easy to build and use

The input impedance, Zin (looking into the quarter

wave long section of lossless ZS impedance line

terminated in a resistive load RL):




Quarter-wave transformer.

ljRZ

ljZRZZ

LS

SLSin

tan

tan

, 24

2

l

l tan

LS RZZ 0

Transmission Lines


Quarter Wave Transformer

Example

Calculate the position and characteristic impedance of

a quarter wave transformer that will match a load

impedance, RL = 15Ω; to a 50 Ω line.

The transformer’s impedance

the position of quarter-wave transformer from the

load:

d = 0.25 λ




Quarter-wave transformer.

39.27)15)(50(0 LS RZZ

Transmission Lines



Example

A transistor has an input impedance of ZL = 25 Ω, at

an operating frequency of 500 MHz.

Assume:

Thickness of the dielectric is d = 1 mm,

The relative dielectric constant εr is 4.

The surface resistance, R and shunt conductance, G, are

be neglected.

Find:

The length, l

The width, w

The characteristic impedance of the quarter-wave

parallel plate line transformer for which matching is

achieved.


Transmission Lines Smith Chart: Impedance Matching


Example

ZL = 25 Ω, f = 500 MHz, d= 1mm, εr = 4, and R=G=0

The characteristic of the line is

From the table,

The line length l follows from the condition

The input impedance of the combined transmission line and

the load is:

where d = l = λ/4 and the reflection coefficient is


355.35)25)(50(0 LS RZZ

w

d

C

LZ

p

Line mmZ

dw

rline

p329.5

0

0

mnHw

dL

p/8.235

mpFd

wC

p

/6.188

o = 8.85 x 10-12 F/m

µo = 4 x 10-7 H/m

mmLCf

l 967.744

1

4

)(1

)(1

tan

tan

d

dZ

djZZ

djZZZZ line

Lline

lineLlinein

d

v

fj

ZZ

ZZed

plineL

lineLdj 22exp)( 2

0

Transmission Lines


Stub Matching

Shunt stub matching network

The matching network has to

transform the real part of load

impedance, RL to Z0 and reactive

part, XL to zero Use two adjustable

parameters – e.g. shunt-stub. Thus,

The main idea of shunt stub matching

network is to:

Find length d and l in order to get yd

and yl .

Ensure total admittance ytot = yd + yl =

1 for complete matching network.


YLY0

d

lY=1/Z

Y0

Open

or

shorted

stub(a)

YLZ0

d

lZ=1/Y

Z0

Open

or

shorted

stub(b)

Z0

Single-stub tuning circuits. (a) Shunt stub. (b) Series stub.

Transmission Lines


Stub Matching

Shunt stub matching network using

Smith Chart

Locate the normalized load impedance ZNL.

Draw constant SWR circle and locate YNL.

Move clockwise (WTG) along circle to

intersect with 1 ± jB value of yd.

The length moved from YNL towards yd is the

through line length, d.

Locate yl at the point jB .

Depends on the shorted/open stub, move

along the periphery of the chart towards yl

(WTG).

The distance traveled is the length of stub, l .


YLY0

d

lY=1/Z

Y0

Open

or

shorted

stub(a)

YLZ0

d

lZ=1/Y

Z0

Open

or

shorted

stub(b)

Z0

Single-stub tuning circuits. (a) Shunt stub. (b) Series stub.

j

L e


Shunt stub matching network using Smith Chart

Example

Construct the shorted shunt stub matching network for a

50Ω line terminated in a load ZL = 20 – j55Ω

Locate the normalized load impedance,

ZNL = ZL/Z0 = 0.4 – j1.1Ω


Locate YNL. (0.112λ at WTG)

Moving to the first intersection with the ± jB circle, which

is at 1 + j2.0 yd

Get the value of through line length, d from 0.112λ to

0.187λ on the WTG scale, so d = 0.075λ

Locate the location of short on the Smith Chart (note: when

short circuit, ZL = 0, hence YL = ∞) on the right side of the

chart with WTG=0.25λ.

Move clockwise (WTG) until point jB, which is at 0 - j2.0,

located at WTG= 0.324λ yl

Determine the stub length, l 0.324λ – 0.25λ = 0.074 λ




Figure 6-28bc (p. 302) (b) Adding shunt admittances. (c) Using the Smith Chart to find through line and

stub lengths. Values on the chart apply to Example 6.7.

j

L e

Thus, the values are:

d = 0.075 λ

l = 0.074 λ

yd = 1 + j2.0 Ω

yl = -j2.0 Ω

Where

YTOT = yd + yl = (1 + j2.0) + (-j2.0) = 1



Example

Construct the open ended shunt stub matching network

for a 50Ω line terminated in a load ZL = 150 + j100Ω

Locate the normalized load impedance,

ZNL = ZL/Z0 = 3.0 + j2.0Ω


Locate YNL. (0.474λ at WTG)

Moving to the first intersection with the ± jB circle, which

is at 1 + j1.6 yd

Get the value of through line length, d from 0.474λ to

0.178λ on the WTG scale, so d = 0.204λ

Locate the location of open end on the Smith Chart (note:

when open circuit, ZL = ∞, hence YL = 0) on the left side of

the chart with WTG=0.00λ.

Move clockwise (WTG) until point jB, which is at 0 – j1.6,

located at WTG= 0.339λ yl

Determine the stub length, l 0.339λ – 0.00λ = 0.339 λ


j

L e

Thus, the values are:

d = 0.204 λ

l = 0.339 λ

yd = 1 + j1.6 Ω

yl = -j1.6 Ω

Where

YTOT = yd + yl = (1 + j1.6) + (-j1.6) = 1



Example

50-Ω transmission 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.

The normalized load impedance is

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).

First matching points, C.

At C, is at 0.178λ on WTG scale.

Distance B and C is d= (0.178λ- 0.115λ)= 0.063λ

Normalized input admittance

at the juncture is:

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

Normalized admittance of −j 1.58 at F and position 0.34λ on the

WTG scale gives: l1= (0.34λ- 0.25λ)= 0.09λ


jj

Z

Zz

5.0

50

5025

0

LL

(located at A).

8.04.0L jy

58.11d jy

58.1

58.1101

s

s

sin

jy

jyj

yyy d



Example

50-Ω transmission 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.

The normalized load impedance is

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).

Second matching points, D.

At D, is at 0.322λ on WTG scale.

Distance B and D is d= (0.322- 0.115λ)= 0.207λ

Normalized input admittance at the juncture is: at G

Rotating from point E to point G, we get


jj

Z

Zz

5.0

50

5025

0

LL

(located at A).

8.04.0L jy

58.11 jyd

58.1s jy

41.016.025.02 l

ANTENNAS and MICROWAVES ENGINEERING …...Voltage Standing Wave Ratio (VSWR): It is the ratio of the...

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Transcript of ANTENNAS and MICROWAVES ENGINEERING …...Voltage Standing Wave Ratio (VSWR): It is the ratio of the...