ch2

EE 330

1 transmission lines

Transmission Lines

V)10sin(50)( 5 ttvs π=

l

What if this is 5 km ???

+vC

−2.5 H

0.1 F

6 Ω

3 Ω)(tvs+−~

6 Ω

+vC

−2.5 H

0.1 F 3 Ω+−~)(tvs

EE 330


EE 330


Lumped Element Model

Δ ΔΔ

Δ Δ

Δ Δ Δ Δ Δ

Δ Δ Δ Δ Δ Δ

EE 330


Transmission lines transmit energy and signals from a source (generator) to a load

Distinguishing characteristics of a transmission line:-

the devices to be connected are separated by distances comparable to or larger than the signal wavelength-

the parameters of the circuit are distributed and are evaluated on a per-unit length basis: R' –

resistance per unit length, Ω/mL' –

inductance per unit length, H/mG' –

conductance per unit length, S/mC' –

capacitance per unit length, F/m-

transmission lines are circuit elements that have complex impedances, which are functions of line length and signal frequency

TEM transmission lines-

two-wire line-

coaxial cable-

parallel-plate line

01.0≥λl+

−∼Vg

Rg

ZL

generator transmission line load

fu p=λ

l

01.0≥λl+

−∼Vg

Rg

ZL


fu p=λ 01.0≥

λl+

−∼+−∼Vg

Rg

ZL


fu p=λ

l

EE 330


coaxial line

two wire line

parallel plate

microstrip

EE 330


Notice: εμ=''CL

EE 330


Power loss: due to non-ideal conductors and dielectrics

+−∼Vg

Rg

ZL

zL Δ' zL Δ' zL Δ' zL Δ'

zC Δ' zC Δ' zC Δ' zC Δ'

z – l 0

Lossless transmission line model (R' = 0, G' = 0)

Dispersion: due to frequency dependent properties of the materials

EE 330


Examples:

1. Compute the line parameters for the following lossless transmission lines:

A. A coaxial cable with diameters of the concentric conductors 2a = 0.4 mm and 2b = 3 mm employs polyethylene as insulator between the conductors.

μH/m 4.04.0

3ln2

104ln2

'7

0 =⎟⎠⎞

⎜⎝⎛×

=⎟⎠⎞

⎜⎝⎛=

−

ππ

πμμ

abL r

pF/m 9.9

6.010ln

10854.8

ln'

120 =

⎟⎠⎞

⎜⎝⎛

××=

⎟⎠⎞

⎜⎝⎛

=−πεεπ

ad

C r

12

2

>>⎟⎠⎞

⎜⎝⎛

ad

B. The conducting wires of a two-wire line have 1.2-mm diameter and are a distance of 1 cm from one another. The two wires are in air.

Assume

pF/m 62

4.03ln

10854.825.22

ln

2'12

0 =⎟⎠⎞

⎜⎝⎛

×××=

⎟⎠⎞

⎜⎝⎛

=−πεεπ

ab

C r

μH/m 1.16.0

10ln104ln'7

0 =⎟⎠⎞

⎜⎝⎛×

=⎟⎠⎞

⎜⎝⎛=

−

ππ

πμμ

adL r

0'=R

0'=G

0'=R

0'=G

EE 330


Examples:

2. Compute the line parameters at 1 MHz for a coaxial air line with

an inner conductor diameter of 0.6 cm and an outer conductor diameter of 1.2 cm. The conductors are made of copper. Assume the air has zero conductivity.

Ω×=×

×××== −

−4

7

760 106.2

108.510410 ππ

σμμπ

c

rcs

fR

/m1008.2006.01

003.01

2106.211

2' 2

4

Ω×=⎟⎠⎞

⎜⎝⎛ +

×=⎟

⎠⎞

⎜⎝⎛ += −

−

ππ baRR s

μH/m 14.06.02.1ln

2104ln

2'

70 =⎟

⎠⎞

⎜⎝⎛×

=⎟⎠⎞

⎜⎝⎛=

−

ππ

πμμ

abL r

pF/m 26.80

6.02.1ln

10854.82

ln

2'12

0 =⎟⎠⎞

⎜⎝⎛

××=

⎟⎠⎞

⎜⎝⎛

=−πεεπ

ab

C r

0 todue 0' == σG

EE 330


Generalized Transmission Line

For AC currents and voltages, using phasor

notations and applying Kirchhoff’s voltage and current laws to a section of the line:

+

v(z,t)

–

zL Δ'

zC Δ'

Δz

+

v(z + Δz,t)

–

i(z,t) i(z + Δz,t)

Phasor

notations:

( ) ( ) ]~Re[, tjezVtzv ω=

( ) ( ) ]~Re[, tjezItzi ω=

( ) ( ) ( )

( ) ( ) ( )zVCjGtdzId

zILjRtdzVd

~''~

~''~

ω

ω

+=−

+=−

telegrapher’s equations

Transmission line equations:

KVL, KCL

zR Δ'

zG Δ'

Δ ΔΔ

Δ Δ

Δ Δ Δ Δ Δ

Δ Δ Δ Δ Δ Δ

Δ ΔΔ

Δ Δ

Δ Δ Δ Δ Δ

Δ Δ Δ Δ Δ Δ

EE 330


Upon uncoupling the telegrapher’s equations, two second order differential equations are obtained: ( ) ( )

( ) ( ) 0~~

0~~

22

2

22

2

=−

=−

zItd

zId

zVtd

zVd

γ

γ

wave equations

( )( )'''' CjGLjRj ωωβαγ ++=+=

]Im[γβ =

– complex propagation constant

– phase constant, rad/m

The general solution to the wave equation is a sum of two waves,

one traveling from the generator toward the load and another one traveling from the load toward the generator:

( )( ) zz

zz

eIeIzI

eVeVzVγγ

γγ

−−+

−−+

+=

+=

00

00~

~

]Re[γα = – attenuation constant, Np/m

EE 330


Traveling waves parameters:

amplitude:

frequency:

phase velocity on the line:

wavelength on the line:

zz eVeV αα −−+00 ,

πω2

=f

βω

=pu

fu p=λ

( ) ( ) ( )−−+−+ ++++−= φβωφβω αα zteVzteVtzv zz coscos, 00

( ) zjzjzjzj eeeVeeeVzV βαφβαφ −+ −−−+ += 00~

Phasor

notation :

incident wave,

reflected wave,

Instantaneous function (time domain):

0 z

Zg

+−∼ ZLgV~

TL

( )tv+ ( )tv−

( )tv+ ( )tv−

( ) zz eVeVzV γγ −−+ += 00~

EE 330


Current in terms of voltage (using the solutions to the wave equations into the telegrapher’s equations):

( ) ( )

zz

zz

eZVe

ZV

eVeVLjR

zI

γγ

γγ

ωγ

0

0

0

0

00''~

−−

+

−−+

−=

=−+

=

where

0

0

0

0

0

''''

''

ZCjGLjR

LjRIV

IV

=++

=

=+

=−

= −

−

+

+

ωω

γω

– characteristic impedance of the line, Ω

EE 330


A lossless coaxial cable with diameters of the concentric conductors 2a = 0.4 mm and 2b = 3 mm employs polyethylene as insulator between the conductors.

μH/m 4.0'=L

12

2

>>⎟⎠⎞

⎜⎝⎛

ad

The conducting wires of a two-wire lossless line have 1.2-mm diameter and are a distance of 1 cm from one another. The two wires are in air.

Assume

pF/m 62'=CFrom before: 0'=R 0'=G

Ω==++

= 3.80''

''''

0 CL

CjGLjRZ

ωω

μH/m 1.1'=L pF/m 9.9'=CFrom before: 0'=R 0'=G

Ω==++

= 333''

''''

0 CL

CjGLjRZ

ωω

Examples:

3. Compute the characteristic impedance of the transmission lines from examples 1 and 2.

EE 330


Signal frequency of 1 MHz, a coaxial air line with an inner conductor diameter of 0.6 cm and an outer conductor diameter of 1.2 cm. The conductors

are made of copper. Assume the air has zero conductivity.

μH/m 14.0'=L pF/m 26.80'=CFrom before: Ω×= −21008.2'R 0'=G

Ω°−∠=×

=×

+×=

=×××

×××+×=

++

=

°−

°

−

−

−

−−

35.14210588.0

10588.01008.2

1026.801021014.01021008.2

''''

904

65.88

4

2

126

662

0

j

j

ee

jj

jj

CjGLjRZ

ππ

ωω

EE 330


Lossless Transmission Line

''''2 CLjCL ωωγ =−=

εμωωγβ === '']Im[ CL

Complex propagation constant:

Phase constant:

Attenuation constant: 0]Re[ == γα

''

0 CLZ =Characteristic impedance:

εμβω 1

''1

===CL

u pPhase velocity:

εμωπ

ωπ

βπλ 2

''22

===CL

Wavelength:

0'=R 0'=G

(real quantity)

EE 330


H/m]['L

F/m]['C

][''

0 Ω=CLZ

parameter

coaxial line two wire line parallel plate

( )ad 2>>

⎟⎠⎞

⎜⎝⎛

abln

2πμ

( )abln2 επ

wdη

wdμ

dwε

⎟⎠⎞

⎜⎝⎛

abln

2πη

⎟⎠⎞

⎜⎝⎛

adln

πμ

( )adlnεπ

⎟⎠⎞

⎜⎝⎛

adln

πη

εμη = – intrinsic impedance of the dielectric

material between the two conductors εμ=''CL

Lossless TEM Transmission Lines

EE 330


Examples:

4. Design a parallel-plate transmission line according to the following requirements: (a) characteristic impedance of 50 Ω; (b) capacitance per unit length of 100 pF/m. You have at your disposal three dielectric materials, namely, teflon, polyethylene, and polystyrene, in the form of 1.9-mm thick strips. Assume the insulating materials are perfect dielectrics and the metal plates are perfect conductors.

μH/m 25.05010100'''' 2122

00 =××==⇒= −ZCLCLZ

wide.mm 9.6 bemust plates theand nepolyethyle bemust material dielectric The2.6 :epolystyren 2.25 : nepolyethyle 2.1 :teflon === rrr εεε

mm6.91025.0

0019.0104'

' 67

00 Ldw

wdL r =

××==⇒= −

−πμμμ

25.20096.010854.8

0019.010100'' 12

12

00 =

××××

==⇒= −

−

wdC

dwC rr ε

εεε

wd

CLZ η==

''

0 wdL μ='

dwC ε='

0εεε r=00 μμμμ == r

1material?

size?

?=rε

?=w

EE 330


Generalized impedance on the line:

( ) ( )( ) ⎥

⎦

⎤⎢⎣

⎡−+

== −−+

−−+

zjzj

zjzj

eVeVeVeVZ

zIzVzZ ββ

ββ

00

000~

~

At z = 0, load impedance: ( ) 000

000 ZVVVVZZL ⎟⎟

⎠

⎞⎜⎜⎝

⎛−+

== −+

−+

At z = –

l, input impedance: ( )lZZin −=

+−⎟⎟⎠

⎞⎜⎜⎝

⎛+−

= 00

00 V

ZZZZV

L

L

Reflected wave amplitude in terms of incident wave amplitude:

Note: No reflected wave ( ), when !!!00 =−V 0ZZL =

( )

( ) zjzj

zjzj

eZVe

ZVzI

eVeVzV

ββ

ββ

0

0

0

0

00

~

~

−−

+

−−+

−=

+=

0 z

+−∼

z = –

l

( )zZ LZinZ

EE 330


Relating the reflection coefficient to the load impedance through and using Euler’s identity and some trigonometry:

( )( )⎥⎦

⎤⎢⎣

⎡++

=lZjZlZjZZZ

L

Lin β

βtantan

0

00

+−⎟⎟⎠

⎞⎜⎜⎝

⎛+−

= 00

00 V

ZZZZV

L

L

( ) ( )( )⎥⎦

⎤⎢⎣

⎡−+−+

=zZjZzZjZZzZ

L

L

ββ

tantan

0

00

A distance from the load:zl −=

( ) ⎥⎦

⎤⎢⎣

⎡

−+

= −−+

−−+

zjzj

zjzj

eVeVeVeVZzZ ββ

ββ

00

000

EE 330


–

reflection coefficient in terms of load impedance

0

0

ZZZZ

L

L

+−

=Γ

Generally, the load impedance is a complex number, therefore the reflection coefficient is also a complex number:

ΓΓ=Γ θje

Note: 1≤Γ

+−⎟⎟⎠

⎞⎜⎜⎝

⎛+−

= 00

00 V

ZZZZV

L

L

Voltage reflection coefficient at the load:

+

−

=−+

−

==Γ0

0

00

0

VV

eVeV

zzj

zj

β

β

EE 330


( ) ( )( ) ( )zjzjzjzj

zjzjzjzj

eeZVe

ZVe

ZVzI

eeVeVeVzV

ββββ

ββββ

Γ−=−=

Γ+=+=

−+−

−+

−+−−+

0

0

0

0

0

0

000

~

~

( ) ( )( )

( )( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛Γ−Γ+

=⎥⎦

⎤⎢⎣

⎡Γ−Γ+

== −+

−+

zj

zj

zjzj

zjzj

eeZ

eeVeeVZ

zIzVzZ β

β

ββ

ββ

2

2

00

00 1

1~~

Voltage and current in terms of reflection coefficient:

Impedance in terms of reflection coefficient:

At z = – l, ( ) ( )( ) ⎥

⎦

⎤⎢⎣

⎡Γ−Γ+

=−−

=−= −

−

lj

lj

in eeZ

lIlVlZZ β

β

2

2

0 11

~~

EE 330


Examples:

5. A 75-Ω

transmission line is terminated in two different loads: (a) ZL = 75 Ω, and (b) open circuit. Find the corresponding reflection coefficients

at the load.

line matched075757575a)

0

0

⇒=+−

=Γ

+−

=ΓZZZZ

L

L

111b)

0

0 ⎯⎯ →⎯+−

=Γ ∞→LZ

L

L

ZZZZ

EE 330


Examples:

6. The reflection coefficient at the load is measured when a 75-Ω

transmission line is terminated in three different loads. The corresponding values of

Γ are as follows: (a) Γ

= –

j0.5, (b) Γ

= –

1, (c) Γ

= 0.5. Find the load impedances.

Ω=−+

×= 2255.015.0175c) LZ

Γ−Γ+

=11

0ZZL

Ω−==+−

×= °− 6045755.015.0175a) 13.53 je

jjZ j

L

circuitshort 0111175b) ⇒Ω=

+−

×=LZ

0

0

ZZZZ

L

L

+−

=Γ

EE 330


Power flow:

( ) ( )( ) ( )zjzj

zjzj

eeZVzI

eeVzV

ββ

ββ

Γ−=

Γ+=

−+

−+

0

0

0

~

~

Instantaneous incident power at the load (z = 0):

( ) ( ) ( ) [ ] ( )+++

+++ +=⎥⎥⎦

⎤

⎢⎢⎣

⎡==

++

φωωφωφ tZ

Vee

ZV

eeVtitvtP tjjtjji 2

0

2

0

0

00 cosReRe

Instantaneous reflected power at the load (z = 0):

( ) ( ) ( ) [ ] =⎥⎥

⎦

⎤

⎢⎢

⎣

⎡Γ−Γ==

+Γ

+Γ

++−− tjjjtjjjr ee

Z

VeeeVetitvtP ωφθωφθ

0

00 ReRe

( )+Γ

+

++Γ−= φθωtZ

V2

0

202 cos

0 z

Zg

+−∼ ZLgV~ ( )tPi ( )tPr

– l

EE 330


Time-average incident power at the load (z = 0):

( )0

20

0 21

Z

VtdtP

TP

Tii

av

+

== ∫

Time-average reflected power at the load (z = 0):

( ) iav

Trr

av PZ

VtdtP

TP 2

0

202

0 21

Γ−=Γ−==+

∫

The same result can be obtained easier using phasors. The average value of a product of two time-harmonic quantities is:

*~~Re21 IVPav ⋅=

Power delivered to the load: rav

iavav PPP +=

Percentage of the incident power delivered to the load:

211 Γ−=+=+

= iav

rav

iav

rav

iav

iav

av

PP

PPP

PP

10 ≤≤ iav

av

PP

Note:

EE 330


Examples:

7. A load impedance ZL = 80 –

j100 Ω

terminates a 50-Ω

line. The operating frequency is 100 MHz and the wavelength on the line is 2 m. Find (a) the phase velocity of the wave on the line, and (b) the percentage of power absorbed by the load.

%4.59594.0637.011637.0

637.050100805010080b)

22

7.35

0

0

==−=Γ−=⇒=Γ

=+−−−

=+−

=Γ °−

iav

av

j

L

L

PP

ejj

ZZZZ

8. Consider the transmission line from the previous example and find what percentage of the incident power is delivered to the load when the line is terminated in a load impedance ZL = Z0

=

50 Ω.

%100110 2

0

0 ==Γ−=⇒=+−

=Γ iav

av

L

L

PP

ZZZZ

m/s 10221010022a) 86 ×=××==== λ

λππ

βω ffu p

EE 330


Examples:

9. A voltage generator with internal impedance Zg = 100 Ω

is connected to a λ/8-long 50-Ω

transmission line. A load impedance ZL = 50 + j50 Ω

terminates the line. Find

(a)

the input impedance,

( ) ( )V 90120sin220 °−−= ttvg π

( )( )

( ) ( ) Ω−=⎥⎦

⎤⎢⎣

⎡++++

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛ ×++

⎟⎠⎞

⎜⎝⎛ ×++

=

=⎥⎦

⎤⎢⎣

⎡++

=

5010050505050505050

82tan505050

82tan505050

50

tantan

0

00

jjjjj

jj

jj

lZjZlZjZZZ

L

Lin

λλπ

λλπ

ββ

Zin

0 z

Zg

+−∼ ZLgV~

– l

Z0

EE 330


(b)

the reflection coefficient at the load,

4.02.0505050505050

0

0 jjj

ZZZZ

L

L +=++−+

=+−

=Γ

(c)

the phasor

of the input voltage

( )

( ) V3.11950100100

50100220

~~~

5.12 °−=−+

−=

=+

=−=

j

ing

ingi

ej

j

ZZZV

lVV

Zin

0 z

Zg

+−∼ ZLgV~

– l

−

+

iV~

EE 330


(d)

the average power delivered to the load.

*~~Re21

LLav IVP ⋅= ( ) ( ) ( ) ( )Γ−==Γ+==+

+ 10~~,10~~

0

00 Z

VIIVVV LL

( ) ljljiljlj

i eeVVeeVV ββ

ββ−

+−+

Γ+=⇒Γ+=

~~00

2.

Plug in into the expressions about and +0V LI~LV~

W571.161Re215.14.107Re

21 457631 === °°°− jjj

av eeeP

Analysis:

Computation:

4.

Plug in and into the expression about the average powerLV~ ∗LI~

3.

Take the complex conjugate of LI~

1.

Compute +0V

( )V2.85

4.13.119

4.02.0

3.119~4.49

9.36

5.12

44

5.12

0°−

°

°−

−

°−

−+ ==

++

=Γ+

= jj

j

jj

j

ljlji e

ee

eje

eee

VV ππββ

( ) ( )

( ) ( ) A5.150

4.02.012.851~

V4.1074.02.012.851~

764.49

0

0

314.490

°−°−+

°−°−+

=−−

=Γ−=

=++=Γ+=

jj

L

jjL

ejeZVI

ejeVV

A5.1~ 76* °= jL eI

EE 330


Standing WavesThe interference of the incident and the reflected voltage waves

gives rise to a pattern called standing wave, that is, a wave-like spatial distribution of the voltage magnitude on the transmission line.

( )zV~

z

2λ

4λ

minl

maxl

2λ

( )max

~ zV

( )min

~ zV

( )zI~

z

2λ

4λ

maxl

minl

2λ

( )max

~ zI

( )min

~ zI

This “wave”

does not travel along the transmission line, it is “frozen”.Two neighboring maximums (or minimums) are spaced by a half-

wavelength.

A similar pattern is formed by the interfering current waves.

The two patterns are in phase opposition.

EE 330


disturbance amplitude

Standing wave on a string

EE 330


( ) ( ) ( ) ( )Γ+∗ +Γ+Γ+== θβ zVzVzVzV 2cos21~~~ 2

0

Mathematically, the voltage standing wave pattern is described by

when( ) ( )Γ+= + 1~0max

VzV ( ) 12cos =+ Γθβ zπθβ nz 22 −=+ Γ

⎩⎨⎧

<=≥=

+=+

==−Γ

ΓΓΓ

0if...2,10if...2,1,0

,242

2max θ

θλπθλ

βπθ

nnnnlz

when( ) ( )Γ−= + 1~0min

VzV ( ) 12cos −=+ Γθβ z( )πθβ 122 ±−=+ Γ nz

⎩⎨⎧

≥−<+

±=±+==− Γ

4if""4if""

,4424 max

maxmaxmin λ

λλλλπθλ

ll

lnlz

Voltage standing wave ratio (VSWR): ( )( ) Γ−

Γ+==

11

~

~

min

max

zV

zVS

The extrema

in the pattern are related to the magnitude of Γ, and their positions are related to the phase of Γ. Therefore, from the standing wave pattern, the reflection coefficient can be constructed and then the load impedance and the power delivered to the load can be found.

1≥S

EE 330


Computing Γ from the standing wave pattern:1.

Compute the VSWR by taking the ratio of the maximum and minimum values of the voltage2.

Compute the magnitude of Γ from VSWR.3.

Determine the distance from the load in wavelengths of the nearest maximum or minimum.4.

Find the phase of Γ, θΓ

, using the result from step 3.5.

Construct the reflection coefficient in polar form from the results from steps 2 and 4.

Plotting the standing wave pattern from Γ:1.

Compute VSWR from the magnitude of Γ. If the amplitude of the incident wave is given, compute the actual values of minimum and maximum voltage on the line.

2.

Determine the distance from the load in wavelengths of the nearest maximum or minimum from the phase of Γ. If the signal wavelength on the line is given or can be determined from the data, compute the actual physical distance from the load.

3.

Draw a coordinate system in which the load is placed at z = 0 and the transmission line occupies the negative values of z.

4.

Plot the standing wave pattern starting from the load and moving

to the left where

z < 0. In steps of λ/4 mark the positions of several minimums and maximums starting from the position determined in step 2 and altering maximums and minimums. Make sure your pattern goes up to the maximum voltage value at the positions of

the maximums and down to the minimum voltage value at the positions of the minimums. Note: If you do not know the amplitude of the incident voltage wave and the wavelength on the

line, draw a normalized pattern, i.e., in terms of incident voltage amplitude and wavelength.

EE 330


Examples:

10. Determine the reflection coefficient at the load from the measured voltage standing wave pattern.

–33.5

4

6

V [V]

z[cm]–83.5

( )( )

5.146

~

~

min

max ===zV

zVS

2.051

15.115.1

11

==+−

=+−

=ΓSS

m2m5.04

m835.0 m,335.0 minmax =⇒=⇒== λλll

°==⎟⎠⎞

⎜⎝⎛ −=⇒+= Γ

Γ 6.12067.02

424 maxmax πλ

λπθλ

πθλ nlnl

πθ 67.02.0 jj ee =Γ=Γ Γ

EE 330


Examples:11. Plot the normalized voltage standing wave pattern if a 50-Ω

line is terminated in a load ZL = 40 + j50 Ω.

rad 26.1,495.0495.0505040505040 26.1

0

0 ==Γ⇒=++−+

=+−

=Γ Γθj

L

L ejj

ZZZZ

( ) ( ) ++ =Γ+= 00max495.11~ VVzV ( ) ( ) ++ =Γ−= 00min

505.01~ VVzV

K

K

,2,1,0,2

35.025.0

,2,1,0,2

1.024

26.124

maxmin

max

=⎟⎠⎞

⎜⎝⎛ +=+=

=⎟⎠⎞

⎜⎝⎛ +=⎟

⎠⎞

⎜⎝⎛ +=⎟⎟

⎠

⎞⎜⎜⎝

⎛+= Γ

nnll

nnnnl

λλ

λλπ

λπ

θ

–0.1

0.505

1.495

–0.35

( )+

0

~

V

zV

λz–0.85 –0.6

EE 330


Examples:

12. When a transmission line is terminated in load A, the voltage standing wave ratio is S = 5.8. When the same line is terminated in load B, it is S = 1.5. What percentage of the incident power is delivered to load in these two cases?

22

1111 ⎟

⎠⎞

⎜⎝⎛

+−

−=Γ−=SS

PP

iav

av

%505.018.518.51 :A Load

2

==⎟⎠⎞

⎜⎝⎛

+−

−=iav

av

PP

%9696.015.115.11 :B Load

2

==⎟⎠⎞

⎜⎝⎛

+−

−=iav

av

PP

EE 330


A transmission line is used to deliver signals to a load.

The reflection coefficient at the load is

What is the voltage standing wave ratio on the line?

2

31 πj

e−=Γ

A.

2B.

1C.

1/2D.

0.8 + j0.6E.

0.8 –

j0.6

Probing Question

2311311

11

31

=−+

=Γ−Γ+

=⇒=Γ S

EE 330


Special Cases

Short-circuited line: 0=LZ0Z

z2λ

−4λ

−

+02V

( ) ( ) ,21~00max

++ =Γ+= VVzV

( ) ( ) ,01~0min

=Γ−= +VzV

24maxmaxλλ nlz +==−

2minminλnlz ==−

∞=Γ−Γ+

=11

S

10

0 −=+−

=ΓZZZZ

L

L πθ ==Γ Γ,1

01 2 =Γ−=iav

av

PP

All the power is reflected back toward the generator.

EE 330


( ) ( ) XjzZjzZ sc =−= βtan0

The impedance at any point on the line is pure imaginary,i.e., reactive (either inductive or capacitive):

X

X < 0, capacitive

X > 0, inductiveX = 0, short circuit

X = ∞, open circuit

EE 330


Open-circuited line: ∞=LZ0Z

( ) ( ) ,21~00max

++ =Γ+= VVzV

( ) ( ) ,01~0min

=Γ−= +VzV24minλλ nz +=−

2maxλnz =−

∞=Γ−Γ+

=11

S

111

0

0 =+−

=ΓL

L

ZZZZ 0,1 ==Γ Γθ

01 2 =Γ−=iav

av

PP

All the power is reflected back toward the generator.

z2λ

−4λ

−

+02V

EE 330


The impedance at any point on the line is pure imaginary,i.e., reactive (either inductive or capacitive):

X

X < 0, capacitive

X > 0, inductive

X = 0, short circuit

X = ∞, open circuit

( ) ( ) Xjz

ZjzZ oc =−

−=βtan

0

EE 330


Matched line: 0ZZL =0Z

111

=Γ−Γ+

=S

00

0 =+−

=ΓZZZZ

L

L 0,0 ==Γ Γθ

11 2 =Γ−=iav

av

PP

All the power is absorbed by the load.

( ) 0ZzZZin == –

the impedance at any point on the line is resistiveand equals the characteristic impedance.

( ) ( ) +Γ

+ =+Γ+Γ+= 02

0 2cos21~ VzVzV θβ

z

+0V

–

There is no reflected wave.

EE 330


Examples:

13. The input impedance of a 75-Ω

open circuited transmission line is Zin = j25 Ω. Determine the length of the line in terms of the signal wavelength.

( ) ⎟⎠⎞

⎜⎝⎛−=−=

λπβ lZjlZjZ oc

in 2cotcot 00

The line length can be any length that satisfies the above expression, the shortest being 0.3λ for n = 1.

K,2,1,0,2

2.023

1cot21

27525cot

21

2cot

21

cot2

11

0

1

0

1

=+−=+⎟⎠⎞

⎜⎝⎛−=+⎟

⎠⎞

⎜⎝⎛=+⎟⎟

⎠

⎞⎜⎜⎝

⎛=

+⎟⎟⎠

⎞⎜⎜⎝

⎛=

−−−

−

nnnnjjnZZjl

nZZjl

ocin

ocin

πππλ

πλ

π

EE 330


Examples:

14. The input impedance of a transmission line is j25 Ω when it is short circuited and –

j100 Ω

when it is open circuited. Determine the characteristic impedance line.

( )( )lZjZ

lZjZocin

scin

β

β

cot

tan

0

0

−=

=

( ) Ω=−== 50100250 jjZZZ ocin

scin

( )[ ] ( )[ ] 2000 cottan ZlZjlZjZZ oc

inscin =−= ββ

EE 330


Examples:

15. An air-filled 2-m long lossless coaxial cable is short circuited. Determine what

type of load (capacitive, inductive, resistive, short circuit, open circuit) this cable is to a source generating signals of frequency (a) 18.75 MHz, (b) 37.5 MHz, (c) 56.25 MHz, (d) 75 MHz.

( )

( ) inductive 4

tan103

1075.1822tanMHz75.18

2tan2tantan

008

6

0

000

⇒=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛

××××

=

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛==

ZjZjZjZ

cflZjlZjlZjZ

scin

scin

ππ

πλπβ

(a)

(b)

(c)

(d)

( ) circuitopen 2

tan103

105.3722tanMHz5.37 08

6

0 ⇒∞=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛

××××

=ππ ZjZjZ sc

in

( ) capacitive 4

3tan103

1025.5622tanMHz25.56 008

6

0 ⇒−=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛

××××

= ZjZjZjZ scin

ππ

( ) ( ) circuitshort 0tan103

107522tanMHz75 08

6

0 ⇒==⎟⎟⎠

⎞⎜⎜⎝

⎛

××××

= ππ ZjZjZ scin

EE 330


Smith’s ChartWhat is Smith’s chart? A graphical tool.

Purpose of introducing Smith’s chart: To avoid heavy math and speed up the process of analyzing and designing transmission line circuits.

How is it constructed? On the basis of the reflection coefficient plane.

Γ=ΓΓ=ΓΓ+Γ=Γ=Γ Γ Im and Re where, irirj je θ

Γr

Γi

1–1

1

–1

Only a circle of radius 1 has a meaning since

Any point within the unit circle represents certain reflection coefficient.

Smith’s chart works with normalized impedances and admittances:

1≤Γ

( ) ( )0ZzZzz =

EE 330


LLL

L xjrZZz +=

Γ−Γ+

==11

0

The normalized load impedance has positive resistive part and reactive part that can be either positive or negative:

Equating the real and the imaginary parts on both sides of the upper equation, two parametric equations are obtained:

22

2

11

1 ⎟⎟⎠

⎞⎜⎜⎝

⎛+

=Γ+⎟⎟⎠

⎞⎜⎜⎝

⎛+

−ΓL

iL

Lr rr

r

( )22

2 111 ⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛−Γ+−Γ

LLir xx

These two parametric equations describe two families of circles:

The first family consists of circles centered on the Γr –

axis and represent the value of the normalized resistance, rL .

Γr

Γi

1–1

1

–1The second family consists of circles centered on vertical line at Γr = 1 and represent the value of the normalized reactance, xL .

EE 330


The load impedance can be represented on the Smith’s chart by a point, at which two circles intersect: the circle corresponding to rL and the circle corresponding to xL .

Examples:15. Determine the load impedance denoted by the red star on the chart if this load terminates a 50-Ω

transmission line.

Ω−== 40200 jZzZ LL

The point representszL = 0.4 –

j0.8 Then

EE 330


Probing Questi

Which one of the points on the Smith's chart represents the load impedance of an open-circuited transmission line?

A

B

C

D E

EE 330


|Γ|

If a generalized reflection coefficient at an arbitrary point on

the transmission line is defined

as

, then the normalized impedance at that point can be expressed through the generalized

reflection coefficient the same way the normalized load impedance is expressed in terms of the reflection coefficient at the load:

( ) ( )zj rez βθ 2+Γ=Γ

( ) ( ) ( )( )zz

ee

ZzZzz zj

zj

Γ−Γ+

=Γ−Γ+

==11

11

2

2

0β

β

This observation suggests that Smith’s chart can be used to represent the impedance at any point on the line.The reflection coefficient along a particular transmission line maintains its magnitude; only its phase changes. Therefore, the normalized impedance at any point on the transmission line must lie on a circle with radius

.Also, it is obvious that the impedance on the line repeats itself every λ/2 length of the line. Therefore, the Smith chart represents a λ/2-long section of a transmission line.

Γ

EE 330


How to find the impedance at a point on the line that is a certain distance Δl from the load?

1.

Normalize the load impedance.2.

Normalize the distance, i.e., express Δl in terms of λ: ln = Δl/λ 3.

Mark on the chart the point representing the load impedance.4.

Draw the |Γ|-circle through that point. 5.

Draw a line through the load impedance point and the center of the chart.6.

Read the relative position of the load on the outer scale of the

chart.7.

Move along the outer scale in clockwise direction (toward the generator) a distance of ln and draw a line through that point on the outer scale and the center of the chart.

8.

Mark the point at which the line intersects the |Γ|-circle.9.

Read the resistive and the reactive parts of the impedance using

the corresponding families of circles. The result is the normalized impedance a distance of Δl from the load.

10.

To obtain the actual impedance value, multiply by the characteristic impedance.

How to find the impedance at a point on the line that is a certain distance Δl from the line input?

Perform the same steps as above with the following differences:1.

Start from the input impedance instead of the load impedance.2.

Move along the second outer scale in counterclockwise direction (toward the load).

EE 330


Examples:16. Consider the load impedance from the previous example. Determine the impedance 20 cm away from the load if the signal wavelength on the line is 1 m.

|Γ|

lnzL

z(– ln )

relative position of the load

relative position of the point of interest

zL = 0.4 –

j0.8

2.012.0cm 20 ==

Δ=⇒=Δ

λlll n

Rel. pos. z = 0.385 + 0.2 = 0.585

( ) ( ) Ω+=−=Δ− 5.275.150 jlzZlZ n

z(–ln ) = 0.31 + j0.55

Rel. pos. load = 0.385

EE 330


How to find the admittance that corresponds to certain impedance?

( ) ( )n

n

lj

lj

lj

lj

n ee

ee

ZlZlz π

π

β

β

4

4

2

2

0 11

11

−

−

−

−

Γ−Γ+

=Γ−Γ+

=−

=−

Consider the normalized impedance and admittance a distance l from the load:

( ) ( )( )

( ) ⎟⎠⎞

⎜⎝⎛ +=

Γ−

Γ+=

Γ−Γ+

=Γ+Γ−

=−

=−⎟⎠⎞

⎜⎝⎛ +−

⎟⎠⎞

⎜⎝⎛ +−

+−

+−

−

−

41

1

111

111

414

414

2

2

2

2

nlj

lj

lj

lj

lj

lj

nn lz

e

eee

ee

lzly

n

n

π

π

πβ

πβ

β

β

Notice that the admittance at certain point on the line equals the impedance a quarter wavelength away. Then, admittances also can be represented by points on Smith’s chart by simply adding a quarter wavelength to the actual position of interest. That is, the impedance and the admittance at the same position on the transmission line are represented by two points that are diametrically opposite to each other on the |Γ|-circle.

To summarize, Smith’s chart can be used to display normalized admittances. In that case, the r-circles play the role of normalized conductance circles (g-circles), and the x-

circles are considered normalized susceptance

circles (b-circles).

EE 330


zL

yL

Examples:17. Consider a 50-Ω

transmission line. Determine the load admittance if the load impedance is denoted by the red star on the chart.

1

00

02.001.0505.0

15.0

−Ω+=+

=

===

+=

jjZ

yYyY

jy

LLL

L

EE 330


Single-Stub Impedance Matching

The power absorbed by the load is maximum (100% of the incident power) when the transmission line is matched, i.e., the load impedance equals the characteristic impedance, ZL = Z0

.Unfortunately, oftenFortunately, there are means to match the load to the transmission line via impedance-matching network. One of the possibilities is: Single-stub matching using a short-circuited section of a transmission line.Essentially, a short-circuited section of the same type transmission line as the feeding (main) line is connected in parallel to the main line at

a position that is relatively close to the load. The point of connection and the length of the short-

circuited stub can be chosen in such a way so that to ensure an input impedance Z(–

d) = Z0

of the matching network that contains the load.

0ZZL ≠

LZ0Z

d

l

( )dZ −

matching network

feeding line

EE 330


Single-stub matching network: Analysis

To use Smith’s chart, we must work with normalized quantities. Then, what is needed is z(–

d) = 1.

Because the stub is connected in parallel to the feeding line, it is easier to work with admittances instead of impedances. The required normalized input admittance is also equal to one:

y(– d) = 1

Note that the required admittance is entirely real and a short-circuited line has a pure imaginary admittance. Then, the shorted stub can be used to compensate for the non-zero susceptance

of the feeding network.

What must be done? –

Find a point on the feeding line where yd = 1 + jb. Then, at that point, connect in parallel a shorted piece of line

that has input admittance ystub = –

jb . Eventually, the total admittance at the point of connection becomes

y(–

d) = yd + ystub = 1

EE 330


Single-stub matching network: Analytical solution

Examples:18. Design a single-stub matching network for a 50-Ω

transmission line transmitting signals with λ

= 2 m on the line to a (20 –

j40)-Ω

load.

jjz

yjjzL

LL +=−

==⇒−=−

= 5.08.04.0

118.04.050

4020

( ) ( )( ) ( )

( )( )( ) ( )

( )( )[ ] ( ) ( )[ ]( ) ( )( ) ( ) ( )( ) bj

djddjddjddj

djddj

djj

djj

dyjdjydy

L

L

+=−−+−

−−++=

=+−++

=⎟⎠⎞

⎜⎝⎛ ×++

⎟⎠⎞

⎜⎝⎛ ×++

=+

+=−

1tan5.0tan1tan5.0tan1

tan5.0tan1tan15.0

tan5.0tan1tan15.0

22tan5.01

22tan5.0

tan1tan

πππππππ

πππ

π

π

ββ

( ) ( ) ( ) ( )

( ) ( ) cm 8.7m 087.0279.0tan1 cm, 37.4m 374.0387.2tan1279.0tan,387.2tan05.0tan2tan75.0

12

11

212

======

==⇒=+−

−−

ππ

ππππ

dd

dddd

( ) ( )( ) ( ) ( )( )( )( ) ( )

1tan5.0tan1

tan1tan5.0tan15.0Re222 =

+−++−

=−dd

ddddyππ

πππ

EE 330


∞=stubLy ,

( ) ( )( ) ( )

( )( )( ) ( )

( )( )[ ] ( ) ( )[ ]( ) ( )( ) ( ) ( )( ) bj

djddjddjddj

djddj

djj

djj

dyjdjydy

L

L

+=−−+−

−−++=

+−++

=

=⎟⎠⎞

⎜⎝⎛ ×++

⎟⎠⎞

⎜⎝⎛ ×++

=+

+=−

1tan5.0tan1tan5.0tan1

tan5.0tan1tan15.0tan5.0tan1

tan15.02

2tan5.01

22tan5.0

tan1tan

πππππππ

πππ

π

π

ββ

( ) ( ) ( )( ) ( )( )( )( ) ( )

bdd

ddddy =+−

−++−=−

πππππ

222

2

tan5.0tan1tan1tan1tan5.0Im

( ) ( ) ( ) 58.1,For ,58.1,For 22,211,1 jlydjlydbjly stubstubstub −=−=−⇒−=−

( ) ( )( ) ( ) ( )⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−=⇒

−=

⎟⎠⎞

⎜⎝⎛ ×

=+

+=− −

lyjl

lj

ljlyjljy

lystubstubL

stubLstub

1

,

, tan1tan

22tan

1tan1tan

πππββ

cm 18m 18.0,For ,cm 82m 82.0,For 2211 ==== lblb

To summarize, there are two distinct solutions to the line-matching problem:(1)

A shorted stub of length 82 cm connected in parallel to the feeding line 37.4 cm away from the load.(2)

A shorted stub of length 18 cm connected in parallel to the feeding line 8.7 cm away from the load.

The above extremely involving math analysis can be avoided by an

approximate graphical solution using Smith’s chart.

EE 330


Single-stub matching network design: Step by step using Smith’s chart

1.

Mark the normalized load impedance, zL , on Smith’s chart2.

Draw the |Γ|-circle.3.

Find the relative position of the normalized load admittance, yL , on Smith’s chart.4.

Mark the two points where the |Γ|-circle intersects the (g = 1)-circle. 5.

Choose one of the above points to be your point of connection. Usually, the better choice is the point that is closer to the load.

6.

Find the relative position of the point of connection.7.

Compute its normalized distance from yL .8.

Compute the actual distance of the connection point from the load, d. 9.

Read the susceptance

at the point of connection, b. The stub must have an admittance (–

jb).10.

Find the circle corresponding to susceptance

(–

b).11.

Mark the point where the (–

b)-circle intersects the |Γ|-circle. This point is the input of the stub, ystub .

12.

Read the relative position of ystub .13.

The stub is short-circuited, therefore its load admittance is yL, stub = ∞. Find its relative position on the chart.

14.

Determine the normalized distance from yL, stub to ystub . This is the normalized length of the stub.

15.

Compute the actual length of the stub.

EE 330


zL

yL

Examples:18. Design a single-stub matching network for a 50-Ω

transmission line transmitting signals with λ

= 2 m on the line to a (20 –

j40)-Ω

load.

yL, stub

ystub

y(– d)

ln

dn

8.04.050

4020 jjzL −=−

=

Relative positions of the two possible connection points:0.1787 and 0.3213.

Choosing the point at 0.1787

25.0rel.pos. , ,, =∞= stubLstubL yy

cm 8.8m 088.02044.0044.01347.01787.0

==×===−=

λn

n

ddd

cm 8.17m 178.02089.0089.025.0339.0

==×===−=

λn

n

lll

339.0rel.pos. ,6.1 =−= stubstub yjy

1347.0rel.pos. ,5.0 =+= LL yjy