Lect_6_2014-RF Circuit Design (ECE321/521)

7/25/2019 Lect_6_2014-RF Circuit Design (ECE321/521)

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Indraprastha Institute ofInformation Technology Delhi ECE321/521

Lecture 6 Date: 21.08.2014 Lossy Transmission Line Introduction to Smith Chart: The complex plane Transformations on the complex plane Mapping Z to Smith Chart Construction Smith Chart Geography


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Lossy Transmission Lines Recall that we have been approximating low-loss transmission lines as

lossless (R =G = 0):

0 LC

But, long low-loss lines require a better approximation:

00

12

RGZ

Z

LC

Now, if we have really long transmission lines (e.g., long distance

communications), we can apply no approximations at all: Re Im

For these very long transmission lines, we find that = isa function of signal frequency . This results in an extremely

serious problem signal dispersion .


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Lossy Transmission Lines (contd.) Recall that the phase velocity (i.e., propagation

velocity) of a wave in a transmission line is: pv

Im Im R j L G j C

Thus, for a lossy line, the phase velocity is a function offrequency (i.e., ( ) )this is bad !

Any signal that carries significant information must has some non-zerobandwidth . In other words, the signal energy (as well as the information itcarries) is spread across many frequencies.

If the different frequencies that comprise a signal travel at differentvelocities, that signal will arrive at the end of a transmission line distorted .We call this phenomenon signal dispersion.

Recall for lossless lines, however, the phase velocity is independent offrequency no dispersion will occur!


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Lossy Transmission Lines (contd.)

For lossless line: 1 pv

LC however, a perfectly lossless line isimpossible, but we find phase

velocity is approximately constantif the line is low-loss.

Therefore, dispersion distortion on low-loss lines ismost often not a problem.

Q : You say most often not aproblem that phrase seems toimply that dispersion sometimes

is a problem!


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Lossy Transmission Lines (contd.)A: Even for low-loss transmission lines, dispersion can be a problem if thelines are very long just a small difference in phase velocity can result insignificant differences in propagation delay if the line is very long!

Modern examples of long transmission lines include phone lines and cableTV. However, the original long transmission line problem occurred with the

telegraph , a device invented and implemented in the 19 th century. Early telegraph engineers discovered that if they made their telegraph lines

too long , the dots and dashes characterizing Morse code turned into amuddled, indecipherable mess . Although they did not realize it, they hadfallen victim to the heinous effects of dispersion !

Thus, to send messages over long distances, they were forced to implementa series of intermediate repeater stations, wherein a human operatorreceived and then retransmitted a message on to the next station. Thisreally slowed things down!


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Q: Is there any way to preventdispersion from occurring?

A: You bet! Oliver Heaviside figured outhow in the 19 th Century!

Heaviside found that a transmission line would bedistortionless (i.e., no dispersion) if the line parameters

exhibited the following ratio :

R G L C

Lets see why this works. Note the complex propagation constant can beexpressed as:

/ / R j L G j C LC R L j G C j


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Then IF: R G L C

we find:

/ / /

C LC R L j R L j R L j LC R j LC

L

Thus: Re C R L

Im LC

The propagation velocity of the wave is thus: 1 pv LC

The propagation velocity is independent of frequency! This lossytransmission line is not dispersive!


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Q: Right. All the transmission lines I usehave the property that > . Ive

never found a transmission line with thisideal property = !

A: It is true that typically > . But, we can reduce the ratio (until itis equal to ) by adding series inductors periodically along thetransmission line.

This was Heavisides solution and it worked! Long distancetransmission lines were made possible.

Q: Why dont we increase G instead ?

A:


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Smith Chart

Smith chart what? The Smith chart is a very convenient graphical tool for analyzing TLsstudying their behavior.

It is mapping of impedance in standard complex plane into a suitablecomplex reflection coefficient plane.

It provides graphical display of reflection coefficients. The impedances can be directly determined from the graphical display (ie,

from Smith chart) Furthermore, Smith charts facilitate the analysis and design of

complicated circuit configurations.


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The Complex - Plane

Let us first display the impedance Z on complex Z-plane

30 40 Z j

60 30 Z j

InvalidRegion

InvalidRegion

Re (Z)

Im (Z)

Note that each dimension is defined by a single real line: the horizontal line (axis) indicates the real component of Z , and the vertical line (axis)indicates the imaginary component of Z Intersection of these linesindicate the complex impedance


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The Complex - Plane (contd.)

How do we plot an open circuit (i.e, = ), short circuit (i.e, = 0 ), andmatching condition (i.e, = = 50 ) on the complex Z-plane

Re (Z)

Im (Z)

Z = Z0

Z = 0

= somewhere over there!!

It is apparent that complex is not very useful


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The Complex -Plane (contd.)

The limitations of complex Z-plane can be overcome by complex -plane We know Z (i.e, if you know one , you know the other ). We can therefore define a complex -plane in the same manner that we

defined a complex Z-plane. Let us revisit the reflection coefficient in complex form:

Real part of 0

Imaginary part of 0

In the special terminated conditions of pure short-circuit and pure open-circuit conditions the corresponding 0 are -1 and +1 located on the realaxis in the complex -plane.

000 0 0 0

0

j Lr i

L

Z Z e

Z Z

Where, 1 00

0

tan ir


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0r

0i

000 0 0 0

0

j Lr i

L

Z Z e Z Z

Representation of reflectioncoefficient in polar form

0

0

0 0

Observations:

A radial line is formed by the locusof all points whose phase is 0

A circle is formed by the locus of allpoints whose magnitude is | 0|

It means the reflection coefficient has a valid regionthat encompasses all the four quadrants in the complex

-plane within the -1 to +1 bounded region

In complex Z-plane the valid region was unbounded on the right half of theplane as a result many important impedances could not be plotted


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0r

0i

Validity Region

Invalid Region| 0| > 1

Valid Region| 0| < 1

| 0| = 1

1

-1

-1

1


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The Complex -Plane (contd.) We can plot all the valid impedances (i.e R > 0) within this bounded region.

0r

0i

(short)0

1.0 je (matched)0 0

(open)

00 1.0

je

| 0| = 1

Z = jX purely reactive


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A TL with a characteristic impedance of Z0 = 50 is terminated into

following load impedances:(a) ZL = 0 (Short Circuit)(b) ZL (Open Circuit)(c) ZL = 50

(d) ZL = (16.67 j16.67) (e) ZL = (50 + j50) Display the respective reflection coefficients in complex -plane

Example 1

Solution: We know therelationship between Z and :

000 0 0 0

0

j Lr i

L

Z Z e

Z Z

(a) 0 = -1 (Short Circuit)(b) 0 = 1 (Open Circuit)(c) 0 = 0 (Matched)(d) 0 = 0.54


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(a) Short Circuit (b)Open Circuit

(c) Matched

(e)

(d)

Example 1 (contd.)


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Transformations on the Complex -Plane

At z =0 , the reflection coefficient is called load reflection coefficient ( 0) this actually describes the mismatch between the load impedance (ZL) andthe characteristic impedance ( ) of the TL.

The move away from the load (or towards the input/source) in thenegative z-direction (clockwise rotation) requires multiplication of 0 by afactor exp (+ 2 ) in order to explicitly define the mismatch at location z known as (z).

This transformation of 0 to (z) is the key ingredient in Smith chart as agraphical design/display tool.

The usefulness of the complex -plane will be evident when we considerthe terminated, lossless TL again.

, Z 0 , Z 0 Z Ll

in

0


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Transformations on the Complex -Plane (contd.) Graphical interpretation of 2

0( ) j z z e

0

( ) 1 z

0r

0i

0( 0) z 0

( ) in z l 0 2 l


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Transformations on the Complex -Plane (contd.)

It is clear from the graphical display that addition of a length of TL to aload 0 modifies the phase 0 but not the magnitude 0, we trace acircular arc as we parametrically plot (z) ! This arc has a radius 0 andan arc angle 2 l radians.

We can therefore easily solve many interesting TL problems

graphically using the complex -plane! For example , say we wish todetermine in for a transmission line length l = / 8 and terminated with ashort circuit.

in , Z 0 , Z 0 0= -1l = /8

z = -l z = 0


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Transformations on the Complex -Plane (contd.) The reflection coefficient of a short circuit is 0 = 1 =1*e(j ), and

therefore we begin at the leftmost point on the complex -plane. Wethen move along a circular arc 2 l = 2(/ 4) = / 2 radians (i.e., rotateclockwise 90 ).

When we stop, we find weare at the point for in; inthis case in = 1*e(j /2 )

0r

0i

( ) z

/21* jin e

0 1* je


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Transformations on the Complex -Plane (contd.) Now let us consider the same problem, only with a new transmission line

length l = / 4. Now we rotate clockwise 2 l = radians.

( ) z

In this case the inputreflection coefficient is in = 1*e(j0) = 1

The reflection coefficientof an open circuit

The short circuit load has beentransformed into an open circuit

with a quarter-wave TL

0r

0i

0 1* je

01* jin e


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Transformations on the Complex -Plane (contd.) We also know that a quarter-wave TL transforms an open-circuit into

short-circuit graphically it can be shown as:

0r

0i

1* jin e

00 1*

je ( ) z


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Transformations on the Complex -Plane (contd.) Now let us consider the same problem again, only with a new

transmission line length l = / 2. Now we rotate clockwise 2 l = 2 radians (360 )

( ) z 0

0 1* je

( ) 1 z

We came clear around towhere we started!

Thus we conclude that in = 0

It comes from the fact thathalf-wavelength TL is a

special case, where we knowthat Zin = ZL eventually it

leads to

in =

0

0r

0i

1* jin e


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Transformations on the Complex -Plane (contd.) Now let us consider the opposite problem. Say we know that the input

reflection coefficient at the beginning of a TL with length l = / 8 is: = . .

What is the reflection coefficient at the load ? In this case we rotate counter-clockwise along a circular arc (radius =0.5)

by an amount 2 l = /2 radians (90 ). In essence, we are removing the phase associated with the TL.

0r

0i

600.5* jin e

in

1500 0.5*

je

0 2in l

( ) 1 z

The reflection coefficient atthe load is:

1500 0.5*

je

0.5

f


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Mapping Z to We know that the line impedance and reflection coefficient are equivalent

either one can be expressed in terms of the other.

0

0

( )( )

( ) Z z Z

z Z z Z

0

1 ( )( )

1 ( ) z

Z z Z z

The above expressions depend on the characteristic impedance Z

0 of the

TL. In order to generalize the relationship, we first define a normalized impedance value z as:

0 0 0

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

Z z R z X z z z j r z jx z

Z Z Z

therefore

00

0 0

( ) / 1( ) ( ) 1( )

( ) ( ) / 1 ( ) 1

Z z Z Z z Z z z z

Z z Z Z z Z z z

1 ( )( )

1 ( )

z z z

z

I d h I i f


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Mapping Z to (contd.)

00

0 0

( ) / 1( ) ( ) 1( )( ) ( ) / 1 ( ) 1

Z z Z Z z Z z z z Z z Z Z z Z z z

1 ( )( ) 1 ( ) z z z z

These equations describe a mapping between z and . That meansthat each and every normalized impedance value likewise corresponds

to one specific point on the complex -plane For example, we wish to indicate the values of some common normalized

impedances (shown below) on the complex -plane and vice-versa.Case Z z

1 1

2 0 0 -1

3 Z0 1 0

4 jZ0 j j

5 -jZ0 -j -j

I d h I i f


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r

i

Invalid Region| | > 1

| | = 1

= j(z = j)

= -1(z = 0)

= -j(z = -j)

= 0(z = 1)

= 1 (z =)

The five normalized impedances map five specific points on the complex -plane.

I d th I tit t f


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Invalid Region

r

x

( = 0)z = 1

( = -1)z = 0

( = -j)z = -j

( = j)z = j

The five complex- map onto five points on the normalized Z-plane

It is apparent that the normalized impedances can be mapped on complex -plane and vice versa

It gives us a clue that whole impedance contours (i.e, set of points) can be

mapped to complex -plane

I d th I tit t f


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Mapping Z to (contd.)Case-I: Z = R impedance is purely real

0 z r j 11

r r

11r

r r

0i

r

i


(

i = 0)x = 0

I n v a l i d R e g i o n

r

x

( i = 0)x = 0

r

Indraprastha Instit te of


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Mapping Z to (contd.)Case-II: Z = jX impedance is purely imaginary

0 z jx Purely reactive impedance results in a

reflection coefficient with unity magnitude1

r

i

| |= 1r = 0


I n v a l i d R e g i o n

x j

x j

r

x

These cases (I and II) demonstrate thateffectively any complex impedance can be

mapped to complex -plane Smith Chart

| |= 1r = 0

Indraprastha Institute of


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The Smith Chart

In summary A vertical line r = 0 on complex Z-plane maps to a circle | | = 1 on thecomplex -plane

A horizontal line x = 0 on complex Z-plane maps to the line i = 0 onthe complex -plane

Very fascinating in an academic sense, but are not relevant consideringthat actual values of impedance generally have both a real and imaginary

component

Mappings of more general impedance contours (e.g,r = 0.5 and x = - 1.5 corresponding to normalized

impedance 0. 5 j1.5) can also be mappedSmith Chart



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The Smith Chart (contd.)

Therefore, the normalized impedance can be formulated as:

Let us revisit the generalized reflection coefficient formulation:

0 20( )

j j z r i z e e j

0

1( ) 1 ( )( ) 1 ( ) 1

r i

r i

j Z z z z z r jx Z z j

The separation of real and imaginary part results in:

1 1r i r r x

1 r i i x r

Real

Imaginary

1 1r i r i j r jx j



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The Smith Chart (contd.) Simplification and then elimination of reactance ( x ) from these two give:

Multiplying through by 1 r

2 21- r + 1+ r = 1+ 1-r i r r

ir i r r

1- r + 1+ r = 1+

1-

2 2 21 1 1r i r r r

2 21 2 1 1 0r r ir r r r



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2 21 2 1 1r r ir r r r

2 2 1

2 1 1r r ir r

r r

2 22

r i

r 1- r r - + = +

1+ r 1+ r 1+ r

222

2

1 1

1 1

r i

r r r r r

r



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22

2

11 1

r i

r r r

2 2 2, :

r i

l

p q

p q l

Similar equation to circle of radius ,

centered at

This is equation of a circle

, ,01

r p q

r

center:

radius: 11

l r

and

Observations: For r =0: p2 + q2 = 1; (p, q) = (0, 0) and l = 1 For r =1/2: (p - 1/3) 2 + q2 = (2/3) 2; (p, q) = (1/3, 0) and l = 2/3 For r =1: (p - 1/2) 2 + q2 = (1/2) 2; (p, q) = (1/2, 0) and l = 1/2 For r =3: (p 3/4) 2 + q2 = (1/4) 2; (p, q) = (3/4, 0) and l = 1/4

Circles ofdistinct

centre andradii



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0r

r

i 1r 3r

1/ 2r

r

1

1 p l Note:

Therefore the resistance circles on the complex -plane are:

Because of(q 0)2

term, all theconstant

resistance (r)circles havecenters on

this line


This approach enables mapping of anyrealizable vertical line (representing r) in the

complex

-plane



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For the mapping of horizontal lines of the normalized impedance plane to -plane, let us simplify and eliminate resistance ( r ) from the following:

1 1r i r r x

1 r i i x r

Real

Imaginary

1

1 1r ir i r i

x x

2 21 1 1 0r i r i i r x x

2 2

1 2 0r i i x x




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2 221 0r i i x

Observations: For x =1: (p 1)2 + (q 1)2 = (1)2; (p, q) = (1, 1) and l = 1 For x =-1: (p 1)2 + (q + 1) 2 = (1)2; (p, q) = (1, -1) and l = 1 For x =1/2: (p 1)2 + (q 2)2 = (2)2; (p, q) = (1, 2) and l = 2 For x =-1/2: (p 1)2 + (q + 2) 2 = (2)2; (p, q) = (1, -2) and l = 2

Circles of

distinctcentre andradii

2 2

2 1 11r i x x

, 1,1/ p q xcenter:

radius:1

l x

q l Note:

Indraprastha Institute of ECE321/521


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0r


r

i1 x

1 x

0 x

0.5 x

0.5 x

3 x

3 x

x

q l Note:

All constant reactance( x ) circles have theirorigins along this linep=1 because of the

term (p 1)2

This approach enables mapping of any realizable horizontal line

(representing x ) in the complex

-plane



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The Smith Chart (contd.) Combination of these constant resistance and reactance circles define the

mappings from normalized impedance ( z) plane to -plane and is called asSmith chart.

( ) 1( )

( ) 1 z z

z z z

1 ( )( )

1 ( ) z

z z z

r

i

0r

0 x

jx

r

z r jx

1r

Positive(Inductive)Reactance

Negative(Capacitive)

Reactance



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pInformation Technology Delhi ECE321/521

The Smith Chart (contd.) Important Features

It is apparent: for 0 , we get | (z)| 1. This condition is easily met forpassive networks (i.e, no amplifiers) and lossless TLs (real Z 0)

Consequently, the standard Smith chart only shows only the inside of the

unit circle in the -plane. That is, | (z)| 1 which is bounded by the = 0 circle described by:

2 2 1r i

1. By definition:

( ) 1 1( )

( ) 1 1 z z r jx

z z z r jx

2 2

2 2

1( )

1

r x z

r x



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2. Notice that in the upper semi-circle of the Smith chart, 0 which is aninductive reactance . Consequently, the generalized reflection coefficients (z) r + j i in the upper semi-circle are associated with normalized TLimpedances + that are inductively reactive.

Conversely, the lower semi-circle of the Smith chart represent capacitivereactive impedances

3. If z(z) is purely real (ie, x = 0) then the reactance term:

2 21 2 0

r i i x x

suggests

i = 0 except possibly at r = 1

Consequently, purely real z(z) values are mapped to (z) values on the r axis.



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4. If z(z) is purely imaginary (ie, r= 0) then the impedance term:


Consequently, purely imaginary z(z) values are mapped to (z) valueson the unit circle in -plane.

222

2

1 1

1 1r i

r r r r r r

suggests

2 2 1r i

Unit Circle on -plane

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