Chapt.3-Transmission Line, Imped Transformation, 2-Port Net

8/2/2019 Chapt.3-Transmission Line, Imped Transformation, 2-Port Net

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Chapter 3

Transmission Lines

Transmission Line Model

Transmission lines are characterized by series inductance, L, per meter, series resistance,R per meter, capacitance, C, per meter between the two wires and G, conductance per

meter between the two wires.

Consider a small length of the transmission line. The series inductance is L and its

impedance is . The series resistance is

z z

zLj zR . The capacitance between the two

wires is and its admittance is . The conductance between the two wires iszC zCj

zG

Fig.3-1

Kirchoffs Voltage Law (KVL) can be written as:

I(z)zR)z(IzLj)z(V-)zz(V =+ (3.1)

Kirchoffs Current law can be written as

V(z)zG)z(VzCj)z(I-)zz(I =+ (3.2)

ividing (3.1) by

z D

)z(RI)z(LIjz

)z(V-)zz(V=

+

(3.3)

ividing (3.2) by

z D

3-1


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V(z))()(-)(

VzIzzI

=+

GzCjz

(3.4)

Taking limits as 0z

I)LjR(RILIjdz

dV

+== .5)(3

V)CjG(GVCVjdz

dI+== (3.6)

Differentiating (3.5) with respect to z,

dz

dI)LR(

dz

Vd2

+= j2

SubstitutingdzdI from (3.6),

VV)CjG)(LjR(Vd2

dz

2

2=++= ,

e,i. VVd2

dz2

where G)(LjR(2 ++=

2=

smitting

formation as telegraphs.

he solution of this second order ordinary differential equation is well known

(3.7)

tituting in (3.5),

)Cj

This equation is called the telegraphists equation as it was used cables for tran

in

T

zz eVeVV ++ +=

Subs

I)LjR()eVeV( zz +=+ +

)eVeV(LjR

I z +

+= z+ Or,

Or, )eV-e(Z1I zz-0

++= (3.8)V

CjG)(LjR( ++= is called the propagation constantObviously, is a complex

nt with unit radians/m

(3.9)

number, j+= where is called the attenuation constant

with unit Nepers/m and is called the phase consta

3-2


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CjG

LjR

LjRZ0 +

+=

+= is c edance. (3.10)

oximati

s are small we ignore the losses represented by R and G. They

can be taken into accoun

Ignoring R

alled the characteristic imp

This is also a complex quantity. Its unit is ohms.

Lossless appr on

In circuits, the line length

t during circuit simulation.

and G, i.e., setting R=0=G in (3.9)

=)Cj)(Lj(= j=LCj , where LC= (3.11)

Setting R=0=G in (3.10)

C

L=Z0 , which is pure real (like resistance) (3.12)

Thus, we can write (3.7) and (3.8) aszjzj-

eVeVV ++ += (3.13)

zjzj- Ve

VI +

+=

00

The first terms represent a wave traveling in +z direction and is called the forward orincident wave. The second terms represent a wave traveling in the z direction.

How can we find +V and

V ?

eZZ

(3.14)

The values are determined by the impedance/admittance connected at the two ends of the

transmission line (or what is called boundary conditions in Maths.). But we are notinterested in this at the moment

Fig 3-2

The phase velocity of the wave is

Phase velocity

LC

1=

LC

=

=vp (3.15)

3-3


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From (3.12) and (3.15)

pv

0Z=L (3.16)

p0vZ

1=C

(3.17)

d C

ined.

ie

ote that we can define z=0 anywhere on the line. A common practice is to have z=0 ay

the load end as shown in Fig.3-3. Note that values of z to the left are negative as indicted

by a point at a distance from the load impedance, .

Thus if the characteristic impedance, 0Z and the phase velocity, pv are known, L an

can be determ

More about phase velocity when we deal with different types of transmission lines.

Reflection coeffic nt

N

l LZ

Fig.3-3

The ratio of the reflected to incident voltage is called the reflection coefficient, .

At the load (z=0), the load reflection coefficient is

+

=

+

+=

V

V

eV

eV

0z

zj

zj

(3.18)=L

00z0zjzj

zjzj

0zL

Z/)VV(

VV

Z/)eVeV(

eVeV

I

VZ

+

+

=++

++

= +

+=

+== But

0)( Z

Z

VV

VV L=+

+

+

Hence,

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0L

0L

ZZ

ZZ

VHence

V=

+

+

m (3.18),

0L

0L Z-Z= (3.19)Fro LZZ +

complex, is complex (although s real).

Matched line

As LZ can be L 0Z i

Suppose that LZ = 0Z . Then0L

0LL

ZZ

Z-Z

+= =0 which means that 0=V . So there is no

flected wave. Thus the characteristic impedance is that impedance which when

said to be

How does the reflection coefficient vary with the distan

re

connected to the transmission line produces no reflected wave. Such a line ismatched all the power is delivered to the load as there is no reflection.

cel

from the load?l=)( l

l

l

je

VeV +++=

== (3.20)l

jj

zj

Ve

eV

2-

-

+

3.21)

hase shift. Thus, along the lossless line,

Input impedance at distance

zj VeV +

z

From (3.18) and (3.20),

=)( l lj-2

Le (

This is just the load reflection coefficient with a p

he reflection coefficient remains constant in magnitude but changes in phase by l2 t

radians.This will be useful when we discuss Smith Chart.

, l

The input impedance at distance from the load isl

l

l

ll

ll

l

2-

2-

0

00)/(-1

)/(1

/)-( j

j

z

jj

jj

z

ineVV

eVVZ

ZeVeV

eVeV

I

VZ == +

+

=

++

++

=

+=

+

From (3.18)

l

ll

ll

j2L

j2

00ine1)(1

(3.22)

This can be expressed in terms of and by substituting for from (3.19)

+=

+= L

e1Z

)(1Z)(Z

LZ 0Z L

l

l

l

l

l

j

LL

j

LL

j

L

L

j

L

L

ineZZZZ

eZZZZZ

eZZ

ZZ

eZZ

ZZ

ZZ2-

0

2-

0

02-

0

0

2-

0

0

0)-(-)(

)-()(

--1

-1

)(

0

0

+

++=

+

++

=

ll

ll

l

j

L

j

L

je-)L

j

L

ineZZeZZ

ZZeZZZZ

-

0

0

0)-(-)(

-()()(

0

0

+

+

+

++=

3-5


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W )sin(j)cos(e j lll = r anditing )sin(j+)cos(=e j+ lll

)sin(jZ+)cos(Z

Z=)(ZL0

0L0in

lll (3.23)

Thus voltage current and input impedance change with distance along the line. Therefore

at high frequencies, it is not enough to state the value of impedance,

)sin(jZ+)cos(Z ll

it must be specifiedwhere it is measured. We will see the importance of this when we deal with multiport

etworks.

Standing waves

n

Short circuited line

If the line is short circuited,

From (3.19),

0=ZL

0L

0LL

ZZ += =

Z-Z1

Z0 0

0 =+

Z-0

The voltage at distance from the load isj

Lj-jj-jj- llllll +++++++ +=+=+ (3.24)

For

l+ Including the time variation,

l

)ee(Ve)V/V(e(VeVeVV =1L =

jV2)e-e(VV jj- ll == ++ )sin(

)2/+tcos()sin(V2)sin(jV2=V ++ ll

It is seen that for n=l , n = 0, 1, 2 ., the voltage is zero at all time, i.e., the wave

does not travel at all. Such a wave is called a

=)tcos(

pure standing wave. The minimum value of

he amplitude is zero and the maximum

value is . The variation of the voltage with

/mini is a s

e

+

V2tdistance and time is shown in Fig.3-4.

The distance between two consecutive maximum mum nd the di tance

between a maximum and the next minimum is )2/( , As /2= , wher

/

is the

wavelength, the distance between two consecutive maximum/minimum is and the

distance between a maximum and the next mi

2/

nimum is 4/

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Fig.3-4

Voltage standing wave ratio (VSWR)

The VSWR is given by

min

max

V

V=S

(3.25)

Where Vmax is the maximum value of the voltage and minV is the minimum value of theoltage.

or a short circuit, because

v

=S minV = 0.F

For a matched load, we may consider maxV = minV because there is no reflection and

o the values of S always lie between 1 and infinity.

VSWR is sometimes expressed in dB

VSWR in dB =

hence no standing wave. Hence S=1S

min10 V

maxVog20l (3.26)

o the VSWR in dB lies between 0 (for S=1) to infinity.

o find the magnitude f

(3.27)

S

T rom dB,20/)dBinVSWR(

10=S

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What about the input impedance?

From (3.25) with

0in ll

Consider , then and hence

0=ZL

Z )tan(jZ=)(

1


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A short circuited line of length less than is often used as an inductor.

pen circuited line

4/

O

cuited line,From (3.19),

=ZL For an open cir

0L

L

0L

ZZ

Z-Z

+= = 1

Z/Z1 L0+

Z/Z-1 L0 +=

For we get from (3.24)

l+ Including the time variation,

1L +=

V2)ee(VV jj- ll =+= ++ )cos(

)tcos()cos(V2=V + l

It is seen that for 2/n=l , n = 0, 1, 2 ., the voltage is zero at all time, i.e., the wave

does not travel at all. Once again we have a pure standing wave. The minimum value ofthe amplitude is zero and the maximum value is . The variation of the voltage with

istance and time is shown in Fig.3-6.

/mini is a s

e

+V2

d

The distance between two consecutive maximum mum nd the di tance

between a maximum and the next minimum is )2/( , As /2= , wher

/

is the

wavelength, the distance between two consecutive maximum/minimum is and the

distance between a maximum and the next mi

2/

nimum is 4/

Fig.3-6

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For an open circuit, because=S minV = 0.

ut impe ance?What about the inp d

From (3.23) with =ZL

)sin(jZ+)cos(Z

)sin(jZ+)cos(Z

Z=)(Z L0

0L

0 )sin(j+)cos()Z/Z(

)sin()Z/Z(j+)cos(

Z L0

L0

0 ll

ll

ll

ll

=in l

= )t( l cojZ0Consider , then and hence1


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For a mismatched load which is neither a short or open circuit, 0Vmin as shown inFig.3-8. The wave is partly standing and partly travelling.

Fig.3-8

Relationship between VSWR and reflection coefficient

Consider (3.24)

( )Lmax 1VV +=+ and ( )Lmin 1VV =

+

Hence,

L

L

1

1S

+= (3.30)

from which,1

1

+

=S

SL (3.31)

Some practical transmission lines

Two wire line

3-11


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Fig.3-9

Coaxial cable

Fig.3-10

Microstrip line and strip line

3-12


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Fig.3-11: (a) Microstrip line (b) Strip line

Parameters of a two-wire and co-axial transmission lines.

Table 3.1: Distributed parameters of a two-wire transmission line with wire radius a and

separated by a distanceD, and a coaxial transmission line with a centre conductor radiusofa and outer conductor radius b. '0= , where 0 is the permittivity of free space and

is the equivalent conductivity and is the permeability of the dielectric. Thesurface resistivity is given by

=

/s c cR f= ,where c is the skin depth and is the

relative permeability of the conductor.

c

Parameter Two-wire line Coaxial line

R (/m) sRa

1 1

2

sR

a b

+

L (H/m) 1cosh ( / 2 )D a

ln( / )2

b a

G (S/m)1cosh ( / 2 )D a 2

ln( / )b a

C (F/m)1cosh ( / 2 )D a

2

ln( / )b a

Parameters of microstrip lines

In the case of microstrip, the wave does not propagate in a homogeneous dielectric. For

this reason, the microstrip line does not support a pure TEM wave, since the phase

velocity of TEM fields would be / rc , but the phase velocity in the air would be c. A

phase match at the dielectric interface would then be impossible. For the quasi-TEMfields, we find that

effp

c=v and 0 0 eff 0 eff k = = (3.32)

where is the effective dielectric constant of the microstrip line. In general,

, and it is dependent on the substrate thickness, d, and the conductor width W.eff

eff1 r< <

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If is the real part of the relative permittivity of the dielectric, the effective dielectric

constant of a microstrip line is given approximately byr

eff

1 1 1

2 2 1 12 /

r r

d W

+ = +

+(3.33)

Given the dimensions of the microstrip, the characteristic impedance can be calculated as

eff

0

eff

60 8ln / 1

4

120/ 1

[ / 1.393 0.667 ln( / 1.444)]

d WW d

W dZ

W dW d W d

+

+ + +

(3.34)

Alternatively, given the characteristic impedance 0Z , we can calculate the appropriate

line width from

2

8/ 2

2/

12 01 ln(2 1) ln( 1) 0.39 / 2

2

A

A

r

r r

eW d

eW d

B B B W d

.61(3.35)

where 01 1 0.11

0.2360 2 1

r r

r r

ZA

+ = + +

+ and

0

377

2 r

BZ

=

.

Smith Chart

Why Smith chart?

The Smith chart was first devised by P.H.Smith as an aid for calculations involvingtransmission lines. The first publication appears in a nonacademic journal (P.H.Smith,

Transmission line calculator, Electronics, Vol.12, January, 1939). Over the years, it has

been refined. Although there are various extensions of the chart, such as charts including

negative resistance, it is the basic Smith chart that has wide acceptance.There are two attitudes to Smith chart. Engineers with microwave background

consider it indispensible. Traditional analog designers sometimes do not bother about it

(Razavis book given in your reference does not even mention it). Students wonder outloud why they have to learn it for calculations that are rough at best, when so much

circuit simulation software available for accurate analysis.

This is what Tom Lee writes:IC designers accustomed to working with lower frequency circuits tend to have, at best, only apassing familiarity with two staples of traditional RF design: Smith charts and S-parameters(scattering parameters). Although Smith charts today are less relevant as a computational aidthan they once were, RF instrumentation continues to present data in Smith chart form.Furthermore these data are often S-parameter characterization of two ports, so it is important,even in the modern era, to know something about Smith charts and S-parameters

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(Lee TH, The design of CMOS Radio-Frequency integrated circuits, 2nd

Edition, CambridgeUniversity Press, 2006).

To this, an RF engineer will add that Smith chart gives us a visual representation of the

design process and helps us in getting insights.

Smith Chart for impedance

Recall (3.19)

Load reflection coefficient imre0L

0LL j

ZZ

Z-Z+=

+= , where re and im are the real

and imaginary parts of .L

For all passive impedances, LLL jXRZ += with positive ,LR 1L

(This can be shown easily,0LL

0LLL

ZjXR

Z-jXR

++

+= 1

X)ZR(

X)ZR(

2L

20L

2L

20L2

L ++

+= for

positive ).LR

Consider now the complex plane of L as shown in Fig.3-12. All passive impedancescan be represented by points inside the circle of radius 1 because their reflection

coefficients are always less than or equal to 1.

Fig.3-12

To make representation independent of , we divide impedances by and write0Z 0Z

1Z/Z

1Z/Z

ZZ

Z-Z

0L

0L

0L

0LL +

=+

= ,

i.e.,1z

1-zL +

= (3.36)

where z = is called the normalized load impedance.0L Z/Z

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For any z, we can find plot the corresponding L in Fig.3-12. So if we write z = r+jx,(3.36) represents mapping of points from the complex z plane to complex plane as

shown in Fig.3-13.L

Fig.3-13 Notice that we are considering only positive r (passive loads)

If we use the r-x co-ordinates, we require an infinitely large paper to plot all possible z.However if we use Smith chart, we can plot all points inside the circle of radius 1, which

is called the unit circle.

Now if we want to plot normalized impedance in the unit circle, it will be good to havegrid lines for r and x. Fig.3-14 shows how the grid lines for r are obtained.

Fig3-14: Grid lines of r are the green straight lines in the z (r-x) plane as shown on the

left. They are transformed to circles in the plane as shown in the right.

Notice the radii of the circles get smaller and their centres shift right as r increases. =r is just the point to the right. It is possible to find expressions for the radii and centres

from (3.36) considering r to be a constant. But we dont want to get into this trouble asthe circles are provided.

3-16


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To make the use of the chart more convenient, the angle of the reflection coefficient is

given along the periphery of the unit circle (Notice that the angles are given from 00

to+180

0and 0

0to -180

0). Further, the square grid lines are removed. The chart is shown in

Fig.3-17.

You will notice another outermost scale wavelengths towards generator andwavelengths towards load. These scales allow us to find the normalized impedance

after a length along the line. To understand this, refer to P.3-5 equation (3.21). It is seen

that the magnitude of the reflection coefficient remains same, but the angle decreases by

(where

l

)/(4/222 == lll is the wavelength in the transmission line )as onemoves by a length towards the generator. The scale wavelength towards the generator

means towards the generator. The direction of the arrow ensures that the angle is

decreasing. Similarly, the angle will increase if we move towards the load. Hence the

other scale - towards the load. To use these scales, we have to calculate first ,

where is the phase velocity and f is the frequency. Then we divide the length by

l

/l

f/vp=

pv

and move by this amount to find the impedance at the distance l .

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Fig.3-17

Smith chart for admittance

L0

L0

0L

0L

0L

0LL

YY

YY

Y

1

Y

1

Y

1

Y

1

ZZ

Z-Z

+

=

+

=+

=The load reflection coefficient , where,

andLL Y/1Z = 00 Y/1Z =

Or,y1

y1j imreL +

=+= , where, 0L Y/Yy = is the normalized admittance (3.37)

We can write this as1y

1y)(j)( imre +

=+ ,i.e., as

1y

1yj 'im

're +

=+ , where

andre're = im

'im = (3.38)

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(3.38) is the same as (3.36) except y has replaced z.

We write, , where g is the normalized conductance and b is the normalized

susceptance. So if we plot the constant g and constant b grids in and

jbgy +='re

'im axes, the

results will be the same as those for constant r and constant b grids in andre

im

axes.

Thus the result is the same as Fig.3-16 with r replaced by g, x replaced by b, replaced

by and

re're im replaced by

'im . This is shown in Fig.3-19.

This means that the same chart Fig.3-17 can be used for both normalized

impedance/admittance. This is why it is sometimes called immittance (impedance oradmittance) chart. There are two things to take care of.

(1)r=0 and x=0 represents short circuit, but g=0 and b=0 represents open circuit(2)r= and x= represents open circuit, but g= and b= represents short circuit.(3)The reflection coefficient co-ordinates are given for the impedance chart. So if the

reflection coefficient is plotted in the chart, it is considered to be impedance chart.

For the same reason, if normalized admittance is given in the chart, we shouldmultiply the reflection co-efficient read from the chart by -1, i.e., je ..

Fig.3-19

Finding normalized admittance from normalized impedance and vice versa

The reversal of axes provides a method for this. Suppose a point in the chart represents anormalized impedance. To obtain the corresponding location of the reflection coefficientin the admittance chart we have to multiply the reflection coefficient by -1, which means

rotating the point by radians, i.e., 180 0. This point is just the diametrically oppositepoint. The value read at this point from the grid is the normalized admittance.

The same argument applies for finding the normalized admittance. We read the values atthe diametrically opposite point.

3-20


21/46

Examples of using the Smith Chart

Example 3.1

Consider a characteristic impedance of 50 with the following impedances:

Z1 = 100 +j50 Z2 = 75 j100 Z3 =j200 Z4 = 150

Z5 = (an open circuit) Z6 = 0 (a short circuit) Z7 = 50 Z8 = 184 j900

The normalized impedances shown below are plotted in Figure 5.6.

z1 = 2 +j z2 = 1.5 j2 z3 =j4 z4 = 3z5 = z6 = 0 z7 = 1 z8 = 3.68 j18It is also possible to directly extract the reflection coefficient on the Smith chart ofFig.3-20. Once the impedance point is plotted (the intersection point of a constant

resistance circle and of a constant reactance circle), simply read the rectangular

coordinates projection on the horizontal and vertical axis. This will give , the real part

of the reflection coefficient, andre

im , the imaginary part of the reflection coefficient.

Alternatively, the reflection coefficient may be obtained in polar form by using the scalesprovided on the commercial Smith chart.

1 = 0.4 + 0.2 j=0.45 27

2 = 0.51 0.4 j= 0.65 38

3 = 0.875 + 0.48j= 0.998 29

4 = 0.5= 0.5 0

5 = 1= 1 0

6 = 1= 1 180

7 = 0= 0

8 = 0.96 0.1 j= 0.97 6

Fig. 3-20: Points plotted on the Smith chart for Example 3.1.

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Example 3.2

Consider the network of Fig.3-21. The elements are normalized withZ0 = 50 . We wishto find the impedanceZas seen from the input terminals.

x = 0.9

b = 1.1

x = -1.4 x = 1

b = -0.3 r = 1z = ?

Fig.3-21: A multi-element circuit

The series reactance (x) is positive for inductance and negative for capacitance. The

susceptance (b) is positive for capacitance and negative for inductance.

Starting at the right side, where there is a resistor and an inductor with a reactance value

of 1, we plot a point at the intersection of the where the r = 1 circle and thex = 1 circle.This becomes point A. As the next element is an element in shunt (parallel), we switch to

an admittance chart. To do this, we need to convert the previous point into an admittanceby rotating through 180. This becomes A'. We are now in the admittance mode.

The shunt element can be added by going along the conductance circle by a distance

corresponding to 0.3. This is done in a counter-clockwise direction (negativesusceptance) and gives point B. Then we have another series element. We therefore have

to switch back to the impedance Smith chart. This is accomplished by rotating through

180 to get back to the impedance mode.

After the conversion, we can determine B'. The series element is added by following

along the resistance circle by a distance corresponding to 1.4 and marking point C. Thisneeds to be done counter-clockwise (negative reactance value). For the next element, the

same operation is performed (conversion into admittance and plane rotation). Then move

the prescribed distance of 1.1 in a clockwise direction (because this is a positivesusceptance), along the constant conductance circle. We mark this as D.

Finally, we reconvert back to impedance mode and add the last element (the series

inductor). We then determine the required value, z, located at the intersection of resistor

circle 0.2 and reactance circle 0.5. Thus, z = 0.2 + j0.5. If the system characteristic

impedance is 50 , thenZ= 10 +j25 (see Fig.3-22).

3-22


23/46

A

A

B

B

C

C

D

D

Z

Fig.3-22: Network elements plotted on the Smith chart for Example 3.2.

Example 3.3:

Use the Smith chart to find the impedance of a short-circuited section of a lossless 50 co-axial transmission line that is 100 mm long. The transmission line has a dielectric of

relative permittivity between the inner and outer conductor, and the frequency

under consideration is 100 MHz.

9= r

3-23


24/46

For the transmission line, we find that 2875.600 == r rad/m and

m. The transmission line of length19993.0/2 == 100z = mm is therefore

wavelengths long./ 0.z = 1 Since 0=Lz , enter the Smith chart at a point Psc.

Move along the perimeter of the chart ( 1= ) by 0.1 wavelengths towards thegenerator in a clockwise direction to point P1. At P1 , read 0=r and 725.0x , or 725.0jzi = . Then == 3.3650725.0 jjZi .

1P

3P

'3P

scP

ocP

2P

'2P

MPO

Fig.3-23: Smith chart calculations for Example 3.3 and Example 3.4.

Example 3.4: A lossless transmission line of length 0.434 and characteristic impedance100 is terminated in an impedance 260 + j180 . Find the voltage reflectioncoefficient, the standing-wave ratio, the input impedance, and the location of a voltage

maximum on the line.

Given , and0.434z = =1000Z += 180260 jZL . Then

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Enter the Smith chart at 8.16.2/ 0 jZZz LL +== shown as point P2 in Fig.3-23

With the centre at the origin, draw a circle of radius 6.02 == LOP .

Draw the straight line 2OP and extend it to 2P on the periphery. Read 0.220 onwavelengths towards generator scale. The phase angle of the load reflection may

either be read directly from the Smith chart as 21 on the "Angle of ReflectionCoefficient" scale. Therefore 21 /180 0.120.6 0.6 = .=j jL e e The 6.0= circle intersects the positive real axis scOP at 4== Sr . Therefore the

voltage standing-wave ratio is 4.

The find the input impedance, move 2P at 0.220 by a total of 0.434 wavelengthstoward the generator first to 0.500 (same as 0.000) and then further to 0.434(0.5000.220)=0.154 to 3P .

Join O and 3P by a straight line which intersects the 6.0= circle at 3P . Here69.0=r and 2.1=x , or 2.169.0 jzi += . Then +=+= 12069100)2.169.0( ji jZ .

In going from 2P to 3P , the 6.0= circle intersects the positive real axis at M where there is a voltage maximum. Thus the voltage maximum appears at

0.2500.220=0.030 wavelengths from the load.

P

Impedance Transformation

Transformer

Fig.3-24 shows an ideal transformer with a turns ratio (ratio of primary turns to

secondary turns), n. Then if an impedance is connected to the secondary, the

impedance seen at the primary side is . This is useful for matching purelyresistive load and source impedances, and as shown in the figure. For matching

therefore, . Thus given and , the turns ratio required for match

is

LZ

L2in ZnZ =sR LR

L2

ins RnZR == sR LR

L

s

R

Rn = (3.39)

Fig.3-24

3-25


26/46

Unfortunately, ideal transformers are not practical. The simplest model is that of two loss

less coils with no losses and with all the magnetic flux coupled to them.

Fig.3-25: , - Primary voltage and current, - Secondary voltage and current,

- Primary and secondary turns, n the turns ratio =

pV pI sV , sI

sp N,N pssp I/IV/V = , m is the

mutual flux. Notice that all the flux is common to both coils.

Fig.3-26; Simplest equivalent circuit of a practical transformer. and are the

inductances of the primary and secondary coils respectively.1L 2L

A transformer is rarely used for impedance match at high frequencies. But there are

reports of integrated circuit transformers.

The concept of transformer is still important because as we see later, many impedance

matching circuits still use the turns ratio as defined by (3.39)

3-26


27/46

Inductor tap and capacitor tap

The inductor and capacitor tap circuits are widely used with resonant circuits requiring ahigher resistance than the load. The reason may be to increase the quality factor of the

circuit or it may be to satisfy the condition for oscillation in an oscillator. The method isvery convenient because the inductor or capacitor of the resonant circuit can be tapped

We consider expressions in terms of reactance as the results can be applied to capacitor

or inductor tap. The problem is shown in Fig. 3-27, where the resistance R is required to

be transformed to the equivalent resistance tR

Fig.3-27

A very simple result can be obtained if . In this case, we can apply voltage

division to get the voltage across R as

2L XR >>

VXX

XV

21

2R +

=

Equating the power in the resistance and its equivalentR tR

t

22

RR

V

R

V

= , giving

2

2

212R

t

X

XX

V

V

R

R

+=

=

Thus the circuit behaves as a transformer.

2

212t

X

XXnwithn

R

R +==

With , we may neglect R and get2XR>> 21t XXX +=

Approximate design procedure

(1)Calculate n=R

Rt

(2)n

XX t2 =

(3) 22t1 X)1n(XXX ==

3-27


28/46

For an accurate calculation, we have to use the parallel-series transformation consideredin page 2-8 of chapter 2. The calculation is easier if we convert both parallel circuit to the

series circuit in between and then equate resistances and reactances as shown in Fig.3-28

Fig.3-28

Exact design procedure

(1)Calculate n=R

Rt

(2)Calculate

t

tt

X

RQ =

(3)Calculate

2/1

2

2t

p 1n

Q1Q

+= (this is obtained from the first equation)

(4) p2 Q/RX =

(5)

+=

2p

2p

2t1Q!

QXXX

It is easy to show that these reduce to the approximate formulae for large , say

>10 ( is always greater than ).

pQ

pQ tQ pQ

(3) approximates ton

QQ tp = , i.e., n

Q

Q

p

t =

ButR

R

X

X

Q

Q t

t

2

p

t = , or 2

t

2 nX

Xn = , or,

n

XX t2 = which is the same as step 2 of the

approximate procedure.

3-28


29/46

Further (5) approximates to 2t1 XXX = which is the same as step 3 of the approximateprocedure.

The quality factor of the circuit is and the 3dB bandwidth istQ t0 Q/ .A very important point to note is that for inductor tapped network, the two inductors

should not have mutual inductance (no transformer like coupling by magnetic fields). Ifthey do the theory described here can not be used.

L-matching

We discussed how to increase the resistance of a parallel resonant circuit. We now

consider matching two resistive loads at some frequency 0 . We can do this by L-network.

One of the two resistances (source or load) is larger. We call it . The smaller

resistance (load or source) is called . Then the L-network must be of the type

shown in Fig. 3-29. must be in series with a reactance , because when

converted to the parallel form shown on the right of the figure, its value increases to

earglR

smallR

smallR sX

( 2SsmallP Q1RR += first. So the rule is the smaller resistance must be in series with areactance and the larger resistance must be in parallel with another reactance.

Fig.3-29

For matching, we set ( 2SsmallPeargl Q1RRR +== and = -p'p XX = 2S

2S

SQ

Q1X

+

Design procedure

(1)Calculate the transformation ratiosmall

eargl

R

Rn =

(2)Calculate 1nQ 2s =

(3)Calculate smalls s RQX =

(4)Calculate 'pX = - 2S

2S

SQ

Q1X

+

3-29


30/46

Fig3-30: Two possible implementations of Fig3-29

Matching with multiple L networks

With the L network, there is no control of the quality factor and hence the bandwidth.

Multiple L-networks are used for design when 0 bandwidth/quality factor and the tworesistances to be matched, and , are specified. The interested student may

consult the text by Besser and Gillmore for design. It should be noted that bandwidth can

not be decreased or increased indefinitely by increasing the number of section becauselosses of the capacitors and inductors if not other parasitics sets a limit to the bandwidth

from these designs.

earglR smallR

Matching using transmission lines

Quarter wavelength line

Recall equation (3.23))sin(jZ+)cos(Z)sin(jZ+)cos(ZZ=)(Z

L0

0L0inll

lll

Also, and/2=)f/v/(2=v/= pp =

ll 2

If the length is a quarter wavelength, i.e., 4/=l , or 4/1/ =l , 2/=l

Substituting,L

20

inZ

ZZ = and hence

L

20

inY

YY = (3.40)

3-30


31/46

Thus a quarter wave line inverts impedance/admittance. This property can be used for

matching two resistances, say and as illustrated in Fig.3-31.sR LR

Fig.3-31: sLLsL20in RR/RRR/ZZ ===

As before, perfect matching is correct at only one frequency.

There are many other types of matching. But we will consider only single stub matchingusing a Smith chart.

Single stub matching

In matching of impedances, we are only allowed to use reactive components (i.e.

equivalent to inductors and capacitors no resistors). Recall that for short-circuited and

open-circuited lossless transmission line sections of length l, the input impedance wasgiven by

, (3.41), 0 0tan tan(2 / )i sZ jZ l jZ l= =

and

, (3.42), 0 0cot cot(2 / )i oZ jZ l jZ l= =

where is purely real. The impedances in (3.41) and (3.42) are purely reactive

(imaginary), and therefore these transmission line sections act as inductors or capacitors,depending on the line length. We are going to make use of these elements (called

transmission line stubs) to design matching circuits. In practice, it is more convenient to

use short-circuited stubs. Short-circuited stubs are usually used in preference to open-

circuited stubs because an infinite terminating impedance is more difficult to realize thana zero terminating impedance. Radiation from the open end of a stub makes it appear

longer than it is, and compensation for these effects makes the use of open-circuited stubs

more cumbersome. A short-circuited stub of an adjustable length is much easier to

construct than an open-circuited stub.

00 RZ =

It is also more common to connect these stubs in parallel with the main line. For parallel

connections, it is convenient to use admittances rather than impedances. In thee cases, weuse the Smith chart as an admittance chart to design the matching networks.

A single-stub matching circuit is depicted in Fig.3-32. Note that the short-circuited stub isconnected in parallel with the main line. In order to match the complex load impedance

3-31


32/46

LZ to the characteristic impedance of the lossless main line, , we need to

determine the lengths dand l.00 RZ =

Fig.3-32: Impedance matching by single stub method.

For the transmission line to be matched at the point B B , the basic requirement is

0

0

1.

i B sY Y Y

YR

= +

= =(3.43)

In terms of normalized admittances, (3.42) becomes

1i B sy y y= + = . (3.44)

where for the load section and0/YYjbgy BBBB =+= 0/ YYy ss = for the short-circuited

stub. Note that is purely imaginary. It can therefore only contribute to

the imaginary part of . The position of

)/2cot( = ljys

iy B B (or, in other words, the length d) must be

chosen such that , i.e.1Bg =

1By Bjb= + . (3.45)

Next, the length l is chosen such that

s By jb= , (3.46)

which yields 1)()1( =++=+= BBsBi jbjbyyy . The circuit is therefore matched at

B B , and at any point left ofB B as well.

If we use the Smith chart, we would rotate on a -circle in a clockwise direction

(towards the generator) when transforming the normalized load admittance to the

3-32


33/46

admittance . However, according to (5.25), must also be located on theBy By 1=g

circle.

The use of the Smith chart for the purpose of designing a single-stub matching network isbest illustrated by means of an example.

Example 3.5: A 50 transmission line is connected to a load impedance. Find the position and length of a short-circuited stub required to match

the load at a frequency of 200 MHz. Assume that the transmission line is a co-axial line

with a dielectric for which .

= 5.3735 jZL

9= r

Given and== 5000 RZ = 5.3735 jZL . Therefore 75.0j7.0Z/Zz 0LL == .

Enter the Smith chart at Lz shown as point 1P in Figure 5.16.

Draw a -circle centred at O with radius 1OP .

Draw a straight line from 1P through O to point 2P on the perimeter, intersecting the

-circle at 2P , which represents Ly . Note 0.109 at 2P on the wavelengths toward

generator scale.

Note the two points of intersection of the -circle with the 1=g circle:

o At 3P : 11 12.11 BB jbjy +=+=

o At 4P : 22 12.11 BB jbjy +==

Solutions for the position of the stub:o For 3P (from 2P to 3P ) == 059.0)109.0168.0(1d

o For 4P (from 2P to 4P ) == 223.0)109.0332.0(2d

Solutions for the length of the short-circuited stub to provide Bs jby = :

o For 3P (from scP on the extreme right of the admittance chart to 3P , whichrepresents 2.11 jjby Bs == ): == 111.0)250.0361.0(1l

o For 4P (from scP on the extreme right of the admittance chart to 4P , which

represents 2.12 jjby Bs == ): =+= 389.0)250.0139.0(2l

To compute the physical lengths of the transmission line sections, we need to calculatethe wavelength on the transmission line. Therefore

m5.0//1

=

==f

c

ff

urp .

Thus:

mm5.29059.01 ==d mm5.55111.01 ==l

mm5.111223.02 ==d mm5.194389.02 ==l

Note that either of these two sets of solutions would match the load. In fact, there is a

whole range of possible solutions. For example, when calculating , instead of going

straight from to , we could have started at1d

2P 3P 2P , rotated clockwise around the Smith

3-33


34/46

chart n times (representing an additional length of 2/n ) and continued on to 3P ,

yielding . The same argument applies for , and ...,2,1,0,2/059.01 =+= nnd 2d 1l 2l

scPO

1P4P

'4P"3P

3P

'3P

2P

"4P'2P

1=g

Figure 3-33: Single-stub matching on an admittance chart (Example 3.5).

3-34


35/46

Multiport Networks

Port refers to the terminals of a network. Thus a one port network has a pair of terminals.

Two port network has two pairs of terminals and an N port network has N pairs ofterminals.

Fig, 3-34

We can just imagine two wires connected to the pair of terminals. At low frequencies, the

length of the line does not matter. However, we just saw that the wires are actually

transmission lines. At high frequencies, the length of the wires may be comparable towavelength and transmission line effect must be considered. We now know that in a

transmission line, voltage current and impedance vary along the length of the line. So wemust specify exactly where they are measured. These locations are called referenceplanes. In Fig.3-34 they are indicated by the dotted lines and the ports lie on these lines. It

is not enough to say what the impedance/ admittance is, we must specify the location of

the reference planes.Notice that the ports are numbered 1,2, N. The voltages and currents are indicated at

the ports. Most importantly notice the direction of the currents they flow into the ports.

We consider linear circuits for which output (voltage or current) is proportional to theinput (voltage or current). Thus the outputs can be written as linear combinations of

inputs.

Z and Y parameters

For the one port, we can write,

1111 IZ=V ( is input and is output, is impedance in ), or as1I 1V 11Z

1111 VY=I ( is input and is output, is admittance in Siemens)1V 1I 11Y

For the two port, we can write ( and are inputs and and are outputs)1I 2I 1V 2V

3-35


36/46

+=

+=

2221212

2121111

IZIZV

IZIZV(3.47)

or as ( and are inputs and and are outputs),1

V2

V1

I2

I

+=

+=

2221212

2121111

VYVYI

VYVYI(3.48)

In matrix form,

]I][Z[=]V[ (3.49)

]V][Y[=]I[ (3.50)

Where, , , ,

=

2

1

V

V]V[

=

2

1

I

I]I[

=

2221

1211

ZZ

ZZ]Z[

=

2221

1211

YY

YY]Y[

This can be generalized to N ports.

[Z] is called the impedance matrix and [Y] is called the admittance matrix. The meanings

of these parameters can be established easily.

Physical meaning of Z parameters

For the one port, is simply the impedance.11Z

From equations (3.47), we get the following:

Fig.3-35

Fig.3-36

Thus the Z parameters can be measured or calculated by applying a current at one port

and measuring the voltage at all ports with all other ports open circuited.

3-36


37/46

Reciprocity condition for Z parameters

Consider the two port where we apply a current at port 1 and measure the voltage at opencircuited port 2. Next we apply the same current at port 2 and measure the voltage at open

circuited port 1.

Fig.3- 37

If the measured voltages are equal, the circuit is said to be reciprocal. Hence for a

reciprocal circuit, , i.e.,IZ=V=IZ=V 2121212112 Z=Z (3.51)

Physical meaning of Y parameters

For one port, is the admittance.11Y


Fig. 3-38

Fig.3-39

Thus the Y parameters can be measured or calculated by applying a voltage at one port

and measuring the current at all ports with all other ports short circuited.

3-37


38/46

Reciprocity condition for Y parameters

Consider the two port where we apply a voltage at port 1 and measure the current at short

circuited port 2. Next we apply the same voltage at port 2 and measure the current atshort circuited port 1.

Fig.3-40

If the measured currents are equal, the circuit is said to be reciprocal. Hence for a

reciprocal circuit, , i.e.,IY=I=IY=I 1222112112 Y=Y (3. 52)

Scattering or S-parameters

Consider a two port with transmission lines of characteristic impedances and

connecting the two ports. We consider these characteristic impedances to be real.

From transmission line theory, we can write the voltages at the two ports as

01Z

02Z

+ += 111 VVV , , (3.53)-222 VVV +=

+

Where the + and signs refer to the forward and reflected voltages

The corresponding currents can be written as

01

1

01

1-111

Z

V-

Z

VIII

++ =+= ,

02

2

02

2222

Z

V-

Z

VIII

++ =+= (3.54)

Fig.3-41

The scattering parameters relate the forward and reflected voltages as:

+=

+=++

++

2221212

2121111

VSVSV

VSVSV(3.55)

Or in matrix form

3-38


39/46

]V][S[]V[ + = (3.56)

Where,

= +

++

2

1

V

V]V[ ,

=

2

1

V

V]V[ ,

=

2221

1211

SS

SS]S[

As with Z and Y parameters, this can be generalized to N port networks.

So far we have been calling and as reflected voltages. The scattering parameter

equation (3.55) however shows that they are the sum of voltage reflected from the port

and another term due to voltage incident on the other port. Thus I prefer calling and

reverse voltages rather than reflected voltages. Henceforth, we will use this term.

1V

2V

1V

2V

Physical meaning of S parameters


Fig. 3-42

Fig.3-43

3-39


40/46

In figures (3-42 ) and (3-43 ), it is understandable that we connect load impedances,

and to kill the reflected voltage. But why do we have to connect sources with

internal impedances and

01Z 02Z

01Z 02Z ?

Recall that the terminals have transmission lines of characteristic impedances and

. If the source internal impedances are not the same, additional reflections will becaused by the mismatch of the source and the transmission lines and these will affect the

measured/calculated S-parameters.

01Z

02Z

Why S-parameters?

We saw that to measure Z-parameters, we need open circuits and to measure Y-parameters, we need short circuits. On the other hand, to measure S-parameters, we need

matched loads.

Like old books, your text states When we attempt to create a short circuit with a wire:

the wire itself possesses a inductance that can be of substantial magnitude at highfrequency. Also, the open circuit leads to capacitive loading at the terminal In otherwords, the argument is that it is difficult to produce open and short circuits at high

frequencies. As matched loads might be produced easily, it is easier to measure S-

parameters rather than Z and Y parameters.

This argument has some truth, but vector network analyzers do not measure the

parameters by the simple ways considered while dealing with the physical meanings ofparameters. High quality open and short circuits may be difficult to produce, but they can

be produced. They are costly but are employed for calibration of vector network

analyzers used to measure S-parameters!

The proper reasons are as follow:

Recall that we stated the importance of reference planes when we introducedmultiport networks. In other words, it is not enough to give values of parameters,it must be stated where they are measured. These reference planes can be

established easily using S-parameter measurement techniques using a vector

network analyzer.

Note that the determination of S-parameters requires measuring forward andreverse voltages. At high frequencies, it is easier to sample and measure these

voltages rather than measure voltages and currents this is what is done by vectornetwork analyzers.

As correctly pointed out in your text, active devices (like transistors) can getunstable with open/short circuit loads or may be damaged by open/short circuits .Scattering parameter measurements use matched loads and active devices are

usually stable for such loads.

3-40


41/46

Normalized S-parameters

You might be wondering why we did not consider reciprocity when discussing S-

parameters. It can be proved that for the scattering parameters we have defined,for a reciprocal circuit unless2112 SS 0201 ZZ = , even though and

.

2112 ZZ =

2112 YY =One motivation for defining normalized scattering parameters is to have , where

the small s refers to normalized scattering parameters. To obtain the normalized s-

parameters, we define normalized voltages and currents as:

2112 ss =

01

11

Z

Vv = ,

02

22

Z

Vv = ..; 1011 IZi = , 2022 IZi = .. (3.57)

Consider the equation from (3.53). Dividing both sides by+ += 111 VVV 01Z ,

01

1

01

1

01

1

Z

V

Z

V

Z

V + += , or + += 111 vvv .

Similarly, dividing both sides of from (3.53) by222 VVV +=+

02Z ,

and so on.

+ += 222 vvv

Multiplying both sides of01

1

01

1-111

Z

V-

Z

VIII

++ =+= from (3. ) by 01Z ,

++

+ ==+= 1101

1

01

1-111 vv

Z

V-

Z

Viii .

Similarly, and so on.++ =+= 22-222 vviii

In matrix form, we can write

=

+=+

+

]v[]v[]i[

]v[]v[]v[(3.58)

Where, ,

=

2

1

v

v]v[

= +

++

2

1

v

v]v[ ,

=

2

1

v

v]v[ ,

=

2

1

i

i]i[

The normalized scattering parameters are defined in terms of the normalized forward and

reverse voltages as

+=

+=++

++

2221212

2121111

vsvsv

vsvsv(3.59)

Or in matrix form

3-41


42/46

]v][s[]v[ + = , (3.60)

=

2221

1211

ss

ss]s[

It can be proved that for normalized scattering parameters, s12 is always equal to s21for a reciprocal network.

Now you are going to have two objections.

Objection 1: These normalized voltages are fictitious. So what do we do in practice?

Measure scattering parameters and do some mathematical conversions?

The answer is no. Scattering parameters are measured with 00201 ZZZ == (=50 ) by avector network analyzer. It will measure the unnormalized scattering matrix [S}.

Consider the original scattering parameter equation as given by (3.55)

+=

+=

++

++

2221212

2121111

VSVSV

VSVSV

Dividing both sides by 00201 ZZZ == and remembering the expression for

normalized voltage,

+=

+=++

++

2221212

2121111

vSvSv

vSvSv

The normalized and unnormalized scattering parameters are the same when

00201 ZZZ == !

The measurement gives us normalized/unnormalized scattering parameters. When giving

scattering parameters, we must state the characteristic impedance. This impedance hasbeen standardized as 50 . So it is common practice for RF engineers to give scattering

parameters without stating whether it is normalized/unnormalized and what the

characteristic impedance is.

For unequal characteristic impedances, normalized scattering parameters, [s], and

unnormalized scattering parameters, [S], are not the same. For other characteristicimpedances other than 50 normalized scattering parameters can be obtained by matrix

methods we consider later.

Objection 2: The scattering concept is difficult to visualize when forward and reversevoltages do not exist for example as in the lumped element L matching networks

discussed earlier.

In any case we can define the forward and reverse voltages from voltages and currents

using transmission line relationship. Consider (3.53) and (3.54).

3-42


43/46

+ += 111 VVV and01

1

01

11

Z

V-

Z

VI

+=

Solving,2

ZIVV 01111

+=+ ,

2

ZIVV 01111

=

The corresponding normalized voltages will be01

01111

Z2ZIVv +=+ , and

01

01111

Z2ZIVv =

These normalized voltage can be defined at any port and we can choose the impedances

whatever we like, not necessarily the characteristic impedances of

transmission lines. This is why, the general name for these impedances is normalizing

01Z , 01Z

impedance.

Further generalization for normalized S-parameters

( Although this generalization has given rise to many research papers and complex

algebraic techniques, we will not use them in this course. It is discussed here becausemany books, including your text, abruptly introduce it and create confusion you may

omit this discussion.)

We have so far considered these impedances as real because we were thinking of them as

characteristic impedances. But now they can be anything. Suppose they are complex. We

now have to modify the definition as)ZRe(2

ZIVv

01

*0111

1+

=+ , and)ZRe(2

ZIVv

01

*0111

1

=

Note the differences from the earlier formula. We use the complex conjugate of n

the numerator and Re( ) in the denominator. The transmission line relationships

remain unchanged. The reason for using the complex conjugate is that if the port sees aconjugate match, there is no reflected wave. This is explained in Fig.3-44

01Z i

01Z

Fig.3-44

The definition can be further generalized to include complex normalizing impedanceswith negative real parts. The generalized definition is then

3-43


44/46

)ZRe(2

ZIVv

01

*0111

1+

=+ , and)ZRe(2

ZIVv

01

*0111

1

=

These quantities now lose their physical meaning as voltage and current waves.

Notice that the ratio of the normalized voltage and current is 1. So the incident power is

=++ *11 )i(v2

1v+ and similarly

2

1v is the reverse power. Thus these may be considered

as forward and reverse power waves.

Some definitions

Reflection or Return loss

Consider a two port and normalized scattering parameters. is the reflected voltage

and the corresponding reflected power is

+111vs

2

12

11 vs+ , while the incident power is

2

1v+ .

Thus the power loss in dB due to reflection is 111021

211

2

1

10 sog20

vs

vog10 ll =

+

+

dB

This is called Reflection or Return loss

Insertion Gain/Loss

Consider a two port matched to output and input. The power in the load is2

22

21 vs+ and

the incident power is2

1v+ .

The power gain in dB is 2111021

2

12

21

10 sog20

v

vsog10 ll =

+

+

This is called insertion

gain. If the two port is lossy, the insertion loss is 2111021

221

2

1

10 sog20

vs

vog10 ll =

+

+

While lossy networks, such as filters and attenuators are usually matched at both outputs

and inputs, it may not be possible to do so with amplifiers (we will see this later). In this

case, the insertion gain will not be 21110 sog20l , it will be the available power gain. We

will also see this later.

3-44


45/46

Two ports in series

For two 2-port is series, the overall Z matrix is given by the sum of the Z matrices of the

2-ports.

Fig. 3-55: ]Z[]Z[]Z[ 21 +=

Two ports in parallel

For two 2-port is parallel, the overall Y matrix is given by the sum of the Y matrices of

the 2-ports.

Fig.3-56: ]Y[]Y[]Y[ 21 +=

The formulae for two ports in series and parallel can be proved easily. These formulae

can not be used for certain networks. An example is given in Fig.4-5. p.155 of your text.

However the proper explanation is that for the formulae to be applicable, the circuit must

satisfy Brunes tests. These tests are given in books on electrical networks.

Two ports in cascade

For two 2-ports in cascade, we can not use matrix operations on Z, Y and S matrices

because these matrices do not relate output to input. Two types of matrices are used ABCD matrix and the transmission matrix. The latter involves travelling waves like those

used for defining S parameters and we will not deal with it.

3-45


46/46

The two port with current and voltage directions is shown in Fig.3-57

Fig.3-57

Notice that current flows into port 1 and out of port 2. For defining Z and Y parameters,currents flow into both ports. The ABCD parameters are defined through:

+=

+=

221

221

DICVI

BIAVV(3.61)

(3.61 ) can be written in matrix form as

=

2

2

1

1

I

V

DC

BA

I

V(3.62)

Now the input is expressed in terms of the output.Now consider the two 2-ports in cascade as shown in Fig.3-58.

Fig.3-58

=

'2

'2

11

11

1

1

I

V

DC

BA

I

Vand

=

2

2

22

22

'1

'1

I

V

DC

BA

I

V

As =

'2

'2

I

V

'1

'1

I

V

=

11

11

1

1

DC

BA

I

V

2

2

22

22

I

V

DC

BA

Thus the overall ABCD parameters of two cascaded 2-ports is the product of the

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Chapt.3-Transmission Line, Imped Transformation, 2-Port Net

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