EE 481 SDSMT

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K. W. Whites EE 481 Course Syllabus Page 1 of 3

South Dakota School of Mines and Technology Revised 9/2/08

EE 481Microwave EngineeringFall 2008

Instructor: Dr. Keith W. WhitesOffice: 317 Electrical Engineering/Physics (EEP) Building

Email: [email protected]: http://whites.sdsmt.eduOffice hours: MWF 3:00-4:00 PM

To contact the instructor, please use e-mail rather than the telephone. All e-mail will beanswered. The instructor will be available for assistance during the hours listed above, as well asother times when the office door is open.

Catalog Description: (3-1) 4 credits. Prerequisite: 382 completed or concurrent. Presentation ofbasic principles, characteristics, and applications of microwave devices and systems.Development of techniques for analysis and design of microwave circuits.

Time and Location: The lectures for this course will meet Monday, Wednesday, and Fridayfrom 10:00-10:50 AM in room 251B EEP. Laboratory work will be performed in 230 EEP.There is no common laboratory time for this course.

Course Reference Materials: The required materials for this course are

D. M. Pozar, Microwave Engineering, New York: John Wiley & Sons, third ed., 2004,which is available at the SDSMT Bookstore.

Additionally, the lecture notes K. W. Whites, EE 481 Microwave Engineering LectureNotes, 2008, are available from the course web page.

Grading: 30 % Two exams

30 % Laboratory20 % Homework20 % Final exam

Homework Policy: One homework set will generally be assigned each week. These homeworkassignments are to be turned in at the beginning of the class period on the due date. Latehomework will be assessed a 10% per calendar day reduction in points.

Labwork Policy: Near the middle of the semester, we will begin the first of approximately fourto five labs for the course. These will involve the design, construction, and measurement ofpassive and active microwave circuits. The labs will also require the simulation of your circuits

using Advanced Design System (ADS) from Agilent Technologies. Measurements will beperformed in theLaboratory for Applied Electromagnetics and Communications (LAEC) locatedin room 230 EEP using Agilent 8753ES vector network analyzers. Laboratory work will beperformed in pairs of students and open lab hours will be posted. Late lab reports will beassessed a 10% per calendar day reduction in points.

Exam Policy: The exams will be closed book and closed notes with no formula sheets. Using orreferring to equations stored in a calculator is not allowed, even if these equations come pre-programmed into the calculator. If you feel an exam problem was graded incorrectly, it must be

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EE 481Microwave Engineering

Lecture Notes

Keith W. WhitesFall 2008

Laboratory for Applied Electromagnetics and CommunicationsDepartment of Electrical and Computer Engineering

South Dakota School of Mines and Technology

2008 Keith W. Whites

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Whites, EE 481 Lecture 1 Page 1 of 5


Lecture 1: Introduction. Overview of

Pertinent Electromagnetics.

In this microwave engineering course, we will focus primarily

on electrical circuits operating at frequencies of 1 GHz and

higher. In terms of band designations, we will be working with

circuits above UHF:

Band Frequency

RFRegion HF 3 MHz-30 MHz

VHF 30 MHz-300 MHz

UHF 300 MHz-1 GHz

Microwav

eRegion

(=30cm

to8mm)

L 1-2 GHz

S 2-4 GHz

C 4-8 GHz

X 8-12 GHz

Ku 12-18 GHz

K 18-27 GHz

Ka 27-40 GHz

Millimeter

Wave

Region V 40-75 GHz

W 75-110 GHz

mm 110-300 GHz

RF, microwave and millimeter wave circuit design and

construction is far more complicated than low frequency work.

So why do it?

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Advantages of microwave circuits:

1. The gain of certain antennas increases (with reference toan isotropic radiator) with its electrical size. Therefore,

one can construct high gain antennas at microwave

frequencies that are physically small. (DBS, for example.)

2. More bandwidth. A 1% bandwidth, for example,provides more frequency range at microwave frequencies

that at HF.

3.

Microwave signals travel predominately by line of sight.Plus, they dont reflect off the ionosphere like HF signals

do. Consequently, communication links between (and

among) satellites and terrestrial stations are possible.

4. At microwave frequencies, the electromagnetic propertiesof many materials are changing with frequency. This is

due to molecular, atomic and nuclear resonances. This

behavior is useful for remote sensing and other

applications.

5. There is much less background noise at microwavefrequencies than at RF.

Examples of commercial products involving microwave circuits

include wireless data networks [Bluetooth, WiFi (IEEE Standard

802.11), WiMax (IEEE Standard 802.16), ZigBee], GPS,

cellular telephones, etc. Can you think of some others?

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Lecture 10: TEM, TE, and TM Modes forWaveguides. Rectangular Waveguide.

We will now generalize our discussion of transmission lines by

considering EM waveguides. These are pipes that guide EM

waves. Coaxial cables, hollow metal pipes, and fiber optical

cables are all examples of waveguides.

We will assume that the waveguide is invariant in the z-

direction:

x

y

z a

b

,

Metal walls

and that the wave is propagating in z as j ze . (We could also

have assumed propagation in z.)

Types of EM Waves

We will first develop an extremely interesting property of EM

waves that propagate in homogeneous waveguides. This will

lead to the concept of modes and their classification as

Transverse Electric and Magnetic (TEM),

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Transverse Electric (TE), or Transverse Magnetic (TM).

Proceeding from the Maxwell curl equations:

x y z

x y z

E j H j H x y z

E E E

= =

or x :y

z x

EEj Hy z

=

y : xz yEE

j Hx z

=

z :y x

z

E Ej H

x y

=

However, the spatial variation in z is known so that

( )( )

j z

j ze

j ez

=

Consequently, these curl equations simplify to

zy x

E j E j H

y

+ =

(3.3a),(1)

zx y

E j E j H x

=

(3.3b),(2)

y xz

E Ej H

x y

=

(3.3c),(3)

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We can perform a similar expansion of Ampres equation

H j E = to obtain

z

y x

H j H j E

y

+ =

(3.4a),(4)

zx y

H j H j E

x

=

(3.4b),(5)

y xz

H Hj E

x y

=

(3.5c),(6)

Now, (1)-(6) can be manipulated to produce simple algebraicequations for the transverse (x and y)components ofE and H.

For example, from (1):

zx y

j E H j E

y

= +

Substituting for Ey from (5) we find

2

2 2

1z zx x

z zx

j E H H j j H y j x

j E j H H

y x

= +

= +

or,2

z zx

c

j E H H

k y x

=

(3.5a),(7)

where2 2 2

ck k and 2 2k = . (3.6)

Similarly, we can show that

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2

z zy

c

j E H H

k x y

= +

(3.5b),(8)

2z z

x

c

j E H Ek x y

= + (3.5c),(9)

2

z zy

c

j E H E

k y x

= +

(3.5d),(10)

Most important point: From (7)-(10), we can see that all

transverse components of E and H can be determined fromonly theaxial components zE and zH . It is this fact that allows

the mode designations TEM, TE, and TM.

Furthermore, we can use superposition to reduce the complexity

of the solution by considering each of these mode types

separately, then adding the fields together at the end.

TE Modes and Rectangular Waveguides

A transverse electric (TE) wave has 0zE = and 0zH .

Consequently, all E components are transverse to the direction

of propagation. Hence, in (7)-(10) with 0zE = , then all

transverse components ofE and H are known once we find a

solution for only zH . Neat!

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For a rectangular waveguide, the solutions for xE , yE , xH , yH ,

and zH are obtained in Section 3.3 of the text. The solution and

the solution process are interesting, but not needed in this

course.

What is found in that section is that2 2

,

, 0,1,

( 0)c mn

m nm nk

m na b

= = + =

(11)

Therefore,

2 2

,mn c mnk k = =

(12)

These m and n indices indicate that only discrete solutions for

the transverse wavenumber (kc) are allowed. Physically, this

occurs because weve bounded the system in the x and y

directions. (A vaguely similar situation occurs in atoms, leading

to shell orbitals.)

Notice something important. From (11), we find that 0m n= =

means that ,00 0ck = . In (7)-(10), this implies infinite field

amplitudes, which is not a physical result. Consequently, the

0m n= = TE or TM modes are not allowed.

One exception might occur if 0z zE H= = since this leads toindeterminate forms in (7)-(10). However, it can be shown that

inside hollow metallic waveguides when both 0m n= = and

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Lecture 11: Dispersion. Stripline andOther Planar Waveguides.

Perhaps the biggest reason the TEM mode is preferred over TE

or TM modes for propagating communication signals is that

ideally it is notdispersive. That is, the phase velocity of a TEM

wave is not a function of frequency [ ( )p

v g ] if the material

properties of the waveguide are not functions of frequency.

To see this, recall for a TEM wave that LC = . Therefore,1pv

LC

= =

which is not a function of frequency, as conjectured, provided

neither L nor Care functions of frequency.

However, for either TE or TM modes, pv is a function of

frequency regardless of the material properties of the

waveguide.

Take the rectangular waveguide as an example. In the last

lecture, we found that

2 2

,mn c mnk k = =

and

2 2

2

,c mn

m n

k a b

= +

where , 0,1,2,m n = ( )0m n= for TE modes, while, 1,2,m n = for TM modes.

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For a CW signal carried by one of these modes, the phase

velocity is

, 2 2,

p mnc mn

vk

=

which is clearly a function of frequency. Consequently, we have

confirmed that TE and TM modes in a rectangular waveguide

are dispersive.

One special case is 0m n= = . Since,00

0c

k = , then ( )p

v g f

which means this is not a dispersive mode. However, the

0m n= = mode is the TEM mode, which cannot exist in a

hollow conductor waveguide.

The problem with (temporally) dispersive modes is that they can

severely distort signals that have been modulated onto them as

the carrier. As the signal propagates down the waveguide:

t t t t

In communications, such distortion is often unacceptable.

Therefore, the TEM mode is the one commonly used inmicrowave engineering. (For high power applications, hollow

waveguides made be required; hence, one would need to

somehow work around the distortion.)

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Since we prefer to work with the TEM mode of wave

propagation, it is important that we use waveguides in our

microwave circuits that will support TEM or quasi-TEM

modes. Examples of such structures are:

Microstrip and covered microstrip, Stripline, Slotline, Coplanar waveguide.

In this course, we will work primarily with microstrip. Actually,in the lab we will exclusively use microstrip.

Before delving into microstrip, however, lets quickly overview

some of these other TEM waveguides, beginning with stripline.

Stripline

Stripline is a popular, planar geometry for microwave circuits.

As shown in Fig. 3.22:

W

b r

Metal planes

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Lecture 12: Microstrip. ADS and LineCalc.

One of the most widely used planar microwave circuit

interconnections is microstrip. These are commonly formed by a

strip conductor (land) on a dielectric substrate, which is backed

by a ground plane (Fig. 3.25a):

rWd

t

We will often assume the land has zero thickness, t.

In practical circuits there will be metallic walls and cover to

protect the circuit. We will ignore these effects, as does the text.

Unlike the stripline, there is more than one dielectric in which

the EM fields are located (Fig 3.25b):

r

E

H

This presents a difficulty. Notice that if the field propagates as a

TEM wave, then

0p

r

cv

=

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But which r do we use?

The answer is neither because there is actually no purely TEM

wave on the microstrip, but something that closely approximates

it called a quasi-TEM mode. At low frequency, this mode is

almost exactly TEM. Conversely, when the frequency becomes

too high, there are appreciable axial components ofE and/or H

making the mode no longer quasi-TEM. This property leads to

dispersive behavior.

Numerical and other analysis have been performed on microstrip

since approximately 1965. Some techniques, such as the method

of moments, produce very accurate numerical solutions to

equations derived directly from Maxwells equations and

incorporate the exact cross-sectional geometry and materials of

the microstrip.

From these solutions, simple and quite accurate analytical

expressions for 0Z , pv , etc. have been developed primarily by

curve fitting.

The result is that at relatively low frequency, the wave

propagates as a quasi-TEM mode with an effective relativepermittivity, ,r e :

,

1 1 1

2 2 1 12

r rr e

d W

+ = +

+(3.195),(1)

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The phase velocity and phase constant, respectively, are:

0

,

p

r e

cv

= (3.193),(2)

0 ,r ek = (3.194),(3)

as for a typical TEM mode.

In general,

,1 r e r (4)

The upper bound occurs if the entire space above the microstrip

has the same permittivity as the substrate, while the lower bound

occurs if in this situation the material is chosen to be free space.

The characteristic impedance of the quasi-TEM mode on the

microstrip can be approximated as

,

0

,

60 8

ln 14

1201

1.393 0.667ln 1.444

r e

r e

d W W

W d d

Z W

dW W

d d

+ =

> + + +

(3.196),(5)

Alternatively, given a desired 0Z and r , the necessary W d can

be computed from (3.197).

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Again, (1) and (5) were obtained by curve fitting to numerically

rigorous solutions. Equation (5) can be accurate to better than

1%.

Example N12.1. Design a 50- microstrip on RogersRO4003C laminate with 1/2-oz copper and a standard thickness

slightly less than 1 mm.

Referring to the attached RO4003C data sheet from RogersCorporation, we find that 3.38 0.05r = and 0.032"d = . Wewill ignore all losses (dielectric and metallic).

What does 1/2-oz copper mean? Referring to the attached

technical bulletin from the Rogers Corporation, copper foil

thickness is more accurately measured through an areal mass.

The term 1/2-oz copper actually means 1/2 oz of copper

distributed over a 1-ft2 area.

For 1-oz copper, 34t = m. For 2-oz copper, double thisnumber and for -oz copper divide by 2.

We will use (3.197) to compute the required W d to achieve a

50- characteristic impedance:

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( ) ( )

2

82

2

2 1 0.611 ln 2 1 ln 1 0.39 22

A

A

r

r r

e W

e dW

Wd B B B d

= + + >

(3.197),(6)

To apply this equation, we first need to compute the constants A

and B:

0 1 1 0.11

0.2360 2 1

r r

r r

Z

A

+

= + + + 1.376= (7)

0

377

2 rB

Z

= 6.442= (8)

Next, we will arbitrarily assume that 2W d < and use thesimpler equation in (6). We find that

1.376

2 1.376

82.317

2

W e

d e

= =

.

Is this result less than 2? The answer is no. So, we need to

recompute W d using the bottom equation in (6). We find here

that 2.316W d = , which is greater than 2 as assumed.

So, with this result and 0.032"d = , then 2.316 0.032"W = 0.0741"= .

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A more common unit for width and thickness dimensions in

microwave circuits is mil where

1 mil1

" 25.41000

= = m

Therefore,

74.10.0741" "

14 1

0007 .W = == mils ( 1.88= mm).

This completes the design of the 50- microstrip.

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Lecture 13: Simple Quasi-Static MomentMethod Analysis of a Microstrip.

Computational electromagnetics (CEM) can provide accurate

solutions for Z0 and other microstrip properties of interest

including plots ofE and H everywhere in space and sJ and s

on the land or the ground plane. This can be accomplished

regardless of the cross-sectional geometry of the microstrip, the

thickness of the land or its conductivity.

The method of moments (MM) is a very popular CEM

technique. It is particularly useful for planar geometries such as

microstrip, stripline, conformal antennas, etc.

The MM was popularized by R. F. Harrington in 1965 with his

book Field Computations by Moment Methods. Today, it is

one of the most widely used CEM techniques.

Well illustrate the MM technique with a solution to a quasi-

static microstrip immersed in an infinite dielectric as shown:

x

y

w d

Land

Ground plane

That is, there is no substrate,per se.

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Integral Equation

Well imagine that a time harmonic voltage source has been

applied across the two conductors:

x

y

Line chargedensity [C/m]

+++ + + + +

++

- ---------- -- - - - - - - - -

V+

-

This causes a charge accumulation as shown.

Next, the image method will be employed to create an

equivalent problem for the fields in the upper half space ( 0y ):

x

y

e=0 naturally

satisfied+++ + + + +

++

V+

-

-- - - - - -

--

-V+

-

O.b. pointr

rd

d

In a previous EM course, youve likely learned that the electricpotential e at a point r in a homogeneous space produced by a

line charge density ( )l r is given by

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( ) ( )

( ) ( ) ( )( )2 2

1 1ln

2

1 ln2

e l

C

l

C

r r dlr r

r x x y y dx

=

= +

(1)

(For example, see J. Van Bladel, Electromagnetic Fields. New

York: Hemisphere Publishing, 1985.)

It is very important to realize that this contour C must include

all charge densities in the space, which means we must include

both conductors in this integral.

To develop an equation from which we can solve for the charge

density, well apply the boundary condition

( ) { }upper stripe r V r = (2)

Now, using (2) in (1) and accounting for both the +l and l

strips yields

( ) ( ) ( )2 21

ln2

lV r x x d d

= +

( ) ( ) ( )

0

top

2 2

bottom

lnl

dx

r x x d d dx

+ +

or ( ) ( ) ( )( )2 2

0

1 ln ln 42

w

lV r x x x x d dx

= + (3)

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Recall that the unknown in (3) is the line charge density l. But

how do we solve for this function? It varies along the strip so we

cant simply pull it out of the integral.

Actually, (3) is called an integral equation because the unknown

function is located in an integrand. You most likely havent

encountered such equations before. Integral equations are very

difficult to solve analytically. Well use a numerical solution

method instead.

Basis Function Expansion

In the moment method, we first expand l in a set of basis

functions. For a simple MM solution, here well use pulse basis

functions and divide the strips intoNuniform sections:

1

2

x 12

Nx

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where ( ) ( )11

; ,N

l n n n n

n

r P x x x =

(4)

What is not known in (4) are the amplitudes n of the line

charge density expansion. These are just numbers. So, instead of

directly solving for the spatial variation ofl in (3), now well

just be computing theseNnumbers, n. Much simpler!

However, we need to allow enough degrees of freedom in this

basis function expansion (4) so that an accurate solution can befound. This is accomplished by choosing the proper type of

expansion functions, a large enoughN, etc.

The next step in the MM solution is to substitute (4) into (3)

( ) ( )110

1; ,

2

w N

n n n n

n

V P x x x G x x dx

=

=

(5)

where ( ) ( ) ( )( )2 2ln ln 4G x x x x x x d = + (6)and is called the Greens function.

We can interchange the order of integration and summation in

(5) since these are linear operators, except perhaps when x x= .

In this case, the integrand becomes singular. Well consider thissituation later in this lecture.

Then, (5) becomes

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( ) ( )11 0

1; ,

2

wN

n n n n

n

V P x x x G x x dx

=

=

or ( )1

1

1

2

n

n

xN

n

n xV G x x dx

= = (7)

Testing the Integral Equation

In (7), we haveNunknown coefficients n to solve for, but only

a single equation. We will generate a total ofN equations by

evaluating (7) at Npoints along the (top) strip. This process is

called testing the integral equation. Well test (7) at the

centers of each of theNsegments,xm, giving

( )1

1

11, ,

2

n

n

xN

n m

n x

V G x x dx m N

=

= = (8)

This is the final system of equations that we will use to solve for

all the coefficients n.

Matrix Equation

It is helpful to cast (8) into the form of a matrix equation

[ ]

[ ]

[ ]

1 1 N N N N

V Z

= (9)

wherem

V V= (10a)

n n = (10b)

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( )1

1

2

n

n

x

mn

x

Z G x x dx

= (10c)

The numerical solution to (9) is accomplished by filling or

populating [V] and [Z], then solving a system of linear,

constant coefficient equations. In particular, for

[V] choose V= 1 V in (10a), for example. [Z] compute (10c) analytically, if possible, or by

numerical integration.

The filling of [V] is very simple, while filling [Z] is a bit more

difficult. In this quasi-static microstrip example, though, it is

possible to evaluate all of the terms analytically since a simple

anti-derivative is available.

In particular, with the center of the strip located at the origin asshown:

-/2

x

y

/2

(x,y) O.b. point

l

r

then the electrostatic potential at point r produced by a strip ofwidth supporting a constant line charge density l is given by

( ) ( )2

2 2

2

ln2

le r x x y dx

= + (11)

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Lecture 14: Impedance andAdmittance Matrices.

As in low frequency electrical circuits, a matrix description for

portions of microwave circuits can prove useful in simulations

and for understanding the behavior of the subcircuit, among

other reasons.

Matrix descriptions are a very convenient way to integrate the

effects of the subcircuit into a circuit without having to concernyourself with the specific details of the subcircuit.

We will primarily be interested in ABCD and S matrices in this

course, though Z and Y matrices will also prove useful. The

ABCD and S parameters are probably new to you. As well see,

using these matrix descriptions is very similar to other two-port

models for circuits youve seen before, such as Zand Ymatrices.

Z Matrices

As an example ofZmatrices, consider this two-port network:

1V

+

-

[ ]Z

1I

2V

+

-

2I

The Z-matrix description of this two-port is defined as

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[ ]

1 11 12 1

2 21 22 2

Z

V Z Z I

V Z Z I

=

(1)

where

0,k

iij

j I k j

VZ

I=

= (4.28)

As an example, lets determine the Z matrix for this T-network

(Fig. 4.6):

1V

+

-

AZ

2V

+

-

CZ

BZ1I 2I

Applying (1) repeatedly to all four Zparameters, we find:

2

111

1 0

A C

I

V Z Z Z

I =

= = + ( inZ at port 1 w/ port 2 o.c.)

1

112 1 2

2 0

C

I

V Z V I Z

I=

= = (think of 2I as source) 12 CZ Z=

2

221 2 1

1 0

C

I

V Z V I Z

I=

= = (think of 1I as source) 21 CZ Z=

1

222

2 0

B C

I

V Z Z Z I

=

= = + ( inZ at port 2 w/ port 1 o.c.)

Collecting these calculations, then for this T-network:

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[ ] A C C

C B C

Z Z Z Z

Z Z Z

+ = +

Notice that this matrix is symmetric. That is,ij jiZ Z= for i j .

It can be shown that [ ]Z will be symmetric for all reciprocalnetworks.

Whats the usefulness of an impedance matrix description? For

one thing, if a complicated circuit exists between the ports, one

can conveniently amalgamate the electrical characteristics intothis one matrix.

Second, if one has networks connected in series, its very easy to

combine the Zmatrices. For example:

1V

1I

2V

+

-

+

-

2I

[ ]Z

[ ]Z+-

+

-1

V

1I 2I

2V

[ ]Z

1V

1I

2I

2V

+

-

+

-

By definition

[ ]1 1

2 2

V IZ

V I

=

and [ ]1 1

2 2

V IZ

V I

=

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From the figure we see that1 1

I I = ,2 2

I I = , and that

1 1 1V V V = + , 2 2 2V V V = + . So, summing the above two matrix

equations gives

[ ] [ ]1 1 1 1

2 2 2 2

V V I I Z Z

V V I I

+ = +

+

Also from the figure, note that1 1

I I= and2 2

I I = . Therefore,

[ ] [ ]{ }[ ]

1 1

2 2Z

V IZ Z

V I

= +

(2)

From this result, we see that for a series connection of two-port

networks, we can simply add the Z matrices to form a single

super Zmatrix

[ ] [ ] [ ] Z Z Z = + (3)

that incorporates the electrical characteristics of both networks

and their mutual interaction.

YMatrices

A closely related characterization is the Y-matrix description of

a network:

1V

+

-

[ ]Y

1I

2V

+

-

2I

By definition:

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Lecture 15: S Parametersand the Scattering Matrix.

While Z and Y parameters can be useful descriptions for

networks, S and ABCD parameters are even more widely used in

microwave circuit work. Well begin with the scattering (or S)

parameters.

Consider again the multi-port network from the last lecture,

which is connected to Ntransmission lines as:

1 1,V I

+ + 1t

1 1,V I

2 2,V I

+ + 2t

2 2

,V I

3 3,V I

+ +3t

3 3,V I

,N N

V I+ +N

t

,N N

V I

[ ]S

0Z

0Z

0Z

0Z

Rather than focusing on the total voltages and currents (i.e., the

sum of + and - waves) at the terminal planes 1t ,., nt , the S

parameters are formed from ratios of reflected and incident

voltage wave amplitudes.

When the characteristic impedances of all TLs connected to thenetwork are the same (as is the case for the network shown

above), then the S parameters are defined as

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1 11 1 1

1

N

N N NN N

V S S V

V S S V

+

+

=

or [ ]V S V + = (4.40),(1)

where [ ]S is called the scattering matrix.

As we defined in the last lecture, the terminal planes are the

phase = 0 planes at each port. That is, with( ) ( ) ( )n n n n n n j z t j z t n n n nV z V e V e

+ = + 1, ,n N=

then at the terminal plane nt

( )n n n n n nV z t V V V + = = +

Each S parameter in (1) can be computed as

0,k

iij

j V k j

VSV

+

+

=

= (4.41),(2)

Notice in this expression that the wave amplitude ratio is defined

from port j to port i:

i jS

Lets take a close look at this definition (2). Imagine we have a

two-port network:

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1V

+1

t

1V

0Z [ ]S

2t

0Z

2V

+

2V

Then, for example,

2

111

1 0V

VS

V +

+

=

=

Simple enough, but how do we make 2 0V+ = ? This requires that:

1. There is no source on the port-2 side of the network, and2. Port 2 is matched so there are no reflections from this

port.

Consequently, with 2 11 110V S+ = = , which is the reflection

coefficient at port 1.

Next, consider

2

2

21

1 0V

VS

V +

+

=

=

Again, with a matched load at port 2 so that 2 0V

+ = , then

21 21S T=

which is the transmission coefficient from port 1 to port 2.

It is very important to realize it is a mistake to say 11S is the

reflection coefficient at port 1. Actually, 11S is this reflectioncoefficient only when 2 0V

+ = .

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Lecture 16: Properties ofS Matrices.Shifting Reference Planes.

In Lecture 14, we saw that for reciprocal networks the Z and Y

matrices are:

1.Purely imaginary for lossless networks, and2.Symmetric about the main diagonal for reciprocal

networks.

In these two special instances, there are also special properties

of the S matrix which we will discuss in this lecture.

Reciprocal Networks and S Matrices

In the case ofreciprocal networks, it can be shown that

[ ] [ ]

t

S S=

(4.48),(1)where [ ]

tS indicates the transpose of [ ]S . In other words, (1) is

a statement that [ ]S is symmetric about the main diagonal,which is what we also observed for the Zand Ymatrices.

Lossless Networks and S Matrices

The condition for a lossless network is a bit more obtuse for S

matrices. As derived in your text, if a network is lossless then

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[ ] [ ]{ }1

* tS S

= (4.51),(2)

which is a statement that [ ]S is a unitary matrix.

This result can be put into a different, and possibly more useful,

form by pre-multiplying (2) by [ ]t

S

[ ] [ ] [ ] [ ]{ } [ ]1

*t t tS S S S I

= = (3)

[ ]I is the unit matrix defined as

[ ]

1 0

0 1

I

=

Expanding (3) we obtain

[ ]

* * *11 21 1 11 12 1

* *12 22 21 22

* *1 1

1 0

0 1

t

N N

N NN N NN

S

i j

k

S S S S S S

S S S S

S S S S

=

=

(4)

Three special cases

Take row 1 times column 1:* * *

11 11 21 21 1 1 1N NS S S S S S+ + + = (5)

Generalizing this result gives

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*

1

1N

ki ki

k

S S=

= (4.53a),(6)

In words, this result states that the dot product of any column

of[ ]S with the conjugate of that same column equals 1 (for alossless network).

Take row 1 times column 2:* * *

11 12 21 22 1 2 0N NS S S S S S+ + + =

Generalizing this result gives

( )*1

0 , ,N

ki kj

k

S S i j i j=

= (4.53b),(7)

In words, this result states that the dot product of any column

of [ ]S with the conjugate of another column equals 0 (for alossless network).

Applying (1) to (7): If the network is also reciprocal, then [ ]S is symmetric and we can make a similar statement concerningthe rows of[ ]S .

That is, the dot product of any row of [ ]S with the conjugateof another row equals 0 (for a lossless network).

Example N16.1 In a homework assignment, the S matrix of a

two port network was given to be

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[ ]0.2 0.4 0.8 0.4

0.8 0.4 0.2 0.4

j jS

j j

+ = +

Is the network reciprocal? Yes, because [ ] [ ]t

S S= .

Is the network lossless? This question often cannot be answered

simply by quick inspection of the S matrix.

Rather, we will systematically apply the conditions stated above

to the columns of the S matrix:

*C1 C1 : ( )( ) ( )( )0.2 0.4 0.2 0.4 0.8 0.4 0.8 0.4 1 j j j j+ + + = *C2 C2 : Same = 1 *C1 C2 : ( )( ) ( )( )0.2 0.4 0.8 0.4 0.8 0.4 0.2 0.4 0 j j j j+ + + = *C2 C1 : Same = 0Therefore, the network is lossless.

As an aside, in Example N15.1 of the text, which we saw in thelast lecture,

[ ]0.1 0.8

0.8 0.2

jS

j

=

This network is obviously reciprocal, and it can be shown that

its also lossy.

Example N16.2 (Text Example 4.4). Determine the S

parameters for this T-network assuming a 50- system

impedance, as shown.

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1V

+

1V

050Z =

[ ]S

2V

+

2V

050Z =

1V

AV

2V

inZ

First, take a general look at the circuit:

Its linear, so it must also be reciprocal. Consequently, [ ]S must be symmetric (about the main diagonal).

The circuit appears unchanged when viewed from eitherport 1 or port 2. Consequently, 11 22S S= .

Based on these observations, we only need to determine 11S and

21S since 22 11S S= and 12 21S S= .

Proceeding, recall that 11S is the reflection coefficient at port 1

with port 2 matched:

2

2

111 11 0

1 0

V

V

VSV

+

+

+=

=

= =

The input impedance with port 2 matched is

( )in 8.56 141.8 8.56 50 50.00Z = + + = which, not coincidentally, equals 0Z ! With this Zin:

in 0

i 0

11

n

0Z Z

Z Z

S =

=

+

which also implies22 0S = .

Next, for 21S we apply 1V+ with port 2 matched and measure 2V

:

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Lecture 17: S Parameters and Time AveragePower. Generalized S Parameters.

There are two remaining topics concerning S parameters we will

cover in this lecture.

The first is an important relationship between S parameters and

relative time average power flow. The second topic is

generalized scattering parameters, which are required if the port

characteristic impedances are unequal.

S Parameters and Time Average Power

There is a simple and very important relationship between S

parameters and relative time average power flow. To see this,

consider a generic two-port connected to a TL circuit:

1V

+

1V

2V

+

2V

1V

2V1 2

By definition,1 11 1 12 2V S V S V

+ += + (1)

2 21 1 22 2V S V S V + += + (2)

At port 1, the total voltage is

1 1 1V V V+ = + (3)

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and the total time average power at that port is comprised of the

two terms (see 2.37):2

1

inc

02

V

PZ

+

= and

2

1

ref

02

V

PZ

= (4),(5)

Further, since port 2 is matched the total voltage there is

22 20V

V V+

== (6)

Consequently, for this circuit the transmitted power is

2

2

2

trans0 02

V

VP

Z

+

=

= (7)

Using the results from (4), (5), and (7), we will consider ratios

of these time average power quantities at each port and relate

these ratios to the S parameters of the network.

At Port 1. Using (4) and (5), the ratio of reflected andincident time average power is:

2 2

1ref 1

2

inc 11

VP V

P VV

++= = (8)

From (1) and noticing port 2 is matched so that

2

1

11

1 0V

V

S V +

+==

then in (8):

2

2ref11

inc 0V

PS

P + =

= (9)

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This result teaches us that the relative reflected time average

power at port 1 equals2

11S when port 2 is matched.

At Port 2. Using (7) and (4), the ratio of transmitted andincident time average power is:

2

2

trans 20

2

inc 1

VP V

P V

+

=

+= (10)

However, from (2) and with 2 0V+ = , then

2

2trans21

inc 0V

P SP + =

= (11)

This result states that the relative transmitted power to port 2

equals2

21S when port 2 is matched.

Equations (9) and (11) provide an extremely useful physical

interpretation of the S parameters as ratios of time averagepower. Note that this interpretation is valid regardless of the loss

(or even gain) of the network.

However, if the network is lossless we can use (9) and (11) to

develop other very useful relationships. Recall that for a lossless

network,

[ ] 11 12

21 22

S SS

S S

=

must be unitary. As a direct result of this

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Lecture 18: Vector Network Analyzer.

A network analyzer is a device that can measure S parameters

over a range of frequencies. There are two types:

1.Scalar network analyzer. Measures only the magnitude ofthe S parameters.

2.Vector network analyzer (VNA). Measures both themagnitude and phase of the S parameters.

The latter is generally a much more expensive piece of

equipment.

The VNA is basically a sophisticated transmitter and receiver

pair with vast signal processing capabilities. Here is a block

diagram of a typical vector network analyzer (text p. 183):

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We will examine the basic subsystems of a VNA in this lecture

and some important topics concerning the sources of error and

calibration of the VNA.

The following pages are from Network Analyzer Basics,

Agilent Product Note E206. (Agilent Technologies is the

company that was formed when Hewlett Packard Corporation

spun-off its test and measurement business.)

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Lecture 19: Proper MicrowaveLaboratory Practices.

Microwave circuit measurements are very different than

electrical measurements at lower frequencies. Here are four key

differences:

1.We often use a network analyzer to make S parametermeasurements rather than traditional instruments for

voltage or current measurements.

2.Precision connections are necessary for accurate andrepeatable measurements.3.Special tools are used to tighten coaxial connectors.4.Electrostatic protection is absolutely necessary.

You will be using an Agilent 8753ES Vector Network Analyzer

(VNA) to make all of your measurements this semester. This

VNA operates from 30 MHz to 6 GHz.

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A VNA measures both the magnitude and phase of S

parameters. Note that a VNA is intrinsically making frequency

domain measurements, i.e., sinusoidal steady state.

Types of Coaxial Connectors

There are nine types of coaxial connectors that you may

encounter in most RF and microwave engineering laboratories:

1.BNC2.Type F3.Type N4.SMA

5.APC-76.APC-3.57.2.92 mm8.2.4 mm9.1.85 mm

The right-hand column lists metrology-grade connectors.

The photograph on p. 130 of your text shows Type N, TNC,

SMA, APC-7 and 2.4 mm connectors:

In this course, you will primarily be using the SMA connector.

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Proper Microwave Coaxial Connections

We will use only SMA connectors in the EE 481 laboratory. Do

NOT finger tighten these, or any other, precision microwave

connectors. Instead, use a black-handled torque wrench, which

provides the proper 5 in-lbs of torque required for SMA.

If you over-tighten a connection, it is possible you will damage

the VNA connector, the cable or your microwave circuit. There

may be different types of torque wrenches lying about the lab.Be certain you are using the BLACK-handled torque wrench.

Standard operating procedures for making microwave coax

connections include:

1) When initially making coaxial connections: Make sure you are electrostatically grounded. Be certain there are no metal filings or other debris inside

the connectors.

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Lecture 2: Telegrapher Equations

For Transmission Lines. Power Flow.

Microstrip is one method for making electrical connections in a

microwave circuit. It is constructed with a ground plane on one

side of a PCB and lands on the other:

Microstrip is an example of a transmission line, though

technically it is only an approximate model for microstrip, as we

will see later in this course.

Why TLs? Imagine two ICs are connected together as shown:

A

B When the voltage at A changes state, does that new voltage

appear at B instantaneously? No, of course not.

If these two points are separated by a large electrical distance,

there will be a propagation delay as the change in state

(electrical signal) travels to B. Not an instantaneous effect.

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In microwave circuits, even distances as small as a few inches

may be far and the propagation delay for a voltage signal to

appear at another IC may be significant.

This propagation of voltage signals is modeled as a

transmission line (TL). We will see that voltage and current

can propagate along a TL as waves! Fantastic.

The transmission line model can be used to solve many, many

types of high frequency problems, either exactly orapproximately:

Coaxial cable.

Two-wire.

Microstrip, stripline, coplanar waveguide, etc.

All true TLs share one common characteristic: the E and H

fields are all perpendicular to the direction of propagation,

which is the long axis of the geometry. These are called TEM

fields for transverse electric and magnetic fields.

An excellent example of a TL is a coaxial cable. On a TL, the

voltage and current vary along the structure in time t and

spatially in the z direction, as indicated in the figure below.There are no instantaneous effects.

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E

E

z

H

Fig. 1

A common circuit symbol for a TL is the two-wire (parallel)

symbol to indicate any transmission line. For example, the

equivalent circuit for the coaxial structure shown above is:

Analysis of Transmission Lines

On a TL, the voltage and current vary along the structure in time

(t) and in distance (z), as indicated in the figure above. There are

no instantaneous effects.

( ),i z t

( ),i z t

z( ),v z t+

-

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How do we solve for v(z,t) and i(z,t)? We first need to develop

the governing equations for the voltage and current, and then

solve these equations.

Notice in Fig. 1 above that there is conduction current in the

center conductor and outer shield of the coaxial cable, and a

displacement current between these two conductors where the

electric field E is varying with time. Each of these currents

has an associated impedance:

Conduction current impedance effects:oResistance, R, due to losses in the conductors,

o Inductance, L, due to the current in the conductors and

the magnetic flux linking the current path.

Displacement current impedance effects:

oConductance, G, due to losses in the dielectric

between the conductors,

oCapacitance, C, due to the time varying electric field

between the two conductors.

To develop the governing equations for ( ),V z t and ( ),I z t , wewill consider only a small section z of the TL. This z is so

small that the electrical effects are occurring instantaneously and

we can simply use circuit theory to draw the relationshipsbetween the conduction and displacement currents. This

equivalent circuit is shown below:

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( ),v z t

( ),i z t L z ( ),i z z t +

( ),v z z t + +

-

+

-

z

C z G z

R z

Fig. 2

The variables R, L, C, and G are distributed (or per-unit length,

PUL) parameters with units of /m, H/m, F/m, and S/m,

respectively. We will generally ignore losses in this course.

In the case of a lossless TL where R = G = 0, a finite length of

TL can be constructed by cascading many, many of these

subsections along the total length of the TL:

+

-

L z

C z

z

L z

C z

L z

C z

L z

C z

RL

Rs

vs(t)

z z z

This is a general model: it applies to any TL regardless of its

cross sectional shape provided the actual electromagnetic field is

TEM.

However, the PUL-parameter values change depending on the

specific geometry (whether it is a microstrip, stripline, two-wire,

coax, or other geometry) and the construction materials.

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Transmission Line Equations

To develop the governing equation for ( ),v z t , apply KVL inFig. 2 above (ignoring losses)

( )( )

( ),

, ,i z t

v z t L z v z z t t

= + +

(2.1a),(1)

Similarly, for the current ( ),i z t apply KCL at the node

( )( )

( ),

, ,v z z t

i z t C z i z z t

t

+ = + +

(2.1b),(2)

Then:

1.Divide (1) by z :

( ) ( ) ( ), , ,v z z t v z t i z t L

z t

+ =

(3)

In the limit as 0z , the term on the LHS in (3) is the

forward difference definition of derivative. Hence,

( ) ( ), ,v z t i z t L

z t

=

(2.2a),(4)

2.Divide (2) by z :

( ) ( ) ( )0 ,, , v z z t i z z t i z t Cz t

+ +

= (5)

Again, in the limit as 0z the term on the LHS is the

forward difference definition of derivative. Hence,

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Lecture 20: Transmission (ABCD) Matrix.

Concerning the equivalent port representations of networks

weve seen in this course:

1.Z parameters are useful for series connected networks,2.Y parameters are useful for parallel connected networks,3.S parameters are useful for describing interactions of

voltage and current waves with a network.

There is another set of network parameters particularly suitedfor cascading two-port networks. This set is called the ABCD

matrix or, equivalently, the transmission matrix.

Consider this two-port network (Fig. 4.11a):

1V

+

-

A B

C D

1I

2V

+

-

2I

Unlike in the definition used for Z and Y parameters, notice that

2I is directed away from the port. This is an important point and

well discover the reason for it shortly.

The ABCD matrix is defined as

1 2

1 2

V VA B

I IC D

=

(4.63),(1)

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It is easy to show that

2

1

2 0I

VA

V=

= ,

2

1

2 0V

VB

I=

=

2

1

2 0I

IC

V=

= ,

2

1

2 0V

ID

I=

=

Note that not all of these parameters have the same units.

The usefulness of the ABCD matrix is that cascaded two-port

networks can be characterized by simply multiplying theirABCD matrices. Nice!

To see this, consider the following two-port networks:

1V

+

-

1 1

1 1

A B

C D

1I

2V

+

-

2I

2V

+

-

2 2

2 2

A B

C D

2I

3V

+

-

3I

In matrix form

1 1 1 2

1 1 1 2

V A B V

I C D I

=

(4.64a),(2)

and32 2 2

32 2

2

VV A B

IC DI

=

(3)

When these two-ports are cascaded,

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Lecture 21: Signal Flow Graphs.

Consider the following two-port network (Fig 4.14a):

1a

1t

1b

0Z [ ]S

2t

0Z

2a

2b

A signal flow graph is a diagram depicting the relationships

between signals in a network. It can also be used to solve for

ratios of these signals.

Signal flow graphs are used in control systems, power systems

and other fields besides microwave engineering.

Key elements of a signal flow graph are:

1.The network must be linear,2.Nodes represent the system variables,3.Branches represent paths for signal flow.

For example, referring to the two-port above, the nodes and

branches are (Fig 4.14b):

1a

2a

2b

1b

21S

12S

22S

11S

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4.A signal ky traveling along a branch between nodes ka and jb is multiplied by the gain of that branch:

ka jbjkS

That is,

j jk k b S a=

5.Signals travel along branches only in the direction of thearrows.

This restriction exists so that a branch from ka to jb denotes a

proportional dependence ofj

b onk

a , but not the reverse.

Solving Signal Flow Graphs

Signal flow graphs (SFGs) can form an intuitive picture of the

signal flow in a network. As an application, we will develop

SFGs in the next lecture to help us calibrate out systematic

errors present when we make measurements with a VNA.

Another useful characteristic is that we can solve for ratios of

signals directly from a SFG using a simple algebra.

There are four rules that form the algebra of SFGs:

1. Series Rule. Given the two proportional relations

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2 21 1V S V= and 3 32 2V S V=

then ( )3 32 21 1V S S V = (4.69),(1)

In a SFG, this is represented as (Fig. 4.16a):

V1

V2

V3

S21

S32

V1

V3

S21

S32

In other words, two series paths are equivalent to a single

path with a transmission factor equal to a product of the two

original transmission factors.

2. Parallel Rule. Consider the relation:

( )2 1 1 1a b a bV S V S V S S V = + = + (4.70),(2)

In a SFG, this is represented as (Fig 4.16b):

V1 V2

Sa

Sb

V1

V2

Sa

+ Sb

In other words, two parallel paths are equivalent to a single

path with a transmission factor equal to the sum of the

original transmission coefficients.

3. Self-Loop Rule. Consider the relations

2 21 1 22 2V S V S V = + (4.71a),(3)

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and 3 32 2V S V= (4.71b),(4)

We will choose to eliminate 2V . From (3)

( ) 212 22 21 1 2 122

11

SV S S V V V S

= =

Substituting this into (4) gives

32 213 1

221

S SV V

S=

(4.72),(5)

In a SFG, this is represented as (Fig 4.16c):

V1

V2

V3

S32

V1

V2

V3

S21

S32

S22

V1

V3

21

221

S

S

32 21

221

S S

S

In other words, a feedback loop may be eliminated by

dividing the input transmission factor by one minus the

transmission factor around the loop.

4. Splitting Rule. Consider the relationships:4 42 2V S V= (6)

2 21 1V S V= (7)

and 3 32 2V S V= (8)

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The SFG is

V1

V2

V3

V4

S21

S32

S42

From (6) and (7) we find that 4 42 21 1V S S V = . Hence, if we use

the Series Rule in reverse we can define:

4 42 4V S V= and 4 21 1V S V

=

In a SFG, this is represented as (Fig 4.16d):

V1

V2

V3

V4

V4'

S21

S32

S21

S42

V1

V2

V3

V4

S21

S32

S42

In other words, a node can be split such that the product of

transmission factors from input to output is unchanged.

Example N21.1. Construct a signal flow graph for the network

shown below. Determine in and LV using only SFG algebra.

1a

1b

2a

2b

LVin LS

1t

2t

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Lecture 22: Measurement Errors.TRL Calibration of a VNA.

As we discussed in Lecture 18, a VNA measures both the

magnitude and phase ofS parameters.

However, there will invariably be significant errors in these

microwave measurements that must be removed somehow if

we are to obtain accurate results.

There are three general types of errors:

1. Systematic: repeatable errors due to imperfections in

components, connectors, test fixture, etc.

2. Random: vary unpredictability with time and cannot be

removed. From noise, connector repeatability, etc.

3. Drift: caused by changes in systems characteristics after a

calibration has been performed due to temperature, humidity

and other environmental variables.

Using well-designed and maintained equipment in an

unchanging environment is about all we can do to minimize

random errors. A similar environment helps minimize drift

errors, or the network analyzer can be recalibrated.

The effects ofsystematic errors can be largely removed from the

S parameters using calibration. (In the context of microwave

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measurements, calibration has a much different meaning than

calibrating low-frequency equipment.)

To do this calibration, we need to assume a general model for

the effects of systematic errors, such as that shown in Fig 4.20:

Well definem

S as the S parameters that are actually

measured by the VNA. These include all of the errors we

mentioned earlier.

The error boxes (with parameters [ ]S ) and how these arespecifically connected to the DUT form the model of the

systematic errors. The parameters [ ]S are those we desire toknow. These are the S parameters of the DUT, which also,

unfortunately, contain random errors.

The purpose of network analyzer calibration is to determine the

numerical values of all the S (or ABCD) parameters in the error

model at each frequency of interest.

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For coaxial measurements, we often use precision Short, Open,

Load and Thru (SOLT) standards as loads connected to the test

ports. With these known standards as loads, we make several S-

parameter measurements to construct enough equations from

which we can numerically determine the error parameters.

Thru-Reflect-Line (TRL) Calibration

SOLT standards are difficult to implement for VNAmeasurements of microstrip and similar circuits. Instead, the

Thru-Reflect-Line (TRL) method is more commonly used.

The TRL calibration method is very cleverly designed. It doesnt

rely on preciselyknown standards and it uses only three simple

connections to completely characterize the error model.

The three connections for TRL calibration of microstrip are:

1. Thru. Directly connect port 1 to 2, at the desired reference

planes, using matched microstrip.

2. Reflect. Terminate a microstrip connected to each port with a

load that produces a large reflection, say an open or short.

These can be imperfect loads.

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3. Line. Connect the two ports together through a microstrip

approximately /4 longer than the Thru (at the center

frequency).

We will step through each of these connections and outline the

solutions for the S parameters using signal flow diagrams.

1. Thru Standard. The configuration for this measurement is

shown in Fig 4.21a. The measured S matrix is defined as [ ]T :

Notice that:

a. The [ ]S matrices for the two error boxes are assumed to beidentical. This simplifies things for us right now, though

this is not assumed in actual VNA TRL cal kits.b. The reflection planes for the DUT are coincident.

Consequently, this is called a zero length Thru. Youll

use this in the lab.

c. 21 12S S= for a reciprocal error box.

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Lecture 23: Basic Properties ofDividers and Couplers.

For the remainder of this course were going to investigate a

plethora of microwave devices and circuits both passive and

active.

To begin, during the next six lectures we will focus on different

types of power combiners, power dividers and directional

couplers.

Such circuits are ubiquitous and highly useful. Applications

include:

Dividing (combining) a transmitter (receiver) signal tomany antennas.

Separating forward and reverse propagating waves (canalso use for a sort of matching).

Signal combining for a mixer.As a simple example, a two-way power splitter would have the

form (Fig 7.1a):

1P

Divider

orCoupler

2 1P P=

( )3 11P P=

where and 0 1 . The same device can often be used

as a power combiner:

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1 2 3P P P= +

Divider

or

Coupler

2P

3P

We see that even the simplest divider and combiner circuits are

three-port networks. It is common to see dividers and couplers

with even more than that.

So, before we consider specific examples, it will be beneficial

for us to consider some general properties of three- and four-port

networks.

Basic Properties of Three-Port Networks

As well show here, its not possible to construct a three-port

network that is:

1. lossless,

2. reciprocal, and

3. matched at all ports.

This basic property of three-ports limits our expectations for

power splitters and combiners. We must design around it.

To begin, a three-port network has an S matrix of the form:

[ ]11 12 13

21 22 23

31 32 33

S S S

S S S S

S S S

=

(7.1),(1)

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If the network is matched at every port, then 11 22 33 0S S S= = = .

(It is important to understand that matched means 1 , 2 and

30 = when all other ports are terminated in 0Z .)

If the network is reciprocal, then 21 12S S= , 31 13S S= and

32 23S S= . Consequently, for a matched and reciprocal three-port,

its S matrix has the form:

[ ]12 13

12 23

13 23

0

0

0

S S

S S S

S S

=

(7.2),(2)

Note there are only three different S parameters in this matrix.

Lastly, if the network is lossless, then [ ]S is unitary. Applying(4.53a) to (2), we find that

22

12 13 1S S+ = (7.3a),(3)

2212 23 1S S+ = (7.3b),(4)

2 2

13 23 1S S+ = (7.3c),(5)

and applying (4.53b) that:*

13 23 0S S = (7.3d),(6)*

23 12 0S S = (7.3e),(7)*

12 13 0S S = (7.3f),(8)

From (6)-(8), it can be surmised that at least two of the three S

parameters must equal zero. If this is the case, then none of the

equations (3), (4) or (5) can be satisfied. [For example, say

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13 0S = . Then (6) and (8) are satisfied. For (7) to be satisfied and

230S , we must have 12 0S = . But with S12 and S13 both zero,

then (3) cannot be satisfied.]

Our conclusion then is that a three-port network cannot be

lossless, reciprocal and matched at all ports. Bummer. This

finding has wide-ranging ramifications.

However, one can realize such a network if any of these three

constraints is loosened. Here are three possibilities:1.Nonreciprocal three-port. In this case, a lossless three-port

that is matched at all ports can be realized. It is called a

circulator (Fig 7.2):

Port 1

Port 2

Port 3

[ ]

0 0 1

1 0 00 1 0

S

=

Notice that

ij jiS S .

2.Match only two of the three ports. Assume ports 1 and 2are matched. Then,

[ ]12 13

12 23

13 23 33

0

0

S S

S S S

S S S

=

(7.7),(9)

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Lecture 24: T-Junction and ResistivePower Dividers.

The first class of three-port network well consider is the T-

junction power divider. We will look at lossless, nearly lossless

and lossy dividers in this and the next lecture.

A simple lossless T-junction network is shown in Fig. 7.6:

1V

+

1V

3V

+

3V

2V

+

2V

0Z

1Z

2Z

1

2

3

inY

There are two basic constraints we need to incorporate into thispower splitter:

1.The feedline should be matched.2.The input time average power Pin should be divided

between ports 2 and 3 in a desired ratio.

In the text, this ratio is defined as X:Y where:

( )/ 100% X X Y + of the incident power is deliveredto one output port, and

( )/ 100%Y X Y+ of the incident power is delivered tothe other.

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For example:

1:1 means 50% of the incident time average power isdelivered to each output port.

2:1 means 67% of the incident time average power isdelivered to one output port and the remaining to the

other.

Referring to the circuit above, in order to enforce the first

constraint on the power splitter requires that

in

1 2 0

1 1 1Y Z Z Z

= + = (7.25),(1)

Consequently, to divide the incident power between the two

output ports, we simply need to adjust the characteristic

impedances of the two TLs.

Because port 1 is matched, the input time average power is

simply:2

0

in

0

1

2

VP

Z= (2)

where V0 is the phasor voltage at the junction.

The output powers can be computed similarly as2

0

1

1

1

2

VP

Z= and

2

0

2

2

1

2

VP

Z= (3),(4)

Dividing (3) and (4) by (2) we find

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01 1

in 0 1

1/

1/

ZP Z

P Z Z= = and 02 2

in 0 2

1/

1/

ZP Z

P Z Z= = (5),(6)

Because the network is lossless:

1 2

in in

1P P

P P+ =

Substituting (5) and (6) into this expression gives

0 0

1 2

1Z Z

Z Z+ = so that

1 2 0

1 1 1

Z Z Z + =

Consequently, not only have we split the power between theoutput ports, but in light of (1) we have also ensured that the

feedline is matched.

So, once we have specified the desired ratios for the output port

powers, we can use (5) and (6) to compute the required

characteristic impedances of these TLs:

01

1 in

ZZ

P P= and 02

2 in

ZZ

P P= (7),(8)

Thats basically it for the design of a simple T-junction power

divider. An example of this design process is given in Example

7.1 of the text, which well cover later.

From a practical standpoint, there are two important points that

arise with T-junction power splitters:

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1.Junction effects. At the junction of the TLs, there is likelyto be an accumulation of excess charge. Take a microstrip

junction for example:

+++

++

+++

++

These charges attract oppositely-signed charges on the

ground plane:

+ +++

- ---

E

This time-varying electric field is a displacement current,

of course. We can model this effect as a lumped capacitor

connected to ground, as shown in Fig. 7.6.

2.Characteristic impedance of the output lines. It is not toopractical to have these Z1 and Z2 characteristic impedances

in the system. We generally like to work with just one

system impedance, Z0.

To compensate for this, we can use QWTs for matching:

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1V

+

1V

3V

+

3V

2V

+

2V

0Z

0Z

0Z

inY

in,1Z

in,2Z

Using QWTs makes this power splitter narrow-banded,

unfortunately.

Here, instead ofZ1 and Z2, the impedances of interest in thepower splitter design are Zin,1 and Zin,2. From (1), the match

condition now becomes

in,1 in,2 0

1 1 1

Z Z Z + = (9)

and from (5) and (6), the power division constraints

become01

in in,1

ZP

P Z= and 02

in in,2

ZP

P Z= (10),(11)

Example N24.1 (text example 7.1). Design a 1:2, T-junction

power divider in a 50- system impedance.

Well choose to use the network in Figure 7.6 with B = 0:

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Lecture 25: Wilkinson Power Divider.

The next three port network we will consider is the Wilkinson

power divider (Fig 7.8b):

0Z

0Z

0Z

Port 1

Port 2

Port 3

0,QZ

0,QZ

/4

/4

R

This is a popular power divider because it is easy to construct

and has some extremely useful properties:

1.Matched at all ports,2.Large isolation between output ports,3.Reciprocal,4.Lossless when output ports are matched.

There is much symmetry in this circuit which we can exploit to

make the S parameter calculations easier. Specifically, we will

excite this circuit in two very special configurations

(symmetrically and anti-symmetrically), then add these two

solutions for the total solution.

This mathematical process is called an even-odd mode

analysis. It is a technique used in many branches of science

such as quantum mechanics, antenna analysis, etc.

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We will now show that for a 1:1 Wilkinson power divider,

0, 02QZ Z= and 02R Z= . To simplify matters, as in the text, we

will:

1.Normalize all impedances to 0Z ,2.Not draw the return line for the TL.

For example, a TL with characteristic impedance 02Z will be

delineated as

2

Hence, the Wilkinson power divider shown in the first figure

above and with matched terminations can be drawn as

0,Qz

0,Qz

Even-Odd Mode Analysis of the

Wilkinson Power Divider

In the even-odd mode analysis for the S parameters, we will first

excite this network symmetrically at the two output ports,

followed by an anti-symmetrical excitation.

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Symmetric excitation (even mode):

0,Qz

0,Qz

Notice that 0I= because we have symmetric excitation. Hence,

2 3V V= and we can bisect this circuit as shown to simplify theanalysis (Fig. 7.10a):

0,Qz

/4

r/22

1

eV

2

eV

o.c.

o.c.

in

ez

x =- /4

x =0

1

22gV V=

1

We can recognize this circuit as a QWT. Consequently,

2

0,

in2

Qez

z = (7.33),(1)

or 0, in2e

Qz z= (2)

We want the output ports to be matched. Therefore,in 1e

z = 0, 2Qz = (3)

Since in 1e

z = , then by voltage division at the output port

in2 2 2

in

1

1 2

ee

g ge

zV V V V

z= = =

+(4)

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Since the circuit is fed anti-symmetrically, 3 2V V= and the

voltage = 0 at points A and B. Hence, to simplify the analysis,

we can bisect the circuit with grounds as shown (Fig 7.10b):

2

1

oV

2

oV

in

oz

2V

For in

oz , notice that the load is a short circuit and the TL is 4

long (1/2 rotation around the Smith chart). This means inoz = .Therefore, to match port 2 (and 3) for odd mode excitation,

select

12

r= 2r= [/] (9)

Further, becausein

oz = , then with 2r= and port 2 matched:

( )2

9

22

2 1

o rV V V

r= =

+(10)

Even and odd solutions are eigenvectors. Any solution can be

determined by summing appropriately weighted eigenvectors.

With this information, well be able to deduce most of the S

parameters. But first, lets determine in,1z so we can compute

11S . Terminating ports 2 and 3 gives the circuit in Fig. 7.11a:

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Lecture 26: Quadrature (90) Hybrid.

Back in Lecture 23, we began our discussion of dividers and

couplers by considering important general properties of three-

and four-port networks. This was followed by an analysis of

three types of three-port networks in Lectures 24 and 25.

We will now move on to (reciprocal)directional couplers, which

are four-port networks. As in the text, we will consider these

specific types of directional couplers:1.Quadrature (90) Hybrid,2.180 Hybrid,3.Coupled Line, and4.Lange Coupler.

We will begin with the quadrature (90) hybrid. Fig 7.21 shows

this coupler implemented with microstrip as a 1:1 power divider:

Because of symmetry, we can simplify the analysis of this

circuit considerably using even-odd mode analysis. This process

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is similar to what we did in the last lecture with the Wilkinson

power divider.

Even-Odd Mode Analysis of the

Quadrature Hybrid

The normalized (wrt 0Z ) TL circuit is shown in Fig 7.22, minus

the return lines:

A symmetric (even mode) excitation of this circuit is shown in

Fig. 7.23a:

1

eA =

4

eA =

1

1

/ 8

and an anti-symmetric (odd mode) excitation is shown in Fig.7.23b:

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1

1

1

1

1

oA =

4

oA =

/8

Observe that the circuit and its boundary conditions remain the

same in both the even and odd mode configurations. It is only

the excitation that changes. Because of this and the circuit being

linear, the total solution is simply the sum of the even and odd

mode solutions.

Each solution (even and odd) is simpler to determine than the

complete circuit, which is why we employ this technique.

Even mode. Because the voltages and currents must be thesame above and below the line of symmetry (LOS) in Fig7.23a, then 0I = at the LOS open circuit loads at the ends

of/8 stubs, as shown.

Referring to the definition of iB ( 1, ,4i = ) in Fig 7.22, we

can write from Fig 7.23a that for the even mode excitation:

1 1

e e

eB A= , 2 1e e

e B T A= (1a)

3 2 1

e e e

e B B T A= = , 4 1 1

e e e

e B B A= = (1b)

where 1 1 2eA = , and e and eT are the reflection and

transmission coefficients for the even mode configuration.

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Odd mode. Because the voltages and currents must haveopposite values above and below the LOS in Fig 7.23b, then

0V = along the LOS short circuit loads at the ends of/8

stubs, as shown.

Then, 1 1o o

oB A= , 2 1o o

o B T A= (2a)

3 2 1

o o o

o B B T A= = , 4 1 1o o o

o B B A= = (2b)

where 1 1 2oA = and o and oT are reflection and transmission

coefficients for the odd mode configuration.

Total solution. The total solution is the sum of the voltages inboth circuits. From this fact, we can deduce that the total

iB

coefficients will be the sum of (1) and (2):

1 1 1

1 1

2 2

e o

e o B B B= + = + (7.62a),(3)

2 2 2

1 1

2 2

e o

e o B B B T T = + = + (7.62b),(4)

3 3 3

1 1

2 2

e o

e o B B B T T = + = (7.62c),(5)

4 4 4

1 1

2 2

e o

e o B B B= + = (7.62d),(6)

Likewise, the incident wave coefficients are

1 1 1 1 1 12 2

e o A A A= + = + =

4 4 4

1 10

2 2

e o A A A= + = =

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These match the assumed excitation in the original circuit on

p. 2.

To finish the calculation of the S parameters for the quadrature

hybrid, we need to determine the reflection and transmission

coefficients for the even- and odd-mode configurations.

Your text shows that the solutions for e and eT are

0e = and ( )1

1

2

eT j

= + (7.64),(7),(8)

Here well derive solutions foro and oT .

From Fig 7.23b:

oT

o

1 21 1

1 2

1 1

We have three cascaded elements, so well use ABCD

parameters to solve for the overall S parameters of this circuit.

Elements 1 and 3. These are short circuit stubs of length 8 ,which appear as the shunt impedance

in 0 tan Z jZ l= where 28 4

l

= =

Therefore, in

0

Zj

Z= , or NY j=

From the inside flap of your text:

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Lecture 27: The 180 Hybrid.

The second reciprocal directional coupler we will discuss is the

180 hybrid. As the name implies, the outputs from such a

device can be 180 out of phase.

There are two primary objectives for this lecture. The first is to

show that the S matrix of the 180 hybrid is

[ ]

0 1 1 0

1 0 0 11 0 0 12

0 1 1 0

jS

=

(7.101),(1)

with reference to the port definitions in Fig. 7.41:

1

4

2

3

()

()180

Hybrid

The second primary objective is to illustrate the three common

ways to operate this device. These are:

1.In-phase power splitter:1

4

2

3

Input

Isolation

Through

Coupled

180

Hybrid

With input at port 1 and using column 1 of [S], we can deduce

that port 1 is matched, the outputs are ports 2 and 3 (which are

in phase) and port 4 is the isolation port.

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2.Out-of-phase power splitter:1

4

2

3Input

Isolation Through

Coupled

180

Hybrid

With input at port 4 and using column 4 of [S], we can deduce

that port 4 is matched, the outputs are ports 2 and 3 (which are

completely out of phase) and port 1 is the isolation port.

3.Power combiner:1

4

2

3Difference ()

Sum () Input A

Input B

180

Hybrid

With inputs at ports 2 and 3 and using columns 2 and 3 of [S],

we can deduce that both ports 2 and 3 are matched, port 1 will

provide the sum of the two input signals and port 4 willprovide the difference.

Because of this, ports 1 and 4 are sometimes called the sum

and difference ports, respectively.

There are different ways to physically implement a 180 hybrid,

as shown in Fig. 7.43. Well focus on the ring hybrid and

specifically consider the first two applications described above.

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Ring Hybrid

The ring hybrid (aka the rat race) is shown in Fig. 7.43a:

Well analyze this structure using the same even-odd mode

approach we applied to the Wilkinson power divider and the

branch line coupler in the previous two lectures. In the present

case, the physicalsymmetry plane bisects ports 1 and 2 from 3

and 4 in the figure above.

1. In-phase power splitter. Assume a unit amplitude voltage

wave incident on port 1:

1B

1Port 1

Port 3

Port 4

Port 2

3B

4B

2B

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As in Lecture 26, proper symmetric and anti-symmetric

excitations of this device are required to produce the even

and odd mode problems, as shown in Fig. 7.44:

1

1

1 1

1 1

1

1

As we derived in Lecture 26,

1

1 1

2 2e o

B = + (7.102a),(2)

2

1 1

2 2e o

B T T = + (7.102b),(3)

3

1 1

2 2e o

B = (7.102c),(4)

4

1 1

2 2e o

B T T = (7.102d),(5)

Each of the even and odd solutions fori

B ( 1, ,4i = ) can be

found by cascading ABCD matrices, then converting to S

parameters. Since the ports are terminated by matched loads,

we can directly determineeo

andeo

T from these S parameters.

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Lecture 28: Coupled Line andLange Directional Couplers.

These are the final two directional couplers we will consider.

They are closely related and based on two TLs that interact with

each other, but are not physically connected.

Coupled Line Directional Coupler

When two TLs are brought near each other, as shown in the

figure below (Fig. 7.26), it is possible for power to be coupled

from one TL to the other.

This can be a serious problem on PCBs where lands are close

together and carry signals changing rapidly with time. EMC

engineers face this situation in high speed digital circuits and in

multiconductor TLs.

For coupled line directional couplers, this coupling between TLs

is a useful phenomenon and is the physical principle upon which

the couplers are based.

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Consider the geometry shown in Fig 7.27:

r

+++++

+ ++++

++

- -- - -- - -- - --

1 2

When voltages are applied, charge distributions will be induced

on all of the conductors. The voltages and total charges are

related to each other through capacitance coefficients Cij:

V1

V2

1 2C12

C11 C22

By definition (Q CV= ): 1 11 1 12 2Q C V C V = +

2 21 1 22 2Q C V C V = +

where 11C = capacitance of conductor 1 with conductor 2 present but

grounded.

22C = capacitance of conductor 2 with conductor 1 present butgrounded.

12C = mutual capacitance between conductors 1 and 2. (If theconstruction materials are reciprocal, then 21 12C C

=

.)

By computing only these capacitances and the quasi-TEM mode

wave speed, well be able to analyze these coupled line

problems. Why? Assuming TEM modes, then

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1

p

L LC Z

C C v C = = = (1)

Notice that L doesnt appear here. Hence we

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