EE481 - Microwave Engineering

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

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

EE 481 Microwave Engineering Fall 2008

Instructor: Dr. Keith W. Whites Office: 317 Electrical Engineering/Physics (EEP) Building Email: [email protected] Web: http://whites.sdsmt.edu Office hours: MWF 3:00-4:00 PM To contact the instructor, please use e-mail rather than the telephone. All e-mail will be answered. The instructor will be available for assistance during the hours listed above, as well as other times when the office door is open. Catalog Description: (3-1) 4 credits. Prerequisite: 382 completed or concurrent. Presentation of basic 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 Friday from 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 Lecture

Notes, 2008, are available from the course web page. Grading: 30 % – Two exams 30 % – Laboratory 20 % – Homework 20 % – Final exam Homework Policy: One homework set will generally be assigned each week. These homework assignments are to be turned in at the beginning of the class period on the due date. Late homework 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 four to five labs for the course. These will involve the design, construction, and measurement of passive and active microwave circuits. The labs will also require the simulation of your circuits using Advanced Design System (ADS) from Agilent Technologies. Measurements will be performed in the Laboratory for Applied Electromagnetics and Communications (LAEC) located in room 230 EEP using Agilent 8753ES vector network analyzers. Laboratory work will be performed in pairs of students and open lab hours will be posted. Late lab reports will be assessed a 10% per calendar day reduction in points. Exam Policy: The exams will be closed book and closed notes with no formula sheets. Using or referring 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

EE 481 Microwave Engineering

Lecture Notes

Keith W. Whites Fall 2008

Laboratory for Applied Electromagnetics and Communications Department of Electrical and Computer Engineering

South Dakota School of Mines and Technology

© 2008 Keith W. Whites

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

RF

Reg

ion HF 3 MHz-30 MHz

VHF 30 MHz-300 MHz UHF 300 MHz-1 GHz

Mic

row

ave

Reg

ion

(λ =

30

cm to

8 m

m)

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

Mill

imet

er

Wav

e R

egio

n 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?


Advantages of microwave circuits:

1. The gain of certain antennas increases (with reference to an 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 don’t 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 properties of 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 microwave frequencies 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?



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:

ε

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.


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 or approximately: • 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.


E

E

zΔ

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+

-


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:

o Resistance, 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:

o Conductance, G, due to losses in the dielectric between the conductors,

o Capacitance, 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 , we will consider only a small section zΔ of the TL. This zΔ is so small that the electrical effects are occurring instantaneously and we can simply use circuit theory to draw the relationships between the conduction and displacement currents. This equivalent circuit is shown below:


( ),v z t

( ),i z t L zΔ ( ),i z z t+ Δ

( ),v z z t+ Δ+

-

+

-

zΔ

C zΔ G zΔ

R zΔ

Fig. 2 The variables R, L, C, and G are distributed (or per-unit length, PUL) parameters with units of Ω/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 zΔ

C zΔ

zΔ

L zΔ

C zΔ

L zΔ

C zΔ

L zΔ

C zΔ RL

Rs

vs(t)

zΔ zΔ zΔ

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.


Transmission Line Equations

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

( ) ( ) ( ),, ,

i z tv z t L z v z z t

t∂

= Δ + + Δ∂

(2.1a),(1)

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

( ) ( ) ( ),, ,

v z z ti z t C z i z z t

t∂ + Δ

= Δ + + Δ∂

(2.1b),(2)

Then: 1. Divide (1) by zΔ :

( ) ( ) ( ), , ,v z z t v z t i z tL

z t+ Δ − ∂

= −Δ ∂

(3)

In the limit as 0zΔ → , the term on the LHS in (3) is the forward difference definition of derivative. Hence,

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

Lz t

∂ ∂= −

∂ ∂ (2.2a),(4)

2. Divide (2) by zΔ :

( ) ( ) ( )0 ,, , v z z ti z z t i z tC

z t∂ + Δ+ Δ −

= −Δ ∂

(5)

Again, in the limit as 0zΔ → the term on the LHS is the forward difference definition of derivative. Hence,



Lecture 3: Phasor Wave Solutions to the Telegrapher Equations. Termination of TLs

We will continue our TL review by considering the steady state response of TLs to sinusoidal excitation. Consider the following TL in the sinusoidal steady state:

Zs

z

Z0, vp

l

ZL

+

-Vs

We previously derived the wave equations for the voltage and current as

( ) ( )2 2

2 2 2

, ,1

p

v z t v z tz v t

∂ ∂=

∂ ∂ (1)

and ( ) ( )2 2

2 2 2

, ,1

p

i z t i z tz v t

∂ ∂=

∂ ∂ (2)

For sinusoidal steady state, we will employ the phasor representation of the voltage and current as ( ) ( ), j tv z t e V z e ω⎡ ⎤= ℜ ⎣ ⎦ (3)

( ) ( ), j ti z t e I z e ω⎡ ⎤= ℜ ⎣ ⎦ (4) where ( )V z and ( )I z are spatial phasor functions.


Substituting (3) into (1) gives

( ) ( ) ( ) ( )2 2

22 2 2

1

p p

d V zj V z V z

dz v vωω= = − (5)

We define LCβ ω= [rad/m] (2.12a),(6) as the phase constant for reasons that will be apparent shortly. (L and C are the usual TL per-unit-length parameters.)

From (6) 2

2 22p

LCvωβ ω= =

Substituting this into (5) gives

( ) ( )

22

2 0d V z

V zdz

β+ = (7)

Similarly, from (4) and (2) we can derive

( ) ( )

22

2 0d I z

I zdz

β+ = (8)

Equations (7) and (8) are the wave equations for V and I in the frequency domain (i.e., the phasor domain). The solutions to these two second-order ordinary differential equations are ( ) j z j z

o oV z V e V eβ β+ − − += + (2.14a),(9) and ( ) j z j z

o oI z I e I eβ β+ − − += + (10) , , ,o o o oV V I I+ − + − are complex constants.


We can confirm the correctness of these two solutions by direct substitution into (7) and (8). For example, substituting j z

oV e β+ − from (9) into (7) gives ( ) ( )2 2 ? 0j z

oV j e V zββ β+ −− + =

or 2 2 ? 0j z j zo oV e V eβ ββ β+ − + −− + =

which is indeed true. Therefore, j zoV e β+ − in (9) is a valid solution

to (7). The constants oI + and oI − in (10) can be expressed in terms of oV + and oV − . In particular, it can be shown that

0

oo

VIZ

++ = (11)

and 0

oo

VIZ

−− = − (12)

If we substitute (11) and (12) into (10) we find that

( )0 0

j z j zo oV VI z e eZ Z

β β+ −

− += − (2.14b),(13)

and ( ) j z j zo oV z V e V eβ β+ − − += + (2.14a),(14)

Both of these equations should be committed to memory. They are the general form of phasor voltages and currents on transmission lines.


The first terms in (13) and (14) are the phasor representation of waves propagating in the +z direction along the TL. The second terms in both equations represent waves propagating in the -z direction.

Discussion • As stated above, the first terms in (13) and (14) are the phasor

representation of waves traveling in the +z direction. To see this, convert the first term in (14) to the time domain:

( ) ( )

( )

,

cos cos

cos

j t zj z j t jo o

o o

op

v z t e V e e e V e e

V t z V t z

zV tv

ω ββ ω φ

βω β φ ω φω

ω φ

+ −+ − +

+ + + +

+ +

⎡ ⎤⎡ ⎤= ℜ =ℜ⎣ ⎦ ⎣ ⎦⎡ ⎤⎛ ⎞= − + = − +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

⎡ ⎤⎛ ⎞= − +⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

We can clearly see in this last result that we have a function of time with argument / pt z v− . From our previous discussions with TLs we recognize that this is a wave that is propagating in the +z direction with speed vp.

• Similarly, we can show that j z

oV e β− + (and j zoI e β− + ) are waves

propagating in the -z direction.



Lecture 4: TL Input Impedance, Time Average Power, Return and Insertion Losses.

VSWR. Example N4.1: Determine an expression for the voltage at the input to the TL assuming Rs = Z0:

Rs

z

Z0, β

z = 0

+

-Vs

+

-Vg

l

Zin To calculate the input voltage gV , we’ll first determine the effective impedance seen at the TL input terminals seen looking towards the load at z = 0. This is called the input impedance

inZ . Forming the ratio of (19) and (20) from the previous lecture gives

( )( )

( )

( )( )in 0

0

2 coscot

2 sin

o

o

V l V lZ jZ l

j VI l lZ

ββ

β

+

+

− −= = = −

− − − [Ω]

In other words, the input impedance is purely reactive in inZ jX= where ( )in 0 cotX Z lβ= − (2.46c) A plot of this reactance is shown in Fig. 2.8c of the text.


An equivalent circuit can now be constructed at the input to the TL by using sR and inZ as

Using voltage division,

( )( )

0in

in 0 0 0

cotcotg s s

jZ lZV V VZ Z jZ l Z

ββ

−= =

+ − +

This circuit voltage gV is also the voltage on the TL at z l= − . That is, from (19) in the previous lecture ( ) ( )2 cosoV z l V lβ+= − = − Since ( )gV V z l= = − , we can equate these two voltages giving

( ) ( )( )

0

0 0

cot2 cos

coto sjZ l

V l VjZ l Z

ββ

β+ −

=− +

More often than not, expressions of this type are used to determine oV + in terms of sV and sR . We’ll see more on this topic in Lecture 5.

Input Impedance of a Transmission Line In problems like the one in the last example, it is helpful to have an analytical expression for the input impedance of an arbitrarily terminated TL.


As we saw in the last lecture, the voltage and current everywhere on a homogeneous TL are ( ) j z j z

o oV z V e V eβ β+ − − += + (2.34a),(1)

and ( )0 0

j z j zo oV VI z e eZ Z

β β+ −

− += − (2.34b),(2)

We can readily construct an input impedance expression for a TL of length l by dividing (1) and (2) for some arbitrary load reflection coefficient ΓL at z = 0:

( ) ( )j l j l j l j loo o L

o

VV l V e e V e eV

β β β β−

+ + − + + −+

⎛ ⎞− = + = + Γ⎜ ⎟

⎝ ⎠ (3)

( ) ( )0 0

j l j l j l j lo o oL

o

V V VI l e e e eZ V Z

β β β β+ − +

+ − + −+

⎛ ⎞− = − = −Γ⎜ ⎟

⎝ ⎠ (4)

such that

( )( )

( )( )

2

in 0 2

0

11

j l j l j lo L L

j lj l j lo L

L

V e eV l eZ ZVI l ee eZ

β β β

ββ β

+ + − −

+ −+ −

+ Γ− + Γ≡ = =

− −Γ−Γ (2.43)

Substituting for LΓ and simplifying gives

( )( )

0in 0

0

tantan

L

L

Z jZ lZ Z

Z jZ lββ

+=

+ [Ω] (2.44),(5)

This is the input impedance for a lossless TL of length l and characteristic impedance Z0 with an arbitrary load ZL. Three special cases are: 1. With an open circuit load ( LZ = ∞), (5) yields


( )in 0 cotZ jZ lβ= − [Ω] (2.46c),(6) as we derived in the last lecture.

2. With a short circuit load ( 0LZ = ), (5) yields ( )in 0 tanZ jZ lβ= [Ω] (2.45c),(7)

A plot of this input reactance is shown in Fig. 2.6c.

3. With the resistive load 0LZ Z= , (5) yields in 0Z Z= [Ω]

The input impedance is Z0 regardless of the length of the TL.

All of these last three expressions should be committed to memory. You will use them often in microwave circuits. Note that both input impedances (6) and (7) are purely reactive, which is expected since neither type can dissipate energy, assuming lossless TLs.

Time Average Power Flow on TLs A hugely important part of microwave engineering is delivering signal power to a load. Examples include efficiently delivering power from a source to an antenna, or maximizing the power delivered from a filter to an amplifier.


Often, the “power” we are ultimately concerned with is the time average power Pav, expressed as

( ) ( ) ( )av12

P z e V z I z ∗⎡ ⎤= ℜ ⎣ ⎦ (8)

This expression is similar to that used in circuit analysis. Substituting V(z) and I(z) from (1) and (2) into (8) gives

( )2

2* 2 2av

0

1 12

o j l j lL L L

VP z e e e

Zβ β

+− +⎡ ⎤= ℜ −Γ + Γ − Γ⎣ ⎦ (9)

Notice that the second and third terms are conjugates so that ( )2 2 22j l j l j l

L L Le e j m eβ β β∗+ + +⎡ ⎤− Γ + Γ = ℑ Γ⎣ ⎦

The real part of this sum is zero. Consequently, (9) simplifies to

( )2

2av

0

1 12

oL

VP

Z

+

= − Γ [W] (2.37),(10)

Since this power is not a function of z (true for a lossless and homogeneous TL), a z-dependence is no longer indicated for Pav. It is important to reiterate that we’re assuming a lossless TL throughout this analysis. These results are not valid for lossy TLs. Equation (10) is very illuminating. It shows that the total time average power delivered to a load is equal to the incident time


average power ( )2

02oV Z+ minus the reflected time average power ( )2 2

02oV Z+ Γ . The relative reflected time average power from an arbitrary load on a lossless TL is the ratio of the two terms in (10) = 2

LΓ . From (10) we see that if the load is entirely reactive so that

1LΓ = , then av 0P = and no time average power is delivered to the load, as expected. For all other passive loads, av 0P > . The time average power that is not delivered to the load can be considered a “loss” since the signal from the generator was intended to be completely transported – not returned to the generator. This return loss (RL) is defined as ( ) ( )2

10 10RL 10log 20logL L= − Γ = − Γ dB (2.38),(11)

The two extremes for return loss with a passive load are: 1. A matched load where 0LΓ = and RL = ∞ (no reflected

power), and 2. A reactive load where 1LΓ = and RL 0= (all power

reflected).



Lecture 5: Generator and Load Mismatches on TLs.

Up to this point, we have focused primarily on terminated transmission lines that lacked a specific excitation. That is, the TL was semi-infinite and terminated by a load impedance. In this lecture, we’ll complete our review of TLs by adding a voltage source together with an arbitrary load (Fig. 2.19):

This TL model is very useful and applicable to a wide range of practical engineering situations. Quantities of interest in such problems include the input impedance (for matching purposes) and signal power delivered to the load. We will first consider the computation of the latter quantity assuming the TL is lossless. Proceeding, the voltage on the TL is expressed by


( ) j z j zo oV z V e V eβ β+ − −= +

or ( ) ( )j z j zo LV z V e eβ β+ −= + Γ (2.69),(1)

where 0

0

LL

L

Z ZZ Z

−Γ =

+ (2.68),(2)

We’ll assume that the physical properties of the TL, the source and the load are known. This leaves the complex constant oV + as the only unknown quantity in (1). Generally speaking, we compute oV + by applying the boundary condition at the TL input. (Recall that we have already applied boundary conditions at the load.) This is accomplished by applying voltage division at the input:

inin

ing

g

ZV VZ Z

=+

(3)

Observe that inV is an electrical circuit quantity. However, at the input to the TL, voltage must be continuous from the generator to the TL. This implies that inV must also equal ( )V z l= − on the TL. Proceeding, then from (1) at the input ( ) ( )j l j l

o LV z l V e eβ β+ + −= − = + Γ (4)

Equating (3) and (4) to enforce the boundary condition at the TL input we find


( ) in

in

j l j lo L g

g

ZV e e VZ Z

β β+ + −+ Γ =+

or ( ) 1in

in

j l j lo g L

g

ZV V e eZ Z

β β −+ −= + Γ+

[V] (2.70),(5)

Maximum Power Because the TL is lossless, the time average power avP delivered to the input of the TL must equal the time average power delivered to the load. Therefore,

*

* inav in in in *

in

1 1[ ]2 2

VP e V I e VZ

⎡ ⎤= ℜ = ℜ ⎢ ⎥

⎣ ⎦

or 2

inav *

in

12

VP e

Z⎡ ⎤

= ℜ ⎢ ⎥⎣ ⎦

[W] (2.74),(6)

Now, substituting (3) into (6) gives

2 2

inav *

in in

12g

g

V ZP eZ Z Z

⎡ ⎤= ℜ ⎢ ⎥+ ⎣ ⎦

(2.74),(7)

If we define in in inZ R jX= + and g g gZ R jX= + , (7) becomes

( ) ( )

2

inav 2 2

in in2g

g g

V RPR R X X

=+ + +

(2.75),(8)


Employing this last result, we’ll consider three special cases for avP in an effort to maximize this quantity. We will assume that gZ is both nonzero and fixed:

(1.) Load is matched to the TL: 0LZ Z= .

From (2), 0LΓ = in this situation, which also implies that

in 0Z Z= . [This should be intuitive. If not, see (2.43).] Consequently, from (8) with in 0R Z= and in 0X = :

( ) ( )

2

0av,1 2 2

02g

g g

V ZPZ R X

=+ +

(2.76),(9)

(2.) Generator is matched to an arbitrarily loaded TL:

in gZ Z= and 0LΓ ≠ .

Specific values for lβ , 0Z , and LZ would need to be chosen so that in gZ Z= . Then from (8) and with in gR R= and in gX X= :

( ) ( )

2

av,2 2 22g g

g g g g

V RP

R R X X=

+ + +

or 2

av,2 2 28g g

g g

V RP

R X=

+ (2.78),(10)



Lecture 6: The Smith Chart The Smith chart began its existence as a very useful graphical calculator for the analysis and design of TLs. It was developed by Phillip H. Smith in the 1930s. The Smith chart remains a useful tool today to visualize the results of TL analysis, oftentimes combined with computer analysis and visualization as an aid in design. The development of the Smith chart is based on the normalized TL impedance ( )z z defined as

( ) ( ) ( )( )0

11

Z z zz z

Z z+ Γ

≡ =−Γ

(1)

where ( ) ( ) ( )/Z z V z I z= is the total TL impedance at z and ( ) 2j z

Lz e β−Γ = Γ (2) is the generalized reflection coefficient at z. The real and imaginary parts of the generalized reflection coefficient ( )zΓ will be defined as ( ) ( ) ( )r iz z j zΓ ≡ Γ + Γ . Substituting this definition into (1) gives

( ) ( )( )

11

r i

r i

jz z

j+ Γ + Γ

=− Γ + Γ

(3)

Now, we will define ( )z z r jx≡ + and separate (3) into its real and imaginary parts


( ) ( )

( )( )( )

( )

*

*

2 2

2 2

1 11 1

1 21 2

r i r i

r i r i

i r i

r r i

j jz z r jx

j j

j

+ Γ + Γ − Γ + Γ≡ + = ⋅

− Γ + Γ − Γ + Γ

+ Γ − Γ + Γ=

− Γ + Γ + Γ

Equating the real and imaginary parts of this last equation gives

( )

( )

2 2

2 2

1

1r i

r i

r− Γ + Γ

=Γ − + Γ

and ( )2 2

21

i

r i

x Γ=

Γ − + Γ (2.55)

Rearranging both of these leads us to the final two equations

2 2

2 11 1r i

rr r

⎛ ⎞ ⎛ ⎞Γ − + Γ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ (2.56a),(4)

and ( )2 2

2 1 11r i x x⎛ ⎞ ⎛ ⎞Γ − + Γ − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(2.56b),(5)

We will use (4) and (5) to construct the Smith chart. Definition: The Smith chart is a plot of normalized TL resistance and reactance functions drawn in the complex, generalized reflection coefficient [ ( )zΓ ] plane. To understand this, first notice that in the Γr-Γi plane:

1. Equation (4) has only r as a parameter and (5) has only x as a parameter.

2. Both (4) and (5) are families of circles. Consequently, we can plot (4) and (5) in the Γr-Γi plane while keeping either r or x constant, as appropriate.


Plot (4) in the Γr-Γi plane: • For 0r = : 2 2 21r iΓ + Γ =

• For 1r = : 2 2

21 12 2r i

⎛ ⎞ ⎛ ⎞Γ − + Γ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

• For 12

r = : 2 2

21 23 3r i

⎛ ⎞ ⎛ ⎞Γ − + Γ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Plot these curves in the Γr-Γi plane:

( )Rer zΓ = Γ⎡ ⎤⎣ ⎦

( )Imi zΓ = Γ⎡ ⎤⎣ ⎦

1

1

-1

-1

r=0

r=1/2r=1

1/2

Complex Γ(z)plane

Plot (5) in the Γr-Γi plane: • For 1x = : ( ) ( )22 21 1 1r iΓ − + Γ − = • For 1x = − : ( ) ( ) ( )22 21 1 1r iΓ − + Γ + = −

• For 100x = : ( )2 2

2 1 11100 100r i

⎛ ⎞ ⎛ ⎞Γ − + Γ − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

• For 1100

x = : ( ) ( )22 21 100 100r iΓ − + Γ − =


Plot these curves in the Γr-Γi plane:

( )Rer zΓ = Γ⎡ ⎤⎣ ⎦

( )Imi zΓ = Γ⎡ ⎤⎣ ⎦

1

1

-1

-1

r=0

x=-0.01

x=100

x=0.01

Complex Γ(z)plane

x=-100

x=1

x=-1

Combining both of these curves (or “mappings”), as shown on the next page, gives what is called the Smith chart. As quoted from the text (p. 65):

“The real utility of the Smith chart, however, lies in the fact that it can be used to convert from reflection coefficients to normalized impedances (or admittances), and vice versa, using the impedance (or admittance) circles printed on the chart.”

Additionally, it is very easy to compute the generalized reflection coefficient and normalized impedance anywhere on a homogeneous section of TL.


Notice that the Γr and Γi axes are missing from the “combined” plot. This is also the case for the Smith chart.


Important Features of the Smith Chart

1. By definition ( ) ( )( )

( )( )

1 11 1

z z r jxz

z z r jx− + −

Γ = =+ + +

. Therefore

( ) ( )( )

2 2

2 2

11 11 1 1

r xr jx r jxzr jx r jx r x

− ++ − − +Γ = ⋅ =

+ + − + + +

From this result, we can show that if 0r ≥ then ( ) 1zΓ ≤ . This condition is met for passive networks (i.e., no amplifiers) and lossless TLs (real 0Z ). Consequently, the standard Smith chart only shows the inside of the unit circle in the Γr-Γi plane. That is, ( ) 1zΓ ≤ which is bounded by the 0r = circle described by 2 2 1r iΓ + Γ = .

2. If ( )z z is purely real (i.e., 0x = ), then since

( )2 2

21

i

r i

x Γ=

Γ − + Γ

we deduce that 0iΓ = (except possibly at 1rΓ = ). Consequently, purely real ( )z z values are mapped to ( )zΓ values on the ( )r e zΓ =ℜ Γ⎡ ⎤⎣ ⎦ axis.

3. If ( )z z is purely imaginary (i.e., 0r = ) then from (4) 2 2 21r iΓ + Γ =

which is the unit circle in the Γr-Γi plane.


Consequently, purely imaginary ( )z z values are mapped to ( )zΓ values on the unit circle in the Γr-Γi plane.

Example N6.1: Using the Smith chart, determine the voltage reflection coefficient at the load and the TL input impedance.



Lecture 7: Transmission Line Matching Using Lumped L Networks

Impedance matching (or simply “matching”) one portion of a circuit to another is an immensely important part of microwave engineering. Additional circuitry between the two parts of the original circuit may be needed to achieve this matching. Why is impedance matching so important? Because:

1. Maximum power is delivered to a load when the TL is matched at both the load and source ends. This configuration satisfies the conjugate match condition.

2. With a properly matched TL, more signal power is transferred to the load, which increases the sensitivity of the device.

3. Some equipment (such as certain amplifiers) can be damaged when too much power is reflected back to the source.

Factors that influence the choice of a matching network include:

1. The desire for a simple design, if possible. 2. Providing an impedance match at a single frequency is

often not difficult. Conversely, achieving wide bandwidth matching is usually difficult.


3. Even though the load may change, the matching network may need to perform satisfactorily in spite of this, or be adjustable.

We will discuss three methods for impedance matching in this course:

1. L networks, 2. Single stub tuners (using shunt stubs), 3. Quarter wave transformers.

You’ve most likely seen all three of these before in other courses, or in engineering practice.

Matching Using L Networks

Consider the case of an arbitrary load that terminates a TL:

Z0, ZLL

To match the load to the TL, we require 0LΓ = . However, if

0LZ Z≠ additional circuitry must be placed between ZL and Z0 to bring the VSWR = 1, or least approximately so:


For 0LΓ = , this implies in 0Z Z= . In other words, in 0[ ]R e Z=ℜ and in 0X = , if the TL is lossless. Note that we need at least two degrees of freedom in the matching network in order to transform LZ at the load to 0Z seen at the input to the matching network. This describes impedance matching in general. For an L network specifically, the matching network is either (Fig. 5.2): 0LR Z> : 0LR Z< :

Z0,

Zin

ZL

jX

(jB)-1

where L L LZ R jX= + . This network topology gets its name from the fact that the series and shunt elements of the matching network form an “L” shape. There are eight possible combinations of inductors and capacitors in the L network:


0LR Z> : 0LR Z< :

ZL ZL

ZL ZL

ZL ZL

ZL ZL

Notice that this type of matching network is lossless; or at least the loss can potentially be made extremely small with proper component choices. As in the text, we’ll solve this problem two ways: first analytically, then using the Smith chart.

Analytical Solution for L Network Matching Assume 0LR Z> . Using Fig. 5.2(a):

1

in1

L L

Z jX jBR jX

−⎛ ⎞

= + +⎜ ⎟+⎝ ⎠ (5.1),(1)

Through the proper choice of X and B we wish to force in 0Z Z= . Solving (1) for the B and X that produce this

outcome (by equating real and imaginary parts, as shown in the text) we find that


( ) 2

0 02 2

L L L L L

L L

X R Z R R Z XB

R X± − +

=+

(5.3a),(2)

0 01 L

L L

X Z ZXB R BR

= + − (5.3b),(3)

Comments: 1. Since 0LR Z> , the argument is positive in the second

square root of (2). (B must be a real number.) 2. Note that there are two possible solutions for B in (2). 3. X in (3) also has two possible solutions, depending on

which B from (2) is used. Assume 0LR Z< . Using Fig. 5.2(b) with in 0Z Z= , we obtain

1

in 01

L

Z Z jBZ jX

−⎛ ⎞

= = +⎜ ⎟+⎝ ⎠ (5.4),(4)

Solving this equation by equating real and imaginary parts as shown in the text gives

( )0L L LX R Z R X= ± − − (5.6a),(5)

0

0

1 L

L

Z RBZ R

−= ± (5.6b),(6)

Comments: 1. Since 0LR Z< , the argument is positive for the square root

in (5). 2. There are two solutions for both X and B. Use the top

signs in both (5) and (6) for one solution and the bottom signs for the other.


Smith Chart Solution for L Network Matching

L-network matching can also be computed graphically using the Smith chart. This approach is less accurate than the analytical approach. However, more insight into the matching process is often obtained using the Smith chart. For example, the contribution each element makes to the matching is quite clear. The process of using the Smith chart to design the matching network is probably best illustrated by example. Example 5.1 in the text illustrates the design of an L network when 0LR Z> (Fig. 5.2a). Here, we’ll give an example when 0LR Z< . Example N7.1 Design an L network to match the load 25 30j+ Ω to a TL with 0 50Z = Ω at the frequency 1f = GHz. Since 0LR Z< , we’ll use the circuit topology in Fig. 5.2b:


We’ll solve this problem two ways: first with the Smith chart and then analytically. Steps for a Smith chart solution:

1. 0

1 32 5

LL

Zz jZ

= = + p.u.Ω. Mark this point on the chart.

2. The overall concept behind this type of L-network

matching is to add a reactance x to Lz such that the sum of admittances b and ( ) 1

Lz jx −+ yield in in01y j z= + = (the center of the Smith chart). In such a case, the TL sees a matched load.

So, in this case we need to add a normalized impedance 0.1jx j= − p.u.Ω in order to move to 1 jx+ on an admittance chart.

3. Convert this impedance to an admittance value by

reflecting through the origin to the diametrically opposed point on the constant VSWR circle.

4. Add the normalized susceptance 1.0b = p.u.S to reach the

center of the Smith chart. Here 0Γ = and in 1 0y j= + , which means the TL now sees a matched load.


10

20

30

40 50

60

70

80

90

100

110

120

130140

150

160

170

0

20

20

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0 0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.2

1.4

1.6

1.8

2.0

3.0

4.0

5.0

10

0.2

0.4

0.6

0.8

1.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

3.0

50

6.0

7.0

8.0

9.0

50

10

5.0

4.0

20

1.00.90.

8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.2

0.4

0.6

0.8

1.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

3.0

6.0

7.08.09.0

50

10

5.0

4.0

zLStart

End

1+jx admittancecircle (on Z chart)

Constant VSWRcircle

1+jx impedancecircle (on Z chart)

K. W. Whites

Un-normalizing, we find that 0 0.1 50 5.0jX jx Z j j= ⋅ = − ⋅ = − Ω

011.0 0.0250

jB jb Y j j= ⋅ = ⋅ = S


What are the L and C values of these elements? We can identify the type of element by the sign of the reactance or susceptance:

Inductor Capacitor

Impedance LZ j Lω= 1

CjZ

j C Cω ω−

= =

Admittance 1

LjY

j L Lω ω−

= = CY j Cω=

Since 0X < , we identify this as a capacitor. Therefore,

2

5.0jjX jCω−

= = − Ω

For operation at 1 GHz, we need

21 31.8

5 2C

fπ= =

⋅ pF

Since 0B > , we also identify this as a capacitor. Therefore, 1 0.02jB j C jω= = S

For operation at 1 GHz, we need

10.02 3.182

Cfπ

= = pF

The final circuit is:

50 Ωf = 1 GHz

Zin

ZL

C2=31.8 pF

C1=3.18 pF

=25+j30 Ω



Lecture 8: Single-Stub Tuning The second matching network we’ll discuss is the single-stub tuner (SST). The single-stub tuner uses a shorted or open section of TL attached at some position along another TL:

This is an example of a parallel SST, which is the only type we’ll study. (A series SST is shown in Fig. 5.4b of the text.) The shunt-connected section is called the stub. Although not necessary, all sections of TL will be assumed to have the same

0Z and β . Why an open or shorted section? Because these are easy to fabricate, the length can easily be made adjustable and little to no power is dissipated in the stub. (An open stub is sometimes easier to fabricate than a short.)


We will study the SST from two perspectives. First, we will develop an analytical solution, followed by a Smith chart graphical solution. Referring to the figure above, the transformed load impedance at the stub position z = -d is

( ) ( )( )

0 00 0

0 0

tantan

L L

L L

Z jZ d Z jZ tZ z d Z ZZ jZ d Z jZ t

ββ

+ += − = =

+ + (5.7),(1)

where ( )tant dβ≡ . With a shunt connection, it is much simpler to work with admittances than impedances. So, we’ll define the transformed load admittance as 1/Y Z G jB= = + . The distance d is chosen so that ( )0 01/G Y Z= = . As shown in the text, this condition leads to the solutions

( )

1

1

1 tan , 021 tan , 0

2

t td

t t

πλ π

π

−

−

⎧ ≥⎪⎪= ⎨⎪ + <⎪⎩

(5.10),(2)

where

( ) ( )2 20 0

00

00

,

,2

L L L L

LL

LL

X R Z Z R XR Z

R ZtX R ZZ

⎧ ⎡ ⎤± − +⎪ ⎣ ⎦ ≠⎪ −= ⎨⎪− =⎪⎩

(5.9),(3)

and L L LZ R jX= + .


With this location of the stub, the transformed load admittance has a real part = Y0, which is almost a matched state. In general, however, this transformed YL will also have an imaginary part B. The length of the stub, ls, is chosen so that its input susceptance

sB B= − . Consequently, the parallel combination of the stub input susceptance and the transformed load admittance yield an input admittance in 0Y Y= , as seen from the source end of the TL. As shown in the text, this second condition provides the solutions

1 01 tan2

sl YBλ π

− ⎛ ⎞= ⎜ ⎟⎝ ⎠

short-circuit stub (5.11b),(4)

or

1

0

1 tan2

ol BYλ π

− ⎛ ⎞= − ⎜ ⎟

⎝ ⎠ open-circuit stub (5.11a),(5)

where B is the transformed load susceptance at z = -d. Lengths of TL that are integer multiples of λ/2 can be added or subtracted from (2), (4), and (5) without altering the tuning. Example N8.1: Match the load 35 47.5LZ j= − Ω to a TL with

0 50Z = Ω using a shunt, short-circuited single-stub tuner.


Single Stub Tuning Using the Smith Chart We will now solve the single stub tuner problem using the Smith chart. In terms of quantities normalized to the characteristic impedance or admittance, the geometry is



Lecture 9: Quarter-Wave-Transformer Matching

For a TL in the sinusoidal steady state with an arbitrary resistive load (Fig. 2.16)

the input impedance of the right-hand TL is given as

1 1in 1

1 1

tantan

L

L

R jZ lZ ZZ jR l

ββ

+=

+ (2.61),(1)

Now imagine that we have a special length 1 / 4l λ= of TL, as indicated in the figure above. At this frequency and physical length, the electrical length of the TL is

11

1

24 2

l π λ πβλ

= = rad (2)

Consequently, for a / 4λ -length TL, 1tan lβ →∞ . Using this result in (1) gives

2

1in

L

ZZR

= (2.62),(3)

This result is an interesting characteristic of TLs that are exactly λ/4 long. We can harness this characteristic to design a matching network using a λ/4-length section of TL.


Note that we can adjust 1Z in (3) so that in 0Z Z= . In particular, from (3) with in 0Z Z= we find 1 0 LZ Z R= (2.63),(4) In other words, a λ/4 section of TL with this particular characteristic impedance will present a perfect match ( 0Γ = ) to the feedline (the left-hand TL) in the figure above. This type of matching network is called a quarter-wave transformer (QWT). Through the impedance transforming properties of TLs, the QWT presents a matched impedance at its input by appropriately transforming the load impedance. This is accomplished only because we have used a very special characteristic impedance 1Z , as specified in (4). Three disadvantages of QWTs are that:

1. A TL must be placed between the load and the feedline. 2. A special characteristic impedance for the QWT is

required, which depends both on the load resistance and the characteristic impedance of the feedline.

3. QWTs work perfectly only for one load at one frequency. (Actually, it produces some bandwidth of “acceptable” VSWR on the TL, as do all real-life matching networks.)


Real Loads for QWTs Ideally, a matching network should not consume (much) power. In (4) we can deduce that if instead of LR we had a complex load, then the QWT would need to be a lossy TL in order to provide a match. So, QWTs work better with resistive loads. However, if the load were complex, we could insert a section of TL to transform this impedance to a real quantity (is this possible?), and then attach the QWT. But, again, this would work perfectly for only one load at one frequency.

Adjusting TL Characteristic Impedance We see in (4) that the QWT requires a very specific characteristic impedance in order to provide a match 1 0 LZ Z R= With coaxial cable, twin lead and other similar TLs this is often not a practical solution for a matching problem. However, for stripline and microstrip adjusting the characteristic impedance is as simple as varying the width of the trace. Consequently, QWTs find wide use in these applications.


As we’ll see in Lecture 12, the characteristic impedance of a microstrip

εrWd

as a function of W/d is

2 4 6 8 10Wêd

25

50

75

100

125

150

175Z0 @ΩD

To construct this curve, it was assumed that 3.38rε = , which is the quoted specification for Rogers Corporation RO4003C laminate that we’ll be using in the lab. Example N9.1: Design a microstrip QWT to match a load of 100 Ω to a 50-Ω line on Rogers RO4003C laminate. Estimate the fractional bandwidth under the constraint that no more than 1% of the incident power is reflected.


28 High Frequency Electronics

High Frequency Design

MATCHING NETWORKS

The Yin-Yang of Matching:Part 2—Practical MatchingTechniques

By Randy RheaConsultant to Agilent Technologies

The Standard Quarter-Wavelength Transmis-sion Line Transformer

Awell-known dis-tributed matchingnetwork is the

quarter-wavelength longtransmission line trans-former. I will refer to this

network as a type 11. The characteristicimpedance of this line is given by

(41)

For example, a 100 ohm load is matched toa 50 ohm source using a 90° line with charac-teristic impedance 70.71 ohms. The matchablespace of the quarter-wavelength transformeris small, essentially only the real axis on theSmith chart. Nevertheless, it enjoyswidespread use. A quarter-wavelength line isalso used in filter design as an impedanceinverter to convert series resonant circuits toparallel resonance, and vice versa [4].

The General Transmission LineTransformer

Perhaps less well-known is that a singleseries transmission line can matchimpedances not on the axis of reals. Thematchable space of this type 12 network isplotted in Figure 9. The characteristicimpedance of the series line is given by

(42)

where

(43)

(44)

and the electrical length of the line is given by(see text)

(45)

where

(46)a Z XL= 12

θ121

2 242

90= + − + °−tana b b

a

ρρρ

ρmax

Im

ReRe=

[ ]( )[ ] + [ ]load

loadload

2

ρloadL

L

ZZ

= ′ −′ +

11

Z Z12 0

11

= +−

ρρ

max

max

Z R RS L0 =

The conclusion of this article covers transmission

line matching networks,plus a discussion of how

characteristics of the loadaffects matching band-width and the choice of

network topologies

Figure 9 · By allowing line lengths other than90°, the matchable impedance space for asingle, series transmission line extendsbeyond the real axis.

From April 2006 High Frequency ElectronicsCopyright © 2006 Summit Technical Media



Lecture 10: TEM, TE, and TM Modes for Waveguides. 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

μ, ε

Metal walls

and that the wave is propagating in z as j ze β− . (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),


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

Proceeding from the Maxwell curl equations:

ˆ ˆ ˆ

x y z

x y z

E j H j Hx y z

E E E

ωμ ωμ∂ ∂ ∂∇× = − ⇒ = −

∂ ∂ ∂

or x : yzx

EE j Hy z

ωμ∂∂

− = −∂ ∂

y : xzy

EE j Hx z

ωμ∂∂⎛ ⎞− − = −⎜ ⎟∂ ∂⎝ ⎠

z : y xz

E E j Hx y

ωμ∂ ∂

− = −∂ ∂

However, the spatial variation in z is known so that

( ) ( )

j zj z

ej e

z

βββ

−−

∂= −

∂

Consequently, these curl equations simplify to

zy x

E j E j Hy

β ωμ∂+ = −

∂ (3.3a),(1)

zx y

E j E j Hx

β ωμ∂− − = −∂

(3.3b),(2)

y xz

E E j Hx y

ωμ∂ ∂

− = −∂ ∂

(3.3c),(3)


We can perform a similar expansion of Ampère’s equation H j Eωε∇× = to obtain

zy x

H j H j Ey

β ωε∂+ =

∂ (3.4a),(4)

zx y

Hj H j Ex

β ωε∂− − =

∂ (3.4b),(5)

y xz

H H j Ex y

ωε∂ ∂

− =∂ ∂

(3.5c),(6)

Now, (1)-(6) can be manipulated to produce simple algebraic equations for the transverse (x and y) components of E and H . For example, from (1):

zx y

j EH j Ey

βωμ

⎛ ⎞∂= +⎜ ⎟∂⎝ ⎠

Substituting for Ey from (5) we find

2

2 2

1z zx x

z zx

j E HH j j Hy j x

j E j HHy x

β βωμ ωε

β βωμ ω με ω με

⎡ ⎤∂ ∂⎛ ⎞= + − −⎜ ⎟⎢ ⎥∂ ∂⎝ ⎠⎣ ⎦∂ ∂

= + −∂ ∂

or, 2z z

xc

j E HHk y x

ωε β⎛ ⎞∂ ∂

= −⎜ ⎟∂ ∂⎝ ⎠ (3.5a),(7)

where 2 2 2ck k β≡ − and 2 2k ω με= . (3.6)

Similarly, we can show that


2z z

yc

j E HHk x y

ωε β⎛ ⎞∂ ∂

= − +⎜ ⎟∂ ∂⎝ ⎠ (3.5b),(8)

2z z

xc

j E HEk x y

β ωμ⎛ ⎞− ∂ ∂

= +⎜ ⎟∂ ∂⎝ ⎠ (3.5c),(9)

2z z

yc

j E HEk y x

β ωμ⎛ ⎞∂ ∂

= − +⎜ ⎟∂ ∂⎝ ⎠ (3.5d),(10)

Most important point: From (7)-(10), we can see that all transverse components of E and H can be determined from only the axial 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 of E and H are known once we find a solution for only zH . Neat!


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 that

2 2

,

, 0,1,( 0)c mn

m nm nkm na b

π π =⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟ = ≠⎝ ⎠ ⎝ ⎠

… (11)

Therefore, 2 2,mn c mnk kβ β= = − (12)

These m and n indices indicate that only discrete solutions for the transverse wavenumber (kc) are allowed. Physically, this occurs because we’ve 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 to indeterminate forms in (7)-(10). However, it can be shown that inside hollow metallic waveguides when both 0m n= = and



Lecture 11: Dispersion. Stripline and Other 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 not dispersive. That is, the phase velocity of a TEM wave is not a function of frequency [ ( )pv g ω≠ ] if the material properties of the waveguide are not functions of frequency. To see this, recall for a TEM wave that LCβ ω= . Therefore,

1pv

LCωβ

= =

which is not a function of frequency, as conjectured, provided neither L nor C are 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 kβ β= = − and

2 22,c mn

m nka bπ π⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

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


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

, 2 2,

p mn

c mn

vk

ωω με

=−

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 0ck = , then ( )pv 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 in microwave engineering. (For high power applications, hollow waveguides made be required; hence, one would need to somehow work around the distortion.)


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, let’s 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 εr

Metal planes



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):

εrWd

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):

εr

E

H

This presents a difficulty. Notice that if the field propagates as a TEM wave, then

0p

r

cvε

=


But which rε 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 of E 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 Maxwell’s 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 relative permittivity, ,r eε :

,1 1 1

2 2 1 12r r

r e d Wε εε + −

= ++

(3.195),(1)


The phase velocity and phase constant, respectively, are:

0

,p

r e

cvε

= (3.193),(2)

0 ,r ekβ ε= (3.194),(3) as for a typical TEM mode. In general, ,1 r e rε ε≤ ≤ (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 8ln 14

120 11.393 0.667ln 1.444

r e

r e

d W WW d d

Z WdW W

d d

ε

π

ε

⎧ ⎛ ⎞+ ≤⎜ ⎟⎪ ⎝ ⎠⎪⎪= ⎨>⎪ ⎡ ⎤⎛ ⎞⎪ + + +⎜ ⎟⎢ ⎥⎪ ⎝ ⎠⎣ ⎦⎩

(3.196),(5) Alternatively, given a desired 0Z and rε , the necessary W d can be computed from (3.197).


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-Ω microstrip on Rogers RO4003C laminate with 1/2-oz copper and a standard thickness slightly less than 1 mm. Referring to the attached RO4003C data sheet from Rogers Corporation, we find that 3.38 0.05rε = ± and 0.032"d = . We will 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 = μm. For 2-oz copper, double this number and for ½-oz copper divide by 2. We will use (3.197) to compute the required W d to achieve a 50-Ω characteristic impedance:


( ) ( )

2

8 22

2 1 0.611 ln 2 1 ln 1 0.39 22

A

A

r

r r

e We dW

Wd B B Bd

επ ε ε

⎧≤⎪ −⎪= ⎨ ⎧ ⎫⎡ ⎤−⎪ ⎪⎪ − − − + − + − >⎨ ⎬⎢ ⎥⎪ ⎪ ⎪⎣ ⎦⎩ ⎭⎩

(3.197),(6) To apply this equation, we first need to compute the constants A and B:

0 1 1 0.110.2360 2 1

r r

r r

ZA ε εε ε

⎛ ⎞+ −= + +⎜ ⎟+ ⎝ ⎠

1.376= (7)

0

3772 r

BZ

πε

= 6.442= (8)

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

1.376

2 1.376

8 2.3172

W ed 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"= .


A more common unit for width and thickness dimensions in microwave circuits is “mil” where

1 mil 1 " 25.41000

= = μm

Therefore,

74.10.0741" "1

4 1000

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

This completes the design of the 50-Ω microstrip.






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

Computational electromagnetics (CEM) can provide accurate solutions for Z0 and other microstrip properties of interest including plots of E and H everywhere in space and sJ and sρ 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. We’ll illustrate the MM technique with a solution to a quasi-static microstrip immersed in an infinite dielectric as shown:

x

y

w dε

Land

Ground plane

That is, there is no substrate, per se.


Integral Equation

We’ll imagine that a time harmonic voltage source has been applied across the two conductors:

x

y

ε

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

εΦe=0 naturallysatisfied++++ + ++ ++

V+-

-- - - - - - ---V

+

-

O.b. pointr

ε

r′d

d

In a previous EM course, you’ve likely learned that the electric potential Φe at a point r in a homogeneous space produced by a line charge density ( )l rρ ′ is given by


( ) ( )

( ) ( ) ( )( )2 2

1 1ln2

1 ln2

e lC

lC

r r dlr r

r x x y y dx

ρπε

ρπε

′

′

⎛ ⎞′ ′Φ = ⎜ ⎟′−⎝ ⎠

′ ′ ′ ′= − − + −

∫

∫ (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, we’ll apply the boundary condition ( ) upper stripe r V rΦ = ∀ ∈ (2) Now, using (2) in (1) and accounting for both the +ρl and –ρl strips yields

( ) ( ) ( )2 21 ln

2 lV r x x d dρπε

′ ′= − − + −

( ) ( ) ( )

0

top

2 2

bottom

lnl

dx

r x x d d dxρ

⎧ ⎡ ⎤⎪ ′ −⎨ ⎢ ⎥⎪ ⎣ ⎦⎩

⎫⎡ ⎤′ ′ ′− + + ⎬⎢ ⎥⎣ ⎦ ⎭

∫

∫

or ( ) ( ) ( )( )2 2

0

1 ln ln 42

w

lV r x x x x d dxρπε

⎡ ⎤′ ′ ′ ′= − − − − +⎢ ⎥⎣ ⎦∫ (3)


Recall that the unknown in (3) is the line charge density ρl. But how do we solve for this function? It varies along the strip so we can’t 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 haven’t encountered such equations before. Integral equations are very difficult to solve analytically. We’ll use a numerical solution method instead.

Basis Function Expansion In the moment method, we first expand ρl in a set of basis functions. For a simple MM solution, here we’ll use pulse basis functions and divide the strips into N uniform sections:

12

x 12Nx −


where ( ) ( )11

; ,N

l n n n nn

r P x x xρ α −=

′ ′≈∑ (4)

What is not known in (4) are the amplitudes αn of the line charge density expansion. These are just numbers. So, instead of directly solving for the spatial variation of ρl in (3), now we’ll just be computing these N numbers, αn. Much simpler! However, we need to allow enough “degrees of freedom” in this basis function expansion (4) so that an accurate solution can be found. This is accomplished by choosing the proper type of expansion functions, a large enough N, etc. The next step in the MM solution is to substitute (4) into (3)

( ) ( )110

1 ; ,2

w N

n n n nn

V P x x x G x x dxαπε −

=

⎡ ⎤′ ′ ′= − −⎢ ⎥⎣ ⎦∑∫ (5)

where ( ) ( ) ( )( )2 2ln ln 4G x x x x x x d′ ′ ′− = − − − + (6)

and is called the Green’s 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. We’ll consider this situation later in this lecture. Then, (5) becomes


( ) ( )11 0

1 ; ,2

wN

n n n nn

V P x x x G x x dxαπε −

=

′ ′ ′= − −∑ ∫

or ( )1

1

12

n

n

xN

nn x

V G x x dxαπε

−=

′ ′= − −∑ ∫ (7)

Testing the Integral Equation In (7), we have N unknown coefficients αn to solve for, but only a single equation. We will generate a total of N equations by evaluating (7) at N points along the (top) strip. This process is called “testing” the integral equation. We’ll test (7) at the centers of each of the N segments, xm, giving

( )1

1

1 1, ,2

n

n

xN

n mn x

V G x x dx m Nαπε

−=

′ ′= − − =∑ ∫ … (8)

This is the final system of equations that we will use to solve for all the coefficients αn.

Matrix Equation It is helpful to cast (8) into the form of a matrix equation [ ] [ ] [ ]

1 1N N N N

V Z α× × ×

= ⋅ (9)

where mV V= (10a) n nα α= (10b)


( )1

12

n

n

x

mnx

Z G x x dxπε

−

′ ′= − −∫ (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 as shown:

-Δ/2

x

y

Δ/2

(x,y) O.b. point

ρlr

then the electrostatic potential at point r produced by a strip of width Δ supporting a constant line charge density lρ is given by

( ) ( )2

2 2

2

ln2

le r x x y dxρ

πε

Δ

−Δ

⎡ ⎤′ ′Φ = − − +⎢ ⎥⎣ ⎦∫ (11)



Lecture 14: Impedance and Admittance 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 concern yourself 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 we’ll see, using these matrix descriptions is very similar to other two-port models for circuits you’ve seen before, such as Z and Y matrices.

Z Matrices As an example of Z matrices, consider this two-port network:

1V+

-[ ]Z

1I

2V+

-

2I

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


[ ]

1 11 12 1

2 21 22 2

Z

V Z Z IV Z Z I

≡

⎡ ⎤ ⎡ ⎤ ⎡ ⎤= ⋅⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦ (1)

where 0,k

iij

j I k j

VZI

= ∀ ≠

= (4.28)

As an example, let’s determine the Z matrix for this T-network (Fig. 4.6):

1V

+

-

AZ

2V

+

-

CZ

BZ1I 2I

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

2

111

1 0A C

I

VZ Z ZI

=

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

1

112 1 2

2 0C

I

VZ V I ZI

=

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

2

221 2 1

1 0C

I

VZ V I ZI

=

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

1

222

2 0B C

I

VZ Z ZI

=

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

Collecting these calculations, then for this T-network:


[ ] A C C

C B C

Z Z ZZ

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 “reciprocal” networks. What’s 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 into this one matrix. Second, if one has networks connected in series, it’s very easy to combine the Z matrices. For example:

1V ′1I ′

2V ′+-

+-

2I ′

[ ]Z ′

[ ]Z ′′+-

+-1V ′′

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

⎡ ⎤ ⎡ ⎤′′ ′′′′⎢ ⎥ ⎢ ⎥= ⋅

′′ ′′⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦


From the figure we see that 1 1I I′ ′′= , 2 2I 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 IZ Z

V V I I

⎡ ⎤ ⎡ ⎤ ⎡ ⎤′ ′′ ′ ′+′ ′′⎢ ⎥ ⎢ ⎥ ⎢ ⎥= ⋅ + ⋅

′ ′′ ′ ′⎢ ⎥ ⎢ ⎥ ⎢ ⎥+⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Also from the figure, note that 1 1I I ′= and 2 2I 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” Z matrix [ ] [ ] [ ]Z Z Z′ ′′= + (3) that incorporates the electrical characteristics of both networks and their mutual interaction.

Y Matrices A closely related characterization is the Y-matrix description of a network:

1V+

-[ ]Y

1I

2V+

-

2I

By definition:



Lecture 15: S Parameters and 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. We’ll begin with the scattering (or S) parameters. Consider again the multi-port network from the last lecture, which is connected to N transmission 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 NV I+ +Nt

,N NV I− −

[ ]S0Z

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 the network are the same (as is the case for the network shown above), then the S parameters are defined as


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 nj z t j z t

n n n nV z V e V eβ β− − −+ −= + 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

Let’s take a close look at this definition (2). Imagine we have a two-port network:


1V + 1t

1V −

0Z [ ]S

2t

0Z

2V +

2V −

Then, for example, 2

111

1 0V

VSV +

−

+=

=

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, and 2. Port 2 is matched so there are no reflections from this

port. Consequently, with 2 11 110V S+ = ⇒ = Γ , which is the reflection coefficient at port 1. Next, consider

2

221

1 0V

VSV +

−

+=

=

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 reflection coefficient only when 2 0V + = .



Lecture 16: Properties of S 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, and 2. 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 of reciprocal networks, it can be shown that [ ] [ ]tS 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 Z and Y matrices.

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


[ ] [ ] 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 [ ]tS

[ ] [ ] [ ] [ ] [ ]1*t t tS S S S I−

⋅ = ⋅ = (3)

[ ]I is the unit matrix defined as

[ ]1 0

0 1I

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

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 SS 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


*

11

N

ki kik

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 a lossless network).

Take row 1 times column 2:

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

Generalizing this result gives

( )*

10 , ,

N

ki kjk

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 a lossless network).

Applying (1) to (7): If the network is also reciprocal, then [ ]S

is symmetric and we can make a similar statement concerning the rows of [ ]S .

That is, the dot product of any row of [ ]S with the conjugate of 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


[ ] 0.2 0.4 0.8 0.40.8 0.4 0.2 0.4

j jS

j j+ −⎡ ⎤

= ⎢ ⎥− +⎣ ⎦

Is the network reciprocal? Yes, because [ ] [ ]tS 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 1j 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 0j j j j+ + + − − = *C2 C1⋅ : Same = 0

Therefore, the network is lossless. As an aside, in Example N15.1 of the text, which we saw in the last lecture,

[ ] 0.1 0.80.8 0.2

jS

j⎡ ⎤

= ⎢ ⎥⎣ ⎦

This network is obviously reciprocal, and it can be shown that it’s also lossy. Example N16.2 (Text Example 4.4). Determine the S parameters for this T-network assuming a 50-Ω system impedance, as shown.


1V +

1V −

0 50Z = Ω

[ ]S

2V +

2V −

0 50Z = Ω1V AV 2V

inZ

First, take a general look at the circuit: It’s linear, so it must also be reciprocal. Consequently, [ ]S

must be symmetric (about the main diagonal). The circuit appears unchanged when “viewed” from either

port 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 0V

V

VSV+

+

−

+==

= Γ =

The input impedance with port 2 matched is ( )in 8.56 141.8 8.56 50 50.00Z = + + Ω = Ω which, not coincidentally, equals 0Z ! With this Zin:

in 0

i 011

n

0Z ZZ Z

S =−

=+

which also implies 22 0S = . Next, for 21S we apply 1V + with port 2 matched and measure 2V − :



Lecture 17: S Parameters and Time Average Power. 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 2V1Γ 2Γ

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)


and the total time average power at that port is comprised of the two terms (see 2.37):

2

1inc

02V

PZ

+

= and 2

1ref

02V

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

2trans 0

02V

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 and

incident time average power is:

2 2

1ref 12

inc 11

VP VP VV

− −

++= = (8)

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

2

111

1 0V

VSV +

−

+=

=

then in (8): 2

2ref11

inc 0V

P SP + =

= (9)


This result teaches us that the relative reflected time average power at port 1 equals 2

11S when port 2 is matched. At Port 2. Using (7) and (4), the ratio of transmitted and

incident time average power is:

2

2trans 20

2inc 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 equals 2

21S when port 2 is matched. Equations (9) and (11) provide an extremely useful physical interpretation of the S parameters as ratios of time average power. 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



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 of the S parameters.

2. Vector network analyzer (VNA). Measures both the magnitude 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):


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.)









Lecture 19: Proper Microwave Laboratory 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 parameter measurements rather than traditional instruments for voltage or current measurements.

2. Precision connections are necessary for accurate and repeatable 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.


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. BNC 2. Type F 3. Type N 4. SMA

5. APC-7 6. APC-3.5 7. 2.92 mm 8. 2.4 mm 9. 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.


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.



Lecture 20: Transmission (ABCD) Matrix. Concerning the equivalent port representations of networks we’ve 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 suited for 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 BC 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 we’ll discover the reason for it shortly. The ABCD matrix is defined as

1 2

1 2

V VA BI IC D⎡ ⎤ ⎡ ⎤⎡ ⎤

= ⋅⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦ (4.63),(1)


It is easy to show that

2

1

2 0I

VAV

=

= , 2

1

2 0V

VBI

=

=

2

1

2 0I

ICV

=

= , 2

1

2 0V

IDI

=

=

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 their ABCD matrices. Nice! To see this, consider the following two-port networks:

1V+

-

1 1

1 1

A BC D⎡ ⎤⎢ ⎥⎣ ⎦

1I

2V+

-

2I

2V ′+

-

2 2

2 2

A BC D⎡ ⎤⎢ ⎥⎣ ⎦

2I ′

3V+

-

3I

In matrix form

1 1 1 2

1 1 1 2

V A B VI C D I⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= ⋅⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

(4.64a),(2)

and 32 2 2

32 22

VV A BIC DI

⎡ ⎤′ ⎡ ⎤⎡ ⎤⎢ ⎥ = ⋅ ⎢ ⎥⎢ ⎥′⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦

(3)

When these two-ports are cascaded,


2V ′+

-

2 2

2 2

A BC D⎡ ⎤⎢ ⎥⎣ ⎦

2I ′

3V+

-

3I

1V+

-

1 1

1 1

A BC D⎡ ⎤⎢ ⎥⎣ ⎦

1I

2V+

-

2I

it is apparent that 2 2V V′ = and 2 2I I′ = . (The latter is the reason for assuming 2I out of the port.) Consequently, substituting (3) into (2) yields

31 1 1 2 2

31 1 1 2 2

VV A B A BII C D C D⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= ⋅ ⋅ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

(4.65),(4)

We can consider the matrix-matrix product in this equation as describing the cascade of the two networks. That is, let

3 3 1 1 2 2

3 3 1 1 2 2

A B A B A BC D C D C D⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= ⋅⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

(5)

so that 3 3 31

3 3 31

A B VVC D II⎡ ⎤ ⎡ ⎤⎡ ⎤

= ⋅⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

(6)

where

1V+

-

3 3

3 3

A BC D⎡ ⎤⎢ ⎥⎣ ⎦

1I

3V+

-

3I

In other words, a cascade connection of two-port networks is equivalent to a single two-port network containing a product of the ABCD matrices. It is important to note that the order of matrix multiplication must be the same as the order in which the two ports are



Lecture 21: Signal Flow Graphs. Consider the following two-port network (Fig 4.14a):

1a1t

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

22S11S


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 kb S a=

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

arrows.

This restriction exists so that a branch from ka to jb denotes a proportional dependence of jb on ka , 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


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

S21S32

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)


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

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

( ) 212 22 21 1 2 1

22

11

SV S S V V VS

− = ⇒ =−

Substituting this into (4) gives

32 213 1

221S 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

221S

S−

32 21

221S 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)


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

V4V4'

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 inΓ and LV using only SFG algebra.

1a

1b

2a

2b

LVinΓ LΓSΓ

1t 2t



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

As we discussed in Lecture 18, a VNA measures both the magnitude and phase of S 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 of systematic errors can be largely removed from the S parameters using “calibration.” (In the context of microwave


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:

We’ll define mS⎡ ⎤⎣ ⎦ 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 are specifically connected to the DUT form the model of the systematic errors. The parameters [ ]S′ are those we desire to know. 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.


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 VNA measurements 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 doesn’t rely on precisely known 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.


3. Line. Connect the two ports together through a microstrip approximately λ/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 be identical. 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.” You’ll use this in the lab.

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



Lecture 23: Basic Properties of Dividers and Couplers.

For the remainder of this course we’re 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 to

many antennas. • Separating forward and reverse propagating waves (can

also 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):

1PDivider

orCoupler

2 1P Pα=

( )3 11P Pα= −

where α ∈ and 0 1α≤ ≤ . The same device can often be used as a power combiner:


1 2 3P P P= +Divider

orCoupler

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 we’ll show here, it’s 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 SS S S S

S S S

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

(7.1),(1)


If the network is matched at every port, then 11 22 33 0S S S= = = . (It is important to understand that “matched” means 1Γ , 2Γ and

3 0Γ = 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

00

0

S SS 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)

2 213 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


13 0S = . Then (6) and (8) are satisfied. For (7) to be satisfied and 23 0S ≠ , 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 11 0 00 1 0

S⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

Notice that ij jiS S≠ .

2. Match only two of the three ports. Assume ports 1 and 2

are matched. Then,

[ ]12 13

12 23

13 23 33

00

S SS S S

S S S

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

(7.7),(9)



Lecture 24: T-Junction and Resistive Power Dividers.

The first class of three-port network we’ll 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

1Γ2Γ

3ΓinY

There are two basic constraints we need to incorporate into this power 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 delivered

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

the other.


For example: 1:1 means 50% of the incident time average power is

delivered to each output port. 2:1 means 67% of the incident time average power is

delivered 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

in1 2 0

1 1 1YZ 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

0in

0

12

VP

Z= (2)

where V0 is the phasor voltage at the junction. The output powers can be computed similarly as

2

01

1

12

VP

Z= and

20

22

12

VP

Z= (3),(4)

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


01 1

in 0 1

1/1/

ZP ZP Z Z

= = and 02 2

in 0 2

1/1/

ZP ZP Z Z

= = (5),(6)

Because the network is lossless:

1 2

in in

1P PP P

+ =

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

0 0

1 2

1Z ZZ Z

+ = so that 1 2 0

1 1 1Z Z Z

+ =

Consequently, not only have we split the power between the output 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

ZZP P

= and 02

2 in

ZZP P

= (7),(8)

That’s 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 we’ll cover later. From a practical standpoint, there are two important points that arise with T-junction power splitters:


1. Junction effects. At the junction of the TLs, there is likely to 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 too

practical 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:


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 of Z1 and Z2, the impedances of interest in the power splitter design are Zin,1 and Zin,2. From (1), the match condition now becomes

in,1 in,2 0

1 1 1Z Z Z

+ = (9)

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

01

in in,1

ZPP Z

= and 02

in in,2

ZPP Z

= (10),(11)

Example N24.1 (text example 7.1). Design a 1:2, T-junction power divider in a 50-Ω system impedance. We’ll choose to use the network in Figure 7.6 with B = 0:



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

λ/4

λ/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.


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.


• 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 the analysis (Fig. 7.10a):

0,Qz

/4

r/22

1eV

2eV

o.c.o.c.

inez

x =- /4

x =0

Γ

1

2 2gV V=

1

We can recognize this circuit as a QWT. Consequently,

20,

in 2Qe z

z = (7.33),(1)

or 0, in2 eQz z= (2)

We want the output ports to be matched. Therefore, in 1ez = ⇒ 0, 2Qz = (3) Since in 1ez = , then by voltage division at the output port

in2 2 2

in

11 2

ee

g ge

zV V V Vz

= = =+

(4)


Next, to find 1eV we’ll use the TL equation

( ) ( )j x j xV x V e eβ β+ −= + Γ

so that ( ) ( ) 10 1 eV V V+= + Γ = (5) Therefore,

( )

( )

2 24 4

2(4)

4

1

j j

e

V V e e

jV V V

π λ π λλ λλ⋅ − ⋅+

+

⎛ ⎞− = + Γ⎜ ⎟

⎝ ⎠= −Γ = =

(7.34),(6)

Γ is the reflection coefficient at port 1 seen looking towards the normalized load of 2 Ω/Ω. Therefore,

( )

0,

30,

2 2 22 2 2

Q

Q

zz

− −Γ = =

+ + (7)

Substituting V + from (6) into (5) and using (7) we find that

( ) ( )

( )1

7

1 21

e VV jVj

= ⋅ + Γ = −−Γ

(8)

• Anti-symmetric excitation (odd mode):

0,Qz

/4

0,Qz

/4

r/22

r/22B A

V2

V3

1

2V

1

1

-2V1


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):

21oV

2oV

inoz

Γ

2V

For in

oz , notice that the load is a short circuit and the TL is 4λ long (1/2 rotation around the Smith chart). This means in

oz = ∞ . Therefore, to match port 2 (and 3) for odd mode excitation, select

12r= ⇒ 2r = [Ω/Ω] (9)

Further, because in

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

( )

29

2 22 1

o rV V Vr

= ⋅ =+

(10)

Even and odd solutions are eigenvectors. Any solution can be determined by summing appropriately weighted eigenvectors. With this information, we’ll be able to deduce most of the S parameters. But first, let’s determine in,1z so we can compute

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



Lecture 26: Quadrature (90º) 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 (90º) Hybrid, 2. 180º Hybrid, 3. Coupled Line, and 4. Lange Coupler.

We will begin with the quadrature (90º) 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


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:

1eA =

4eA =

1

1/8λ

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


1

1

1

1

1oA =

4oA− =

/8λ

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 the

same above and below the line of symmetry (LOS) in Fig 7.23a, then 0I = at the LOS ⇒ open circuit loads at the ends of λ/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 1e e

eB A= Γ , 2 1e e

eB T A= (1a) 3 2 1

e e eeB B T A= = , 4 1 1

e e eeB B A= = Γ (1b)

where 1 1 2eA = , and eΓ and eT are the reflection and transmission coefficients for the even mode configuration.


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

0V = along the LOS ⇒ short circuit loads at the ends of λ/8 stubs, as shown.

Then, 1 1

o ooB A= Γ , 2 1

o ooB T A= (2a)

3 2 1o o o

oB B T A= − = − , 4 1 1o o o

oB B A= − = −Γ (2b) where 1 1 2oA = and oΓ and oT are reflection and transmission coefficients for the odd mode configuration.

• Total solution. The total solution is the sum of the voltages in

both circuits. From this fact, we can deduce that the total iB coefficients will be the sum of (1) and (2):

1 1 11 12 2

e oe oB B B= + = Γ + Γ (7.62a),(3)

2 2 21 12 2

e oe oB B B T T= + = + (7.62b),(4)

3 3 31 12 2

e oe oB B B T T= + = − (7.62c),(5)

4 4 41 12 2

e oe oB B B= + = Γ − Γ (7.62d),(6)

Likewise, the incident wave coefficients are

1 1 11 1 12 2

e oA A A= + = + =

4 4 41 1 02 2

e oA A A= + = − =


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 eΓ and eT are

0eΓ = and ( )1 12eT j−

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

Here we’ll derive solutions for oΓ and oT . From Fig 7.23b:

oT

oΓ1 2

1 1

1 2

1 1

We have three cascaded elements, so we’ll use ABCD parameters to solve for the overall S parameters of this circuit. Elements 1 and 3. These are short circuit stubs of length 8λ ,

which appear as the shunt impedance

in 0 tanZ jZ lβ= where 28 4

l π λ πβλ

= ⋅ =

Therefore, in

0

Z jZ

= , or NY j= −

From the inside flap of your text:



Lecture 27: The 180º Hybrid. The second reciprocal directional coupler we will discuss is the 180º hybrid. As the name implies, the outputs from such a device can be 180º out of phase. There are two primary objectives for this lecture. The first is to show that the S matrix of the 180º hybrid is

[ ]

0 1 1 01 0 0 11 0 0 120 1 1 0

jS

⎡ ⎤⎢ ⎥−− ⎢ ⎥=⎢ ⎥⎢ ⎥−⎣ ⎦

(7.101),(1)

with reference to the port definitions in Fig. 7.41: 1

4

2

3(Δ)

(Σ) 180ºHybrid

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

Coupled180º

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.


2. Out-of-phase power splitter:

1

4

2

3Input

Isolation Through

Coupled180º

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 (Δ)

Sum (Σ) Input A

Input B180º

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 will provide 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 180º hybrid, as shown in Fig. 7.43. We’ll focus on the ring hybrid and specifically consider the first two applications described above.


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

We’ll 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 physical symmetry 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

3B4B

2B


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,

11 12 2e oB = Γ + Γ (7.102a),(2)

21 12 2e oB T T= + (7.102b),(3)

31 12 2e oB = Γ − Γ (7.102c),(4)

41 12 2e oB T T= − (7.102d),(5)

Each of the even and odd solutions for iB ( 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 determine e

oΓ and e

oT from these S parameters.



Lecture 28: Coupled Line and Lange 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.


Consider the geometry shown in Fig 7.27:

εr

++++++ ++++++

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

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 but

grounded. • 12C =mutual capacitance between conductors 1 and 2. (If the

construction materials are reciprocal, then 21 12C C= .) By computing only these capacitances and the quasi-TEM mode wave speed, we’ll be able to analyze these coupled line problems. Why? Assuming TEM modes, then


1

p

L LCZC C v C

= = = (1)

Notice that L doesn’t appear here. Hence we only need pv and C , as conjectured. This is a widely used approach in all TL problems, not just microstrip or coupled lines.

Even-Odd Mode Characteristic Impedances To simplify the problem analysis, we’ll assume the two strips are identical and located on top of a dielectric. Because of the symmetry, we can use an even-odd mode solution approach. • Even mode. The voltages and currents are the same on both

strips, as shown in Fig 7.28a:

Hence, the electric field has even symmetry about the plane of symmetry (POS). This plane is called an “H-wall.”

Notice that no E field lines from conductor 1 (2) terminate on conductor 2 (1). Consequently, the two halves are decoupled, as we expect. This leads us to the equivalent circuit:


V1 V2

1 2

C11 C22

We define the effective capacitance to ground for either conductor in this configuration as

11 22eC C C= = (7.68),(2) Then using (1)

0,,

1e

p e e

Zv C

= (7.69),(3)

which is the characteristic impedance of either TL when both are operated in the even mode. This is called the even mode characteristic impedance.

• Odd mode. The voltages and currents are opposite on each

line as shown in Fig 7.28(b):

Notice here that the electric field lines are perpendicular to the POS. Therefore, similar to image theory, we can consider the POS as an equipotential surface (or an “E wall”). That is, as a ground plane where 0V = .


This approach leads to the equivalent capacitance circuit:

V1 V2

1 22C12

C11 C22

2C12

Sum = C12

The effective capacitance to ground of either conductor in this configuration is then

12 11 11 122 2oC C C C C= = + (7.70),(4) so that, from (1)

0,,

1o

p o o

Zv C

= (7.71),(5)

This is the odd mode characteristic impedance. It is the characteristic impedance seen when a voltage wave is launched on the structure with odd symmetry, as shown in Fig. 7.28b.

The computation of eC and oC required in (3) and (5) is a bit involved. Alternatively, the text presents two examples of graphical design data for specific geometries. Fig 7.29 contains design data for striplines and Fig 7.30 for microstrips with

10rε = .



Lecture 29: Microwave Filter Design by the Insertion Loss Method.

The next major topic we’re going to cover in this course is microwave filter design. Its theoretical basis is exactly the same as low frequency analog filters, as you saw in your electronics courses. For example, you’ll see Butterworth and Chebyshev filters, but designed to operate at microwave frequencies. Implementation of these filters is different, however. For example, we won’t use discrete inductors and capacitors.

Insertion Loss Method We will begin this process with the design of analog filters, but perhaps with more detail than you’ve seen before. There are different methods for systematically designing filters, but the insertion loss method is probably the most prominent. In this technique, the relative power loss due to a lossless filter with reflection coefficient ( )ωΓ :

[ ]S( )ωΓ


is specified in the loss ratio PLR defined as:

( )

incLR 2

load 1o

o

P PPP P ω

= =⎡ ⎤− Γ⎣ ⎦

or ( )12

LR 1P ω−

⎡ ⎤= − Γ⎣ ⎦ (8.49),(1)

In Section 4.1, the text shows that ( ) 2

ωΓ is an even function of ω. This implies that ( ) 2

ωΓ can be expanded in a polynomial series in 2ω . In particular, for a linear and time invariant system, ( ) 2

ωΓ is a rational function meaning It can be expressed as a quotient of real polynomials ( )2M ω and ( )2N ω . We’ll choose:

( ) ( )( ) ( )

22

2 2

M

M N

ωω

ω ωΓ =

+ (8.51),(2)

Using (2) in (1) ( )( )

2

LR 21

MP

N

ω

ω= + (8.52),(3)

This is valid for any linear, time invariant system that is an even function of ω.

Types of Low Pass Filters There are four types of low pass filters discussed in the text that are all based on (3):


1. Maximally Flat, Butterworth, Binomial Filter. For this type of low pass filter:

2

2LR 1

N

c

P k ωω⎛ ⎞

= + ⎜ ⎟⎝ ⎠

(8.53),(4)

where N =filter order and cω =cutoff frequency.

If 1k = , then LR 2P = at cω ω= , which is the 3-dB frequency:

oP

2oP

1LRP−

ω

Flat

Pass band Stop bandcω

For large ω and with 1k = , then

2

2LR

N

c

P k ωω⎛ ⎞

≈ ⋅⎜ ⎟⎝ ⎠

=2

21N

c

ωω⎛ ⎞⋅⎜ ⎟⎝ ⎠

(5)

From this result we learn that the insertion loss IL, defined as ( )LRIL 10log P= , (8.50)

increases by 20N dB/decade in the stop band for the maximally flat low pass filter.


2. Equal Ripple or Chebyshev Filter. For this type of low pass filter:

2 2LR 1 N

c

P k T ωω⎛ ⎞

= + ⎜ ⎟⎝ ⎠

(8.54),(6)

where ( )NT x is the Chebyshev polynomial. A typical plot of 1

LRP− in (6) is

oP

21oPk+

1LRP−

ω

Equal ripple


Generally, N is chosen to be an odd integer when the source and load impedances are equal (two-sided filters).

For large cω ω and using the large argument form of NT , (6) becomes

22

LR2

4

N

c

kP ωω

⎛ ⎞≈ ⎜ ⎟

⎝ ⎠ (7)

As with the Butterworth filter, (7) also increases at 20N dB/decade, but with the extra factor

224

N

(8)

compared to (5). Consequently, there is more roll off with the Chebyshev low pass filter. For example,



Lecture 30: Scaling of Low Pass Prototype Filters. Stepped Impedance Low Pass Filters. In the last lecture, we discussed the design of prototype low pass filters where 1s LR R= = Ω and 1cω = rad/s. Of course, one generally is not going to implement the prototype filter. So what good is it? It is possible to scale and transform the low pass prototype filter to obtain a low pass, high pass, band pass and band stop filters for any impedance “level” ( s LR R= ) and cutoff frequency. Nice! The process of filter design has three basic steps as discussed in the last lecture: (1) collect the filter specifications, (2) design the low pass prototype filter, (3) scale and convert the prototype. The first two steps were performed in the previous lecture. We’ll now consider the last step, beginning with scaling. There are two types of scaling for low pass prototype circuits, impedance scaling and frequency scaling: 1. Impedance Scaling. Since the filter is a linear circuit, we can

multiply all the impedances (including the terminating resistances) by some factor without changing the transfer function of the filter. Of course, the input impedances will change.


If the desired source and load impedances equal 0R , then • ( )0 0L LX R X R Lω′ = = . Therefore, 0L R L′ = . (8.64a),(1)

• 00

1C C

RX R XCω

⎛ ⎞′ = = − ⎜ ⎟⎝ ⎠

. Therefore, 0

CCR

′ = . (8.64b),(2)

• 0 01sR R R′ = ⋅ = . (8.64c),(3)

• 0 0L L LR R R R R′ = ⋅ = . (8.64d),(4) 2. Frequency Scaling. As defined for the prototype 1cω = rad/s.

To scale for a different cutoff frequency, we substitute

c

ωωω

→ (8.65),(5)

Applying this to the inductive and capacitive reactances in the prototype filter we find

• c

Lc

LX L ωωω

ω ωω→

⎛ ⎞′ = = ⎜ ⎟⎝ ⎠

. Therefore, c

LLω

′ = . (8.66a),(6)

• 1 1

c

cCX

C Cωωω

ωω ω→

⎛ ⎞′ = = ⎜ ⎟⎝ ⎠

. Therefore, c

CCω

′ = . (8.66b),(7)

For a one-step impedance and frequency scaling, we can combine (1)-(4), (6) and (7) to obtain

• 0 kk

c

R LLω

′ = (8.67a),(8)

• 0

kk

c

CCRω

′ = (8.67b),(9)


• 0sR R′ = (10)

• 0L LR R R′ = (11) where 1, ,k N= … as in Fig. 8.25. For example, in the circuit of Fig. 8.25a, 1 1C g= , 2 2L g= , 3 3C g= , etc. Example N30.1. Design a 3-dB, equi-ripple low pass filter with a cutoff frequency of 2 GHz, 50-Ω impedance level and at least 15-dB insertion loss at 3 GHz. The first step is to determine the order of the filter needed to achieve the required IL at the specified frequency. From equation (7) in the previous lecture for cω ω

22

LR2

4

N

c

kP ωω

⎛ ⎞≈ ⎜ ⎟

⎝ ⎠ (12)

(This is just an approximation here since / 1.5cω ω = .) What value do we use for k? From Fig. 8.21

1

2

11 k+

1LRP−

ω

Equal ripple



we see that the passband ripple equals 21 k+ . So, with A = ripple in dB, then ( )210log 1 k A+ =

so that /1010 1Ak = − (13) Consequently, for 3A = dB then 0.998 1k = ≈ . Therefore, equation (12) becomes 2

LR 3 / 4NP ≈ so that

N 10logPLR Fig. 8.27b w/ |ω/ωc|-1=0.5 1 3.5 dB 6 dB 3 22.6 dB 19 dB 5 41.7 dB 35 dB

The third column is the more accurate number since it originates from the plot in Fig. 8.27b. The second column is less accurate because we used (12) with / 1.5cω ω = , which is not 1. For this filter, we’ll choose N = 3 to meet the IL specification. From Table 8.4 (3.0-dB ripple), we find the immitance values to be 1 3.3487g = , 2 0.7117g = , 3 1g g= and 4 1g = . Using (8)-(11) with 0 50R = Ω, 2cf = GHz and arbitrarily choosing the prototype circuit having the fewest inductors

+

-Vs

sR ′

1C ′3C ′

2L ′

LR ′


then, • 0 50S LR R R′ ′= = = Ω

• 1 11 3 9

0 0

3.3487 5.332 2 10 50c c

C gC CR Rω ω π

′ ′= = = = =⋅ ⋅ ⋅

pF

• 0 2 0 22 9

50 0.7117 2.832 2 10c c

R L R gLω ω π

⋅′ = = = =⋅ ⋅

nH

The response of this filter was computed in ADS and is shown below. This filter was also designed in ADS using the Filter DesignGuide feature for automatic filter design.

S_ParamSP1

Step=0.01 GHzStop=4 GHzStart=0.5 GHz

S-PARAMETERS

TermTerm2

Z=50 OhmNum=2

PortP2Num=2

PortP1Num=1

TermTerm1

Z=50 OhmNum=1

LL1

R=L=2.83 nH

CC2C=5.33 pF

CC1C=5.33 pF


1.5 2.5 3.50.5 4.0

-20

-10

-30

0

freq, GHz

dB(S

(1,1

))dB

(S(2

,1))

1.5 2.5 3.50.5 4.0

-100

0

100

-200

200

freq, GHz

phas

e(S

(1,1

))ph

ase(

S(2

,1))

Using SmartComponents in ADS can greatly speed up the filter design process. Here using the low pass filter from the “Filter DG – All Networks” palette. The filter is designed using the Filter DesignGuide, which is activated by pointing to DesignGuide -> Filter.

DA_LCLowpassDT1_lpdesign1DA_LCLowpassDT1

Rl=50 OhmRg=50 OhmResponseType=ChebyshevN=3As=15 dBAp=3 dBFs=3 GHzFp=2 GHz

DT



Lecture 31: Stub Synthesis. Kuroda’s Identities. Stub Low Pass Filters.

Another method for synthesizing microwave filters without lumped elements is to use shorted and opened stubs to realize the inductances and capacitances of the filter. However, the separations between the stubs are generally not negligible and thus will degrade the filter performance if they are neglected. Kuroda’s identities are transformations that prove useful with this type of problem.

Stub Synthesis

The text calls this stub synthesis process “Richard’s transformation.” Here we’ll show a simpler approach. In Section 2.3 of text, we derived the stub input impedances:

• Short circuit TL: in 0 tanZ jZ lβ′= (2.45c),(1)

• Open circuit TL: in 0 tanY jY lβ′= (2.46c),(2) To be consistent with Section 8.4, the prime indicates an impedance-scaled (i.e., normalized) value.


From (1) with 4l λ< (or 2lβ π< ), the input impedance for a short circuit stub is a positive reactance (i.e., an effective inductance) while from (2), the open circuit stub presents a negative reactance (i.e., an effective capacitance). We’ll use these two properties to construct effective inductances and capacitances for filters. From (1), it is apparent that we cannot express the input impedance in the form Lω since ω appears in the tangent function. We can conclude that this effective inductance varies with frequency. If this is the case, then which f would one choose? One choice is

8l λ= ( 4lβ π= ), which is halfway between 0 and 4λ (beyond this, the reactance changes sign). So, with 8l λ= thentan 1lβ = and (1) becomes in 0Z jZ ′= (3) Now, for an inductor at cω ω= L cZ j Lω= (4) Equating (3) and (4) we see that by choosing 0 cZ Lω′ = (5) then this λ/8-long TL has the same input impedance as an inductor with inductance L.


We will employ (5) in the design of stub filters. In such an application, the filter coefficients gk will be associated with unscaled component values (i.e., the unprimed Lk and Ck values in Section 8.4). So, from (5)

00

(5)

1c

c c

ZZ L Lωω ω′

≡ = = (6)

where the unprimed quantity indicates an unscaled coefficient. This relationship in (6) is very useful. It shows us that we can realize an effective inductance with filter coefficient Lk by using a short circuit TL with an unscaled characteristic impedance 0 kZ L= (7) that is λ/8-long at cω ω= . Similarly, one can show that a filter coefficient Ck can be effectively realized by an open circuit stub with unscaled characteristic impedance

01

k

ZC

= (8)

that is 8λ -long at cω ω= . These relationships are shown in Fig. 8.34:


These effective L and C values for the stubs change with frequency. This affects, and generally degrades, the filter performance for cf f≠ .

Kuroda’s Identities Now that we can construct stubs to perform as effective inductors and capacitors in a filter [but with ( )L f and ( )C f ], we must next address the effects that occur when the stubs are separated from each other. This is an effect we ignored in the low pass prototype filter. We assumed it contained lumped elements that were interconnected without any time delay between them. In other words, all the elements existed at a point in space.


In microwave circuits, this restriction may be difficult to realize in the physical construction. Hence, the distances between stubs may not be electrically small. The four Kuroda identities allow us to add redundant TLs to the microwave filter circuit and transform it into a more practical form. By definition, a redundant TL is a matched TL of 8λ length. Because it is matched at both ends, these “unit elements” have no effect on the filter at the center frequency. However, they will have an effect at other frequencies. The four Kuroda identities are shown in Table 8.7:

Each box represents a unit element with the indicated characteristic impedance and length 8λ . The lumped elements


represent short- or open-circuit stubs acting as normalized (i.e., unscaled) series or shunt TLs. For example, the first figure in entry (a) represents:

1Z1Z2

1Z

2Z

8λ 8

λ

The text shows a proof of the first Kuroda identity entry in Table 8.7. We’ll prove the second one in the following example. Example N31.1. Prove the second Kuroda identity in Table 8.7. The left hand circuit is

1Z

2Z

8λ

8λ

inZ

From this circuit in 1 1tanZ jZ l jZβ= = Ω where tan lβΩ ≡ . Zin is an un-normalized value because of Z1. Cascading ABCD matrices for this circuit:



Lecture 32: High Pass and Bandpass Microwave Filters. Resonant Stub Filters.

As has been stated in recent lectures, the low pass prototype can also be used to design high pass, bandpass, and bandstop filters, in addition to low pass filters. To achieve this, the prototype filter must be “converted,” in addition to impedance and frequency scaled. We will discuss the design process for high pass and bandpass filters in this lecture.

High Pass Filter Transformation The frequency substitution

cωωω

−→ (8.68),(1)

in a transfer function converts a low pass filter response to a high pass one. Referring to the low pass prototype filters in Fig. 8.25, we will use (1) to convert the impedances of the series inductances and the shunt capacitances. • Series Inductance. With LZ j Lω= and substituting (1):


1ck k

k

j L j Lj C

ωωω ω

→ − =′ (2)

where 1k

c k

CLω

′ ≡ (8.69a),(3)

We can deduce from this result that the series inductances of the low pass prototype filter are converted to series capacitances. This is indicative of a high pass filter. The purpose of the negative sign in (1) is also apparent from (3): it yields physically realizable capacitor values.

• Shunt Capacitance. With CY j Cω= and substituting (1):

1ck k

k

j C j Cj L

ωωω ω

→ − =′

(4)

where 1k

k

LCω

′ ≡ (8.69b),(5)

Here we see that the substitution in (1) transforms the shunt capacitances in the low pass prototype to shunt inductances. This is also indicative of a high pass filter.

Including impedance scaling, the complete conversion of a low pass prototype circuit to a high pass filter is accomplished using

0

1k

c k

CR Lω

′ = (8.70a),(6)

0k

c k

RLCω

′ = (8.70b),(7)


Bandpass Filter Transformation

The conversion of the low pass prototype to bandpass or bandstop filters is only slightly more involved than for high pass filters. The conversion to a bandpass filter is accomplished with the substitution

0

0

1 ωωωω ω

⎛ ⎞→ −⎜ ⎟Δ ⎝ ⎠

(8.71),(8)

where 2 1

0

ω ωω−

Δ = (8.72),(9)

is the fractional bandwidth of the passband:

oP

2oP

P

ω

cω2ω1ω

0ω

Notice that ω0 is not the center frequency ωc. The text defines 0 1 2ω ω ω= (8.73) which is the geometric mean of ω1 and ω2, rather than the more common arithmetic-mean definition. This was done to make the design equations simpler.


As we did with the high pass filter, we’ll apply the transformation (8) to the series inductors and shunt capacitors in the low pass prototype circuit.

• Series Inductance. With LZ j Lω= and substituting (8):

0 0

(8) 0 01/

1 1

kk

k kk k

CL

L Lj L j L jj

ω ωωω ωω ω ω ω

′′

⎛ ⎞→ − = +⎜ ⎟Δ Δ Δ⎝ ⎠

(10)

From this result we see that the series inductors in the low pass prototype are transformed to a series LC combination with elements

0

0

kk

L RLω

′ =Δ

(series element 1 of 2) (8.74a),(11)

and 0 0

kk

CL RωΔ′ = (series element 2 of 2) (8.74b),(12)

where we’ve also included the impedance scaling.

• Shunt Capacitance. With CY j Cω= and substituting (8):

0 0

(8) 0 01/

1 1

kk

k kk k

LC

C Cj C j C jj

ω ωωω ωω ω ω ω

′′

⎛ ⎞→ − = +⎜ ⎟Δ Δ Δ⎝ ⎠

(13)

From this result we see that the shunt capacitors in the low pass prototype are transformed to a parallel LC combination with elements

0 0

kk

CCRω

′ =Δ

(shunt element 1 of 2) (8.74d),(14)


0

0k

k

RLCωΔ′ = (shunt element 2 of 2) (8.74c),(15)

where we’ve also included impedance scaling. Table 8.6 in the text succinctly summarizes these transformations:

How do we implement these filters in microstrip? Consider a second order bandpass filter:

1 1

1L ′1C ′

2L ′2C ′

We could use the first Kuroda identity to transform 1L ′ to a shunt capacitance, but what about the series capacitance 1C ′? Kuroda’s fourth identity transforms a series capacitance to a series capacitance. That’s no help here. We’re stuck!



Lecture 33 – Active Microwave Circuits: Two-Port Power Gains.

We are going to focus on active microwave circuits for the remainder of the semester. There are many types of active circuits such as amplifiers, oscillators, and mixers. We will concentrate only on amplifiers. It is often a much more involved process to design and construct active circuits that operate correctly than passive ones. Reasons for this include: • A bias network is required, • The devices are nonlinear, • Unintended oscillations produced by circuit instability.

More care, patience, and experience are often required in the design of active RF and microwave circuits than purely passive ones. The analysis of such circuits is usually very difficult given the nonlinear behavior of the devices. For linear amplifiers, though, a linear analysis is applicable which helps simplify matters. For this reason, we will focus on linear, small signal amplifiers. Furthermore, we will use measured (or given) S parameters for the devices (transistors) rather than detailed device parameters (β, Cπ, rπ, etc.). Consequently, we can treat the transistor as a


two port, but possibly one with gain. This approach works well for the steady state analysis of linear, small signal amplifiers. For other types of active circuits, such as oscillators, mixers, or power amplifiers, the nonlinear behavior of the circuit devices must be explicitly accounted for, which precludes the use of S parameters. Much more difficult. One big difference with active devices is that the magnitude of the S parameters may be greater than one. Often it is only S21 that has this characteristic, with port 1 serving as the input and port 2 the output. With passive devices, S parameters with magnitudes greater than unity are physically impossible.

Types of Power Gains

Referring to a generic two-port network circuit such as

inP LP

0Z0Z

there are three commonly used definitions for power gain.

1. Operating Power Gain: in

LPGP

= (1)


This is the ratio of the time-average power dissipated in a load to the time-average power delivered to network.

2. Available Gain: ,

,

av nA

av S

PG

P= (2)

This is the ratio of the maximally available time-average power from the network to the maximally available time-average power from the source.

3. Transducer Gain: ,

LT

av S

PGP

= (3)

This is the ratio of the time-average power dissipated in the load to the maximally available time-average power from the source.

It is this latter transducer gain that we used in EE 322 Electronics II – Wireless Communication Electronics to characterize the performance (i.e. gain) of the active devices in circuits.

Among other applications, these three definitions of power gain are used to design different types of amplifiers:

1. Operating Power Gain, G. Maximum linear output power amplifiers.

2. Available Power Gain, GA. Low Noise Amplifiers (LNAs).


3. Transducer Power Gain, GT. Simultaneously conjugate matched input and output ports (leads to maximum linear gain).

Power Gain Expressions

We will now derive analytical expressions for these power gains in terms of the S parameters of the network, as well as the source and load impedances. These will prove central to the design of linear microwave amplifiers. Referring to this generic two-port circuit (Fig. 11.1):

ZL[S]

(wrt Z0)

1V +

1V −

2V +

2V −

inΓ LΓ

ZS

+

-VS SΓ

1t 2t

TLs are infinitesimally short, with characteristic impedance Z0.

outΓ

inZ

+

-1V

outZ

then by the definition of the S parameters we can write 1 11 1 12 2LV S V S V− + −= + Γ (11.2a),(4) and 2 21 1 22 2LV S V S V− + −= + Γ (11.2b),(5) In these equations we have used the relationship 2 2LV V+ −= Γ . As we showed in Lecture 21 using signal flow graphs


1 12 21in 11

1 221L

L

V S SSV S

−

+

ΓΓ = = +

−Γ (11.3a),(6)

Similarly, it can be show that

12 212out 22

2 111S

S

S SV SV S

−

+

ΓΓ = = +

−Γ (11.3b),(7)

Next, by voltage division at the source

( )in1 1 1 1 in

in

1SS

ZV V V V VZ Z

+ − += = + = + Γ+

(8)

so that in1

in in1S

S

VZVZ Z

+ =+ + Γ

(9)

Now, using ( ) ( )in in 0 in 0Z Z Z ZΓ = − + and after some algebra, (9) can be reduced to

1in

11 2

S S

S

VV + − Γ= ⋅

− Γ Γ (11.4),(10)

There are four different time-average power quantities we need to determine in order to compute (1)-(3): 1. Pin: Time-average power provided by the source

( )2

21in in

0

| | 1 | |2VP

Z

+

= − Γ (11.5),(11)

Substituting for 1V + from (10) gives

( )2 2

2in in2

0 in

| | |1 | 1 | |8 |1 |

S S

S

VPZ

−Γ= − Γ

−Γ Γ (11.5),(12)



Lecture 34 – Amplifier Stability.

You’ve seen in EE 322 that a simple model for a feedback oscillator has an amplifier and a feedback network connected as:

Oscillation occurs at the output power P0 and frequency f0 where:

Anytime a portion of a circuit has gain, the circuit may oscillate, even though that is not the intended function of the circuit. For example, an amplifier circuit has gain and if it’s not properly designed, it may oscillate. This concept is referred to as circuit stability. A circuit that is not stable may oscillate. This “oscillation” may not be easily detected. That is, you will probably not measure nice sinusoids at the circuit nodes.


Instead, the circuit may just not “work,” or you will see DC voltages that “are not possible,” or the waveforms are incredibly “noisy”, etc. A very frustrating experience, especially if you don’t understand circuit stability. Even a brief period of oscillation could permanently damage a circuit because of large voltages and power levels that might be generated. A thorough stability analysis requires a large signal analysis of the nonlinear circuit. Very difficult. Here, we will perform a much simpler two-port, S-parameter analysis. This is sufficient only for linear, small signal circuit applications. This analysis can be considered accurate as a test for startup instabilities. It can also provide a starting point for large signal analysis.

Negative Resistance Consider the generic linear, small-signal transistor amplifier circuit we saw in the last lecture:


inΓ LΓSΓ outΓ

inZSZ LZoutZ

We will imagine the transistor is characterized by the S matrix:

[ ] 11 12

21 22t

S SS

S S⎡ ⎤

= ⎢ ⎥⎣ ⎦

(1)

Further, we will assume the matching networks are passive so that 1SΓ < and 1LΓ < (2) Oscillation is possible in this circuit if a signal “incident” on the input or output port of the transistor is “reflected” with a gain > 1. That is, if in 1Γ > or out 1Γ > (3) the circuit may become unstable and oscillate. This could occur when noise in the circuit is incident on either port, is reflected with gain, is reflected again at the corresponding matching network, and so on. It is then possible that at some frequency, such noise could be amplified repeatedly to such a level that the device is forced into nonlinear operation and instability.


In order for 1Γ > , the real part of the impedance seen looking into the port must be negative. That is, in 0R < or out 0R < (4) for instability. This region lies outside the unit circle on a Smith chart. As we derived in Lecture 21

12 21in 11

221L

L

S SSS

ΓΓ = +

−Γ (11.3a),(5)

12 21out 22

111S

S

S SSS

ΓΓ = +

−Γ (11.3b),(6)

Using these in (3), we find that for the circuit to be unconditionally stable:

12 21in 11

22

11

L

L

S SSS

ΓΓ = + <

−Γ (11.19a),(7)

and 12 21out 22

11

11

S

S

S SSS

ΓΓ = + <

−Γ (11.19b),(8)

If 1SΓ > or 1LΓ > , then (7) and (8) are requirements only for conditional stability.

Stability Circles

It can be very helpful to generate a graphical depiction of the range of SΓ and LΓ values that may lead to instability. Stability circles are one way to do this. They are particularly helpful


because of the extra information we gain because they are drawn on the Smith chart. Stability circles define the boundary between stable and potentially unstable LΓ or SΓ . To determine these boundaries, we will set in 1Γ = (or out 1Γ = ) and draw these curves in the LΓ (or SΓ ) plane. For the load stability circle, from (5) we find

12 2111

22

11

L

L

S SSS

Γ+ =

−Γ (11.20),(9)

After some manipulation, as shown in the text, (9) can be rearranged to L L LC RΓ − = (10) which is an equation for a circle in the complex LΓ plane.

In (10) ( )**

22 112 2

22| | | |L

S SC

S− Δ

=− Δ

(11.25a),(11)

is the center of the circle in the complex LΓ plane, and

12 212 2

22| | | |LS SR

S=

− Δ (11.25b),(12)

is the radius, where 11 22 12 21S S S SΔ = − (11.21),(13)


For a conditionally stable circuit, the load stability circle may look something like this on the extended Smith chart:

LR

LC

( )Re LC

( )Im LC

( )Re LΓ

( )Im LΓ

Again, this circle defines those LΓ values where in 1Γ = . Inside the load stability circle (and with 1LΓ ≤ ) are those LΓ that produce a stable or unstable circuit. We don’t know which yet. The converse can be said for those LΓ outside of the circle (and with 1LΓ ≤ ). So how do we identify the stable region? Very easily, as it turns out. We will choose a special load ZL that will quickly uncover the region of stability. Specifically, we’ll choose 0LZ Z= ⇒

0LΓ = so that from (5) in 11| |SΓ = (14)

Consequently, if 11 1S < , then the region containing the origin in the LΓ plane (and 1LΓ ≤ ) is the stable region. Otherwise, if

11 1S > , then the region inside the stability circle (and 1LΓ ≤ ) is the stable region. These two situations are illustrated below.



Lecture 35 – Single Stage Amplifier: Design for Maximum Gain.

Amplifiers must be designed for different performance requirements that depend on the application. Examples of these different requirements are maximum gain, maximum output power, specific gain, circuit stability with varying load impedance, wide bandwidth, and low noise. Often, the first amplifier in a microwave-frequency receiver will be a low noise amplifier (LNA). After the signal level is raised well above the noise level, gain often becomes more important than noise. We will first consider the design of amplifiers that are needed for large gain rather than for their noise or bandwidth characteristics. To realize maximum gain, the input and output matching networks are simultaneously conjugate matched to the transistor. We also need the entire amplifier system (the transistor and matching networks) to be matched to the system impedance. We will use the transducer gain approach to design the amplifier of Figure 11.2 for maximum gain.


inΓ LΓSΓ outΓ

In this approach, we begin with the transducer gain we developed in Lecture 33: 0T S LG G G G= (1)

where 2

2in

1 | ||1 |

SS

S

G − Γ=

−Γ Γ (11.16a),(2)

20 21G S= (11.16b),(3)

and 2

222

1 | ||1 |

LL

L

GS

− Γ=

− Γ (11.16c),(4)

Maximum power transfer from the input matching network (IMN) to the transistor occurs when the two are conjugate matched so that *

in SZ Z= . For a real-valued system impedance, this leads to the requirement *

in SΓ = Γ (11.36a),(5) Similarly, maximum power is transferred from the transistor to the output matching network (OMN) when *

out LΓ = Γ (11.36b),(6) When (5) and (6) are realized in the amplifier circuit, maximum transducer gain will be obtained (for lossless matching


networks). The value of this maximum transducer gain, maxTG , is

found by substituting (5) into (1)

max

22

212 222

1 1 | || |1 | | |1 |

LT

S L

G SS

− Γ=

− Γ − Γ (11.37),(7)

In order to achieve this gain, the IMN and OMN must be simultaneously designed so that (5) and (6) are satisfied. To develop these design equations, we substitute the expressions for

inΓ and outΓ derived in Lecture 33 into (5) and (6) giving

* 12 2111

221ML

MSML

S SSSΓ

Γ = +− Γ

(11.38a),(8)

* 12 2122

111MS

MLMS

S SSSΓ

Γ = +− Γ

(11.38b),(9)

where the ‘M’ subscript indicates values obtained with conjugate matching. These are two equations from which we can solve for the two unknowns MSΓ and MLΓ . As shown in the text, these solutions are

2 2

1 1 1

1

4 | |2MS

B B CC

− −Γ = (11.40a),(10)

and 2 2

2 2 2

2

4 | |2ML

B B CC

− −Γ = (11.40b),(11)

(not ‘+’ as in the text – see Gilmore and Besser, vol. II, p. 79). In (10) and (11):


2 2 21 11 221 | | | | | |B S S= + − − Δ (11.41a),(12)

2 2 22 22 111 | | | | | |B S S= + − − Δ (11.41b),(13)

*1 11 22C S S= − Δ (11.41c),(14)

*2 22 11C S S= − Δ (11.41d),(15)

and 11 22 12 21S S S SΔ = − (11.21),(16) These design equations will produce physically realizable, passive matching networks provided the transistor is unconditionally stable (or has been “stabilized”). The final step is to construct these lossless passive matching networks. This will be illustrated in the following example. Example N35.1 (Text example 11.3). For the amplifier circuit shown earlier in this lecture, design input and output matching networks to achieve maximum transducer gain at 4 GHz. The S parameters for the transistor, referenced to 50 Ω, are: f(GHz) S11 S21 S12 S22 3.0 0.80 89°− 2.86 99° 0.03 56° 0.76 41°− 4.0 0.72 116°− 2.60 76° 0.03 57° 0.73 54°− 5.0 0.66 142°− 2.39 54° 0.03 62° 0.72 68°− The first step in this design process is to ensure that the device is unconditionally stable at 4.0 GHz.



Lecture 36 – Single Stage Amplifier: Design for Specific Gain.

Maximum gain amplifiers discussed in the previous lecture are designed to provide the maximum gain possible from the circuit. If a specific gain value less than the maximum is required, then another design approach must be used, which is the topic of this lecture. Additional motivation for amplifiers with less than maximum gain is related to the gain-bandwidth product concept. Increased gain often occurs at the expense of bandwidth and vice versa. Generally speaking, more bandwidth from an amplifier may be obtained if the gain is reduced.

Unilateral Transistor Before introducing the method of design for specific gain, we will first discuss the concept of a unilateral transistor. This is an approximation in which 12 0S = is assumed for a transistor. This is in contrast to a bilateral transistor where 12 0S ≠ . The unilateral approximation can substantially simplify the design of an amplifier circuit while, in some cases, introducing acceptable error in the gain calculation.


The assumption that 12 0S = has some important consequences for the input and output reflection coefficients in a two-port network:

inΓ LΓSΓ outΓ

inZSZ LZoutZ

In particular, from (11.3a) for the input reflection coefficient and

12 0S = :

12 21in 11 11

221L

L

S SS SS

ΓΓ = + =

−Γ (11.3a),(1)

while from (11.3b) for the output reflection coefficient

12 21out 22 22

111S

S

S SS SS

ΓΓ = + =

−Γ (11.3b),(2)

Using (1) and (2) in the expression for GT (from (11.13) in the text) provides what is called the unilateral transducer gain, GTU, given as

2 2

2212 2

11 22

1 | | 1 | || ||1 | |1 |

S LTU

S L

G SS S

− Γ − Γ=

−Γ − Γ (11.15),(3)

The error in gain calculated by the unilateral approximation is sometimes quantified through the ratio of GTU to the regular (or


“bilateral”) transducer gain GT. As given in the text, this ratio is bounded as

( ) ( )2 2

1 11 1

T

TU

GGU U

< <+ −

(11.45),(4)

where U is the unilateral figure of merit defined as

( )( )

11 21 12 222 2

11 221 1

S S S SU

S S=

− − (11.46),(5)

The unilateral assumption is often found acceptable when the error is less than approximately 0.5 dB or so. We have found that this expression in (4) is actually not correct. You will be investigating this in your homework.

Constant Gain Circles The unilateral transducer gain in (3) is the product of three terms 0TU SU LUG G G G= (6)

where 2

211

11

SSU

S

GS− Γ

=− Γ

, 2

222

11

LLU

L

GS− Γ

=− Γ

(7),(8)

and 20 21G S= (9)

The S parameters of the transistor are fixed by the device itself as well as the particular biasing.


To achieve a specific gain, which is less than the maximum, from the transistor amplifier circuit, mismatch will be purposefully designed into the input and/or output “matching” networks to yield the gain we seek. In other words, the source and load gain factors in (7) and (8) will be adjusted downward from their maximum values

maxSUG and

maxLUG in order to achieve the specified gain. The maximum GS occurs when there is a conjugate match at the transistor input, so that for a unilateral device * *

in 11(1)

S SΓ = Γ = (10)

Likewise, the maximum GL occurs when there is a conjugate match at the transistor output so that * *

out 22(2)

L SΓ = Γ = (11)

Applying (10) in (7) and (11) in (8) leads to an expression for the maximum unilateral transducer gain,

maxTUG , given as

max max max

221 02 2

11 22

1 11 1TU SU LUG S G G G

S S= ≡

− − (12)

where the maximum source and load gain factors for the unilateral device are

max 2

11

11SUG

S=

− (11.47a),(13)


and max 2

22

11LUG

S=

− (11.47b),(14)

We will define normalized gain factors gS and gL to quantify the amount of reduction in the source and load gain factors needed to achieve the desired gain. These factors are given as

( )max

22

11211

11

1SS

SSU S

Gg SG S

− Γ≡ = −

− Γ (11.48a),(15)

and ( )max

22

22222

11

1LL

LLU L

Gg SG S

− Γ≡ = −

− Γ (11.48b),(16)

so that 0 1Sg≤ ≤ and 0 1Lg≤ ≤ . As shown in the text, (15) and (16) can be cast in the forms S S SC RΓ − = and L L LC RΓ − = , (17),(18) respectively, where

( )

*11

2111 1

SS

S

g SCg S

=− −

(11.51a),(19)

( )

( )

211

211

1 1

1 1S

SS

g SR

g S

− −=

− − (11.51b),(20)

and ( )

*22

2221 1

LL

L

g SCg S

=− −

(11.52a),(21)

( )

( )

222

222

1 1

1 1L

LL

g SR

g S

− −=

− − (11.52b),(22)

Documents

EE481 - Microwave Engineering