EE481 - Microwave Engineering

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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

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

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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

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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 Region HF 3 MHz-30 MHz VHF 30 MHz-300 MHz UHF 300 MHz-1 GHz Microwave Region ₍λ = 30 cm to 8 mm) 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 Millimete r Wave _Region V 40-75 GHz W 75-110 GHz mm 110-300 GHz

RF, microwave and millimeter wave circuit design and construction is far more complicated than low frequency work. So why do it?

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Whites, EE 481 Lecture 1 Page 2 of 5

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?

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Lecture 2: Telegrapher Equations

For Transmission Lines. Power Flow.

Microstrip is one method for making electrical connections in a microwave circuit. It is constructed with a ground plane on one side of a PCB and lands on the other:

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

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In microwave circuits, even distances as small as a few inches may be “far” and the propagation delay for a voltage signal to appear at another IC may be significant.

This propagation of voltage signals is modeled as a

“transmission line” (TL). We will see that voltage and current can propagate along a TL as waves! Fantastic.

The transmission line model can be used to solve many, many types of high frequency problems, either exactly 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.

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E E z Δ H Fig. 1

A common circuit symbol for a TL is the two-wire (parallel) symbol to indicate any transmission line. For example, the equivalent circuit for the coaxial structure shown above is:

Analysis of Transmission Lines

On a TL, the voltage and current vary along the structure in time (t) and in distance (z), as indicated in the figure above. There are no instantaneous effects.

( )

, i z t

( )

, i z t z

( )

, v z t +

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-Whites, EE 481 Lecture 2 Page 4 of 12

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 Δz of the TL. This Δz 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:

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( ), v z t ( ), i z t L zΔ i z( + Δz t, ) ( , ) v z+ Δz t + -+ -z Δ 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 z Δ L zΔ C zΔ L zΔ C zΔ L zΔ C zΔ R_L R_s v_s(t) z Δ Δz Δz

This is a general model: it applies to any TL regardless of its cross sectional shape provided the actual electromagnetic field is TEM.

However, the PUL-parameter values change depending on the

specific geometry (whether it is a microstrip, stripline, two-wire, coax, or other geometry) and the construction materials.

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Transmission Line Equations

To develop the governing equation for v z t , apply KVL in

( )

, Fig. 2 above (ignoring losses)

( )

, i z t

( ) (

, ,

)

v 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

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:

_{( ) ( )}

_{( )}

2 ₂ 2 2 2 2 1 p p d V z j V z V z dz v v ω ω = = − (5) We define LC β 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.)

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

( )

= I e_o+ −j zβ + I e_o− +j zβ (10) , , , o o o o

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We can confirm the correctness of these two solutions by direct substitution into (7) and (8). For example, substituting V e_o+ −j zβ from (9) into (7) gives

(

)

2 ₂

( )

? 0 j z o V + −jβ e− β +β V z = or −β2V e_o+ −j zβ + β2V e_o+ −j zβ = ? 0

which is indeed true. Therefore, V e_o+ −j zβ in (9) is a valid solution to (7).

The constants I_o+ and I_o− in (10) can be expressed in terms of V_o+ and V_o−. In particular, it can be shown that

0 o o V I Z + + ₌ (11) and 0 o o V I Z − − _{= −} (12) If we substitute (11) and (12) into (10) we find that

( )

0 0 j z j z o o V V I z e e Z Z β β + − − + = − (2.14b),(13) and V z

( )

=V e_o+ −j zβ +V e_o− +j zβ (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.

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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 z j z j t j o o o o o p v z t e V e e e V e e V t z V t z z V t v ω β β ω φ β ω β φ ω φ ω ω φ + ₋ + − + + + + + + + ⎡ ⎤ ⎡ ⎤ = ℜ _⎣ _⎦ = ℜ _⎣ _⎦ ⎡ ⎛ ⎞ ⎤ = − + = _⎢ _⎜ − _⎟ + _⎥ ⎝ ⎠ ⎣ ⎦ ⎡ ⎛ ⎞ ⎤ = ⎢ ⎜_⎜ − ⎟_⎟+ ⎥ ⎢ _⎝ _⎠ ⎥ ⎣ ⎦

We can clearly see in this last result that we have a function

of time with argument t − z v/ _p. 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

o

V e− + β (and I e_o− +j zβ ) are waves propagating in the -z direction.

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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 R_s = Z₀:

R_s z Z₀, β z = 0 + -V_s + -V_g l Z_in

To calculate the input voltage V_g, 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

in Z .

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

( )

(

₍

)

₎

( )

in 0 0 2 cos cot 2 sin o o V l V l Z jZ l j V I l l Z β _β β + + − − = = = − − ₋ ₋ [Ω]

In other words, the input impedance is purely reactive

in in

Z = jX where X_in = −Z₀cot

( )

βl (2.46c)

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

Using voltage division,

( )

0 in in 0 0 0 cot cot g s s jZ l Z V V V Z Z jZ l Z β β − = = + − +

This circuit voltage V_g is also the voltage on the TL at z = − . l That is, from (19) in the previous lecture

(

)

2 _o cos

(

)

V z = − =l V+ −βl

Since V_g =V z

(

= − , we can equate these two voltages giving l

)

( )

0

_{( )}

( )

0 0 cot 2 cos cot o s jZ l V l V jZ l Z β β β + ₌ − − +

More often than not, expressions of this type are used to

determine V_o+ in terms of V_s and R_s. 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.

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

( )

j z j z o o V z =V e+ − β +V e− + β (2.34a),(1) and

( )

0 0 j z j z o o V V I z e e Z Z β β + − − + = − (2.34b),(2)

(

₍

)

Z = jZ βl [Ω] (2.45c),(7)

A plot of this input reactance is shown in Fig. 2.6c. 3. With the resistive load Z_L = Z₀, (5) yields

in 0

Z = 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.

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

( )

( ) ( )

av

1 2

P z = ℜ ⎣e V z I z⎡ ∗⎤_{⎦ (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 2 av 0 1 1 2 o _j _l _j _l L L L V P z e e e Z β β + − + ⎡ ⎤ = ℜ _⎣ − Γ + Γ − Γ _{⎦ (9)}

Notice that the second and third terms are conjugates so that

(

2

)

2 2 2 j l j l j l Le Le j m Le β ∗ β β + + _⎡ + _⎤ − Γ + Γ = ℑ _⎣Γ _⎦

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

(

)

2 2 av 0 1 1 2 o L V P 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

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Whites, EE 481 Lecture 4 Page 6 of 12 average power V_o+ 2

(

2Z₀

)

minus the reflected time average power V_o+ 2 Γ 2

(

2Z₀

)

.

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

From (10) we see that if the load is entirely reactive so that 1

L

Γ = , then P_av = 0 and no time average power is delivered to the load, as expected. For all other passive loads, P_av >0.

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

RL= −10log₁₀

( )

Γ_L 2 = −20log₁₀

( )

Γ dB _L (2.38),(11) The two extremes for return loss with a passive load are:

1. A matched load where Γ =_L 0 and RL= ∞ (no reflected power), and

2. A reactive load where Γ = and RL 0_L 1 = (all power

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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.

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( )

j z j z o o V z =V e+ − β +V e− β or V z

( )

=V_o+

(

e−j zβ + Γ_Lej zβ

)

(2.69),(1) where 0 0 L L L Z Z Z 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 V_o+ as the only unknown quantity in (1).

Generally speaking, we compute V_o+ 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:

in in in g g Z V V Z Z = + (3)

Observe that V_in 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 V_in must also equal (V z = − on the TL. l)

Proceeding, then from (1) at the input

(

)

(

j l j l

)

o L

V z = − =l V+ e+ β + Γ e− β (4)

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

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(

)

in in j l j l o L g g Z V e e V Z Z β β + + _{+ Γ} − ₌ + or in

(

)

1 in j l j l o g L g Z V V e e Z Z β β − + ₌ _{+ Γ} − + [V] (2.70),(5)

Maximum Power

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

* * in av in in in * in 1 1 [ ] 2 2 V P e V I e V Z ⎡ ⎤ = ℜ _{= ℜ ⎢} _⎥ ⎣ ⎦ or 2 in av * in 1 2 V P e Z ⎡ ⎤ = _{ℜ ⎢ ⎥} ⎣ ⎦ [W] (2.74),(6)

Now, substituting (3) into (6) gives

2 2 in av * in in 1 2 g g V _Z P e Z Z Z ⎡ ⎤ = _{ℜ ⎢ ⎥} + _⎣ _⎦ (2.74),(7)

If we define Z_in = R_in + jX_in and Z_g = R_g + jX_g, (7) becomes

(

) (

)

2 in av 2 2 in in 2 g g g V _R P R R X X = + + + (2.75),(8)

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Employing this last result, we’ll consider three special cases for

av

P in an effort to maximize this quantity. We will assume that

g

Z is both nonzero and fixed:

(1.) Load is matched to the TL: Z_L = Z₀.

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

in 0

Z = Z . [This should be intuitive. If not, see (2.43).] Consequently, from (8) with R_in = Z₀ and X_in = 0:

(

) ( )

2 0 av,1 2 2 0 2 g g g V _Z P Z R X = + + (2.76),(9)

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

in g

Z = Z and Γ ≠_L 0.

Specific values for βl, Z₀, and Z_L would need to be chosen so that Z_in = Z_g. Then from (8) and with R_in = R_g and X_in = X_g:

(

) (

)

2 av,2 2 2 2 g g g g g g V R P R R X X = + + + or 2 av,2 2 2 8 g g g g V R P R X = + (2.78),(10)

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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 1 1 Z z z z z Z z + Γ ≡ = − Γ (1)

where Z z

( )

=V z

( ) ( )

/I z is the total TL impedance at z and

( )

j2 z

L

z e− β

Γ = Γ (2)

is the generalized reflection coefficient at z.

₍

)

₎

1 r i r i j z z j + Γ + Γ = − Γ + Γ (3)

Now, we will define z z

( )

≡ +r jx and separate (3) into its real and imaginary parts

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( )

(

₍

₎

)

(

)

(

( )

z ] 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

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Plot (4) in the Γr-Γi plane: • For 0r = : Γ + Γ =2_r 2_i 12 • For r =1: 2 2 2 1 1 2 2 r i ⎛_{Γ −} ⎞ _{+ Γ =} ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ • For 1 2 r = : 2 2 2 1 2 3 3 r i ⎛_{Γ −} ⎞ _{+ Γ =} ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

Plot these curves in the Γr-Γi plane:

( ) Re r z Γ = ⎡_⎣Γ ⎤_⎦ ( ) Im i z Γ = ⎡_⎣Γ ⎤_⎦ 1 1 -1 -1 r=0 r=1/2 r=1 1/2 Complex Γ(z) plane

Plot (5) in the Γr-Γi plane:

2 = −1 2 • For 100x = :

(

( ) Re r z Γ = ⎡_⎣Γ ⎤_⎦ ( ) Im i z Γ = ⎡_⎣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.

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Notice that the Γr and Γi axes are missing from the “combined”

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Important Features of the Smith Chart

1. By definition

( )

(

)

1 1 1 1 z z r jx z z z r jx − + − Γ = = + + + . Therefore

( )

(

)

(

)

2 ₂ 2 ₂ 1 1 1 1 1 ₁ r x r jx r jx z r jx r jx _r _x − + + − − + Γ = ⋅ = + + − + ₊ ₊

From this result, we can show that if r ≥ then 0 Γ

( )

z ≤1. This condition is met for passive networks (i.e., no amplifiers) and lossless TLs (real Z₀).

Consequently, the standard Smith chart only shows the inside of the unit circle in the Γr-Γi plane. That is, Γ

( )

z ≤ which is 1

bounded by the r = circle described by 0 Γ + Γ =2_r 2_i 1. 2. If z z

( )

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

(

)

2 ₂ 2 1 i r i x = Γ Γ − + Γ

we deduce that Γ =_i 0 (except possibly at Γ =_r 1).

Consequently, purely real z z

( )

values are mapped to Γ

( )

z

values on the Γ = ℜ Γ_r e⎡_⎣

( )

z ⎤_{⎦ axis.}

3. If z z

( )

is purely imaginary (i.e., r = ) then from (4) 0

2 2 2

1

r i

Γ + Γ =

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Consequently, purely imaginary z z

( )

values are mapped to

( )

z

Γ values on the unit circle in the Γ_r-Γ_i plane.

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

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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.

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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, L ZL

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

0

L

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

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For Γ =_L 0, this implies Z_in = Z₀. In other words, R_in = ℜe Z[ ₀]

and X_in =0, if the TL is lossless.

Note that we need at least two degrees of freedom in the matching network in order to transform Z_L at the load to Z₀ 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):

0 L R > Z : R_L < Z₀: Z0, Zin ZL jX (jB)-1 where Z_L = R_L + jX_L.

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

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0 L R > Z : R_L < Z₀: Z_L Z_L Z_L Z_L Z_L Z_L Z_L Z_L

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

y Assume R_L > Z₀. Using Fig. 5.2(a):

1 in 1 L L Z jX jB R jX − ⎛ ⎞ = +_⎜ + _⎟ + ⎝ ⎠ (5.1),(1)

Through the proper choice of X and B we wish to force

in 0

Z = 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

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(

)

2 0 0 2 2 L L L L L L L X R Z R R Z X B R X ± − + = + (5.3a),(2) 0 0 1 _L L L X Z Z X B R BR = + − (5.3b),(3) Comments:

1. Since R_L > Z₀, 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.

y Assume R_L < Z₀. Using Fig. 5.2(b) with Z_in = Z₀, we obtain

1 in 0 1 L Z Z jB Z jX − ⎛ ⎞ = =_⎜ + _⎟ + ⎝ ⎠ (5.4),(4)

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

(

0

)

L L L X = ± R Z −R − X (5.6a),(5) 0 0 1 _L L Z R B Z R − = ± (5.6b),(6) Comments:

1. Since R_L < Z₀, 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.

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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 R_L > Z₀ (Fig. 5.2a). Here, we’ll give an example when R_L < Z₀.

Example N7.1 Design an L network to match the load 25+ j30 Ω to a TL with Z₀ =50 Ω at the frequency f = GHz. 1

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

Steps for a Smith chart solution:

1. 0 1 3 2 5 L L Z z j Z

= = + 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 z_L such that the sum of admittances b and

(

z_L + jx

)

−1 yield y_in = +1 j0 = z_in (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 jx= − j0.1 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 Γ = and 0 y_in = +1 j0, which means the TL now sees a matched load.

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10 20 30 40 50 60 70 8₀ 90 1 0 0 11 0 120 13 0 14 0 15 0 16 0 1 7 0 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. 0 0.9 0 .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 7.0 8.0 9.0 50 10 5.0 4.0 z_L Start End 1+jx admittance circle (on Z chart)

Constant VSWR circle

1+jx impedance circle (on Z chart)

K. W. Whites

Un-normalizing, we find that

0 0.1 50 5.0 jX = jx Z⋅ = −j ⋅ = − j Ω 0 1 1.0 0.02 50 jB = jb⋅ =Y j ⋅ = j S

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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 Z_L = j Lω C 1 j Z j Cω ωC − = = Admittance L 1 j Y j Lω ωL − = = C Y = j Cω y Since 0X < , we identify this as a capacitor. Therefore,

2 5.0 j jX j C ω − = = − Ω

For operation at 1 GHz, we need

2 1 31.8 5 2 C f π = = ⋅ pF

y Since 0B > , we also identify this as a capacitor. Therefore,

1 0.02

jB = j Cω = j S For operation at 1 GHz, we need

1 0.02 3.18 2 C f π = = pF

The final circuit is:

50 Ω f = 1 GHz Z_in Z_L C₂=31.8 pF C₁=3.18 pF =25+j30 Ω

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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

0

Z 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.)

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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

( )

_{( )}

0 0 0 0 0 tan tan L L L L Z jZ d Z jZ t Z z d Z Z Z jZ d Z jZ t β β + + = − = = + + (5.7),(1)

where t ≡ tan

( )

βd . With a shunt connection, it is much simpler to work with admittances than impedances. So, we’ll define the transformed load admittance as Y =1/Z = +G jB.

The distance d is chosen so that G =Y₀

(

=1/Z₀

)

. As shown in the text, this condition leads to the solutions

(

)

1 1 1 tan , 0 2 1 tan , 0 2 t t d t t π λ _π π − − ⎧ _≥ ⎪⎪ = ⎨ ⎪ ₊ _< ⎪⎩ (5.10),(2) where

(

) (

)

2 ₂ 0 0 0 0 0 0 , , 2 L L L L L L L L X R Z Z R X R Z R Z t X R Z Z ⎧ _± _⎡ ₋ ₊ _⎤ ⎪ _⎣ _⎦ _≠ ⎪ ₋ = ⎨ ⎪ − = ⎪ ⎩ (5.9),(3) and Z_L = R_L + jX_L.

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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 Y_L will also have an imaginary part B.

The length of the stub, l_s, is chosen so that its input susceptance

s

B = −B. Consequently, the parallel combination of the stub input susceptance and the transformed load admittance yield an input admittance Y_in =Y₀, as seen from the source end of the TL. As shown in the text, this second condition provides the solutions 1 _tan 1 0 2 s l Y B λ π − ⎛ ⎞ = _{⎜ ⎟} ⎝ ⎠ short-circuit stub (5.11b),(4) or 1 0 1 tan 2 o l B Y λ 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 Z_L =35− j47.5 Ω to a TL with

0 50

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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

(45)

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 1 in 1 1 1 tan tan L L R jZ l Z Z Z jR l β β + = + (2.61),(1)

Now imagine that we have a special length l =λ₁/ 4 of TL, as indicated in the figure above. At this frequency and physical length, the electrical length of the TL is

1 1 1 2 4 2 l π λ π β λ = = rad (2)

Consequently, for a / 4λ -length TL, tanβ₁l → ∞. Using this result in (1) gives 2 1 in L Z Z R = (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.

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Note that we can adjust Z₁ in (3) so that Z_in = Z₀. In particular, from (3) with Z_in = Z₀ we find

1 0 L

Z = Z R (2.63),(4)

In other words, a λ/4 section of TL with this particular

characteristic impedance will present a perfect match (Γ = ) to 0 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 Z₁, 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.)

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Real Loads for QWTs

Ideally, a matching network should not consume (much) power. In (4) we can deduce that if instead of R_L 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 L

Z = 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.

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As we’ll see in Lecture 12, the characteristic impedance of a microstrip ε_r W d as a function of W/d is 2 4 6 8 10 Wêd 25 50 75 100 125 150 175 Z0 @ΩD

To construct this curve, it was assumed that ε_r =3.38, 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.

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28 High Frequency Electronics

High Frequency Design

MATCHING NETWORKS

The Yin-Yang of Matching:

Part 2—Practical Matching

Techniques

By Randy Rhea

Consultant to Agilent Technologies

The Standard Quarter-Wavelength Transmis-sion Line Transformer

A

well-known

dis-tributed matching network is the quarter-wavelength long transmission line trans-former. I will refer to this network as a type 11. The characteristic impedance of this line is given by

(41) For example, a 100 ohm load is matched to a 50 ohm source using a 90° line with charac-teristic impedance 70.71 ohms. The matchable space of the quarter-wavelength transformer is small, essentially only the real axis on the

Smith chart. Nevertheless, it enjoys

widespread use. A quarter-wavelength line is also used in filter design as an impedance inverter to convert series resonant circuits to parallel resonance, and vice versa [4].

The General Transmission Line Transformer

Perhaps less well-known is that a single series transmission line can match impedances not on the axis of reals. The matchable space of this type 12 network is plotted in Figure 9. The characteristic impedance 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 X₁₂ _L θ12 1 2 2 4 2 90 =_tan− a +b −b+ ° a ρ ρ ρ ρ max Im Re Re =( [ ]) [ ]loadload + [ ] load 2 ρload L L Z Z = ′ − ′ + 1 1 Z₁₂ Z₀ 1 1 = + − ρ ρ max max Z₀= R R_S _L

The conclusion of this article covers transmission

line matching networks, plus a discussion of how characteristics of the load affects matching band-width and the choice of network topologies

Figure 9 · By allowing line lengths other than 90°, the matchable impedance space for a single, series transmission line extends beyond the real axis.

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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 e−j zβ . (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

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• 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 H x y z E E E ωμ ∂ ∂ ∂ ωμ ∇ × = − ⇒ = − ∂ ∂ ∂ or ˆx : z y x E E j H y z ωμ ∂ ∂ ₋ _{= −} ∂ ∂ ˆy : z x y E E j H x z ωμ ∂ ∂ ⎛ ⎞ −_⎜ − _⎟ = − ∂ ∂ ⎝ ⎠ ˆz : y x z E _E j H x y ωμ ∂ _∂ − = − ∂ ∂

However, the spatial variation in z is known so that

(

j z

)

₍

₎

j z e j e z β β β − − ∂ = − ∂

Consequently, these curl equations simplify to

z y x E j E j H y β ωμ ∂ ₊ _{= −} ∂ (3.3a),(1) z x y E j E j H x β ωμ ∂ − − = − ∂ (3.3b),(2) y _x z E _E j H x y ωμ ∂ _∂ − = − ∂ ∂ (3.3c),(3)

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We can perform a similar expansion of Ampère’s equation H jωεE ∇ × = to obtain z y x H j H j E y β ωε ∂ ₊ ₌ ∂ (3.4a),(4) z x y H j H j E x β ∂ ωε − − = ∂ (3.4b),(5) y _x z H _H j E x 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):

z x y j E H j E y β ωμ ⎛ ∂ ⎞ = _⎜ + _⎟ ∂ ⎝ ⎠

Substituting for E_y from (5) we find

2 2 2 1 z z x x z z x j E H H j j H y j x j E j H H y x β β ωμ ωε β β ωμ ω με ω με ⎡∂ ⎛ ∂ ⎞⎤ = _⎢ + _⎜− − _⎟_⎥ ∂ _⎝ ∂ _⎠ ⎣ ⎦ ∂ ∂ = + − ∂ ∂ or, _x ₂ z z c j E H H k ωε y β x ⎛ ∂ ∂ ⎞ = _⎜ − _⎟ ∂ ∂ ⎝ ⎠ (3.5a),(7) where k_c2 ≡ k2 − β2 and k2 =ω με2 . (3.6)

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Whites, EE 481 Lecture 10 Page 4 of 10 2 z z y c j E H H k ωε x β y ⎛ ∂ ∂ ⎞ = − _⎜ + _⎟ ∂ ∂ ⎝ ⎠ (3.5b),(8) 2 z z x c j E H E k β x ωμ y ⎛ ⎞ − ∂ ∂ = _⎜ + _⎟ ∂ ∂ ⎝ ⎠ (3.5c),(9) 2 z z y c j E H E k β y ωμ x ⎛ ∂ ∂ ⎞ = _⎜− + _⎟ ∂ ∂ ⎝ ⎠ (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 E_z and H_z. 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 E_z =0 and H_z ≠ 0. Consequently, all E components are transverse to the direction

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

transverse components of E and H are known once we find a solution for only H_z. Neat!

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For a rectangular waveguide, the solutions for E_x, E_y, H_x, H_y, and H_z 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 n m n k m n a b π π = ⎛ ⎞ ⎛ ⎞ = _⎜ _⎟ +_⎜ _⎟ = ≠ ⎝ ⎠ ⎝ ⎠ … (11) Therefore, β β= _mn = k2 −k_{c mn}2_, (12)

These m and n indices indicate that only discrete solutions for the transverse wavenumber (k_c) 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 m= =n 0

means that k_c_,00 = 0. In (7)-(10), this implies infinite field amplitudes, which is not a physical result. Consequently, the

0

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

One exception might occur if E_z = H_z =0 since this leads to indeterminate forms in (7)-(10). However, it can be shown that

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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 [v_p ≠ g( )ω ] if the material properties of the waveguide are not functions of frequency.

To see this, recall for a TEM wave that β ω= LC . Therefore,

vp 1

LC

ω β

= =

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, v_p 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 k kc mn β β= = − and 2 2 2 , c mn m n k a b π π ⎛ ⎞ ⎛ ⎞ =_⎜ _⎟ +_⎜ _⎟ ⎝ ⎠ ⎝ ⎠

where m n, =0,1, 2,…

(

m = ≠ for TE modes, while n 0

)

, 1, 2,

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For a CW signal carried by one of these modes, the phase velocity is , ₂ ₂ , p mn c mn v k ω ω 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 m = = . Since n 0 k_c_,00 = , then 0 v_p ≠ g f( ) which means this is not a dispersive mode. However, the

0

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

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.)

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Since we prefer to work with the TEM mode of wave propagation, it is important that we use waveguides in our

microwave circuits that will support TEM or “quasi-TEM”

modes. Examples of such structures are:

• Microstrip and covered microstrip,

• Stripline,

• Slotline,

• Coplanar waveguide.

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

Before delving into microstrip, however, 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

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Lecture 12: Microstrip. ADS and LineCalc.

One of the most widely used planar microwave circuit interconnections is microstrip. These are commonly formed by a strip conductor (land) on a dielectric substrate, which is backed by a ground plane (Fig. 3.25a):

ε_r W

d 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):

ε_r

E

H

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

0 p r c v ε =