EE3104 Introduction to RF and Microwave

Systems and Circuits

Introduction

Lecturer: Prof Yeo Tat SoonConsultation location: E1-05-05

Office phone: 6516-2119E-mail: eleyeots@nus.edu.sg

1

Course Syllabus1. Transmission lines 2. Scattering parameters3. Antennas4. Amplifiers5. Mixers6. Oscillators7. Transceiver architectures8. Radar systems9. GSM/CDMA systems10.RFID systems11.Short range wireless communication systems12.Electromagnetic Compatibility (EMC)

Part 1(Components)

Part 2(Systems)

2

3

Course Syllabus

Course Syllabus

generation

modulation

amplification

radiation

propagation

reception

amplificationdemodulation4

Course OverviewContinuous Assessment Components:

• Patch antenna design (individual) 5%•Paper design (group) 5%•Two experiments (pt1) 5%• Quiz (pt1) 5%• Tutorials/Projects/Quiz (pt2): 20%• Final Exam.: 60%

5

Part 1 Lecture Schedule (Guide)Week Monday (1200-1400 @

E1-06-04)Friday* (1700-1800 @ E1-06-04)

1 (12-16 Aug) House keeping and Transmission Lines, Scattering Parameters

Introduction to EMC

2 (19-23 Aug) Scattering Parameters and Antennas

Antenna

3 (26-30 Aug) Antennas and Amplifier Tutorial 1

4 (2-6 Sept) Amplifier Class Quiz5 (9-13 Sept) E-learning week

Mixer** and Oscillator** Paper design

6 (16-20 Sept) Amplifier, Discussion on Quiz, Tutorials 2 and 3

Tutorial 4

7 (23-27 Sept) Mid-term Break (Re-test, Revision, if necessary)

Mid-term Break6

Part 1 Patch Antenna• A printed antenna around 2.5 GHz (different center

frequencies will be assigned individually) designed usingcommercial EM software. One of the design will beselected for fabrication, and the S parameters of theprototype will be measured at the Microwave laboratory.

• The parameters of the PCB are listed as follows:• Material: Rogers RO4003• Substrate thickness: 0.8 mm• Dielectric constant: 3.38• Metal thickness: 35 μm. (1 oz.)

• The designated frequency shall start from 2.0 GHz andincrease by 20 MHz according to class roll

• Submit the final design online.7

8

Part 1 Paper Design• Each group of five (5) students will be assigned to do a

paper design of a particular RF front-end (e.g., RFID,synthetic aperture radar, mobile phone) with certainspecifications/requirements.

• The students need to source the internet for componentsto build up the particular front-end.

• A short report (not more than 5 pages, including figures,etc) is to be submitted after the mid-term break. Thereport shall briefly discussed how you think theassembled components shall meet the specifications.

• Please note:- even if you copy from the internet, do notcopy words for words. I will run a check and if foundcopied words for words, the whole team will get 0 mark.

Part 1 Experiments• Familiarization on VNA Calibration and S-parameter

Measurement• Two-tone Third-order Inter-Modulation Distortion IMD3

Measurement

9

• D. M. Pozar, Microwave Engineering, 3rd Ed., Wiley, 2005.• C.A. Balanis, Antenna Theory: Analysis and Design, 3rd Edition

Recommended Text Books

10

Outline

A. Introduction

B. Transmission lines

C. Transmission line theory

11

For efficient point-to-point transmission, the source energy must be guided. Note: Plan wave propagation also follow transmission line characteristics –the “guide” is just of infinite extent.Evolution of guided structure:1) Simplest – two-wire line (improvement – twisted wires)2) Coaxial cables – lower loss3) Semi-rigid coaxial cables – much lower loss4) Metallic waveguides – very low loss, high power capacity

1. Transmission Lines

Coaxial line- Support a TEM wave;- High bandwidth, convenient fortest;- Not suitable for IC/MMIC

Waveguide-High power-handling capability-Low loss-Bulky and expensive

12

1. Transmission Lines

TEM modes can only exist in two-conductor waveguides such astwo-wire transmission lines, co-axial lines, parallel-platewaveguides, etc, but not in single-conductor waveguides such asrectangular waveguides and circular waveguides. A closedconductor (such as a rectangular waveguide) cannot support TEMwaves since the corresponding static potential would be zero (orpossibly a constant).

Waveguide theory is more difficult to handle than coaxial lines.

New type of transmission lines – planar, easy to analyse, easy tointegrate with printed circuit, light weight, low cost, easy tofabricate, but low power handling capacity.

13

Types of Planar Transmission Lines

14

–Basic structure: one dielectric substrate with conducting metals on both or either sides

• Main advantages– simple and inexpensive to fabricate by printed circuit techniques –easy integration with other passive and active devices for microwave integrated circuit (MIC) and Monolithic microwave integrated circuit (MMIC)–small size & light weight

• Major disadvantages–low efficiency & low power–also, losses and coupling

Hybrid MIC

Why Planar Transmission Lines

MMIC 15

Stripline vs Microstrip LineStripline Microstrip line

This is made by sandwiching two boards together. All fields are within material of εr .

Hence velocity ~ .rc /

In microstrips the fields are partly in air.

So for wide lines, the fields are almost all within dielectric, while narrower lines will have proportionally more field energy in air.Wide lines: rcv /

16

• For finite-width (W) microstrip lines, this leads to the idea of an equivalent permittivity

• Effective dielectric constant can be interpreted as the dielectric constant of a homogenous medium that replaces the air and dielectric regions of the microstrip, as shown Fig. (c) below

Note: z-axis referred as to “longitudinal direction”xy-plane referred as to “transverse plane”, transverse to the direction of wave propagation

Equivalent Dielectric Constant

Wide line: εe => εrNarrow line: εe => (εr + 1)/2

17

• The microstrip line cannot support a pure TEM wave, since the phase velocity of TEM fields in the dielectric region and in the air would be different. In most applications, the dielectric substrate is electrically very thin (h<<λ) and so the fields are quasi-TEM.• Phase velocity and propagation constant can be expressed as

Quasi-TEM

18

• Given the dimensions of the microstrip line, characteristic

impedance can be calculated as

For a fixed thickness (h) of the selected substrate,

(a)wide line (large W): large εe & small Z0;

(b)narrow line (small W): small εe & large Z0

Why??

Characteristic Impedance

There is no need to memorize these equations. Can be found in any handbook. If needed in exam, they will be given.

19

• For given Z0 and εr, W/h ratio can be found

where

Microstrip-Line Synthesis

There is no need to memorize these equations. Can be found in any handbook. If needed in exam, they will be given.

20

Calculate width and length of a microstrip line with Z0=50 Ω and 90 degree phase shift (φ)

at 2.5 GHz using the substrate with h = 0.05’’ and εr = 2.20.

Solution We first find W/h for Z0= 50 Ω, and initially guess that W/h > 2.

, B = 7.985

W/h = 3.081 (otherwise we would use the expression for W/h < 2)

For line length L, for a 90 degree phase shift

Example 1Too m

uch iterations and computations,

not much theoretical insights, best leave

it to computer softw

are.

21

Transmission-line analysis if circuit lengths comparable with λ

Schematic for TL of short length (Δz → 0) Lumped-element equivalent circuit

Kirchhoff’s voltage law:

Kirchhoff’s current law:

Dividing the above two equations by and taking the limit as gives:

Transmission Line Theory (re-visited)

0),(),(),(),(

tzzv

ttzizLtzziRtzv

0),(),(),(),(

tzzit

tzzvzCtzzzvGtzi

z 0z

ttzvCtzGv

ztzi

ttziLtzRi

ztzv

),(),(),(

),(),(),(This slide is for self-reading/self interest only.

22

For the sinusoidal steady-state condition, i.e.,

Then

Wave equations on TL simplified as

Complex propagation constant

Solutions of the Wave equations

Then

tjV(z)et) v(z, )(),( tjezItzi

0)()( ,0)()( 22

22

2

2

zIdz

zIdzVdz

zVd

))(( CjGLjRj

zz

zz

eIeIzI

eVeVzV

00

00

)(

)(

CjGLjR

IV

IVZ

0

0

0

00

Transmission Line Theory (re-visited)

)()()( ),()()( zVCjGdz

zdIzILjRdz

zdV

)()(][)()(

0000 zILjReVeV

dzeVeVd

dzzdV zz

zz

][)( 00zz eVeV

LjRzI

Characteristic Impedance

This slide is for self-reading/self interest only.

23

24

Any doubt?

CjGLjRZ

0

1) R = G = 0 => lossless. But isn’t R = 1/G, thus R = 0 => G = ∞ and vice versa?

2) Lossless line => Zo = √(L/C) is real. But shouldn’t real load (e.g., a resistor) lossy?

Lossless TL (R=G=0, α=0, no attenuation)

Characteristic Impedance

Solutions of the Wave equations

Converting back to the time domain, the voltage waveform:

where is the phase angle of the complex voltage

Phase velocity: a fixed phase point on the wave travels.

Wavelength: two successive maxima on the wave at a fixed instant of time.

jLCjj

zjzj

zjzj

eIeIzI

eVeVzV

00

00

)(

)(

CL

IV

IVZ

0

0

0

00

Transmission Line Theory (re-visited)

zjzj

zjzj

eZVe

ZVzI

eVeVzV

0

0

0

0

00

)(

)(

)cos(||)cos(||),( 00 ztVztVtzv

0V

dtdzvp

/2 ,2)]([][ ztzt

You are expected to know the information in this slide.

25

Terminated lossless TL

Forward (+z):Reverse (-z):

0000

0000

, ,

, ,

IVeIeV

IVeIeVzjzj

zjzj

)tan()tan(

)sin)(cos()sin)(cos()sin)(cos()sin)(cos()(

0

00

00

000 ljZZ

ljZZZljlZZljlZZljlZZljlZZZlZ

L

L

LL

LLin

||1||1

min

max

L

L

VVSWR

Transmission Line Theory (re-visited)

zjzj

zjzj

eZVe

ZVzI

eVeVzV

0

0

0

0

00

)(

)(

000

00

)0()0( Z

VVVV

IVZL

00

00 V

ZZZZV

L

L

ZZZ-Z/

0L

0L00L(V) VVReflection Coefficient at z=0:

)()()(

)()()(

0

0

0

0

0

0

00

00

zjL

zjzjzj

zjL

zjzjzj

eeZVe

VVe

ZVzI

eeVeVVeVzV

zjL

lj

zjL

lj

in eeeeZ

lIlVlZ

0)()()(

)( 2L

0

0 ljlj

lj

eeVeVl

26

What is ГL(I)= I-o/I+o ?

Special cases: Terminated lossless TL

Short circuit

Open circuit

Matched load

lj

in

L

eljZZlz

SWRz 2

0

L

);tan(;@

;1;0 Z,0@

lj

in

L

eljZZlz

SWRz 2

0

L

);cot(;@

;1; Z,0@

0;;@1;0; Z,0@

0

0L

ZZlzSWRZz

in

L

Questions: At high frequency, can a “opened” circuit be “open circuit’? How short can be considered a “short circuit”?

Special Cases

ZZZ-Z

0L

0L)(L VL

||1||1

L

LSWR

)( 2L

ljel

)tan()tan()(

0

00 ljZZ

ljZZZlZL

Lin

er transformimpedance wave-Quarter , Z,4/ If20

inLZ

Zl

27

calculate ΓL, SWR, Zin

9.527.69)6.0tan()2040(75)6.0tan(75)2040(75

6.03.02)(

05.2345.01345.01

11

)(

345.07.116

31.4075)2040(75)2040( )( 39.140

87.9

26.150

0

0

jjjjjZ

lc

SWRb

ee

ejj

ZZZZ

a

in

jj

j

L

LL

o

o

o

Example 2

28

Power Flow and Return loss

Now consider the time-average power flow along the line at the point z:

zjzj

zjzj

eZVe

ZVzI

eVeVzV

0

0

0

0

00

)(

)(

||1Re||21])()(Re[

21 222*

0

20*

zjzjav ee

ZVzIzVP

Thus smission.power tran no imaginary,purely ),Im(2 that Note * AjAA

)||1(||21 2

0

20

ZVPav

Incident power: Reflected power: 0

20

2||

ZV

2

0

20 ||

2||

ZV

29

)()()(

)()()(

0

0

0

0

0

0

00

00

zjL

zjzjzj

zjL

zjzjzj

eeZVe

VVe

ZVzI

eeVeVVeVzV

30

Power Flow and Return lossWhen the load is mismatched, then not all of the available power from the generator is delivered to the load. This loss is called return loss (RL), and is defined (in dB) as:

dB ||log20 RL

Notes:1) RL is defined as a positive value (for lossy termination) as Г < 1.2) RL, although is called a return loss, is not power (thus its unit is dB and not

dBm or dBW) but a power ratio = reflected power/incident power.3) In dB terms, 10log(Preflected) = 10log(Pin) – RL, if the powers are in dBm, then

Preflected (dBm) = Pin (dBm) – RL (dB).4) In other words, the higher the value will indicate a better matched network.

Insertion loss

The voltage for z<0 is 0z ),()( 0 zjzj eeVzV

No reflection for z>0, the voltage for z>0 is outgoing only and can be written as

Equating these voltages at z=0 gives the transmission coefficient, T, as0z ,)( 0 zjTeVzV

zjzj

zjzj

eZVe

ZVzI

eVeVzV

0

0

0

0

00

)(

)(

01

1

01

01 211ZZ

ZZZZZT

31

32

Insertion losstransmission coefficient between two pints in a circuit is often expressed in dB as the insertion loss, IL,

dB ||log20 TIL

Notes:1) IL is defined as a positive value (for lossy termination) as T < 1.2) IL, although is called a insertion loss, is not power (thus its unit is dB and

not dBm or dBW).3) It other words, the lower the value will indicate the better power transfer.

Case #1 – Matched load

We have ZL = Zo and Zin = Rin= Zo, and Xin = 0.

Then

0L

XXRR

RV

ginging

inP

22

2

21

XRZZV

ggog

o22

2

21

Zg

ZoVg

No standing wave

Special Cases

33

Zg

ZL

Vg

Case #2(a) – Matched generator

Zg = Zo

We have Z’in (looking into line from load side) = Zo.

Where Zo

ZLVL

VVV ooL

VZZ

ZZV o

oL

oLo

eeZZ

ZVV

LjLjing

ingo

1

Note: Standing wave – single reflection

Special Cases

34

ZL

VgZg

Case #2(b) – Matched generator

Zg = Zin

Half of the power delivered by the source goes into the transmission line without reflection at the input end.

Where Zg

ZinVg

Note: Standing wave – multiple reflections

XXRR

RV

ginging

inP

22

2

21

)(421

22

2

XR

RV

gg

gg

Special Cases

35

Zg

ZL

Vg

Case #3 – Matched generator – Conjugate match

This is the case for maximum power transfer.

0

Rin

P 0)(21

)()()()( 22 222 XXRR

RRRXXRR gingingingin

ginin

0)( 222 XXRR gining

0X in

P

0)(2)()( 22 2

XXRR

XXX

gingin

ginin

0)( 2 XXX ginin

Solving, Rg = Rin , Xg = - Xin , or Zg = Z*in

Special Cases

36

Note: Standing wave – multiple reflectionsMaximum power !!

XXRR

RV

ginging

inP

22

2

21

RV

gg 4

121 2

Special Cases

37

But then why is it that, in earlier years, the impression was matched load gives maximum power transfer?Hints: This is because, at that time, Zg is always assumed to be equal to Zo. Insert Zg = Rg = Zo and Xg = 0 into the result from slide 32 (repeated below) and you will get the same result as above.

XRZZV

ggog

o22

2

21

RVg

g 41

21 2

Special Cases

ZL = Zo No reflection

Zg = Zo Single reflection

Zg = Zin Multiple reflections

Conjugate match

Multiple reflections

ZL = Zg = Zo No reflectionSpecial case of conjugate match

XRZRV

ggog

in22

2

21

)(421

22

2

XRRV

gg

gg

RVo

g 41

21 2

XRZRV

ininog

in22

2

21

38