Post on 09-Feb-2016
description
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
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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)
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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%
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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
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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
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• D. M. Pozar, Microwave Engineering, 3rd Ed., Wiley, 2005.• C.A. Balanis, Antenna Theory: Analysis and Design, 3rd Edition
Recommended Text Books
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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
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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.
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Types of Planar Transmission Lines
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–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 /
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• 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
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• 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
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• 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.
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• 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
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