Lecture 1
마이크로파공학(1)
판서용 강의록
교재: Microwave Engineering, by Pozar
홍익대학교 전자전기공학부
오이석
2018년 1학기
Welcome to this class
^^
Ver. 1: 2018.03.02
캔버스에 아크릴, 2013.02.
1
Microwave Engineering (1)
Electromagnetics:
Engineering
Mathematics:
(Vector Analysis)
Coordinate Systems
Experimental Laws:
Coulomb’s Law
Ampere’s Law
Theories:
Gauss’s Law
Potential Theory
Differentiation,
Integral in space
Gradient, Divergence,
Curl operator
line, surface Integrals
Divergence Theorem,
Stoke’s Theorem
Concepts:
Electric Fields,
Magnetic Fields
Electric Flux,
Magnetic Flux
Densities
Electromagnetic Fields:
Voltage  Conduction Current : R
Voltage  Charge storage : C
Current  Magnetic Flux Storage : L
Maxwell’s equations
(time-varying)
Faraday’s Law
(Lenz’s law)
Displacement
current
Boundary
Conditions
Maxwell’s equations (statics)
Electric
Circuits(1):
Waves:
Wave Propagation
Wave Reflection
Wave Refraction
Transmission Line
Theory
Wave
Guidance
Ch. 1, 2
Resistor,
Capacitor,
Inductors,
Voltage,
Current
TIME DOMAIN
SPACE DOMAIN
Microwave Engineering (1):
Electromagnetic Theory
Transm.
Line
Theory
Ch.2
Electric
Circuits(2):
Ch.1
Wave Guidance:
Two plates, rectangular,
circular Waveguides,
Coaxial lines, Microstrip
Ch.3
Network
Theory
Microwave Network Theory
(S-parameters)
Ch.4
Impedance Matching
2017.03.02, 오이석 교수
Ch.5
Microwave Engineering (2)
Microwave Passive Circuits:
Microwave Resonators
Power dividers, Hybrids,
Filters, Circulators, etc.
Laser Eng.
Electro-Optics,
Fiber Optics
Ch. 6, 7, 8, 9
Microwave Active Circuits:
Detectors, Mixers,
Oscillator, Amplifiers
(Wireless communication)
Ch. 10,11,12,13
Antenna
Design
Radiation,
Antennas
Electronic Circuits: Diodes, Transistors
2
Ch. 1 Electromagnetic Theory
Lecture 2
1.1 Introduction to Microwave Engineering
Why do we study Microwave Engineering?
Because, microwave circuits are core parts of the wireless
communication systems such as mobile communication,
satellite communication, and radar systems.
An Example of Wireless Communication Systems:
Mixer
-. Mobile phones
-. Satellite comm.
-. Radars
Microwave
Modulator
(signal with
information)
D.C.
Power
Oscillators
Microwave Circuits
Antenna
Amplifier
Waveguides,
Transmission
lines
Filters,
Matching circuits,
Power dividers,
Resonators, etc.
3
Microwaves:
Frequencies: 0.3 GHz ~ 300 GHz
(wavelengths: 1 m ~ 1 mm; λ=c/f, c=3x108 m/s)
TV(UHF; 14~ ): 473~806 MHz (analog) 470~698 MHz (digital)
4
Why Microwaves?
• Antenna size ∝ λ (wavelength)


• Information capacity (data speed) ∝ BW (bandwidth)  frequency → the higher, the better.

• Radar target detectability ∝ 1/λ

Windows
XCL P
5
Short History
-. 1785: Coulomb (French): Electrical Force between two charges
 Coulomb’s Law: :
-. 1800: Volta (Italian): Developed the first electric battery.
-. 1820: Oersted (Danish): interconnection between electricity and
magnetism (compass needle)
-. 1831: Faraday (English): Built the first electric generator.
A changing magnetic field  an electromotive force (voltage source)
-. 1864: Maxwell (Scottish): Proposed that
a changing electric  (induces) a magnetic field.
-. 1873: Maxwell’s equations
 (1885-1887: modern form (vectors) by Heaviside)
-. 1887-1891: Experiments (Fig. 1.2) by Herz (German)
-. 1888: Tesla: Invented the ac electric motor.
-. 1896: Marconi (Italian): Radio telegraphy across Atlantic Ocean.
-. 1905: Albert Einstein (1879~1955): Adopted the quantum concept of
energy (E=mc2)
(bridged between the classical and modern eras of electromagnetics.)
-. 1935: Watson-Watt (Scottish): invented radar
6
1.2 Maxwell’s Equations
∂B
∇× E = −
−M
∂t
∂D
∇×=
+J
H
∂t
∇⋅D =ρ
E = E ( x, y , z , t )
∇ ⋅ B =0
Symbols
Descriptions
E:
Electric Field Intensity
H:
Magnetic Field Intensity
D:
Electric Flux Density
B:
Magnetic Flux Density
J:
Electric Current Density
M:
Magnetic Current Density
ρ:
electric charge density
Units
Related
V/m

V (V)
A/m

I (A)
E, D=εE
C/m2 
T (Wb/m2)  H, B=µH
I (A)
A/m2 
V/m2 (fictitious)
C/m3
E = −∇V
D=εE
ε:
Permittivity
F/m
 C (F)
B=µH
µ:
Permeability
H/m
 L (H)
(in free space)
µ0=4π x 10-7 H/m
ε0 =8.854 x10-12 F/m,
v = c, velocity of EM wave, c2 = 1/ε0 µ0 , c =3x 108 m/s (light velocity)
7
∂ρ
∂t
∇ ⋅ M =0
Continuity equation: ∇ ⋅ J = −
No free magnetic charge:
∂
∇⋅D +∇⋅ J
∂t
∂
∇ ⋅∇× E = 0 = − ∇ ⋅ B − ∇ ⋅ M
∂t
∇ ⋅ ∇ × H = 0=
→ ∇ ⋅ D= ρ
→ ∇⋅B = 0
}
: dependent eq.
E = E ( x, y , z , t )
Assuming time-harmonics,
ε (r , t ) = Re{E (r )e jωt }
Function of position r
Function of
jω t

e
∂
∂

{ E ( r ) e jω t } =  E ( r )
∂t
∂t

time (sinusoidal)

jω t
 = { jω E ( r ) e }


∂
j
ω
→



 ∂t
In phasor form,
∇ × E = − jω B − M
∇⋅D =ρ
∇ ×=
H jω D + J
∇ ⋅ B =0
E = E ( x, y , z )
8
1.3 fields in Media and Boundary Conditions
Lecture 3
-. Dielectric material
P: polarization vector (volume density of electric dipole moment)
D=εo E + Pe
( C/m2 )
where D : electric displacement, or electric flux density
εo : permittivity of free-space
Pe : polarization vector = εo χe E, where χe =electric susceptibility
D = ε o (1 + χ e )E
= ε oε r E
=ε E
Permittivity
Dielectric Constant, or relative Permittivity
-. Magnetic Material:
M: magnetization vector (volume density of magnetic dipole moment)
B=µo (H + Pm) ( C/m2 ), where B : magnetic flux density
µo : permeability of free-space
Pm : magnetization vector = χm H, where χm =magnetic susceptibility
B = µ o (1 + χ m )H = µ o µ r H = µ H (Wb / m 2 )
Permeability
relative Permeability
9
∇ × H = jωD + J
= jωεE + σE
= jωε ′E + ωε ′′E + σE
= jω (ε ′ − jε ′′) E + σE
= jωε ′E + (ωε ′′ + σ ) E
σ
= jω (ε ′ − jε ′′ − j ) E
ω
= jωε c E
εc =
ε ′(1 − j tan δ ) =
ε 0ε r (1 − j tan δ )
Microstrip substrate
ε ′′ σ
δ
+
Loss tangent: tan =
ε ′ ωε ′
Dielectric damping loss: ε ′′
Conduction loss: σ / ω
 are not distinguishable. 
δ
tan
=
′′
ε eff
σ eff
=
ε′
ωε ′
Anisotropic Media: ε or μ depends on the field direction.
For example,
Ferrimagnetic material (Ch. 9):
B = [µ ] H ,
 µ0
[ µ ] =  0
 0
0
µ
− jκ
0
jκ 
µ 
10
nˆ2 × ( E1 − E2 ) =
−M s
Boundary Conditions:
nˆ2 × ( H1 − H 2 ) =
Js
ρs
nˆ2 ⋅ ( D1 − D2 ) =
nˆ2 ⋅ ( B1 − B2 ) =
0
Derived from the integral forms of Maxwell equations.
For example, from
C
∂D
⋅ ds
H ⋅ dl = J ⋅ ds +
c
s
s ∂t
∫
∫
∫
H1
dl1 = tˆdl
n̂2
l3
∂D
∫S ∂t ⋅ ds → 0 as ∆h → 0
H2
ˆˆ
as ∆h → 0
⋅
=
⋅
H
dl
H
l
1
1∆l + H 2 ⋅ l2 ∆l
∫
C
I=
∫
s
J ⋅ ds → 0 as ∆h → 0
using
n̂
n̂
l4
∆h
l2
∆l
unless the surface current exists.
If the surface current density Js exists, then
n̂2
Therefore,
lˆ1
tˆ
dl2 = −tˆdl
n̂1
S
l1
lˆ1= nˆˆ× n2
(H
1
− H 2 ) ⋅ lˆ1∆l = J s ⋅ nˆ ∆l
→ nˆ2 × ( H1 − H 2 ) =
Js
11
(1) No conductors (Both are dielectrics)
J s = 0, M s = 0, ρ s = 0 on the boundary
nˆ2 × ( E1 − E2 ) =
0
nˆ2 × ( H1 − H 2 ) =
0
nˆ2 ⋅ ( D1 − D2 ) =
0
nˆ2 ⋅ ( B1 − B2 ) =
0
(2) Medium 2 is a electric perfect conductor (electric wall)
E=
2 H=
2 D=
2 B=
2 0
nˆ2 × E1 =
0
Ms = 0
nˆ2 × H1 =
Js
nˆ2 ⋅ D1 =
ρs
nˆ2 ⋅ B1 =
0
(3) Medium 2 is a magnetic conductor (magnetic wall)
E=
2 H=
2 D=
2 B=
2 0
ˆ
=
J s 0,=
ρs 0
−M s
n2 × E1 =
0
nˆ2 × H1 =
0
nˆ2 ⋅ D1 =
0
nˆ2 ⋅ B1 =
Radiation condition: A field cannot be infinite at an infinite distance.
12
1.4 Wave Equation and basic Plane Wave Solutions
Lecture 4
What is the EM wave?
What makes the EM wave to propagate?
Water Wave:
Snap
shot:
Wave Height
X (Spatial Displacement)
Wave Height
Float
movement:
T (Time)
(Show EM waves radiating from a horn antenna)
View of
te_horn.m,
tm_open.m,
tm_box.m
13
Assumptions:
-. Source-free region (Ji=0, M=0,
-. Time-harmonics
∇ × H = jωεE
∇ × E = − jωµH
ρv = 0 )
∇ × ∇ × E = ∇∇ ⋅ E − ∇ 2 E
∇ ⋅ E = 0,
∇ E −γ E = 0
2
∇⋅H = 0
2
Solutions
(Wave Propagation
Wave Guidance)
γ 2 = ( jωε c )( jωµ ) = −ω 2ε c µ
→ γ = jω ε c µ
γ = jω µε
where
'
1− j
= jk
σ
ωε
'
= α + jβ
γ = propagation constant
α = attenuation constant,
α = Re{γ }
β = phase constant,
β = Im{γ }
14
Plane Waves in a Lossless Medium:
d 2 Ex
dz 2
− k 2 Ex =
0
=
E x ( z ) E + e− jkz + E − e+ jkz
E x ( z , t ) = Re{E x ( z ) e jωt }
E=
( z , t ) E + cos(ωt − kz ) + E − cos(ωt + kz )
dz d ωt − constant
ω
=
=
vp =
(
)=
dt dt
k
k
1
µε
vp
2π 2π
→λ =
=
vp =
k
ω
f
15
E ( z , t ) = E0 cos(ωt − kz )xˆ
Magnitude
(source,
distance,
etc.)
Ex
Sinusoidal
Wave
Time
Variation
2π
ω=
T
T
Ex
Z-directed
propagation
2π
k=
λ
λ
Time
In a lossy medium:
E ( z, t )
Vector
(Polarization)
z, distance
=
E x ( z ) E + e −γ z + E − e + γ z
E0 e −α z cos (ω t − β z ) xˆ
16
Lecture 5
Magnetic Field:
∇× E
− jωµ
∇ × E = − jωµ H → H =
→
1 ˆ
=
H
k×E
for plane wave:
η
Intrinsic Impedance
ηc =
µ
=
ε
µ
ε ' (1 − j
σ
)
'
ωε
=
µ
ε'
1
1− j
σ
ωε '
Good Conductors: σ >> 1
ωε
γ = jω µε ' 1 − j
= j − j ω µε ′
σ
σ
≈ jω µε ' − j
ωε ′
ωε ′
σ
= ωµσ e
ωε ′
j
π
2e
−j
1
(1 + j ) = α + jβ → α = β ≅
= ωµσ
2
1
Skin depth: e −αz = e −1
→
=
δ
s
z =δ
s
α
π
ωµσ ∠
4 =
ωµσ
2
π
4
= πfµσ
17
1.5 General plane wave solution
Ex
(∇ 2 + k02 ) E = 0 (∇ 2 + k02 ) E y = 0 , E x , E y , E y : Same form
Ez
E x ( x, y , z )
∂ 2 Ex ∂ 2 Ex ∂ 2 Ex
+
+
+ k02 E x = 0
∂ x2
∂ y2
∂ z2
Let E x ( x, y, z ) = f ( x) g ( y )h( z ) (Separation of variables − Ch.3)
∂2
∂x
2
∴ gh
( f gh) = gh
d2 f
d x2
+ fh
∂2 f
∂ x2
d 2g
d y2
+ fg
d 2h
d z2
+ k02 fgh = 0
1 d 2 f 1 d 2 g 1 d 2h
+
+
+ k02 = 0
f d x2 g d y2 h d z 2
= Constant for x , let ≡ k x2
18
∴
+ xˆ
2
d f
d x2
− xˆ direction
+ k x2 f = 0 → f ( x) = Ae − jk x x + Be jk x x
choose
≡ k y2
1 ∂ 2 g 1 ∂ 2h
+
+ k 02 − k x2 = 0
g ∂ y2 h ∂ z2
∂2g
∂ y2
+ k y2 g = 0
(
→ g ( y) = C e
− jk y y
)
1 ∂ 2h
+ k 02 − k x2 − k y2 h = 0 → h ( y ) = D e − jk z z
h ∂ z2
let ,
k02 − k x2 − k y2 ≡ k z2
→∴ k x2 + k y2 + k z2 = k02
19
− jk x x − jk y y − jk z z
∴ E x = E xo e
= E xo e
e
e
k = k x xˆ + k y yˆ + k z zˆ,
 → k = kkˆ
r = x xˆ + y yˆ + z zˆ 
− j (k x x + k y y + k z z )
∴E = E0 e − jk ⋅r , E0 = E xo xˆ + E yo yˆ + E zo zˆ
Plane wave
k ⋅ r → kkˆ ⋅ r
∇ × E = − jωµH
→
k ˆ
1 ˆ
H=
k×E = k×E
ωµ
η
For Plane Wave
(TEM wave)
CD Module (Ulaby’s book)
-. EM waves
20
Lecture 6
Polarization
Time-varying behavior of E at a given point. E (0, t )
usually z = 0
E ( z ) = xˆE1 ( z ) + yˆ E2 ( z )
= xˆE x e
− jkz
+ yˆ E y e
− jkz
= E x ( xˆ +
(1) Linear polarization:
Ey
Ey
Ex
yˆ ) e − jkz
= real
Ex
ωt =
Ey
π
2
y
(2) Circular Polarization:
=±j
Ex
− jkz
E ( z ) = ( xˆ − jyˆ ) e
{
→ E ( z, t ) = Re ( xˆ − jyˆ ) e − jkz e jωt
π
ωt =
4
wt = 0
}
z
x
E (0, t ) = xˆ cos ωt + yˆ sin ωt
(3) Elliptical polarization:
RHCP
( Right-hand Circularly polarized wave )
Ey
Ey
Ex
≠ real
Ex
≠±j
21
1.6 Energy and Power
Pout
Pin
We
Wm
Poynting’s Theorem:
Ps = Po + Pl + j 2ω (Wm − We)
pl
v s
Dissipated power loss
where
1
Ps = − ∫v ( E ⋅ J s* + H * ⋅ M s )dv : Power delivered by sources
2
1
Po = ∫ E × H * ⋅ ds
: Power flow out of the surface S
2s
Pl =
σ
2
Wm =
We =
2
∫v E dv +
µ
4
ε
4
ω
2
Re ∫v H 2 dv
2
Re ∫v E dv
Power absorbed
2
2
∫v (ε ′′ E +µ ′′ H )dv : Power dissipated in V
Stored work ½ from W=1/2 (Q1V1+Q2V2)
Time averaged  ½
R
Pav = s ∫S | H |2 ds
by a good conductor:
2 o
Surface resistivity:
Rs = Re(η ) =
1
ωµ
=
2σ σδ s
22
1.7 Plane Wave Reflection
ε 0 , µ0
x
ˆ 0 e − jk0 z
Ei = xE
ε , µ ,σ
(1) H i =
Ei
η0
η0
µ0
ε0
xˆˆΓE0 e − jk0 kr ⋅r =
x ΓE0 e jk0 z ,
(2) Er =
ˆ
Et
1
(3) H r =kˆr × Er
η0
Er
(5) H t =
Using the boundary conditions,
1− Γ
η0
=
− yˆ
1
η0
Γ E0 e jk0 z
ˆ 0 e −γ z
(4) Et = xTE
z
1+ Γ = T,
E
1 ˆ
ki × Ei = yˆ 0 e − jk0 z , η0=
=
T
η
→
Γ=
TE0 −γ z
1 ˆ
kt × Et = yˆ
e
η
η
η − η0
2η
, T=
η + η0
η + η0
23
−
Power density (z<0): S = E × H
Time-average power density: =
P−
*
{
}
1
Re E × H *
2
−
+
Lossless media: P = P + P = P = P
i
r
t
Lossy media:
Perfect conductor: P + = 0
P− ≠ P+
24
Lecture 7
1.8 Oblique incidence
z
Ei
Hi
Hi
⊗
•
x
E
z
Infinite
plane
i
x
Perpendicular Polarization
Parallel Polarization
Electric field is perpendicular
to the incidence plane
Electric field is parallel
to the incidence plane
Using the Maxwell’s equations and the boundary conditions,
η2cosθi − η1cosθ t
,
Γ⊥ =
η2cosθi + η1cosθ t
2η2cosθi
Τ⊥ =
η2cosθi + η1cosθ t
η cosθ t − η1cosθi
Γ // = 2
,
η2cosθ t + η1cosθi
Τ // =
2η2cosθi
η2cosθ t + η1cosθi
25
-. Nonmagnetic media: µ=
1
µ=
µ0
2
cos θi − (ε 2 / ε1 ) − sin 2 θi
Γ⊥ =
cos θi + (ε 2 / ε1 ) − sin 2 θi
−(ε 2 / ε1 ) cos θi + (ε 2 / ε1 ) − sin 2 θi
Γ // =
(ε 2 / ε1 ) cos θi + (ε 2 / ε1 ) − sin 2 θi
26
The Brewster angle :
→ tanθ B = ε 2 / ε1
Γ0
=
//
The critical angle:
|Γ|⊥=1,Γ| |1
What happens, if
=
//
θi > θ c ?
Et , H t ~ e −αz
ε2
→ sin θ c =
ε1
e − jβ z
Attenuation along z
CD Module (Ulaby’s book)
-. EM waves
Propagation along x (interface)
“nonuniform plane wave”, Surface Wave
27
1.9 Image theory
J
air
J
Conductor
air
air
p.e.c
electric wall
M
M
air
Conductor
M
p.e.c
air
air
J
p.m.c (perfect magnetic conductor)
: magnetic wall
28
Ch.2 Transmission Line Theory
Lecture 8
Why do we study Transmission Line Theory?
Because microwave circuits are usually composed with transmission line
sections instead of lumped elements.
Examples of Microwave Circuits
Low Pass Filter
Microstrip
Transmission
lines
Using
Microstrip
lines
Matching
Circuit
29
Transmission Line Theory
Circuit Theory
Field Theory
I, V [ i(t),v(t) ]
E(r), H(r) [ E(z,t),H(z,t) ]
Z, Y (R, L, C, G)
α , β , η (ε , µ , σ )
Network Theory
(z-, y- matrices)
Reflection, Transmission
( Γ, T )
Transmission Line Theory
I(z), V(z) [ i(z,t),v(z,t) ]
Z , α , β
Microwave Network Theory
(S-parameters)
E
H
direction
Z , α , β
30
2.1 Lumped element circuit model for a transmission line
i ( z + ∆z , t)
i(z,t)
σ =∞
R
R=0
L
G
v(z,t)
C
v( z + ∆z , t)
σ =0
G=0
∆Z
∂i(z, t)
+ v( z + ∆z , t) = 0 - - - (1)
∂t
v ( z + ∆z , t ) − v ( z , t )
∂i(z, t)
⇒ - lim
= Ri ( z , t ) + L
∆z
∂t
∆z → 0
∂v(z, t)
∂i ( z , t )
- - - (2)
⇒ = Ri ( z , t ) + L
∂z
∂t
∂v(z + ∆z, t)
- i(z, t) + G∆z v(z + ∆z, t) + C∆z
+ i ( z + ∆z , t) = 0 - - - (3)
∂t
i ( z + ∆z , t ) − i ( z , t )
∂v( z + ∆z , t )
]
⇒ - lim
= lim [Gv( z + ∆z , t ) + C
∆z
∂t
∆z → 0
∆z → 0
KVL ; - v(z, t) + R∆z i(z, t) + L∆z
(1)
1
∆z
KCL ;
(3)
1
∆z
31
⇒ -
∂v( z, t )
∂i(z, t)
= Gv( z , t ) + C
- - - (4)
∂z
∂t
Compare with
∇× H = J +
∂D
∂E
= σE + ε
∂t
∂t
32
Lecture 9
Transmission Line Equations:
-
∂v(z, t)
∂i ( z , t )
∂i(z, t)
∂v( z, t )
, = Ri ( z , t ) + L
= Gv( z, t ) + C
∂z
∂t
∂z
∂t
Assuming harmonic time dependence,
v(z, t) = Re{ V(z)exp[jωt]},
dV(z)
⇒ = ( R + jωL) I ( z ),
dz
dI(z)
= (G + jωC )V ( z )
dz
−I ( z)
1
d
− [(
)
]=
( R + jω L) I ( z )
dz G + jωC dz
2
d 2 I ( z)
d
V ( z)
2
=
→
= γ 2V ( z )
I ( z ) and
γ
dz 2
dz 2
where γ =α + j β = ( R + jω L)(G + jωC )
Propagation
Phase Constant
Constant
Attenuation (rad/m)
Constant (Np/m)
i(z, t) = Re{ I(z)exp[jωt]}
Compare with
 ∇2 E − γ 2 E =
0

 ∇2 H − γ 2 H =
0

γ =α + j β =ω µε (1 − j
σ
)
ωε
33
2.2 Field Analysis of Transmission Lines
Transmission line parameters: C, L, G, R (Table 2.1)
34
2.3 Terminated Lossless Transmission Line
Load:
( another Tr. Line,
or, a resistor,
or, a microwave
circuit )
V ( z ) = V0+ e − jβ z + V0− e jβ z
dV ( z )
1
I ( z) =
− ( R + jωL) dz
0
0
γ = α + jβ , β = ω LC
(
1
V0+ (− jβ )e − jβz + V0− ( jβ )e − jβz
=
− j ωL
=
ω LC + − jβz jβ − jβz
V0 e
V0 e
−
j ωL
ωL
V0+ − jβz V0− jβz
e
e
=
−
Z0
Z0
)
C
1
ω LC
=
=
L Z0
ωL
35
V (0)
ZL =
I (0)
=
→
→
V0+ + V0−
V0+ − V0−
(Ohm' s Law)
Z , β
l
Z0
Z LV0+ − V0+ Z 0 = V0− Z 0 + Z LV0−
V0− ( Z L + Z 0 ) = V0+ ( Z L − Z 0 )
Γ ≡ voltage reflection coefficient
V0−
+ Γe
jβ z
Γ(l )
Γ
Z in − Z 
Γ(l ) ≡
= Γ e − 2 jβ l
Z in + Z 
Return Loss (RL) = - 20log10 | Γ | ( dB )
Z L − Z0
= Γ,
=
=
+
Z L + Z0
V0
V ( z ) = V0+ [e − jβz
ZL
Example > RL = 20 dB
],
→ 1%
V0+ − jβz
[e
I ( z) =
− Γ e jβ z ]
Z0
36
Lecture 10
Time-averaged Power:
V ( z ) = V0+ [e − jβz + Γe jβz ],
V0+ − jβz
− Γ e jβ z ]
I ( z) =
[e
Z0
{
1
Pav = Re V ( z ) I ( z )*
2
}
1 | V0+ |2
Re{1− | Γ |2 +Γe 2 jβz − Γ*e − 2 jβz }
=
2 Z0
A − A* = a + jb − (a − jb) = 2 jb
{
}
→ Γe 2 jβz − Γ*e − 2 jβz = 2 j Im Γ e 2 jβ z : ( pure imaginary )
(
1 | V0+ |2
1− | Γ |2
Pav =
2 Z0
)
Standing Wave Ratio:
(
| V ( z ) |=| V0+ || e − jβz | 1+ | Γ | e + jθ Γ e 2 jβz
)
=| V0+ | 1+ | Γ | e + j (θ Γ + 2 jβz )
θ Γ − 2 β l ( z = −l )
37
Vmax =| V0+ | (1+ | Γ |)
Vmin =| V0+ | (1− | Γ |)
Vmax 1+ | Γ |
=
=ρ
SWR (VSWR) =
Vmin 1− | Γ |
= Voltage Standing Wave Ratio
ZL = 0
l
1 ≤ SWR ≤ ∞
| Γ |= 0
| Γ |= 1
2 βl = 2π
Reflection at a point:
Γ(l ) =
V − ( z)
V + ( z)
=
V0− e − jβl
V0+ e jβl
⇒l=
λ
2π
2π
⇒l=
=
2π
2β
2
2
λ
= Γ(0)e − 2 jβl
Phase for a round trip
Input Impedance:
V (−l ) V0+
=
Z in =
I (−l ) V0+
(e jβl + Γe− jβl ) Z0 =  = Z0 Z L + jZ0 tan βl
Z 0 + jZ L tan βl
(e jβl − Γe− jβl )
38
Special cases:
Short
Γ=
ZL = 0
Z in = Z 0
Z L − Z0
= −1
Z L + Z0
V ( z ) = V0+ (e − jβz − e + jβz ) = −2 jV0+ sin β z
0 + jZ 0 tan βl
= jZ 0 tan βl
Z0 + j0
SWR = ∞
Z L − Z0
Γ=
=1
Z L + Z0
Open
ZL = ∞
V ( z ) = V0+ (e − jβz + e + jβz ) = 2 jV0+ cos β z
|V |
1 + j0
Z in = Z 0
= − jZ 0 cot βl
0 + j tan βl
SWR = ∞
39
Analogy
T
Z2
Z1
Γ
Γ=
1
T
Z 2 − Z1
Z 2 + Z1
T = 1+ Γ
Γ
Γ=
η 2 − η1
η 2 + η1
T = 1+ Γ
40
Lecture 11
2.4 Smith Chart
Z L − Zo zL −1
1+ Γ
Γ=
=
→ zL =
Z L + Zo zL + 1
1− Γ
1 + Γr + jΓi
→ rL + jx L =
1 − Γr − jΓi
Arranging for real and imaginary parts,
1 − Γ r2 − Γi2
2Γi
, and xL
rL =
2
2
(1 − Γ r ) + Γi
(1 − Γ r )2 + Γi2
→ (Γr −
1 2
1
1
rL 2
) + Γi2 = (
) , (Γr − 1) 2 + (Γi − ) 2 = ( ) 2
rL + 1
rL + 1
xL
xL
(1) Real Part  Equation of circles:  r
L ,0
Circles with Center,  rL + 1  and Radius,
(2) Imaginary Part  Equation of circles:
 1 
Circles with centers,  1,

 xL 
Print the Smith Chart.
smithchart.pdf
1
rL + 1
1
and radii, x
L
41
Plot circles:
From (1)
Γi
(i) rL =0 , (0,0) ,1
1
(ii) rL =1 , ( ,0) ,
1
2
2
(iii) rL =2 , ( 2 ,0) , 1
3
3
Γ =1
1
Γ ≤1
xL = 1
(iv) rL =0.5, (1/3, 0), 2/3
xL = 0.5
rL = 1
rL = ∞
rL = 2
-1
xL = 0
-1
3
0
rL = 0.5
From (2)
1
3
(i) xL=0 , (0,0) ,1
xL = −1
(ii) xL=1 , (1,1) , 1
1
(iii) xL= -1 , (1,-1) ,
1
(iv) xL=0.5, (1,2), 2
1
2
2
3
1
Γr
Γ=0
( z L = 1, Z L = Z 0 )
(Matched)
-1
42
Use of the Smith Chart
1) For a given z L = rL + x L , read Γ (Γ = Γr + Γi or Γ ∠φ )
2)
(Γ ↔ z L )
r −1 S −1
z L at real axis , z L = rL : Γ = L
=
 ⇒ S = r L ( S ≥ 1)
rL + 1  S + 1 
Example
Z L = 40 + j 30Ω , Z 0 = 50Ω Find Γ ?
(1)Read from the Smith chart
z L = 0.8 + j 0.6 Re ad Γ → Γ ≈ j 0.3
(2) by calculator ,
40 + j 30 − 50 − 1 + j 3 9 − j 3
Γ=
=
⋅
40 + j 30 + 50 9 + j 3 9 − j 3
1
− 9 + 9 + j 30
=
=j
90
3
(3) Read SWR from the Smith Chart :
SWR ≈ 2
(4) Compute the SWR :
1 + 1/3 4 / 3
SWR =
=
=2
1 - 1/3 2 / 3
xL = 0.6 1
Γi
rL = 0.8
0.33
-1
0
1
Γr
2
-1
43
3) Find the input impedance
Example:
l =0.1 λ
4) Find admittance on Smith chart
l
λ
R0 =50
Z in
Γi
1
Toward
generator
-1
ZL
4
Z in = Z L = 40 + j 30Ω
l
0 .1 λ
zL
0
1
Γr
yL
− transformer
Z L + jR0 tan β
λ
Zin
4
=
zin =
R0 R + jZ tan β λ
L
0
4
1
1
= = = yL
Z L zL
R0
π
 2π λ π


⋅
=
=
∞
,
tan


λ
4
2
2


λ
z ′L
-1
zin ( ) → yL
4
“ VSWR circle”
44
Lecture 12
2.5 Quarter-wave Transformer
Length =
λ
Z 0 (R0 )
Z1 = Z 0 RL
RL
Z1 = Rin RL
4
l=
λ
4
π
2π λ
= tan = ∞
tan β l = tan
λ 4
2
Z L + jZ 0 ∞ Z 02
Z in = Z 0
=
Z 0 + jZ L ∞ Z L
Z 0 = Z L Z in
or ,
λ
4
Z0
Z1
ZL
Z in = Z 0 for matching
Z1 = Z 0 Z L
(1) Z1 is usually real (lossless).
 Load impedance should be real !
(2) Reflection coefficient has a bandwidth.
(matched at a single frequency, because l=λ/4 only at a single frequency)
Questions:
Why do we need multi-section quarter-wave transformer?
 Answer in Sec. 5.4 - 5.6.
45
2.6 Generator and Load Mismatches
Γin
Γl
Γg
Iin
Z0 , β
Vin Zin
-l
Zl − Z 0
Γl =
Zl + Z 0
Zl
0
Z g − Z0
Γg =
Z g + Z0
V ( z ) = V0+ [e − jβz + Γl e jβz ]
Γin = Γ =
z
Z in − Z g
Z in + Z g
At the generator end (z=-l),
V ( z = −l )
+ jβ l
= V0 [e
+
Therefore, V0 = Vg
+ Γl e
− jβ l
Z in
Z in
] and V (−l ) = Vg
Z in + Z g
1
Z in + Z g e jβl + Γ e − jβl
l
by voltage
dividing rule
46
Using
(e jβl + Γl e − jβl )
1 + Γl e − j 2 βl
V (−l )
Z in =
= Z0
= Z0
j
l
β
β
−
j
l
I (−l )
(e − Γl e
)
1 − Γl e − j 2 βl
We can compute
Z in
1
1
e − jβ l
+
V0 = Vg
= Vg
j
l
j
l
β
β
−
(1 + Z g / Z in ) (1 + Γl e − j 2 βl )
Z in + Z g e
+ Γl e

= Vg
Z 0 e − jβ l
( Z 0 + Z g )(1 − Γg Γl e − j 2 βl )
Power delivered to the load: Vin = V (−l )
{ }
Z in
1
1
1
1
1
*
2
2
2
P = Re Vin I in = | Vin | Re{ } = | Vg | |
| Re{ }
Z in
2
2
2
Z in + Z g
Z in
Now let Z in = Rin + jX in
and Z g = Rg + jX g
| Z in |2
1
1
2
P = | Vg |
Re{
}
2
Rin + jX in
2
| Z in + Z g |
2
2
Rin
Rin
+ X in
1
2
= | Vg |
2
2
2
+ X in
( Rin + Rg ) 2 + ( X in + X g ) 2 Rin
47
P=
Rin
1
| V g |2
2
( Rin + Rg ) 2 + ( X in + X g ) 2
(1) Load matched to line: Z in = Z 0 = R0
R0
1
2
→ P = | Vg |
2
2
2
( R0 + Rg ) + X g
(2) Generator matched to the loaded line: Z in = Z g
Rg
1
2
→ P = | Vg |
2
2
2
4( Rg + X g )
∂P
= 0, and
(3) Maximum power transfer:
∂Rin
→ Rin = Rg , and X in = − X g
∂P
=0
∂X in
*
→ Z in = Z g
 “Conjugate Matching”
1
2 1
→ P = | Vg |
2
4 Rg
2.7 Lossy Transmission Lines (skip)
48
Lecture 13
Ch.3 Transmission Lines and Waveguides
Heaviside : 1893 
Rayleigh : 1897 
AT&T , MIT : 1936 
think “EM wave propagation in a hollow pipe.”
proved mathematically
experiment confirmation (circular waveguide)
Common Transmission Lines for microwave
-. coaxial cables – convenient (No cutoff frequency)
-. waveguides – bulky
1960’s 
microstrip line (easy fabrication)
3.1 General solution for TEM, TE and TM waves
TEM : Transverse Electromagnetic Wave → E ⊥ zˆ, H ⊥ zˆ ( E z = H z = 0)
TE : Transverse Electric → E ⊥ zˆ ( E z = 0)
TM : Transverse Magnetic → H ⊥ zˆ ( H z = 0)
(Two-wire Tr.line)
(Coaxial cable)
(General case)
(parallel plate waveguide)
(strip line)
(Circular
waveguide)
(Rectangular
waveguide)
(microstrip line)
49
Maxwell ' s Equations ( source − free)
∇ × E = − jωµH − (1)
∇ × H = jωεE
Assuming time − harmonic field with
e + j ωt
− (2)
∇ × (1) ⇒ ∇ × ∇ × E = − jωµ∇ × H
( p.700)
∇∇ ⋅ E − ∇ 2 E
(∇ ⋅ E = 0 : source − free region)
∴ ∇ 2 E + k 2 E = 0 , similarly
→ − jωµ ( jωε ) E
ω 2 µε ≡ k 2
(∇ 2 + k 2 ) H = 0
Assuming wave propagation in + ẑ direction → e − jβz
E ( x , y , z ) = E 0 ( x , y ) e − jβ z
et ( x, y ) + zˆe z ( x, y )
Transverse
electric field component
and infinitely long in ẑ − direction
z
Longitudinal
electric field component
 E x = e x ( x , y ) e − jβ z

(et ( x, y ) = ex xˆ + e y yˆ ) →  E y =
E =
 z
Similarily , H ( x, y , z ) = (ht ( x, y ) + zˆ hz ( x, y ) )e − jβz
50
From Maxwell ' s Eqn. ∇ × E = − jωµH
xˆ
∂
∂x
Ex
yˆ
∂
∂y
Ey
zˆ
∂E z ∂E y
 H x xˆ 
xˆ − comp.
−
= − jωµH x


∂
y
z
∂
∂
= − jωµ  H y yˆ  →
∂z
∂
 H zˆ 
e y ( x , y ) e − jβ z = − j β e y ( x , y ) e − jβ z = − jβ E y
z


Ez
∂z
(
)
∂E z
+ jβ E y = − jωµH x − (1)
∂y
∂E
− jβ E x − z = − jωµH y − (2)
∂x
∂E y ∂E x
−
= − jωµH z − (3)
∂x
∂y
∴
From, ∇ × H = jωεE ( refer to above eqns)
∂H z
+ jβH y = jωεE x − (4)
∂y
∂H z
− jβ H x −
= jωεE y − (5)
∂x
∂H y ∂H x
−
= jωεE z − (6)
∂x
∂y
51
Rearrange E x , E y , H x , H y as functions of E z and H z
For example from (1) & (5) , remove E y
1 ∂H z
− jβ
Hx −
jωε
jωε ∂x
∂E z
− jβ
− 1 ∂H z
(1) →
+ jβ
= − jωµH x
H x + jβ
∂y
jωε
jωε ∂x
∂E
∂H z
ε z + H x β 2 − jβ
= ω 2 µεH x
∂y
∂x
∂E
∂H z
,
H x (k 2 − β 2 ) = jωε z − jβ
∂y
∂x
(5) → E y =
∂H z 
∂E z
j 

 ,
ωε
β
−
2 
∂x 
∂y
kc 
Similarly ,
Hx =
− j
∂E z
∂H z 


ωε
β
+
2 
∂x
∂y 
kc 
− j  ∂E
∂H z 

E x = 2  β z + ωµ
∂y 
k c  ∂x
Hy =
∂E
∂H z
j 
E y = 2  − β z + ωµ
∂y
∂x
kc 
HW-4
(1) Prob. 3.1 (p.157)
2π 

k 2 = ω 2 µε ,  k = ω µε =

λ 

k 2 − β 2 ≡ kc2 ( β 2 ≡ k 2 − kc2 )
β = k 2 − kc2
in general , ε = ε 0ε r (1 − j tan δ )



If we know E z , H z
we can compute E x , E y , H x , H y .
52
TEM Waves
Transverse EM → E z = 0 , H z = 0
H x =  Ez +  H z → 0
H x =  Ez +  H z → 0
unless kc2 = 0,
Hx =
kc2
→
⇒ β 2 = k 2,
β = k = ω µε
 sin x



=
1
 x

x =0


0
0
kc = cutoff wave number
= 2πf c µε = 0 → f c = 0 ( No cutoff frequency )
2
2
 ∂2

∂
∂
2

+ k Ex = 0
+
+
Wave equation ; for E x or E y
 ∂x 2 ∂y 2 ∂z 2



∂ 2 Ex ∂ 2
=
ex ( x, y )e − jβz = (− jβ )2 E x = − β 2 E x = −k 2 E x
∂z 2
∂z 2
2 
 ∂2
∂
 E x = 0,
∴
+
∇t2 E x xˆ + E y yˆ = 0
 ∂x 2 ∂y 2 


∴ ∇t2e ( x, y ) = 0 → Find e ( x, y ) , then E = e ( x, y )e − jβz
(
ẑ
H
kc2 = k 2 − β 2 = 0,
0
Lecture 14
E
Pass band
reflection
HPF
fc
f
)
(
)
53
When Φ ( x, y ) is a scalar potential
e ( x, y ) = −∇ t Φ ( x, y )
and . ∇ ⋅ D = ε∇ t ⋅ e = 0 ( source − free)
∇ t ⋅ (−∇ t Φ ) = 0
→ ∇ t2 Φ = 0
( Laplace equation)
Find Φ ( x, y ) solving the Laplace equation.
For example,
y
V0
y=d
y=0
d 2Φ
dx
2
+
d 2Φ
dy
2
=0
d 2Φ
dy
2
x
∂
= 0 ( No variation in x − direction )
∂x
= 0 → Φ = ay + b
V
V
B.C ⇒ y = d → Φ = ad + b = V0 → a = 0 → Φ = 0 y
d
d
y = 0 →Φ = b = 0
54
V0
d  V0 
e ( x, y )(= e y 0 ( y ) yˆ ) = −  y  yˆ = − yˆ
dy  d 
d
V0 − jβz
E ( x, y, z ) = − e yˆ
d
H ( x, y, z ) = h ( x, y )e − jβz ,
Z TEM =
Ex
Hy
where h ( x, y ) =
V0
V0
E = − yˆ
d
1
zˆ × e ( x, y )
Z TEM
( wave inpedance)
0
E x ωµ
∂E z
ωµ
From (2) − jβE x −
= − jωµH y →
=
=
=
β ω µε
Hy
∂x
∴ Z TEM
µ
=
=η
ε
− Ey

 or Z TEM =
=

Hx

(Intrinsic Impedance)
µ
ε
µ
ε




(Characteristic Impedance)
1
V12 = − ∫ E ⋅ dl
2
I = ∫ H ⋅ dl
c
,
Z0 =
V
I
55
TE Waves (H-waves)
Transverse Electric waves ( E ⊥ zˆ ⇒ E z = 0) and H z ≠ 0
− jβ ∂H z
− jβ ∂H z
Hy =
Hx =
,
,
kc2 ≠ 0 ( H z ≠ 0)
2 ∂x
2 ∂y
kc
kc
Ex =
− jωµ ∂H z
,
2
∂y
kc
Ey =
jωµ ∂H z
kc2 ∂x
kc2 = k 2 − β 2
→ propagation constant : β = k 2 − kc2
∴ Find H z at first . Then compute H x , H y , E x , E y ( E, H )
using wave equation
(∇ 2 + k 2 )H z = 0 , H z = h z ( x, y )e − jβz
2
2
2
 ∂2

∂
∂
∂
−
β
2
j
z

H z = −β 2 H z
+ k  h z ( x , y )e
= 0, where
+
+

 ∂x 2 ∂y 2 ∂z 2
∂z 2


2
2
 ∂2

 ∂2

∂
∂
−
β
2
2
2
j
z



→
+
− β + k hz e
=0 →
+
+ kc hz = 0, where kc2 ≡ k 2 − β 2
 ∂x 2 ∂y 2

 ∂x 2 ∂y 2





S olve this wave equation for a given geometry → hz ( x, y ) → H z (r ) → E , H
Wave Impedance :
E y ωµ
E
∂E z
+ jβE y = − jωµH x − (1) → ZTE = x = −
=
β
∂y
Hy
Hx
=
ωµ
k 2 − kc2
ωµ = ω µε
µ
= kη
ε
56
TM Waves (E-waves)
Transverse Magnetic wave ( H ⊥ zˆ ⇒ H z = 0) and E z ≠ 0
− jωε ∂E z
jωε ∂E z
,
,
Hy =
Hx =
2 ∂y
2
∂
x
kc
kc
Ex =
− jβ ∂E z
,
2 ∂x
kc
Ey =
− jβ ∂E z
kc2 ∂y
Lecture 15
kc2 ≠ 0
kc2 = k 2 − β 2
→ propagation constant :
β = k 2 − kc2
2
2
 ∂2
 − jβ z
∂
∂
2

∴ Find Ez at first ! (∇ + k ) E z = 0 →
+
+
+ k ez e
=0
 ∂x 2 ∂y 2 ∂z 2



2
2
 ∂

∂
2

+ kc e z = 0 → Find e z → E z = e z ( x, y )e − jβz → E , H
+

 ∂x 2 ∂y 2


Wave Impedance :
2
2
k 2 − kc2
Ex
∂H z
β
Using
+ jβH y = jωεE x − (4) → ZTM =
=
=
H y ωε
∂y
ωε
ε k
=
ωε = ω εµ
µ η
If we find kc → β
βη 


 ZTM =
k 

→ ZTM
when we have loss → γ = α + jβ
Boundary condition
( For conducting surface, nˆ2 × E1 = 0)
Boundaries
E, H
(Etan = 0)
57
Attenuation Constant
α = α d + α c ( Np / m)
(i) α c → conduction loss
P ( z ) = P0 e −2α c z
dielectric loss
conduction loss
→ Pl = −
P
∂P
= 2α c P ( z ) → α c = l , (Perturbation Method, Sec. 2.7)
∂z
2 P0
where
air
1
Re E × H * ⋅ ds , assuming lossless
S
2
Pl = Pt (transmitted into lossy medium conductor) (Sec.1.6)
∫
P0 =
1
Re
2
1
= Re
2
=
∫S
∫S
E × H * ⋅ ds =
ηH ⋅ H *ds =



where Rs = Re{η } = Re


=
σd
σc
P0
conductor
Pl
1
1
Re ( E × H * ) ⋅ zˆds = Re ( zˆ × E ) ⋅ H *ds,
S
S
2
2
R
R
1
Re(η ) | H |2 ds = s | H |2 ds = s | J s |2 ds ( W / m 2 )
S
2
2 S
2 S




 µ

µ
σ
1
1



≈
>> 1
Re
,
assuming



ε
ωε
σ
 ε − j σc 
1− j c 

ωε 
ωε 
∫
∫
∫
ωµ
ωµ
1
Re(
)=
, because
σc
2σ c
−j
∫
1
−j
= ∠450 =
∫
1
2
+j
1
2
58
(ii) α d → dielectric loss
γ = jβ = j k 2 − kc2 ,
if
k 2 − kc2 = real → α d = 0
σ
If k (complex) = ω µε 1 − j d
ωε
→ k 2 − kc2 = complex ,
then γ = j k 2 − kc2 = α d + jβ
σ 

γ = j ω 2 µε 1 − j d  − kc2 = j ω 2 µε (1 − j tan δ ) − kc2 ,
ωε 

σ
where ω 2 µε = k 2 (real ), tan δ = d
ωε
= j (k
2
− kc2 ) −
jk tan δ = j k
2
2
− kc2
2

k
1 − j tan δ
2
2

−
k
k
c





59
1
x, where | x |<< 1, using the Taylor series,
2
1
1
f ' (0)
f ′′(0) 2
f ( x) ≈ f (0) +
x+
x + , f ' ( x) = (1 − x) −1 / 2 (-1) → f ' (0) = − }
1
2!
2
2
2
2
2


δ
β
δ
1
tan
tan
tan δ
k
k
k
 = jβ − ( j )( j )
γ ≈ jβ 1 − j
= jβ +
= jβ + α d
2
2


2 β
2β
2β


{(1 − x)1 / 2 ≈ 1 −
k 2 tan δ
( Np / m);
→α d =
2β
TEM waves ; β = k
→ αd =
k
tan δ ( Np / m)
2
60
Lecture 16
3.2 Parallel plate Waveguide
Support TEM waves (two conductors)
Also support TM and TE waves.
D.C → f c = 0 (kc = 0)
( No cutoff freq.)
Fringing fields
(difficult to analyze with these)
 Need numerical computation (computer)
 TEM mode
For your computation
ignore the fringing fields.
(easy)
(difficult)
TEM modes:
Find E, H using Laplace equation
for the electrostatic potential Φ ( x, y )
V0
d
0
 Φ ( x,0) = 0
∇t Φ ( x, y ) = 0, B.V . 
Φ ( x, d ) = V0
No variation in x direction →
∴
d 2Φ ( y )
dy
2
∂Φ
=0
∂x
= 0 → Φ ( y ) = Ay + B
61
Φ (0) = B = 0,
V
Φ( y) = 0 y
d
Φ (d ) = Ad = V0
⇒
⇒
e ( x, y ) = −∇t Φ ( y ) = −
V
E ( x, y, z ) = e ( x, y ) e − jβz = − yˆ 0 e − jβz ,
d
1
xˆ V0 − jkz
H=
zˆ × E =
e
( A / m)
Z TEM
η d
 π 
V
 E   = yˆ 0
d
 k
 π 
V0
ˆ
=
−
H
x



ηd
 k
V
E (0) = − yˆ 0
d
V
H (0) = xˆ 0
ηd
β = k for TEM modes
y
at kz = −π
at z = 0
V
∂Φ
∂Φ
xˆ −
yˆ = − 0 yˆ
∂x
∂y
d
x
z
V2
P
V21 = V2 − V1 = − ∫P2 E ⋅ dl
1
V1
d
d
y =0
y =0
Voltage : V = − ∫ E ⋅ dl = − ∫
−
V0 − jβz
e ⋅ dy = V0 e − jβz (V )
d
62
Current : on top plate
nˆ2 × H1 = J s
H = xˆ
nˆ2 × H1 = J S
− yˆ × xˆ = zˆ
yˆ × xˆ = − zˆ

V0 − jkz 
 J s = zˆ e 
ηd


ηd
e − jkz
H (r , t ) = xˆ
V0
ηd
cos(ωt − kz )
y
y
JS
V0
kz = 0
x x x x x
kz =
x
z
π
2
x
kz = π
C3
C2
x x x x x
C4
dl1 = dx xˆ
C1
I = ∫C + ∫C + ∫C + ∫C
1
2
3
4
w V0 − jkz
= ∫0
e
dx =
ηd
H = xˆ
V0
ηd
wV0 − jkz
( A)
e
ηd
Top plate
e
− jkz
H ⋅ dl1 =
V0
ηd
H ⋅ dl2 = 0
e − jkz dx
H ⋅ dl3 = 0
(assuming no fringing field)
63
Characteristic Impedance for this two − plate waveguide
V0 e − jkz
ηd
d
V
=
=η
Z0 = =
wV0 − jkz
w
w
I
e
ηd
Material
Geometry
parameter parameter
Phase velocity
vp =
ω
ω
1
=
=
β ω µε
µε
=
1
µ 0 µ r ε 0ε r
=
C
µrε r
,
C = 3 × 108 m / s
β = ω µε
L (cm)
d
l ( in λ )
w
Z0 = η
Z0 , β
η=
< Waveguide for TEM >
< Transmission Line >
For TM and TE modes,
we usually don’t call it “transmission line”
we call it “waveguide”
d
W
µ
ε
64
Lecture 17
(3.2 Parallel plate Waveguide)
TM modes
Transverse magnetic waves ( H z = 0) , H ⊥ zˆ,
2
2
(∇ + k ) E z = 0,
H = H x xˆ + H y yˆ ,
where E z ( x, y, z ) = ez ( x, y )e
− jβ z
,
2
 ∂2

∂
2
2
2
2

+
+ k c  e z ( x , y ) e − jβ z = 0
k
−
≡
k
β
c
 ∂x 2 ∂y 2



≠ 0 (not always )
∂2
∂z 2
Find E z
E z = ( − jβ ) 2 E z = − β 2 E z
( β = k 2 − kc2 )
Assume,
No fringing fields
⇒ No variation of e z in x̂ − direction
⇒
 d2
2

∴  2 + kc  ez ( x, y ) = 0
 dy

∂
ez = 0
∂x
(O.D.E ) Ordinary Differential Eq.
ez ( x, y ) = A cos(kc y ) + B sin( kc y )
( general solution)
Standing wave in ŷ − direction
(propagating in ẑ − direction )
why not e ± jk c y form ?
65
nˆ2 × ( H1 − H 2 ) = J S → can not use
Find (A, B), k c using B.C
nˆ2 × ( E1 − E2 ) = 0
y
nˆ2 × E1 = 0
E2 = 0
n̂2
n̂2
⇒ Etan = 0 → etan = 0
etan = ex or ez
E1
E2 = 0
→ use this
 ez ( x, y = 0) = 0 − (1)
B.C ∴ 
 ez ( x, y = d ) = 0 − (2)
x
(1) → ez ( x,0) = a + 0 = 0 → A = 0
(2) → ez ( x, d ) = B sin kc d = 0 → B ≠ 0 , sin kc d = 0
→ kc d = nπ ,
nπ
kc =
d
n = 0,1, 2, 3 , ,
nπ
y
d
nπ
y e − jβ n z ,
∴ E z ( x, y, z ) = An sin
d
∴ ez ( x, y ) = B sin
2
 nπ 
β n = k 2 − kc2 = ω 2εµ −   ,
 d 
kc =
nπ
(For TM n mode)
d
66
jωε ∂E z jωε
=
An cos kc y e − jβ n z
Hx =
kc
kc2 ∂y
− jωε ∂E z
=0
Hy =
2
∂x
kc
Ex =
− jβ ∂E z
=0
2 ∂x
kc
− jβ ∂E z − jβ
=
An cos kc y e − jβ n z
kc2 ∂y
kc
if n = 0 , → E z = 0 → TEM mode
Ey =
, n = 1, 2, 3, 
(TM 0 → TEM )
if k < kc → β n = ± j kc2 − k 2 ≡ ± jα → − jα
e − jβ n z = e − j ( ± jα ) z = e ±αz → e −αz (attenuating )
Therefore, if k < kc ( f < f c ) → evanescent wave (no propagation)
2πf c µε
→ fc =
→
n
2d µε
fc =
kc
2π µε
=
nπ
d
2π µε
for TM n − mode
67
TM 1 → f c =
Example
f
TM 1
c
εr = 4
2d µε
TEM
TM1
TM2
TEM
TM1
f cTM 2 = 30 GHz
f (GHz )
0
D.C ( static )
15
30
45
βn
1
k 2 − kc2
, =
ωε
ωε
 real if k > kc
= 
 imaginary if k < kc ( f < f c )
vp =
0.5 cm
3 ×108
3 ×108
=
=
= 15 GHz
0.02
2 × 0.005 × 4
Only
TEM
ZTM =
1
No propagation
(stored energy)
(standing wave)
ω
ω
ω
=
> =v
k
β
k 2 − kc2
λg =
fc =
2π
β
=
1
2d µε
2π
k 2 − kc2
=
v
2d
>
2π
=λ
k
→
λc =
v
= 2d
fc
λ 

d = c 
2

68
Lecture 18
Time-averaged Power flow:
{
}
y
1
P0 = Re ∫S E × H * ⋅ ds
2
n̂
zˆ × xˆ = + yˆ 

1
w d


= Re ∫0 ∫0 ( E y yˆ + E z zˆ ) × H x xˆ ⋅ zˆ dx dy
ˆ
ˆ
y
⋅
z
=
0


2

− jωε *
1  w d − jβ
= − Re ∫0 ∫0
An cos kc y e − jβ n z ⋅
An cos kc y e + jβ n z ⋅ zˆ dx dy 
2 
kc
kc

{
=
=
ωε
2kc2
ωε
4kc2
[
{
d
Wd | An | Re{ β } , n ≥ 1
E z = A1 sin
2
π
d
ye
}
]
W Re ∫0 β | An |2 cos 2 kc y dy
− jβ1 z
,
s
x
}
cos 2
π 
where β1 = k 2 −  
d 
nπ
1 1
2nπ
y = + cos
,
d
2 2
d
y
2
π 
 jπ y
j
y
−
1  d
d

e
−e

2j 


 j  π y − β z  − j  π y + β z  
1
1
A1   d
d


 = E + E
e
e
Ez =
−
z1
z2

2j 

 2 plane waves
∫ ( period ) = 0
d
0
θ
θ
z
v p = v(cos θ ) −1,
v g = v cos θ ,
v pvg = v 2
69
(3.2 Parallel plate Waveguide)
TE modes E ⊥ ẑ (E = 0)
z
Find H z from H z = hz ( x, y )e − jβz and (∇ 2 + k 2 ) H z = 0,
∂
=0
No fringing field →
∂x
 d2

2

+ kc hz ( y ) = 0
hz ( x, y ) → hz ( y )
 dy 2



H z = hz e − jβz
E x = 0 , E z = 0 on conductor surface
hz ( x, y ) = A sin kc y + B cos kc y
Boundary conditions :
− jωµ ∂H z − jωµ
( A cos kc y − B sin kc y )e − jβz
=
Ex =
∂y
kc
kc2
E x ( y = 0) = 0 → A = 0
E x ( y = d ) = 0 → sin kc d = 0
nπ
y e − jβ n z
H z ( x, y, z ) = Bn cos
d
→ kc =
nπ
,
d
n = 1, 2, 3 ...
( n=0? )
70
if n = 0 , Ex , H y = 0
jωµ
nπ
Bn sin
y e − jβ n z
→ Ex =
kc
d
→ E = 0 ( No power flow)
No TE0 − mode (∴ n = 1,2,3,...)
jωµ ∂H z
Ey =
=0
2 ∂x
kc
Hx =
− jβ ∂H z
=0
2 ∂x
kc
y
nπ
− jβ ∂H z + jβ
Bn sin
y e − jβ n
=
kc
d
kc2 ∂y
for TE n − modes
Hx =
 nπ
βn = k 2 − 
 d
n
fc =
2d µε
2

 ,

same with TM n − mode
,
E x ωµ
TE n
=
ZTE =
H y βn
*
1
P0 = Re ∫s E × H ⋅ ds
2
''
d
x
0z
eˆ × hˆ = zˆ
 E = E y yˆ
TEM mode : 
 H = H x xˆ
 E = E y yˆ + E z zˆ
TM - modes : 
 H = H x xˆ
 E = E x xˆ
TE - modes : 
 H = H y yˆ + H z zˆ

71
TM1 − mode
y
E z = An sin
Ey =
β
kc
π
d
y cos(ωt − β1z )
An cos
π
d
ωt = 0, E z = An sin
d
y sin(ωt − β1z )
π
d
y cos β1z → max at y =
d
2
0
x
z
π
−β
Ey =
An cos y sin β1z → max at y = 0, d
kc
d
TE1 − mode
ωµ
π
Bn sin y sin β1z
kc
d
β
π
H y = Bn sin y sin β1z
kc
d
π
H z = Bn cos y cos β1z
d
Hx =
π
+ ωε
An cos y sin β1 z
kc
d
Re( j (− j )) = −1
− sin( − βz ) = sin β
d

 sign change at y = 
2

sign change at y
Ex =
max at y =
d
2
d

+ y <
2
max at y = 0, d 
 − y > d
2

Refer to Fig. 3.5 (p. 106)
y
d
x
0
x x
x
x
x
x
x
x
z
72
Lecture 19
3.3 Rectangular Waveguide
y
Find E and H inside
No TEM (no statics )
b
(single conductor)
0
0
z
a
x
(∇ + K ) H z (r ) = 0 → (
2
H z = h z e − jβ z
E z = 0, use H z ,
TE-modes
2
∂2
∂x 2
Separation of variables
hz ( x, y ) = X ( x )Y ( y ) → Y
+
∂2 X
∂x 2
∂2
∂y 2
+X
+ k c 2 )hz ( x, y ) = 0, P.D.E.
∂ 2Y
∂y 2
+ k c 2 XY = 0,
1
XY
1 ∂ 2 X 1 ∂ 2Y
+
+ kc 2 = 0
X ∂x 2 Y ∂y 2
2
→
let
k
Constant for X(x)
x
73
1 d2X
d 2Y
2
+ K c = 0 and
+ ( K c 2 − K x 2 )Y = 0
X dx 2
dy 2
k y 2 → k c2 = k x 2 + k y 2
Y ( y ) = C cos k y y + D sin k y y
X ( x) = A cos k x x + B sin k x x
where
Then,
hz ( x, y ) = X ( x)Y ( y ) = ( A cos k x x + B sin k x x)(C cos k y y + D sin k y y )
H z ( r ) = h z ( x , y ) e − jβ z
B.C :
nˆ2 × ( E1 − E2 ) = 0 → use this
nˆ2 × ( H1 − H 2 ) = J s
(can' t use)
E2 = 0
xˆ × E1 = 0 →
E1
E1 y = 0, at x = 0 plane
n̂
74
E x ( x, y ) = 0 at y = 0, b
ex ( x,0) = ex ( x, b) = 0
E y ( x, y ) = 0 at x = 0, a
e y (0, y ) = e y (a, y ) = 0
Find E x , E y
− jωµ ∂H z
= 0 → k y ( A cos k x x + B sin k x x) ⋅ (−C sin k y y + D cos k y y ) = 0
Ex =
2
∂y
kc
Ey =
jωµ ∂H z
= 0 → k x (− A sin k x x + B cos k x x) ⋅ (C cos k y y + D sin k y y ) = 0
2 ∂x
kc
Applying B.C;
e x ( x,0) = 0
→ − C ⋅ 0 + D ⋅1 = 0 → D = 0
nπ
, n = 0, 1, 2,.....
e x ( x, b) = 0 → sin k y b = 0 → k y b = nπ → k y =
b
e y (0, y ) = 0 → − A ⋅ 0 + B ⋅ 1 = 0 → B = 0
e y (a, y ) = 0
→ sin k x a = 0
→ k x a = mπ → k x =
mπ
, m = 0,1,2,3.....
a
75
∴ H z (r ) = hz ( xy )e − jβz = Amn cos
β mn = k 2 − k 2 mn ,
mπ
nπ
x cos
y e − jβ mn z
a
b
kc 2 mn
2
 mπ   nπ 
=

 +
 a   b 
2
76
Find E x , E y , H x , H y , from H z
− jωµ ∂H z + jωµ nπ
mπ
nπ − jβz
ye
Ex =
=
Amn cos
x sin
2
2
b
∂y
b
a
kc
kc
jωµ ∂H z − jωµ mπ
mπ
nπ − jβz
=
Ey = 2
Amn sin
x cos
ye
2
kc
a
a
b
kc ∂x
Hx =
Lecture 20
kc → kcmn
β → β mn
− jβ ∂H z + jβ nπ
mπ
nπ − jβz
sin
cos
=
A
x
ye
mn
2
2
kc a
a
b
kc ∂x
− jβ ∂H z + jβ nπ
nπ − jβz
mπ
= 2
x sin
ye
Hy = 2
Amn cos
kc b
a
b
kc ∂y
 mπ  2  nπ  2 
2
β mn = k 2 − k cmn
= k 2 − 
 
 +
 a   b  
k > k c ⇒ propagation
k < k c ⇒ Evanescent
77
fc
Dominant mode : lowest
2πf cmn µε = kc → f cmn =
 mπ   nπ 

 +

 a   b 
2
1
2π µε
2
∂H z ∂H z
m = 0, n = 0 →
=
= 0 → E = 0 ? (null field ) → No TE00
∂x
∂y
If a > b , then m=1, n=0
→ TE10
TE10 mode
1 1
 < 
a b
π 
→ β10 = k 2 −  
a
a
π
2π π
1
fc =
,
=
→ λc = 2a
λc a
2π µε a
kc =
π
(dominant mode)
Guided Wavelength :
ω
ω
vp = =
β
k 2 − kc2
2
λg =
2π
β
> vp =
ω
k
=
2π
k 2 − kc2
>λ =
2π
( planewave)
k
( planewave)
78
Example
For an X-band WG
Appendix I: Standard Rectangular Waveguide Data (p.688)
X-band WG: a=2.286 cm, b=1.016 cm
π
3 × 108
=
= 6.56 GHz
f cTE10 =
2π µε a 2 ⋅ 0.02286
1
fc
TE10
= 6.56,
f cTE
20
= 13.12
 Next cutoff
frequency
∴ use 6.56 < f < 13.12 for the dominant-mode transmission
→ X − band ( 8 ≤ f ≤ 12GHz )
79
Waveguide sets
80
Lecture 21
Attenuation
for the Dominant mode
TE10
E = E x xˆ + E y yˆ + E z zˆ
E z = 0, E x = 0, H y = 0
H = H x xˆ + H y yˆ + H z zˆ
E y ≠ 0, H x ≠ 0, H y ≠ 0
E y = A10
Assuming
π
− jωµ π
sin x e − jβz
a
kc2 a
π
jβ π
sin x e − jβz ,
H x = A10
a
kc2 a
lossless WG
H z = A10 cos
π
a
xe
− jβ z
Lossy 
e − α z e − jβ z
α = αc + αd
(1)
αc =
Pl
2P10
where
1
Re ∫0a ∫0b E × H ∗ ⋅ zˆ dxdy
=
P10
2
1
= Re ∫0a ∫0b E y H *x dydx
2
81
R
2
Pl = s ∫c J s dl
2
nˆ 2 × ( H1 − H 2 ) = J s
H2
=0
c1
c4
nˆ 2 H1 n̂2
H 2 = 0 c2
c3
C1
C4
C2
C3
∫c = ∫c1 + ∫c 2 + ∫c3 + ∫c 4
→ J s = nˆ 2 × H
xˆ × H = − yˆ H z x = 0 = J s
yˆ × H = yˆ × ( xˆH x + zˆH z ) = − zˆH x + xˆH z y = 0 = J s
2
Rs b
R
2
2
Pl=
J sy dy + s 2 ∫xa 0 { J sx + J sz }dx
2 ∫ y 0=
2
2
k 2 tan δ
( 2) α d =
( N p / m) ( In air , α d = 0)
2β
82
Separation of variables
TM-modes
B.C
(
∂2
∂x 2
+
∂2
∂y 2
+ k c 2 )e z = 0 → e z ( x, y ) = Amn sin
mπ
nπ
x sin
y
a
b
TM 00 , TM 01,TM 01 → e z = 0 → No such modes
π 
2
π 
Lowest f c → TM 11 → k cTM
,11 =   +  
a
b
2
> k cTE
,10
83
Table 3.2: Summary (p.113)
Figure 3.9: Field distributions (p.114)
84
3.4 Circular Waveguides
Lecture 22
Find E , H In the W.G
x
z
(I) Use cylindrical coordinate.
( ρˆ , φˆ, zˆ )
(II) Solve wave equations
E, H
ρ
TE - mode , use
φ
y
TM - mode , use
∂H z 
− j  ∂E z
β
,
+ ωµ
Eρ =
2 
ρ∂φ 
kc  ∂ρ
∂H z 
− j  ∂E z
β
 ,
− ωµ
Eφ =
2  ρ∂φ
∂ρ 
kc 
H z = h z ( ρ , φ ) e − jβ z
E z = e z ( ρ , φ ) e − jβ z
j 
 ωε
2
kc 
− j
 ωε
Hφ =
2 
kc 
Hρ =
∂E z
∂H z 
 ,
−β
ρ∂φ
∂ρ 
∂E z
∂H z 

+β
∂ρ
ρ∂φ 
TE – modes:
(∇ 2 + k 2 ) H z = 0 → (∇ t 2 + kc2 )hz ( ρ ,φ ) = 0
kc2 = k 2 − β 2
∂hz
1 ∂
1 ∂ 2 hz
(ρ
)+
+ kc2 hz = 0
∂ρ
ρ ∂ρ
ρ 2 ∂φ 2
β = k2 − β 2
85
Separation of variables hz ( ρ , φ ) = R ( ρ ) P (φ )
∂R
∂2P
1 ∂
1
+ k c2 RP = 0
(ρ
)+
P
R
∂ρ
ρ ∂ρ
ρ 2 ∂φ 2
ρ2
RP
∂
∂R
1
1 ∂2P
ρ
+ k c2 ρ 2 = 0
(ρ
)+
R ∂ρ
P ∂φ 2
∂ρ
 separated
1 ∂
∂R
( ρ ) + kc2 ρ 2 ≡ kφ2
ρ
R ∂ρ ∂ρ
(constant for φ)
Let
d 2P
+ kφ 2 P = 0 → Find P(φ )
dφ
d
dR
ρ
(ρ
) + (k c2 ρ 2 − kφ2 ) R = 0 → Find R( ρ )
dρ
dρ
Then,
hz ( ρ )
d 2R
dR
or ρ
+ρ
+
2
dρ
dρ
2
86
d 2P
dφ 2
+ kφ2 P = 0 → P (φ ) = A sin kφ φ + B cos kφ φ
kφ = n,
(Integer) because of the periodicity in
→ P (φ ) = A sin nφ + B cos nφ
φ
sin 0.2(φ ) ≠ sin 0.2(2π + φ )
sin 2(φ ) = sin 2(2π + φ )
2
d
R
dR
2
+ρ
+ ( k c2 ρ 2 − n 2 ) R = 0 : Bessel’s Equation of order n
ρ
dρ
dρ
 We don’t have a closed-form solution for this equation!
gives R ( ρ ) = CJ ( k ρ ) + DY ( k ρ )
n c
n c
1st kind
2 nd kind Bessel functions of order n.
87
p. 684 (App. C) Fig. C.1 
Yn (0) = −∞
J 0 (0) = 1, J1 (0) = 0
Yn (kc ρ ) → −∞ at ρ = 0 is not acceptable in this case.
∴ hz ( ρ ,φ ) = ( A sin nφ + B cos nφ ) J n (kc ρ ) ; General solution
H z ( r ) = h z ( ρ , φ ) e − jβ z
88
Lecture 23
Apply B.C
Etan = 0 on surface → Eφ = 0 at ρ = a
From Maxwell’s Eqns ∇ × E = − jωµ H , ∇ × H = jωε E
⇒ Eφ =
Eφ =
→
jωµ
kc2
( A sin φ + B cos φ )
− j  β ∂E z
∂H z

− ωµ
∂ρ
kc2  ρ ∂φ
 jωµ ∂H z
 =
 kc2 ∂ρ
d ( J n (kc ρ )) − jβz
⋅e
= 0 at ρ = a
dρ
d ( J n (kc ρ ))
= 0 at ρ = a
dρ
Using the chain rule, d ( J n (k c ρ )) d J n (k c ρ ) d (kc ρ )
=
⋅
= kc J n' (kc ρ )
dρ
d (kc ρ )
dρ
Eφ ( ρ = a ) = 0 → J n' (kc a ) = 0 → gives many roots
89
First root for
J 0 (k c a )
1.0
′
J 0' (kc a ) → k ca = 3.832 → p01
′ = 7.016
2 nd root → p02
kc a
J 0' (k c a )
′
pnm
: m-th root of n-th order Bessel function for TE mode
J1′ (k c a )
J1 (k c a )
1.841
st
1 root
2 nd root
5.331
kc a
→ P11' = 1.841
P12' = 5.331
90
'
Find kc such that J n' ( kc a) = 0, ( pnm
= kc a ) → kcnm
where β = k
f cnm =
1
2π
2
− kc2
'
pnm
=
a
2


 , k = 2πf µε

= k −
c
c
 a 


2
'
pnm
′
pnm
µε a
Dominant mode , TE11 , because
'
p11
= 1.841 is the lowest root
We can find E ρ , H ρ , and H φ
91
TM-modes
(∇ 2 + k 2 ) E z = 0 → (∇ t 2 + kc2 )e z ( ρ ,φ ) = 0
∴ E z ( ρ ,φ ) = ( A sin nφ + B cos nφ ) J n (kc ρ ) e − jβz ; General solution
E tan = 0 on surface → E z = 0 at ρ = a
→ J n (kc a ) = 0 → gives many roots
→ P01 = 2.405
92
Review for Mid-Term Exam
Lecture 24
Review of EM theory
-. Course description
-. Maxwell’s equation, Wave equation
-. EM Wave Propagation, EM Wave Reflection
Review of transmission line theory
-. Equivalent Circuit Model: R, L, G, C
-. Voltage and Current Wave
-. Smith Chart
Transmission lines and waveguides
-. TEM, TE, TM waves
-. Parallel plate waveguide
-. EM Fields (Wave equation +B.C.)
-. Modes
-. Rectangular waveguide
-. EM Fields (Wave equation +B.C.)
-. Modes
-. Circular Waveguides
Q&A
93
Mid-Term Exam
중간고사 일정표
과목명
마이크로파공학(1)
시험일
(요일)
시간
4/22
오후7시
(화)
-8시30분
장소
P202, P203
시험
범위
1.1-3.4
94
Lecture 25
3.5 Coaxial line
y
ρ
v0
a
z
(i) TEM-mode can exist  Dominant mode
Use Static field analysis (Laplace’s Equation)
∇ t2 Φ ( ρ ) = 0
φ
x
b
(ii) Solve Laplace eqn in cylindical coord.
dΦ
1 d  dΦ ( ρ ) 
 = 0 because
 ρ
=0
dρ 
ρ dρ 
dφ
→ρ
dΦ
= C1 → Φ = C1 ln ρ + C 2 ( general sol )
dρ
(iii ) B.C Φ ( ρ = a ) = V0
Φ ( ρ = b) = 0
ln
ρ
V
∴ Φ ( ρ ) = 0 (ln ρ − ln b) = V0 b
a
a
ln
ln
b
b
→ c1 ln a + c2 = V0
c1 ln b + c2 = 0
a
c1 (ln ) =
b
V0 → c2 = −c1 ln b
95
1


V
0
V0
∂  (ln ρ − ln b) 
ρ
V0
 = − ρˆ
= ρˆ
(iv) et ( ρ , φ ) = −∇Φ ( ρ ) = − ρˆ
b
a
a
∂ρ 

ρ ln
ln
ln


a
b
b


(v) E ( ρ , φ , z ) = ρˆ
H (r ) =
1
η
V0
b
ρ ln
a
e − jβ z
zˆ × et ⋅ e − jβz = φˆ
Time-varying fields
V0
ηρ ln
b
a
e − jβ z
Static field
TEM- mode
Higher order modes exist!
For TE-mode, solve, (∇ t2 + k c2 ) hz ( ρ , φ ) = 0 , k c ≠ 0
ρ =0
Separation of variables
hz ( ρ , φ ) = ( A sin nφ + B cos nφ )(CJ n (kc ρ ) + DYn (kc ρ ))
NO concern
Apply B.C Eφ ( ρ , φ ) = 0 at ρ = a and b ⇒ gives transcendental eqn.
(3.159)
p.128
96
3.6 Surface Waves on a Grounded Dielectric Slab.
Consider a Grounded Dielectric Slab
TM Modes
In air, the surface wave exists:
x
Air
d
Dielectric
0
εo
ε 0ε r
Let k02 − β 2 = (± jα )2 ≡ −h 2
E ZI
E ZII
e−α x e− j β z
In dielectric:
z
Let=
kc2 ε r k02 − β 2
In air, (i) by the Radiation Condition
In the dielectric, (ii ) ez ( =
x 0)= 0
II
=
(iii ) E z I E=
z at x d
I
II
=
(iv) H tan
H=
tan at x d
After mathematical manipulation,
(kc d ) tan(=
kc d ) ε r (hd ) − − − (1)
(kc d )2 + (hd=
)2 ( ε r − 1k0 d )2 − − − (2)
97
Find kc d and hd from (1) & (2)
h>0 using B.C(R.C)
Graphically
hd
 1 
y = 
x  (tan x )
 εr 
x2 + y2 = r 2
where
y ≡ hd , x ≡ k c d ,
r = ε r − 1 k0 d
ε r − 1k0 d → nπ
k0 = 2πf c
fc =
h0
π
2
ε r − 1k0 d
for TM n
k0 = kc
(3)
kc d
π
3
π
2
kc d
(4)
ε 0 µ0
nπ
2π ε 0 µ 0
nC
=
ε r − 1d 2 ε r − 1d
n=0,1,2...
TM0 mode exists, but kc can not be zero. Instead, kc  0.
98
Lecture 26
3.7 Stripline
Dominant mode: TEM (no longitudinal fields E z = H z = 0)
y
TEM waves : v p =
Ground plane
ε
b
c
ω
ω
1
=
=
=
,
β ω εµ
ε 0 µ0 ε r µr
εr
x
c=
w
z
center conductor
1
ε 0 µ0
εr 1
1
L
LC
=
=
=
C
C
v pC
c C
z0 (characteristic Imp.)
β = k0 ε r , Z 0 =
Compute C approximately  then obtain
Computation of Capacitance
(1) Conformal mapping 
d
⇒C =ε
A
d

(2) Approximate Electrostatic Solution
(p. 140: Skip)
∇ t2Φ = 0 + B.C → Φ → E → ρ s = nˆ ⋅ ( D1 − D2 ) → Q → C =
Q
→ Z0
V
(3) Empirical formula
where We = W for W > 0.35b
30π
b
,
Z0 =
⋅
W
ε r We + 0.441b
W − b(0.35 − ) 2 , for W < 0.35b
b
99
Most popular for microwave circuits
3.8 Microstrip
y
d
w
εr
x
z
TEM-mode?
No Because , not homogenous between conductors; air + dielectric
in air → v p = c
in dielectric → v p =
Assume
c
Phase-match is impossible → hybrid TM-TE wave
εr
d << λ →" quasi − TEM " (Very similar with TEM mode)
ε e : effective dielectric constant
εe =
ε r +1 ε r −1
2
+
TEM – mode with
vp =
c
εe
2
εe
1
d
1 + 12
w
, β = k0 ε e
1 < εe < εr
ε0
ε 0ε r
equivalent
εe
(TEM - mode)
100
Characteristic Imp.
Z0
(1) Conformal mapping:
A+
A-
B
C
C
D
C
D
A+
B
Z0 =
(2) Approximate Electrostatic solution
∇ 2Φ = 0
+ B.C
Φ →V → E
A-
1
v pC
Conducting side wall (B.C)
ρs → Q
V
εe
Q
1
C = → Z0 =
=
V
v pC c C
(3) Empirical formulae (easy, quite accurate)
 60  8d
w
+
ln

 for w ≤ d

 ε e  w 4d 
Zo = 
−1
120
w
w
π



⋅
+ 1.393 + 0.667 ln( + 1.444)
 ε  d
d

e

for w ≥ d
101
Sometimes, need to find W for given Z 0 , d , ε e
 8e A
, w < 2d

W  e2 A − 2
=
d  2
ε r −1 
0.61
[ B − 1 − ln(2 B − 1) +
ln( B − 1) + 0.39 −
,
 π
2ε r 
εr 
Z0 ε r + 1 ε r −1
0.11
where A =
+
(0.23 +
),
60
2
εr +1
εr
w > 2d
B=
377π
2Z 0 ε r
3.9 Transverse Resonance Technique (skip).
3.10 Wave velocities and Dispersion (Skip)
ω
ω
=
: phase velocity
2
2
β
k − kc
1
group velocity : v g =
dβ
dω
vp =
3.11 summary (skip)
102
Lecture 27
Ch .4 Microwave Network Analysis
4.1 Impedance and Equivalent voltages and currents
Using Maxwell’s Equation → gives exact solution ( E (r ), H (r )) (Ch.1)
But, for complicate geometry →impossible, or very difficult to solve
Therefore, use transmission line theory for microwave circuits (Ch.2)
v + ( z) + v − ( z)
E, H
Med. 1
η2 T
E1 = E i + E r η1 Γ
E2 = E t
η −η
Γ= 2 1
η 2 + η1
Impedances;
- Intrinsic Impedance:
η=
→ v + ( z)
Med. 2
µ
ε
I
← v − ( z)
⇔ z
01
η ↔ Z0
for planewave
Γ
z02
→1
←
Γ=
→ T
Z 02 − Z 01
Z 02 + Z 01
β
k
ZTEM = η , ZTE = η , ZTM = η
β
k
R + j ωL
(=
)
G + j ωC
103
Ey
Ex
=−
;
- Wave Impedance: Z w =
Hy
Hx
- Characteristic Impedance: Z o = V
V(Z),I(Z)
Microwave circuits; Dimension is large comparing with
λ
Phase change is not negligible
→ use distributed circuit
Transmission lines (Ch. 2)
Use Network Analysis
Low Pass Filter
Port: Input/output through ports
I1+
I +, I −
Z in
→I
+
V
−
V + ,V −
→
← I1−
One –port
Network.
E, H
← I 2+
→ I1
+
Two –port
V1 Network.
−
+
E, H
V1 →
← V1−
← I2
+
V
−2
I 2− →
← V2 +
V2 − →
104
4.2 Impedance and Admittance matrices
Electric circuits
Vi
Z ij =
Ij
I k = 0 for k ≠ j
Example;
V1 = Z11I1 + Z12 I 2 + ...... + Z1n I n
Z12 =
V1
V2
where I1 = I 3 = I 4 ..... = I n = 0
Vn = Vn+ + Vn−
I n = I n+ + I n−
105
In Network Analysis,
Impedance matrix ;
[V ] = [Z ][I ]
Admittance matrix ; [Y ] = [Z ]−1
[I ] = [Y ][V ]
Reciprocal Networks ; Z ij = Z ji → symmetric matrices
Lossless Networks; Z ij , Yij → Pure Imaginary
(Yij = Y ji )
106
Lecture 28
4.3 The Scattering Matrix (S-parameters)
[Z ][Y ] describe the network
Microwave circuits;
Voltage waves →
incident waves
reflected waves
transmitted waves
for N − ports, ( with currents, voltages )
Figure 4.7 (p. 175)
Coaxial
cable
#2
#1
→ Incident
← reflection
DUT
→ Tran
Network analyzer measures S-parameters (scattering matrix)
V1−   S11
 − 
V2  =  S 21
    
 − 
VN   S N 1
S12
S 22

S N 22
 S1N  V1+ 
 
 S 21  V2+ 

   
 + 
 S NN  VN 
[V ] = [S ][V ]
−
∴ S ij =
+
Vi−
V j+
Vk+ =0, for k ≠ j
107
For example,
V1− = S11V +1 + S12V2 + +  + S1N VN
+
→ S12 =
V1−
V2+ when V + =0,V + =V + =0
1
3
N
S12 = T ( port 2 to port1) when Γ1 = 0
← V1
−
→ V2−
Γ = 0 → V1+ = 0
V1+ = 0 means " matched at port 1"
Sii = reflection cofficient at port , ( when all other ports are matched )
Sij = transmission coefficient from port j to port i
( when all other ports are matched )
108
Example 4.4 (p. 179)
Find s-parameters
8.56Ω
#1 8.56Ω
+
Z0
141.8Ω
−
8.56Ω
Z0
and
V1−   S11
sol.  −  = 
V2   S 21
+
V1
Γ(1)
#2
V2
Z0
V1− = S11V1+ + S12V2 +
−
→ V1+
← V2+
← V1−
→ V2−
8.56Ω
141.8Ω
S12  V1+ 
 
S 22  V2+ 
V2− = S 21V1+ + S 22V2 +
Z0 =
8.56 + 141.8 //(8.56 + Z 0 )
(8.56 + Z 0 )141.8
8.56
=
+
Z0
Z 0 + 8.56 + 141.8
→ ( Z 0 − 8.56)( Z 0 + 8.56 + 141.8) = (8.56 + Z 0 )141.8
Z 02 8.56(8.56 + 141.8 + 141.8)
→=
= 2500
50 Ω
→ Z0 =
(1)
(2)
Zin
=
Zin
=
8.56 + 141.8 //(8.56 + 50)
=
8.56 + 41.44 =
50 Ω
(1)
V1−
Zin
− Z0
S11 =
=
Γ(1) V + =0 =
+
(1)
2
V1 V + =0
Zin
+ Z0
2
=
50 − 50
= 0 = S 22 (by symmetry )
50 + 50
109
S21 =
V2−
V1+ V + =0
2
V2 = V2+ + V2− = V2−
V
V1− = 0 ( S11 = 0) ⇒ V1 = V1+ + V1− = V1+ ⇒ S 21 = 2
V1
8.56Ω
+
V1 → V1+
- ← V1−
V2 = (V2− =)V1
8.56Ω
141.8Ω
+
V2
-
50
→ V2−
← V2+
41.44
50
⋅
8.56 + 41.44 8.56 + 50
V2− = (0.8288)(0.8538) = 0.707 = S 21 (= S12 , by symmetry )
0.707 
 0
∴ [S ] = 

0
.
707
0


1
3dB attenuator ↔ P2 = P1
2
-3dB
110
Lecture 29
4.3 The Scattering Matrix (-continued-)
[S]  [Z]
Consider the normalized impedance, Zon=1
[V ] = [ Z ][ I ] = [ Z ][V + ] − [ Z ][V − ] = [V + ] + [V − ]
→ ([ Z ] + [U ])[V − ] = ([ Z ] − [U ])[V + ],
+
−
V=
n Vn + Vn
I n = I n+ − I n− = Vn+ − Vn−
( Z on = 1)
[U ] : unit matrix, or identity matrix
→ ([ Z ] + [U ]) −1 ([ Z ] + [U ])[V − ] = ([ Z ] + [U ]) −1 ([ Z ] − [U ])[V + ]
→ [V − ] = ([ Z ] + [U ]) −1 ([ Z ] − [U ])[V + ] ↔ [V − ] = [ S ][V + ]
 We can compute [S] from [Z]
→ [ S ] = ([ Z ] + [U ]) −1 ([ Z ] − [U ])
([ Z ] + [U ])[ S ] = ([ Z ] − [U ]) → [ Z ][ S ] + [U ][ S ] = [ Z ] − [U ]
→ [ Z ]([U ] − [ S ]) = [ S ] + [U ]
→ [ Z ] = ([ S ] + [U ])([U ] − [ S ]) −1
Table 4.2 (p. 192)
 We can compute [Z] from [S]
 Useful conversions
[S ]
A
C

B
D

[Z ]
[Y ]
111
Reciprocal Network;
[Z ]
: Symmetric for a reciprocal network
[ Z ] = symmetric ; zij = z ji → [ Z ]t = [ Z ], t : transpose
[S ] =
([ Z ] + [U ])−1 ([ Z ] − [U ])
→ [ S ]t = ([ Z ] − [U ])t {([ Z ] + [U ])−1}t = ([ Z ] − [U ])([ Z ] + [U ])−1 = (4.47) p.181
→ [ S ] [ S ]t , because =
A]t , [U ] [U ]t , =
([ A][ B])t [ B]t [=
and [ Z ] [ Z ]t
 [S]: Symmetric for reciprocal networks
Lossless Network;
[ Z ], [Y ] = pure imaginary
1
Pav = Re{[V ]t [ I ]*}
2
I* 
V1 
 I1 
*
*
t *
1 =
Two-port network  [V ] =
,[
]
[
]
[
]
I
=
→
V
I
=
V
V
V
I
+
V
I
[
]


1
2
1
1
2
2
V2 
 I 2 
*
 I 2 
1
Pav = Re{([V + ]t + [V − ]t )([V + ]* − [V − ]* )}
2
1
Re{([V + ]t [V + ]* − [V + ]t [V − ]* + [V − ]t [V + ]* − [V − ]t [V − ]* )}
2
112
Pav
1
Re{([V + ]t [V + ]* − [V − ]t [V − ]* )},
2
because − [V + ]t [V − ]* + [V − ]t [V + ]* ⇒ − A + A* = − j 2 Im( A)
For a lossless network:
[V + ]t [V + ]* = [V − ]t [V − ]* (input power = output power)
0 for lossless networks
→ Pav =
Using [V − ] = [ S ][V + ]
→ [V + ]t [V + ]* = [V − ]t [V − ]* = ([ S ][V + ])t ([ S ][V + ])* = [V + ]t [ S ]t [ S ]*[V + ]*
Therefore,
[ S ]t [ S ]* = [U ]
([ S ]* = {[ S ]t }−1 )
[S]: Unitary matrix for lossless networks
N
1 if
S ki S kj ∗ = δ ij = 
0 if

k =1
∑
i= j
i≠ j
*
*
 s11 s21 s31 ....  s11 s 12
 s12
  s*
s
  21
 13
  s* 31
 .
 .
 .

 .
 .

 .
s*13 .... 1 0 0 ....


 0 1 0


 0 0 1


= .

 .





 .

113
Example 4.5 (p. 183)
 0.15∠0
=
S
[ ] 

0.85∠45
0.85∠ − 45 

0.2∠0 
S12 ≠ S21 ⇒ [ S ]:not symmetric
→ (i) Network is not reciprocal
Network is lossless ? No, because
N
2
2
2
∑ Sk1 = 0.15 + 0.85 = 0.745 ≠ 1 ⇒ (ii) Network is lossy.
k =1
(iii) Return loss when port #2 is matched
V1−
V2+
+
+
−
V1 = S11V1 + S12V2 → Γ ≡
= S11 + S12
+
V1
V1+
(iv) Port #2: short
If
+
V=
0, =
Γ S=
2
11 0.15
Return Loss RL= − 20log10 Γ =
16.5dB
 0.15
0.85∠ − 450 


0
0.2
0.85∠45

Short
V2+ + V2− ==
V2 0 → V2+ =
−V2−
V2−=
Γ =
S21V1+
V1−
V1+
Re turn
+ S22V2+=
S21V1+
− S22V2−
→ V2− (1 + S22 )=
S21V1+
S21
≠ S11 , not matched ⇒ Γ 1 ≠ S11
1 + S22
Loss RL =
6.9dB (Γ =
−20log10 Γ =
−0.452)
V2−
S21
→ =
V1+ 1 + S22
= S11 − S12
114
4.3 The Scattering Matrix (-continued-)
A shift in Reference planes
+
'+
V
1
V
1
V1'−
Lecture 30
V1−
[S]
[S’]
Vn'+
Vn+
Vn'−
Vn−
[V ] = [S ][V ] [V − ] = [S ][V + ]
'−
'
'+
'
S nn
= e −2 jθ n S nn
; phase of S nn → shifted by e − 2 jθ n
→ travel twice over the length
115
Generalized Scattering parameters
Z 01
V1+ , a1
V1− , b1
[S]
.
.
.
Vn+ , an
Z 0n
Vn− , bn
V2
Z 01 ≠ Z 02 , power ∝
2Z 0
+
V
Let an ∆ n
Z on
Wave amplitudes
Vn−
bn ∆
Z on
b
Sij = i
aj
=
a k = 0 for k ≠ j
Vi− Z oj
V j+
Z oi
[b] = [S ][a ]
if Z oi = Z oj → Sij =
Vk =0 , for k ≠ j
Vi−
V j+
Vk =0 ( k ≠ j )
116
4.4 Transmission Matrix (=ABCD Matrix)
V1   A B  V2 
I  = 
 I ;

 1  C D   2 
for 2 − port network
V1 = AV2 + BI 2
I1 = CV2 + DI 2
A=
V1
V2
B=
V1
I 2 =0
I2
V2 =0
ABCD –matrix is good for “cascade two-port networks”
I3
→
V1   A B  V2   A1
 I  = C D   I  = C
 2   1
 1 
B1   A2
D1  C2
v3 3
B2  V3   A B  V3 
=



D2   I 3  C D   I 3 
117
Example 4.6
A B
Find 
 of
C
D


+
I1
V1
Z
I2
+
V2
−
V1
A=
V2
V
B= 1
I2
I
C= 1
V2
I
D= 1
I2
−
No voltage drop in Z→ V2 = V1 → A = 1
I 2 =0
+
Z
I2
I2 =
V1
V2 = 0
−
I1
+
Z
I 2 = 0 → I1 = 0 → C = 0
V2
−
I 2 =0
I1
Z
I2
I 2 = I1 → D = 1,
V2 = 0
Network is reciprocal →
V1
→B=Z
Z
AD − BC = 1
[S ]
 A B  1 Z 
C D  = 0 1 

 

[Z ]
Table 4.2
Useful conversions
A
C

B
D

[Y ]
118
Lecture 31
4-5 Signal Flow Graphs
Useful tool for network analysis
port1
→ a1
← b1
Nodes;
a1 , a2 ,
[S]
a
 a2
 b2 port2
b1 , b2
S 21 b
S11
2
S 22
b1 S12
Incident, Reflected wave
Branches; signal flow , S-parameters or
⇔
1
a2
Γ
< one-port network >
< A voltage source >
119
Decomposition of signal Flow Graphs.
1) Series rule : branches in series → signal branch.
V3 = S32V2 = S 21S32V1
2) Parallel rule: branches in parallel→ signal branch
V2 = S aV1 = SbV1 = (S a + Sb )V1
3) Self - loop rule: A branch that begins and ends on the same node
V2 = S 21V1 + S 22V2 ⇒ (1 − S 22 )V2 = S 21V1
V2 = S 21 /(1 − S 22 ) V1
V3 = S 32V2
4) Splitting rule. : a node can be splitted
V3 =
S 21S32
V1
1 − S 22
V4 = S 21S 42V1
120
Find Γin =
Example 4.7
By the splitting rule
a1
S 21
b1
a1
Flow graph
By the self-loop rule
b2
a1
S 22 Γl Γl
b1
S12
Vs
b1
S12
By the parallel rule
S11
b1
a1
Vs
a1
Γs
Γl
S11
a2
By the series rule
S 21
1 − S 22 Γl
S 21S12 Γl
1 − S 22 Γl
Γs
S11 +
b1
S 21S12 Γl
= Γin
1 − S 22 Γl
121
Network Analyzer Calibration
4.6 Discontinuities and modal Analysis (Skip).
4.7 Excitation of waveguides (Skip)
122
Lecture 32
Ch. 5. Impedance Matching
Purpose of impedance matching
-. To deliver maximum power
-. To improve signal-to-noise (SNR) ratio
-. To reduce amplitude and phase errors
Zin = Z0
Z0
Load
ZL
Single Sections:
-. Lumped Elements
-. Single Stub Matching
-. Double Stub Matching
-. Quarter-Wave Transformer
Z0
Matching
Network
Load
ZL
Multi-Sections
with Quarter-Wave Transformers
-. Binomial Method
-. Chebyshev Method
Tapered Lines
123
5.1 Matching with Lumped Elements
Point of Interest: P. 228
(Resistors, inductors,
Capacitors)
L-section matching networks:
jX
Z0
jX
jB
ZL
Networks for zL inside the 1+jx circle.
Z0
jB
ZL
Networks for zL outside the 1+jx circle.
jX = jωL,
jB =
Find X and B for Matching.
1
j ωL
,
jX =
1
,
j ωC
jB = jωC
-. Analytic Solution
-. Smith Chart Solution
124
Analytic Solution
jX
jB
Z0
Zin = Z0
ZL
Z in = Z 0 = jX +
Two equations for X and B
(real and imaginary)
1
1
jB +
RL + jX L
X and B (or L and C)
125
jX
Z0
jB
Zin = Z0
ZL
1
1
Yin = Y0 =
= jB +
Z0
jX + RL + jX L
Two equations for X and B
(real and imaginary)
X and B (or L and C)
126
Lecture 33
Smith Chart Solution
Example 5.1
zl=2-j (ZL=200-j 100) 
inside of 1+jx circle
jX
Z0
(1)
(2)
(3)
(4)
(5)
(6)
jB
ZL
ZL
zL =
Z0
Point of normalized load impedance
Transfer to load admittance (180deg. rotation)
Find jB
Transfer to impedance
Find jX
To get the input impedance=1 (Gamma=0)
127
Smith Chart Solution
Example 5.1
yl
Example 5.1
Use
Admittance Chart for jB
and
Impedance Chart for jX
+j0.29 =jb
b=0.29 B=bY0 =b/Z0
x=1.2 X=x Z0
Find L, C from
B=2 pi f C
X =2 pi f L
zl=2-j (ZL=200-j 100)
(Start point)
+j1.2 =jx
For 2nd solution
128
5.2 Single-Stub Tuning
(lumped elements-not preferred)
l
l
Zo
Short
Zo
Z L + jZ 0 tan β
Z in = Z 0
Z L + jZ L tan β Z
Z in = Z 0
L =0
= jZ 0 tan β → − ∞ ~ + ∞
shunt stubs
easier
-. open stub : microstrip
-. short stub : coaxial cable, waveguides
Open
 Z0 
1

= − j 
j tan β
 tan β 
tan β
0
π
2
π
β
(1) Analytical Solution
Z in = Z 0
Yin = Yd + Ys
for matching
where Yin =
1
1
1
, Yd =
, Ys =
Z in
Zd
Zs
129
Impedance matching
Z0

Z0
 2 equations for 2 unknowns
Zd
(real, imag.) (tan β d ,
open
or short
∴ Gd − Y0 = 0
Bd + Bs = 0
Then, find d, l
tan β)
( d , )
Yd + Ys = Y0 , Y0 =
Unknowns:
Z L + jZ 0 tan βd
Z 0 + jZ L tan βd
Z0
Zs = − j
(open stub)
tan β
(= jZ 0 tan β for short stub)
ZL
Z0
Zs
Zd = Z0
1
, Yd = Gd + jBd , Ys = jBs
Z0
− (1)
− ( 2)
jBs =
1
Zs
tan βd ≡ t d , tan βl ≡ tl
2π
1
−1
=
β
=
tan t d ,
t≥0
For example, β d = tan t d ,
λ
λ 2π
d
1
−1
if t d < 0, βd = π + tan t d , ⇒ =
π + tan −1 t d
λ 2π
−1
d
[
]
130
(2) Smith Chart Solution (Example 5-2, p. 229)
Lecture 34
131
yd
d
Z0
Z0
ZL
Z0
ys

yd +ys=yin=y0=1
132
d1
yd = 1 + j1.47
Shorted load
(y=infinity)
l1
ys = - j1.47
133
“Smith Chart 따라 하기”
134
Lecture 35
5.3 Double-Stub Tuning
good for an adjustable tuner
(1) Analytical solution
Z + jZ 0 tan βL
Z d1 = Z 0 L
(known)
Z 0 + jZ L tan β L
⇒ Yd1 =
Y1 = Yd1 + Ys1
1
Z1 =
Y1
Yd 2 =
1
Zd 2
,
1
Z d1
Ys 2 Yd 2
Ys1 =
1
, Z s1 = jZ 0 tan β 1 ( short ),
Z s1
≡ t1 (unknown)
Z + jZ 0 tan βd
Zd 2 = Z0 1
Z 0 + jZ1 tan βd
Yin = Yd 2 + Ys 2 = Y0
where
Ys1 Yd 1
or Z s1 =
Z0
(open)
j tan β 1
λ
2π


, d=
β =

λ
8

known 


π
2π λ
β
d
⇒
tan
=
tan
=
tan
=
1


λ 8
4


 jZ 0 tan β 2

Z s2 =  Z0
 j tan β ≡ t
2

2
 2 equations for 2 unknowns (t1,t2)
Real  ①
1  2
Imag  ②
( 1 ,  1 ) ( 2 ,  2 )
'
'
 1 ,  2
⇒  '
'
 1 ,  2
135
(2) Smith Chart Solution
136
137
138
Lecture 36
5.4 Quarter-wave (λ/4) Transformer
= λ0 4
Resistive load
Γ
π
 2π λ 
=
=∞
tan
⋅
tan


(1)  =
at f 0
2
 λ 4
4
Z L + jZ1 tan β
Z L ∞ + jZ1
= Z1
Z in = Z1
Z1 + jZ L tan β
Z1 ∞ + jZ L
λ0
Z in
Z12
= Z0
Z in =
ZL
Z − Z0
Γ = in
=0
Z in + Z 0
⇒ Z1 = Z 0 Z L
at
for Impedance matching
λ
f 0 ( = 0 )
4
(2) At other frequencies   ≠
λ
4
π λ0
π f
2π λ0
check tan β  = tan(
) = tan(
) = tan(
) ≡ t (≠ ∞)
λ 4
2 λ
2 f0
139
Γ
Z L + jZ1t
− Z0
Z1
Z in − Z 0
Z1 + jZ Lt
Γ=
=
Z in + Z 0 Z Z L + jZ1t + Z
0
1
Z1 + jZ Lt
f1
= function of frequency,
known
f0
f
(Γ ≠ 0)
unknown
where Z1 = Z 0 Z L , t = tan β
Γ = 0 at f = f 0 ( =
λ0
4
)
λ0 = wave length of center frequency ( f 0 )
∆θ = π − θ m − θ m = π − 2θ m
Maximum tolerable Γ
140
5.5 Theory of small reflections
Γ1 , Γ2 , Γ3 : partial reflection coefficient
Τ21 , Τ12 , Τ32 , : partial transmission coefficient
Single-section transformer
Z 2 − Z1
Γ1 =
Γ2 = −Γ1
Z 2 + Z1
Τ21 = 1 + Γ1
Τ12 = 1 + Γ2
ZL − Z 2
Γ3 =
ZL + Z 2
Γ = Γ1 + Τ21e − jθ Γ3e − jθ Τ12 + Τ21Γ3Γ2 Γ3e − j 4θ Τ12 + 
∞
(
= Γ1 + Τ21Τ12 Γ3e − 2 jθ ∑ Γ2 Γ3e − 2 jθ
n =0
n
)
141
∞
1
using ∑ x =
1− x
n =0
x <1
n
Τ21Τ12 Γ3e −2 jθ
Γ = Γ1 +
1 − Γ2 Γ3e
= Γ1 +
Γ=
where Γ2 = −Γ1 , Τ12 = 1 − Γ1 , Τ21 = 1 + Γ1
− 2 jθ
(1 − Γ1 )(1 + Γ1 )Γ3e − 2 jθ
1 + Γ1Γ3e − 2 jθ
Γ1 + Γ12 Γ3e − 2 jθ + Γ3e − 2 jθ − Γ12 Γ3e − 2 jθ
=
1 + Γ1Γ3e − 2 jθ
Γ1 + Γ3e − 2 jθ
1 + Γ1Γ3e − 2 jθ
if
Z1 − Z 2 ,
Γ ≈ Γ1 + Γ3e −2 jθ
Z1 − Z 2
are small, Γ1Γ3 << 1
(Γ1 , Γ3
Phase delay
dominant term)
142
Lecture 37
Multi-section Transformer
ZL
(P.243) Fig 5.12  small Z  wide Δf
0
Z n − Z n −1
: small  instead, use multi-sections
Same length
using equal length (θ)
Z n +1 − Z n
Γn =
Z n +1 + Z n
(
Z n : increase (or decrease) monotonically)
Γ(θ ) = Γ0 + Γ1e − j 2θ + Γ2 e − j 4θ +  + ΓN e − jNθ
(considering Γ ≈ Γ1 + Γ3e −2 jθ
Compare with other
Functions for design purpose
for 1 section transformer) -. Binomial series
-. Chebyshev Polyn.
143
Maximally flat response
Chebyshev (equal ripple)
(Flat at passband)
(Broader Bandwidth)
Optimize bandwidth
↔
passband ripple
Trade off
144
5.6 Binomial Multisection Transformers
Binomial Series
(1 + x) n = 1 + nx +
= C0n
n(n − 1) 2
x +  ,
2!
+ C1n x + C2n x 2
(1 + x) 2 = 1 + 2 x + x 2
+  where C2n =
Use N-section Binomial Transformer
(Binomial coefficients)
Γ(θ ) = A(1 + e − j 2θ ) N
π
π

θ
=
 θ=
Γ = 0 at center frequency f0 
2
2

π 

2
j
−
⋅
π

2  = 1 + ( −1) = 0
⇒ 1 + e
θ=

2



'
− j 2θ N −1
Γ (θ ) = AN (1 + e
∴
d n Γ(θ )
dθ
n
)
n!
n(n − 1)
=
(n − 2)! 2!
2!
⇒
λ
4
− Transformer
(− j 2)
= 0 at θ =
π
2
for
n = 1, 2, , N − 1
145
compare
Γ(θ ) = A(1 + e − j 2θ ) N = AC0N + AC1N e − j 2θ +  + AC NN e − j 2 Nθ
with Γ(θ ) ≈ Γ0 + Γ1e − j 2θ + Γ2 e − j 4θ +  + ΓN e − j 2 Nθ
∴ Γn = ACnN
for Binomial multisection
146
Lecture 38
Binomial Multisection Transformers:
Design Procedure
(i) Find A
From Γ(θ ) = A(1 + e − j 2θ ) N ,
let θ = 0, ⇒ e − j 0 = 1
∴ Γ ( 0) ≈ A 2 N
and
Z0
Γ(0)
Logarithmic Series ;
ZL
Z L − Z0
Γ(0) =
Z L + Z0
∴ A≈2
−N
Z L − Z0
Z L + Z0
 x − 1 1  x − 1 3

ln x = 2 
+ 
 +   if
 x + 1 3  x + 1 

Z n +1 − Z n
is small ,
147
Z
Z / Z −1
Z
− Zn
ln n +1 ≈ 2 n +1 n
= 2 n +1
Zn
Z n +1 / Z n + 1
Z n +1 + Z n
Z L − Z0 1 Z L
Z
1
1 Z
ln L
→
= ln
⇒ A ≈ 2 − N ln L =
Z L + Z0 2 Z0
2 Z 0 2 N +1 Z 0
(ii) Find Z n ,
n = 1, 2,  , N
Z n +1 − Z n 1 Z n +1
Γn =
= ln
Z n +1 + Z n 2
Zn
and
Γn =
A CnN
=2
−N
CnN
1 ZL
ln
2 Z0
 (1)
 (2)
Z
Z
Z
(1) = (2) ⇒ ln n +1 = 2 − N CnN ln L → ln Z n +1 = ln Z n + 2 − N CnN ln L
Zn
Z0
Z0


Z
Z 
Z 
( or , ⇒ n +1 = exp 2 − N CnN ln L  → Z n +1 ≅ Z n exp 2 − N CnN ln L  )
Zn
Z0 
Z0 


148
Example 5.6
(Binomial Transformer Design)
100 Ω
π
π
π
2
2
2
Z1
Z2
Z3
(1) Find Z1, Z2, Z3
(2) Compute bandwidth for Γm=0.05
(1) Find Zn
ln Z1 = ln Z 0 + 2 − N C03 ln
Z L = 50Ω
Z n +1
ZL
−N N
ln
= 2 Cn ln
Zn
Z0
50
100
= 4.5184 → Z1 = 91.7 Ω
1 3
C1 ( −0.693) = 4.2586 → Z 2 = 70.7 Ω
8
1
ln Z 3 = ln 70.7 + 3 (−0.693) = 3.9986 → Z 3 = 54.5 Ω
8
ln Z 2 = ln Z1 +
 3 section of λ/4 –transformers [100Ω-91.7Ω-70.7Ω-54.5Ω-50Ω]
149
(2) bandwidth for Γm=0.05
(i) Find A:
ZL
50
ZL
1
A≈
ln
= ln
= −0.693
ln
N +1
Z
Z0
100
2
0
(ii) Find BW:
Γ(θ ) = Ae − j 2θN (e jθ + e − jθ ) N
N
→ Γ = A 2 N cosθ m
0.05 = 0.0433 ⋅ 23 cosθ m 3
θ m = cos −1 (0.5246) = 58.360
0
0
∆f 2(90 − 58.4 )
=
= 0.702 → 70.2%
f0
90
− 0.693
A=
= −0.0433
4
2
|Γ|
BW
Γm
θm
π
2
π − θm
θ
150
5.7 Chebyshev Multisection Matching Transformers
Binomial method :
Γ(θ )
Chebyshev method : Γ(θ )
Binomial polynomial
Lecture 39
Γ(θ ) = A(1 + e − j 2θ ) N
Chebyshev polynomial
Chebyshev polynomial
n = 3, Τ3 ( x) = 4 x 3 − 3 x
n = 2, Τ2 ( x) = 2 x 2 − 1
n=even
n = 1, Τ1 ( x) = x
All pass this point(1,1)
n=odd
Τn ( x) = 2 x Τn −1 ( x) − Τn − 2 ( x)
Let
x = cosθ
→ Τn (cosθ ) = cos nθ
Example : Τ2 (cosθ ) = 2 cos 2 θ − 1 = (1 + cos 2θ ) − 1 = cos 2θ
151
Chebyshev polynomial:
x = cosθ
∴ Τn ( x) = cos[n cos −1 x]
Τn ( x) = cosh[n cosh −1 x]
→ θ = cos −1 x
x ≤1
for
for
x >1
Τn ( x) = 1 ⇔ Γ(θ ) = Γm
at x = 1
at θ = π − θ m




Τn (−1) = 1

⇒ Γ(θ m ) = Γm 
Τn (θ = θ m ) = 1
 cos θ
when using Τn 
 cos θ m






cos θ
−1  cos θ 
,
≤1
for
cos n cos 


cos θ m

 cos θ  
 cos θ m 
=
∴ Τn 


 cos θ m  
cos θ
−1  cos θ 
cosh n cosh  cos θ  , for cos θ ≥ 1


m 
m

Compare with Γ(θ ) ≈ Γ0 + Γ1e − j 2θ + Γ2 e − j 4θ +  + ΓN e − j 2 Nθ
152
Lets use symmetrical form, Γ = Γ ,
N
0
Γ1 = ΓN −1 , 
Γ(θ ) = Γ0 (1 + e − j 2 Nθ ) + Γ1 (e − j 2θ + e − j 2 ( N −1)θ ) + Γ2 () + 
[
]
= e − jNθ Γ0 (e jNθ + e − jNθ ) + Γ1 (e j ( N − 2 )θ + e − j ( N − 2 )θ ) + Γ2 
 N : odd ⇒ ( N + 1) − even


 N : even ⇒ ( N + 1) − odd


→ ΓN −1 (e jθ − e − jθ )
2
→ ΓN
N+1=even
+ ΓN −1 cos θ : odd ( N ) 


 
2
− jNθ
Γ(θ ) = 2e
Γ0 cos Nθ + Γ1 cos( N − 2)θ +   1


+ ΓN : even( N )
 
 2 2
 

And let Γ(θ ) = A e
− jNθ
2
cosθ
)
ΤN (
cosθ m
N+1=odd
153
Design Procedure
(i) Find A :
Γm ≡ Γ(θ m ) , Τ N (
(ii) Find
θ m : when θ = 0
1
)
Γ(θ = 0) = A ΤN (
cos θ m
cosθ
) =Τ N (1) = 1 → A = Γm
cosθ m
Z0
Γ
ZL
Z L − Z0
Γ=
Z L + Z0
Z L − Z0
=
Z L + Z0
1
1 Z L − Z0
1
ZL
)=
ln
⇒ ΤN (
≈
cos θ m
A Z L + Z0 A ⋅ 2 Z0

ZL
1
1  1
−1 
 =
) = cosh  N cosh 
ln
and ΤN (
cosθ m

 cosθ m  2 A Z 0
1
Z L 
1
−1  1
= cosh  cosh
ln
 ⇒ θm


cosθ m
 N
 2 A Z 0 
154
Lecture 40
Chebyshev multisection transformer Design Procedure (continue)
Γn and Z n
 cosθ 

Compare 2(Γ0 cos Nθ + Γ1 cos( N − 2)θ +  ) = A ΤN 
 cosθ m 
→ find Γ0 , Γ1 , 
→ compute Z1 , Z 2 ,
1 Z n +1
using Γn = ln
→ ln Z1 = ln Z 0 + 2Γ0 , 
2
Zn
(iii) Find
155
Example 5.7
Chebyshev Transformer Design
same problem with Example 5.6
50 Ω
Γ
Z1
λ0 / 4
Z2
Z3
Z L = 100Ω
λ0 / 4 λ0 / 4
Find Z1, Z2, Z3 for matching
with N(=3) sections.
Sol.
Γ(θ ) = Γ0 + Γ1e − j 2θ + Γ2 e − j 4θ + Γ3e − j 6θ
[
= e − j 3θ Γ0 e j 3θ + Γ3e − j 3θ + Γ1e jθ + Γ2 e − jθ
= 2e − j 3θ [Γ0 cos(3θ ) + Γ1 cos θ ]
Find A
compare with
Γ(θ ) = Ae
 cos θ 

Τ3 
 cos θ m 
⇒ Γ(θ m) = 0.05 Τ3 (1) = 1
]
− j 3θ
at θ = θ m
A = 0.05
156
Z L − Z0
1 ZL
Γ(0) =
= A Τ3 (sec θ m ) ≈ ln
Z L + Z0
2 Z0
x −1 


 ln x ≈ 2
x +1

ZL
1
1
100
Τ3 (sec θ m ) =
=
= 6.9315
ln
ln
2 A Z 0 2 × 0.05 50
[
]
Τ3 (sec θ m ) = cosh 3 cosh −1 (sec θ m ) ≈ 6.9315
1

sec θ m = cosh  cosh −1 (6.9315) = 1.407 → θ m = 44.7°
3

∆f 2(90 − 44.7)
=
⇒ 101% (larger than 70.2% of binomial type)
90
f0
157
Find
Γn
and
Zn :
Τ3 ( x) = 4 x 3 − 3x
[
⇒ 2[Γ0 cos 3θ + Γ1 cos θ ] = A 4(sec θ m cos θ )3 − 3(sec θ m cos θ )
]
= 0.05(sec3 θ m ⋅ 3 cos θ + sec3 θ m cos 3θ − 3 sec θ m cos θ )
Z1 − Z 0 1 Z1
1
3
⇒ Γ0 = × 0.05 × sec θ m = 0.0696 ⇒ Γ0 =
≈ ln
2
Z1 + Z 0 2 Z 0
⇒ Z1 = e jΓ0 Z 0 = 57.5Ω,
Z 2 = 70.7Ω, Z 3 = 87.0Ω
cos 3 θ =
1
(3 cos θ + cos 3θ )
4
158
π
5.8 Tapered Lines
2
From Example 5.6
With microstrip lines
Lecture 41
(λ / 4) at f 0
100 Ω
91.7Ω
100 Ω
91.7Ω
70.7Ω
54.5Ω
50 Ω
and broader bandwidth
n=1
n=3
70.7Ω
54.5Ω
50 Ω
Γ=0
Γ=0
How about tapering?
100 Ω
50 Ω
Triangular tapering Exponential tapering
Exponential taper :
0<Z <L
Z ( z ) = Z 0e az , → Z ( L) = Z L = Z 0e aL ,
a=
1 ZL
ln
L Z0
159
Triangular Taper ;
d  ZL 
 ln
 : Triangular Shape
2
dz  Z 0 
1 − jβL Z L  sin( βL 2) 
Γ(θ ) Exp = e
ln


2
Z 0  βL 2 
Klopfenstein Taper ;
cos ( βL) 2 − A2
Γ(θ ) Klopf = Γ0 e
,
cosh A
where cosh A = Γ0 / Γm
− jβ L
for βL > A
2
2
β
−
A
L
cosh
(
)
If βL < A, Γ(θ ) Klopf = Γ0 e − jβL
cosh A
160
Example 5.8
Design  triangular taper,

 exponential taper,
 klopfenstein taper

161
Exponential taper : BW  widest (good)
|Γ| at long taper  high (bad)
Triangular taper : Need longer taper to get lower Γ
(good in passband)
Klopfenstein : optimum  short taper is OK and can get lower |Γ|
162
Lecture 42
Review of Microwave Engineering (1)
Microwave Engineering (1):
Electromagnetic Theory
Ch.1
Transm.
Line
Theory
Ch.2
Wave Guidance:
Two plates, rectangular,
circular Waveguides,
Coaxial lines, Microstrip
Ch.3
Microwave Network Theory
(S-parameters)
Ch.4
Microwave Engineering (2)
Microwave Passive Circuits:
Microwave Resonators
Power dividers, Hybrids,
Filters, Circulators, etc.
Ch. 6, 7, 8, 9
Microwave Active Circuits:
Detectors, Mixers,
Oscillator, Amplifiers
(Wireless communication)
Ch. 10,11,12,13
Impedance Matching
Ch.5
163
Preview of Microwave Engineering (2)
Applications of Microwave Resonators
-. Microwave Filters
-. Microwave Oscillators
-. Measurement of the Dielectric Constant using Microwaves
An Example of Microwave Filters
|S21|
Port #1
BW
Port #2
fo
Circular
Waveguide
Cavities
Iris
(small aperture)
164
Antenna
Amplifier
Mixer
Modulator
(signal with
information)
Microwave
Waveguides,
Transmission
lines
Filters,
Matching circuits,
Power dividers,
Resonators,
etc.
D.C.
Power
Oscillators
An Example of
Wireless Communication Systems
Use One of Microwave
Resonators
Microwave Circuits
Matching
circuit
Resonating
circuit
Microwave
Feedback
circuit
Microwave Oscillator
165
Measurement of the Dielectric Constant
Dielectric (e.g., Soil, rock, glass, etc)
with unknown dielectric constant
Mictostrip
Resonator
(Ring type)
Contact
surface
BW1
|S21|
|S21|
BW2
f01
f02
Dry Soil
Wet Soil
166
Lecture 43
Questioning and Answering for Final Exam
167
Final Exam
기말고사 일정표
과목명
마이크로파공학(1)
시험일
(요일)
시간
6/12
오후7시
(목)
-8시30분
장소
P202, P203
시험
범위
3.5-5.8
168