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UNIT V
WAVEGUIDES AND RESONATORS
2
Transmission Lines and Waveguides• Waveguide and other transmission lines for
the low-loss transmission of microwave power.• Early microwave systems relied on waveguide
and coaxial lines for transmission line media.• Waveguide: high power-handling capability,
low loss, but bulky and expensive• Coaxial line: high bandwidth, convenient for
test applications, difficult medium in which to fabricate complex microwave components.
• Planar transmission lines: stripline, microstrip, slotline, coplanar waveguide compact, low cost, easily integrated with active devices
3
3.1 General Solutions for TEM, TE and TM waves
• General solutions to Maxwell’s equations for the specific cases of TEM, TE and TM wave propagation in cylindrical transmission lines or waveguides.
• Uniform in z direction and infinitely long
Figure 3.1 (p. 92)(a) General two-conductor transmission line and (b) closed waveguide.
4
• Assume ejωt dependence
where e(x,y) and h(x,y): transverse (x,y) E & H field components, ez and hz: longitudinal E & H field components.
• Assume source free,
ˆ( , , ) [ ( , ) ( , )]
ˆ( , , ) [ ( , ) ( , )]
j zz
j zz
E x y z e x y ze x y e
H x y z h x y zh x y e
E j H
H j E
5
,
,
zy x
zx y
y xz
Ej E j H
y
Ej E j H
xE E
j Hx y
,
,
zy x
zx y
y xz
Hj H j E
y
Hj H j E
xH H
j Ex y
2
2
z zx
c
z zy
c
j E HH
k y x
j E HH
k x y
2
2
z zx
c
z zy
c
j E HE
k x y
j E HE
k y x
2 2 2ck k
6
where kc: cutoff wavenumber,
: the wavenumber of the material filling with the transmission line or waveguide region.
TEM Waves
• TEM waves are characterized by Ez = Hz = 0.
• From (3.3a) and (3.4b)
the cutoff wavenumber kc = 0 for TEM waves.
• Helmholtz equation for Ex from (1.42)
2 /k
2 2y yE E k
2 2 22
2 2 20xk E
x y z
7
• For e-jβz dependence,
• Similar result also applies to Ey ( )
Transverse fields e(x,y) of a TEM wave satisfy Laplace’s equation.
• Similarly, • In the electrostatic case, E field can be expressed
as
22 2
2 x x xE E k Ez
2 2
2 20xE
x y
2 ( , ) 0t e x y
2 2 2 2 2/ /t x y
2 ( , ) 0t h x y
ˆ ˆ( , ) ( , )( ( / ) ( / )t te x y x y x x y y
8
• In order for (3.13) to be valid, the curl of e must vanish:
• The voltage between 2 conductors and current flow on a conductor:
• TEM waves can exist when 2 or more conductors are present. (ex: Plane waves)
ˆ 0t ze j h z 20 0 ( , ) 0t tD e x y
2
12 1 2 1,
CV E dl I H dl
9
• The wave impedance of a TEM modex
TEMy
EZ
H
yTEM
x
EZ
H
1ˆ( , ) ( , )
TEM
h x y z e x yZ
10
• The procedure for analyzing a TEM line:– Solve Laplace equation, (3.14) for Φ(x,y)– Find these constants by applying the B.C. for
the known voltages on the conductors– Compute e and E form (3.13) & (3.1a).
Compute h and H from (3.18) and (3.1b).– Compute V from (3.15), I from (3.16).
– The propagation constant is given by (3.8), Z0 is given by Z0 = V/I.
11
TE Waves• Characterized by Ez = 0, Hz ≠ 0.
• In this case, kc ≠ 0, and the propagation constant
is generally a function of frequency
and the geometry of the line or guide.
2
2
zx
c
zy
c
j HH
k x
j HH
k y
2
2
zx
c
zy
c
j HE
k y
j HE
k x
2 2ck k
12
• The Helmholtz equation
• Since Hz(x,y,z) = hz(x,y)e-jβz, and kc2 = k2 –
β2
• TE wave impedance can be
2 2 22
2 2 20zk H
x y z
2 22
2 20c zk h
x y
yxTE
y x
EE kZ
H H
13
TM Waves
• Characterized by Hz = 0, Ez ≠ 0.
2
2
zx
c
zy
c
j EH
k y
j EH
k x
2
2
zx
c
zy
c
j EE
k x
j EE
k x
2 2 22
2 2 20zk E
x y z
2 22
2 20c zk e
x y
yxTM
y x
EEZ
H H k
14
• The procedure for analyzing TE and TM waveguides
1. Solve the reduced Helmholtz equation, (3.21) or (3.25) for hz or ez. The solution will contain several unknown constants, and the unknown cutoff wavenumber, kc.
2. Use (3.19) or (3.23) to find the transverse fields from hz or ez.
3. Apply the B.C. to the appropriate field components to find the unknown constants and kc.
4. The propagation constant is given by (3.6) and the wave impedance by (3.22) or (3.26).
15
Attenuation due to Dielectric Loss
• Using the complex dielectric constant
• In practice, most dielectric materials have a very small loss (tan δ <<1). Using the Taylor expansion,
for x << a
2 2 2
2 2 2 20 0 (1 tan )
c
d c c r
k k
j k k k j
1/ 22 2 22 2 1 1
1 12 2
x x xa x a a a
a a a
16
• (3.27) reduces to
• For TE or TM wave
• For TEM line, kc = 0, k = β
22 2 2 2 2
2 2
2
tantan
2
tan
2
c c
c
jkk k jk k k
k k
kj
2 tan
2d
k
tan
2d
k
17
3.2 Parallel Plate Waveguide
Figure 3.2 (p. 98)Geometry of a parallel plate waveguide.
18
• The simplest type of guide that can support TM and TE modes; can also support a TEM mode.
TEM Modes• TEM mode solution can be obtained by
solving Laplace’s equation.
• Assume• Since there is no variation in x,
• The transverse E-field from (3.13),
2 ( , ) 0 for 0 ,0t x y x W y d
0( ,0) 0, ( , )x x d V
( , )x y A By 0( , ) /x y V y d
0ˆ( , ) ( , ) /te x y x y yV d
0ˆ( , , ) ( , ) jkz jkzVE x y z e x y e y e
d
19
01ˆˆ( , , ) ( , , ) jkzV
H x y z z E x y z x ed
The voltage of the top plate with respect to the bottom plate
00
d jkzyy
V E dy V e
The total current on the top plate
0
0 0 0ˆˆ ˆ( )
w w w jkzs xx x x
wVI J zdx y H zdx H dx e
d
0
V dZ
I w
1pv
20
TM Modes
• Hz = 0, Ez satisfies (3.25) with
• B.C.
/ 0x 2
22
( , ) 0c zk e x yy
( , ) sin cosz c ce x y A k y B k y
( , ) 0 at 0,ze x y y d
0 and cB k d n
2 2 2 2( / )ck k k n d
( , ) sin , ( , , ) sin j zz n z n
n y n ye x y A E x y z A e
d d
21
cos , cos ,
0
j z j zx n y n
c c
x y
j n y j n yH A e E A e
k d k d
E H
TM0 mode = TEM mode
The cutoff frequency fc
2 2c
c
k nf
d
TM1 mode is the lowest TM mode with a cutoff frequency
1
2cf
d
22
• At frequencies below the cutoff frequency of a given mode, the propagation constant is purely imaginary, corresponding to a rapid exponential decay of the fields. cutoff or evanescent modes.
• TMn mode propagation is analogous to a high-pass filter response.
• The wave impedance
pure real for f > fc, pure imaginary for f < fc.• The guide wavelength is defined the distance between equiphase
planes along the z-axis.
yTM
x
EZ
H k
2g
23
• λg > λ = 2π/k, the wavelength of a plane wave in the material.
• The phase velocity and guide wavelength are defined only for a propagation mode, for which β is real.
• A cutoff wavelength for the TMn mode may be defined as
• Poynting vector
2c
d
n
0 0 0 0
2 22 0
1 1ˆRe Re
2 2Re( )
| | cos2
w d w d
o y xx y x y
d
n yc
P E H zdydx E H dydx
w n yA dy
k d
24
22
22
Re( )| | for 0
4
Re( )| | for 0
2
nc
nc
w dA n
k
w dA n
k
Consider the dominant TM1 mode, which has a propagation constant,
1 1
221
[ / ] [ / ]11
/ ,
sin2
j y d z j y d zj zz
k d
y AE A e e e
d j
2 plane waves traveling obliquely in the –y, +z and +y, +z directions.
25
Figure 3.3 (p. 102)Bouncing plane wave interpretation of the TM1 parallel plate waveguide mode.
1sin , cosk kd
2
2 21 k
d
f fc: β1 0: 2 plane waves up and down, no real power flow.
26
• Conductor loss can be treated using the perturbation method.
where Po: the power flow down the guide in the absence of conductor loss given by (3.54), Pl: the power dissipated per unit length in the 2 lossy conductors
2l
co
P
P
2 222
202 | | | |
2
2 2 for 0
for 0
ws s
l s nxc
s sc
s
R R wP J dx A
k
R kRn
d d
Rn
d
27
TE Modes
• Ez = 0, Hz satisfies (3.21) with
• B.C.
/ 0x 2
22
( , ) 0c zk h x yy
( , ) sin cosz c ch x y A k y B k y
( , ) 0 at 0,xe x y y d
0 and cA k d n
( , , ) [ cos sin ] ,
( , ) cos
j zx c c
c
j zz n
jE x y z A k y B k y e
k
n yH x y B e
d
28
sin , sin ,
0
j z j zx n y n
c c
y x
j n y j n yE B e H B e
k d k d
E H
The cutoff frequency fc
2 2c
c
k nf
d
xTE
y
EZ
H k
The wave impedance
29
0 0 0 0
22
1 1ˆRe Re
2 2Re( )
| | for 04
w d w d
o x yx y x y
nc
P E H zdydx E H dydx
wB n
k
If n = 0, Ex = Hy = 0, P0 = 0 no TE0 mode.
2 22 2c s c sc
k R k R
d k d
30
Figure 3.4 (p. 105)Attenuation due to conductor loss for the TEM, TM, and TE1 modes of a parallel plate waveguide.
31
Figure 3.5 (p. 106)Field lines for the (a) TEM, (b) TM1, and (c) TE1 modes of a parallel plate waveguide. There is no variation across the width of the waveguide.
32
3.3 Rectangular Waveguide
TE Modes
• Ez = 0
• Hz must satisfy the reduced wave equation (3.21)
• Can be solved by separation of variables
2 22
2 2( , ) 0c zk h x y
x y
( , ) ( ) ( )zh x y X x Y y
2 22
2 2
1 10c
d X d Yk
X dx Y dy
33
Figure 3.6 (p. 107)Photograph of Ka-band (WR-28) rectangular waveguide components. Clockwise from top: a variable attenuator, and E-H (magic) tee junction, a directional coupler, an adaptor to ridge waveguide, an E-plane swept bend, an adjustable short, and a sliding matched load.
34
Figure 3.7 (p. 107)Geometry of a rectangular waveguide.
35
• We define separation constant kx and ky
• Boundary conditions
• Using (3.19c) and (3.19d)
2 22 2 2 2 2
2 20, 0,x y x y c
d X d Yk X k Y k k k
dx dy
( , ) ( cos sin )( cos sin )z x x y yh x y A k x B k x C k y B k y
( , ) 0 at 0,
( , ) 0 at 0,x
y
e x y y b
e x y x a
2
2
( cos sin )( sin cos )
( sin cos )( cos sin )
x y x x y yc
y x x x y yc
je k A k x B k x C k y D k y
k
je k A k x B k x C k y D k y
k
36
• From B.C , D = 0, and ky = nπ/b, B = 0 , and kx = mπ/a
• The transverse field components of TEmn mode
( , , ) cos cos j zz mn
m x n yH x y z A e
a b
2
2
2
2
cos sin ,
sin cos ,
sin cos ,
cos sin .
j zx mn
c
j zy mn
c
j zx mn
c
j zy mn
c
j n m x n yE A e
k b a b
j m m x n yE A e
k a a b
j m m x n yH A e
k a a b
j n m x n yH A e
k b a b
37
2 22 2 2
c
m nk k k
a b
2 2
c
m nk k
a b
is real when
2 21
2 2c
cmn
k m nf
a b
The mode with the lowest cutoff frequency is called the dominant mode;
10
1
2cf
a
38
• For f < fc, all field components will decay exponentially cutoff or evanescent modes
• If more than one mode is propagating, the waveguide is overmoded.
• The wave impedance
• The guide wavelength (λ: the wavelength of a plane wave in the filling medium)
yxTE
y x
EE kZ
H H
2 2g k
1/pvk
39
• For the TE10 mode
10
10
10
cos ,
sin ,
sin ,
0.
j zz
j zy
j zx
x z y
xH A e
aj a x
E A ea
j a xH A e
aE E H
2 2/ , ( / )ck a k a
40
• The power flow down the guide for the TE10 mode:
• Attenuation can occur because of dielectric loss or conductor loss.
0 0 0 0
2 2210
2 0 0
3 210
2
1 1ˆRe Re
2 2
Re( ) | |sin
2
Re( ) | |
4
a b a b
o y xx y x y
a b
x y
P E H zdydx E H dydx
a A xdydx
a
a b A
2| |2
sl sC
RP J dl
41
• There are surface currents on all 4 walls.
• The surface current on the x = 0 wall is
• The surface current on the y = 0 wall is
100 00ˆ ˆ ˆ ˆˆ j z
s z zx xxJ n H x z H y H yA e
0 00
10 10
ˆ ˆ ˆ ˆ
ˆˆ sin cos
s x zy yy
j z j z
J n H y x H z H
j a x xz A e xA e
a a
2 2 2
0 0
2 32
10 2
| | | | | |
2 2
b a
l s sy s sx szy x
s
P R J dy R J J dx
a aR A b
42
• The attenuation due to conductor loss for TE10 mode 2 2 3
3 210
2 3 23
2
2 2 2
(2 ) /
l sc
s
P R a ab
P a b
Rb a k Np m
a b k
43
TM Modes
• Hz = 0
• BC: ez(x,y) = 0 @ x = 0, a and y = 0, b
2 22
2 2( , ) 0c zk e x y
x y
( , ) ( cos sin )( cos sin )z x x y ye x y A k x B k x C k y D k y
0, /
0, /x
y
A k m a
C k n b
( , , ) sin sin j zz mn
m x n yE x y z B e
a b
44
2
2
2
2
cos sin ,
sin cos ,
sin cos ,
cos sin
j zx mn
c
j zy mn
c
j zx mn
c
j zx mn
c
j m m x n yE B e
ak a b
j n m x n yE B e
bk a b
j n m x n yH B e
bk a b
j m m x n yH B e
ak a b
2 22 2 2
c
m nk k k
a b
2 2
11
1,
2yx
c TMy x
EEf Z
a b H H k
45
Figure 3.8 (p. 112)Attenuation of various modes in a rectangular brass waveguide with a = 2.0 cm.
46
Figure 3.9 (p. 114)Field lines for some of the lower order modes of a rectangular waveguide.
47
Ex 3.1
• a = 1.07 cm, b = 0.43 cm, f = 15 GHz
• Solution: for Teflon εr = 2.08, tan δ = 0.0004 2 2
2cmn
r
c m nf
a b
Mode m n fc (GHz)
TE 1 0 9.72
TE 2 0 19.44
TE 0 1 24.19
TE, TM 1 1 26.07
TE, TM 2 1 31.03
48
• At 15 GHz2
2
2
0
2 3 23
345.1
tan0.119 / 1.03 /
2
0.0322
(2 ) 0.050 / 0.434 /
d
s
sc
ka
knp m dB m
R
Rb a k np m dB m
a b k
49
Figure 3.10 (p. 115)Geometry of a partially loaded rectangular waveguide.
50
51
3.4 Circular Waveguide
Figure 3.11 (p. 117)Geometry of a circular waveguide.
52
2 2
2 2
,
,
z z z z
c c
z z z z
c c
j E H j E HE E
k k
j E H j E HH H
k k
TE Modes
Ez = 0 2 2 0z zH k H 2 2
22 2 2
1 1( , ) 0c zk h
( , ) ( ) ( )zh R P
53
2 22
2 2 2
2 2 22 2
2 2
1 1 10,
1
c
c
d R dR d Pk
R d R d P d
d R dR d Pk
R d R d P d
2 22 2
2 2
10
d P d Pk k P
P d d
2
2 2 2 22
0c
d R dRk k R
d d
The general solution is
( ) sin cosP A k B k
54
• Since
• BC: Since Ez = 0,
( , ) ( , 2 )z zh h n
( ) sin cosP A n B n
2
2 2 2 22
0c
d R dRk n R
d d
( ) ( ) ( )n c n cR CJ k DY k But, D = 0
( , ) sin cos ( )z n ch A n B n J k
( , ) 0 at E a
( , , ) sin cos ( ) j zn c
c
jE z A n B n J k e
k
55
( ) 0n cJ k a
• If the roots of Jn' (x) are defined as p'nm, so that Jn'(p'nm) = 0, where p'nm is the mth root of Jn', then kc must have the value.
• See Table 3.3
• The Temn modes are defined by the cutoff wavenumber, kcmn = p'nm/a, where n refers to the number of circumferential (φ) variations, and m refers to the number of radial (ρ) variations.
nmc
pk
a
56
22 2 2 ,
2 2nm c nm
nm c cnm
p k pk k k f
a a
TE11 mode: dominant mode
2
2
cos sin ( )
sin cos ( )
sin cos ( )
cos sin ( )
j zn c
c
j zn c
c
j zn c
c
j zn c
c
j nE A n B n J k e
k
jE A n B n J k e
k
jH A n B n J k e
k
j nH A n B n J k e
k
57
The wave impedance
TE
E E kZ
H H
Because of the azimuthal symmetry of the circular waveguide, both sin nφ and cos nφ are terms are valid solutions, and can be present in a specific problem to any degree.
The actual amplitudes of these terms will be dependent on the excitation of the waveguide.
Consider the dominant TE11 mode with an excitation such that B = 0.
58
1
1 12
1 12
sin ( ) , 0
cos ( ) , sin ( )
sin ( ) , cos ( )
j zz c z
j z j zc c
c c
j z j zc c
c c
H A J k e E
j n jE A J k e E A n J k e
k k
j j nH A J k e H A n J k e
k k
The power flow down the guide
59
2
0 0
2
0 0
2 2 2 2 2 2 21 14 20 0
22 2 2
1 14 20
1ˆRe
21
Re [ ]2
| | Re( ) 1cos ( ) sin ( )
2
| | Re( ) 1( ) ( )
2
| |
a
o
a
a
c c cc
a
c c cc
P E H z d d
E H E H dydx d d
AJ k k J k d d
k
AJ k k J k d
k
A
2
2 211 14
Re( )( 1) ( )
4 cc
p J k ak
60
2 2
0
2 2 2
0
2 22 2 2 2
14 20
2 22
14 2
| |2
| | | |2
| |cos sin ( )
2
| |1 ( )
2
sl s
sz
sc
c
sc
c
RP J ad
RH H ad
A RJ k a ad
k a
A R aJ k a
k a
4 2 2 22
2 211 11
( )( )
2 ( 1) 1l s c s
c co
P R k a R kk
P k a p k a p
61
TM Modes2 2
22 2 2
1 1( , ) 0c zk e
( , ) sin cos ( )z n ce A n B n J k
( , ) 0 at zE a
( ) 0n cJ k a nmc
pk
a
22 2 2 ,
2 2nm c nm
nm c cnm
p k pk k k f
a a
62
2
2
sin cos ( )
cos sin ( )
cos sin ( )
sin cos ( )
j zn c
c
j zn c
c
j zn c
c
j zn c
c
jE A n B n J k e
k
j nE A n B n J k e
k
j nH A n B n J k e
k
jH A n B n J k e
k
TM
E EZ
H H k
63
Figure 3.12 (p. 123)Attenuation of various modes in a circular copper waveguide with a = 2.54 cm.
64
Figure 3.13 (p. 123)Cutoff frequencies of the first few TE and TM modes of a circular waveguide, relative to the cutoff frequency of the dominant TE11 mode.
Ex 3.2
65
Figure 3.14 (p. 125)Field lines for some of the lower order modes of a circular waveguide.
66
3.5 Coaxial Line
TEM Mode
• Boundary conditions
2
2 2
1 ( , ) 1 ( , )0
2 ( , ) 0
0( , ) , ( , ) 0a V b
( , ) ( ) ( )R P 2
2
10
d dR d P
R d d P d
22 2 2 2
2
1, , 0
d dR d Pk k k k
R d d P d
67
Figure 3.15 (p. 126)Coaxial line geometry.
68
• The general solution to (3.148)
• Since the boundary conditions do not vary with φ, the potential Φ should not vary with φ. n = 0 kρ = 0.
( ) cos sinP A n B n
0d dR
d d
( ) ln ( , ) lnR C D C D
0( , ) ln
( , ) 0 ln
a V C a D
b C b D
0 ln /( , )
ln /
V b
b a
69
Higher Order Modes• The coaxial line also support TE & TM waveguide
modes in addition to a TEM mode.• In practice, these modes are usually cutoff
(evanescent), only a reactive effect near discontinuities or sources, where they are excited.
• For TE modes, Ez = 0, and
• The general solution (from Sec. 3.4),
2 22
2 2 2
1 1( , ) 0c zk h
( , ) ( sin cos )( ( ) ( ))z n c n ch A n B n CJ k DY k
70
• Boundary conditions:
• Nontrivial solution for C & D
Characteristic equation for kc
( , , ) 0 for ,E z a b
sin cos ( ) ( )
z z
c
j zn c n c
c
j E HE
k
jA n B n CJ k DY k e
k
( ) ( ) 0
( ) ( ) 0n c n c
n c n c
CJ k a DY k a
CJ k b DY k b
( ) ( ) ( ) ( )n c n c n c n cJ k a Y k b J k b Y k a
71
Figure 3.16 (p. 129)Normalized cutoff frequency of the dominant TE11 waveguide mode for a coaxial line.
72
Figure 3.17 (p. 129)Field lines for the (a) TEM and (b) TE11 modes of a coaxial line.
73
Ex 3.3
• a = 0.035”, b = 0.116”, εr = 2.2. What is the highest usable frequency, before the TE11 waveguide mode starts to propagate?
74
Photograph on Page 134.
75
Figure 3.18 (p. 131)Geometry of a grounded dielectric slab.
76
Figure 3.19 (p. 133)Graphical solution of the transcendental equation for the cutoff frequency of a TM surface wave mode of the grounded dielectric slab.
77
Figure 3.20 (p. 135)Graphical solution of the transcendental equation for the cutoff frequency of a TE surface wave mode. Figure depicts a mode below cutoff.
78
Figure 3.21 (p. 136)Surface wave propagation constants for a grounded dielectric slab with εr = 2.55.
79
80
3.7 Stripline
• A planar-type of transmission line that lends itself well to microwave integrated circuitry and photolithographic fabrication.
• Since stripline has 2 conductors and a homogeneous dielectric, it can support a TEM wave.
• The stripline can also support higher order TM and TE modes, but these are usually avoided in practice.
81
Figure 3.22 (p. 137)Stripline transmission line. (a) Geometry. (b) Electric and magnetic field lines.
82
Figure 3.23 (p. 138)Photograph of a stripline circuit assembly, showing four quadrature hybrids, open-circuit tuning stubs, and coaxial transitions.
83
Formulas for Propagation Constant, Characteristic Impedance and Attenuation• The phase velocity of a TEM mode:
• The propagation constant of the stripline is
• The characteristic impedance of a transmission line is
• Laplace’s equation can be solved by conformal mapping to find the capacitance per unit length of the stripline. complicated special function
0 01/ /p r rv c
0 0 0r rp
kv
0
1
p
L LCZ
C C v C
84
• For practical computations simple formulas have been developed by curve fitting to the exact solution.
where We is the effective width of the center conductor
• Given the characteristic impedance, the strip width is
where
0
30
0.441er
bZ
W b
2
0 for / 0.35
(0.35 / ) for / 0.35e
W bW W
W b W bb b
r 0
r 0
for 120
0.85 0.6 for 120
x ZW
b x Z
0
300.441
r
xZ
85
• The attenuation due to the dielectric loss is the same as (3.30).
• The attenuation due to the conductor loss
with
• Ex 3.5
-30
r 0
r 00
2.7 10 for 120
30 ( )
0.16 for 120
s r
c
s
R ZA Z
b t
RB Z
Z b
2 1 2 1 ln
0.414 1 41 0.5 ln
(0.5 0.7 ) 2
W b t b tA
b t b t t
b t WB
W t W t
86Figure 3.24 (p. 141)Geometry of enclosed stripline.
• Modified the geometry truncating the plates beyond some distance and placing metal walls on the sides.
An Approximate Electrostatic Solution
87
3.8 Microstrip
• Microstrip line is one of the most popular types of transmission lines, primarily because it can be fabricated by photolithographic process and is easily integrated with other passive and active microwave devices.
• Microstrip line cannot support a pure TEM wave.
• In most practical applications, the dielectric substrate is electrically very thin (d<<λ), and so the fields are quasi-TEM.
0, , 1p e e r
e
cv k
88
Figure 3.25 (p. 143)Microstrip transmission line. (a) Geometry. (b) Electric and magnetic field lines.
89
Figure 3.26 (p. 145)Equivalent geometry of quasi-TEM microstrip line, where the dielectric slab of thickness I and relative permittivity εr has been replaced with a homogeneous medium of effective relative permittivity, εe.
90
Formulas for Propagation Constant, Characteristic Impedance and Attenuation• The effective dielectric constant of a
microstrip line:
• The characteristic impedance of a microstrip line is
1 1 1
2 2 1 12 /r r
ed W
0
60 8ln for / 1
4
120 for / 1
/ 1.393 0.667 ln( / 1.444)
e
e
d WW d
W dZ
W dW d W d
91
• Given Z0, and εr, the strip width is
where
• The attenuation due to dielectric loss
2
8 for / 2
2
2 1 0.611 ln(2 1) ln( 1) 0.39 for / 2
2
A
A
r
r r
eW d
eW
dB B B W d
0
0
1 1 0.110.23
60 2 1
377
2
r r
r r
r
ZA
BZ
0 ( 1) tan
2 ( 1)r e
d
e r
k
92
• The attenuation due to the conductor loss
where is the surface resistivity of the conductor.
• Ex 3.7
0
sc
R
Z W 0 / 2sR
93Figure 3.27 (p. 146)Geometry of a microstrip line with conducting sidewalls.
An Approximate Electrostatic Solution
94
• The potential Φ(x,y) satisfying Laplace’s equation:
with BC
• Since there are 2 regions defined by air/dielectric interface, with a charge discontinuity on the strip, we will have separate Φ(x,y).
( , ) 0 for / 2, 0t x y x a y
( , ) 0 at / 2, 0,x y x a y
2 2
2 2
2 2
( , ) ( ) ( ) 0
0 ,
( , ) ( cos sin )( cosh sinh )
x x
x x x x
x y X x Y yx y
X Y X Yk k
X Y X Yx y A k x B k x C k y B k y
95
• Applying BC & even function on x & y,
• Φ must be continuous at y = d
0, 0, cos 02x
x
k a nB C A k
a
1, ;
/
1, ;
cos sinh for 0
( , )
cos for
nn n odd
n y an
n n odd
n x n yA y d
a ax y
n xB e d y
a
/sinh n d an n
n dA B e
a
96
1, ;
( ) /
1, ;
cos sinh for 0
( , )
cos sinh for
nn n odd
n y d an
n n odd
n x n yA y d
a ax y
n x n dA e d y
a a
1, ;
( ) /
1, ;
cos cosh for 0
cos sinh for d
nn n odd
yn y d a
nn n odd
n n x n yA y d
a a aE
y n n x n dA e y
a a a
97
• The surface current density at y = d,
• By a good guess,
• Taking the orthogonalization for both sides of ρs,
0 0
01, ;
( , ) ( , )
( , ) ( , )
cos sinh cosh
s y y
y r y
n rn n odd
D x y d D x y d
E x y d E x y d
n n x n d n dA
a a a a
1 for / 2( )
0 for / 2s
x Wx
x W
/ 2 / 2 20/ 2 / 2
/ 2
0/ 2
cos cos sinh cosh
sin sinh cosh4
a a
n ra a
W
n rW
n x n n x n d n ddx A dx
a a a a a
a n x n a n d n dA
n a a a a
98
2
0
4 sin / 2
sinh coshn
r
a n W aA
n d n dn
a a
01
/ 2
/ 2
1
0 0
( 0, ) sinh2
( ) , sinh
2
1, /
d
y nn
W
sW
nn
ee
p
n bV E x y dy A
a
Q WQ x dx W C
n bVA
a
LZ C C
C v C cC
HW2. Use MATLAB to calculate the Z0 with εr = 2.2, a = 100d, for W/d = 0 to 10 (step = 0.01) and compare these results with (3.195) & (3.196)
C = capacitance per unit length of the microstrip line with εr ≠ 1
C0 = capacitance per unit length of the microstrip line with εr = 1
99
Figure 3.28 (p. 150)A rectangular waveguide partially filled with dielectric and the transverse resonance equivalent circuit.
100
3.10 Wave Velocities and Dispersion• The speed of light in a medium is the velocity
at which a plane wave would propagate in that medium.
• The phase velocity is the speed at which a constant phase point travels.
• The phase velocity is different for different frequencies, then the individual frequency components will not maintain their original phase relationships as they propagate down the transmission line or waveguide, and signal distortion will occur. dispersion
101
Group Velocity
• The velocity at which a narrow band signal propagates.
( ) ( )
1( ) ( )
2
j t
j t
F f t e dt
f t F e d
102
• For a lossless, matched transmission line or waveguide,
• The time-domain output signal,
• Now if |Z(ω)| = A is a constant, and the phase of ψ is a linear function of ω (ψ = aω),
dispersionless (not distortion for a lossless TEM wave)
( ) ( )j z jZ Ae Z e
( )0
1( ) ( ) ( )
2j tf t F Z e d
( )0
1( ) ( ) ( )
2j t af t AF e d Af t a
103
• For a lossy TEM line, consider a narrow band input signal of the form representing an AM modulated carrier wave of frequency ω0. 0
0
0
0
( ) ( ) cos Re ( )
( ) ( ) ( )
j t
j t j t
s t f t t f t e
S f t e e dt F
0
( )
( ) ( )
1 1( ) ( ) ( )
2 2
o m
o m
j zo
j t j t zo o o
S AF e
s t S e d AF e d
104
• In general, β may be a complicated function of ω. But if F(ω) is narrowband (ωm<<ωo),
• Then, y = ω- ω0,
a time-shifted replica of the original modulation envelope f(t).
0 0 0
0 0
( ) ( ) ( ) ( )
where ( ),
o
o
o o
d
d
d
d
0 0 0
0 0
( ) ( )
( )0 0 0 0
( ) Re ( )2
Re ( ) ( ) cos( )
j t z j t z yo
j t z
As t e F y e dy
A f t z e Af t z t z
105
• The velocity of this envelope is the group velocity, vg:
• Ex 3.9
0
1
0
1g
dv
d
106
Figure 3.31 (p. 155)Cross section of a ridge waveguide.
107
Figure 3.32 (p. 155)Dielectric waveguide geometry.
108
Figure 3.33 (p. 156)Geometry of a printed slotline.
109
Figure 3.34 (p. 156)Coplanar waveguide geometry.
110
Figure 3.35 (p. 157)Covered microstrip line.