text stringlengths 1 1k ⌀ | source stringclasses 12
values |
|---|---|
∂2
∂x2
˜Ex + ∂2
∂y2
˜Ex + ∂2
∂z2
˜Ex + β2 ˜Ex = 0 (6.148)
∂2
∂x2
˜Ey + ∂2
∂y2
˜Ey + ∂2
∂z2
˜Ey + β2 ˜Ey = 0 (6.149)
∂2
∂x2
˜Ez + ∂2
∂y2
˜Ez + ∂2
∂z2
˜Ez + β2 ˜Ez = 0 (6.150)
In
general, we expect the total field in the waveguide
to consist of unidirectional waves propagating in the
+ˆz and −ˆz directions. W e may analyz... | Electromagnetics_Vol2.pdf |
˜Hz = 0; i.e., is transverse (perpendicular) to the
direction of propagation. Thus, the TM component is
completely determined by ˜Ez.
Equations 6.136–6.139 simplify to become:
˜Ex = −jkz
k2ρ
∂˜Ez
∂x (6.151)
˜Ey = −jkz
k2ρ
∂˜Ez
∂y (6.152)
˜Hx =
+jωµ
k2ρ
∂˜Ez
∂y (6.153)
˜Hy = −jω
µ
k2ρ
∂˜Ez
∂x (6.154)
where
k2
ρ ≜ β2 − k... | Electromagnetics_Vol2.pdf |
∂2
∂x2 ˜ez + ∂2
∂y2 ˜ez − k2
z˜ez + β2˜ez = 0 (6.157)
The
last two terms may be combined using
Equation 6.155, yielding:
∂2
∂x2 ˜ez + ∂2
∂y2 ˜ez + k2
ρ˜ez = 0 (6.158)
This
is a partial differential equation for ˜ez in the
variables xand y. This equation may be solved using
the technique of separation of variables. In t... | Electromagnetics_Vol2.pdf |
112 CHAPTER 6. W A VEGUIDES
Substituting this expression into Equation 6.158, we
obtain:
Y ∂2
∂x2 X+ X ∂2
∂y2 Y + k2
ρXY = 0 (6.160)
Next dividing through by XY, we obtain:
1
X
∂2
∂x2 X+ 1
Y
∂2
∂y2 Y + k2
ρ = 0 (6.161)
Note
that the first term depends only on x, the second
term depends only on y, and the remaining term ... | Electromagnetics_Vol2.pdf |
and Y, respectively , we find:
∂2
∂x2 X+ k2
xX = 0 (6.165)
∂2
∂y2 Y + k2
yY = 0 (6.166)
These
are familiar one-dimensional differential
equations. The solutions are: 4
X = Acos (kxx) + Bsin (kxx) (6.167)
Y = Ccos (kyy) + Dsin (kyy) (6.168)
where A, B, C, and D– like kx and ky – are
constants to be determined. At this po... | Electromagnetics_Vol2.pdf |
tangent to a perfectly-conducting wall must be zero.
Note that the ˆz component of ˜E is tangent to all four
walls; therefore:
˜Ez(x= 0) = 0 (6.170)
˜Ez(x= a) = 0 (6.171)
˜Ez(y= 0) = 0 (6.172)
˜Ez(y= b) = 0 (6.173)
Referring to Equation 6.169, these boundary
conditions in turn require:
X(x= 0) = 0 (6.174)
X(x= a) = 0 (... | Electromagnetics_Vol2.pdf |
Each positive integer value of mand nleads to a
valid expression for ˜Ez known as a mode. Solutions
for which m= 0 or n= 0 yield kx = 0 or ky = 0,
respectively . These correspond to zero-valued fields,
and therefore are not of interest. The most general | Electromagnetics_Vol2.pdf |
6.9. RECT ANGULAR W A VEGUIDE: TE MODES 113
expression for ˜Ez must account for all non-trivial
modes. Summarizing:
˜Ez =
∞∑
m=1
∞∑
n=1
˜E(m,n)
z (6.186)
where
˜E(m,n)
z ≜ E(m,n)
0 sin
(mπ
a x
)
sin
(nπ
b y
)
e−jk(m
,n )
z z
(6.187)
where E(m,n)
0 is an arbitrary constant (consolidating
the constants Band D), and, sinc... | Electromagnetics_Vol2.pdf |
“TM mn. ” For example, the mode TM 12 is given by
Equation 6.187 with m= 1 and n= 2.
Finally , note that values of k(m,n)
z obtained from
Equation 6.188 are not necessarily real-valued. It is
apparent that for any given value of m, k(m,n)
z will be
imaginary-valued for all values of ngreater than some
value. Similarly ... | Electromagnetics_Vol2.pdf |
propagation of waves. Rectangular waveguide is
commonly used for the transport of radio frequency
signals at frequencies in the SHF band (3–30 GHz)
and higher. The fields in a rectangular waveguide
consist of a number of propagating modes which
depends on the electrical dimensions of the
waveguide. These modes are broad... | Electromagnetics_Vol2.pdf |
waveguide which is free of sources. Expressed in
phasor form, the magnetic field intensity within the
waveguide is governed by the wave equation:
∇2 ˜H + β2 ˜H = 0 (6.189)
where
β = ω√
µǫ (6.190)
Figure 6.7: Geometry for analysis of fields in a rect-
angular
waveguide. | Electromagnetics_Vol2.pdf |
114 CHAPTER 6. W A VEGUIDES
Equation 6.189 is a partial differential equation. This
equation, combined with boundary conditions
imposed by the perfectly-conducting plates, is
sufficient to determine a unique solution. This
solution is most easily determined in Cartesian
coordinates, as we shall now demonstrate. First we... | Electromagnetics_Vol2.pdf |
∂z2
˜Hx + β2 ˜Hx = 0 (6.196)
∂2
∂x2
˜Hy + ∂2
∂y2
˜Hy + ∂2
∂z2
˜Hy + β2 ˜Hy = 0 (6.197)
∂2
∂x2
˜Hz + ∂2
∂y2
˜Hz + ∂2
∂z2
˜Hz + β2 ˜Hz = 0 (6.198)
In
general, we expect the total field in the waveguide
to consist of unidirectional waves propagating in the
+ˆz and −ˆz directions. W e may analyze either of
these waves; then... | Electromagnetics_Vol2.pdf |
˜Ez = 0; i.e., is transverse (perpendicular) to the
direction of propagation. Thus, the TE component is
completely determined by ˜Hz.
Equations 6.136–6.139 simplify to become:
˜Ex = −jωµ
k2ρ
∂˜Hz
∂y (6.199)
˜Ey =
+jωµ
k2ρ
∂˜Hz
∂x (6.200)
˜Hx = −jkz
k2ρ
∂˜Hz
∂x (6.201)
˜Hy = −jkz
k2ρ
∂˜Hz
∂y (6.202)
where
k2
ρ ≜ β2 − k2... | Electromagnetics_Vol2.pdf |
∂2
∂x2
˜hz + ∂2
∂y2
˜hz − k2
z˜hz + β2˜hz = 0 (6.205)
The
last two terms may be combined using
Equation 6.203, yielding:
∂2
∂x2
˜hz + ∂2
∂y2
˜hz + k2
ρ˜hz = 0 (6.206)
This
is a partial differential equation for ˜hz in the
variables xand y. This equation may be solved using
the technique of separation of variables. In t... | Electromagnetics_Vol2.pdf |
6.9. RECT ANGULAR W A VEGUIDE: TE MODES 115
Substituting this expression into Equation 6.206, we
obtain:
Y ∂2
∂x2 X+ X ∂2
∂y2 Y + k2
ρXY = 0 (6.208)
Next dividing through by XY, we obtain:
1
X
∂2
∂x2 X+ 1
Y
∂2
∂y2 Y + k2
ρ = 0 (6.209)
Note
that the first term depends only on x, the second
term depends only on y, and the... | Electromagnetics_Vol2.pdf |
ρ (6.212)
Now multiplying Equations 6.210 and 6.211 by X
and Y, respectively , we find:
∂2
∂x2 X+ k2
xX = 0 (6.213)
∂2
∂y2 Y + k2
yY = 0 (6.214)
These
are familiar one-dimensional differential
equations. The solutions are: 5
X = Acos (kxx) + Bsin (kxx) (6.215)
Y = Ccos (kyy) + Dsin (kyy) (6.216)
where A, B, C, and D– li... | Electromagnetics_Vol2.pdf |
this case, it is required that any component of ˜E that
is tangent to a perfectly-conducting wall must be
zero. Therefore:
˜Ey(x= 0) = 0 (6.218)
˜Ey(x= a) = 0 (6.219)
˜Ex(y= 0) = 0 (6.220)
˜Ex(y= b) = 0 (6.221)
Referring to Equation 6.217 and employing
Equations 6.199–6.202, we obtain:
∂
∂xX(x= 0) = 0 (6.222)
∂
∂xX(x= ... | Electromagnetics_Vol2.pdf |
Equations 6.231 and 6.233 reduce to:
sin (kxa) = 0 (6.234)
sin (kyb) = 0 (6.235)
This in turn requires:
kx = mπ
a , m = 0,1,2... (6.236)
ky = nπ
b , n= 0,1,2... (6.237) | Electromagnetics_Vol2.pdf |
116 CHAPTER 6. W A VEGUIDES
Each positive integer value of mand nleads to a valid
expression for ˜Hz known as a mode. Summarizing:
˜Hz =
∞∑
m=0
∞∑
n=0
˜H(m,n)
z (6.238)
where
˜H(m,n)
z ≜ H(m,n)
0 cos
(mπ
a x
)
cos
(nπ
b y
)
e−jk(m
,n )
z z
(6.239)
where H(m,n)
0 is an arbitrary constant (consolidating
the constants Aan... | Electromagnetics_Vol2.pdf |
“TE mn. ” For example, the mode TE 12 is given by
Equation 6.239 with m= 1 and n= 2.
Although Equation 6.238 implies the existence of a
TE00 mode, it should be noted that this wave has no
non-zero electric field components. This can be
determined mathematically by following the
procedure outlined above. However, this is... | Electromagnetics_Vol2.pdf |
value. Similarly , it is apparent that for any given value
of n, k(m,n)
z will be imaginary-valued for all values of
mgreater than some value. This phenomenon is
common to both TE and TM components, and so is
addressed in a separate section (Section 6.10).
Additional Reading:
• “W aveguide (radio frequency)” on Wikiped... | Electromagnetics_Vol2.pdf |
6.10. RECT ANGULAR W A VEGUIDE: PROP AGA TION CHARACTERISTICS 117
6.10 Rectangular W aveguide:
Propagation Characteristics
[m0224]
In this section, we consider the propagation
characteristics
of TE and TM modes in rectangular
waveguides. Because these modes exhibit the same
phase dependence on z, findings of this sectio... | Electromagnetics_Vol2.pdf |
k(m,n)
z obtained from Equation 6.242 are not
necessarily real-valued. For any given value of m,
(k(m,n)
z )2 will be negative for all values of ngreater
than some value. Similarly , for any given value of n,
(k(m,n)
z )2 will be negative for all values of mgreater
than some value. Should either of these conditions
occ... | Electromagnetics_Vol2.pdf |
of the wave decreases exponentially with increasing
z. Such a wave does not effectively convey power
through the waveguide, and is said to be cut off.
Since waveguides are normally intended for the
efficient transfer of power, it is important to know the
criteria for a mode to be cut off. Since cutoff occurs
when (k(m,n... | Electromagnetics_Vol2.pdf |
if the frequency is high enough to meet this criterion.
Thus, it is useful to make the following definition:
fmn ≜ vpu
2
√ (m
a
)2
+
(n
b
)2
(6.253)
The cutoff frequency fm n (Equation 6.253) is the
lowest frequency for which the mode (m,n) is
able to propagate (i.e., not cut off). | Electromagnetics_Vol2.pdf |
118 CHAPTER 6. W A VEGUIDES
Example 6.4. Cutof f frequencies for WR-90.
WR-90 is a popular implementation of
rectangular waveguide. WR-90 is air-filled with
dimensions a= 22.86 mm and b= 10.16 mm.
Determine cutoff frequencies and, in particular,
the lowest frequency at which WR-90 can be
used.
Solution. Since WR-90 is a... | Electromagnetics_Vol2.pdf |
The lowest-order TM mode that is non-zero and
not cut off is TM 11 (f11 = 16.145 GHz).
Phase velocity. The phase velocity for a wave
propagating within a rectangular waveguide is greater
than that of electromagnetic radiation in unbounded
space. For example, the phase velocity of any
propagating mode in a vacuum-filled ... | Electromagnetics_Vol2.pdf |
Equation 6.253 and also noting that ω= 2πf,
Equation 6.256 may be rewritten in the following
form:
vp = vpu√
1 − (fmn/f)2
(6.257)
F
or any propagating mode, f >fmn; subsequently ,
vp >vpu. In particular, vp >c for a vacuum-filled
waveguide.
How can this not be a violation of fundamental
physics? As noted in Section 6.1,... | Electromagnetics_Vol2.pdf |
velocity vg. In unbounded space, vg = vp, so the
speed of information is equal to the phase velocity in
that case. In a rectangular waveguide, the situation is
different. W e find:
vg =
(
∂k(m,n)
z
∂ω
)−1
(6.258)
= vpu
√
1 − (fmn/f)2 (6.259)
which
is always less than vpu for a propagating mode.
Note that group velocity ... | Electromagnetics_Vol2.pdf |
6.10. RECT ANGULAR W A VEGUIDE: PROP AGA TION CHARACTERISTICS 119
The speed of a signal within a rectangular waveg-
uide
is given by the group velocity of the as-
sociated mode (Equation 6.259). This speed is
less than the speed of propagation in unbounded
media having the same permittivity and perme-
ability . Speed d... | Electromagnetics_Vol2.pdf |
next-lowest cutoff frequency is
f20 = 13.114 GHz. Therefore, only the TE 10
mode is available for this signal. The group
velocity for this mode at the frequency of
interest is given by Equation 6.259. Using this
equation, the speed of propagation is found to be
∼= 2.26 × 108 m/s
, which is about 75.5% of c.
[m0212] | Electromagnetics_Vol2.pdf |
120 CHAPTER 6. W A VEGUIDES
Image Credits
Fig. 6.1: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Geometry for analysis of fields in parallel plate waveguide.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/), modified.
Fig. 6.2: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wi... | Electromagnetics_Vol2.pdf |
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 6.5: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:TM component of electric field in parallel plate waveguide.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/), modified. | Electromagnetics_Vol2.pdf |
Chapter 7
T ransmission
Lines Redux
7.1 Parallel Wire T ransmission
Line
[m0188]
A parallel wire transmission line consists of wires
separated
by a dielectric spacer. Figure 7.1 shows a
common implementation, commonly known as “twin
lead. ” The wires in twin lead line are held in place by
a mechanical spacer comprised ... | Electromagnetics_Vol2.pdf |
line lacks the self-shielding property of coaxial cable;
i.e., the electromagnetic fields of coaxial line are
isolated by the outer conductor, whereas those of
c⃝ Sp inningSpark, Inductiveload CC BY SA 3.0 (modified)
Figure 7.1: T win lead, a commonly-encountered form
of parallel wire transmission line.
c⃝ C. W ang CC BY... | Electromagnetics_Vol2.pdf |
applications where the signal sources and/or loads are
also differential; common examples are the dipole
antenna and differential amplifiers. 2
Figure 7.2 shows a cross-section of parallel wire line.
Relevant parameters include the wire diameter, d; and
the center-to-center spacing, D.
1 The references in “ Additional R... | Electromagnetics_Vol2.pdf |
122 CHAPTER 7. TRANSMISSION LINES REDUX
c⃝ S. Lally CC BY SA 4.0
Figure 7.3: Structure of the electric and magnetic
fields for a cross-section of parallel wire line. In this
case, the wave is propagating away from the viewer.
The associated field structure is transverse
electromagnetic (TEM) and is therefore completely
d... | Electromagnetics_Vol2.pdf |
Z0 =
√
R′ + jωL′
G′ + jωC′ (7.1)
where R′, G′, C′,
and L′ are the resistance,
conductance, capacitance, and inductance per unit
length, respectively . This analysis is considerably
simplified by neglecting loss; therefore, let us assume
the “low-loss” conditions R′ ≪ ωL′ and G′ ≪ ωC′.
Then we find:
Z0 ≈
√
L′
C′ (lo w los... | Electromagnetics_Vol2.pdf |
µ0/ǫ0 ≜ η0, we obtain
Z0 ≈ 1
π
η0
√ǫr
ln (2D/d) (7.8)
The characteristic impedance of parallel wire
line,
assuming low-loss conditions and wire spac-
ing much greater than wire diameter, is given by
Equation 7.8.
Observe that the characteristic impedance of parallel
wire line increases with increasing D/d. Since this
r... | Electromagnetics_Vol2.pdf |
calculating transmission line parameters, since
the jacket and spacer have only a small effect on
the fields. For these values, Equation 7.8 gives
Z0 ≈ 298 Ω, as expected.
Under the assumption that the wire jacket/spacer
material has a negligible effect on the electromagnetic | Electromagnetics_Vol2.pdf |
7.2. MICROSTRIP LINE REDUX 123
fields, and that the line is suspended in air so that
ǫr ≈ 1, the phase velocity vp for a parallel wire line is
approximately that of any electromagnetic wave in
free space; i.e., c. In practical twin-lead, the effect of
a plastic jacket/spacer material is to reduce the phase
velocity by a... | Electromagnetics_Vol2.pdf |
transmission line on a printed circuit board, and so
accounts for an important and expansive range of
applications. The reader should be aware that
microstrip is distinct from stripline, which is a
distinct type of transmission line; see “ Additional
Reading” at the end of this section for disambiguation
of these terms... | Electromagnetics_Vol2.pdf |
length in the direction perpendicular to the direction
of propagation. Despite this difference, the parallel
plate waveguide provides some useful insight into the
operation of the microstrip line. Microstrip line is
nearly always operated below the cutoff frequency of
3 The reference in “ Additional Reading” at the end... | Electromagnetics_Vol2.pdf |
124 CHAPTER 7. TRANSMISSION LINES REDUX
Figure 7.5: Appr oximate structure of the electric and
magnetic fields within microstrip line, assuming TM 0
operation. The fields outside the line are possibly sig-
nificant, complicated, and not shown. In this case, the
wave is propagating away from the viewer.
all but the TM 0 mo... | Electromagnetics_Vol2.pdf |
The limited width W of the trace results in a “fringing
field” – i.e., significant deviations from TM 0 field
structure in the dielectric beyond the edges of the
trace and above the trace. The fringing fields may
play a significant role in determining the
characteristic impedance Z0. Since Z0 is an important
parameter in th... | Electromagnetics_Vol2.pdf |
with propagating waves lies directly underneath the
trace, and Figure 7.5 provides a relatively accurate
impression of the fields in this region. The
characteristic impedance Z0 may be determined using
the “lumped element” transmission line model using
the following expression:
Z0 =
√
R′ + jωL′
G′ + jωC′ (7.9)
where R′,... | Electromagnetics_Vol2.pdf |
the capacitance Cis given by
C ≈ ǫA
d (parallel plate capacitor) (7.11)
In terms of wide microstrip line, A= Wl where lis
length, and d= h. Therefore:
C′ ≜ C
l ≈ ǫW
h (W ≫ h) (7.12)
T
o determine L′, consider the view of the microstrip
line shown in Figure 7.6. Here a current source
applies a steady current I on the le... | Electromagnetics_Vol2.pdf |
7.2. MICROSTRIP LINE REDUX 125
c⃝ C. W ang CC BY SA 4.0
Figure 7.6: V iew from the side of a microstrip line,
used to determine L′.
where H is the magnitude of H. Next, recall that:
L≜ Φ
I (7.15)
So,
we may determine Lif we are able to obtain an
expression for I in terms of H. This can be done as
follows. First, note t... | Electromagnetics_Vol2.pdf |
I = JsW ≈ HW (7.17)
Returning to Equation 7.15, we find:
L≜ Φ
I ≈ µ0Hhl
HW = µ0hl
W (7.18)
Subsequently
,
L′ ≜ L
l ≈ µ0h
W (W ≫ h) (7.19)
No
w the characteristic impedance is found to be
Z0 ≈
√
L′
C′ ≈
√
µ0h/W
ǫW/h =
√ µ0
ǫ
h
W (7.20)
The
factor
√
µ0/ǫis recognized as the wave
impedance η. It is convenient to express th... | Electromagnetics_Vol2.pdf |
see.
Narrow case. Figure 7.5 does not accurately depict
the fields in the case that W ≪ h. Instead, much of
the energy associated with the electric and magnetic
fields lies beyond and above the trace. Although these
fields are relatively complex, we can develop a
rudimentary model of the field structure by
considering the ... | Electromagnetics_Vol2.pdf |
126 CHAPTER 7. TRANSMISSION LINES REDUX
Figure 7.8: Noting the similarity of the fields in nar-
ro
w microstrip line to those in a parallel wire line.
Figure 7.9: Modeling the fields in narrow microstrip
line
as those of a parallel wire line, now introducing
the ground plane.
Figure 7.8. Note that the fields above the die... | Electromagnetics_Vol2.pdf |
plane. Thus, we see that the parallel wire transmission
line provides a pretty good guide to the structure of
the fields for the narrow transmission line, at least in
the dielectric region of the upper half-space.
W e are now ready to estimate the characteristic
impedance Z0 ≈
√
L′/C′ of low-loss narrow
microstrip line.... | Electromagnetics_Vol2.pdf |
properties can be useful for making fine adjustments
to a microstrip design.
The characteristic impedance of “narrow” (W ≪
h) microstrip
line can be roughly approximated
by Equation 7.22.
Intermediate case. An expression for Z0 in the
intermediate case – even a rough estimate – is difficult
to derive, and beyond the scop... | Electromagnetics_Vol2.pdf |
7.2. MICROSTRIP LINE REDUX 127
Figure 7.10: Z0 for FR4 as a function of h/W, as
determined by the “wide” and “narrow” approxima-
tions, along with the Wheeler 1977 formula. Note that
the vertical and horizontal axes of this plot are in log
scale.
the wide and narrow expressions (blue and green
curves, respectively) for... | Electromagnetics_Vol2.pdf |
h/W ∼ 1. Here it is:
Z0 ≈ 42.4 Ω
√ǫr + 1 ×
ln
[
1
+ 4h
W′
(
K+
√
K2 + 1 + 1/ǫr
2 π2
)]
(7.23)
where
K ≜ 14
+ 8/ǫr
11
(4h
W′
)
(7.24)
and W′ is W adjusted
to account for the thickness t
of the microstrip line. T ypically t≪ W and t≪ h,
for which W′ ≈ W. Although complicated, this
formula should not be completely surpris... | Electromagnetics_Vol2.pdf |
having h∼= 1.575 mm and ǫr ≈ 4.5. Figure 7.10
shows Z0 as a function of h/W, as determined
by the wide and narrow approximations, along
with the value obtained using the Wheeler 1977
formula. The left side of the plot represents the
wide condition, whereas the right side of this
plot represents the narrow condition. Th... | Electromagnetics_Vol2.pdf |
Also worth noting from Figure 7.10 is the
important and commonly-used result that
Z0 ≈ 50 Ω is obtained for h/W ≈ 0.5. Thus, a
50 Ω microstrip line in FR4 has a trace width of
about 3 mm.
A useful “take away” from this example is that the
wide and narrow approximations serve as useful
guides for understanding how Z0 ch... | Electromagnetics_Vol2.pdf |
128 CHAPTER 7. TRANSMISSION LINES REDUX
method; e.g., using the Wheeler 1977 formula or from
measurements.
FR4 circuit board construction is so common that the
result from the previous example deserves to be
highlighted:
In FR4 printed circuit board construction (sub-
strate
thickness 1.575 mm, relative permittivity
≈ ... | Electromagnetics_Vol2.pdf |
W avelength in microstrip line. An accurate general
formula for wavelength λin microstrip line is
similarly difficult to derive. A useful approximate
technique employs a result from the theory of uniform
plane waves in unbounded media. For such waves,
the phase propagation constant βis given by
β = ω√
µǫ (7.25)
It
turns... | Electromagnetics_Vol2.pdf |
(for “effective relative permittivity”). Then:
β ≈ ω√
µ0 ǫr,ef f ǫ0 (low-loss microstrip)
= β0
√ǫr,ef f (7.26)
In other words, the phase propagation constant in a
microstrip line can be approximated as the free-space
phase propagation β0 ≜ ω√µ0ǫ0 times a correction
f
actor √ǫr,ef f.
Next, ǫr,eff is crudely approximated... | Electromagnetics_Vol2.pdf |
relationship
vp = ω
β = c√ǫr,ef f
(7.29)
i.e., the phase velocity in microstrip is slower than c
by a factor of √ǫr,ef f.
Example 7.3. W a velength and phase velocity in
microstrip in FR4 printed circuit boards.
FR4 is a low-loss fiberglass epoxy dielectric that
is commonly used to make printed circuit boards
(see “ Add... | Electromagnetics_Vol2.pdf |
7.3. A TTENUA TION IN COAXIAL CABLE 129
• “Printed circuit board” on Wikipedia.
• “Stripline” on Wikipedia.
• “Single-ended signaling” on Wikipedia.
• Sec. 8.7 (“Differential Circuits”) in
S.W . Ellingson, Radio Systems Engineering,
Cambridge Univ . Press, 2016.
• H.A. Wheeler, “Transmission Line Properties of
a Strip ... | Electromagnetics_Vol2.pdf |
outer conductors, whereas G′ represents loss due to
current flowing directly between the conductors
through the spacer material.
The parameters used to describe the relevant features
of coaxial cable are shown in Figure 7.11. In this
figure, aand bare the radii of the inner and outer
conductors, respectively . σic and σo... | Electromagnetics_Vol2.pdf |
which the current flows. The latter is equal to the
circumference 2πatimes the skin depth δic of the
ϵr
σ sσ ic
σ oc
b
a
Figure 7.11: Parameters defining the design of a coax-
ial
cable. | Electromagnetics_Vol2.pdf |
130 CHAPTER 7. TRANSMISSION LINES REDUX
inner conductor, so:
R′
ic ≈ 1
(2πa· δic) σic
for δic ≪ a (7.30)
This
expression is only valid for δic ≪ abecause
otherwise the cross-sectional area through which the
current flows is not well-modeled as a thin ring near
the surface of the conductor. Similarly , we find the
resista... | Electromagnetics_Vol2.pdf |
√
2/ωµ0
[ 1
a√σic
+ 1
b√σoc
]
(7.35)
At
this point it is convenient to identify two particular
cases for the design of the cable. In the first case,
“Case I, ” we assume σoc ≫ σic. Since b>a, we
have in this case
R′ ≈ 1
2π
√
2/ωµ0
[ 1
a√σic
]
= 1
2πδicσic
1
a (Case I) (7.36)
In the second case, “Case II, ” we assume σoc... | Electromagnetics_Vol2.pdf |
frequency , at least to the extent that σs is independent
of frequency .
Attenuation. The attenuation of voltage and current
waves as they propagate along the cable is
represented by the factor e−αz, where zis distance
traversed along the cable. It is possible to find an
expression for αin terms of the material and
geom... | Electromagnetics_Vol2.pdf |
7.3. A TTENUA TION IN COAXIAL CABLE 131
where Z0 is the characteristic impedance
Z0 ≈ η0
2π
1√ǫr
ln b
a (lo w loss) (7.46)
and where KR is a unitless constant to be determined.
The justification for Equation 7.45 is as follows: First,
αR must increase monotonically with increasing R′.
Second, R′ must be divided by an im... | Electromagnetics_Vol2.pdf |
conductor is difficult to quantify because it
consists of a braid of thin metal strands.
However, σoc ≫ σic, so we may assume Case I;
i.e., σoc ≫ σic, and subsequently C = 0.
Figure 7.12 shows the components αG and αR
computed for the particular choice
KR = KG = 1/2. The figure also shows
αG + αR, along with αcomputed us... | Electromagnetics_Vol2.pdf |
Figure 7.12: Comparison of α = Re { γ} to αR, αG,
and αR + αG for KR = KG = 1/2. The result for
α has been multiplied by 1.01; otherwise the curves
would be too close to tell apart.
may consider αR and αG independently . Let us first
consider αG:
αG ≜ 1
2G′Z0
≈ 1
2 · 2πσs
ln (b/a) · 1
2π
η0
√ǫr
ln (b/a)
= η0
2
σs
√ǫr
(7... | Electromagnetics_Vol2.pdf |
132 CHAPTER 7. TRANSMISSION LINES REDUX
Here we see that αR is minimized by minimizing
ǫr/σic. It’s not surprising to see that we should
maximize σic. However, it’s a little surprising that we
should minimize ǫr. Furthermore, this is in contrast to
αG, which is minimized by maximizing ǫr. Clearly
there is a tradeoff to... | Electromagnetics_Vol2.pdf |
on aand b. This implies the existence of a
generally-optimum geometry . T o find this geometry ,
we minimize αR by taking the derivative with respect
to a, setting the result equal to zero, and solving for a
and/or b. Here we go:
∂
∂aαR = 1
2
√
2 · η0
√ ωµ0ǫr
σic
· ∂
∂a
[1
/a+ C/b]
ln (b/a)
(7.51)
This
derivative is wor... | Electromagnetics_Vol2.pdf |
can be solved by plotting the function, or by a few
iterations of trial and error; either way one quickly
finds
b/a∼= 3.59 (Case II) (7.55)
Summarizing, we have found that αis minimized by
choosing the ratio of the outer and inner radii to be
somewhere between 2.72 and 3.59, with the precise
value depending on the relat... | Electromagnetics_Vol2.pdf |
attenuation is less for dielectric-filled cables than it is
for air-filled cables. For example, let us once again
consider the RG-59 from Example 7.4. In that case,
ǫr ∼= 2.25 and C = 0, indicating Z0 ≈ 39.9 Ω is
optimum for attenuation. The actual characteristic
impedance of Z0 is about 75 Ω, so clearly RG-59 is
not opt... | Electromagnetics_Vol2.pdf |
7.4. POWER HANDLING CAP ABILITY OF COAXIAL CABLE 133
and
∂
∂a ln
(b
a
)
= ∂
∂a [ln (b) − ln (a)]
= − ∂
∂a ln (a)
= − 1
a (7.59)
So:
∂
∂a
[
a2 ln
(b
a
)]
= [2a] ln
(b
a
)
+ a2
[
− 1
a
]
= 2
aln
(b
a
)
− a (7.60)
This
result is substituted for a2 ln(b/a) in
Equation 7.51 to obtain Equation 7.52.
7.4 Power Handling Capabi... | Electromagnetics_Vol2.pdf |
one design a coaxial cable to maximize Pmax for a
given Epk?
W e begin by finding the electric potential V within
the cable. This can be done using Laplace’s equation:
∇2V = 0 (7.61)
Using the cylindrical (ρ,φ,z ) coordinate system with
the zaxis along the inner conductor, we have
∂V/∂φ = 0 due to symmetry . Also we set... | Electromagnetics_Vol2.pdf |
134 CHAPTER 7. TRANSMISSION LINES REDUX
The electric field intensity is given by:
E = −∇V (7.67)
Again we have ∂V/∂φ = ∂V/∂z = 0, so
E = −ˆρ ∂
∂ρV (7.68)
= −ˆρ ∂
∂ρ
[ −V0
ln (b/a) ln ρ+ V0 ln
(b)
ln (b/a)
]
(7.69)
=
+ˆρ V0
ρln (b/a) (7.70)
Note
that the maximum electric field intensity in the
spacer occurs at ρ= a; i.e.,... | Electromagnetics_Vol2.pdf |
Now let us consider if there is a value of awhich
maximizes Pmax. W e do this by seeing if
∂Pmax/∂a = 0 for some values of aand b. The
derivative is worked out in an addendum at the end of
this section. Using the result from the addendum, we
find:
∂
∂aPmax =
π
E2
pk
η0/√ǫr
[2 aln (b/a) − a] (7.76)
For the above expressi... | Electromagnetics_Vol2.pdf |
example, a material with higher ǫr may also have
higher σs, which means more current flowing through
the spacer and thus more ohmic heating. This
problem is so severe that cables that handle high RF
power often use air as the spacer, even though it has
the lowest possible value of ǫr. Also worth noting is
that σic and σ... | Electromagnetics_Vol2.pdf |
identified above. For air-filled cables, we obtain 30 Ω.
Since ǫr ≥ 1, this optimum impedance is less for
dielectric-filled cables than it is for air-filled cables.
Summarizing:
The power handling capability of coaxial trans-
mission
line is optimized when the ratio of radii
of the outer to inner conductors b/ais about 1.6... | Electromagnetics_Vol2.pdf |
7.5. WHY 50 OHMS? 135
Using the chain rule, we find:
∂
∂a
[ 1/a+ C
/b
ln (b/a)
]
=
[ ∂
∂a
(1
a + C
b
)]
ln−1
(b
a
)
+
(1
a + C
b
)[ ∂
∂a ln−1
(b
a
)]
(7.79)
Note
∂
∂a
(1
a + C
b
)
= − 1
a2 (7.80)
T
o handle the quantity in the second set of square
brackets, first define v= ln u, where u= b/a. Then:
∂
∂av−1 =
[ ∂
∂vv−1
][ ... | Electromagnetics_Vol2.pdf |
commonly specified as the port impedance for signal
sources, amplifiers, filters, antennas, and other RF
components. So, what’s special about 50 Ω? The
short answer is “nothing. ” In fact, other standard
impedances are in common use – prominent among
these is 75 Ω. It is shown in this section that a broad
range of impedan... | Electromagnetics_Vol2.pdf |
values to accommodate the smaller number of
applications where there may be specific compelling
considerations.
So, the question becomes “what makes characteristic
impedances in the range of 10s of ohms particularly
useful?” One consideration is attenuation in coaxial
cable. Coaxial cable is by far the most popular type... | Electromagnetics_Vol2.pdf |
having typical ǫr ≈ 2.25. Thus, 50 Ω is clearly a
reasonable choice if a single standard value is to be
established for all such cable.
Coaxial cables are often required to carry high power
signals. In such applications, power handling | Electromagnetics_Vol2.pdf |
136 CHAPTER 7. TRANSMISSION LINES REDUX
capability is also important, and is addressed in
Section 7.4. In that section, we find the power
handling capability of coaxial cable is optimized
when the ratio of radii of the outer to inner conductors
b/ais about 1.65. For the air-filled cables typically
used in high-power appl... | Electromagnetics_Vol2.pdf |
V alues of 50 Ω and 75 Ω also offer some convenience
when connecting RF devices to antennas. For
example, 75 Ω is very close to the impedance of the
commonly-encountered half-wave dipole antenna
(about 73 + j42 Ω), which may make impedance
matching to that antenna easier. Another
commonly-encountered antenna is the qua... | Electromagnetics_Vol2.pdf |
half-wave dipole has an impedance of about 300 Ω
and is balanced (not single-ended); thus, there is a
market for balanced transmission line having
Z0 = 300 Ω. However, it is very easy and
inexpensive to implement a balun (a device which
converts the dipole output from balanced to
unbalanced) while simultaneously steppi... | Electromagnetics_Vol2.pdf |
7.5. WHY 50 OHMS? 137
Image Credits
Fig. 7.1: c⃝ SpinningSpark, Inductiveload,
https://commons.wikimedia.org/wiki/File:T win-lead cable dimension.svg,
CC
BY -SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/).
Minor modifications.
Fig. 7.2: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Parallel w... | Electromagnetics_Vol2.pdf |
Fig. 7.6: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:V iew from the side of a microstrip line.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 7.7: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:Electric and Magnetic Fields for Microstrip.svg,
CC
BY -SA 4... | Electromagnetics_Vol2.pdf |
Chapter 8
Optical
Fiber
8.1 Optical Fiber: Method of
Operation
[m0178]
In its simplest form, optical fiber consists of
concentric
regions of dielectric material as shown in
Figure 8.1. A cross-section through the fiber reveals a
circular region of transparent dielectric material
through which light propagates. This is su... | Electromagnetics_Vol2.pdf |
cal fiber.
material is commonly quantified in terms of its index
of refraction. Index of refraction is the square root of
relative permittivity , and is usually assigned the
symbol n. Thus, if we define the relative
permittivities ǫr,f ≜ ǫf/ǫ0 for the fiber and
ǫr,c ≜ ǫc/ǫ0 for the cladding, then
nf ≜ √
ǫr,f (8.1)
nc ≜ √ǫr... | Electromagnetics_Vol2.pdf |
greater than the wavelength. Thus, from the
perspective of the light ray , the fiber appears to be an
unbounded half-space sharing a planar boundary with
the cladding, which also appears to be an unbounded
half-space.
Continuing under this presumption, the criterion for
total internal reflection is (from Section 5.11):
θ... | Electromagnetics_Vol2.pdf |
8.1. OPTICAL FIBER: METHOD OF OPERA TION 139
c⃝ S. Lally CC BY -SA 4.0
Figure 8.2: T otal internal reflection in optical fiber.
the fiber, and is reflected onward. Otherwise, power is
lost into the cladding.
Example 8.1. Critical angle for optical fiber.
T ypical values of nf and nc for an optical fiber
are 1.52 and 1.49, re... | Electromagnetics_Vol2.pdf |
bend. This will occur even for light rays which are
traveling perfectly parallel to the axis of the fiber
before they arrive at the bend.
Note that the cladding serves at least two roles. First,
it determines the critical angle for total internal
reflection, and subsequently determines the minimum
radius of curvature for... | Electromagnetics_Vol2.pdf |
140 CHAPTER 8. OPTICAL FIBER
8.2 Acceptance Angle
[m0192]
In this section, we consider the problem of injecting
light
into a fiber optic cable. The problem is
illustrated in Figure 8.3. In this figure, we see light
incident from a medium having index of refraction n0,
with angle of incidence θi. The light is transmitted
... | Electromagnetics_Vol2.pdf |
nf
(8.9)
Squaring
both sides, we find:
cos2 θ2 ≥ n2
c
n2
f
(8.10)
c⃝ S. Lally CC BY -SA 4.0
Figure 8.3: Injecting light into a fiber optic cable.
Now invoking a trigonometric identity:
1 − sin2 θ2 ≥ n2
c
n2
f
(8.11)
so:
s
in2 θ2 ≤ 1 − n2
c
n2
f
(8.12)
No
w we relate the θ2 to θi using Snell’s law:
sin θ2 = n0
nf
sin θi (... | Electromagnetics_Vol2.pdf |
cone having half-angle θa with respect to the axis
of the fiber.
The associated cone of acceptance is illustrated in
Figure 8.4.
It is also common to define the quantity numerical
aperture NA as follows:
NA ≜ 1
n0
√
n2
f − n2c (8.17)
Note
that n0 is typically very close to 1
(corresponding to incidence from air), so it i... | Electromagnetics_Vol2.pdf |
8.3. DISPERSION IN OPTICAL FIBER 141
Example 8.2. Acceptance angle.
T ypical values of nf and nc for an optical fiber
are 1.52 and 1.49, respectively . What are the
numerical aperture and the acceptance angle?
Solution. Using Equation 8.17 and presuming
n0 = 1, we find NA ∼= 0.30. Since sin θa = N A,
we find θa = 17.5 ◦. ... | Electromagnetics_Vol2.pdf |
digital signals sent over fiber optic cable. In this
section, we analyze this dispersion and its effect on
digital signals.
Figure 8.5 shows the variety of paths that light may
take through a straight fiber optic cable. The nominal
path is shown in Figure 8.5(a), which is parallel to the
axis of the cable. This path has ... | Electromagnetics_Vol2.pdf |
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