Solucionário Notaros - Capítulo 11

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Solucionário Notaros -
capitulo 11
Engenharia Elétrica
Universidade Federal do Rio Grande do Norte (UFRN)
14 pag.
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P11 SOLUTIONS TO PROBLEMS FIELD
ANALYSIS OF TRANSMISSION LINES
Section 11.6 Attenuation Coefficients for Line
Conductors and Dielectric
PROBLEM 11.1 Circuit/field quantities in the time domain for a coax-
ial cable. (a) From Eqs.(11.16), (9.47), and (11.48), the phase coefficient and
characteristic impedance of the cable amount to β = 2πf
√
εr/c0 = 31.4 rad/m and
Z0 = 50 Ω, respectively. Having in mind Eqs.(11.45), (11.49), (11.63), and (9.84),
the instantaneous voltage and current along the cable are given by
v(z, t) = V0
√
2 e−αz cos(ωt − βz) = 1.414 e−0.0184z cos(6.28 × 109t − 31.4z) V ,
i(z, t) =
V0
√
2
Z0
e−αz cos(ωt − βz) = 28.28 e−0.0184z cos(6.28 × 109t − 31.4z) mA
(−∞ < z < ∞ ; z in m ; t in s) . (P11.1)
(b) Based on Eqs.(11.50) and (11.51), the instantaneous electric and magnetic field
intensities in the cable dielectric are (vectors E and H are shown in Fig.11.5)
E(r, z, t) =
V0
√
2
r ln(b/a)
e−αz cos(ωt−βz) = 1.13
r
e−0.0184z cos(6.28×109t−31.4z) V/m ,
H(r, z, t) =
V0
√
2
2πrZ0
e−αz cos(ωt−βz) = 4.5
r
e−0.0184z cos(6.28×109t−31.4z) mA/m
(a < r < b ; −∞ < z < ∞ ; z in m ; t in s) . (P11.2)
(c) Eqs.(11.56) tell us that the instantaneous surface charge densities on the inner
and outer conductor of the cable (Fig.11.5) are
ρs1(z, t) = εrε0E(a
+, z, t) = 22.5 e−0.0184z cos(6.28 × 109t − 31.4z) nC/m2 ,
ρs2(z, t) = −εrε0E(b−, z, t) = 6.43 e−0.0184z cos(6.28 × 109t − 31.4z) nC/m2 ,
(P11.3)
and, by means of Eqs.(11.58), the corresponding surface current densities on the
conductors come out to be
Js1(z, t) =
c0√
εr
ρs1(z, t) = 4.5 e
−0.0184z cos(6.28 × 109t − 31.4z) A/m ,
Js2(z, t) =
c0√
εr
ρs2(z, t) = 1.29 e
−0.0184z cos(6.28 × 109t − 31.4z) A/m . (P11.4)
(d) From Eq.(8.161), the instantaneous Poynting vector in the cable dielectric is
P(r, z, t) = E(r, z, t) × H(r, z, t) = V
2
0
πr2 ln(b/a)Z0
e−2αz cos2(ωt − βz) ẑ
=
5.08
r2
e−0.0368z cos2(6.28 × 109t − 31.4z) ẑ mW/m2 . (P11.5)
301
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302 Branislav M. Notaroš: Electromagnetics (Pearson Prentice Hall)
(e) The total power transported by the TEM wave along the cable equals the flux of
P through an arbitrary cross section Sz of the cable (defined by the coordinate z)
for the reference direction coinciding with the direction of wave travel. Computed
as in Eq.(8.172), this flux turns out to be
P (t) =
∫
Sz
P · dSz =
∫ b
r=a
P(r, z, t) 2πr dr
︸ ︷︷ ︸
dSz
=
2V 20
Z0
e−2αz cos2(ωt − βz) = v(t)i(t) ,
(P11.6)
as expected.
PROBLEM 11.2 Three different optimizations of a coaxial cable. From
Eqs.(11.85), (11.89), and (11.93), the attenuation coefficient (α = αc), breakdown
rms voltage, and maximum permissible (breakdown) time-average transferred power
of the cable are the following functions of the outer to inner conductor radii ratio
(x):
αc(x) =
Rs
2bZTEM
1 + x
lnx
, |V (x)|cr =
Ecrb√
2
lnx
x
, Pcr(x) =
πE2crb
2
ZTEM
lnx
x2
, x =
b
a
(P11.7)
(ZTEM = η0/
√
εr = 251.3 Ω), where the surface resistance (Rs) of copper is com-
puted using Eq.(10.80).
(a) If a = b/3.59 or x = 3.59 (for which αc is minimum), Eqs.(P11.7) give αc =
0.00217 Np/m, |V |cr = 102 kV, and Pcr = 202 MW.
(b) For a = b/ e or x = e (for which |V |cr is maximum), we obtain αc =
0.00224 Np/m, |V |cr = 105 kV, and Pcr = 276 MW.
(c) Finally, substituting a = b/
√
e or x =
√
e (for which Pcr is maximum) in
Eqs.(P11.7), the three parameters come out to be αc = 0.0032 Np/m, |V |cr =
86.7 kV, and Pcr = 376 MW.
Section 11.8 Evaluation of Primary and Secondary
Circuit Parameters of Transmission Lines
PROBLEM 11.3 Circuit parameters of a nonsymmetrical two-wire line.
(a) The capacitance per unit length of the nonsymmetrical thin two-wire line in air
is obtained in Problem 2.32, and we now substitute ε0 by ε in the result to obtain
C′ for this present case. The inductance and conductance p.u.l. of the line are then
computed from Eqs.(11.41) and (3.171), respectively, and we can write
C′ =
πεrε0
ln(d/
√
ab)
, L′ =
µ0
π
ln
d√
ab
, G′ =
πσd
ln(d/
√
ab)
. (P11.8)
In finding the high-frequency p.u.l. line resistance using Eq.(11.66), the only
difference with respect to the corresponding analysis of a symmetrical thin two-wire
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P11. Solutions to Problems: Field Analysis of Transmission Lines 303
line, which is carried out in Example 11.8, is that we now have b in place of a in the
integral in Eq.(11.66) along the contour of the second wire (of radius b). Therefore,
in place of the expression in Eq.(11.103), R′ of the nonsymmetrical line comes out
to be
R′ =
Rs
2πa
+
Rs
2πb
=
Rs
2π
(
1
a
+
1
b
)
, where Rs =
√
πµ0f
σc
. (P11.9)
(b) Using Eq.(11.37), the characteristic impedance of the line is
Z0 =
√
εrε0µ0
C′
=
ZTEM
π
ln
d√
ab
(
ZTEM =
√
µ0
εrε0
)
. (P11.10)
Eqs.(11.16) and (11.17) tell us that the phase coefficient, phase velocity,
and wavelength of the line are β = 2πf
√
εrε0µ0, vp = 1/
√
εrε0µ0, and λz =
1/(
√
εrε0µ0f), respectively.
The attenuation coefficient for the losses in the line conductors, Eq.(11.77),
amounts to
αc =
R′
2Z0
=
Rs
4ZTEM
1/a + 1/b
ln(d/
√
ab)
, (P11.11)
and that for the losses in the dielectric, Eq.(11.80), to αd = σdZTEM/2 =
σd
√
µ0/(εrε0)/2. The total attenuation coefficient of the line is α = αc + αd.
PROBLEM 11.4 Maximum power transfer along a two-wire line. Having
in mind Eq.(2.224), we realize that dielectric breakdown in the line occurs when the
peak-value (amplitude) of the electric field intensity on the surface of the thinner
wire, that of radius b (b < a), at the beginning (at generator terminals) of the
line reaches the critical field value (dielectric strength), Ecr, for the dielectric. This
peak-value equals the corresponding rms field intensity times
√
2, as in Eq.(11.88),
and therefore, in place of Eqs.(2.224)-(2.226), the maximum breakdown rms voltage
of the line is obtained as
Ecr = Epeak =
|Q′|cr
√
2
2πεb
−→ |Q′|cr =
√
2πεrε0bEcr = 7.08 µC
−→ |V |cr =
|Q′|cr
C′
= 259 kV , (P11.12)
where C′ = 27.3 pF/m is the capacitance per unit length of the line (previous
problem).
As in Eq.(11.93), we then use the characteristic impedance of the line, Z0 =
211.5 Ω (previous problem), to find the time-average power carried by the TEM
wave along the cable corresponding to the breakdown rms voltage of the cable, at
the beginning of the line,
Pcr =
|V |2cr
Z0
= 318 MW , (P11.13)
and this is the maximum time-average power that the line can receive from the
generator for the safe operation of the structure, i.e., prior to an eventual dielectric
breakdown.
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304 Branislav M. Notaroš: Electromagnetics (Pearson Prentice Hall)
The attenuation coefficient of the line is α = 7.27 × 10−4 Np/m [previous
problem; note that the coefficient for the losses in the dielectric is computed as
αd = (β tan δd)/2, Eq.(11.80)], so that the corresponding maximum time-average
power delivered to the load, at the other end of the line, Eq.(11.96), amounts to
Pl = Pcr e
−2αl = 295 MW , (P11.14)
PROBLEM 11.5 Charge and current distributions on the ground plane.
Similarly to the computation of the electric field in Fig.1.48 and Eqs.(1.220), the
vector E at the point M in Fig.11.9 is obtained as
E = Eoriginal + Eimage = 2Ewire cos θ (−n̂) =
Q′h
πεR2
(−n̂)
(
Ewire =
Q′
2πεR
)
,
(P11.15)
where n̂ is the normal unit vector on the conducting plane directed upward and
Q′ = Q′(z) = I(z)/c, from Eq.(11.29), c = 1/
√
εµ0 being the intrinsic phase velocity
of the dielectric above the plane. As expected, E is normal to the plane. The vector
H at the same point is given in Eq.(11.108). With the use of boundary conditions
for PEC surfaces in Eqs.(8.33), the associated complex rms surface charge density
(ρ
s
) and current density vector (Js) on the plane are
ρ
s
(x, z) = εn̂ ·E = −
Q′(z)h
π(x2 + h2)
, Js(x, z) = n̂×H = −
I(z)h
π(x2 + h2)
ẑ . (P11.16)
It is important to note that these expressions approximately hold also for the ground
plane made of an imperfect but good conductor (with finite but very large σc), such
as the plane in Example 11.9. Of course, J s = cρs, as in Eq.(11.24).
PROBLEM 11.6 Satisfaction of the continuity equation on the ground
plane. (a) Since the current density vector on the PEC plane, computed in the
previous problem, is of the form Js = J s(x, z) ẑ, the surface divergence of Js at
the point M in Fig.11.9 can, having in mind Eqs.(8.43), (11.33), and (11.17), be
evaluated as follows:
∇s · Js(x, z) =
∂Js
∂z
= − h
π(x2 + h2)
dI(z)
dz
= − h
π(x2 + h2)
[−jβI(z)]
= − h
π(x2 + h2)
[
−jβcQ′(z)
]
= −jω
[
−
Q′(z)h
π(x2 + h2)
]
= −jωρ
s
(x, z) , (P11.17)
so it turns out to equal −jω times the surface charge density (from the previous
problem) at the same point, which proves that the continuity equation for surface
currents (for plates), Eq.(10.14), is satisfied on the PEC plane.
(b) By integrating, respectively, the charge and current distributions obtained in
the previous problem along the conducting plane contour in Fig.11.9(a), like in
Eq.(11.25), the total charge on the plane per unit length of the line (Q′
2
) and the
total current (I2) come out to be
Q′
2
=
∫
plane contour
ρ
s
(x) dx = −
∫
∞
x=−∞
Q′h
πR2
dx = −
Q′
π
∫ π/2
θ=−π/2
dθ = −Q′ = −I
c
,
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P11. Solutions to Problems: Field Analysis of Transmission Lines 305
I2 =
∫
plane contour
J s(x) dx = −
∫
∞
x=−∞
Ih
πR2
dx = −I
(
dx
R2
=
dθ
h
)
, (P11.18)
where, as in Eq.(11.109), the length to angular coordinate transformation given
by Eq.(1.55) is used to simplify the integration. So, Q′
2
and I2 indeed equal the
negative of the p.u.l. charge and current intensity of the upper conductor of the line.
PROBLEM 11.7 Three-wire transmission line. (a) Since σc/(2πfε0) =
3.5× 109 ≫ 1 (σc = 58 MS/m – copper), conditions in Eqs.(11.61) are met (σd = 0
– air), for the line to be treated as having small losses. The capacitance per unit
length of the line being that in Eq.(2.145), the use of Eq.(11.41) or (7.13) gives the
per-unit-length line inductance,
L′ =
ε0µ0
C′
=
3µ0
4π
ln
d
a
= 1.174 µH/m . (P11.19)
With the TEM wave on the line, the currents of the three wires are distributed
in the same way as the charges in Fig.2.23: I, −I/2, and −I/2 for wires 1, 2,
and 3, respectively. Hence, the magnitudes of the corresponding (entirely tangen-
tial) magnetic field vectors on the surfaces of the wires are |H1| = |I|/(2πa) and
|H2| = |H3| = (|I|/2)/(2πa). From Eq.(11.66), the high-frequency resistance per
unit length of the line is
R′ =
1
|I|2
[
∮
Cc1
Rs
( |I|
2πa
)2
dl +
∮
Cc2
Rs
( |I|/2
2πa
)2
dl +
∮
Cc3
Rs
( |I|/2
2πa
)2
dl
]
=
Rs
2πa
(
1 +
1
4
+
1
4
)
=
3Rs
4πa
= 1.08 Ω/m , (P11.20)
where Cc1, Cc2, and Cc3 stand for contours of the wires, and Rs for their surface
resistance, given in Eq.(10.80) for copper. The dielectric of the line is air, and thus
the conductance G′ is zero.
(b) Combining Eqs.(11.37), (2.145), and (9.23), the characteristic impedance of the
line is
Z0 =
√
ε0µ0
C′
= 90 Ω ln
d
a
= 352 Ω . (P11.21)
As αd = 0 (air line), using Eqs.(11.75), (P11.20), (P11.21), and (9.89), the line
attenuation coefficient comes out to be α = αc = R
′/(2Z0) = 0.00153 Np/m =
0.0133 dB/m.
PROBLEM 11.8 Maximum permissible power delivered to a load. The
electric field on the surface of wire 1 in Fig.2.23 is twice as strong as the field on the
surface of either one of the other two wires, so, for a high enough voltage of the line,
the breakdown occurs near wire 1, when the peak-value of the corresponding field
intensity, at the beginning of the line, reaches the dielectric strength of air, Eq.(2.53).
Having in mind Eqs.(2.224), (2.225), and (11.88), this breakdown condition can be
written as ∣
∣Q′
∣
∣
cr
√
2
2πε0a
= Ecr0 = 3 MV/m , (P11.22)
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306 Branislav M. Notaroš: Electromagnetics (Pearson Prentice Hall)
from which the critical rms p.u.l. charge of the line at breakdown amounts to
∣
∣Q′
∣
∣
cr
=√
2πε0aEcr0 = 118 nC/m. Eqs.(11.36), (11.60), and (11.96) then yield the following
for the maximum (critical) time-average power that can be delivered to a load (L)
at the end of the line:
|V |cr =
∣
∣Q′
∣
∣
cr
C′
= 12.45 kV −→ (PL)cr =
|V |2cr
Z0
e−2αl = 437.6 kW , (P11.23)
where C′ = 9.48 pF/m, Z0 = 352 Ω, and α = 1.53× 10−3 Np/m (from the previous
problem).
PROBLEM 11.9 Four-wire transmission line. (a) The capacitance per unit
length of the line amounts to C′ = 9.86 pF/m (Problem 2.35), and hence, using
Eq.(11.41), the p.u.l. line inductance is L′ = 1/(c20C
′) = 1.13 µH/m (c0 = 3 ×
108 m/s). The conductance is G′ = 0, since the dielectric is air. As the TEM wave
is established on the line, with the line current I(z), the current I(z) flowingin the
upper pair of wires in Fig.2.44 is, because of symmetry, distributed equally between
the two wires, and the same holds true for the current −I(z) (with respect to the
same reference direction) in the lower pair. This means that the magnitudes of
currents in the four wires in Fig.2.44 are all the same (in any given cross section of
the line), equal to |I|/2. This also means that the magnitudes of the corresponding
(entirely tangential) magnetic field vectors on the surfaces of the wires are all the
same, given by
|H1| = |H2| = |H3| = |H4| =
|I|/2
2πa
. (P11.24)
By means of Eq.(11.66), the high-frequency resistance per unit length of the line
comes out to be
R′ =
1
|I|2 4
∮
Cc1
Rs|H1|2 dl =
Rs
4π2a2
∮
Cc1
dl =
Rs
2πa
= 0.587 Ω/m , (P11.25)
with Cc1 standing for the contour of wire 1 (or any other wire) in Fig.2.44 and Rs
for the surface resistance of copper [Eq.(10.80)].
(b) From a combination of Eqs.(11.37) and (11.75), the attenuation coefficient of
the line is α = αc = R
′/(2Z0) = c0R
′C′/2 = 8.69 × 10−4 Np/m.
PROBLEM 11.10 Two-wire line and a foil. (a) The capacitance per unit
length of the line, evaluated in Problem 2.36, equals C′ = 26.5 pF/m, inductance is
L′ = 420 nH/m [Eq.(11.41)], and conductance G′ = 0. As the losses in the foil can
be neglected and the two wires in Fig.2.45 carry currents of the same magnitude,
|I|/2, so that the magnitude of the magnetic field vector on the surfaces of both
wires is |H1| = |H2| = (|I|/2)/(2πa), the high-frequency resistance per unit length
of the line, Eq.(11.66), amounts to
R′ =
1
|I|2 2
∮
Cc1
Rs
( |I|/2
2πa
)2
dl =
Rs
4πa
= 0.656 Ω/m . (P11.26)
(b) The induced surface charge density (ρs) at the central point O on the foil in
Fig.2.45 is obtained in Problem 2.36 – in the electrostatic regime on the line. In
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P11. Solutions to Problems: Field Analysis of Transmission Lines 307
the dynamic case (line with the TEM wave), for the induced rms surface charge
density at the point O in an arbitrary cross section of the line (for arbitrary z), the
expression for |ρ
s
| has exactly the same form as in the electrostatic case, the only
difference being that the p.u.l. charge of the line is a function of z, Q′ = Q′(z):
|ρ
s
(z)| =
4|Q′(z)|h
π(d2 + 4h2)
=
4Q′0h
π(d2 + 4h2)
e−αz = 33.95 e−0.00261z nC/m2 , (P11.27)
where Q′0 = I0/c0 = 3.33 nC (c0 = 3 × 108 m/s) is the rms p.u.l. charge at the
beginning of the line (for z = 0) and α = R′/(2Z0) = c0R
′C′/2 = 2.61×10−3 Np/m
is the attenuation coefficient of the line. The induced rms surface current density
at the point O is
|J s(z)| = c0|ρs(z)| =
4I0h
π(d2 + 4h2)
e−αz = 10.19 e−0.00261z A/m . (P11.28)
Of course, this current density can alternatively be obtained invoking the boundary
condition for the magnetic field vector (H) in Eqs.(8.33), Js = n̂ × H, with H, in
turn, being found using image theory for current, as in Eq.(11.108).
PROBLEM 11.11 Wire-corner transmission line. (a) The capacitance per
unit length of the transmission line in Fig.1.57 is given by C′ = 2πε0/ ln(h
√
2/a)
(Problem 2.37), and – for the given numerical data in the present case – it amounts to
C′ = 11.8 pF/m. Neglecting the contribution of the losses in the corner screen, the
high-frequency resistance per unit length of the line is R′ = Rs/(2πa) = 2.07 Ω/m
[a half of the expression in Eq.(11.103)], so that the attenuation coefficient of the
line comes out to be α = αc = R
′/(2Z0) = c0R
′C′/2 = 0.00365 Np/m [Eqs.(11.75)
and (11.37)].
A
B
Q'
a
I z( )
1
2
3
4
h
h
h h
d
-Q'
-Q'
Q'
nn
E2
E1
E4
E3
z
Figure P11.1 Evaluation of the induced surface charge (and current) on the corner
screen in the transmission line of Fig.1.57 – by virtue of image theory for charge
(applied twice).
(b) In the equivalent structure in Fig.P11.1, obtained (in Problem 1.89) by virtue
of image theory for charge, the resultant complex electric field intensity vector at
the point A (one of the two points on the screen in the original structure, Fig.1.57,
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308 Branislav M. Notaroš: Electromagnetics (Pearson Prentice Hall)
that are closest to the wire) is
E = 2
Q′
2πε0h
(−n̂) + 2
Q′
2πε0d
cosα n̂ = −
4Q′
5πε0h
n̂ −→ ρ
s
= ε0n̂ · E = −
4Q′
5πh
(P11.29)
(cosα = h/d, d =
√
5h), where Q′ = Q′(z) = I(z)/c0 [Eq.(11.29)]. Then, using
Eq.(11.24), the complex rms surface current density vector at both points A and B
in Fig.P11.1 equals
Js(z) = c0ρs ẑ = −
4I(z)
5πh
ẑ . (P11.30)
Of course, obtaining this current density directly from Js = n̂ × H, with the use
of image theory for current (Fig.11.9) to find the resultant magnetic field vector at
the point A (or B), is also possible.
Section 11.9 Transmission Lines with
Inhomogeneous Dielectrics
PROBLEM 11.12 Planar TEM line with a continuously inhomogeneous
dielectric. As in Problem 2.54, the total capacitance of the transmission line in
Fig.3.37 can be obtained evaluating the following integral for the corresponding
parallel-plate capacitor with continuously inhomogeneous dielectric:
1
C
=
∫ d
x=0
dx
ε(x)wl
=
1
ε0wl
∫ d
0
dx
4 + 3x/d
=
d ln(7/4)
3ε0wl
−→ C = 3ε0wl
d ln1.75
,
(P11.31)
so that the capacitance per unit length of the line turns out to be
C′ =
C
l
=
3ε0w
d ln1.75
. (P11.32)
From the analysis in Problem 3.22 and the leakage current per unit length of
the line, I ′d, the p.u.l. line conductance equals
G′ =
I ′d
V
=
I ′d
E =
2σ0w
11d
. (P11.33)
Since the dielectric in Fig.3.37 is nonmagnetic, both the inductance and the
high-frequency resistance per unit length of the line (L′ and R′) are the same as
those of the same line if air-filled (L′0 and R
′
0) or if with a homogenous nonmagnetic
substrate, as in the microstrip line in Fig.11.10; therefore, L′ and R′ are given by
Eqs.(11.114) and (11.113), respectively,
L′ = L′0 = µ0
h
w
, R′ = R′0 =
2Rs
w
(
Rs =
√
πµ0f
σc
)
. (P11.34)
Using Eqs.(11.42) and (11.43), the characteristic impedance and phase coeffi-
cient of the line are
Z0 =
√
L′
C′
=
η0
w
√
hd ln1.75
3
, β = ω
√
L′C′ =
2πf
c0
√
3h
d ln1.75
(P11.35)
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P11. Solutions to Problems: Field Analysis of TransmissionLines 309
(η0 =
√
µ0/ε0 , c0 = 1/
√
ε0µ0), and then Eqs.(11.17), (11.77), and (11.79) result
in the following expressions for the phase velocity, wavelength, and attenuation
coefficients for line conductors and dielectric:
vp =
ω
β
= c0
√
d ln1.75
3h
, λz =
vp
f
=
c0
f
√
d ln1.75
3h
,
αc =
R′
2Z0
=
Rs
η0
√
3
hd ln1.75
, αd =
G′Z0
2
=
σ0η0
11
√
h ln1.75
3d
. (P11.36)
PROBLEM 11.13 Quasi-TEM wave on a coaxial cable with two dielec-
tric layers. (a) The capacitance per unit length of the cable (in Fig.2.50) is found
in Problem 2.48, and – for numerical data given in Problem 3.17 – it amounts to
C′ = 2πε0
(
1
εr1
ln
b
a
+
1
εr2
ln
c
b
)
−1
= 217 pF/m . (P11.37)
The p.u.l. conductance of the cable (Fig.3.34) is (Problem 3.17)
G′ = 2π
(
1
σ1
ln
b
a
+
1
σ2
ln
c
b
)
−1
= 6.84 pS/m . (P11.38)
Based on Eqs.(7.12) and (11.70), the p.u.l. cable inductance and resistance equal
L′ = L′0 =
µ0
2π
ln
c
a
= 241 nH/m , R′ = R′0 =
Rs
2π
(
1
a
+
1
c
)
= 0.0254 Ω/m ,
(P11.39)
with the surface resistance (Rs) of copper being calculated using Eq.(10.80).
(b) Since the p.u.l. capacitance of the cable if air-filled is C′0 = 2πε0/ ln(b/a) =
65.7 pF/m [Eq.(2.123) with ε = ε0], the effective relative permittivity of the cable,
Eq.(11.120), amounts to
εreff =
C′
C′0
= 3.3 . (P11.40)
(c) From Eqs.(11.122), the phase velocity of the propagating wave in the cable is
vp =
c0√
εreff
= 1.65 × 108 m/s
(
c0 = 3 × 108 m/s
)
. (P11.41)
Finally, Eqs.(11.42), (11.77), and (11.79) give the following for the attenuation
coefficient of the cable:
Z0 =
√
L′
C′
= 33.3 Ω −→ α = αc + αd =
R′
2Z0
+
G′Z0
2
= 3.82 × 10−4 Np/m .
(P11.42)
PROBLEM 11.14 Power capacity of a coaxial cable with a two-layer
dielectric. (a) The breakdown voltage of the cable in the electrostatic regime is
evaluated in Problem 2.77, and we only need here, in the dynamic case (line with the
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310 Branislav M. Notaroš: Electromagnetics (Pearson Prentice Hall)
TEM wave), to divide that result by
√
2 [see Eq.(11.88)], to obtain the maximum
breakdown rms voltage of the cable:
|V |cr =
εr2bEcr2√
2
(
1
εr1
ln
b
a
+
1
εr2
ln
c
b
)
= 27.4 kV . (P11.43)
The capacitance per unit length of the cable is C′ = 115 pF/m (also computed
in Problem 2.77) and the p.u.l. cable inductance, using Eq.(7.12) with c in place
of b as the inner radius of the outer conductor of the cable, amounts to L′ =
L′0 = 277 nH/m, so that the characteristic impedance of the cable turns out to be
Z0 =
√
L′/C′ = 49.2 Ω [Eq.(11.42)]. Having in mind Eq.(11.93), the maximum
time-average power, limited by an eventual dielectric breakdown in the structure,
that the cable can receive from a generator is given by
Pcr =
|V |2cr
Z0
= 15.3 MW . (P11.44)
(b) From Eq.(11.70) with b replaced by c, the high-frequency resistance p.u.l. of
the cable is R′ = R′0 = 5.1 Ω/m, and its attenuation coefficient, Eq.(11.77), is thus
α = αc = R
′/(2Z0) = 0.0519 Np/m. Eq.(11.96) then tells us that the corresponding,
maximum permissible, power delivered to a load at the other end of the cable equals
Pl = Pcr e
−2αl = 5.43 MW . (P11.45)
PROBLEM 11.15 Quasi-TEM analysis of coaxial cables with dielectric
sectors. (a) The capacitance per unit length of the cable in Fig.2.33 is evaluated
in Problem 2.81, and it is given by
C′ =
εrε0α + ε0(2π − α)
ln(b/a)
. (P11.46)
When air-filled, C′0 = 2πε0/ ln(b/a), and hence the effective relative permittivity,
Eq.(11.120), of the cable
εreff =
C′
C′0
=
εrα + (2π − α)
2π
= 1 + (εr − 1)
α
2π
. (P11.47)
Note that α = 0 and α = 2π result in εreff = 1 and εreff = εr, respectively, as
expected.
The p.u.l. cable inductance and resistance are given in Eqs.(7.12) and (11.70),
respectively, and the attenuation coefficient of the cable, using Eqs.(11.77) and
(11.42), amounts to α = αc = R
′/(2Z0) = R
′
√
C′/L′/2.
(b) The capacitance p.u.l. of the cable in Fig.2.51, found in Problem 2.50, and its
effective relative permittivity are as follows:
C′ =
π (εr1 + εr2 + εr3 + εr4) ε0
2 ln(b/a)
−→ εreff =
C′
C′0
=
εr1 + εr2 + εr3 + εr4
4
(P11.48)
(of course, for εr1 = εr2 = εr3 = εr4 = εr, we have εreff = εr). Using this capacitance,
α is computed as in (a).
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P11. Solutions to Problems: Field Analysis of Transmission Lines 311
PROBLEM 11.16 Coated wire-ground plane quasi-TEM transmission
line. (a) Applying image theory for charge to the line in Fig.11.16, to free charges on
the surface of the metallic wire and bound charges on the two surface of the dielectric
coating, we remove the ground plane and obtain the equivalent transmission line
with two symmetric dielectrically coated wires, as the one in Fig.2.31. Having
in mind Eq.(2.146), the capacitance per unit length of the original line (C′), in
Fig.11.16, equals twice that of the equivalent two-wire line (C′e), which, in turn, can
be obtained modifying Eqs.(2.182) and (2.183) for the case in Fig.11.16, where the
thickness of the coating is b, and not a, and the dielectric between the coated wire
and the ground plane is not air. With this, the respective limits in the integrals for
the equivalent two-wire line are a + b and 2h − (a + b) (the distance between wire
axes is d = 2h) in place of 2a and d − 2a, which results in
C′ = 2C′e = 2 πε0
(
1
εr1
ln
a + b
a
+
1
εr2
ln
2h
a + b
)
−1
. (P11.49)
By the same token, the conductance per unit length of the line in Fig.11.16
equals twice G′e of the equivalent two-wire line with imperfect dielectric coatings in
Fig.3.23, where the same modifications in Eq.(3.186) regarding the thickness of the
coatings now being b instead of a are in order, which yields
G′ = 2G′e = 2 π
(
1
σd1
ln
a + b
a
+
1
σd2
ln
2h
a + b
)
−1
. (P11.50)
Alternatively, this expression for G′ can be obtained by substituting the permittiv-
ities in the expression for C′ in Eq.(P11.49) by the corresponding conductivities,
because the inhomogeneities of the two systems in terms of ε and σ, respectively,
are of the same form (are dual).
The dielectric materials in Fig.11.16 being nonmagnetic, both the inductance
and the high-frequency resistance per unit length of the line (L′ and R′) are the
same as those of the same line if air-filled (L′0 and R
′
0) or if with a homogenous non-
magnetic substrate, which means that L′ and R′ are actually those in Eqs.(11.106)
and (11.110),
L′ = L′0 =
µ0
2π
ln
2h
a
, R′ = R′0 =
1
2π
(
Rs1
a
+
Rs2
h
)
. (P11.51)
Combining Eqs.(11.77), (11.79), and (11.42), the attenuation coefficient of the
line is α = αc + αd = R
′/(2Z0) + (G
′Z0)/2 = R
′
√
C′/L′/2 + G′
√
L′/C′/2.
(b) Distributions of surface charge and current on the ground plane for the same line
but with a homogeneous dielectric, the one in Fig.11.9, are evaluated inProblem
11.5. As the inhomogeneous dielectric in Fig.11.16 is nonmagnetic, the surface
current density on the plane is the same in the two structures, and we simply take
the result from Problem 11.5,
Js(x, z) = n̂ × H = −
I(z)h
π(x2 + h2)
ẑ . (P11.52)
For the surface charge distribution on the ground plane in Fig.11.16, we invoke,
because the dielectric is inhomogeneous, the electric flux density vector, D, instead
of dealing with the vector E in Problem 11.5, and write [also see Eqs.(2.181)]
D = Doriginal + Dimage =
Q′h
πR2
(−n̂) −→ ρ
s
(x, z) = n̂ ·D = −
Q′(z)h
π(x2 + h2)
,
(P11.53)
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312 Branislav M. Notaroš: Electromagnetics (Pearson Prentice Hall)
which turns out to be, again, the same result as in Problem 11.5. However, the
p.u.l. charge Q′(z) here is
Q′(z) = C′V (z) = C′Z0I(z) = C
′
√
L′
C′
I(z) =
√
L′C′ I(z) , (P11.54)
which, based on Eqs.(11.43), can be written using the phase velocity for the line as
Q′(z) = I(z)/vp, and this is a generalization, valid for transmission lines with inho-
mogeneous dielectrics as well, of the current-charge proportionality in Eq.(11.29).
Consequently, J s(x, z) = vpρs(x, z), a generalization (for lines with inhomogeneous
dielectrics) of the relationship in Eq.(11.24).
PROBLEM 11.17 High-frequency internal inductance of three different
lines. The high-frequency internal inductances per unit length of the two-wire line
from Problem 11.4 (L′i1), coaxial cable from Problem 11.13 (L
′
i2), and microstrip
line from Example 11.12 (L′i3) are obtained using Eq.(11.100) and the respective
expressions and values of their high-frequency p.u.l. resistances, R′, as follows:
L′i1 =
R′1
ω
=
Rs
4π2f
(
1
a
+
1
b
)
= 0.53 nH/m , L′i2 =
R′2
ω
=
Rs
4π2f
(
1
a
+
1
c
)
= 2.02 nH/m , L′i3 =
R′3
ω
=
Rs
πfw
= 0.328 nH/m
(
Rs =
√
πµ0f
σc
)
. (P11.55)
The corresponding values of per-unit-length external inductances of these lines are
L′1 = 1.22 µH/m, L
′
2 = 241 nH/m, and L
′
3 = 118 nH/m, and we see that L
′
i ≪ L′
in all cases, as expected.
Section 11.10 Multilayer Printed Circuit Board
PROBLEM 11.18 Microstrip lines with different strip width to height
ratios. (a)-(g) At the frequency of f = 3 GHz, the effective relative permittivity,
εreff , characteristic impedance, Z0, phase coefficient, β, and attenuation coefficients
for conductors and dielectric, αc and αd, of the microstrip line are computed for
the given w/h ratios, cases (a)-(g), with the fringing effects taken into account –
following the procedure from Example 11.15 [note that the phase velocity, vp, and
wavelength, λz, along the line can be obtained from β using Eqs.(11.17)]. Then, the
per-unit-length capacitance, inductance, resistance, and conductance of the line, C′,
L′, R′, and G′, are found as in Example 11.16. All these results are tabulated in
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P11. Solutions to Problems: Field Analysis of Transmission Lines 313
Table P11.1.
Table P11.1 Primary and secondary circuit parameters of a microstrip line taking into account fringing effects,
for different strip width to substrate height ratios (analysis as in Examples 11.15 and 11.16).
case w/h C′ (pF/m) L′ (nH/m) R′ (Ω/m) G′ (µS/m) εreff Z0 (Ω) β (rad/m) αc (Np/m) αd (Np/m)
(a) 0.05 29.1 1015 286 45.5 2.65 187 102 0.765 0.00425
(b) 0.1 34.1 876 143 53.8 2.68 160 103 0.446 0.00431
(c) 0.5 56.3 556 28.6 91.3 2.81 99.4 105 0.144 0.00453
(d) 1 76.9 422 14.3 127 2.92 74.1 107 0.0965 0.0047
(e) 2 114 298 7.15 194 3.07 51 110 0.07 0.00495
(f) 10 405 96.5 1.43 727 3.51 15.4 118 0.0463 0.00562
(g) 20 765 53.6 0.715 1401 3.69 8.37 121 0.0427 0.00586
(h) Table P11.2 shows the results for the circuit parameters of the line in cases (d)-
(g) obtained neglecting the fringing effects, using the equations from Example 11.10.
Comparing these results with the corresponding values in Table P11.1, we realize
that only for w/h ≥ 10, the analysis with no fringing effects taken into account
makes some sense.
Table P11.2 Circuit parameters of the microstrip line in cases (d)-(g) obtained neglecting the fringing effects
(analysis as in Example 11.10).
case w/h C′ (pF/m) L′ (µH/m) R′ (Ω/m) G′ (µS/m) εreff Z0 (Ω) β (rad/m) αc (Np/m) αd (Np/m)
(d) 1 35.4 1.26 14.3 66.8 4 188 126 0.0379 0.00629
(e) 2 70.8 0.628 7.15 133 4 94.2 126 0.0379 0.00629
(f) 10 354 0.126 1.43 668 4 18.8 126 0.0379 0.00629
(g) 20 708 0.0628 0.715 1335 4 9.42 126 0.0379 0.00629
(i) Finally, for cases (a)-(d), presented in Table P11.3 are the results of the analysis
of a wire-plane transmission line [Fig.11.9(a)] with the conducting strip in Fig.2.20
replaced by a thin wire of an equivalent radius equal to a = w/4 and the entire
half-space above the ground plane filled with the line dielectric – carrying out the
solution procedure from Example 11.9. When compared to the corresponding rows
of Table P11.1, these data indicate that some orientational qualitative results and
conclusions can be obtained with a wire-plane approximation of the microstrip line,
but, of course, these results are not accurate enough for any quantitative evaluations
of the line parameters or realistic designs.
Table P11.3 Results in cases (a)-(d) for a wire-plane transmission line, in Fig.11.9(a), with a wire of an
equivalent radius a = w/4 (analysis as in Example 11.9).
case w/h C′ (pF/m) L′ (µH/m) R′ (Ω/m) G′ (µS/m) εreff Z0 (Ω) β (rad/m) αc (Np/m) αd (Np/m)
(a) 0.05 43.8 1.01 92.1 82.6 4 152 126 0.303 0.00629
(b) 0.1 50.8 0.876 46.6 95.7 4 132 126 0.177 0.00629
(c) 0.5 80.3 0.554 10.2 151 4 83.1 126 0.0616 0.00629
(d) 1 107 0.416 5.69 202 4 62.3 126 0.0456 0.00629
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314 Branislav M. Notaroš: Electromagnetics (Pearson Prentice Hall)
PROBLEM 11.19 Primary circuit parameters of a strip line with fring-
ing. We start with the values for the secondary circuit parameters of the strip line,
in particular, for Z0, αc, and αd, obtained in Example 11.18. From Eqs.(11.37) and
(9.47), the capacitance per unit length of the line comes out to be
Z0 =
1
cC′
=
√
εr
c0C′
−→ C′ =
√
εr
c0Z0
= 203.9 pF/m
(
c0 = 3 × 108 m/s
)
.
(P11.56)
On the other side, Eq.(11.40) gives the p.u.l. line inductance:
Z0 = cL
′ =c0L
′
√
εr
−→ L′ =
√
εr Z0
c0
= 218 nH/m . (P11.57)
As in Eqs.(11.135), the high-frequency resistance and conductance p.u.l. of the line
are
R′ = 2Z0αc = 7.13 Ω/m , G
′ =
2αd
Z0
= 384.2 µS/m . (P11.58)
PROBLEM 11.20 Design of microstrip and strip lines. (a) The charac-
teristic impedance of the coaxial cable from Example 11.3 is Z0 = 50.08 Ω, and to
design a microstrip line that has this same characteristic impedance for the same
relative permittivity of the dielectric as the cable, εr = 2.25, a use of Eqs.(11.129)
gives w/h = 3.067 from the expression for w/h ≤ 2, which is contradictory, whereas
w/h = 3.027 from the other expression. Hence, this latter result is the required w/h
ratio in this case. Eq.(11.127) then yields εreff = 1.905, and a check in Eqs.(11.128)
confirms that Z0 = 50.36 Ω ≈ 50.08 Ω, as desired.
(b) To design a strip line with the same characteristic impedance for the same
relative permittivity of the dielectric as the two-wire line from Problem 11.4, so Z0 =
211.48 Ω for εr = 3, we have that
√
εrZ0 = 366.3 Ω > 0.316η0 ≈ 120 Ω, meaning
that Eq.(11.137) applies. It results in w/h = −0.0705, which is an impossible
w/h ratio (w/h must be positive). This is illustrative of the general fact that not
every characteristic impedance can be achieved for a given transmission line choice.
Namely, a quite high impedance (Z0 = 211.48 Ω) that is easily realizable with a
two-wire line is impossible to obtain in a strip-line design.
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