Answer Key
RES.6-001: Solutions to Chapter 14
-
University:
Massachusetts Institute of Technology -
Course:
RES.6-001 | Electromagnetic Fields and Energy Academic year:
2023
-
Views:
413
Pages:
20
Author:
Marin Schmitt
SOLUTIONS TO CHAPTER 14
14.1 DISTRIBUTED PARAMETER EQUIVALENTS AND
MODELS
14.1.1
The fields are approximated as uniform in each of the dielectric regions. The
integral of E between the electrodes must equal the applied voltage and D is con
tinuous at the interface. Thus,
(1)
and it follows that
Ea
V
(2)
[a + b(€a/€b
=-.,.------,-------:---:-:
so that the charge per unit length on the upper electrodes is
(3)
where C is the desired capacitance per unit length. Because the permeability of the
region is uniform, H = I/w between the electrodes. Thus,
>.
= (a
+ b)J.LoH =
LI;
L == J.Lo(a
where L is the inductance per unit length. Note that LC
unless €a
= €b·
14.1.2
+ b)/w
t
(4)
J.L€
(which permittivity)
The currents at the node must sum to zero, with that through the inductor
related to the voltage by V = Ldiconductor/ dt
L
D.z
a
at [I(z) -
I(z + D.z)] = V
(1)
and C times the rate of change of the voltage drop across the capacitor must be
equal to the current through the capacitor.
C
D.z
In the limit where D.z line equations.
a
at [V(z)
- V(z
+ D.z)] = I
(2)
0, these become the given backward-wave transmission
14.2 TRANSVERSE ELECTROMAGNETIC WAVES
1 Solutions to Chapter 14
14-2
14.2.1
(a) From Ampere's integral law, (1.4.10),
H",
= 1/21ff'
(1)
and the vector potential follows by integration
H",
= _-!:.. aaA•
IJo
r
=> A.(r) - A.(a)
= -IJ201'n(r/a)
(2)
1t'
and evaluating the integration coefficient by using the boundary condition on
A. on the outer conductor, where r = a. The electric field follows from Gauss'
integral law, (1.3.13),
Er
= ).,f21t'Er
(3)
and the potential follows by integrating.
E r = - alb => lb(r) :- lb(a) =
ar
~'n(~)
21t'E
(4)
r
Using the boundary condition at r = a then gives the potential.
(b) The inductance per unit length follows from evaluation of (2) at the inner
boundary.
! = A.(b) -
A.(a) = IJo In(a/b)
(5)
1
1
21t'
Similarly, the capacitance per unit length follows from evaluating (5) at the
inner boundary.
0== A, =
21t'E
(6)
V
In(a/b)
L ==
14.2.2
The capacitance per unit length is as given in the solution to Prob. 4.7.5. The
inductance per unit length follows by using (8.6.14), L = 1/0c2 •
14.3 TRANSIENTS ON INFINITE TRANSMISSION LINES
14.3.1
(a) From the values of Land 0 given in Prob. 14.2.1, (14.3.12) gives
Zo= ~ln(a/b)/21t'
(b) From IJ
=
IJo
=
41t'
X
10- 7 and E
=
2.5Eo
=
(2.5)(8.8.5
X
10- 12), Zo
=
(37.9)ln(a/b). Because the only effect of geometry is through the ratio alb
and that is logarithmic, the range of characteristic impedances encoutered in
practice for coaxial cables is relatively small, typically between 50 and 100
ohms. For example, for the four ratios of alb, Zo = 26,87,175 and 262 Ohms,
respectively. To make Zo = 1000 Ohms would require that alb = 2.9 x lOll! Solutions to Chapter 14
14.8.2
14-3
The characteristic impedance is given by (14.3.13). Presuming that we will
find that 1/R ::> 1, the expression is approximated by
Zo = ..;;[iln(21/ R)/fr
and solved for 1/R.
= iexp[frZo/..;;[iJ = exp[fr(300)/377J
Evaluation then gives 1/R = 6.1.
l/R
14.8.8
The solution is analogous to that of Example 14.3.2 and shown in the figure.
v
I
Figure 814.8.8
14.3.4
From (14.3.18) and (14.3.19), it follows that
V::I:
= Vo exp(- z2 /2a 2 ) /2
Then, from (9) and (10)
V
1
= iVo{exp[-(z
-
ct)2/2a2J + exp[-(z + ct)2/2a2]} Solutions to Chapter 14
14-4
14.3.5
In general, the voltage and current can be represented by (14.3.9) and (14.3.10).
From these it follows that
14.3.6
By taking the ao/at and ao/az of the second equation in Prob. 14.1.2 and
substituting it into the first, we obtain the partial differential equation that plays
the role played by the wave equation for the conventional transmission line
(1)
Taking the required derivatives on the left amounts to combining (14.3.6). Thus,
substitution of (14.3.3) into (I), gives
By contrast with the wave-equation, this expression is not identically satisfied.
Waves do not propagate on this line without dispersion.
14.4 TRANSIENTS ON BOUNDED TRANSMISSION LINES
14.4.1
When t = 0, the initial conditions on the line are
I
=0
for 0 < z < I
From (14.4.4) and (14.4.5), it follows that for those characteristics originating on
the t = 0 axis of the figure
For those lines originating at z = I, it follows from (14.4.8) with RL = oo(rL = 1)
that
V_ =V+
Similarly, for those lines originating at z = 0, it follows from (14.4.10) with
and Vg = 0 that
V+ =0
rg
= 0
Combining these invarlents in accordance with (14.1.1) and (14.1.2) at each location
gives the (z, t) dependence of V and I shown in the figure. Solutions to Chapter 14
14-5
:II
"
V
I
Vj
~
V.
1, = 0
/
V+ =0
J
= -V"/2Z,,
Figure 814.4.1
14.4.2
When t = 0, the initial conditions on the line are
V=Oj
!II
.I ==
1= Vo/Zo for 0 < z < l
v./z.
V. == V./2 ..
1 == V./2Z.
Figure 814.4.2
From (14.4.4) and (14.4.5), it follows that for those characteristics originating on 14-6
Solutions to Chapter 14
the t = 0 axis of the figure
V+ = Vo /2;
V_ = -Vo /2
For those lines originating at z = " it follows from (14.4.8) with R L = 0 (r L
that
V_ = -V+
Similarly, for those lines originating at z = 0, it follows from (14.4.10) that
= -1)
V+ =0
Combining these invarients in accordance with (14.4.1) and (14.4.2) at each location
gives the (z, t) dependence of V and I shown in the figure.
14.4.S
H the voltage and current on the line are initially zero, then it follows from
(14.4.5) that V_ = 0 on those characteristic lines z + ct = constant that originate
on the t = 0 axis. Because RL = Zo, it follows from (14.4.8) that V_ = 0 for all of
the other lines z + ct = constant, which originate at z = ,. Thus, at z = 0, (14.4.1)
and (14.4.2) become
V =V+;
I=V+/Zo
and the ratio of these is the terminal relation V/ 1= Zo, the relation for a resistance
equal in value to the characteristic impedance. Implicit to this equivalence is the
condition that the initial voltage and current on the line be zero.
14.4.4
z.
~-,....--~---....,.--'''--------.. t
• t
lie
Figure 814.4.4
21lc 14-7
Solutions to Chapter 14
The solution is constructed in the z - t plane as shown by the figure. Because
the upper transmission line is both terminated in its characteristic impedance and
free of initial conditions, it is equivalent to a resistance R a connected to the ter
minals of the lower line (see Prob. 14.4.3). The values of V+ and V_ follow from
(14.4.4) and (14.4.5) for the characteristic lines originating when t = 0 and from
(14.4.8) and (14.4.10) for those respectively originating at z = I and z = o.
14.4.5
When t < 0, a steady current flows around the loop and the initial voltage
and current distribution are uniform over the length of the two line-segments.
v., -
Ra Vo
L' _
•
'-Ra+Rb'
Vo
'-Ra+Rb
In the upper segment, shown in the figure, it follows from (14.4.5) that V_ = O.
Thus, for these particular initial conditions, the upper segment is equivalent to a
termination on the lower segment equal to Za = R a. In the lower segment, V+ and
V_ originating on the z axis follow from the initial conditions and (14.4.4) and
(14.4.5) as being the values given on the z - t diagram. The conditions relating the
incident to the, reflected waves, given respectively by (14.4.8) and (14.4.10), are also
summarised in the diagram. Use of (14.4.1) to find V(O, t) then gives the function
of time shown at the bottom of the figure.
1'_
=0
/
z.
\' =!:! (R., -
.12
Ro)
(P.... Ro)
z.
\' =!:! 1R., -
2
Ro)
(R.,.,. Ro)
1'.1
1
_ _.!...---I-",(t)
+
",
--:'
v.. =0
':"""'"
.~
t
I
d
1'(0,1)
:
~-.L~____
_ _ 1~_ _
1
t=JI
I'. fR., - R.)
TIR.,.,. R,}
l/e
\'.IR. -R.}'
.. t
21fe
T (R., - R,l'
Figure 81-&.4.5
14.4.6
From (14.4.4) and (14.4.5), it follows from the initial conditions that V+ and
V_ are zero on lines originating on the t = 0 axis. The value of V+ on lines coming Solutions to Chapter 14
14-8
from the z = 0 axis is determined by requiring that the currents at the input
terminal sum to zero.
or
ig
(l/Zo - 1/ Rg )
= (1/ R g + l/Zo ) + V_ (l/Zo + 1/R g )
It follows that for 0 < t < T, V+ = l o Ru/2 while for T < t, V+ = o. At z = I,
(14.4.8) shows that V+ = -V_. Thus, the solution is as summarized in the figure.
V+
v;=-v+=o
z~
f::::::::t:=lt:=====::::;::=========:::P
•
i
g
I o Rg/2
V(o. t)t
-q-J...,----2l-/e-ci:Jr--2l-/e-+-T- -...
I~/2
t
l/e
Fleur. 814.4.8
14..4..7
(a) By replacing V+/Zo -+ 1+, V_/Zo -+ -L, the general solutions given by
(14.4.1) and (14.4.2) are written in terms of currents rather than voltages.
1
V=Y (I+-L)
o
(1)
where Yo == l/Zo. When t = 0, the initial conditions are zero, so on char
acteristic lines originating on the t = 0 axis, 1+ and L are zero. At z = "
it follows from (lb) that 1+ = L. At z = 0, summation of currents at the
terminal gives
(2)
i g = (G g jYo )(l+ - L) + (1+ + L)
which, solved for the reflected wave in terms of the incident wave gives
(3) 14-9
Solutions to Chapter 14
where
(4)
Feom these relations, the wave components 1+ and L are constructed as
summarized in the figure. The voltage at the terminals of the line is
(5)
.
+
4.
I Q (l+fQI
:t,(ItIqtIql
2
3
N=
iDL
t
Figure 814.4.7'
It follows that during this same interval, the terminal current is
1(0, t)
=1 (1 0
rN - 1 )
(1 + ;"o/Og)
(6)
(b) In terms of the terminal current I, the circuit equation for the line in the limit
where it behaves as an inductor is
ig
= lLOg ~~ + I
Solution of this expression with i g
1(0, t)
=1
= 10 (1 -
0
(7)
and 1(0)
e- t/T );
l'
= 0 is
== lLO g
(8) Solutions to Chapter 14
14-10
(c) In the limit where Gg/Yo is very large
(9)
Thus,
I
2Nl
2(N-l)- i
Bx x
where
_
Bq>i
By 'Y
Vl~-:t)2+y,]
q> -
V In [ Vlv'P-R'+:t)'+Y' (e-i~z + ei~z) e3'wt
-Re - g
2
In[:k+V(l/R)2_1]
(ei~l+ei~l)
Using (14.2.2) A z follows from this potential.
where
II
V:
A z -- -Re !:!!.......J!..l n
21rZo
(e-i~z _ ei~z) .wt
3
y(v'l2 _ R2 + x)2 + y2 ( e3'~l + e 3'~l) e
J(v'l2 - R2 - x)2 + y2
In these expressions, f3 and Zo are evaluated from (14.7.8) and (14.7.11) using the
values of C and L given by (4.6.27) and (4.6.12) with R = 0 and G = (u/€)C.
14.8.4
(a) The integral of E around the given contour is equal to the negative rate of
change of the magnetic flux linked. Thus,
and in the limit
~z -+
0,
BEa
Bz
a--
BE b
+ b-=
Bz
BH
Bt
-J1.o(a + b)--Y
(2)
Because €aEa = €bEb, this expression becomes
(a
€a
BEa
+ -b)-B
=
4
z
BH
-J1.o(a+ b)-B
Y
t
(3)
If Ea and H y were to be respectively written in terms of V and I, this would
be the transmission line equation representing the law of induction (see Prob.
14.1.1).
(b) A similar derivation using the contour closing at the interface gives
(4) 14-19
Solutions to Chapter 14
and in the limit ti.z -+ 0,
E.
aH"
= -aJ.'0ljt -
a
aEa
az
(5)
With the use of (3), this expression becomes
H
E. = - [aJ.'o(Ea - l)b/(a + Ea b)] aa "
Eb
t
Eb
(6)
Finally, for a wave having a z dependence exp(-j,8z), the desired ratio follows
from (6) and (3).
IE.I = b(,8a) 11 lEal a+b
Ea
I
(7)
Eb
Thus, the approximation is good provided the wavelength is large compared
to a and b and is exact in the limit where the dielectric is uniform.
14.9 QUASI-ONE-DIMENSIONAL MODELS
14.9.1
From (14.9.11)
2
R=-
1rWU
while, from (4.7.2) and (8.6.12) respectively
c-
211"E
•
- In[(l/a) + y(l/a) -II'
L = e'n[('/a) + Y(I/a)2 - 1]
11"
To make the skin depth small compared to the wire radius
6
=
V
2
WJ.&U
:> R => w <: 2/a 2 J.W
For the frequency to be high enough that the inductive reactance dominates
2
wL = wJ.'ua In [(l/a)
+ Y(I/a)2 _ 1]
R
2
Thus, the frequency range over which the inductive reactance dominates but the
constant resistance model is still appropriate is
2
-==-:-:-::-:--:---;:;:;;::;::::;::===:
2
J.'uR Inl(l/a) + y(l/a)2 -
11
2
< w < -2
a J.'u
For this range to exist, the conductor spacing must be large enough compared to
their radii that
1 <: In[( i) + Y(I/a)2 - 1]
a
Because of the logarithmic dependence, the quantity on the right is not likely to be
very large. 14-20
14.9.2
Solutions to Chapter 14
From (14.9.11),
1
R
1
1
1
1
= 0'27l'a~ + 0'7l'b2 = 'frO' [a~ + b2 ]
while, from Prob. 14.2.1
L=
~;ln(a/b)j
C=
2'frE
In(a/b)
For the skin depth to be large compared to the transverse dimensions of the con
ductors
0==
2 ::> ~ or b => W -< 2/b 21J0' and 2/~21J0'
V
WIJO'
This puts an upper limit on the frequency for which the model is valid. To be
useful, the model should be valid at sufficiently high frequencies that the inductive
reactance can dominate the resistance. Thus, it should extend to
wL
R
= WlJoO' In(a/b)/[..!.... + -!.]
>
2
a~
2
1
b
For the frequency range to include this value but not exceed the skin depth limit,
2[~+;,\]
1J00In(a/b)
2
d
-< W -< b21J0' an
2
~21J0'
which is possible only if
1
In(a/b)
-< [a~ + ;,\ ](b2and~2)
Because of the logarithms dependence of L, this is not a very large range.
14.9.3
Comparison of (14.9.18) and (10.6.1) shows the mathematical analogy between
the charge diffusion line and one-dimensional magnetic diffusion. The analogous
electric and magnetic variables and parameters are
H;s
+-+
V,
Kp
+-+
Vp ,
IJO' +-+
RC,
b +-+ I,
:z: +-+ %
Because the boundary condition on V at % = 0 is the same as that on H;s at :z: = 0,
the solution is found by following the steps of Example 10.6.1. From 10.6.21, it
follows that the desired distribution of V is
V =
co
(-l)n. n1r%
-v.P,- - '"
2V. - - S i n (_)e- t /.,.....
LJ P n7l'
I
'
n=l
%
This transient response is represented by Fig. 10.6.3a where H;s/K p
-
V /Vp and
z/b - %/1.
14.9.4
See solution to Prob. 10.6.2 using analogy described in solution to Prob. 14.9.3.
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RES.6-001: Solutions to Chapter 14
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