p5-23 Some of the tedium of the ray-tracing can be removed by using a so-called covering space (see
E. F. Kuester and D. C. Chang, IEEE Trans. Ant. Prop., vol. 31, pp. 27-34, 1983). This method
will not be used here, however. We will (at least temporarily

p6-4 To show that (6.34) and (6.35) are equivalent, use the trigonometric identity
tan1 (x) =
tan1
2
1
x
for x = p2 /h1 , and the result follows immediately. Next, let m be even and divide (6.34) by 2,
then take the tangent of both sides:
tan
h1 a
2
= ta

p5-20 Since the only change in solving this problem as compared with the corresponding problem for
the full circular waveguide will lie in the boundary conditions, we follow the method of section 5.5
exactly, up until equation (5.70), except that Ez Hz :

p3-9 From (3.29) we have for this case
d[V ]
= j[x][I];
dz
where
d[I]
= j[b][V ]
dz
l11
2
1 2 /c
[x] =
0
0
l22
while [b] = [c], and [c] is given by (3.28). Assuming the mode solutions
[V ] = [Vm ]em z ;
[I] = [Im ]em z
gives us
m [Vm ] = j[x][Im ];
m [I

p4-2 Pulses will overlap if their envelopes touch, and envelopes travel at the group velocity. Consider a
pulse (call it the mth pulse) emitted by the generator. On line N , the pulse arrives at the output
earliest (with a delay time of d/vg,N later than

p5-5 For TE modes, we have from (5.8) and (5.18),
Hz =
1
m
1 Ey
mx
uz T ET =
= 2jE1
cos
j
j x
ja
a
2
By (5.23), we have (with kc = 2 + 2 = (m/a)2 ):
HT =
a
m
2
m T Hz =
2jE1
mx
ux sin
TEm
a
Then from the normalization condition (5.27) for a forward mo

p4-10 (a) From (1.14) with g = 0, we have
b r2 + x2 xb
=
=
2
lc
=
1+x1=
2
c r2 + 2 l2 2 lc
2
x
c 2
1+x+1
and from (1.15) we get
=
rb
rc
r
=
=
2
2
2
1
c
l
=
c 2
1+x1
1+x+1
[Note the subtle dierence in font between capacitance c and velocity c, and betw

p1-20 (a) We compute
Zc (z)
1
=
ln
2Zc (z)
2d
ZL
Zc0
+
A
cos
d
2z
d
From (1.106), we have that the rst term gives the result for the exponential line in (1.112),
while the second term contributes
A
d
d
e2jz cos
0
2z
d
dz =
d sin d jd
Ae
2 (d)2
to the re

p3-16 Using the product rule for dierentiation, we have
d
dV (z)
dI (z)
[V (z)I (z)] =
I (z) + V (z)
dz
dz
dz
and by (3.6) this becomes
d
[V (z)I (z)] = [zI(z) + Eext (z)] I (z) + V (z) [yV (z) + Hext (z)]
dz
= z |I(z)|2 y |V (z)|2 + I (z)Eext (z) + V (z

p1-22 We can break up the solution by repeated application of formula (1.55). On the right side, the
semi-innite section of line with Zc1 can be replaced with a lumped impedance equal to Zc1 , since
with no reection on this line there is only a forward wa

p1-23 The geometry of this problem (shown in the gure below)
0 ,
0
z = 0
2
,
2
z = d
0 ,
0
has the equivalent circuit shown in Figure 1.10, where from Table 1.1 we have for the TM case
Zc1 = 0 cos ;
Zc2 = cos 2 ;
1 = jk0 cos ;
2 = jk cos 2
where
0 =
0
;
0

p2-10 We must have V2 + V3 = 0 for any combination of currents I1 , I2 and I3 . Then add the second
and third rows of the impedance matrix equations to get
0 = V2 + V3 = (Z21 + Z31 )I1 + (Z22 + Z32 )I2 + (Z23 + Z33 )I3
Assume reciprocity: Z23 = Z32 , etc.

p1-21 (a) A possible structure that realizes this distributed circuit might be a wire over a ground plane,
periodically interrupted in series to give s and periodically shorted to ground to give rm , with
the lengths of interrupted wire accounting for the

p2-20 (a) Equation (2.12) expands to
V1 = AV2 BI2 ;
()
CV2 DI2
Solve the second of these for I2 :
CV2 I1
D
Direct comparison of () with the second of (2.19) gives
I2 =
H21 =
1
;
D
()
H22 =
C
D
A
BC
D
Now plug () into the rst of (), resulting in
V1 = AV2