1) Page 3
read: J-band as: (5.38 – 8.17) GHz
read: C-band as: (3.94 – 5.99) GHz
2) Page 4
Replace: existing Table 1.1 by:
3) Page 16
read:  0  120  377 Ω
4) Page 30
Eq. (2.20f) should be: H1t – H2t = Js
Eq. (2.20g) should be: D1n – D2n = ρs
5) Page 34
2 1
Eq. (2.31b) should be: T 
1   2
Same correction applies to expression of T in page 35 (in solution of Example
2.7).
In page 34, read: Eq. 2.13a as: Eq. 2.31a
Solution of Example 2.7 read: T 
2 2
 0.000196
 2  1
6) Page 38
read: Eq. 2.32b as: Eq. 2.33a
7) Page 44 and 45
replace: Raleigh by: Rayleigh
8) Page 46
read: Xin = 42.5 Ω
Fig. 2.13 read: dΩ (instead of dω) = (Rdθ) (R sinθ dφ)
Caption of Fig. 2.13 read: (dω) as: (dΩ)
9) Page 47
Eq. (2.55) read: dω (in the denominator) as: dΩ
In the line following this equation replace: dω = sinθ dθ dφ by: dΩ = sinθ dθ dφ
Following this line replace: (Fig. 2.12) by: (Fig. 2.13)
10) Page 48
replace: 0.79 ω by: 0.79 Ω
replace: Rrad = 73ω by: Rrad = 73 Ω
11) Page 50
Solution of Example 2.8, replace: Eq. 2.48 by: Eq. 2.58
replace Eq. 2.46 by: Eq. 2.56
12) Page 52
read:  0  120  377 Ω
13) Page 54
2 1
read: T 
1   2
14) Page 55
replace: Raleigh by Rayleigh
In the last line read: dω (in the denominator) as: dΩ
15) Page 56
replace: dω = sinθ dθ dφ by: dΩ = sinθ dθ dφ
16) Page 58, Review question no. 2.38
replace: Raleigh by: Rayleigh
17) Page 59, Numerical problem 2.1, line 4


read: Aaz sin pxsin(t  z) as: az Asin px sin(t  z)
18) Page 59, Numerical problem 2.3
Ignore the answer part given; it is irrelevant
19) Page 59, Numerical problem 2.4



H ( z )  3.34(ax  ja y )e j 3.92
read: [Ans:



H ( z )  3.34(ax  ja y )e j 3.92
as: [Ans:
μA-m-1]
20) Page 60, Numerical problem 2.6, line 1–2
replace: and the other has µr2 = 2
by: and the other having µr2 = 2
line 5
replace: (b) the magnetic field intensity in the material 2
by: (b) the magnetic flux density in material 2
21) Page 60, Numerical problem 2.9, 2nd line
replace: cos1 (v / c) by: sin 1 (v / c)
22) Page 60, Numerical problem 2.10
read: [Ans: Aet = 1.175, Aer = 1.074, Pr = 2.243 mW]
as: [Ans: Aet = 1.175, Aer = 1.074, Pr = 2.243 μW]
23) Page 67, 2nd line after Eq. 3.11
replace: Ω by: ω
24) Page 70
replace: Eq. 3.10b by: Eq. 3.11
25) Page 71, Eq. 3.18
replace: (the last part of the equation) L e j  2 jz by: L e j (  2 z )
replace: Eq. 3.10b by: Eq. 3.11
26) Page 72, Eq. 3.19b
1st line, replace: e  j ( 2 z ) by: e j ( 2 z )
2nd line, replace: ( L sin ) by: ( L sin )2
replace: where   ( L  2z ) by: where   (  2z )
27) Page 73
replace: Using Eq. 3.17 by: Using Eq. 3.18
28) Page 74, last diagram of Fig. 3.5
replace: d by: l
29) Page 88
replace: Eq. 3.17 by: Eq. 3.18
30) Page 104, last line
replace: Eqs 3.46a by: Eqs 3.45a
31) Page 105, 1st line
replace: 3.46b by: 3.45b
32) Page 102, solution of Example 3.7, last line
replace: eff   /  eff , thus  eff
 


 eff
2
  1 2
 
 6.44
  0.394 

2
    1 2
 
by: eff   /  reff , thus  reff  
 6.44
    0.394 
 eff 
33) Page 116–117, Numerical problem 3.1
1st line, read: j150 as: j150 Ω
5th line, read: – j150 as: – j150 Ω
34) Page 117, Numerical problem 3.3, delete the answer given and read as:
Ans: (a) Vmax = 133.3 V and Vmin = 99.99 V (note Vmin = VL = 100 V, as RL < Z0);
(b) The minima occurs at: z = 0, 0.5λ, 1.0λ, 1.5λ … while the maxima
occurs at: z = 0.25λ, 0.75λ, 1.25λ,1.75λ … (c) Zmin = 75.18 Ω and Zmax =
133 Ω; (d) PL = 48.98 W].
35) Page 118, Numerical problem 3.7
o
read: [Ans:   0.56e j 24.5 , ρ =3.7, Zi = 12.5 – j2.5 , a voltage maximum occurs
at 0.034λ from the load]
as: [Ans:   0.6224.5o , ρ =4.3, Zi = 12.5 – j2.5 , a voltage maximum occurs at
0.034λ from the load]
36) Page 118, Numerical problem 3.8
read: [Ans: W = 0.095 inch, εeff = 1.891, vpms = 2.18 × 1010 cm/s]
as: [Ans: W = 0.095 inch, εreff = 1.891, vpms = 2.18 × 1010 cm/s]
37) Page 121




replace: [  H  j E ] by: [  H  j E ]
38) Page 123, preceding Eq. 4.6b in the text
replace: Eq. 4.6 by: Eq. 4.6a
39) Page 124, third paragraph
replace: Eq. 4.6 by: Eq. 4.6a
40) Page 125
replace: Eq. 4.3a by: Eq. 4.3
replace: Eq. 4.8c by: Eq. 4.7f
41) Page 127, in the text following Eq. 4.9b
replace: Above the cut-off frequency, the phase constant can be written as
by: Above the cut-off frequency (   cTEmn ), the phase constant of the
propagating wave can be written as
42) Page 127, in the text following Eq. 4.9b
replace: Eq. 4.9 by: Eq. 4.8
43) Page 127, in Eq. (4.9c)
2
replace:  gTEmn
 m   n 
    
 

 a   b 
2
2
2
2
 m    n 
by:  gTE mn      
 

 a   b 
44) Page 128, 2nd paragraph
replace: Eq. 4.9 by: Eq. 4.8
45) Page 129, in the 2nd paragraph following Eq. 4.12
replace: Eq. 4.12 by: Eq 4.11
46) Page 129, corresponding to Eq. 4.13b
replace: Case II: For f < fc, by: Case II: For f > fc,
47) Page 130, Section 4.2.3, 1st paragraph, penultimate line
replace: TE/TM models by: TE/TM modes
48) Page 130, 1st paragraph after Fig. 4.5, 1st line
replace: Eq.4.11b by: Eq. 4.10b
49) Page 136, Example 4.1, in (iv)
replace: 864 ω by: 864 Ω
50) Page 136, Section 4.2.5
1st line, replace: Eq. 4.10c by: 4.9c
2nd line, replace: Eq. 4.10d by: Eq. 4.9d
2
51) Page 137, paragraph before Eq. 4.15b, penultimate line
replace: Eq. 4.16a by: 4.15a
replace: Eq. 4.10c by: Eq. 4.9c
52) Page 137, paragraph after Eq. (4.15b), 2nd line
replace: Eq. 4.10d by: 4.9d
replace: Eq. 4.16b by: Eq. 4.15b
53) Page 138, 2nd paragraph, last line
replace: Eq. 4.16b by: Eq. 4.15b
replace: Eq. 4.10d by: Eq. 4.9d
54) Page 138, paragraph after Eq. (415c), 2nd line
replace: Eq. 4.10c by: Eq. 4.9c
55) Page 139, 3rd paragraph, last line just preceding Table 4.1
replace: xxx/1000 by: xxx/100
56) Replace Table 4.1 by:
57) Page 143, paragraph after Eq. (4.16b), 1st line
replace: Eq. 4.16 by: Eq. 4.16b
58) Page 145, in the solution of Example 4.4, last line
replace:
2
2
2
 3  103  1
2.5  1.0
E0 ab
 2 
 
 

1 
Ppeak 
1    = 
 = 0.646 MW
2
 2
 2 2 
 2a 
 2  377
by:
2
2
2
2
 3  103 
E0 ab
1 2.5  1.0
  
 2 
 
Ppeak 
1    = 

1 
 = 6.837 kW
 2
2
 2a 
 2  2.5 
 2  377
59) Page 148, Caption of Fig. 4.12, 2nd line
replace: s = 5.8 × 107 S/m by: σ = 5.8 × 107 S/m
60) Page 161, last paragraph, 1st line
replace: Fig. 4.18b by: Fig. 4.17b
61) Page 162, Caption of Fig. 4.17, 2nd line
replace: s = 5.8 × 107 S/m by: σ = 5.8 × 107 S/m
62) Page 168, Eq. 4.45

 m 
 n 
 p 
replace: Ez  E0 cos
 x sin 
 y cos
z
 a 
 b 
 l 
by: E z  E0 sin k x x sin k y y cos k z z
63) Page 169, Eq. 4.46
delete: Ez  E0 sin kx x sin k y y coskz z
64) Page 170, Eq. (4.48), last line


a2 


a2 
abl  1  2  H 0
abl 1  2  H 02
16
l 
16 
l 

replace:
by:
70) Page 179, paragraph after Fig. 4.22, 4th line from bottom
read: 1750 as: 1750 GHz
71) Page 180, Eq. 4.68 a and b
replace: kcnm(TM ) by: kcTMnm
72) Page 184, paragraph after Fig. 4.26, 3rd line
replace: Fig. 12.24 by: Fig. 12.25
73) Page 186
replace:  gTE   2   ( m / a )2  ( n / b)2
by:  gTE   2   ( m / a ) 2  ( n / b) 2
74) Page 194, Numerical problem 4.1
Read answer as: [Ans: Allowed mode is TEmm with cut-off frequency
mc
fm 
]
2a
75) Page 194, Numerical problem 4.6, 3rd line
read: 2.18 as: 2.16
76) Page 195, Numerical Problem 4.7, 1st line
replace: Fig. 4.18a by: Fig. 4.17a
replace: Fig. 4.18b by: Fig. 4.17b
77) Page 195, Numerical Problem 4.10, last line
replace: Fig. 4.23 by: Fig. 4.22
78) Page 198, Eq. 5.1
replace (in the last term of last set of equations): Smn by: Snn
79) Page 199, Eq. 5.4f
replace (in the LHS of the equation): b1 by: b2
80) Page 209
Fig. 5.7d and e: arrow sign of S22 to be reversed in direction
Fig. e (i) and e (ii): the straight connecting line from b2 to a2 to be deleted
81) Page 203, 214
replace: Meson by: Mason
82) Page 236, 3rd paragraph
replace: Fig. 6.19c(ii) by: Fig. 6.20c(ii)
83) Page 241, last paragraph, last line
replace: Eq. 4.11b by: Eq. 4.10b
84) Page 253, text following Fig. 6.31
replace: Fig. 4.15 by: Fig. 4.15e
85) Page 295
replace: Fig. 7.8 by:
86) Page 295, penultimate line
replace: l1a = 0.266 and l1b = 0.456λ by: l1a = (0.25 + 0.126)λ = 0.376λ and
l1b = (0.25 + 0.206)λ = 0.456λ
Same change in the last line of text of Step 4
87) Page 301
replace: i  ( Z i 1  Z i )( Z i 1  Z i ) by: i  ( Z i 1  Z i ) /( Z i 1  Z i )
88) Page 331, Example 7.10, 1st line of Solution
replace: z L
zL
Z
by: z L  L
z0
Z0
89) Page 332, penultimate line
replace: 75 Ω lin by: 75 Ω line
90) Page 345, 1st line of the paragraph following Eq. 8.7b
replace: Eqs 8.5a and 8.5b by: Eqs 8.7a and 8.7b
91) Page 361, 2nd paragraph following Eq. 8.24, 2nd line
replace: identical series tuned resonators by: identical parallel tuned
resonators
replace: identical parallel tuned resonators by: identical series tuned
resonators
replace: Fig. 8.11 (b) and (c) by:
Fig. 8.11 (a) A section of band-pass filter (b) Conversion to all parallel resonators with
K-inverter (c) Conversion to all series resonators with J-inverter
92) Page 363, Solution of Example 8.2, Method I
replace: (Fig. 8.5b) by: (Fig. 8.4b)
93) Page 368, Section 8.5.2, 2nd line of
replace: It consists of series and parallel resonant circuits by: It consists of
parallel and series resonant circuits
94) Page 369, Eq. 8.30


J

replace: k  1, k  

Y0
 2c gn gk gk  1 
1/ 2

J k 1,k 


by:

Y0
 2 c g k g k 1 
1/ 2
95) Page 371, Step 3, first expression
replace:
J 10 J 6,5


Y0
Y0

2 g 0 g1


2 g5 g6

  0.108
2  1  1.3394
 0.356
J 10 J 6 , 5


  0.108




 0.356
Y0
Y0
2 g 0 g1
2 g5 g6
2  1.3394 1
96) Page 375, Section 8.5.3, 2nd line
replace: It consists of parallel and series resonant circuits
by: It consists of series and parallel resonant circuits
97) Page 388, last line
repalce: Wmainline  2.1  0.707 = 1.48 mm by: Wmainline  2.1  0.7874 =
1.653 mm
replace: W stub line  0.44  0.707 = 0.31 mm by: W stub line  0.44  0.7874 =
0.346 mm
98) Page 392, Table 9.2, last line
replace: (Amplifier frequency standard) by: (Amplifier/frequency standard)
99) Page 399, 1st paragraph, last line
replace: Fig. 9.5 by: Fig. 9.4
100) Page 400, Fig. 9.6
replace existing figure by:
by:
101) Page 401
replace line 7 and 8 by:
voltage and hence an inductive impedance is presented across the cavity;
raising the oscillation frequency of a particular mode above the resonant
value.
102) Page 401
replace line 11 and 12 by:
Negative voltage for a particular mode, then a capacitive impedance is
presented to the cavity; lowering the oscillation frequency with respect to the
resonant value for the particular mode of
103) Page 407, Fig. 9.12, caption
interchange the position of position of (ii) and (i)
104) Page 416, Eq. 9.5a, denominator
replace: fp by: fc
105) Page 419, Example 9.4, 1st numerical equation
replace: ms by: m/s
106) Page 420, Fig. 9.24
read: Double-coil helial electron beam as: double-coil helical electron beam
107) Page 428, Fig. 9.30c
replace: Lp by: Ls
108) Page 429
 A2  1 
A2  1
 Z 0
replace: RS 
by: RS  
2A
 2A 
109) Page 430, 4th paragraph, last line
replace: takes place form port 1 to port 2 by: takes place from port 1 to port 2
110) Page 432, 3rd paragraph, 16th line
Replace: IU >> IL by: IU << IL
111) Page 433, Fig. 9.32a
reverse the battery terminals
112) Page 446, Eq. 9.10, RHS, denominator of first expression
replace: t by: τ
113) Page 474
read: Eq. 9.43 as: Eq. 9.18
read: Eq. 9.44 as: Eq. 9.19
114) Page 482, Fig. 9.65, the lowest power supply
should not be VDS, it should be VGS
115) Page 493, penultimate line at the end of the paragraph
read: bandwidth of 25 MHz as: bandwidth of 250 MHz
116) Page 545, 1st line following Eq. 11.2a
replace: (Eq. 2.47) by: (Eq. 2.57)
117) Page 550, Example 11.3, last line
replace: 109 by: 10–9
118) Page 554, 4th line from the bottom
replace: W by: fm
119) Page 563
replace: Raleigh by: Rayleigh
120) Page 586, last paragraph, 8th line
replace:  K s cost [ K d cos(  t ) ] by:  K d cost [ K d cos(  t ) ]
121) Page 638
replace: the contents of Eq. (12.26) by:
nf1 = fH = fs and (n – 1)f2 = fH = fs, whence:
fs 
f1 f 2
f 2  f1
…….. (12.26)
122) Page 639, Section12.6, 2nd paragraph, 5th line from the bottom
replace: (RAM), Section 13.5) by: RAM, (Section 14.5)
123) Page 695
read: Eq. (14.1c) as:
0
 20 log10 et /   20 log10 1  e  2t / 
4 sm
[RdB]
[AdB]
[MRdB]
SE  20 log10
…. (14.1c)
Page 709, same correction
124) Page 719, Eq. (15.2a)
read: D1n – D2n = 0 as: D1n = D2n
125) Page 721, Eq. (15.2)
replace: nreff   reff  reff by: nreff    reff  reff
126) Page 734, Fig. 15.15
replace: +βc by: –βc (on the left side)
replace: –βc by: +βc (on the right side)
127) Page 743, 2nd paragraph after Eq. (15.16b)
read: (also known as surface polariton) as: (also known as polariton)
128) Page 749, paragraph following Table 15.6,
replace: Section 15.3.2 by: Section 15.3.2.2
129) Page 765, paragraph preceding Section 16.2.2
read: Q-band as: V-band
130) Page 768, Fig. 16.3, in the triangular box marked as IF amp
replace: i with: 1
131) Page 769
read: radar altimeter as: radio altimeter.
132) Page 780, 3rd paragraph, 7th line
replace: but should be higher with: but should not be higher
133) Page 785, 2nd paragraph 4th line
replace: define a 2 × 2 transmission by: define a 2 × 2 transmission matrix
134) Page 826, Chapter 9, 7
replace: Holtman by: Holzman