Solution Manual
Solution Manual
Solution Manual
by
S R M Prasanna
Department of Electronics and Electrical Engineering
Indian Institute of Technology Guwahati
Contents
1 Introduction to Communication Systems 1
2 Noise 2
3 Amplitude Modulation Techniques 5
4 Angle Modulation Techniques 9
5 Pulse Modulation Techniques 13
6 Digital Modulation Techniques 14
7 Radio Transmitters and Receivers 15
8 Television Broadcasting 17
9 Transmission Lines 18
10 Radiation and Propagation of Waves 26
11 Antennas 28
12 Waveguides, Resonatorsand Components 30
13 Microwave Tubes and Circuits 36
14 Semiconductor Microwave Devices and Circuits 37
15 Radar System 38
16 Broadband Communication Systems 41
17 Introduction to Fiber Optics Technology 42
18 Information Theory, Coding and Data Communication 43
Acknowledgements
I would like to thank all my students and project engineers working in the Electromedical and Speech
Technology (EMST) laboratory, Dept. of Electronics and Electrical Engineering, Indian Institude of
Technology Guwahati for helping me in developing this solution manual. A special mention goes
to the following persons: Gayadhar Pradhan, Debmalya Chakrabarty, Biswajit Dev Sarma, Syed
Shahnawazuddin, Haris B.C., Govind. D and Kukil Khanikar. Thank you Gayadhar Pradhan for
coordinating the whole effort and finally compiling the entire solution manual into one document.
1
Introduction to
Communication Systems
No Review Problems
Noise
2
(No. 1)
vn = 4kT D f R
Vn0 µ Df
Hence,
D f2 5 ¥ 103
V¢n0 = ¥ Vn 0 = ¥ 1 ¥ 10-3 V
3
D f1 20 ¥ 10
1
= ¥ 10-3 V = 0.5 mV
2
(b)
2
Vn0
Vn0 = A 4kT D f R =
A2 ¥ 4kT D f
(1 ¥ 10-3 ) 2
= W
602 ¥ 4 ¥ 1.38 ¥ 10-23 ¥ (80 + 273) ¥ 20 ¥ 103
= 0.7 MW
(No. 3)
Given,
Q = 20
Noise 3
At resonance,
XC = QRS
XC
RS =
Q
Thus,
XC = wC = 2p ¥ 200 ¥ 106 ¥ 10 ¥ 10–12 W
wC
RS =
Q
= 20p ¥ 10–5W
Therefore,
RP = Q2RS = 400 ¥ 20p ¥ 10–5 W
Also,
Resonant frequency
Q=
Df
Therefore,
Resonant frequency
Df =
Q
200
= MHz = 10 MHz
20
Now,
v = Voltage across the capacitor
= 4kT D f RP
R3 = 470 kW
4 Kennedy’s Electronic Communication Systems
R2 R3
Req = R1 + +
A12 A12 A22
15.7 ¥ 103 470 ¥ 103
= 800 + + W
152 152 202
= 875 W
(No. 6)
Here,
A1 = 25
Now,
R1 = 1 kW + 2 kW = 3 kW
R2 = 125 kW
R2
Req = R1 +
A12
125
= 3 + kW
252
= 3200 W
Hence,
Vn = 4kT D f Req
(No. 1)
Given carrier frequency, fc = 1000 kHz
Modulating frequencies, f1 = 300 Hz
f2 = 800 Hz
f3 = 2 kHz
Frequencies present in the output will be,
(fc + f1), (fc – f1), (fc + f2), (fc – f2), (fc + f3), and (fc – f3),
fc + f1 = 1000 + 0.3 kHz = 1000.3 kHz
fc – f1 = 1000 – 0.3 kHz = 999.7 kHz
fc + f2 = 1000 + 0.8 kHz = 1000.8 kHz
fc – f2 = 1000 – 0.8 kHz = 999.2 kHz
fc + f3 = 1000 + 2 kHz = 1002 kHz
fc – f3 = 1000 – 2 kHz = 998 kHz
(No.2)
Given, Radiated carrier power, Pc = 50 kW
Modulation index, m = 0.85
Total radiated power by the AM transmitter will be,
PAM = Pc (1 + m2/2)
= 50 (1 + (0.85)2/2) kW
= 68.0625 kW
(No. 3)
Given, Total radiated power, PAM = 10 kW
Modulation index, m = 0.75
6 Kennedy’s Electronic Communication Systems
PAM
Carrier power, Pc =
(1 + m 2 /2)
10
=
(1 + (0.75) 2 /2)
= 7.804 kW
Power in one sideband is = Pc ¥ m2/4
= 7.804 ¥ (0.75)2/4
(Carrier power + one sideband power )
Percentage power saving is = ¥ 100%
Total power
(7.804 + 1.0975)
= ¥ 100%
10
= 89.015%
(No. 4)
Given, carrier power, Pc = 360 W
Modulation indexes are, m1 = 0.55
m2 = 0.65
= 0.552 + 0.652
= 0.851469
mt2
Total side band power = Pc ¥ W
2
0.8514692
= 360 ¥ W
2
= 130.4999 W
(No. 5)
Given, Collector dissipation = 20 W
Collector efficiency = 75% W
(a) Modulation index, m1 = 0.90
(i) If maximum carrier power is Pc
0.25 ¥ Pc = 20
Pc = 80 W
m2
(ii) Maximum side band power Ps = Pc ¥
2
Amplitude Modulation Techniques 7
(0.9) 2
= 80 ¥
2
= 32.40 W
I t1
Unmodulated current, Ic =
1 + m12 /2
12
=
1 + 0.52 /2
= 11.31 A
New antenna current with modulation index (m2 = 0.9)
m22
will be, It2 = Ic ¥ 1+
2
0.92
= 11.31 ¥ 1+
2
= 13.40 A
(No. 7)
Given, Antenna current, It1 = 1.5A
Modulation index, m1 = 0.60
I t1
Unmodulated current, Ic =
1 + m12 /2
1.5
=
1 + 0.62 /2
= 1.3808 A
8 Kennedy’s Electronic Communication Systems
After modulating the generator with an additional audio wave with modulation index (m2 = 0.7), total modula-
tion index will be, mt = m12 + m22
= 0.62 + 0.7 2
= 0.92
mt2
New total current, It = Ic ¥ 1 +
2
0.922
= 1.3808 ¥ 1+
2
= 1.6472 A
I c2 Ê m2 ˆ
= ¥ 1 + ¥ 100 %
I t2 ÁË 4 ˜¯
1.38082 Ê 0.922 ˆ
= ¥ 1 + ¥ 100 %
1.64722 ÁË 4 ˜¯
= 85.13%
Angle Modulation
4
Techniques
(No. 1)
Given, Modulating frequency, fm1 = 500 Hz
Frequency deviation, dp1 = 2.25 kHz
Hence, Modulation index, mp = dp1 = 2.25 kHz
d p1
Proportionality constant, kp =
Vm1
New modulating frequency, fm2 = 6 kHz
New modulating voltage is not changed. Hence, Vm2 = Vm1
d p2 d p1
kp = =
Vm 2 Vm1
which implies, dp2 = dp1 = 2.25 kHz
Thus, the change in modulating frequency makes no difference to the deviation.
(No. 2)
Given, Modulating frequency, fm1 = 400 Hz
Modulating voltage, Vm1 = 2.4 V
Modulation index, mf1 = 60
Maximum frequency deviation, df1 = mf1 ¥ fm1 = 60 ¥ 400 = 24 kHz
d f1 24
Proportionality constant, kf = = = 10 kHz
Vm1 2.4
New modulating frequency, fm2 = 250 Hz
And Modulating voltage, Vm2 = 3.2 V
New maximum frequency deviation, df2 = kf ¥ Vm2 = 10 ¥ 3.2 = 32 kHz
df2
New modulation index, mf2 = = 32 /0.25 = 128
fm2
10 Kennedy’s Electronic Communication Systems
(No. 3)
Given, Angle modulated voltage is, v = 10 sin (108t + 3 sin 104t)
The voltage can be both frequency or phase modulated
104
where, modulating frequency, fm = = 1591.54 Hz
2p
108
where, carrier frequency, fc = = 15.9 MHz
2p
Modulation index m = 3
2
Vrms (10/ 2 ) 2
Power dissipated in a 100-ohm resistor will be, P = = = 0.5W
R 100
If we consider it as frequency-modulated voltage, the maximum frequency deviation will be
df = mf ¥ fm = 3 ¥ 1591.54 = 4774.64 Hz
If we consider it as phase-modulated voltage, the maximum frequency deviation will be,
dp = mp = 3
(No. 4)
Given, Center frequency of the LC oscillator, f = 70 MHz
Reactance of bias capacitor is 10 times the resistance of bias resistor.
Hence, n = 10
Capacitance of fixed tuning capacitor, C = 25 pF
gmmin = 1 mS
gmmax = 2 mS
Therefore, minimum eq. capacitance of reactance FET,
gmmin 1 ¥ 10-3
Cn = = = 0.22736 pF
2p fn 2p ¥ 7 ¥ 107 ¥ 10
And, maximum eq. capacitance of reactance FET,
gmmax 2 ¥ 10-3
Cx = = = 0.4572 pF
2p fn 2p ¥ 7 ¥ 107 ¥ 10
Ratio of maximum and minimum frequency,
1
L(C + Cn )
fx C + Cx
= p
2 =
fn 1 C + Cn
L(C + C x )
2p
= 1.0045
fx f +d
Again, =
fn f -d
where d is maximum frequency deviation.
f + d = 1.0045 ¥ (f – d)
2.0045 d = 0.0045 f
0.0045 f 0.0045 ¥ 70 ¥ 106
d= =
2.0045 2.0045
d = 0.157 MHz
(No. 5)
Given, Center frequency of the LC oscillator, f = 10 MHz
Reactance of bias capacitor is 10 times the resistance of bias resistor.
Hence, n = 10
gmmin = 0 mS
gmmax = 0.628 mS
Therefore, minimum eq. capacitance of reactance FET,
gmmin 0
Cn = = =0
2p f n 2p ¥ 10 ¥ 106 ¥ 10
And, maximum eq. capacitance of reactance FET,
f x2 - fn2 C x - Cn
2
=
fn C + Cn
( f x + f n )( f x - f n ) C x - Cn
2
=
fn C + Cn
12 Kennedy’s Electronic Communication Systems
4fd C x - Cn
2
=
fn C + Cn
(C x - Cn ) f n 2
C + Cn =
4fd
Now, let, fn ª f
(C x - Cn ) f 2
C + Cn =
4fd
(C x - Cn ) f
C= – Cn
4d
1 ¥ 10-12 ¥ 10 ¥ 106
=
4 ¥ 0.05 ¥ 106
= 50 pF
Again,
1
f=
2p L(C + Ca v)
1
ª
2p LC
1
L=
4p 2 f 2 C
1
=
4p ¥ 100 ¥ 1012 ¥ 50 ¥ 10-12
2
= 5.06 ¥ 10–6 H
5
Pulse Modulation
Techniques
No Review Problems
6
Digital Modulation
Techniques
No Review Problems
Radio Transmitters
7
and Receivers
1. Given
fs = 555 kHz
f0 = 1010 kHz
fi = f0 – fs = 455 kHz
a= 1 + Q 2 r 2 = 76.61
3. Given
fi = 450 kHz
a = 120
fs = 15 MHz
fsi = fs + 2fi = 15.9 MHz
f si f
r= - s = 0.116
fs f si
Since
a= 1 + Q2 r 2
or
a2 -1
Q2 =
r2
or
Q = 1034
4. Given
fi1 = 2 MHz
fi2 = 200 kHz
Q = 75
fs = 30 MHz
Therefore,
fsi1 = fs + 2fi1 = 34 MHz
f si1 f
r1 = - s = 0.251
fs f si1
a1 = 1 + Q 2 r12 = 18.85
fsi2 = fs + 2fi2 = 30.4 MHz
f si 2 f
r2 = - s = 0.0265
fs f si 2
a2 = 1 + Q 2 r22 = 2.224
a = a1a2 = 41.93
8
Television Broadcasting
No Review Problems
Transmission Lines
9
(No. 1)
L
Z0 =
C
4 ¥ 10-6
= W
100 ¥ 10-12
2
= ¥ 103 = 200 W
10
(No. 2)
138 D
Z0 = log ÊÁ ˆ˜
a Ë d¯
138 6
50 = log ÊÁ ˆ˜
1.6 Ëd¯
6 50 ¥ 1.6
log ÊÁ ˆ˜ =
Ëd¯ 138
d = 2.008 mm
(No. 3)
Z0 = 300 W
Vmax = 7.5 mV
Vmin = 5 mV
Vmin will be across the load resistance.
Vmin
ZL =
I
V
I = max
Z0
Transmission Lines 19
7.5 ¥ 10-6
= = 2.5 ¥ 10–8 A
300
5 ¥ 10-6
Hence, ZL = W = 200 W
2.5 ¥ 10-8
(No. 4)
Z02
ZS =
ZL
75 ¥ 75
= = 112.5 W
50
\ the source impedance is matched to the characteristic impedance of matching transformer.
Hence, characteristic impedance = 112.5 W
(No. 5)
Characteristics impedance = Z0 = 150 W
Load impedance = Zl = (225 – j75) W
Zl - Z0
Reflection coefficient = t =
Zl + Z0
225 - j 75 - 150 75 - j 75 106.006–-45∞
t= = =
225 - j 75 + 150 375 - j 75 382.426–-11.31∞
t = 0.277 – – 33.69
Voltage Standing Wave Ratio (VSWR)
1 + | t | 1 + 0.277
VSWR = =
1 - | t | 1 - 0.277
VSWR = 1.767
By Smith Chart
(a) Calculate the normalized load impedance
Z l 225 - j 75
zl = = = 1.5 – j0.5
Z0 150
Locate zl on a Smith Chart at the point where the r = 1.5 circle and x = –0.5 circle meet. To get t at zl,
extend OP to meet the r = 0 circle at Q and measure OP and OQ. Since OQ corresponds to |t |, then at P
OP 2.15 cm
|t | = = = 0.2866
OQ 7.5 cm
Note that OP = 2.15 cm and OQ = 7.5 cm were taken from the Smith chart. The angle between OS and
OP, that is –POS = – 34∞.
20 Kennedy’s Electronic Communication Systems
at B. Here, r = 0.56 and convert this normalized resistance into an actual resistance by multiplying by Z0
of the line. Here, R = 0.56 ¥ 150 = 84 W
l
(d) 84 W is the resistance which the transformer will have to match to the 150 W line.
4
Therefore,
z0¢ = Z 0 Z g = 150 ¥ 84
= 112.25 W
(No.7)
Normalized load impedance,
Zl 50 + j 50
zl = = = 1 + j1
Z0 50
Reflection coefficient,
OP 3.4
|t | = = = 0.4533
OQ 7.5
Qr = –POS = 63.5∞
t = 0.4533–63.5∞
Voltage standing wave ratio = VSWR = r = 2.63
(b) Moving towards the generator, i.e. clockwise, find the nearest point at which the line impedance is purely
resistive. Around the rim of the chart, the measure of the distance from the load to the point (B); the dis-
244.5
tance = 0.339l. Read the normalized resistance at B. Hence, r = 0.38 and convert this normalized
720
resistance into an actual resistance by multiplying by Z0 of the line. Hence, R = 50 ¥ 0.38 = 19 W. 19 W
l
is the resistance which the transformer will have to match to the 50 W line. Therefore,
4
(No. 8)
Normalized load impedance,
Transmission Lines 21
Zl 100
zl = = =2
Z0 50
Plotting P on the Smith chart gives SWR = 0.5
Q(admittance of load) is plotted. Point of intersection with r = 1 circle, R is plotted. Distance from load im-
110∞
pedance, Q-R is found to be = 0.152 l. This is the distance of the stub.
720∞
At R, Zline = 1 – j0.7, hence Sstub = j0.7. Plotting S and measuring the distance of S from •, j• gives the length
290∞
of stub as = 0.402l
728∞
f ≤ is 110% of f ¢
l
So, l¢ =
1.1
Thus, the distance of the stub is 0.152 ¥ 1.1 = 0.1672 l¢ and the length of the stub is
0.402 ¥ 1.1 = 0.4422 ¥ l¢
0.1672 l¢ = 0.1672 ¥ 720∞ = 120.384∞
and 0.1672 l¢ = 0.442 ¥ 720∞ = 318.384∞
(No. 9)
Load impedance, Zl = 180 + j120
Characteristics impedance, Z0 = 300 W
180 + j120
Normalized load impedance zl = = 0.6 + j0.4
300
Circle plotted around SWR = 2.0
Yl = 1.15 – j0.75 from the chart
To obtain Yl, extend PO to POP¢ and note P¢ where the S-Circle meets POP¢
Obain Yl = 1.15 – j0.75
Nearest point of Y = 1 ± b is 1 + j0.75. This is found from the Smith chart and marked as R. The distance from
178∞
P’ To R is found along the rim of the chart and given by = 0.177l
720∞
Therefore, the stub will be placed 0.177l from the load and will have to be tuned to b = + 0.75. Thus the
stub must have the susceptance – 0.75.
Starting from •, j• and traveling clockwise around the rim of the chart, reach the point 0 – j0.75 and mark
S¢ on the chart. From the chart, the distance of this point from the short-circuit admittance point is the stub
75∞
length = = 0.104 l.
720∞
22 Kennedy’s Electronic Communication Systems
50 45
55 40
60 Q 35
65
30
70
75
25
80
20
P
85
ANG
ANG
90
LE OF
LE OF
170
TRANSMISS
REFLECTIO
63.5°
N COEFFIENTION D
ION COEFFICIENT IN
B
180
CONDUCTANCE COMPONENT S
39°
170
EGRE
DEGR
ES
EES
–90
–20
AD
LO
5
RD
–8
WA
–25
TO
G TH
R
LE
0
244.5° –8
= 0.339 l
E
0
AV
720 –3
W
5
–7
5
–3
0
–7
0
–4
–65
–45
–60
–50
–60
–55
244.5°
)
.P
AN W. PL W L TE
SW BS S [ EF , E
P T
TR S. R S AT
E
] P rI
F,
1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.05 0.03 0 0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.5 15•
EF
3 4 5
CO
1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 1 0.99 0.95 0.9 0.6 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
SM
AN
CENTER
TR
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2
ORIGIN
50 45
55 40
60
0
35
18 0 2°
11 0 = 0.40
65
29
30
720 tub
s
70
New ngth
Le
75
25
80
20
85
ANG
ANG
90
LE OF
LE OF
170
TRANSMISS
REFLECTIO
N COEFFIENTION D
ION COEFFICIENT IN
P
180
CONDUCTANCE COMPONENT
39° S
B
170
EGRE
DEGR
ES
–90
EES
–20
AD
LO
5
–8
RD
WA
–25
TO
TH
RG
LE
0
–8
E
0
AV
–3
W
5
–7
5
–3
0
–7
0
–4
–65
–45
–60
–50
–60
–55
110°
= 0.152°
720
TOWARD LOAD
.P
AN RAN .W. PL W L TTE
TOWARD GENERATOR
F, NS ]
RT R FL
5W DE S [ EF , E
P T
R AS dB] , I or
LO C E
•60 38 20 15 10 10 20 •
R
S O FF
1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.05 0.03 15•
EF
0 0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.5 3 4 5
T S
CO
1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 1 0.99 0.95 0.9 0.6 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
SM
I
CENTER
TR
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2
ORIGIEN
50 45
55 40 Distance
of stub
60
35 128
= 0.177l
65 720
30
70
75
25
80
20
85
P R
ANG
ANG
90
LE OF
LE OF
170
TRANSMISS
REFLECTIO
2.0 (VSWK)
N COEFFIENTION D
ION COEFFICIENT IN
P
180
2.0
CONDUCTANCE COMPONENT S
170
EGRE
DEGR
ES
EES
–90
P –20
D
OA
DL
5
–8
AR
–25
OW
THT
RG
LE
0
–8
E
0
AV
–3
W
5
–7
5
–3
0
–7
0
–4
–65
Stub length
–45
75
–60
= 0.104l
–50
–60
–55
720°
)
.P
TOWARD LOAD
AN W. PL W L TE
F, NS ]
TOWARD GENERATOR
RT R FL
5W E S [ EF , E
P T
TR S. R S AT
R AS dB] , I or
D
LO C E
•40 38 10 20 •
R
S O FF
EF
1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.05 0.03 0 0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.5 3 4 5 15•
CO
1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 1 0.99 0.95 0.9 0.6 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
SM
I
AN
CENTER
TR
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2
ORIGIN
(No. 1)
Pt
PD =
4p r 2
Pt
200 ¥ 10–6 =
4p ¥ 202
Pt = 4p ¥ 202 ¥ 200 ¥ 10–6
Hence, PD at 25 km away from the source,
(No. 2)
500 10-2
PD = = = 159.2 mW
4p ¥ 500 ¥ 500 4p ¥ 5
3 ¥ 103 3
PD = = = 0.1843 mW
4p ¥ 36000 ¥ 36000 4p ¥ 36 ¥ 36000
(No. 3)
37 ¥ 10-18
Given, PD = W/m2
8400
37 ¥ 10-18 Pt
=
8400 4p ¥ 800 ¥ 106 ¥ 800 ¥ 106
(No. 4)
sin 20∞ v
=
sin 30∞ 3 ¥ 108
fc 7 ¥ 106
MUF = = = 7.143 MHz
cos q 0.98
(No. 7)
d = dt + dr
= 4 ht + 4 hr
d= 4 h+4 h
Since ht = hr = h
40 = 8 h
h =5
h = 25 km
Antennas
11
(No. 1)
60p Le I sinq
We Know, E =
l
In direction of maximum radiation, q = 90°
60p Le I
So, E=
l
60p10 ¥ 10-2 ¥ 2
=
3 ¥ 108
6
¥ 20 ¥ 103
10 ¥ 10
= 2p ¥ 10–5ms/m
(No. 2)
Ê P2 ˆ
Given, A = 10log10 Á ˜
Ë P1 ¯
2400 ˆ
= 10log10 ÊÁ
Ë 180 ˜¯
= 19.03 dB
(No. 4)
Given, Rrad = 72 W
Rt = 8 W
Rrad
Hence, Antenna efficiency, = ¥100%
( Rrad + RD )
72
= ¥100%
(72 + 8)
= 90%
Power gain, = 16
Directivity, = 10log10 (Power Gain)
Antennas 29
= 10log10(16)
= 12 dB
(No. 5)
Given, D = 64 m
3 ¥ 108
m l=
1430 ¥ 106
= 0.21 m Hence, beam width between half power point,
70 ¥ l
f=
D
70 ¥ 0.21
= = 0.22∞
64
Beamwidth between nulls
= 2f = 2 ¥ .22∞ = 0.44∞
Power Gain,
2
D
Ap = 6 ÊÁ ˆ˜
Ë l¯
0.60 ˆ 2
= 6 ÊÁ
Ë 0.21 ˜¯
= 163265.306
(No. 6)
Given, D = 5 m
l = 10 cm
We know that,
D 2
Ap = 6 ÊÁ ˆ˜
Ë l¯
2
Pr ad Ê 5 ˆ
6
Pf ed = ÁË 10 ¥ 10-2 ˜¯
2
Pr ad Ê 5 ˆ
= 6Á -2 ˜
20 ¥ 103 Ë 10 ¥ 10 ¯
Prad = 300 MW
Waveguides, Resonators
12
and Components
(No. 1)
For dominant mode of the rectangular waveguide, m = 1
2a
The cutoff wavelength, l0 =
m
= 2a = 2 ¥ 10 cm
= 20 cm
v 3 ¥ 108
Free space wavelength, =
f 2.5 ¥ 109
= 1.2 ¥ 10–1 m = 12 cm
l
Hence, guide wavelength, lp =
2
Ê lˆ
1- Á ˜
Ë l0 ¯
12
=
2
12
1- Ê ˆ
Ë 20 ¯
= 15 cm
Phase velocity, vp = flp = 2.5 ¥ 109 ¥ 15–2 m/s
= 3.75 ¥ 108 m/s
vc2 9 ¥ 1016
Group velocity, vg = = m/s
vp 3.75 ¥ 108
= 2.4 ¥ 108 m/s
377 377
Z0 = = = 471.25 W
Ê lˆ
2 0.8
1- Á ˜
Ë l0 ¯
Waveguides, Resonators and Components 31
(No. 2)
Given, f = 6 GHz
a = 7.5 cm
3 ¥ 108
l= m
6 ¥ 109
1
= ¥ 10–1 m = 5 cm
2
2a
l1,0 = = 2a = 2 ¥ 7.5 cm = 15 cm
m
2a
l2,0 = = a = 7.5 cm
m
2a 2 ¥ 7.5
l3,0 = = = 5 cm
m 3
377
Z1,0 =
2
Ê lˆ
1- Á ˜
Ë l0 ¯
377 377
= = = 399.87 W
0.5 ˆ 2 0.94
1- Ê
Ë 15 ¯
377 377
Z2,0 = = = 505.79 W
5 ˆ 2 0.745
1- Ê
Ë 7.5 ¯
377 377
Z3,0 = = =•
5 2 0
1- Ê ˆ
Ë 5¯
For TM1,1,
2
l0 =
2 2
Ê mˆ + Ê nˆ
Ë a ¯ Ë b¯
2
= = 6.7 cm
2 2
Ê 1 ˆ +Ê 1 ˆ
Ë 7.5 ¯ Ë 3.75 ¯
32 Kennedy’s Electronic Communication Systems
5 ˆ 2 = 0.666
r= 1 - ÊÁ
Ë 6.7 ˜¯
Z0 = 120p ¥ r
vc 3 ¥ 108
l= = m = 5 cm
f 6 ¥ 109
2
Ê lˆ
vg = vc 1- Á ˜
Ë l0 ¯
2 2
Ê lˆ Ê vg ˆ
1 – Á ˜ = Á ˜ = 0.81
Ë l0 ¯ Ë vc ¯
l
= 0.19
l0
l
l0 =
0.19
l 5
l0 = = = 11.4 cm
0.19 0.19
For dominant mode, m = 1
2a
l0 =
m
l0 11.4
a= = cm = 5.7 cm
2 2
377 377
Z0 = = = 418.89 W
Ê lˆ
2 0.9
1- Á ˜
Ë l0 ¯
(No. 4)
Given f = 12 GHz
Z0 = 450 W
Waveguides, Resonators and Components 33
3 ¥ 108
l= m
3 ¥ 109
1
= ¥ 10–1 m = 0.25 ¥ 10–1 m = 2.5 cm
4
377
Z0 =
2
Ê lˆ
1- Á ˜
Ë l0 ¯
2
Ê lˆ 377
1- Á ˜ =
Ë l0 ¯ 450
2
Ê lˆ
ÁË l ˜¯ = 0.298
0
2.5
l0 = = 4.578 cm
0.546
For TE1,0 mode, m = 1
2a
l0 = = 2a
m
l0 4.578
a= = = 2.289 cm
2 2
l
lp =
2
Ê lˆ
1- Á ˜
Ë l0 ¯
2.5
lp = = 2.98 cm
0.84
Given, distance = 30 cm
nlp = 30
30
n= = 10
2.98
vp = 12 ¥ 109 ¥ 2.98 ¥ 10–2 m/s = 3.57 ¥ 108 m/s
vc 2 9 ¥ 1016
vg = v = m/s = 2.52 ¥ 108 m/s
p 3.57 ¥ 108
d = vg ¥ t
30 ¥ 10-2
t= s = 11.9 ¥ 10–10 s
2.82 ¥ 108
34 Kennedy’s Electronic Communication Systems
(No. 5)
For WR 128, Frequency range = 2.6 – 3.95 GHz
a = 7.62 cm
Hence, l0 = 2a = 15.24 cm
Range of free space wavelength,
3 ¥ 108
l1 = m = 11.5 cm
2.6 ¥ 109
3 ¥ 108
l2 = m = 7.5 cm
3.9 ¥ 109
l1 11.5
lp1 = = = 17.5 cm
2
Êl ˆ 11.5 ˆ 2
1- Á 1 ˜ 1- Ê
Ë l0 ¯ Ë 15.24 ¯
l2
lp2 = = 8.61 cm
2
Êl ˆ
1- Á 2 ˜
Ë l0 ¯
3 ¥ 108
fp1 = Hz = 1.71 GHz
17.5 ¥ 10-2
3 ¥ 108
fp2 = Hz = 3.484 GHz
8.61 ¥ 10-2
Hence, frequency range = 1.71 – 3.484 GHz
(No. 6)
Given d = 5 cm, r = 2.5 cm
Cut off wavelengths
2p r
l0 (TE1,1) = = 8.53 cm
1.84
2p r
l0(TM0,1) = = 6.54 cm
2.4
2p r
l0(TE0,1) = = 4.09 cm
3.83
Cutoff frequencies,
3 ¥ 108
fc(TE1,1) = = 3.51 GHz
8.53 ¥ 10-2
Waveguides, Resonators and Components 35
3 ¥ 108
fc(TM0,1) = = 4.58 GHz
6.541 ¥ 10-2
3 ¥ 108
fc(TE0,1) = = 7.32 GHz
4.009 ¥ 10-2
(No. 6)
f = 4 GHz
AdB = 120 dB
3 ¥ 108
l= = 7.5 cm
4 ¥ 109
For WR28 waveleguide,
a = 7.62 cm
2a
l0 = = 2a (For dominant mode)
m
= 15.24 cm
Waveguide is not operated well below cutoff.
8
Hence, AdB = 54.5 ¥ does not hold value.
l0
For WR 128, 1.9 dB attenuation is caused at 100 m length of waveguide.
100
Hence, 1 dB attenuation Æ
1.9
100
Hence, 120 dB attenuation Æ ¥ 120 m = 6.315 km
1.9
13
Microwave Tubes and Circuits
No Review Problems
Semiconductor Microwave
14
Devices and Circuits
(No. 1)
Given rs = 500 W, rl = 500 W, r = 200 W, so
1 1
¥4¥
A= 500 500
1 1 Ê 1 ˆ Ê 1 1 ˆ
4¥ ¥ + ¥ -4¥
500 500 Ë 200 ¯ Ë 200 500 ¯
1.6 ¥ 10-5
=
1.6 ¥ 10-5 - 1.5 ¥ 10-5
A = 12 dB
(No. 2)
1
rs = 1000 W, gs =
1000
1
rL = 1000 W, gL =
1000
Let negative resistance be r; then negative conductive is g.
1 1
¥4¥
23 = 10 log 1000 1000
1 1 1 ˆ
4¥ ¥ + gÊg -4¥
1000 1000 Ë 1000 ¯
4 ¥ 10-6
1023 =
4 ¥ 10-6 + g ( g - 0.004)
g2 – 0.004 g + 4 ¥ 10–6 = 4 ¥ 10–29
g2 – 0.004g + 4 ¥ 10–6 = 0
vc 3 ¥ 108
Wavelength, l = = = 0.24 m
f 1.25 ¥ 109
1
Ê r 4 4pl 2 Pmin ˆ 2
Face area of the antenna, A0 = Á max ˜¯
Ë PS t
1
Ê (185.2 ¥ 103 ) 4 ¥ 4 ¥ p ¥ 0.242 ¥ 2 ¥ 10-13 ˆ 2 2
= Á ˜¯ = 7.53 m
Ë 3 ¥ 106 ¥ 1
(iv)
Operating frequency, f = 5 GHz
Peak power output, Pt = 1 ¥ 106 W
Antenna diameter, D = 3.66 m
Receiver bandwidth, Df = 1.6 ¥ 106 Hz
Noise figure = 11 dB
Target size, S = 1 m2
11
Noise figure (expressed as ratio), F = antilog = 12.59
10
vc 3 ¥ 108
Wavelength, l = = = 0.06 m
f 5 ¥ 109
Pt D 4 S 1 ¥ 106 ¥ 3.664 ¥ 1
Maximum detection range, rmax = 48 = 48 = 47.06 km
Df p 2 ( F - 1) 1.6 ¥ 106 p 2 (12.59 - 1)
(v)
Peak pulse power = 400 kW
Pulse reception frequency, PRF = 1500 pps
Pulse width, PW = 0.8 ms
Pulse reception time, PRT = 666.66 ms
PRT 666.66
Maximum unambiguous range, mur = = = 333.33 miles
2 2
vc 3 ¥ 108
Wavelength, l = = = 0.0375 m
f 8 ¥ 109
(vii)
Operating frequency = 10 GHz
Pulse reception frequency, PRF = 3000 pps
vc 3 ¥ 108
Wavelength, l = = = 0.03 m
f 10 ¥ 109
nl 3000 ¥ 1 ¥ 0.03
Lowest blind speed (corresponds to n = 1), vb = PRF = = 45 m/s
2 2
(vii)
Operating frequency = 3 GHz
Pulse reception frequency, PRF = 500 pps
vc 3 ¥ 108
Wavelength, l = = = 0.1 m
f 3 ¥ 109
nl 500 ¥ 1 ¥ 0.03
Lowest blind speed (corresponds to n = 1), vb = PRF = = 25 m/s
2 2
16
Broadband Communication
System
No Review Problems
Introduction to fiber optics
17
technology
(i) Under the worst case scenario, the ratio of spacing between the repeaters of optic fiber to copper fiber is
35:1.
0.35
(ii) The system bandwidth, BW =
tr
Information theory, coding
18
and data communication