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Web answers for Principles of
Communication Engineering
Chapter 2: Amplitude Modulation
Example 8
An amplitude-modulated wave has a power content of 800W at its carrier frequency. Calculate
the power content of each of the sidebands for 80% modulation.
Solution:
Pc = 800W
m = .8
2
1284
c
LSB USB
m PP P W
Example 10
A carrier wave Vc = A sin c t is amplitude modulated by an audio wave
a a a a
B BV Bsinω t sin3ω t sin5ω t
3 5
Determine the upper and lower bands. Sketch the complete spectrum of the modulated wave.
Estimate the total power in the side bands in terms of the carrier power if the modulated index
B/A = 0.6.
Solution:
After modulated, the frequencies will be c a, c 3a, and c 5a, respectively. The frequency
spectrum is shown below.
2
MA
6
MA
10
MA
c
- 5
a
c
- 3
a
c
-
a
c
a
c
+
a
c
+ 3
a
c
+ 5
a L
A
Total power in side bands,
2 2 2
1 2 3
2 2 2
C C C
SBT
P m P m P mP
m1 = m = 0.6
2
0.60.2
3 3
mm
and 3
0.60.12
5 5
mm
2 2 2
SBT 1 2 3
C
P
P 2
m m m
0.36 0.04 0.0144
2
= 0.2072
% Power = 0.2072 100 = 20.72%
Example 17
Determine the percent increase in the effective value of the antenna current when the
modulation of a transmitter is increased to 65% from 0%.
Solution:
We know
22
12
t
c
I m
I
Where It = modulated current (or total)
Ic = unmodulated current
m = modulation index
2(0.65) 0.42251 1 1.2112
2 2
1.2112 1.1t
c
I
I or It = 1.1 Ic
So increase in antenna current = (It – Ic)
= 1.1 Ic – Ic = 0.1 Ic
Example 19
The ac rms antenna current of an AM transmitter is 6.2 A when unmodulated and rises to 6.7 A
when modulated. Calculate % age modulation.
Solution:
We know
22
12
t
c
I m
I
or
22
12
t
c
Im
I
where It = modulated current = 6.7
Ic = unmodulated current = 6.2
12.6
7.6
2
22
m
]12[(1.08) 2 m
1) - 2[(1.16
%5656.00.32
Example 22
A certain transmitter radiates 9 kW with the carrier unmodulated, and 10.125 kW when the
carrier is sinusoidally modulated. Calculate the modulation index. If another sine wave,
corresponding to 40% modulation, is transmitter
Simultaneously, determine the total radiated power.
Solution:
1st part:
Here, Pt = Total Power (with carrier modulated) 10.125 kW
Pc = Power with carrier unmodulated = 9 kW
From equation, Pt = Pc
21
2m
2 10.1251 1 1.125 -1 0.125
2 9
t
c
Pm
P
m2 = 2 0.125
m = m1 = 50.0125.02
2nd
Part:
40% modulation means m2 = 0.4
mt = 2
2
2
1 mm
So, (Total unmodulation indes)
64.041.016.025.0)4.0()5.0( 2 t
m
2
12
tc
mP Pt
2
)64.0(1109
23
= 9 103 (1 + 0.205)
=10.84 kW.
Example 24
An AM broadcast station operates at its maximum allowable output of 50 kW and at 90%
modulation. How much of its transmitted power is intelligence?
Solution:
Maximum allowable output, Pt = 50 kW
m = 90% = 0.9
We know
2
12
t c
mP P
50 kW =
2
)9.0(1
2
cP
405.01
1050P
3
c
c P 35.58 kW
This will be the intelligence power transmitted.
Example 30
When the modulation percentage is 75, an AM transmitter produces 10 kW (a) How much of
this is carrier power. (b) What would be the percentage power saving if the carrier and one of
the sidebands were suppressed before transmission took place?
Solution:
(a)
2
t c
mP P 1
2
2
c
(0.75)10 P 1
2
cP 1.28
c
10P 7.81kW
1.28
(b) We know,
2 2
t c c c
m mP P P P
4 4 for two side bands
where power due to carrier and one of the sidebands
2
c c
mP P
4
2
c
mP 1
4
4
75.0181.7
2
14.181.7
= 8.9 kW
power suppressed, i.e., saved as per condition.
= 89% [Since, 10 kW considered 100%, so 8.9 kW in 89%]
Example 31
Refer to the following figure. Determine the effect of a phase error in the local oscillator on
synchronous DSB demodulation.
Solution:
Let be the phase error of the local oscillator in the given figure. Then the local carrier is expressed
as ).cos( tc Now
ttmtx cDSB cos)()(
and )cos(]cos)([)( tttmtd cc
)]2(cos)[cos(2
1 ttm c
)2cos()(2
1cos)(
2
1 ttmtm c
The second term on the R. H. S. is filtered out by the low-pass filter, and we get
cos)(2
1)( tmty
When is constant, this output is proportional to m(t). The output is completely lost when
2 . Thus, as long as is constant and not equal to 2 , the phase error in the local carrier
causes attenuation of the output signal without any distortion. If the phase error varies randomly
with time, then the output also will vary randomly and is undesirable.
Example 32
When a broadcast AM transmitter is 50% modulated, its current is 12 A. What will be the
current when the modulation depth is increased to 0.9?
Solution:
We know It = Ic ,2
12m
here I = 12 A
m = 0.5
Ic = ?
2
5.01I 12
2
c
LPF
xDSB(t) d(t) y(t)
cosct ~
Fig: Synchronous demodulator
06.1125.1I 12 c cI
32.1106.1
12 c I
c tNext, m 0.9, I 11.32, I ?
2
)9.0(132.11
2
tI
= A35.1318.132.11
Example 36
Take the example of an analog baseband communication system with additive white noise. The
transmission channel is assumed to be distortionless and the power spectral density of white
noise /2 is 10–9
watt per hertz (W/Hz). The signal to be transmitted is an audio signal with 4
kHz bandwidth. At the receiver end, an RC low-pass filter with a 3dB bandwidth of 8 kHz is
used to limit the noise power at the output. Determine the output noise power.
Solution:
The frequency response H() of an RC low-pass filter with a 3 db bandwidth of 8 kHz is given by
0/1
1
jH
where 800020 . Then the output noise power N0 is
d
dHtnEN
2
0
22
00
/1
1
2
1
2
22
1
9 3
0
1 12 10 2 8 10
4 4
25.2 W
W
Chapter 3: Angle Modulation
Example 6
A modulating signal 35cos 2 (15 10 )t , angle modulates a carrier tA ccos .
a) Find the modulation index and the bandwidth for FM systems.
b) Determine change in the bandwidth and modulation index for FM, if fm is reduced to 5KHz.
Solution:
a) Here 5 ; 15KHzm mE V f
So frequency deviation
KHz75
51015
Ekf
3
mf
515
75mSo f
Hence bandwidth
2( ) 2(75 15) 2 90 180KHzmf f
b) 155
75fm
Therefore, Bandwidth
2( ) 2(75 5) 160mf f KHz
Example 22
A FM wave is represented by voltage equation v = 20 sin (5 108t + 4 sin 1500 t). Find the
carrier, modulating frequency modulation index, and maximum deviation of the FM. What
power will this FM voltage dissipate in a 20 resistor?
Solution:
A = 20 (Emax), c = 5 108, mf = 4
MHz. 6.7928.6
105
2
105 88
c
f
m = 1500
Hz. 8.23828.6
1500
2
1500
mf
mf
fm f
f = mf fm = 4 238.8 = 955 Hz
Power W10220
2020
20
)2/20(
R
E 22
rms
Example 27
A FM signal has deviation of 3 kHz and modulating frequency of 1 kHz. Its total power is
PT = 5W developed across a 50 resistor. The carrier frequency is 160 MHz.
(i) Calculate RMS signal voltage VT.
(ii) If J 0 = - 0.26, J 1 = 0.34, J 2 = 0.49 Find the amplitude of carrier and first two side bands.
(iii) Find power in side bands
(iv) Plot the frequency spectrum.
Solution:
Df = 3 kHz, PT = 5 W, R = 50 , fm = 1 KHz,
FC = 160MHz
3
3
3 103
1 10
f
m
D
f
We have, )] )(22
)(21
2( )(20
[ JJJC
PT
P
)] 49.02(0.34 )26.0[(5 222 C
P
5 = PC [0.0676 + 0.1156 + 0.2401]
5 = PC (0.779)
W418.6
779.0
5CP
We have,
R
CE
CP
2
50418.6 RPE CC
EC = 17.91 V = Amplitude of Carrier
The amplitudes of sidebands can be determined by
En = Jn() EC
= 0.26 + 17.91 = 4.65 V
E1 = J1() EC
= 0.34 17.91 = 6.08 V
E2 = J2 () EC
= 0.49 17.91 = 8.77 V
Power in side bands is given by,
n
2P E /n
R
WREP 43.050/)65.4(/ 22
00
WREP 739.050/)08.6(/2 22
11
WREP 53.150/)77.8(/ 22
22
505; 22 TTT VRPV
V 81.15250 TV
Example 29
In an Armstrong FM transmitter with a carrier frequency of 152 MHz and a maximum
deviation of 15 kHz at a minimum audio frequency of 100 Hz, the primary oscillator is a
100 kHz crystal, oscillator and the initial phase modulation deviation is to be kept less than
12º in order to minimize distortion.
Estimate the frequency multiplication to give the required deviation and specify the required
mixer crystal and multiplier stages.
Solution:
Here, the maximum phase deviation of the modulator is
0
P 12 12 rad 0.2094 mmax 180
Now, f max = max f min = 0.2094 100
= 20.94 Hz = 0.02094 kHz
So, the required frequency deviation is,
(allowed)Maximum frequency deviation allowed max
Maximum frequency deviationmax
f
f
3
3
15 10716.33
0.02094 10
Also 729 = 36, this requires a chain of six tripler stages to give a total deviation of
0.02094 729 = 15.265 KHz at a carrier frequency of 72.9 MHz.
The mixer oscillator signal is
f0 = (152- 72.9) = 79.1 MHz
Here f0 can be contained by employing two tripler stages from a 8.7889 MHz crystal oscillator.
Example 30
The following is a block diagram of an indirect (Armstrong) FM transmitter. Calculate the
maximum frequency deviation f of the output of the FM transmitter and the carrier frequency
fc if f1 =200 KHz, fLO = 10.8 MHz, f1 = 25 Hz, n1 = 64, and n2 = 48.
Solution:
1 1 2( )( )( ) (25)(64)(48) Hz 76.8KHZf f n n
MHz12.8Hz)10(8.12)10)(200)(64( 63
112 fnf
MHz2.0
MHz23.6 Hz)10)(8.108.12( 6
23 LOfff
Thus, if f3 = 23.6 MHz, then
8.1132)6.23)(48(32 fnf c MHz
If f3 = 2 MHz, then
96)2)(48(32 fnf c MHz.
Example 32
The crystal oscillator frequency in an Armstrong-type FM generator of the following figure is
200 KHz. To avoid distortion, the maximum phase deviation is limited to 0.2. fm ranges from
50 Hz to 15 KHz. The carrier frequency at the output is 108 MHz, and the maximum frequency
deviation is 75 KHz. Select multiplier and mixer oscillator frequencies.
Solution:
Referring to the given figure, we have
1 (0.2) (50) 10mf f Hz
3
1 2
1
75 107500
10
fn n
f
~
NBFM Frequency
Multiplier Frequency
Multiplier
f2 = n1f1
m(t)
f1
f1 f
n2
f3
n1
xc(t)
fLO
fc
~
NBFM Frequency
Multiplier Frequency
Multiplier
f2 = n1f1
m(t)
f1
f1 f
n2
f3
n1
xc(t)
fLO
fc
5
2 1 1 1(2 10 )f n f n Hz
Let us assume down conversion. Then, we have
2
2n
fff c
LO
Thus,
5 6
1 1
2 2
7500(2 10 ) 108 10cLO
ff n f
n n
6
2
139210
n Hz
Let n2 = 150. Then, we get
n1 = 50 and fLO = 9.28 MHz.
Example 35
Determine the resonant frequency of series combination of a 0.001 F capacitor and 2 mH
inductor.
Solution:
The resonant frequency is given by
LCf r
2
1
36 10210001.02
1
rf
= 112.53 103 Hz
= 112.53 KHz.
Chapter 4: A/D & D/A Converter
Example 5
Compare the maximum conversion periods of an 8-bit digital ramp ADC and an 8-bit successive
approximation ADC if both utilize a 1 MHz clock frequency.
Solution:
For the digital-ramp converter, the maximum conversion time is
(2N 1) (1 clock cycle) = 255 1 s = 255 s
For an 8-bit successive-approximation converter, the conversion time is always 8 clock periods,
i.e., 8 1 s = 8 s.
Thus, the successive-approximation conversion is about 30 times faster than the digital-ramp
conversion.
Chapter 6: Pulse Modulation
Question 2: What are the disadvantages for digital communication?
Answer:
Disadvantages
1. Digital transmission system are incompatible with all practical analog system.
2. Digital transmission requires Synchronization.
3. Digital Systems are expensive.
Question 7: Compare TDM and FDM systems.
Answer:
1) TDM instrumentation is somewhat simpler than FDM. FDM requires modulators, filters and
demodulators. TDM however needs commutators and distributor only.
2) TDM is invulnerable to the usual sources of FDM inter channel cross talk. In TDM virtually no
cross talk occur when pulse are completely isolated and are non-overlapping.
3) The bandwidth required for TDM system, for multiplexing n signals, each band limited to fm Hz is
nfm and if modulated by a carrier it becomes 2n fm. Now if the ‘n’ signals are multiplexed in
PDM, using (SSB) technique, the bandwidth is nfm.
4) Because of the advantages of TDM over FDM, TDM systems are commonly used in long distance
telephone communication.
Question 9: Calculate SNR for PCM systems. Explain with an example.
Answer:
Pulse-code modulation (PCM) is a digital representation of an analog signal where the magnitude of
the signal is sampled regularly at uniform intervals, then quantized to a series of symbols in a numeric
(usually binary) code. PCM has been used in digital telephone systems and 1980s-era electronic
musical keyboards. It is also the standard form for digital audio in computers and the compact disc
format.
There are two major sources of noise in a PCM system
(I) Transmission noise introduced outside the transmitter
(II) Quantization noise introduced in the transmitter.
The quantized staircase waveform is an approximation to the original waveform. The difference
between the two waveforms amounts to ‘noise’ added to the signal by the quantizing circuit. The
mean square quantization noise voltage has a value of
12
22 S
Enq
Where S is the voltage of each step or the sub range voltage span.
As a result, the number of quantization level must be kept high in order to keep the quantization noise
below some acceptable limit, given by power signal-to-noise ratio, which is the ratio of average signal
power to average noise power for a sinusoidal signal which occupies the full range, the mean square
signal voltages is,
8
(MS)
2
MS
2
1E
2
1E
22
2
peak
2
S
where M = Number of steps
S = Step height voltage
The signal-to-noise ratio is now given by,
2 22
2 2
Signal ( ) 12 3
Noise 8 2
S
nq
E MSM
E S
The number of levels M is related to the number n of bits per level by, M = 2n.
2 2Signal 3 3(2 ) 2
Noise 2 2
n nS
N
n
dbN
Sor 22
2
3log10
ndb02.6761.1
In PCM SNR can be controlled by transmission band width. The same feature is also observed for FM
or PM. But that case it require double of band width to quadruple the SNR So PCM compare to PM or
FM.
0
2
4
6
8
1
3
5
7 m(t)
m7(t)
Example 9
Suppose that a binary channel with bit rate Rb = 36000 bps is available for PCM voice
transmission. What will be the appropriate values of the sampling rate fs, the quantizing level L,
and the binary digits n, assuming fM = 3.2 KHz.
Solution:
It is required that
6400f2f ms and 36000Rnf bs .
So,
6.56400
36000
f
Rn
s
b
Therefore,
72005
36000fs Hz = 7.2 KHz
Example 10
Consider an analog signal to be quantized and transmitted by using a PCM system. When each
system at the receiving end of the system must be known to within 0.5% of the peak-to-peak
full-scale value, calculate the number of binary digits that each sample must contain.
Solution:
Let pm2 be the peak-to-peak value of the signal. The peak error is then .m01.0)m2(005.0 pp
The peak-to-peak error is pp m02.0)m01.0(2 (the maximum step size ).
Therefore, the required number of quantizing levels is
n
p
pp
m
mmL 2100
02.0
22
Thus, the required number of binary digits is .7n
Example 15
Calculate the roll-off factor for a communication channel of bandwidth 75 KHz which will be
required to transmit binary data at a rate of 0.1 Mb/s using raised-cosine pulses.
Solution:
5
6
110
0.1 10bT
s
375 kHz 75 10Bf Hz
Then, we have
5.1)10)(10)(75(221 53
bBTf
Hence we get the roll-off factor as
= 0.5
Chapter 7: Digital Modulation
Question 4: Draw the diagram of Integrate and Dump Receiver.
Answer:
Integrate and Dump Receiver
The circuit diagram of Integrate and Dump filter is shown in figure.
The input to the integrator and output is shown below:
Question 5: What is Matched filter?
Answer:
Matched filter
Matched filter is a type of filter in which after a time delay, the transfer function of the optimum filter
is the same as the complex conjugate of the spectrum of the input signal. The Block diagram of a
matched filter is shown in figure.
+
+ White noise, n(t)
+
x(t)
Integrator
Fig: Integrate and dump type filter
x(t) y(t) Matched filter Decision
device
Fig: Matched filter Receiver
Threshold
Integrator output
v(t)
0 t
Integrator input variations
A
t
x(t)
The figure clearly depicts that it consists of a filter matched to input signal, a sampler and a decision-
device. At time t = T, matched filter output is sampled and amplitude of this sample is compared with
a present threshold . If threshold is exceeded, receiver decides that known signal S(t) is present,
otherwise as shown it will decide that it is absent.
Question 10: Draw the block diagram of Band Pass Binary delta transmission scheme.
Answer:
Band Pass Binary delta transmission scheme
Block diagram of a band bass being data transmission scheme using digital modulation is shown
below.
Transmitter
Channel
Clock pulse
Modulator
Clock pulse
Demodulator
Binary data output
Local carrier
+
+
Node
Binary data
Fig: Band pass binary data transmission system
Chapter 8: Information Theory & Coding
Question 6: Prove that maximum entropy is maximum when all messages are equally likely?
Also calculate the average information per message.
Answer:
Let us consider a memoriless source m emitting messages m1, m2, ……, mn with probabilities P1, P2,
……, Pn respectively.
Conditions, 1P......PP n21
We know that,
n
1iiiIPH
We need to consider only terms –Pi log Pi and Pn log Pn [because Pn is a function of Pi]
i
n
n
n
ni
i
i
nnii
ii
P
Plog
PlogelogP
1PPlogelog
P
1P
PlogPPlogPdP
d
dP
dHSo
when Pn = Pi then 0dP
dH
i
Because this is true for all i so
1 2
1...... (1)nP P P
n
Similarly,
.1
......212
2
nPPPforve
dP
Hdn
i
So maximum entropy occurs when messages are equally likely.
When P1 = 1 and P2 = P3 = … Pn = 0, H(m) = 0, whereas the probabilities in equation (1) gives,
Average information per message = log n.
Example 4
A source generates one of four possible messages during each interval with probabilities
.
8
1PP,
4
1P,
2
1P 4321
Find the information content of each of these messages.
Solution:
We know that,
iP
I1
log
1
2
Example 7
A transmission channel has a bandwidth of 4KHs and signal to noise power ratio of 31.
a) How much should the bandwidth be in order to have the same channel capacity if N
S ratio is
reduced to 15.
b) What will be the signal power if bandwidth is reduced to 3KHz?
Solution:
a) Channel capacity,
.sec/bitsK20
311log104
sec/bitsN
S1logBC
23
2
when
N
S ratio is reduced from 31 to 15 then we can write,
151logB1020 23
4B1020,or 3
Therefore, Bandwidth, B = 5KHz
b) In the second case,
B = 3KHz
So,
So, 3
20
N
S1log 2
So, 4.90N
S
So, required 4.90N
S
Example 10
An ideal receiver receives information from a channel with bandwidth B Hz. Assuming the
message to be band limited to fm Hz, compare
N
S ratio at receiver output to that at its input.
Solution:
According to Shannon Hartley’s law, channel capacity at receiver input is
2log 1 bits / secin
in
SC B
N
Similarly, channel capacity at receiver output is,
2log 1 bits / seco
o
SC B
N
Since at receiver output, B = fm then
2log 1 bits / secom
o
SC f
N
According to the problem, there is no information loss in the receiver.
So,
o
o
m
in
in
N
Sf
N
SB 1log1log 22
o
of
B
in
in
N
S
N
Sor
m
11,
Since
in
in
N
S and
o
o
N
S are very large compared to unity
So we can write,
mfB
in
in
o
o
N
S
N
S/
Example 16
Consider the parity-check code having the following parity check matrix
100110
010011
001101
H
(i) Find the generator matrix G.
(ii) Find the code word that begins 101….
(iii) Let that the received word be 110110. Decode this received word.
Solution:
(i) Since H is a 63 matrix, n = 6 and k = 3.
So,
101
110
011TP
Then the generator matrix G is
101100
110010
011001
][ 3
TPIG
(ii) c = dG =
101100
110010
011001
101
= 110101
(iii) 011011r
100
010
001
101
110
011
011011TrHs 010
As s is equal to the second row of HT, an error is at the second bit, the correct code word is 100110
and the data bits are 100.
Example 17
Show that if Ci and Cj are two code vectors in an (n, K) linear Block code then their sum is also
a code vector.
Solution:
We know that for any code,
CHT = 0
So we can write, 0HCand0HC Tj
Ti
So,
So, ji CC is also a code vector.
Example 22
A signal amplitude X is a randomly variable uniformly distributed in the range (-1,1). This
signal is passed through an amplifier of gain 2. The output y is also a random variable,
uniformly distributed in the range (-2, 2). Determine the entropies H(x) and H(y).
Solution:
We have 1
12
P x x
= 0 otherwise
1
12
P x x
= 0 otherwise
Hence,
1
1
1log 2 1bit
2H x dx
2
2
1log 4 2bits
4H y dx
Chapter 9: Satellite Communication
Cross-talk
All communication channels are of limited bandwidths. Hence a communication channel can be
represented by an RC low pass filter as shown in Figure (a) below whose upper cut-off frequency is
RCfC
2
1
Now, when a pulse is applied to this channel, the output of the channel will be distorted as shown in
Figure (b). This is due to the HF limitations of the channel. It can be seen that the signal pulse allotted
to the time slot 1 extends into the time slot 2, resulting in cross-talk. On an average, we can assume
that the multiplexed signals are equally strong. Hence during the time slot 2, the area corresponding to
the signal, i.e., A2 (not shown in the figure) will be the same as A1. We can define the cross-talk
factor, K, as the ratio of the cross talk signal to the desired signal. Thus
1
12
2
12
A
A
A
AK … (1)
The pulse in time slot 1 is almost rectangular (the figure here is exaggerated for clarity). Hence,
VA1 … (2)
In the above Fig. Cross Talk due to HF Cut-off of Channel; (a) Low pass Filter Representation of the
Channel (b) Output of Channel of (a) for a pulse Input
Now,
/ /
12 (1 )c
g c cA V e e
… (3)
Fig: (b)
V
g Guard time
T.S.1 T.S.2
Time constant
A1
A12 c = RC
V
t
R
C
Fig: (a)
The time constant C must be much smaller than in order to have minimum cross talk. Therefore,
c , and Equation.3 becomes
/
12 c
g cA V e
… (4)
From Equation (1), (2) and (4), we get
/g ccK e
… (5)
The cross-talk factor K should be as low as possible. Equation 5 suggests that g should be much
larger than c . Of course, there are other considerations as well. For example, a large g will result in
less number of channels that can be multiplexed, and/or reduced signal strength.
This type of cross talk is restricted to the neighboring channel because c is very small, and hence the
pulse ends in the neighboring time slot only.
Question 9: What is ISL (inter satellite link)?
Answer:
Occasionally, there is an application where it is necessary to communicate between satellites. This is
done using satellite cross-links or inters satellite links (ISLs).
A disadvantage of using an ISL is that both the transmitter and receiver are space bound.
Consequently, both the transmitter’s output power and the receiver’s input sensitivity are limited.
Sat
1
Sat
2
Cross-link
Up-link/down-link Up-link/down-link
ES 1
ES 2 Earth
Fig: Inter Satellite link (ISL)
Question 15: Define antenna pattern and study how the coverage or footprint depends on
beamwidth?
Answer:
An antenna pattern is a plot of the field strength in the far field of the antenna when the antenna is
driven by a transmitter.
3 dB beam width of the antenna is given by,
3dB
75degrees
D
Thus from then we see that the coverage or footprint depends on beam width. A table below shows
the dependence.
Beam width () Coverage dia (D)
10° 3821 miles
5.7° 2235 miles
2.8° 1117 miles
1.0° 382 miles
0.7° 223 miles
Example 6
A satellite moving in an elliptical eccentric orbit has the semi major axis of the orbit equal to
16000 Km (Fig. Shown below). If the difference between the apogee and the perigee is 30100
Km, determine the orbit eccentricity.
Solution:
Apogee = a (1 + e)
Perigee = a (1 – e)
Where, a = semi-major axis of the ellipse
e = orbit eccentricity
Apogee – Perigee = a(1 + e) – a(1 – e)
= 2ae
or, Eccentricity, Apogee-Perigee
2ae
160002
30100
32000
30100
9406.0
Orbit
Earth
a
b
Example 7
The farthest and the closest points in a satellite’s elliptical eccentric orbit from earth’s surface
are 30,000 Km and 200 Km respectively. Determine the apogee, the perigee and the orbit
eccentricity. Assume radius of earth to be 6370 Km.
Solution:
Apogee = 30000 + 6370 = 36370 Km
Perigee = 200 + 6370 = 6570 Km
Eccentricity a
PerigeeApogee
2
Where a = semi-major axis of the elliptical orbit
Also, 2
PerigeeApogeea
PerigeeApogeea2,or
Therefore, orbit eccentricity
PerigeeApogee
PerigeeApogee
657036370
657036370
693.042940
29800
Example 8
Refer to Fig. showing a satellite moving in an elliptical, eccentric orbit. Determine the apogee
and perigee distances if the orbit eccentricity is 0.5.
Solution:
The distance from center of ellipse (o) to the center of earth (c) is given by (a × e) where (a) is the
semi-major axis and (e) is the eccentricity.
15000 km
Earth
C O
S
Therefore, a × e = 15000
Kma 30000
5.0
15000
Now apogee
(1 )a e 30000(1 0.5) Km45000
Perigee
(1 )a e 30000(1 0.5) 5.030000 Km15000
Example 14
The semi-major axis and the semi-minor axis of an elliptical satellite orbit are
20,000 Km and 16000 Km respectively. Determine the apogee and perigee distances.
Solution:
If (ra)and (rp) are the apogee and perigee distances respectively, then semi-major axis 2
pa rr
Semi-minor axis parr
Kmrr pa
200002
Kmrr pa 40000
16000 pa rr
256000000 pa rr
Now a pr r 40000 ......(1)
)2......(256000000 pa rr
Substituting the value of (rp) from (2) in (1)
(40000 ) 256000000a ar r
or, 0256000000400002
aa rr
2
1024.10101640000 88 ar
2
1076.540000 8
2
104.240000 4
44 106.1,102.3
32000Km, 16000Km
ra = 32000 Km as it cannot be 16000 Km if the semi-major axis is 20,000 Km.
Kmrp 80003200040000
Example 23
What would be the new maximum coverage angle and the slant range if the minimum possible
elevation angle is 5 and not zero as in Example – 22.
Solution:
The maximum coverage angle, (2 max) is given by
mine
e1max Ecos
HR
Rsin22
where Emin = Minimum elevation angle
1
max
1
1
63782 2sin cos5
6378 35786
2sin [0.1512 0.996]
2sin 0.1506
2 8.66
17.32
Example 24
The Fig. (Showing below) a geostationary satellite orbiting earth. Calculate the angle ()
subtended by the arc of the satellite’s footprint at the center of the earth.
Solution:
1 2
1 1 1 190 ( 5 )E E
2 2 2 290 ( 0 )E E
2max
5 5
Therefore,
1 1 2 2
1 2 1 2
1 2
1
1
1
1
2
(90 ) (90 )
180 ( ) ( )
175 ( )
6378sin cos5
6378 35786
6378cos5sin
42164
8.66
6378sin cos 0
6378 35786
8.69
E E
E E
Therefore, 175 (8.66 8.69 ) 157.65 .
Example 28
A satellite is currently in its elliptical transfer orbit (Fig.) with apogee and perigee being at
distances of 35786 Km and 300 Km respectively above the surface of earth. If the transfer orbit
inclination to the equatorial plane is 0, the incremental velocity to be given to the satellite at the
apogee point by the apogee kick motor to circularize the orbit (Assume earth’s radius = 6378
Km).
Solution:
The velocity (V) at any point along an elliptical orbit is given by:
a
1
r
2V
35786 km
6378 km
90
2
1
B
A
Circular
path
Transfer
orbit
Equator
Perigee
300 km
35786
km
Where = GM
r = apogee distance from center of earth
a = semi – major axis of ellipse
Now r = 35786 + 6378 = 42164 Km
Km24421
2
3006378637835786a
The velocity (Va) at the apogee point can then be computed from
s/Km61.11058.2V
1058.2V
102442142164
6678108.39
1024421
110
42164
2108.39V
a
1
r
2V
6a
62a
3
13
313132a
a
For a circular orbit with a radius of 42164 Km,
s/Km07.3
1042164
108.39
rV
3
13
c
The velocity change (V) required to circularize is given by
icosVV2VVV ca
2c
2a
Here i = 0°
Therefore,
caca VVVVV 222
Example 30
Determine the maximum line of sight distance between two communication satellites moving in
a circular orbit at a height (H) above the surface of earth.
Solution:
Maximum line of sight distance between the two satellites would occur when the satellites are so
placed that the line joining the two becomes tangent to the earth’s surface as shown in Fig.
Maximum line of sight distance
= AB = OA + OB = 2 × OA or 2 × OB
as OA = OB
If (R) us the radius of earth and (H) the height of the orbit, then
sinHR
sinACOA
Now,
HR
R1cos
Therefore,
HR
RHROA 1cossin
Maximum line of distance
HR
RHR 1cossin2
Example 31
If the two satellites in Example – 30 have an orbital radius that is twice the radius of earth,
determine the maximum line of sight distance. Repeat the Example for geostationary satellites
having an orbital radius of 42164 Km. (Assume earth’s radius, R = 6370 Km).
Solution:
Maximum line of sight distance
HR
RcossinHR2 1
Orbit A
B
O
Earth
C
It is given that (R + H) = 2R
Therefore, maximum line of sight distance
12 2 sin cos 0.5
4 sin 60
4 3
2
2 3
2 3 6370
22066 Km
R
R
R
R
For geostationary satellite, (R + H) = 42164 Km
Therefore, maximum line of sight distance
1
1
63702 42164sin cos
42164
84328sin(cos 81.3)
84328 0.998
83361Km
Example 43
Fig. shows a cascaded arrangement of three gain blocks. It is given that G1 = 106 and its
associated equivalent noise temperature Te1 = 100 deg Kelvin, G2 = 104 and Te2 = 60 deg K,
G3 = 1000 and Te3 = 20 deg K. Determine the equivalent noise temperature of the cascaded
arrangement.
Solution:
If (Te) is the equivalent noise temperature of the cascaded arrangement,
Then
K100
10201060100
1010
20
10
60100
GG
T
G
TTT
106
466
21
3e
1
2e1ee
G1.Te1
G1.Te2
G1.Te3
Te
Example 44
Fig. Shows the cascaded arrangement of four gain blocks with their gain and noise figures as
G1 = 100, F1 = 2, G2 = 10, F2 = 6 , G3 = 10, F3 = 15, G4 = 10, F4 = 20. Determine the noise figure of
the cascaded arrangement.
Solution:
The overall noise figure (F) is given by
32 41
1 1 2 1 2 3
11 1FF FF F
G G G G G G
6 1 15 1 20 12
100 100 10 100 10 10
2 0.05 0.014 0.0019 2.0659dB
Example 45
Fig. shows the receive side of satellite earth station comprising of earth station antenna followed
by wave guide that connects the antenna feed point to the low noise amplifier input with the
output of the low noise amplifier feeding the down-converter. Assume that the receive antenna
has a gain of 66 dB at the received down link frequency of 11.9 GHz. The other parameters
characterizing the receive chain are:
Antenna noise temperature, TA = 60 K
Loss in the wave guide, L1 = 1.075 (0.3 dB)
Equivalent noise temperature of low noise amplifier Te2 = 160 K
Low noise amplifier gain, G2 = 106
Down converter equivalent noise temperature, Te3 = 10000 K
Ambient temperature, T0 = 290 K
Determine the earth station system noise temperature and (G/T) referred to the input of the low
noise amplifier.
Solution:
Earth station system noise temperature referred to the input of the low noise amplifier is given by:
es TTT
where Ts = Noise temperature measured at the output of the wave guide
Te = Equivalent noise temperature at the input of low noise amplifier
G1=100 F1=2
G2=10 F2= 6
G3=10 F3=15
G4=10 F4=20
Wave guide
(L1, Te1)
Low noise
Amplifier
(G2, Te2)
Down
Converter
(Te3)
TA
Now
1
01
1
1
L
TL
L
TT A
s
K04.76
23.2081.55
290075.1
1075.1
075.1
60
01.160
10
10000160
G
TTT
6
2
3e2ee
dB
G
K
T
7.65
3.066
05.236
01.16004.76
./97.41
73.237.65
05.236log107.65
log10
KdB
TdBinGT
G
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