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