142302 – PRINCIPLES OF COMMUNICATION
DEPARTMENT OF ECE
St JOSEPH’S COLLEGE OF ENGINEERING
DEPARTMENT OF INFORMATION TECHNOLOGY
SUBJECT : PRINCIPLES OF COMMUNICATION
CODE: 142302
UNIT I : FUNDAMENTALS OF ANALOG COMMUNICATION
1. What is the need for modulation?
 It is extremely difficult to radiate low frequency signals through earth’s
atmosphere in the form of electromagnetic energy.
 At low frequency, the antenna size required becomes impractical.
 Information signals often occupy the same frequency band. Signals from two
or more sources would interfere if they are not modulated and translated to a
different frequency band.
2. With reference to AM define modulation index (or) depth of modulation.
E
m m
Ec
E c = peak amplitude of the unmodulated carrier voltage
E m = peak amplitude of modulating signal voltage
3. A broadcast radio transmitter radiates 5 kW power when the modulation
percentage is 60%. How much is the carrier power?
 m2 

Pt  Pc 1 
2 

Pc 
Pt
 m
1 
2

2




5000
 0.6 2 
1 

2 

 4237.28W
4. What is the relationship between total power in AM wave and unmodulated
carrier power?
 m2 

Pt  Pc 1 
2 

Pc  un mod ulated  carrier  power
pt  total  power
m  mod ulation  index
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DEPARTMENT OF ECEPage 1
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5. What is the relationship between total current in AM wave and unmodulated
carrier current?
 m2 

I t  I c 1 
2 

I t  total  current
I c  un mod ulated  carrier  cuurent
m  mod ulation  index
6. An unmodulated carrier is modulated simultaneously by three modulating
signals with coefficients of modulation m1 = 0.2, m2 = 0.4, m3 = 0.5. Determine
the total coefficient of modulation.
mt  m12  m22  m32  0.2 2  0.4 2  0.52  0.67
7. Define selectivity of a radio receiver.
It is the ability of the receiver to accept a given band of frequencies and
reject all others.
11. Define sensitivity of a radio receiver.
 It is the ability of the receiver to pick up weak signals and amplify them.
 It is defined in terms of the voltage that must be applied to the receiver
input terminals to give the standard output power. It is usually expressed
in micro volts.
12.Define the term fidelity of a radio receiver.
Fidelity is a measure of the ability of a receiver to reproduce an exact replica
of the original source information.
i.e., fidelity is the ability of a receiver to reproduce faithfully all frequency
components present in the base band signal
8. Define band width improvement factor and noise figure improvement.
Band width improvement factor ( BI ) is the ratio of RF bandwidth and IF
bandwidth.
RF  BandWidth
BI 
IF  BandWidth
Noise figure improvement = 10 log ( BI )
9. Define Noise figure
Noise figure is a parameter commonly used to indicate the quality of a receiver.
Lower the Noise figure value, better is the performance.
Noise figure = 10 log ( F )
Where, F = noise factor
Noise factor is the ratio of input SNR and output SNR.
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Noise  Factor 
DEPARTMENT OF ECE
SNRinput
SNRoutput
SNR ( signal to noise ratio ) is the ratio of signal power to noise power.
Signal  power
SNR 
Noise  power
10. Define Equivalent Noise Temperature
Equivalent noise temperature ( Te ) is a parameter used to indicate the
quality of a receiver. Lower the value of Te , better the quality.
Te = T ( F – 1 )
T = Temperature in Kelvin
F = Noise factor
16. What is Amplitude modulaltion?
Amplitude Modulation is the process of changing the amplitude of a
relatively high frequency carrier signal in proportion with the instantaneous value
of the modulating signal.
17. What is Frequency modulation?
Frequency of carrier is varied in accordance with amplitude of modulating
signal.
18.What is Phase modulation?
Phase of carrier is varied in accordance with the amplitude of modulating signal.
19.What is Bandwidth of AM wave?
Band width is difference between highest upper side frequency and lowest
lower side frequency. B.W = 2fm(max).
20. What is over,under,critical modulation?
If m >1, has severe distortion.This codn is Over modulation. If m=1, has
greatest output and codn is Critical modulation. If m< 1 ,has no distortion and
codn is Uunder modulation.
21.With reference to FM, define modulation index.
Modulation index is the ratio of frequency deviation and modulating signal
frequency
m = ∆f / fm
∆f = frequency deviation in Hz
fm = modulating signal frequency in Hz
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22.Define deviation ratio.
It is the worst-case modulation index which is the ratio of maximum
permitted frequency deviation and maximum modulating signal frequency.
Deviation ratio = ∆f(max) / fm(max)
23.State Carson’s rule for determining approximate Band Width of FM signal.
Band Width = 2 [ ∆f + fm(max) ] Hz
∆f = frequency deviation in Hz
fm(max) = highest modulating signal frequency in Hz
24.A carrier is frequency is frequency modulated with a sinusoidal signal of 2
KHz resulting in a maximum frequency deviation of 5 KHz. Find the
approximate band width of the modulated signal.
∆f = frequency deviation in Hz = 5 KHz
fm(max) = highest modulating signal frequency in Hz = 2 KHz
Band Width = 2 [ ∆f + fm(max) ] Hz = 14 KHz
25.Determine the modulation index of a FM system with a maximum frequency
deviation of 75 KHz and maximum modulating frequency of 10 KHz.
m = ∆f / fm = 75 KHz / 10 KHz = 7.5
26.Distinguish between narrow band FM and wide band FM.
Narrow band FM
Frequency deviation in carrier
frequency is very small
Band width is twice the
highest modulating frequency
Wide band FM
Frequency deviation in carrier
frequency is large
Band width is calculated as per
Carson’s rule
27.What are the advantages of FM over AM?





The amplitude of FM is constant. Hence transmitter power remains
constant in FM where as it varies in AM.
Since amplitude of FM is constant, the noise interference is minimum
in FM. Any noise superimposing on modulated carrier can be
removed with the help of amplitude limiter.
The depth of modulation have limitation in AM. But in FM, the depth
of modulation can be increased to any value.
Since guard bands are provided in FM, there is less possibility of
adjacent channel interference.
Since space waves are used for FM, the radius of propagation is
limited to line of sight( LOS ) . Hence it is possible to operate several
independent transmitters on same frequency with minimum
interference.
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Since FM uses UHF and VHF ranges, the noise interference is
minimum compared to AM which uses MF and HF ranges.
28. What is the advantage and disadvantage of Angle modulation?
Advantages: 1. Noise Reduction
2. Improved system fidelity
3. More effective use of power
Disadvantage: 1. Require more Bandwidth
2. Use more complex circuits in both transmitter and receiver
29. With respect to PM define modulation index?
m = k Em = Δθ
30. Define peak phase deviation?
k Em = Δθ
where k is deviation sensitivity in rad/volt
and Em is amplitude of modulating signal.
31.Define frequency deviation?
Δf = k1 Em/2п in Hz
Where k1 is deviation sensitivity rad/volt-sec
32. Define percent modulation?
Percent modulation = [actual frerquency deviation/max allowable frequency
deviation] x100
33.What are the types of modulation Index?
1.Low index modulation (m<1)
2.Medium index modulation(m=1)
3.High index modulation(m>1)
34.Obtain total angle modulated wave power?
Pt = Po + P1 +P2 +….Pn
35.With reference to AM define modulation index.
It is defined as the ratio of maximum amplitude of the modulating voltage to the
maximum amplitude of carrier voltage
=Am/Ac
(or)
=ka Am
36.Why is modulation necessary?
 Ease of radiation
 Multiplexing
 To overcome hardware limitation
 To reduce noise and interference
 Frequency Assignment
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37.The carrier of a 100% modulated wave is suppressed Find the percentage of
power saving
Power saving in DSBSC compared to AM is 66.6%
38.Define selectivity and sensitivity of a radio receiver
Selectivity: It is defined as the ability to select a particular frequency from different
frequency
Sensitivity: It is defined as the ability of receiver to give maximum output even though
i/p is very low
39.A Tx supplies 8KW to the antenna when unmodulated. Determine the total
power when amplitude modulate to 30%.
Pt=Pc(1+ma2 /2)
=8x103 (1+0.32/2)
40.What is the main difference b/w frequency modulation and phase modulation
Frequency modulation :It is the form of angle modulation in which instantaneous
frequency fI(t) is varied linearly with the base band signal m(t)
Where,fI (t)=fc+kf m(t) fc unmodulated carrier
kf –Frequency sensitivity of the modulator
m(t)-Base band signal
integrating above equation w. r .t time and multiplying with 2
i(t)= 2fc t+2Kf m(t) dt
s(t)=Ac cos i (t)
s(t)= Ac cos(2fc t+2Kf m(t) dt)
Phase modulation :
It is that form of Angle modulation in which angle i(t) is varied linearly with
the base band signal m(t) as as shown by
i(t)= 2fc t+Kpm(t)
s(t)=Ac cos i (t)
s(t)= Ac cos(2fc t+Kpm(t)
41.Define modulation index for AM and FM
=Am/Ac---------------------------AM
=f/fM
Ratio of frequency deviation and modulating signal frequency
42.With reference to FM define modulation index and frequency deviation
modulation index
=f/fM
Frequency deviation :It represent the maximum departure of instantaneous frequency of
FM wave from the carrier frequency fc .Frequency deviation is proportional to the
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amplitude of modulating wave and is independent of the modulation frequency f=Kf
Am
43.Determine the modulation depth of FM system with a maximum frequency
deviation of 75 KHz and the maximum modulating frequency of 10 KHz
=f /fM
75
x103 /10 x103
=
=7.5
44.Write down the expression for FM signal with sinusoidal modulation
Frequency modulation :It is the form of angle modulation in which
instantaneous frequency fI(t) is varied linearly with the base band signal m(t)
Where,fI (t)=fc+kf m(t) fc unmodulated carrier
kf –Frequency sensitivity of the modulator
m(t)-Base band signal
integrating above equation w. r .t time and multiplying with 2
i(t)= 2fc t+2Kf m(t) dt
s(t)=Ac cos i (t)
s(t)= Ac cos(2fc t+2Kf m(t) dt)
45.What are the main difference b/w AM Rx and FM Rx
The basic difference is AM demodulator replaced by FM demodulator such as limiter and
discriminator.
 Typical freq parameters of commercial AM value
 RF carrier range 0.535-1.605 MHz
 Mid band freq of IF station =.455 MHz
 IF band width =10 KHz
 Typical freq parameters of commercial FM value
 RF carrier range 88-108 MHz
 Mid band freq of IF station =.10.7 MHz
 IF band width =.2 MHz
 In an FM system the message information is transmitted by variation of the
instantaneous frequency of sinusoidal carrier wave and its amplitude is
maintained constant .Therefore any variation of the carrier amplitude at the R x
i/p must result from the noise or interference The amplitude limiter before
discriminator remove amplitude variation.
46.Distinguish b/w direct method and indirect method of producing FM wave
IN indirect method the instantaneous frequency of the carrier wave is varied
directly in accordance with the base band signal by means of a device known as vco e.g
Hartley oscillator
Disadvantage is that carrier frequency is not obtained from a highly stable oscillator
In indirect method the base band signal m(t) is first integrated used to phase modulate a
crystal controlled oscillator .Hence stable carrier frequency is obtained
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47.Why is FM system preferred over AM system.
PM system can provide better discrimination against noise and interference than AM
48.What is the main difference between natural sampling and instantaneous
sampling
Instantaneous sampling:
Consider an arbitrary signal g(t) of finite energy which is specified for all time
as shown in fig a .Suppose that we sample the signal g(t) instantaneously and at uniform
rate once every T seconds .As a result of this sampling process we obtain an infinite
sequence of numbers spaced Ts seconds apart and denoted by g(nTs) where n takes all a
possible integer values .Ts is the sampling period.
Natural sampling: This is applied in PDM(pulse duration modulation ).Samples of
message signal are used to vary the duration of individual pulses .
49.Define Modulation index and percent modulation for an AM wave.
Modulation index is a term used to describe the amount of amplitude
change present in an AM waveform .It is also called as coefficient of modulation.
Mathematically modulation index is
m = Em/Ec
Where m = Modulation coefficient
Em = Peak change in the amplitude of the output waveform voltage.
Ec = Peak amplitude of the unmodulated carrier voltage.
Percent modulation gives the percentage change in the amplitude of the output
wave when the carrier is acted on by a modulating signal.
49.Define direct frequency modulation.
In direct frequency modulation , frequency of a constant amplitude carrier
signal is directly proportional to the amplitude of the modulating signal at a rate
equal to the frequency of the modulating signal.
50.Define indirect frequency Modulation.
In indirect frequency modulation phase of a constant amplitude carrier is
directly proportional to the amplitude of the modulating signal at a rate equal to
the frequency of the modulating signal.
51.Define instantaneous frequency deviation.
The instantaneous frequency deviation is the instantaneous change in the
frequency of the carrier and is defined as the first derivative of the instantaneous
phase deviation.
52.Define frequency deviation.
Frequency deviation is the change in frequency that occurs in the carrier
when it is acted on by a modulating signal frequency.Frequency deviation is
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simply the product of the deviation sensitivity and the peak modulating signal
voltage and i
53.State Carson rule.
Carson rule states that the bandwidth required to transmit an angle
modulated wave as twice the sum of the peak frequency deviation and the highest
modulating signal frequency. Mathematically carson’s rule is
54.Define Deviation ratio.
Deviation ratio is the worst case modulation index and is equal to the
maximum peak frequency deviation divided by the maximum modulating signal
frequency. Mathematically, the deviation ratio is
PART B :
1. Obtain AM wave equation and explain each term with the help of frequency
spectrum and also obtain an expression for its power?
2. Obtain the AM wave equation and draw the AM Envelope. Also explain
modulation index and percent modulation in terms of Vmax and Vmin.
3. An AM modulator has a carrier of 400 KHz with amplitude of 20v; modulating
signal of 8 KHz with amplitude of 8.5v is applied. Determine
(a) Upper and lower side frequencies.
(b) Modulation coefficient and percent modulation
(c) Peak amplitude of the modulated carrier and upper and lower side
frequency voltages.
(d) Maximum and minimum amplitude of the envelope.
(e) Expression of modulated wave.
(f) Sketch the output spectrum and envelope.
4.Write down the expression for FM and PM waves and draw their frequency
spectrum
and explain.
5 .Obtain the mathematical expressions for AM & FM modulated waves & draw
the necessary waveforms in both cases.
6. Compare AM, FM and PM systems.
7.Determine (A)the deviation ratio and bandwidth for the worst case modulation
index for an FM broadcast band transmitter with max frequency of 15khz
(B)the Deviation Ratio and max BW for a equal modulation index with only half
the peak frequency deviation and Modulation signal frequency.
8.Expalin the mathematical analysis of angle modulated wave .
9.Expalin the phase deviation and Modulation index of angle modulated wave
10.Discuss about frequency deviation and Percentage modulation .
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Unit II : DIGITAL COMMUNICATION
1.Obtain the Bandwidth for narrow band FM?
For narrowband FM bandwidth is 2fm.where fm is modulating frequency.
2. For commercial FM broadcast band transmitter frequency deviation is 75KHz. If
the given modulating signal produce 35KHz frequency deviation. What is the
percent modulation?
percent modulation = 35/75 = 46.66
3.Determine peak phase deviation for PM modulator with deviation sensitivity
k = 2.5 rad/v and em = 2cos(2п2000t).
k Em = Δθ
Δθ = 2.5(2) = 5rad/v
4.What is a constellation diagram? Draw the constellation diagram for BPSK.
Constellation diagram is used to show the relative positions of the peaks of the
phasors. It is similar to a phasor diagram except that the entire phasor is not drawn(peak
of the phasors are shown).
5.What is the minimum bandwidth required for an FSK system?
2
band widthrequired  f m  f s 
tb
1
 f b  bit rate ; f m  mark frequency ; f s  space frequency
tb
6.What is the primary advantage of DBPSK and what is its disadvantage?
Advantage: simple implementation. No carrier recovery circuit needed for
detection.
Disadvantage: It requires between 1 dB and 3 dB more signal to noise ratio to
achieve the same BER as that of standard absolute PSK
7.What are the advantages of M-ary signaling schemes?
 M-ary signaling schemes transmit multiple bits at a time.
 Bandwidth requirement of M-ary signaling schemes is reduced.
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8.Compare binary PSK with QPSK.
BPSK
Binary Phase Shift Keying
One bit form a symbol
Two possible symbols
Minimum bandwidth required = f b
where f b is bit rate
QPSK
Quadrature Phase Shift Keying
Two bits form a symbol
Four possible symbols
Minimum bandwidth required = f b / 2
where f b is bit rate
9.What are the advantages of QPSK as compared to BPSK?
 For the same bit rate, the bandwidth required by QPSK is reduced to half
as compared to BPSK.
 Because of reduced bandwidth, the information transmission rate of QPSK
is higher.
10.What happens to the probability of error in M-ary PSK as the value of M
increases?
As the value of M increases, the Euclidean distance between the symbols
reduces. Hence the symbols are closer to each other. This increases the
probability of error in M-ary systems.
11.What is the minimum bandwidth required for BPSK, QPSK, 8-PSK, 8-QAM
and16-QAM systems if the bit rate is 10 Mbps?
system
BPSK
QPSK
8 - PSK
8- QAM
16 - QAM
Minimum band width
required
if f b = bit rate
fb
fb / 2
fb / 3
fb / 3
fb / 4
Minimum band width
required
if f b = 10 Mbps
10 MHz
5 MHz
3.33 MHz
3.33 MHz
2.5 MHz
12.What is difference between coherent and non coherent detection?
Coherent detection
Carrier which is in perfect
coherence with that used in
transmitter
is
used
for
demodulation.
Carrier recovery circuit is
needed for detection
Relatively complex
Non- Coherent detection
No carrier recovery circuit needed
for detection.
Simple implementation
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13.Define Bandwidth efficiency. What is the bandwidth efficiency of BPSK and 8PSK system?
It is the ratio of the transmission bit rate to the minimum bandwidth required for a
particular modulation scheme.
transmissi on rate (bps )
Band width efficiency 
min imum band width ( Hz )
For BPSK , transmission rate = f b and minimum bandwidth = f b
Band width efficiency = 1
For 8-PSK , transmission rate = f b and minimum bandwidth = f b/3
Band width efficiency = 3
14.What is the difference between probability of error P(e) and bit error rate BER?
P(e) Probability of error is a theoretical (mathematical) expectation of the
bit error rate for a given system.
BER is an empirical record of a system’s actual bit error performance.
For Example, if a system has a P(e) of 10 -5 , this mean that, you can expect one
bit error in every 100,000 bits transmitted.
If a system has a BER of 10 -5 , this mean that, there was one bit error for every
100,000 bits transmitted.
BER is measured and then compared to the expected probability of error to
evaluate the system’s performance.
15.What is the probability of error for (i) non-coherent FSK and (ii) coherent FSK?
Compare their error performance.
P (e) 
 E 
1
exp   b 
2
 2N 0 
 probabilit y of error for non  coherent FSK
 Eb 

P(e)  erfc 
 probabilit y of error for coherent FSK
 N 
0


Eb
 energy per bit to noise power density ratio
N0
For a given energy per bit to noise power density ratio, probability of error for noncoherent FSK is greater than that of coherent FSK.
16. Define ( Eb / N0 ) Energy per bit to Noise power density ratio.
Energy per bit to noise power ratio is used to compare two or more digital
modulation systems that uses different bit rates and modulation schemes.
It is the product of carrier to noise power ratio and the noise band width to bit rate
ratio. This is equivalent to signal to noise ratio.
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Eb
 energy per bit to noise power density ratio
N0
Eb  Energy per bit  C Tb
where C  carrier power in watt
Tb  bit duration in sec
N
B
where N  thermal noise power in watt
B  noise band width in Hz
Eb C B
 
N0 N fb
N 0  Noise power density 
PART B :
1. Explain BPSK Transmitter and receiver with a neat diagram.
2. Explain BFSK Transmitter and receiver with a neat diagram.
3. Explain QPSK Transmitter and receiver with a neat diagram.
4. Explain DPSK Transmitter and receiver with a neat diagram.
5. Compare the different types of digital modulation techniques.
6. Explain the error performance of different digital modulation techniques.
7. Explain QAM.
8. Write short notes on the probability error of the following
(i)FSK (coherent and non coherent)
9. Write short notes on the Carrier Recovery of the following
(i)
Coastas loop.
(ii)
Squaring loop.
10.For an 8-PSK modulator with an input data rate equal to 20mbps and a carrier freq of
70Mhz .Determine the minimum double sided Nyquist bandwidth (fn) and baud.
11.For an QPSK modulator with an input data rate (Fb)equal to 20mbps and a carrier freq
of 70Mhz .Determine the minimum double sided Nyquist bandwidth (fn) and baud.
12.For an BPSK modulator with an input data rate equal to 20mbps and a carrier freq of
70Mhz . Determine the minimum and minimum upper and lower side freq double sided
Nyquist bandwidth (fn) and baud.
13.expalin the operation (a)8-PSK (b)QAM.
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UNIT III : DIGITAL MODULATION TECHNIQUES
1.State sampling Theorem.
If a finite energy signal g(t) contains no frequency component higher than W Hz, it is
completely determined by specifying its ordinates at a separation of points spaced 1/
2W seconds apart.
If a finite energy signal g(t) contains no frequency component higher than W Hz, it is
completely recovered from its ordinates at a separation of points spaced 1/ 2W
seconds apart.
2.Methods of Sampling
1. Ideal Sampling (or) Instantaneous Sampling
2. Natural Sampling
3. Flat-top Sampling
3.What is Aliasing or Foldover?
Aliasing occurs when the signal is sampled at a rate less than Nyguist rate(2W
samples/ sec). It is prevented by using
Guard Bands
Pre-alias Filter
2. Define Nyquist rate and Nyquist interval.
According to sampling theorem, a continuous time signal can be completely
represented in its samples and recovered back if the sampling frequency is fS ≥ 2W.
Here fS is sampling frequency and W is the highest frequency component of the signal.
Nyquist rate: The minimum sampling rate of 2W samples per second is called Nyquist
rate.
i.e., fS = 2W → Nyquist rate
Nyquist interval: Reciprocal of 2W is called the Nyquist interval.
Nyquist interval = 1/2W
3. What is aliasing?
When the continuous time signal g(t) is sampled at the rate less than Nyquist
rate, frequencies higher than W takes on the identity of the low frequencies in sampled
signal spectrum . This is called aliasing.
The use of a low pass reconstruction filter, with its pass band extending from –W
to W will not yield an undistorted version of the original signal. Aliasing can be reduced
by sampling at a rate higher than Nyquist rate.
4. Give the practical procedure for the sampling of a signal whose spectrum
is not strictly band limited.
i) Prior to sampling, a low pass pre-alias filter ( anti-alias filter) of high enough
order is used to attenuate those frequency components of the signal that do not contribute
significantly to the information content of the signal.
ii) The filtered signal is sampled at a rate slightly higher than the Nyquist rate 2W,
where W is the cutoff frequency of the pre-alias filter.
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5. Define aliasing error. Give the upper bound for the aliasing error.
Let {g(n/fS)} denote the sequence obtained by sampling an arbitrary signal g(t) at
the rate fS samples per second. Let gi(t) denote the signal reconstructed from this
sequence by interpolation;
That is, gi(t)=∑ g( ) sinc ( fSt - n)
The absolute error ε = | g(t) – gi(t) | is called the aliasing error.
The aliasing error is bounded as
ε≤2
| G(f) | df
6. Given the signal m(t)=10 cos (2000π t) cos (8000 π t ) , what is the minimum
sampling rate based on the low pass uniform sampling theorem?
The equation shows that m(t) is generated by multiplication of two signals. We
know that cosAcosB = ½ [cos (A-B)+cos (A+B)]
There fore, m(t) = (10/2) [ cos (6000π t) + cos (10000π t)]
The two frequencies in m(t) are 3000 Hz and 5000 Hz and the highest frequency present
in m(t) is 5000 Hz.
Minimum sampling rate is 2 (5000) = 10000 samples per second.
7. What is Inter symbol Interference (ISI) ?
ISI arises because of imperfections in the overall frequency response of the
system.
When a short pulse of duration Tb seconds is transmitted through a band limited
system, the frequency components constituting the input pulse are differentially
attenuated and, more significantly, differentially delayed by the system. Consequently the
pulse appearing at the output of the system is dispersed over an interval longer than Tb
seconds.
Thus when a sequence of short pulses are transmitted through the system, one
pulse every Tb seconds, the dispersed responses originating from different symbol
intervals will interfere with each other, there by resulting in ISI.
8. What is eye pattern?
When the received signal in a data transmission system is applied to the vertical
deflection plates of an oscilloscope and saw tooth wave( frequency = symbol rate) is
applied to the horizontal deflection plates of an oscilloscope, the resulting display is
called an eye pattern.
The display pattern is called an eye pattern because of its resemblance to the
human eye. An eye pattern provides a great deal of information about the performance of
the data transmission system. Eye pattern is particularly useful in studying ISI problem.
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9. List out the information provided by eye pattern about the system performance.
i)The width of the eye opening defines the time interval over which the received wave
can be sampled without error from ISI. The preferred time for sampling is the instant of
time at which the eye is open widest.
ii) The sensitivity of the system to timing error is determined by the rate of closure of the
eye as the sampling time is varied.
iii) The height of the eye opening , at a specified sampling time, defines the margin over
noise.
iv) Completely closed eye shows that the effect of ISI is severe.
10. Compare base band binary PAM and base band M-ary PAM.
 In the baseband binary PAM system , the out put of the pulse generator takes on
one of two possible amplitude levels. In a baseband M-ary PAM system, the out
put of the pulse generator takes on one of M possible amplitude levels.
 An M- ary PAM system transmit information at a rate that is log2M faster than
the corresponding binary PAM for a fixed channel band width. That is for a
fixed bit rate, an M-ary PAM system requires less channel bandwidth than a
binary PAM.
 In order to realize the same average probability of symbol error , an M-ary PAM
system requires more transmitted power.
12. State Nyquist criterion for zero ISI.
∑ P( f – n Rb ) = Tb
Tb = bit duration in seconds ; Rb = bit rate = 1/Tb
p(t ) = received pulse
P(f) = Fourier Transform of p(t)
13. What is adaptive equalization?
Equalization is the process of correcting channel induced distortion. This process is said
to be adaptive when it adjusts itself continuously during data transmission by operating
on the input signal.
In adaptive equalization, filters adapt themselves to the channel characteristics.
That is , the filter coefficients are adjusted in such a way that the distortion in the data is
reduced.
14. What is PAM?
It is a process in which amplitudes of regularly spaced pulses are varied in proportion
to the corresponding sample values of continuous message signal.
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15.What do you mean by Aperture Effect?
It is nothing but amplitude distortion occurring at PAM due to the sinc function. It is
overcome by using a Equalizer whose transfer function is
| H(f)| = T –1 sinc(fT)
16. What is PPM?
It is the process in which the position of a pulse relative to its unmodulated time of
occurrence is varied in accordance with message signals.
17. What is PWM?
It is the process in which the samples of message signal are used to vary the duration
of individual pulses in the carrier.
18. What is Quantization?
It is the process in which the analog sample of the original signal is converted into a
digital form.
19. What is Sampling?
It is the process in which the original analog signal is converted into a discrete time
and continuous amplitude signal.
20. Classify Quantizers.
Uniform Quanatizer – Representation levels are uniformly spaced
Non-Uniform Quanatizer – Representation levels are non-uniformly spaced
21. What is Quantization Noise?
The difference between the output analog sample and the discrete output quantized
signal gives raise to an error called Quantization Noise.
22. What are the two types of Quantization Noise?
1. Slope – overload distortion
2. Granular Noise
23. What is Slope- Overload Distortion?
When the message signal varies steeply, the quantized approximation cannot follow
the message signal and this results in slope-overload distortion. It can be reduced by
increasing the step-size. To minimize the distortion, we need to have
 / Ts ≥ max | d(x(t)) / dt |
24. What is Granular Noise?
When the input waveform has a flat segment, then the step-size is larger
when compared to the input. Therefore the approximation hunts around the
segment resulting in Granular noise.
This noise can be reduced by decreasing the step-size.
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25. What is Companding?
It is the combined process of compressing and Expanding used for improving the
dynamic range of signal and also to increase the SNR of low level signals.
26. List the advantages of Digital pulse modulation systems.
Immunity to
3. Transmission distortion.
4. Interference
5. Noise
27.What is meant by hunting with reference to delta modulation?
The up & down movement of the approximated signal m’ (t) above & below the input
signal m (t) because of the step-wise approach during delta modulation is called hunting.
28. What is the principle of adaptive delta modulation?
The principle underlying ADM:
a) If successive errors are of opposite polarity, then the delta modulator is operating
in its granular mode. In this case the step size is reduced.
b) If successive errors are of same polarity, then the delta modulator is operating in
its slope overload mode. In this case the step size is increased.
28. What are the limitations of delta modulation?
The major limitations of delta modulation are:
a) Slope overload error
b) Granular noise
29.State the difference between DPCM and DM.
Parameter
DPCM
DM
Number of Bits
Bits can be more than one
but less than PCM.
It can use 4, 8 or 16 bits per
sample.
Levels, Step size.
Fixed numbers of levels are
used.
Step size is fixed and
cannot be varied.
Bandwidth of
transmission channel.
Bandwidth required is
lesser than PCM.
Lower bandwidth is
required.
30. Define slope overload error?
The signal m (t) over a extended time has a very large slope which cannot be approached
by the approximated signal m’ (t) In short time.as a result the error m (t)-m’ (t) becomes
progressively larger i.e., exceeding s/2 this is known as slope overload error.
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31. Define granular noise?
Even though the signal m(t) remains unvarying the approximated signal m’(t) swinging
up or down above & below m(t) . It is because of step-wise approach of m’ (t) to the
m(t).This process is referred as hunting & it gives the noise known as granular noise.
32. Define Time division multiplexing.
In time division multiplexing common cable/medium is shared by different channels
on time basis Each channel is assigned a fixed rtime slot .Every Channel transmits in its
own time slot.In TDM
33. What are the advantages of digital transmission?
The advantage of digital transmission over analog transmission is noise
immunity. Digital pulses are less susceptible than analog signals to
variations caused by noise.
 Digital signals are better suited to processing and multiplexing than analog
signals.
 Digital transmission systems are more noise resistant than the analog
transmission systems.
 Digital systems are better suited to evaluate error performance.
34. What are the disadvantages of digital transmission?
_ The transmission of digitally encoded analog signals requires
significantly more bandwidth than simply transmitting the original
analog signal.
_ Analog signal must be converted to digital codes prior to
transmission and converted back to analog form at the
receiver, thus necessitating additional encoding and
decoding circuitry.
35. Define pulse code modulation.
In pulse code modulation, analog signal is sampled and converted to fixed length,
serial binary number for transmission. The binary number varies according to
the amplitude of the analog signal.
36. What is the purpose of the sample and hold circuit?
The sample and hold circuit periodically samples the analog input signal and
converts those samples to a multilevel PAM signal.
37. What is the Nyquist sampling rate?
Nyquist sampling rate states that, the minimum sampling rate is equal to twice
the highest audio input frequency.
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38. Define and state the causes of fold over distortion.
The minimum sampling rate(fs) is equal to twice the highest audio input
frequency(fa).If fs is less than two times fa ,distortion will result. The distortion is
Called aliasing or fold over distortion.
The side frequencies from one harmonic fold over into the sideband of another
harmonic. The frequency that folds over is an alias of the input signal hence ,the
names “aliasing” or “fold over distortion”
.
39. Define overload distortion.
If the magnitude of sample exceeds the highest quantization interval, overload
distortion occurs.
40. Define quantization.
Quantization is a process of approximation or rounding off. Assigning PCM codes
to absolute magnitudes is called quantizing.
41. Define dynamic range.
Dynamic range is the ratio of the largest possible magnitude to the smallest
possible magnitude. Mathematically, dynamic range is
DR= Vmax/Vmin
42. Define slope overload. How it is reduced.
The slope of the analog signal is greater than the delta modulator can maintain,
and is called slope overload. Slope overload is reduced by increasing the clock
frequency and by increasing the magnitude of the minimum step size.
43.Define granular noise.How it is reduced.
When the original input signal has relatively constant amplitude, the
reconstructed signal has variations that were not present in the original signal.
This is called granular noise.
Granular noise can be reduced by decreasing the step size.
44. Define peak frequency deviation for FSK.
Peak frequency deviation (_f) is the difference between the carrier rest frequency
and either the mark or space frequency and either the mark or space frequency.
45. Define bit rate.
In digital modulation, the rate of change at the input to the modulator is called the
bit rate (fb) and has the unit of bits per second (bps).
46. Define Baud rate.
The rate of change at the output of the modulator is called baud.
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47. Define QAM.
Quadrature amplitude modulation is a form of digital modulation where the
digital information is contained in both the amplitude and phase of the transmitted
carrier.
48. Write the relationship between the minimum bandwidth required for an
FSK system and the bit rate.
The minimum bandwidth can be approximated as
B=2_f +2fb
Where B=minimum bandwidth (hertz)
_f=minimum peak frequency deviation (hertz)
Fb=bitrate
49.What is PAM?
It is a process in which amplitudes of regularly spaced pulses are varied in proportion to
the corresponding sample values of continuous message signal.
50.What do you mean by Aperture Effect?
It is nothing but amplitude distortion occurring at PAM due to the sinc function. It
is overcome by using a Equalizer whose transfer function is
| H(f)| = T –1 sinc(fT)
51.What is PPM?
It is the process in which the position of a pulse relative to its unmodulated time
of occurrence is varied in accordance with message signals.
52.What is PWM?
It is the process in which the samples of message signal are used to vary the duration of
individual pulses in the carrier.
53. List the advantages of Digital pulse modulation systems.
Immunity to--- Transmission distortion, Interference & Noise
54. What is the minimum bandwidth requirement of a PAM system?
Nfm
N:No.of signals to be multiplexed
fm:Highest frequency of the signal
55. How is synchronization achieved in PAM system?
In a PAM system, marker pulses are introduced in each frame. These pulses are identified
by making its amplitude greater than all possible message pulses and the marker pulse is
identified at the receiver by applying the pulses to the slicer network.
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56.What are the corrective measures taken to avoid aliasing effect?
A low pass anti-alias filter is used to attenuate high frequency of the signal that lie
outside the frequency band of interest prior to sampling. The filtered signal is sampled at
a slightly higher rate than Nyquist rate.
57. Mention the advantages of digital communication over analog communication.
i)It gives ruggedness to transmission noise and interference.
ii) Efficient regeneration of coded signal along transmission path.
iii)Possible to have a uniform format for different kinds of baseband signal.
58. How many bits required to sample audio signals with 128 quantization levels ?
7.
59.What is hunting as related to delta modulation?
The initial large discrepancy between the message signal and its stepwise approach is
called hunting or startup response resulting in redundancy (repetition of same pattern of
bits for almost nil variation in slope in message signal).
60. State the differences between DPCM and DM.
 DM uses only one bit information for transmission.
 Replacement of the prediction filter in DPCM by a single delay element
constitutes DM system.
61.PCM requires larger BW as compared to DM for the same performance. State
true or false. False.
62. What is meant by DPSK method?
It is a non-coherent binary modulation technique. This is a modified technique of binary
phase shift keying. DPSK avoids the need to provide a synchronous carrier required at
the demodulator for detecting BPSK signal.
63. What is meant by Binary – phase- shift keying?
If the transmitted signal is sinusoid of fixed amplitude then it is called as Binary – phaseshift keying. It has one fixed phase when the data is at one level & when the data is at
another level the ‘phase’ is different by 180
64. How the BPSK signal is generated?
The BPSK signal is generated by applying the waveform Cos(uo t) as a carrier to a
‘balanced modulator’ and applying the base band signal b(t) as the modulating waveform,
in this sense BPSK can be thought of as an AM signal.
65. What is the significance of DPCM?
In DPCM instead of quantizing the individual sample values, the difference between the
successive samples is quantized. As the difference between sample values {M(k) – M(k1)} will be smaller than the sample value themselves fewer levels will be required to
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quantize the difference than that required to quantize M(k) and correspondingly fewer
bits will be needed to encode the levels.
PART B
1. What are the limitations of delta modulation? Explain Adaptive and linear delta
modulation system .Discuss its merits over DPCM.
2. Explain the principle of delta modulation.
3. With a relevant block diagram ,describe the operation of a DPCM
4. Write short notes on :
a. Comparison of DPCM with DM
b. Adaptive DELTA Modulation
5. What does slope overloading mean ?Draw the block diagram of DPCM and
explain its
working principle.
6. Compare the performance of PCM ,DPCM and DM.
7. With a neat diagram explain how a DM system works?
8. Obtain the expression for gain in signal to noise ratio for DPCM
9. Explain about ISI and EYE PATTERN
UNIT IV : SPREAD SPECTRUM AND MULTIPLE ACCESS TECHNIQUES
1. What is the primary advantage of spread spectrum communication?
The primary advantage of spread spectrum communication system is its ability to
reject interference whether it be the unintentional interference of another user
simultaneously attempting to transmit through the channel or the intentional interference
of a hostile transmitter attempting to jam the transmission.
2. What are the applications of spread spectrum modulation?



Protection against jammers in military applications
Code division multiple access (CDMA)
Multipath interference suppression in mobile communication
3. Define Pseudo Noise sequence.




A Pseudo Noise (PN) sequence is coded sequence of 1s and 0s with
certain autocorrelation properties.
It is periodic – a sequence of 1s and 0s repeats itself exactly with a known
period
PN sequence is a noise like high frequency signal. The sequence is not
truly random but it is generated by a well defined logic.
This is used in spread spectrum communication.
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4. What are the properties of maximum length PN sequence?
 Balance property: In each period of a maximum length sequence, the
number of 1s is always one more than the number of 0s.
 Run property: Among the runs of 1s and 0s in each period of a
maximum length sequence, one half the runs are of length one, one
fourth are of length two, one eighth are of length three and so on.. as
long as these fractions represent meaningful number of runs
 Correlation property: The auto correlation function of a maximum length
sequence is periodic and binary valued.
5. Define processing gain. What is the processing gain for DSSS?
It is defined as the ratio of spreaded signal bandwidth to unspreaded signal
bandwidth.
6. Define chip rate.
Number of chips per second.
7. What is the difference between slow frequency hopping spread spectrum
technique and fast frequency hopping spread spectrum technique?
 Slow frequency hopping: Symbol rate RS of the MFSK signal is an integer
multiple of the hop rate R h. i.e., several symbols are transmitted in each hop.
 Fast frequency hopping: The hop rate R h is an integer multiple of the
MFSK symbol rate RS. i.e., carrier frequency will change or hop several times
during the transmission of one symbol.
8. What are the different multiple access techniques ( Fixed assignment multiple
access techniques)used in wire less communication?
 TDMA – Time Division Multiple Access
 FDMA- Frequency Division Multiple Access
 CDMA – Code Division Multiple Access.
9. Compare FDMA and TDMA.
FDMA - Frequency Division Multiple
Access
All users access the channel by
transmitting simultaneously but using
disjoint frequency bands
Fixed
assignment
multiple
access
technique
Well suited for analog communication
TDMA - Time Division Multiple Access
All users occupy the same RF band width
of the channel, but they transmit
sequentially in time
Fixed assignment multiple access technique
Well suited for digital communication
10. List the merits of CDMA over TDMA.
 CDMA does not require an external synchronization network which is an
essential feature of TDMA.
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
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CDMA offers a gradual degradation in performance as the number of user is
increased. It is there fore relatively easy to add new users to the system.
CDMA offers an external interference rejection capability.
11. What is CDMA?
CDMA – Code Division Multiple Access. In CDMA , all users transmit
simultaneously and occupy the same RF bandwidth. Each user is assigned a code
which perform the DSSS or FHSS modulation
12. What is meant by spread spectrum?
Spread spectrum is a means of transmission in which the data of interest occupies a
bandwidth in excess of the minimum bandwidth necessary to send the data
13. What are the applications of spread spectrum?
 It is used in military communications systems.
 It allows the transmitter to transmit a message to a receiver without the message
being detected by a receiver for which it is not intended.
 It decreases the transmitted power spectral density so that it lies well below the
thermal noise level of any unfriendly receiver.
 It turns out not to be possible to conceal the transmission.
14. What is DS spread spectrum?
DS spread spectrum signal is one in which the amplitude of an already modulated signal
is AM modulated by a very high rate NRZ binary stream of digits.
15. Define FH spread spectrum.
FH spread spectrum is an FM or FSK technique while DS is a AM or PSK technique. The
signal to be FH is usually a BFSK signal although M-ary FSK, MSK or TFM can be
employed.
16. What is meant by pseudo random noise?
The pseudo random noise is the nose which presents in the DS spread spectrum, if the DS
spread spectrum signal is V(t)=g(t)S(t)=2Psd(t)cosot. Here g(t) is a pseudo random
noise binary sequence having the values 1. A pseudo noise sequence is defined as a
coded sequence of 1s and 0s with certain autocorrelation properties. The class of
sequences used in spread spectrum communication is usually periodic in that a sequence
of 1s and 0s repeats itself exactly with a known period
17. Write the advantages of FH over direct sequence spread spectrum.
 Processing gain is more
 Spread of the transmitted signal is spread sequentially rather than instantaneously
 Greater transmission bandwidth and more chips per bit.
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PART B
1. Explain the Direct sequence spread spectrum technique with neat dig.
2. Explain the Frequency hopping spread spectrum in detail.
3. Explain the properties of PN Sequences.
4. Explain about source coding of Speech for wireless communication
5. Explain the types of Multiple Access techniques.
6. Explain TDMA system with frame structure, frame efficiency and features.
7. Explain CDMA system with its Features .list out various problems in CDMA
systems.
UNIT V : SATELLITE AND OPTICAL FIBRE COMMUNICATIONS
1. State Kepler’s first law.
Kepler’s first law states that the satellite will follow an elliptical path in its orbit
around the primary body. An ellipse has two focal points as F1 and F2.
2. State Kepler’s second & third law.
It states that for equal time intervals the satellite sweeps out equal areas in the orbital
plane, focused at the baycenter.
Third law states that the square of the periodic time of orbit is proportional to the cube
of the mean distance between the two bodies.
3. What is apogee and perigee?
Apogee is the point farthest from the earth. Perigee is the point of closest approach to
the earth.
4. What are the classifications of satellites?
Satellites are generally classified as a low earth orbit (LEO),Medium earth orbit
(MEO),or Geosynchronous earth orbit(GEO) .
5. What is major axis?
The line joining the perigee and the apogee through the centre of earth.
6. What is geo-stationary satellite?
Geo-stationary satellite is one that appears to be stationary relative to the earth.
7. What is ascending node?
The point where the orbit crosses the equatorial plane from south to north. The point
where a polar or inclined orbit crosses the equatorial plane traveling from north to south
is called ascending node.
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8. What are the advantages of optical fibre?
Optical fibers are free from electromagnetic radiation and other electrical interference.
Optical fibres have a very small attenuation .Optical fibers have large bandwidth
capacity.
9. What is meant by the term numerical aperture of an fiber?
Numerical aperture is used as figure merit for the fiber .It is given as
N A =n12 –n22 /no
Where no is refractive index of medium through which light is launched.
n1 is refractive index of fiber core glass.n2 is refractive index of fiber cladding.
10. What is the difference between TDM and FDM?
In TDM ,fixed time slot is given to each channel .In FDM fixed frequency slot is given to
each channel.Thus in FDM ,bandwidth is shared ,whereas in TDM time is shared.
11. What is meant by dispersion and what is it unit of measurement?
The time dispersion of the pulse takes place during transmission. The changes timing
characteristics of the pulse. Because of dispersion ,the pulse is widened and amplitude is
further reduced, This may cause overlapping of adjacent pulses. Dispersion is measured
in nanoseconds per kilometers.
12. Distinguish between graded index and step index multimode fiber?
In step index fiber the refractive index of the fiber core remains same throughout its
bulk. There is sudden transition of refractive index at the interface at the interface of core
and cladding .Because of the constant refractive index of the core material the light rays
travel straight and reflected from the walls of the core. Hence the light path is zig zag
straight segments.
In graded index fiber the refractive index is maximum at the core axis and goes
decreasing gradually towards the wall of the core. The acceptance cone of the graded
index fiber is more than the step index fibers.
13.What is acceptance cone half angle for an optical fiber with core and cladding
refractive index of n1=1.48 and n2=1.46 respectively.
The acceptance cone angle is given by Qo=sin-1 [sqrt(n12 –n22 /no ] (no=1) =14
14. Define satellite.
Satellite is a celestial body that orbits around a planet.In aerospace terms,
a satellite is a space vehicle launched by humans and orbits earth or another
celestial body.
15. State Kepler’s third law.
The third law states that the square of the periodic time of orbit is
proportional to the cube of the mean distance between the primary and the
satellite
.
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16. Define orbital satellite.
Orbital satellites are also called as non synchronous satellite. Non synchronous satellites
rotate around earth in an elliptical or circular pattern. In a circular orbit, the speed or
rotation is constant however in elliptical orbits the speed depends on the height the
satellite is above the earth.
17. Define posigrade orbit.
If the satellite is orbiting in the same direction as earth’s rotation and at an
angular velocity greater than that of earth, the orbit is called a prograde (or)
posigrade orbit.
18. Define retrograde orbit.
If the satellite is orbiting in the opposite direction as the earth’s rotation or
in the same direction with an angular velocity less than that of earth, the orbit is
called a retrograde orbit.
19. Define Geo synchronous satellite.
Geo synchronous or geo stationary satellites are those that orbit in a
circular pattern with an angular velocity equal to that of Erath. Geosynchronous
satellites have an orbital time of approximately 24 hours, the same as earth; thus
geosynchronous satellites appear to be stationary as they remain in a fixed
position in respect to a given point on earth.
20. Define angle of inclination.
The angle of inclination is the angle between the earth’s equatorial plane
and the orbital plane of a satellite measured counterclockwise at the point in the
orbit where it crosses the equatorial plane traveling from south to north.
21. Define Descending node.
The point where a polar or inclined orbit crosses the equatorial plane
traveling from south to north. This point is called descending node.
22. Define line of nodes.
The line joining the ascending and descending nodes through the center of
earth is called line of nodes.
23. Define angle of elevation.
Angle of elevation is the vertical angle formed between the direction of
travel of an electromagnetic wave radiated from an earth station antenna pointing
directly toward a satellite and the horizontal plane.
24. Define Azimuth angle.
Azimuth is the horizontal angular distance from a reference direction,
either the southern or northern most point of the horizon.
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25. What are the advantages of optical fiber communication?
 Greater information capacity
 Immunity to crosstalk
 Immunity to static interference
 Environmental immunity
 Safety & Security

26. Define a fiber optic system.
An optical communications system is an electronic communication system
that uses light as the carrier of information. Optical fiber communication systems
use glass or plastic fibers to contain light waves and guide them in a manner
similar to the way electromagnetic waves are guided through a waveguide.
27. Define refractive index.
The refractive index is defined as the as the ratio of the velocity of
propagation of light ray in free space to the velocity of propagation of a light
ray in a given material.
28. Define critical angle.
Critical angle is defined as the minimum angle of incidence at which a
light ray may strike the interface of two media and result in an angle of refraction
of 90°or greater.
29. Define single mode and multi mode propagation.
If there is only one path for light to take down the cable, it is called single
mode. If there is more than one path, it is called multimode.
30. Define acceptance angle.
It defines the maximum angle in which external light rays may strike the
air/fiber interface and still propagate down the fiber with a response that is no
greater than 10 dB below the maximum value.
31. Define numerical aperture.
Numerical aperture is mathematically defined as the sine of the maximum
angle a light ray entering the fiber can have in respect to the axis of the fiber and
still propagate down the cable by internal reflection.
32. Define modal dispersion.
Modal dispersion or pulse spreading is caused by the difference in the
propagation times of light rays that take different paths down a fiber. Modal
dispersion can occur only in multimode fibers. It can be reduced by using single
mode step index fibers and graded index fibers.
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33.What are the advantages of heterojunction LEDs?
a. The increase in current density generates a more brilliant light spot.
b. The smaller emitting area makes it easier to couple its emitted light into
Fiber.
c. The small effective area has a smaller capacitance, which allows the
planar heterojunction LED to be used at higher speeds.
34. What are the disadvantages of injection laser diode?
_ ILDs are typically on the order of 10 times more expensive than LEDs
_ Because ILDs operate at higher powers, they typically have a much shorter
life time than LEDs.
_ ILDs are more temperature dependent than LEDs.
PART – B
1. Discuss briefly the basic satellite communication system .
2. Write short notes on LEO and GEO orbits.
3. Enumerate the elements of an optical fiber transmission link.
4. With neat diagram of HPA input and output character curve explain each
block of the satellite link model.
5. Describe with neat diagram of satellite antenna radiation pattern or
footprint map.
6. What are satellite orbits and explain the different types of optical Pattern.
7. Explain the Kepler’s law.
8. What are the various optical fiber types .
9. Explain in brief about the losses in optical cable.
10. Explain in brief about the various sources in optical communication .
11. Explain in brief about the various Detectors in optical communication.
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IT 2202 – PRINCIPLES OF COMMUNICATION
DEPARTMENT OF ECEPage 30