Analog and digital communication 2 marks

SUDHARSAN ENGINEERING COLLEGE,
Sathiyamangalam
DEPARTMENT OF COMPUTER SCIENCE AND ENGINEERING
Subject : Analog and Digital Communications Subject Code : CS 2204
Academic Year : 2013 - 2014 Semester : III
Name of the Faculty :G.Dinesh Kumar Question Bank with Answers
1.A.1 Consider an AM signal x(t) =2cos(2π f c t) + 0.5cos(2 π f m t) cos(2 π f c t) .Find the
modulation index used to generate the signal (Nov/Dec 2011,R2008)
General formula for AM
x (t)
m
Given:
Em
Ec
m
=
=
=
=
=
Ec sin (2π f c t) + Em sin (2 π f m t) sin (2 π f c t)
2cos(2π f c t) + 0.5cos(2 π f m t) cos(2 π f c t)
Em / Ec
1.A.2 Draw the frequency spectrum and mention the bandwidth of AM signal.
(Apr/May 2011, R2008)
=
2v
0.5 v
0.250.5 / 2 =
Band width of AM = Max freq- Min freq
= (fc+fm)-(fc-fm)
= 2 fm
1.A.3 For an AM DSBFC modulator with a carrier frequency of 100 KHz and maximum
modulating signal frequency of 5 KHz , Determine Upper and lower side band
frequency and the bandwidth( Nov/Dec 2010 ,R 2008)
UNIT I
PART A
1) AM Band Width calculation( Nov/Dec 2011-R2010 , Apr/May 2011-R2008 , Nov/Dec 2010-
R2008 , April/May 2010)
www.Vidyarthiplus.com

Carrier signal Frequency (fc)
Modulating signal Frequency (fm)
Formula:
Upper Side band Frequency
=
=
100 KHz
5 KHz
Lower side band Frequency
=
=
=
=
=
=
fc +fm
(100 + 5) KHz
105 KHz
fc -fm
100- 5
95 KHz
1.A.4 In a Amplitude modulation modulation system, the carrier frequency is Fc= 100k
Hz. The maximum frequency of the signal is 5 K Hz. Determine the lower and upper
side bands and band width of AM signal.(April/May 2010,R2004)
Given:
Carrier frequency is fc = 100k Hz
Modulating signal frequency (fm) = 5 K Hz
Formula:
Lower side band= fc - fm= 100K - 5K= 95 K Hz
Upper side band
Band width of AM signal
2) Conversion (Nov/Dec 2011-R2010, April/May 2008)
:
= fc + fm
= 2 fm
/
= 100K + 5K
= 2 x5k
=105 k Hz
= 10k Hz
1.A.5 Draw the block diagram of FM signal generator that uses phase modulator in it.( Nov/Dec
2011-R2010)

1

We can generate FM wave by applying the integrated version of x (t) to a phase modulator as shown
in the diagram
1.A.6 How will you convert a frequency modulator into a phase modulator?
(April/May 2008)
A frequency modulator is a circuit in which the carrier is varied in such a way that its
instantaneous phase is proportional to the integral of the modulating signal. Therefore,
with a frequency modulator, if the modulating signal v (t) is differentiated prior to being
applied to the modulator , the instantaneous phase deviation is proportional to the integral
of v(t). An FM modulator that is preceded by a differentiator produces an output wave in
which the phase deviation is proportional to the modulating signal and is, equivalent to a
phase modulator.
PM modulator = differentiator followed by an FM modulator.
3) AM power distribution
(Apr/May 2011, R2008, May/June 2007/R-2004, Nov ‘2006 CSE, AU Nov/Dec 2003, Nov’
2003 CSE)
1.A.7 In an AM transmitter, the carrier power is 10 kW and the modulation index is 0.5. Calculate
the total RF power delivered(Apr/May 2011, R2008)
Given: Pc =10KW and m
Pt = Pc (1 + m2/2)
1.A.8 Determine the relation between the output power of an AM transmitter and depth of
modulation and plot it as a graph for values of modulation index from zero to
maximum.(May/June 2007/R-2004)
In any electrical circuit , the power dissipated is equal to the voltage squared
divided by the resistance .Thus the average power dissipated in a load by an un
modulated carrier is equal to RMS carrier voltage squared divided by the load resistance
.Mathematically power in an un modulated carrier is
Pt = 10K * (1 + .52/2)
Pt = 11250 W
= 0.5
Pc= Ec2/2R.
Pc-Carrier power (watts)
Ec-Peak carrier voltage (volts)
R-Load resistance (ohms)
Upper and lower side band power are expressed as
2

PUSB=PLSB= m2PC/4.
Total power distribution Pt= Pc+ PUSB+ PLSB
Pt = Pc (1+ m2/2).
1.A.9 Determine the powers of the carrier, upper and lower sidebands for an AM DSB-SC
with peak unmodulated carrier voltage of Vc = 20 V and a load resistance RL = 20 Ω.
Assume the modulation index m to be 0.6.(Nov ‘2006 CSE)
2
Pc = VC / RL
= 202 / 20 = 20 watts
Pt = Pc (1 + m2/2)
Pt
Pt
= 20 * (1 + .62/2)
= 23.6 W
1.A.10 A transmitter supplies 8 KW to the antenna when un-modulated. Determine the
total power radiated when modulated to 70%.(AU Nov/Dec 2003)
2
Pt = Pc (1 + m /2)
Pc = total power of the un-modulated carrier
Pt = total power of the modulated signal
1.A.11 A 400W carrier is modulated to a depth of 75 %. Calculate the total power in the
modulated wave(Nov’ 2003 CSE)
2
Pt = Pc (1 + m /2)
Pt = 8 *103 (1 + (0.7)2/2)
= 9.96 KW
Pt = 400 * (1 + .752/2)
Pt = 512.5 W
4) FM Bandwidth and modulation index (Nov/Dec 2011,R2008,Nov/Dec 2010-R 2008,
April/May 2010, Nov/Dec 2009, May/June 2009 , April/May 2008, Nov/Dec‘2006 CSE
,Nov’ 2004 CSE)
2(∆f + fm)
∆f/fm
Band width
Modulation index
=
=
3

1.A.12 State Carson’s rule ( Nov/Dec 2010 ,R 2008)
Carson's bandwidth rule is expressed by the relation
Bandwidth requirement = 2(Δf + fm),
Δf is the peak frequency deviation, and fm is the highest frequency in the
modulating signal.
For example, an FM signal with 5 kHz, peak deviation, and a maximum audio
frequency of 3 kHz, would require an approximate bandwidth 2(5+3) = 16 kHz.
1.A.13 The maximum frequency deviation in an FM is 10 K Hzand signal frequency is
10KHz. Find out the bandwidth using carson’s rule and the modulation index.
(April/May 2010, R2004)
Maximum frequency deviation (∆f) = 10 k Hz
Modulating signal frequency (fm)= 10 K Hz
Formula:
= 2(10K + 10K)
= 40K Hz.
Modulation index (m)
1.A.14 What is the required bandwidth for FM signal, in terms of frequency deviation?
(Nov/Dec 2009, R2004)
Where
=10K/10K
= 1 unit less
Bandwidth=2(∆f+fm)
∆f is the frequency deviation
fm is the modulating frequency
1.A.15 In FM as the modulation index increases, the required bandwidth also increases.
Why? (OR) The required bandwidth for FM transmission depends upon the
modulation index-justify.(May/June 2009 R-2004, Nov/Dec‘2006 CSE)
Theoretically the bandwidth of the FM is infinite. Practically the bandwidth is given by
B.W = 2 fm * no of sidebands
When the modulation index increases the number of sidebands also increases so the
bandwidth also increases.
4

1.A.16 A 20 MHz carrier is frequency modulated by a sinusoidal signal such that the peak
frequency deviation is 100 KHz. Determine the modulation index and approximate
bandwidth of the FM signal if the frequency of the modulating signal is 50 kHz.
(April/May 2008)
Bandwidth = 2(100K +50K)
= 300 KHz
Modulation index (m) =
=
=
100K/50K
2 Unit less
1.A.17 What is the bandwidth required for an FM signal in which the modulating
frequency is 2 KHz and maximum deviation is 10 KHz?(Nov’ 2004 CSE)
Given:
Δφ
B
= 10 KHz and fm = 2 KHz
= 2 * (10 KHz + 2 KHz)
5) AM wave form
1.A.18 Draw the waveform of AM signal and DSB-SC signal
= 24 KHz
(Nov/Dec 2009 ,R2004,Nov/Dec 2008)
(Nov/Dec 2009 ,R2004)
.
5

1.A.19 Draw The Amplitude Modulation Wave Forms With Modulation Index (M) =1,
M<1, M>1.(Nov/Dec 2008,R2004)
M>1
M= 1
6) AM DSBSC and SSBSC(May/June 2009 R-2004, Nov/Dec 2007 )
1.A.20 Write the advantages of SSB-SC-AM.(or) What is the advantage of suppressed
carrier modulation? (or) What are the advantages of SSB-SC modulation?
(May/June 2009 R-2004, Nov/Dec 2007 )
(i)The power of the suppressed carrier and sideband is saved. Hence transmitter
power requirement in SSB is reduced.
(ii)Because of narrow bandwidth of SSB, the effect of noise at the receiver
circuits is reduced. This gives better quality of reception in SSB.
(iii) Fading effect is absent because of SSB. Fading effect arises at the receiver
circuits because of two sidebands and carrier interfere with each other at the
receiver.
(iv) SSB-SC AM requires half of the bandwidth of DSB-SC and considerably less
power is transmitted in SSB-AM transmission.
1.A.21 Compare AM with DSB-SC and SSB-SC.
i.Double sideband suppressed carrier (DSB-SC) is more efficient in
transmitting power but the bandwidth remains the same as DSB-SC.
ii.Single sideband suppressed carrier (SSB-SC) has suppressed carrier and
one of the sidebands removed, hence it requires half of the bandwidth of
M<1
6

DSB-SC and moreover the waveform is not an envelope but it is simply
a sine wave at a single frequency equal to the carrier frequency plus the
modulating signal frequency.
1.A.22 Comparison(April/May 2009 -R-2004, Nov/Dec 2007, May/June 2007, May’ 2006)
a) What is direct and indirect frequency modulation?(April/May 2009 ,R-2004)
Direct FM:
In this type of modulation the frequency of carrier is varied directly by the
modulating signal. An instantaneous frequency deviation is directly proportional
to the amplitude of the modulating signal.
Indirect FM:
FM is obtained by phase modulation of the carrier .Instantaneous phase of the
carrier is directly proportional to the amplitude of the modulating signal.
1.A.23 List the major advantages of angle modulation over amplitude modulation:
(May/June 2007/R-2004)
An important feature of angle modulation is that, it can provide better
discrimination against noise and interference than amplitude modulation. This
improvement in performance is achieved at the expense of increased transmission
bandwidth. Such a trade-off is not possible with amplitude modulation.
1.A.24 Differentiate between an analog signal and a digital signal.(May’ 2006 CSE)
Analog signal is the signal that is defined continuously with respect to time.
Digital signal is the signal that is defined at discrete instants of time.
1.A.25 Distinguish between Narrowband FM and Wideband FM.
(May’ 2005 CSE)
The Bandwidth of an angle modulated wave is a function of the modulating signal
frequency and modulation index. With angle modulation, multiple sets of sidebands are
produced. And consequently the bandwidth is significantly wider than that of an
amplitude modulated wave with the same modulating signal. Angle modulated
waveforms are generally classified as low, medium and high index. In low index angle
modulation most of the information is carried by the first set of sidebands and minimum
bandwidth is required. Hence it is called as narrowband FM. For low index the
modulation index is less than 1 and high index the modulation index is greater than 10.
For low index modulation the frequency spectrum is approximated to double side band
AM, and the minimum bandwidth required is approximated by
B = 2fm Hz
7

8) Modulation index and percent modulation index
(April/may 2008, Nov/Dec 2007, Nov ‘2006 CSE, May’ 2006 CSE)
1.A.26 Define coefficient of modulation and percent modulation for an AM system.
(April/may 2008)
The ratio of maximum amplitude of modulating signal to maximum amplitude of
carrier signal is called modulation index, i.e,
Modulation index, m=Em/Ec
When modulation index is expressed in percentage, it is called Percentage
modulation
1.A.27 What are low level and high level modulation? (Nov/Dec 2007, Nov ‘2006 CSE)
What are low level and high level modulation and how are they related to
bandwidth in angle modulation?
In low level modulation the modulation index is less than 1 and in high level
modulation the modulation index is greater than 10.
1.A.28 Define modulation depth for a FM system and highlight its impact on the spectral
occupation.(May’ 2006 CSE)
For a frequency modulated carrier, the modulation index is directly proportional
to the amplitude and inversely proportional to the frequency of the modulating signal.
The modulation index is mathematically expressed asm = KVm / ωm where K is
deviation sensitivity in radians per volt and Vm in volts and ωm in radians. Here
modulation index is inversely proportional to the frequency
For a low modulation index the bandwidth is small and for a high modulation
index the bandwidth required is high and hence it occupies more spectral space.
1.A.29 The antenna current of an AM transmitter is 8A when only carrier is sent. It
increases to 8.93A when the carrier is modulated by a single sine wave. Find the
percentage modulation.
It= Ic √ 1 + m2/2
Given:
It
Ic
m2
m
= 8.93A,
= 8A, m =?
= ((It / Ic)2-1)*2
= 70.14%
8

1.A.30 Define modulation index for FM and PM.
i) For PM the modulation index is directly proportional to the amplitude of
the modulating signal
m = KVm, where Vm is the peak amplitude of modulating signal and
K is the deviation sensitivity.
ii) For FM the modulation index is directly proportional to the amplitude of
the modulating signal and inversely proportional to the frequency of the
modulating signal the modulation index is mathematically expressed as
m = KVm / fm
Vm is the peak amplitude of modulating signal
fm is the maximum frequency of the modulating signal.
9) Modulation `(Nov/Dec 2007, May/June 2007/R-2001, AUApr/MAY 2004)
1.A.31 What is the need for modulation?(Nov/Dec 2007)
i) To increase the bandwidth of the signal
ii) To multiplex the signals
iii)To reduce the interference made when we transmit the
signals with nearly same frequency in the audio frequency range (20-
20k)Hz.
iv) To favor the complexity of the transmission system
1.A.32 Define frequency modulation and mention its advantages :
(May/June 2007/R-2001)
Frequency modulation is the process of varying the frequency of a constant
amplitude carrier with respect to the instantaneous amplitude of modulating signals
Advantages:
1. Noise immunity
2. Improved SNR
3. Power utilization and efficiency.
1.A.33 Define AM.(AUApr/MAY 2004)
Amplitude modulation is the process of changing the amplitude of a relatively
high frequency carrier signal in proportion with the instantaneous value (amplitude) of
the modulating signal.
1.A.34 Mention the disadvantages of angle modulation.
i)
ii)
High quality angle modulation produces many side frequencies hence
the more bandwidth is necessary for transmission.
PM and FM transmitters and receivers are more complex and costlier
than AM transmitters and receivers.
B = 2 f Hz
9

PART B
1. AM Modulation
(Nov/Dec 2011-R2010 ,Apr/May 2011-R2008, Nov/Dec 2010, R2008, April/May
2008, May/June 2007/R-2001, May’ 2006)
a) Write the equation of single tone amplitude modulation. Derive the Fourier transform of the
signal and plot the signal in time domain and frequency domain for the modulation index
(i) Lesser than unity,
(ii) Equal to unity
(iii) Greater than unity Also calculate the power of each frequency component (Nov/Dec 2011-
R2010)
b) Explain the principle of AM modulation with mathematical analysis. Also draw the AM
wave and explain its power distribution.(Apr/May 2011-R2008)
(Or)
Obtain a relationship between carrier and side band powers in an AM DSBFC wave and
explain how power distribution takes place in AM DSB FC system.
(Nov/Dec 2010, R2008)
(Or)
Define amplitude modulation. Derive the relation between the total transmitted power
and carrier power in an AM system when several frequencies simultaneously
modulate a carrier.(April/May 2008)
(Or)
Explain the principle of amplitude modulation.(May/June 2007/R-2001)
(Or)
Explain the generation and detection of amplitude modulated signals using circuit
diagrams. How is the transmitted power distributed over carrier and side bands in an AM
signal? Explain.(May’ 2006 CSE)
c) For amplitude modulation prove the following statements: (i) Power of the carrier
Pc is unaffected by the modulation process. (ii) Total power in an AM envelope
increases with modulation index. (iii) Modulation index m = Vmax – Vmin / Vmax + Vmin
draw the modulated waveform and show Vmax and Vmin.(Nov’2006 CSE)
d) Define AM. Derive an expression for the AM wave.(AU Apr/May 2005)
e) What is AM? Explain with relevant expressions the Square law modulation.
(AU Apr/May 2005)
f) (i) Define Amplitude Modulation. Derive an expression for the AM wave. (ii) Derive
an expression for the Quantization error.(May’ 2005 CSE)
2. FM (Nov/Dec 2010-R2008, Nov/Dec 2008, Nov/Dec 2003)
a) Briefly discuss the method of deriving wideband FM signal. (Nov/Dec 2010-R2008)
b) Explain the method of generating narrow band FM signal.
(Nov/Dec 2008, Nov/Dec 2003)
10

3. Carson’s rule. (Nov/Dec 2011-R2010 , Apr/May 2011, R2008)
a) State and apply Carson’s rule to calculate the bandwidth occupied by a 3 kHz
message signal frequency modulated with modulation index = 5. (Nov/Dec 2011-R2010)
b) Explain the Bandwidth requirement for FM and define Carson’s rule.
(Apr/May 2011, R2008)
4. FM and PM
(APR /MAY 2011-R2008, Nov/Dec 2010-R 2008, April/May 2010, Nov/Dec 2009)
a) Distinguish between frequency and phase modulation.( APR /MAY 2011-R2008)
(or)
Compare FM with PM.(AU Nov/Dec 2008)
b) Define modulation index for FM and PM and obtain the relation between
modulation index and modulating signal for FM and PM. ( Nov/Dec 2010,R 2008)
c) Distinguish between FM and PM by giving its mathematical analysis (April/May 2010)
d) Define FM and PM modulation. Write down their equation. Describe suitable mechanism that can
produce PM from FM modulator.(Nov/Dec 2009)
5. Problems in AM Modulation
(Apr/May 2011, R2008, May/June 2009, April/May 2008, Nov/Dec2007)
a) The output of a AM transmitter is given by(Apr/May 2011, R2008)
7
Um (t ) = 500 (1 + 0.4 sin 3140t ) sin 6.28 ×10 t .
Calculate
(1) Carrier frequency
(2) Modulating frequency.
(3) Modulation index
(4) Carrier power if the load is 600
(5) Total power.
b) For an AM DSBFC wave with a peak unmodulated carrier voltage Vc=10Vp, a load
resistance RL =10Ω, and a modulation coefficient m=1. determine
(i)Powers of the carrier, upper and lower side bands,
(ii)Total power of the modulated wave
(iii) Total side band power
(iv)Draw the power spectrum
(May/June 2009, April/May 2008)
c) Draw the power spectrum
A carrier frequency of 8MHz with peak value of 6 V is amplitude modulated by a
kHz sine wave signal with amplitude 4 volts. Determine its modulation index and
draw the amplitude spectrum.(Nov/Dec2007)
10
11

d) For an AM DSBFC wave with a peak unmodulated carrier voltage Vc=12Vp, a load
resistance RL =12Ω, and a modulation coefficient m=1. determine
(1) Powers of the carrier, upper and lower side bands,
(2) Total power of the modulated wave
(3) Total side band power
e) (i) Discuss about the sets of side bands produced when a carrier is frequency modulated
by a single frequency sinusoid.
(ii) In an AM Modulator ,500kHz carrier of amplitude 20 V is modulated by 10 K Hz
modulating signal which causes a change in the output wave of +7.5 V. Determine
(1)
(2)
(3)
(4)
Upper and Lower side band frequencies
Modulation index
Peak amplitude of upper and lower side frequency
Maximum and minimum amplitudes of envelope
6. AM frequency spectrum
(Nov/Dec 2010-R2008, April/May 2010,Nov/Dec 2009, Nov/Dec 2003)
a) Derive the expression for a Amplitude Modulated wave and draw
its spectrum.(Nov/Dec 2010, R2008)
(or)
Draw the amplitude-modulated wave equation and explain each term with the help of
frequency spectrum.(AU Nov/Dec 2003)
b) Derive the relationship between the voltage amplitudes of the side band frequencies and
the carrier and draw the frequency spectrum.(April/May 2010)
c) Derive expression for AM wave. Define modulation index and express its value in terms
of maximum and minimum voltage values of signal. Draw the spectrum and time domain
signal of AM wave.(Nov/Dec 2009)
7. Compare the advantages and disadvantages of angle modulation with amplitude
modulation.
12

UNIT II
PART A
1. Shannon’s channel capacity
(Nov/Dec 2011 – R2010, May/June 2009 R-2004, AUNov/Dec
2004, Nov/Dec 2003)
2.A.1 Find the capacity of a channel having 50 kHz bandwidth and produces SNR of 1023 at the
output ( Nov/Dec 2011 – R2010)
I = 3.32 B log10 (1+S/N)
I = Information capacity (bps) B =Bandwidth (Hz) S/N = Signal to Noise ratio.
Given:
B
S/N
I
=
=
=
=
50Hz
1023
3.32 * 50 log10 (1+1023)
500
2.A.2 State Shannon’s fundamental theorem of information theory.
(May/June 2009 R-2004, AU Nov/Dec 2003)
Shannon derived a relationship between Information capacity, Bandwidth and
signal to noise ratio of a communication channel. The higher the signal to noise ratio the
better the performance and higher the information capacity
I = Information capacity (bps) B =Bandwidth (Hz) S/N = Signal to Noise ratio.
2.A.3 Calculate the capacity of a standard 4 KHz telephone channel with a 32 dB signal to
noise ratio.(AU Nov/Dec 2004)
Channel capacity C = B log2 (1 + S/N)
Given: B = 4 KHz, S/N = 32 dB
32 = 10 log10(S/N)
S/N = 10^3.2 = 1584.89
C = 4*103*(1 + 1584.89) = 6.34Mbps
I = B log2 (1 + S/N)
I = 3.32 B log10 (1+S/N)
13

2.A.4 Distinguish between power efficiency and spectral efficiency. (Nov/Dec 2011-R2010)
SI.No
1
Power efficiency Spectral efficiency
The ratio of power obtain to the power Spectralefficiency, spectrum
given that is called power efficiencyefficiency or bandwidth efficiency refers to
the information rate that can be transmitted
over a given bandwidth in a specific
communication system
3. Waveform based problem.
(Apr/May 2011-R2008, April/May 2010, April/May 2005, Nov/Dec 2004)
2.A.5 Draw ASK and PSK waveforms for a data stream 110101.(Apr/May 2011-R2008)
a) Draw ASK and FSK signals for the binary signal s(t) = 1011001 (April/May 2010)
14

2.A.7 Define PSK and draw the PSK waveform for the data 10110.(AU April/May
2005)
In PSK (Phase Shift Keying) the phase of the sinusoidal carrier is changed according to the data
bit to be transmitted. BPSK is the simplest form of PSK. The carrier phase is changed between 0
and 180.
2.A.8 Draw the waveforms for FSK and PSK modulation. (AU Nov/Dec 2004)
PSK Waveforms
15

2.A.9 What are the advantages of QPSK? (April/May 2011-R2008,April/May 2010)
The advantage of the Quadrature Phase Shift Keying (QPSK) modulation versus the Binary
Phase Shift Keying (BPSK) one is well known. It is the possibility to transmit in the same
frequency band twice more information, while the number of errors and the Eb/No relation are
the same. It required minimum Bandwidth as in BPSK
2.A.10 Define channel capacity (or) Define information capacity and bit rate
(Nov/Dec 2010 - R2008, Nov/Dec2009, AU Apr/MAY 2004)
Channel capacity:
Channel capacity is a measure of how much information can be propagate through the
communication system and it is function of bandwidth and transmission time.
Shannon theorem states that if the bandwidth of the channel is B Hz and (S/N) is the
signal to noise ratio then the channel capacity is given by
I (or) C = B log2 (1 + S/N) bps
Information Capacity:
Information Capacity is represents the number of symbols that can be carried through a
system that is called information capacity
It is a measure of how much information can be propagated through communication
system and is function of bandwidth and transmission time.
Where
I=Bxt
B= bandwidth (Hz)
t = Transmission time (sec)
C (or) I = Channel capacity or Information capacity
Bit rate is the number of bits transmitted per second or Rate of change of digital
information signal.
16

6. Phasor Diagram(Nov/Dec 2009,Nov/Dec 2008)
2.A.11 Draw the Phasor diagram of QPSK (Nov/Dec 2009)
2.A.12 What are the advantages of BPSK? (OR)What are the advantages of PSK over FSK
(May/June 2009 R-2004, Nov ‘2006 CSE)
BPSK has a bandwidth which is lower than that of BFSK signal.
BPSK has best performance in presence of noise. It has good noise immunity.
2.A.13. Bandwidth(April/May 2008,Nov/Dec 2007)
a) Determine the bandwidth and baud for the FSK signal with a mark frequency of
49 KHz and a space frequency of 51kHz and a bit rate of 2kbps(April/May 2008)
Bandwidth= ( fs - fm ) +2fb
= 8 KHz
= fb
=2
= (51k - 49 K) +2(2k)
Baud
2.A.14 Why does FSK require more bandwidth when the modulation index is increased?
(Nov/Dec 2007)
Bandwidth-= 2(∆f+fb)
Where,
∆f =mffm
17

Frequency deviation depends on modulation index, when modulation index increases
bandwidth also increases.
9. Comparison
2.A.15 Write the differences between PSK and FSK.
Frequency shift keying(FSK)
1)The binary information signal
modulates the frequency of the analog
signal
2) The output shifts between two
frequencies mark frequency for binary
‘1’ and space frequency for binary ‘0’
(Nov/Dec 2008, April/May 2008)
Phase shift keying (PSK)
modulates the Phase of the analog signal
2) There are limited number of output
phases possible,The input binary
information is encoded into groups of
bits before modulating the carrier.
3) Minimum bandwidth = fb3) Minimum bandwidth = 4 fb
2.A.16 Compare ASK and FSK.
Amplitude shift keying(ASK)
1) The binary information
signal directly modulates
the amplitude of an analog
carrier.
2) The modulated wave will be
Acoswt when the binary
input is ‘1’ and zero if the
binary input is ‘0’.It is also
called as on-off keying.
(AU Nov/Dec 2003)
Frequency shift keying(FSK)
modulates the frequency of the analog
signal
2) The output shifts between two
frequencies mark frequency for binary
‘1’ and space frequency for binary ‘0’
3) Minimum bandwidth = 2 fb 3) Minimum bandwidth = 4 fb
2.A.17 What is digital modulation?(May/June 2007/R-2004)
Digital modulation is the process changing the modulating signal in the form of
binary sequence of 1’s and 0’s .The communication channel used for pass band data
transmission may be a microwave radio link , a satellite channel or optical link . The
modulation process making the transmission possible involves switching, the amplitude
,frequency or phase of a sinusoidal carrier in accordance with incoming data .
18

2.A.18 List the effects of M in M ary digital modulation technique :(May/June 2007)
QPSK offers best tradeoff between power and bandwidth requirement .For M>8
,power requirement becomes excessive . M ary PSK and M ary QAM have similar spectral
and bandwidth characteristics. For M>4, the two schemes have different signal
constellations. For Mary PSK, the constellation is circular .For Mary QAM it is rectangular .
A comparison of these two constellations reveals that the distance between the message
points of Mary PSK is smaller than the distance between the message points of Mary QAM
2.A.19 Define bit rate and baud rate.(Nov ‘2006 CSE)
Baud rate is rate of change of signal on the transmission medium after encoding and
modulation have occurred.
Bit rate is the rate of bits per second.
2.A.20 . Differentiate between Coherent and Non Coherent detection scheme (May’ 2006 CSE)
In Coherent receiver the frequencies generated in the receiver and used for
demodulation are synchronized to oscillator frequencies generated in the transmitter. So
the receiver must have some means of recovering the received carrier and Synchronizing
to it. With Non-coherent receivers, either no frequencies are generated in the receiver or
the frequencies used for demodulation are completely independent from the transmitter
carrier frequency.
2.A.21. What are the major transmission impairments suffered by signal on the transmission
media.(May’ 2006 CSE)
The major transmission impairments suffered by the signal on the transmission media is due to
additive white Gaussian noise and impulse noise which results in bit errors.
2.A.22. State sampling theorem(AU April/May 2005,Dec 2004)
The Sampling theorem states that the sampling rate should be equal to twice the highest
audio frequency.
Fs > 2Fm
Fs = Sampling rate. Fm = Maximum audio frequency
2.A.23 Define the term aliasing.(AU April/May 2005)
When the sampling rate is lesser than twice the maximum audio frequency distortion occurs in
the signal, this distortion is called as aliasing or fold over distortion.
2.A.24. What is ASK?(Nov ‘2004 CSE)
ASK is called amplitude shift keying the amplitude of the carrier is varied according to
the binary symbol logic 1 represents a signal and no signal represents logic 0.
2.A.25. What is the relation between bit rate and baud for a FSK system
Consider the time of one bit = tb
19

So the bit rate is
Number of bits in the symbol (N)
Baud rate is
Baud rate
= 1/tb = fb
=1
= Bit rate / Number of bits in the symbol (N)
= fb/1
= Bit rate
PART B
1. QPSK Modulation
(Nov/Dec 2011 – R2010 , Apr/May 2011 -R2008, Nov/Dec 2010-R2008, Nov/Dec
2009,May/June 2009, Nov/Dec 2008, May’ 2006, Nov/Dec 2004)
a) Draw the block diagram of QPSK modulator, demodulator and explain the principle of
operation. (8)( Nov/Dec 2011 – R2010)
b) Discuss in detail the operation of QPSK modulator and demodulator with its phasor
diagram.(Apr/May 2011 ,R2008)
c) Describe with neat diagram, the operation of a QPSK modulator. Draw its phasor and
constellation diagram.(Nov/Dec 2010, R2008)
or
Explain the QPSK modulation scheme with suitable transmitter and receiver block
diagrams. Also derive the average probability of error in the presence of AWGN.
(May/June 2009, Nov/Dec 2008, May’ 2006 )
or
Explain the QPSK scheme with constellation diagram.(AU Nov/Dec 2004)
d) Explain the bandwidth considerations of QPSK system (Nov/Dec 2010,R2008)
e) Draw the block diagram of QPSK receiver and explain its operation. For QPSK modulator ,
construct the truth table, phasor diagram and constellation diagram.(Nov/Dec 2009)
f) Draw the constellation diagram for QPSK signaling scheme. And for the following
binary sequence sketch the waveforms for the in phase and Quadrature components,
QPSK signal 1100100010.(AU Nov/Dec 2003)
2. Quadrature Amplitude modulation ( Nov/Dec 2011-R2010,Apr/May 2011 -R2008,
Nov/Dec 2008m )
a) With neat constellation diagram, explain the operation of QAM transmitter. List out its
merits over PSK. (8)(Nov/Dec 2011-R2010)
b) Write note on Quadrature amplitude modulation. (Apr/May 2011 ,R2008)
c) Illustrate the concept of 8 QAM transmitter with truth table
(APR/MAY 2010)
d) Explain Quadrature amplitude modulation with the help of relevant diagrams
(Nov/Dec 2008)
20

3. FSK Modulation
(Nov/Dec 2011-R2010 ,Nov/Dec 2009, May/June 2009, Nov/Dec 2007, May’ 2006
CSE, Nov’ 2006 CSE)
a) Explain the operation of the binary FSK modulator and demodulator with its constellation
diagram. (10)(Nov/Dec 2011-R2010)
b) Draw the block diagram of FSK receiver and explain the operation. Determine (i) Peak frequency
deviation (ii) minimum bandwidth (iii) baud for FSK signal with a mark frequency of 49 KHz,
space frequency of 51 KHz, and input bit rate of 2 kbps.
(Nov/Dec 2009)
c) Explain the operation of FSK transmitter and receiver and discuss about the bandwidth
requirements of FSK signals.(May/June 2009, May’ 2006 CSE)
d) (i) Derive the probability of symbol error Pe for a coherent binary FSK and show that Pe
= ½ erfc (√Eb/ 2No). (ii) Draw the block diagram of FSK receiver and explain the
operation(Nov/Dec 2007,Nov’ 2006 CSE)
e) With neat block diagram explain the coherent binary FSK transmitter and receiver.
(AU Nov/Dec 2006)
4. Carrier recovery
(Nov/Dec 2011-R2010, Apr/May 2011 -R2008, APR/MAY 2010, Nov/Dec 2010-
R2008, May’ 2006 CSE)
a) Brief the carrier recovery with squaring loop. (6) (Nov/Dec 2011-R2010)
b) Describe the operation of Costas loop receiver. (Apr/May 2011 ,R2008)
c) What is the need for carrier recovery? Explain the Costas loop method ofcarrier
recovery.(APR/MAY 2010)
Or
What is carrier recovery? Discuss how carrier recovery is achieved by the
squaring loop and Costas loop circuits(Nov/Dec 2010 ,R2008)
d) Write notes on the following: (i) Costos loop method of carrier frequency. (ii)
Differential PSK modulator.(May’ 2006 CSE)
5. Binary Phase shift keying(Apr/May 2011 -R2008, APR/MAY 2010, Nov’ 2006
CSE, May’ 2005 CSE, AU Apr/MAY 2004)
a) Explain the generation and detection of binary phase shift keying.
(Apr/May 2011, R2008)
b) What is known as Binary Phase shift keying? Discuss in detail the BPSK transmitter and
Receiver and also obtain the minimum double sided Nyquist bandwidth.
(APR/MAY 2010)
c) Explain the working of BPSK transmitter and receiver using balanced modulator. How
M-ary encoding and modulation achieved in PSK? Draw a QPSK transmitter block
diagram and explain the working principle.
(Nov’ 2006 CSE)
d) Write an expression for the BFSK and explain the spectrum of BFSK.
(May’ 2005 CSE)
21

e) Draw the block diagram of binary PSK system and explain with signal space diagram.
(AU Apr/MAY 2004)
6. Channel capacity(May/June 2007/R-2001, AU Nov/Dec 2004)
Define channel capacity .Explain Shannon’s channel coding theorem for discrete
memory less channel.(May/June 2007/R-2001)
Or
Derive an expression for channel capacity.(AU Nov/Dec 2004)
7. With neat diagram explain the generation of a DPSK signal and recovering data from
DPSK signal .(May/June 2007/R-2004)
8. Explain differentially encoded PSK system with necessary diagrams.
(AU Apr/May 2005)
9. (i) Explain the terms Average Information and Information Rate. (ii) An analog signal
is band limited 4000Hz, sampled at Nyquist rate and samples are quantized into 4
levels. The levels Q1, Q2, Q3, Q4 are assumed to be independent and occur with
probabilities P1=P4=1/8., P2=P3=3/8, find the information rate of the source.
(May’ 2005 CSE)
10. What is ASK? Draw the waveform (AU Nov/Dec 2004)
22

UNIT III
PART A
1. PCM modulation (Nov/Dec 2011-R2010, May/June 2009,R-2004, Nov/Dec 2007)
3.A.1 A PCM system uses sampling frequency of 16 k. samples/s. Then, find out the maximum
frequency of the signal upto which the signal can be perfectly reconstructed ( Nov/Dec 2011-
R2010 )
Given:
≥2fs
=16kHz
=2*16KHz
=32Khz
So the signal frequency should be greater than or equal to 32 KHz
fm
fs
fm
3.A.2 Explain why the quantization noise cannot be removed completely in PCM ?How do
you reduce this noise?(May/June 2009,R-2004)
A quantizer maps the input random variable of continuous amplitude in to
discrete random variable. For a uniform quantizer ,the quntization error have its sample
values bounded by –Δ/2 ≤ q ≤ Δ/2. If the step size Δ is too small then the quantization
error is uniformly distributed random variable. And hence the quntization error can be
reduced by reducing the value of step size Δ.
Variance of quantization error σQ2= Δ2/12.
3.A.3 In PCM how signal to noise ratio is related to number of bits used in the system?
(Nov/Dec 2007)
SQR(DB) = 10 log {(v2/R)/(q2/12)/R}
R-Resistance
v-rms signal voltage
q-quantization interval
v2/R-average signal power
q2/12-average quantization noise power
2. ISI (Nov /Dec 2011- R2010, Nov/Dec 2007)
3.A.4 What is IST? What is the reason for it occurance? (Nov /Dec 2011- R2010)
Inter Symbol Interference (ISI) is a form of distortion of a signal in which
one symbol interferes with subsequent symbols. This is an unwanted phenomenon as the previous
symbols have similar effect as noise, thus making the communication less reliable. ISI is usually
23

caused by multipath propagation or the inherent non-linear frequency response of a channel
causing successive symbols to "blur" together.
3.A.5 What causes ISI in digital transmission? (Nov/Dec 2007)
a)
b)
c)
d)
Timing inaccuracies
Insufficient bandwidth
Amplitude distortion
Phase distortion
3. Sampling theorem(Apr/May 2011-R2008, April/May 2010)
3.A.6 Define Nyquist sampling theorem.(Apr/May 2011-R2008, April/May 2010)
Nyquist sampling theorem establishes the minimum sampling rate (fs). Each cycle
of the analog input signal (fa) must be sampled at least towice. If fs is less than two times
fa , an impairments called “Aliasing problem or fold over distortion’’.
The minimum Nyquist sampling rate isfs ≥ 2 fa
Where
fs = Minimum Nyquist sampling rate (Hz)
fa = Maximum analog input frequency(Hz)
3.A.7 For the signal m(t) = 3 cos 500лt + 4 sin 1000 лt , Determine the Nyquist sampling
rate.(April/May 2010)
Given:
m(t) = 3 cos 500лt + 4 sin 1000 лt
From the above equation the maximum analog input frequency fa = 1000 Hz .
Hence Nyquist sampling rate fs ≥ 2 fa
fs = 2x 1000 = 2K Hz.
3.A.7 . What is meant by differential pulse code modulation? (Apr/May 2011-R2008)
Differential pulse-code modulation (DPCM): Pulse-code modulation in which an analog
signal is sampled and the difference between the actual value of each sample and its
predicted value, derived from the previous sample or samples, is quantized and converted, by
encoding, to a digital signal.
3.A.8. What are the advantages of digital transmission? (AU Nov/Dec 2010, R 2008)
Following are the Advantages of Digital Transmission over analog transmission.
1. The signal is exact.
For example, Digital formats like PCM (Pulse Code Modulation) can hold the exact
digital format of sounds.
2. Digital signals can checked for errors. Generally, all digital transmission methods use
'head labels' to provide better checking facilities. Head label may contain how many packets
may need to complete the transmission, what is the sequence number of each packet of data,
24

acknowledgment labels etc.
3. A variety of services can afford over one line. For example, IpTV connection can used
to watch cable TV channels while browsing the Internet through a PC using same line. This
line can also used to make a phone call at the same time.
4. Digital data can be compressed and therefore possible to pass over higher
bandwidths. Any digital data; data, image, video, voice can be compressed.
5. More secure. Digital data can be encrypt using an encryption method. Only the dedicated
receiver can decode the message.
6. Supports data integrity. Simple to integrate voice, video and data. Digital transmission
provides easier way to integrate different digital formats.
7. Digital transmission provides higher maximum transmission rates via medium such
as optical fibers. Typically, digital transmissions use less bandwidth
6. Companding(AU Nov/Dec 2010-R 2008 , April/May 2008)
3.A.8 Define companding. (AU Nov/Dec 2010, R 2008)
A process in which compression is followed by expansion; often used for noise reduction
in equipment, in which case compression is applied before noise exposure and expansion after
exposure.
3.A.9 What is µ–law companding?(April/May 2008)
The compression characteristics for µ-law is
Vout = [Vmax ln(1+µVin/Vmax)]/ [ln(1+µ)]
7. Pulse modulation
3.A.10 Draw PWM and PPM
(Nov/Dec 2009, May/June 2007/R-2004, Nov/Dec 2004)
(Nov/Dec 2009)
Pulse Width modulation
25

3.A.11 Define pulse modulation(May/June 2007/R-2004)
In Pulse modulation, some parameter of a pulse train is varied in accordance with
the message signal .In analog pulse modulation a periodic pulse train is used as the carrier
wave and some characteristic feature of each pulse is varied in a continuous manner in
accordance with the corresponding sample value of message signal. In digital pulse
modulation, the message signal is represented in the form that is discrete in both time and
amplitude, thereby permitting its transmission in digital form as a sequence of coded
pulses.
3.A.12 What is PWM?(AU Nov/Dec 2004)
In Pulse Width Modulation PWM) the width of the pulses is varied in accordance
with the modulating signal. The information is represented as width variation. PWM has
more noise immunity than PAM.
8. Eye pattern (Nov/Dec 2009,Nov 2006)
3.A.13 Draw the eye pattern and indicate how ISI is measured from it. (Nov/Dec 2009)
Inter symbol interference
Here
H = ideal vertical opening (cm)
h = degraded vertical opening (cm)
3.A.14 How eye patterns help in measuring ISI?(Nov ‘2006 CSE)
The eye pattern is used to study the effect of ISI in base band digital transmission.
The width of the eye opening defines the interval over which the received wave can be
sampled without error from inter symbol interference. As the ISI increases the eye
opening reduces.
9. Delta modulation
(May/June 2009,R-2004, May/June 2007/R-2004, Nov ‘2006 CSE)
3.A.15 What are the two types of noises present in Delta modulation system?
(May/June 2009,R-2004)
1.Slope overload noise2.Granular noise.
= 20 log (h/H)
26

3.A.16 Distinguish delta modulation & differential pulse code modulation.
(May/June 2007/R-2004)
In delta modulation system one bit (two- level) quantizer is used and prediction
filter of Differential Pulse Code modulation is replaced by, a single delay element. The
principal advantage of delta modulation is the simplicity of its circuitry .In contrast;
differential pulse code modulation employs increased circuit complexity to reduce
channel bandwidth.
3.A.17 Why do you get slope overload error in Delta modulation?(Nov ‘2006 CSE)
Delta modulation is subjected to slope overload distortion. If the slope of the
analog signal x(t) is much higher than that of x1(t) over a long duration then x1(t) will not
be able to follow x(t) at all, this difference between x(t) and x 1(t) is called slope overload
duration.
3.A.18 What is natural sampling? (May/June 2007/R-2001)
Natural sampling is the process when top of the sample pulses retain their natural
shape during the sample interval making it difficult for an ADC to convert the sample to
a PCM code .With natural sampling the frequency spectrum of the sampled output is
different from that of ideal sample. The amplitude of frequency components produced
from narrow finite width sample pulses decreases for the higher harmonics.
1. PCM
(Nov/Dec 2011-R2010 ,Apr/May 2011-R2008, Nov/Dec 2010-R2008,
May/June 2009, Nov/Dec 2008,April/May 2008,Nov/Dec 2007, May/June 2007/R-
2001, May’ 2006 ,Apr/May 2005, Nov/Dec 2004)
a) Draw the block diagram of pulse code modulation system and write Short notes on each
block. (8)(Nov/Dec 2011-R2010)
b) Quantization noise in PCM.(Apr/May 2011-R2008)
c) Draw the block diagram of a PCM transmitter and explain the function of each
block.(Nov/Dec 2010,R2008, Nov/Dec 2007, May/June 2007/R-2001)
or
With neat sketches explain the operation of PCM.
(May/June 2009, April/May 2008, AU Apr/May 2005, AU Nov/Dec 2004)
d) Derive expressions for the quantization noise and signal to noise ratio in a PCM system
using a uniform quantizer. (ii) A sinusoidal signal is to be transmitted using PCM. An
output SNR of 55.8 DB is required. Find the number of representation levels required to
achieve this performance.
PART B
27

(Nov/Dec 2008, May’ 2006 CSE AU Apr/May 2005)
e) In a binary PCM system, the output signal to quantization noise ratio is to be held to a
minimum of 40 dB. Determine the number of required levels, and find the corresponding
out signal to quantization noise ratio.(April/May2008)
2. Delta Modulation
(Nov/Dec 2011-R2010 ,Apr/May 2011-R2008, April/May2010, Nov/Dec 2009,
May/June 2009 R-2004, Nov/Dec 2008, , May/June 2007/R-2001, Nov 2006,May’ 2006
CSE ,AU Apr/May 2005AU Nov/Dec 2004, AU Apr/MAY 2004)
a) Draw the block diagram of delta modulation system and explain. (6) (Nov/Dec 2011-
R2010)
b) With suitable diagram and example, explain the noise associated with DM. (5) (Nov/Dec
2011-R2010)
c) Derive the condition on step size to avoid slope overload error. (5) (Nov/Dec 2011-
R2010)
d) What are the drawbacks of delta modulation? Explain how it is overcome in adaptive
delta modulation. (Apr/May 2011-R2008)
e) Draw the block diagram and describe the operation of a delta modulator. What are its
advantages and disadvantages compared to a PCM system? (Nov /Dec 2010,R2008)
f) Explain in detail the Delta modulation transmitter and Receiver (April/May2010)
g) How does delta modulation differ from PCM? Explain delta modulation transmitter with
help of a block diagram.(Nov/Dec 2009)
h) What are the drawbacks of delta modulation and how are they overcome in adaptive delta
modulation? Explain with the help of a neat block diagram.
(May/June 2009 R-2004)
i) With suitable figures explain a Delta Modulation system. Also discuss the different types
of quantization noise that can arise in the system.
(Nov/Dec 2008, May’ 2006 CSE)
j) (i) What is uniform and non-uniform quantization? Derive the SNR ratio for uniform
quantizer. (ii)Draw the block diagram of Delta modulator and explain the operations.
What are its advantages over PCM?(Nov ‘2006 CSE)
k) Explain delta modulation system with block diagram.
(May/June 2007/R-2001, AU Apr/May 2005AU Nov/Dec 2004, AU Apr/MAY 2004)
3. Sampling and quantization (Nov/Dec 2011- R2010 , Nov/Dec 2010, R2008)
a) Explain in details on the criterions need to be satisfied in its sampling and quantization
blocks. (4+4)(Nov/Dec 2011- R2010)
28

b) What are the types of sampling? Explain the operation of the sample and hold circuit.
(Nov/Dec 2010, R2008)
4. Companding
(Apr/May 2011-R2008, May/June 2009, April/May2010, Nov/Dec 2009)
a) What is Companding? What is the need for Companding?(Apr/May 2011-R2008)
b) What is the need for companding? Explain analog and digital companding
(May/June 2009, R-2004)
c) What is called companding? Briefly discuss the analog companding
(April/May2010)
d) What is companding? Explain analog companding process with the help of block diagram.
(Nov/Dec 2009)
5. Compare analog and digital modulation. (Apr/May 2011-R2008)
6. Eye pattern and ISI
(Apr/May 2011-R2008, April/May2010, May/June 2009/R-2004, Nov/Dec
2008, Nov/Dec 2007, Nov’ 2006 CSE ,April/May2008)
a) Explain about Inter symbol interference & Eye pattern (Apr/May 2011-R2008)
b) Discuss the causes of ISI( April/May2010)
c) What causes ISI in the detection process of a baseband digital system? Explain the effects
of ISI. How ISI can be reduced?
(May/June 2009/R-2004, Nov/Dec 2008, Nov/Dec 2007, Nov’ 2006 CSE)
d) Draw the eye diagram and explain its importance in data transmission.
(April/May2008)
7. DPCM
a) Distinguish between different pulse modulation schemes and delta modulation with neat
diagram(May/June 2007/R-2004,May/June 2007/R2001 )
b) Draw the block diagram of typical DPCM system and explain (May’ 2006 CSE)
(May/June 2007/R-2004,May/June 2007/R2001, May’ 2006 CSE)
8. Explain adaptive delta modulation with block diagram.
(May/June 2007/R-2004)
9. Explain Quantization and determine the quantization error of uniform quantizer
(May/June 2007/R-2001)
10. Pulse Modulation(AU Apr/MAY 2004, AU Nov/Dec 2003)
a) Explain with suitable diagrams the generation of PPM signal.
(AU Apr/MAY 2004)
b) What are the types of pulse modulation systems? Explain each in detail.
(AU Nov/Dec 2003)
29

UNIT IV
PART A
4.A.1 List the layers presented in ISO–OSI reference model.
OSI Model
Data unit Layer
7. Application Network process to application
(Nov/Dec 2011-R2010)
Function
Data
Host
layers
6. Presentation
Data representation, encryption and decryption, convert machine dependent data to machine
independent data
5. Session
Segments 4. Transport
Interhost communication, managing sessions between applications
End-to-end connections, reliability and flow control
Packet/Datagram 3. Network
Media
Frame
layers
Bit
2. Data link
1. Physical
Path determination and logical addressing
Physical addressing
4.A.2 What is modem?(Nov/Dec 2011-R2010)
A modem (modulator-demodulator) is a device that modulates an analog carrier signal to
encode digital information, and also demodulates such a carrier signal to decode the transmitted
information. The goal is to produce a signal that can be transmitted easily and decoded to reproduce
the original digital data. Modems can be used over any means of transmitting analog signals,
from light emitting diodes to radio. The most familiar example is a voice band modem that turns
the digital data of a personal computer into modulated electrical signals in the voice frequency range
of a telephone channel.
30

4.A.3 Define Entropy ( Apr/May 2011-R2008)
Entropy is a measure of the uncertainty associated with a random variable. In this
context, the term usually refers to the Shannon entropy, which quantifies the expected value
of the information contained in a message, usually in units such as bits. In this context, a
'message' means a specific realization of the random variable.
4. A.4 Error Control
(Apr/May 2011-R2008, Nov/Dec 2010, R 2008,April/May 2010, Nov/Dec 09,
May’ 2006 CSE, Nov/Dec 2004, Apr/MAY 2004)
4.A.4 What is the need for error control coding?( Apr/May 2011-R2008)
The goal of error control coding is to encode information in such a way that even if the
channel (or storage medium) introduces errors, the receiver can correct the errors and recover
the original transmitted information
.
4.A.5 What is forward error correction? (AU Nov/Dec 2010, R 2008)
It is the type of error correction scheme, where the errors are detected and corrected without
retransmission but by adding the redundant bit to the message before transmission commences.
Hence the redundant bits are used to detect which bit is in error, whenever an error as found.
Limitation:
Forward error correction is limited to one,two or three bit errors.
4.A.6 Which error detection technique is simple and which one is more reliable?
(April/May 2010)
Parity coding is very simple error detection technique.
Cyclic redundancy checking (CRC) that is convolutional coding scheme is the
most reliable one. CRC approximately 99.99% of all transmission errors are detected.
Hence CRC called as Systematic code.
4.A.7 Compare the merits and demerits of error detection and error correction.
(Nov/Dec 09)
Errors introduced by communications faults, noise or other failures into valid
data, especially compressed data were redundancy has been removed as much as
possible, can be detected and/or corrected by introducing redundancy into the data
stream.
Error Detection is the process of monitoring data transmission and determining when
errors have occurred.
Purpose of error detection is not to prevent errors from occurring but to prevent
undetected errors from occurring. This technique neither corrects nor identifies which
bits are in error and they indicate when an error has occurred.
31

Error Correction: codes are used to determine when an error has occurred and which bit
is in error.
4.A.8 Give typical CRC-16 generating circuit (Nov/Dec 09)
4.A.9 What is your understanding of CRC?(May’ 2006 CSE)
CRC is a systematic code for error detection in transmission. A CRC-16 block
check character is the reminder of a binary division process. A data message polynomial
G(x) is divided by a unique generator polynomial function P(x), the quotient is discarded
and the remainder is truncated to a 16 bit sequence and appended to the original message
as a Block Check Sequence ( BCS).
4.A.10 What are the most common error control techniques?(AU Nov/Dec 2004)
The most common error control techniques are error detection and error
correction. Error detection can be accomplished by techniques like CRC, VRC, HRC and
error correction can be done using ARQ (Automatic repeat request).
4.A.11 What are the two types of ARQ error control schemes?(AU Apr/MAY 2004)
The two types of ARQ error control schemes are stop and wait ARQ and sliding
window ARQ.
4.A. 12 What is data terminal equipment? Give examples. (AU Nov/Dec 2010, R 2008)
Data terminal equipment (DTE) is an end instrument that converts user information into signals
or reconverts received signals. These can also be called tail circuits. A DTE device communicates with
the data circuit-terminating equipment (DCE).
32

4.A.13 What is meant by ASCII code?(April/May 2010)
The American Standard Code for Information Interchange (ASCII) defined by American
National Standards Institute (ANSI). It uses the 7 bit to represent every symbol. ASCII code
represents 27 = 128 various symbols.
a) Example
ASCII code for letter E------ 1000101
Column 4Row 5
4.A.14 Why are synchronous modems required for medium and high speed application?
(April/May 2008)
Synchronous modems can be faster than asynchronous. They depend on timing to
communicate. Data is transmitted in frames with synchronization bits which are used to
be sure the timing of the transmission and reception of data is accurate. Synchronous
modems are normally used on dedicated leased lines. Synchronous modems are one of
binary synchronous communications protocol (bisync), high level data link control
(HDLC), or synchronous data link control (SLDC). Three methods can be used to control
synchronization:
Additional clock signal
Guaranteed state change - Clocking is part of the data signal.
Oversampling - The reciever samples the signal much faster than the data is sent.
The extra samples can be used to be sure the clock is synchronized.
9. Sync and Asyn communication
(May’ 2006, April/May 2005)
4.A.15 What is the major disadvantage of Asynchronous Transmission mode?
(May’ 2006 CSE)
The major disadvantage of Asynchronous transmission is it requires additional
start and stop bits. Asynchronous transmission is slower and timing error will occur while
determining the sampling instants.
4.A.16 Define the term synchronous communication.(AU April/May 2005)
Synchronous communication takes place under the control of a master clock. The
bits transmitted are synchronized to a reference clock. The receiver operates at
exactly the same clock frequency as the transmitter. No start and stop bits are used,
the bits are serially transmitted one after another and the receiver will group the bits
into bytes.
33

10. Communication
4.A.17 Distinguish between half duplex and full duplex communication.
(AU April/May 2005, May 2004 CSE)
Half duplex communication is bidirectional but one cannot transmit and
receive simultaneously. If one is transmitting the other one will be receiving and
vice-versa. An example of half duplex communication is walky talky.
Full duplex communication is bidirectional communication and one can
transmit and receive simultaneously. An example for full duplex communication
is telephone network.
4.A.18 What is full duplex operation of a modem?(AU Nov/Dec 2004)
In full duplex operation, the modem can transmit and receive simultaneously and
the communication is fully bidirectional. There are two carriers simultaneously present in
the communication line, one for incoming and another for outgoing.
4.A.19 What do you mean by simplex transmission?(AU Nov/Dec 2003)
Simplex transmission takes place in only one direction for example television and
radio broadcasting.
. Problem
(May’ 2005 CSE, Nov’ 2004 CSE)
4.A.20 Draw the NRZ and RZ encoding format for the binary data 11010.
(May’ 2005 CSE, Nov’ 2004 CSE)
NRZ wave
1 1 0
1 1 0
1 0
RZ Wave
1 0
34

4.A.21 Given a 4 bit code d= [1 0 1 1] message. If a linear systematic code (7, 4) code is
generated with the redundancy bits r= [100], Form the7 bit code.(May’ 2005
CSE)
d/r = 1 0 1 1 / 1 0 0
=1011100
4.A.22 What is the main drawback of stop and wait algorithm?(AU April/May 2005)
The main drawback of stop and wait algorithm is speed. As the time lapse
between each frame is wasted time. Each frame essentially needs more time to transmit.
13. ARQ
4.A.23 What are three versions of ARQ?
2004)
The three versions of Automatic retransmission request (ARQ) are
a. Stop and wait ARQ system.
c. Selective repeat ARQ.
4.A.24 What do you mean by stop and wait ARQ?
(AU Apr/MAY 2004, Nov/Dec 2003)
(AU Apr/MAY
b. Go back N ARQ system.
With stop and wait flow control, the transmitting station sends one message frame
and then waits for an acknowledgement before sending the next message frame.
4.A.25. What is node-to-node delivery?(AU Nov/Dec 2004)
In node to node delivery communication exists between the transmitting and
receiving nodes using a dedicated connection.
4.A.26. Distinguish between random error and burst error.(Nov’ 2004 CSE)
Random errors are created due to white Gaussian noise in the channel. The error
affecting a particular interval will not affect the subsequent interval hence these noise is
completely uncorrelated. Burst error occurs due to impulse noise due to lightning and
switching transients. The burst errors are dependent on successive message intervals.
(AU Nov/Dec 2003)
PART B
1. Error Control
(Nov/Dec 2011-R2010 ,Apr/May 2011-R2008, Nov/Dec 2010-
R2008,April/May2010, Nov/Dec 2009, April/May2008, May/June 2007/R-2001,
May’ 2006, Apr/May 2004, Nov/Dec 2004)
a) What is CRC? Draw the encoder of the same and explain its operation. (8)
(Nov/Dec 2011-R2010)
b) Write short notes on forward error correction and backward error correction techniques.
(8)(Nov/Dec 2011-R2010)
35

c) Explain error detection and error connection in data communication.
(Apr/May 2011-R2008)
d) Explain how vertical, longitudinal and cyclic redundancy checking is used for detecting
the occurrence of errors in data transmission.(Nov/Dec 2010-R2008)
e) Describe the most common error detection techniques.(April/May2010)
f) Briefly explain the three methods of error correction.(April/May2010)
g) Explain the difference between error detection and error correction. Determine the BCS for the
following data and CRC generating polynomials.(Nov/Dec 2009)
Data G(x) = x
7+x5+ x2+ x1+ x0 CRC P(x) = x5+ x4+ x1+ x0
h) Write short notes on vertical and horizontal check Schemes for error Detection.
(April/May2008)
i) Explain synchronous time division multiplexing. Describe how error and flow control are
accomplished?(May/June 2007/R-2001)
j) Explain the mechanism of error control using stop and wait ARQ.
(May/June 2007/R-2001, May/June 2007/R-2004)
k) What are the types of errors encountered in Data transmission system? Explain the
methods used to overcome the effect of these errors.
(May’ 2006 CSE)
l) Explain in detail about error control schemes with necessary diagrams.
(AU Apr/May 2004, AU Nov/Dec 2004)
2. Modem
(Nov/Dec 2011-R2010, Apr/May 2011-R2008, April/May2010, Nov/Dec 2007)
a) Draw the block diagram of modem and explain the operation with the importance of each block
in the diagram.(Nov/Dec 2011-R2010)
b) Explain the operation of a synchronous modem.(Apr/May 2011-R2008)
c) Compare low speed, medium speed and high speed modems. Mention its application.
(Apr/May 2011-R2008)
d) Discuss the function of a data modem(April/May2010)
e) Explain Asynchronous mod (Nov/Dec 2007)
3. Explain the standard organization for data communication. . (Apr/May 2011-R2008)
4. Interface
(Nov/Dec 2010-R2008, April/May2010, May’ 2006)
a) Describe the features and purposes of serial Interfaces. Describe the mechanical,
electrical and functional characteristics of RS232 interface. (Nov/Dec 2010-R2008)
b) Explain in detail the characteristics of IEEE 488 bus (April/May2010)
c) Write a note on V.24/EIA-232 interface. Draw the timing diagram to show the state of all
EIA-232 leads between two DTE and DCE pairs during the communication of a data call
in a switched telephone network.
(May’ 2006 CSE)
36

5. Asyn/syn Data Transmission
(Nov/Dec 2009, Nov/Dec 2008, May/June 2007/R-2001,Nov/Dec 2003)
a) Discuss UART transmitter and receiver in detail. Under what conditions asynchronous mode of
data transfer is better?(Nov/Dec 2009)
b) Explain the synchronous and asynchronous data transmission techniques.
(Nov/Dec 2008, Nov/Dec 2003)
c) Describe synchronous transmission of data .Compare its advantages over asynchronous
transmission.(May/June 2007/R-2001)
6. Transmission mode
(May/June 2007/R-2001, Apr/May 2005)
a) Discuss the transmission characteristics and applications of three types of guided media
commonly used for data transmission.(May/June 2007/R-2001)
b) Name the guided and unguided transmission media(May/June 2007/R-2001)
c) Distinguish between half duplex and full duplex transmissions
(May/June 2007/R-2001)
d) Explain the different types of transmission modes.
7. Flow control
(AU Apr/May 2005)
(May/June 2007/R-2001, May’ 2006 CSE, AU Apr/May 2005, May’
2005 CSE, May’ 2005 CSE, AU Apr/MAY 2004)
a) Explain with diagrams the stop and wait flow control and sliding window flow control:
(May/June 2007/R-2001)
b) Describe the various flow control schemes with neat sketch.
(May’ 2006 CSE ,AU Apr/May 2005,May’ 2005 CSE, May’ 2005 CSE,
AU Apr/MAY 2004)
c) What are the drawbacks of stop and wait flow control. (AU April/May 2005)
8. Describe the three types of ARQ. (AU April/May 2005, May 2004 CSE )
UNIT V
PART A
5.A.1. Differentiate TDMA and CDMA. ( Nov/Dec 2011 –R2010)
SI.No
1
2
TDMA
Time sharing of satellite
transponder takes place.
Due to incorrect synchronization
there can be interference between
CDMA
Sharing of bandwidth and time both
takes place.
Both types of interferences will be
37

the adjacent times slots.
3
4
5
Synchronization is essential.
Code word is not required.
Guard times between adjacent time
slots are necessary.
present.
Synchronization is not necessary.
Code words are required.
Guard bands and guard times both are
necessary.
5.A.2. Differentiate multiplexing and multiple access. ( Nov/Dec 2011 –R2010)
SI.NoMultiplexingMultiple accessing
Multiplexing is process toMultiple-access is a techniques that
combine multiple signal for transmit it permit many users to simultaneously
over a single channel or media. access a given frequency allocation..
Generally multiplexing combines
severallow-speedsignalsfor
transmission over a single high-speed
connection.
Types:
1
2
Frequency Division Multiplexing Frequency Division Multiple-Access
(FDM), Time Division Multiplexing (FDMA), Time Division Multiple-Access
(TDM), Statistical Time Division (TDMA),
Multiplexing (STDM), Wavelength time/frequencymultiple-access
Division Multiplexing (WDM).randomaccess,
Code Division Multiple-Access (CDMA),
frequency-hopCDMA,
direct-sequence CDMA,
Types:
multi-carrier CDMA (FH or DS)
3. Spread spectrum
5.A.3 What are the applications of spread spectrum modulation? (Apr/May 2011 –
R2008,April/May 2010)
The spread spectrum modulation is used in some applications such as:
(i) Interference rejection
(ii) Secure Communication
(iii)Multiple access capability
(iv)Multipath protection
(v) Low probability of intercept
38

5.A.4 Design processing gain in spread spectrum modulation.
(Apr/May 2011,-R2008, April/May 2010)
Processing gain = BW (Spreaded signal) / BW (unspreaded signal)
Where
Spreaded signal = 1/Tc
Unspreaded signal = 1/Tb
Processing gain = Tb/ Tc ……………….. (1)
One bit period Tb of data signal is equal to N bits periods of spreading pseudo –
noise signal. That is Tb = N Tc putting this in above equation (1) Processing gain = N
5.A.5 Define Processing gain in spread spectrum technique
( May/June 2009/R-2004, Nov/Dec 2007)
i. The processing gain of a DS-SS is the gain achieved by processing a
spread spectrum signal over an un-spread signal
ii. It is the ratio of the bandwidth of the spread spectrum signal to the
bandwidth of the un-spread signal
iii. Processing gain = BW of spread spectrum signal/ BW of un-spread
signal.
5.A.6 What do you mean by spread spectrum techniques?
Spread spectrum is a means of transmission in which the data sequence occupies a
bandwidth in excess of the minimum bandwidth necessary to send it.
5.A.7 What are the different types of SS?
The Spread Spectrum can be classified under two categories
Direct sequence spread spectrum, Frequency hop
4. Frequency Hopping
5.A.7 Define FH spread spectrum technique.( Nov/Dec 2010, R 2008)
It is a spread spectrum technique in which the data is used to modulate a carrier. The data
modulated carrier is then randomly hopped from one frequency to another. So the
spectrum of transmitted signal is spread sequentially rather than instantaneously.What
are the three properties of PN sequence?(Nov ‘2006 CSE)
The three properties of PN sequence are balance property, run property and
correlation property.
5.A.8 What are fast and slow frequency hopping?(Nov ‘2006 CSE,May/June 2009 R-
2004)
i. A slow frequency hopping is multiple symbols transmitted per hop and
each symbol of a slow FH is a chip
39

ii. A fast frequency hop is multiple hops per M-ary symbol and here each
hop is a chip.
5.A.9 What is the basic principle of FH spread spectrum? (Nov/Dec 2008) (Nov/Dec
2009)
In Frequency hop spread spectrum (FH-SS) the data is used to modulate a carrier.
The data modulated carrier is then randomly hopped from one frequency to another. So the
spectrum of transmitted signal is spread sequentially rather than instantaneously.
5.A.10 How the pseudo noise sequences are generated? (Nov/Dec 2008)
Pseudo noise sequence is a periodic noise like sequence with a noise like waveform that
is usually generated by means of a feedback shift register. A feedback shift register is an
ordinary shift register made up of m flip flops.
5.A.11 . State the ‘run property’ of maximum length sequences.
Among the runs of 1s and of 0s in each period of a maximal-length sequence, one
half the runs of each kind are of length one, one-fourth are of length two, one-eighth are
of length three, and so on as these fractions represent meaningful numbers of runs. This
property is called run property.
5.A.12. CDMA(May/June 2007/R-2004)
a) What is the advantage of CDMA as wireless access scheme?
(May/June 2007/R-2004)
A single CDMA radio channel takes up the same bandwidth of approximately
42,30KHz AMPS voice channels . Because of frequency reuse advantage of CDMA,
CDMA offers approximately a 10 to 1 channel advantage over standard analog AMPS and
a 3 to 1 advantage over USDC digital AMPS.
5.A.13 Why purely random sequence cannot be used as a code in CDMA system?
(May/June 2007/R-2004)
In CDMA cellular system, the total radio frequency bandwidth is divided in to a
few broad band radio channels that have a much higher bandwidth than the digitized voice
signal. The digitized voice signal is added to high bit rate signal and transmitted in such a
way that it occupies the entire broad band radio channel. Adding a high bit rate pseudo
random signal to the voice information makes the signal more dominant and less
susceptible to interference, allowing lower power transmission and hence lower number of
transmitters and less expensive repeaters.
8. Time Division Multiple access (Nov/Dec 2007)
(April/May 2008)
5.A.14 What are the advantages of TDMA(Nov/Dec 2007)
i. Digital signals (bits) can be easily encrypted and decrypted,
safeguarding against eavesdropping.
40

ii. The entire telephone system is compatible with other digital formats,
such as those used in computers and computer networks.
iii. Digital systems inherently provide a quieter(less noisy) environment
than their analog counterparts.
5.A.15 What is meant by TDMA & FDMA technique?
i. TDMA: In this type of multiple access the time duration Tf called as
frame duration is subdivided into k non overlapping intervals of
duration
Tf / K. Each user is allotted a particular time slot for transmission.
ii. FDMA: In this technique the channel bandwidth is subdivided into K
non overlapping channels. Each user is allotted a particular channel for
voice and data transmission.
5.A.16 Draw the TDMA frame structure.
Guard time
Time
Slot
1
9. Time
Slot
2
10.
5.A.17. What is frequency division multiplexing?(May’ 2005 CSE, Nov’ 2004 CSE)
To transmit a number of these signals over the same channel, the signals must be
kept apart so that they do not interfere with each other. The technique of separating the
signals in frequency is referred to as frequency division multiplexing.
17. DS Spread Spectrum
5.A.18 Write short notes on DS spread spectrum.
In Direct sequence (DS) spread spectrum technique the data sequence is used to
modulate the pseudo noise sequence directly, the binary sequence is converted into a NRZ
sequence which is used to directly modulate the pseudo noise sequence.
5.A.19 Define processing gain for DS-SS system.
The processing gain of a DS-SS is the gain achieved by processing a spread
spectrum signal over an un-spread signalIt is the ratio of the bandwidth of the spread
spectrum signal to the bandwidth of the un-spread signal
Processing gain = BW of spread spectrum signal/ BW of un-spread signal.
Time
Slot
3
Time Slot
n
11.12. 13.14. 15.
41

5.A.20 Define jamming margin of DS-SS system.
The jamming margin is the ratio of average interference power J to the average
signal power Ps (J/Ps)If it is expressed in dB then,
Jamming margin in dB = Processing gain in dB – 10log10 (Eb/ No)
Where Eb/No is the minimum bit energy to noise density ratio needed to support a
prescribed average error probability.
18. Multiple access techniques
5.A.21 What is the need of multiple access techniques?
The need for multiple access techniques is to accommodate large number or users in
a single communication channel.
5.A.22 What are the different types of multiple access techniques?
The different types of multiple access techniques are
i. Frequency division multiple access (FDMA)
ii. Time division multiple access(TDMA)
iii. Code division multiple access(CDMA)
5.A.23. What does source coding mean?
The wireless communication system such as TDMA and CDMA uses speech
coding for efficient use of channel bandwidth. This speech coding technique removes the
redundancy in speech. LPC (linear predictive coding) is a source coding technique for
speech.
5.A.24. Define Chip duration and chip rate.
The bit period of PN sequence is called chip duration (Tc). Chip rate is the rate at
which the PN sequence is generated.
Chip rate = 1 / Tc
PART B
1. PN sequence
a)
b)
c)
d)
(Nov/Dec 2011 –R2010, Apr/May 2011-R2008, Nov/Dec 2010-R2008,
Nov/Dec 2007, Nov’ 2006)
List three properties of pseudo-noise signal and explain them with an example. (8)
(Nov/Dec 2011 –R2010)
Explain how a pseudo random noise sequence is generated.(Apr/May 2011-R2008)
What is a PN sequence? Explain its important properties. (Nov/Dec 2010,R2008)
Draw the block diagram of a simple PN sequence generator using shift register and obtain
the output sequence. For this output sequence verify the properties of the PN sequence.
(Nov/Dec 2007 ,Nov’ 2006 CSE)
42

2. DS spread spectrum
(Nov/Dec 2011-R2010 ,Apr/May 2011-R2008, Apr/May 2010,Nov/Dec
2009,May/June 2009 R-2004, Nov/Dec 2008, May/June 2007/R-2004,May
2006)
a) With neat block diagram, explain the DS-spread spectrum with coherent BPSK. (8)
(Nov/Dec 2011-R2010)
b) Discuss in detail the operation of DS spread spectrum with coherent binary PSK.
(Apr/May 2011-R2008)
Or
With the help of a neat block diagram, explain DS spread spectrum system with coherent
binary PSK. (Apr/May 2010, May 2006)
Or
Describe with block diagram, DS SS binary PSK spread spectrum system.
c) With help of block diagram explain how DSSS can be implemented. Draw the input and
output waveforms(Nov/Dec 2009)
d) Draw the block diagram of a DS spread spectrum system and explain its working
(May/June 2009 R-2004, Nov/Dec 2008, May/June 2007/R-2004)
e) Explain with neat sketches DS SS with BPSK.
3. TDMA and CDMA
(Nov//Dec 2011-R2010 ,Apr/May 2011-R2008, Nov/Dec 2010/R2008,
May/June 2009, Nov/Dec 2008, April/May 2008, May’ 2006)
a) With neat diagram, describe the operation of CDMA. (8) (Nov//Dec 2011-R2010)
b) Compare the features of FDMA, TDMA and CDMA. (8) (Nov//Dec 2011-R2010)
c) Describe the principles of TDMA and CDMA in wireless communication system.
(Apr/May 2011-R2008)
d) List the advantages of CDMA over TDMA multiple access scheme.(Nov/Dec
2010,R2008)
e) Describe the application of CDMA in wireless communication system.
Or
Describe the operation of a CDMA multiplexing system.(Nov/Dec 2010,R2008)
Compare TDMA with FDMA.(May/June 2009, April/May 2008)
Draw the frame structure of TDMA and explain.(Nov/Dec 2008)
With neat sketches explain the FDMA technique.(Nov/Dec 2008)
Compare the merits and demerits of TDMA and FDMA multiple access schemes.
(May’ 2006 CSE)
4. Write note on coding of speech for wireless communication. (Apr/May 2011-R2008)
5. Explain the near –far problem in spread spectrum modulation.(April/May 2010)
f)
g)
h)
i)
43

6. Frequency hopping
a) Explain fast and slow frequency hopping techniques used in spread spectrum
(May/June2009 R-2004, Nov/Dec 2007)
b) What is frequency hop spread spectrum? Explain the generation of slow frequencyhop
spread M-ary FSK and fast frequency hop spread M-ary FSK with appropriate diagrams.
(April/May 2008)
7. Problems
(April/May 2008, May’ 2006 CSE)
a) A spread spectrum communication system has the following parameters:
Information bit duration Tb =4.095ms, PN chip duration Tc =1µs, the energy
to noise ratio Eb/No =10.Calculate the processing gain and jamming margin.
(April/May 2008)
b) A (15, 5) Linear Cyclic Code has a generator polynomial g(D) = 1 + D + D2 + D4 + D5 +
D8 + D10 (i) Draw block diagrams of an encoder and syndrome calculator for this code.
(ii) Find the code polynomial in systematic form for the message polynomial m (D) = 1 +
D2 + D4. (iii) Is y(D) = 1 + D4 + D6 + D8 + D14 , a doe polynomial? If not, find the
syndrome of y (D).
(May’ 2006 CSE)
(1)c) Consider a rate -1/2 non systematic
convolutional code with g = {1, 0, 1} and g(2)=
{1,1,1}.(i)Determine the encoder output corresponding to the data sequence {1 0 1 0 1}.
(ii) If the first and the fourth its of the encoded sequence are affected during transmission,
demonstrate the error correcting capability of the Viterbi decoding algorithm.
(May’ 2006 CSE)
8. LPC
(Nov/Dec 2007, May/June 2007,Nov’ 2006 )
a) Draw the block diagram of RAKE receiver and explain the operation. (ii) Explain the
working of multi pulse excited LPC and code-excited LPC by suitable diagrams.
(Nov/Dec 2007,Nov’ 2006 )
b) With neat diagram explain the working of Residual excited LPC (RELP) and Code
Excited LPC (CELP).(May/June 2007/R-2004)
9. Draw the trellis diagram to state table for the encoder .
(May/June 2007 R-2001)
10. Consider the convolutional encoder shown in fig below with rate ½ and constraint
length of 3 Determine the encoder output for the message sequences 10011.
(May/June 2007 R-2001)
11. Discuss about linear block code. (ii)Explain the Viterbi decoding algorithm for
convolutional code.(May’ 2004 CSE)
12. Write in detail about multiple access techniques.
44

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