IT 2202 PRINCIPLES OF COMMUNICATION DEPARTMENT OF ECE
142302 PRINCIPLES OF COMMUNICATION DEPARTMENT OF ECE
St JOSEPHS COLLEGE OF ENGINEERING
Department of information technology
SUBJECT : PRINCIPLES OF COMMUNICATION CODE: 142302UNIT I :
FUNDAMENTALS OF ANALOG COMMUNICATION1. What is the need for
modulation?
It is extremely difficult to radiate low frequency signals
through earths 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.
EMBED Equation.3 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?
4. What is the relationship between total power in AM wave and
unmodulated carrier power?
5. What is the relationship between total current in AM wave and
unmodulated carrier current?
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.
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.
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.
SNR ( signal to noise ratio ) is the ratio of signal power to
noise power.
10. Define Equivalent Noise TemperatureEquivalent 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
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 Carsons 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 FMWide band FM
Frequency deviation in carrier frequency is very smallFrequency
deviation in carrier frequency is large
Band width is twice the highest modulating frequencyBand width
is calculated as per Carsons 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.
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 =
30. Define peak phase deviation?
k Em =
where k is deviation sensitivity in rad/volt
and Em is amplitude of modulating signal.
31.Define frequency deviation?
f = k1 Em/2 in Hz
Where k1 is deviation sensitivity rad/volt-sec
32. Define percent modulation?
Percent modulation = [actual frerquency deviation/max allowable
frequency deviation] x10033.What are the types of modulation
Index?
1.Low index modulation (m1)
34.Obtain total angle modulated wave power?
Pt = Po + P1 +P2 +.Pn35.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
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 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
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
typically given as a peak frequency shift in Hertz(f).The peak
to peak frequency
deviation (2f) is sometimes called carrier swing.The peak
frequency deviation is
simply the product of the deviation sensitivity and the peak
modulating signal
voltage and is expressed mathematically as f=K1 Vm Hz
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 carsons rule is
B=2(f +fm) Hz.
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
DR= f (max)/fm(max)
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
.
Unit II : DIGITAL COMMUNICATION1.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 = 46.663.Determine peak phase
deviation for PM modulator with deviation sensitivity k = 2.5 rad/v
and em = 2cos(22000t). k Em =
= 2.5(2) = 5rad/v4.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?
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.
8.Compare binary PSK with QPSK.
BPSK
Binary Phase Shift KeyingQPSK
Quadrature Phase Shift Keying
One bit form a symbolTwo bits form a symbol
Two possible symbolsFour possible symbols
Minimum bandwidth required = f b where f b is bit rate
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 Minimum band width
required
if f b = bit rateMinimum band width
required
if f b = 10 Mbps
BPSKf b10 MHz
QPSKf b / 25 MHz
8 - PSKf b / 33.33 MHz
8- QAMf b / 33.33 MHz
16 - QAMf b / 42.5 MHz
12.What is difference between coherent and non coherent
detection?
Coherent detectionNon- Coherent detection
Carrier which is in perfect coherence with that used in
transmitter is used for demodulation.
Carrier recovery circuit is needed for detectionNo carrier
recovery circuit needed for detection.
Relatively complexSimple implementation
13.Define Bandwidth efficiency. What is the bandwidth efficiency
of BPSK and 8-PSK system?
It is the ratio of the transmission bit rate to the minimum
bandwidth required for a particular modulation scheme.
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 systems 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 systems performance.
15.What is the probability of error for (i) non-coherent FSK and
(ii) coherent FSK? Compare their error performance.
For a given energy per bit to noise power density ratio,
probability of error for non-coherent 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.
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.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.
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 = | g(t) gi(t) | is called the aliasing
error.
The aliasing error is bounded as
2 | G(f) | df
6. Given the signal m(t)=10 cos (2000 t) cos (8000 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 = [cos (A-B)+cos (A+B)]
There fore, m(t) = (10/2) [ cos (6000 t) + cos (10000 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.
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 ) = TbTb = bit duration in seconds ; Rb = bit rate =
1/Tbp(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.
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.
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 noise29.State the difference between DPCM and
DM.Parameter
DPCM
DM
Number of Bits
Levels, Step size.
Bandwidth of transmission channel.Bits can be more than one but
less than PCM.
Fixed numbers of levels are used.
Bandwidth required is lesser than PCM.It can use 4, 8 or 16 bits
per sample.
Step size is fixed and cannot be varied.
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.
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.
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.
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?
NfmN: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.
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 phase- shift 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(k-1)} will be smaller than
the sample value themselves fewer levels will be required to
quantize the difference than that required to quantize M(k) and
correspondingly fewer bits will be needed to encode the levels.
PART B1. 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.
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 AccessTDMA - Time Division
Multiple Access
All users access the channel by transmitting simultaneously but
using disjoint frequency bandsAll users occupy the same RF band
width of the channel, but they transmit sequentially in time
Fixed assignment multiple access techniqueFixed assignment
multiple access technique
Well suited for analog communicationWell 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.
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.
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 Keplers first law.
Keplers 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 Keplers 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.
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 Keplers 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.
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 earths
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
earths 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 earths
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.25.
What are the advantages of optical fiber communication?
Greater information capacity
Immunity to crosstalk
Immunity to static interference Environmental immunity
Safety & Security26. 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 90or 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.
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 Keplers 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 1
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