it 2202 poc

IT 2202 PRINCIPLES OF COMMUNICATION 3 1 0 4

UNIT I FUNDAMENTALS OF ANALOG COMMUNICATION 9 Principles of amplitude modulation, AM envelope, frequency spectrum and bandwidth, modulation index and percent modulation, AM Voltage distribution, AM power distribution, Angle modulation - FM and PM waveforms, phase deviation and modulation index, frequency deviation and percent modulation, Frequency analysis of angle modulated waves. Bandwidth requirements for Angle modulated waves.

UNIT II DIGITAL COMMUNICATION 9Introduction, Shannon limit for information capacity, digital amplitude modulation, frequency shift keying, FSK bit rate and baud, FSK transmitter, BW consideration of FSK, FSK receiver, phase shift keying – binary phase shift keying – QPSK, Quadrature Amplitude modulation, bandwidth efficiency, carrier recovery – squaring loop, Costas loop, DPSK.

UNIT III DIGITAL TRANSMISSION 9 Introduction, Pulse modulation, PCM – PCM sampling, sampling rate, signal to quantization noise rate, companding – analog and digital – percentage error, delta modulation, adaptive delta modulation, differential pulse code modulation, pulse transmission – Intersymbol interference, eye patterns.

UNIT IV SPREAD SPECTRUM AND MULTIPLE ACCESS TECHNIQUES 9 Introduction, Pseudo-noise sequence, DS spread spectrum with coherent binary PSK, processing gain, FH spread spectrum, multiple access techniques – wireless communication, TDMA and CDMA in wireless communication systems, source coding of speech for wireless communications.

UNITV SATELLITE AND OPTICALCOMMUNICATION 9

Satellite Communication Systems-Keplers Law,LEO and GEO Orbits, footprint, Link model-Optical Communication Systems-Elements of Optical Fiber Transmission link, Types, Losses, Sources and Detectors.

TUTORIAL: 15 TOTAL: 45 +15=60

TEXT BOOKS:1. Wayne Tomasi, “Advanced Electronic Communication Systems”, 6/e, Pearson

Education, 2007. 2. Simon Haykin, “Communication Systems”, 4th Edition, John Wiley & Sons., 2001.

REFERENCES:

1. H.Taub,D L Schilling ,G Saha ,”Principles of Communication”3/e,2007.2. B.P.Lathi,”Modern Analog And Digital Communication systems”, 3/e, Oxford

University Press, 20073. Blake, “Electronic Communication Systems”, Thomson Delmar Publications, 2002.4. Martin S.Roden, “Analog and Digital Communication System”, 3 rd Edition, PHI,

2002.5. B.Sklar,”Digital Communication Fundamentals and Applications”2/e Pearson

Education 2007.

IT 2202 1

DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING

Question Bank

SUBJECT: IT2202-PRNCIPLES OF COMMUNICATIONSemester: III Department: IT

UNIT I--Fundamentals Of Analog CommunicationPART-A

1. What is modulation?Modulation is the process of changing any one parameter (amplitude,

frequency or phase) of a relatively high frequency carrier signal in proportion with the instantaneous value of the modulating signal or message signal.

2. Define amplitude Modulation.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.

3. Give the expression for AM modulated wave.Vam =Vc sin ωc t + m Vc cos (ωc - ωm ) t – m Vc cos (ωc + ωm) t

------ ----- 2 2

where, Vam - amplitude of modulated signalVc - amplitude of carrier signalωc = 2 π fc = carrier frequencyωm = 2π fm = modulating signal frequency.

4. 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.

5. Give the bandwidth of AM?Bandwidth (B) of AM DSBFC is the difference between highest upper

frequency and lowest lower side frequency.B= 2fm(max)

fm(max) – maximum modulating signal frequency.

IT 2202 2

6. Draw the spectrum of AM signal. Vc

mVc/2 mVc/2

fm fm

fc-fmfc fc+fm frequency fLSB fUSB

7. Give the expression for modulation index in terms of Vmax and Vmin.

m = Vmax – Vmin / Vmax + Vmin 8. Give the formula for AM power distribution.

Ptotal = Pc [1 + m2 / 2]where, Ptotal – total power

m- modulation indexPc – carrier power

9. Give the expression for total current.Itotal = Ic [1 + m2 / 2] 1/2

where, Itotal – total Currentm- modulation indexIc – carrier current

10. Give the types of AM Modulation.DSBSC-Double sideband suppressed carrier.SSBSC- Single sideband suppressed carrier.DSBFC- Double sideband full carrier.VSBSC-Vestigial sideband suppressed carrier.

11. What are the disadvantages of conventional (or) double side band full carrier system?

In conventional AM, carrier power constitutes two thirds or more of the total transmitted power. This is a major drawback because the carrier contains no information; the sidebands contain the information. Second, conventional AM systems utilize twice as much bandwidth as needed with single sideband systems.

12. Define Single sideband suppressed carrier AM.AM Single sideband suppressed carrier is a form of amplitude modulation

in which the carrier is totally suppressed and one of the sidebands removed.13. Define AM Vestigial sideband.

AM vestigial sideband is a form of amplitude modulation in which the carrier and one complete sideband are transmitted, but only part of the second sideband is transmitted.

14. What are the advantages of single sideband transmission?The advantages of SSBSC are

1.Power conservation2.Bandwidth conservation3.Noise reduction

15. What are the disadvantages of single side band transmission?

IT 2202 3

i. Complex receivers ii. Tuning Difficulties

16. What is the advantage of low-level modulation?An advantage of low-level modulation is that less modulating signal

power is required to achieve a high percentage of modulation.17. Define Low-level Modulation.

In low-level modulation, modulation takes place prior to the output element of the final stage of the transmitter. For low level AM modulator class A amplifier is used. It requires less power to achieve a high percentage of modulation.

18. Define High-level Modulation.In high-level modulators, the modulation takes place in the final element

of the final stage where the carrier signal is at its maximum amplitude. For high level modulator class C amplifier is used. It requires a much higher amplitude modulating signal to achieve a reasonable percent modulation.

19. What is the advantage of low-level modulation?An advantage of low-level modulation is that less modulating signal

power is required to achieve a high percentage of modulation.20. Define image frequency.

An image frequency is any frequency other than the selected radio frequency carrier that, if allowed to enter a receiver and mix with the local oscillator, will produce a cross product frequency that is equal to the intermediate frequency.

21. Define image frequency rejection ratio.The image frequency rejection ratio is the measure of the ability of

preselector to reject the image frequency.Mathematically, IFRR is

IFRR =(1+Q2ρ2)1/2 Where ρ= (fim/fRF)-(fRF/fim)Q – quality factor of preselectorfim-image frequencyfRF- RF frequency

22. Define Heterodyning.Heterodyne means to mix two frequencies together in a nonlinear device

or to translate one frequency to another, using nonlinear mixing.23. 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.

24. Define indirect frequency Modulation.In indirect frequency modulation, phase of a constant amplitude carrier

directly proportional to the amplitude of the modulating signal at a rate equal to the frequency of the modulating signal.

25. Define instantaneous frequency deviation.

IT 2202 4

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.26. 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 (Δf). The peak-to-peak frequency deviation (2 Δf) 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 Δf =K1Vm Hz.

27. State Carson rule.Carson rule states that the bandwidth required to transmit an angle

modulated wave as twice the sum of the peak frequency deviation and the highest modulating signal frequency. Mathematically Carson’s rule is B=2(Δf +fm) Hz.

28. 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= Δf (max) / fm(max)

29. Write down the comparison of frequency and amplitude modulation. Amplitude modulation Frequency modulation1. Noise interference is more Noise interference is less2. 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.

Frequency Modulation is the process of changing the frequency of a relatively high frequency carrier signal in proportion with the instantaneous value of the modulating signal.

3. The depth of modulation has limitation in AM.

But in FM the depth of modulation can be increased to any value by increasing the deviation.

4. Simple circuits used in transmitter and receiver.

Uses more complex circuits in transmitter and receiver.

5. Power varies in AM depending on depth of modulation.

The amplitude of FM is constant. Hence transmitter power remains constant in FM

30. Define Phase modulation. Phase of a constant amplitude carrier is varied directly proportional to the

amplitude of the modulating signal at a rate equal to the frequency of the modulating signal.

31. What are the advantages of angle modulation and also list its disadvantages.Advantages:

i. Noise reduction.ii. Improved system fidelity.

iii. more efficient use of power.Disadvantages:

IT 2202 5

i. wider Bandwidth.ii. uses more complex circuit in receiver and transmitter.

32. What is Phase deviation ?The relative angular displacement (shift) of the carrier phase (rad) in

respect to reference phase is called phase deviation(ΔӨ)33. Give the expression for bandwidth of angle-modulated wave in terms of

Bessel’s table.B= 2(n*fm)n=no. of significant sidebands for m found using Bessel’s table.

34. Define deviation sensitivity for FM and PM and give its units.FM: Change in output frequency occurs when amplitude changes in input signal. Unit K1=(rad/s)/V.PM: Change in output phase occurs when amplitude changes in input signal. Unit K =(rad)/V.

35. If a modulated wave with an average voltage of 20Vp changes in amplitude ±5V, determine the maximum and minimum envelope amplitudes and the modulation coefficients.

Vm = 20VpVc = 5 Vm = Vmax – Vmin / Vmax + Vmin Vmax = Vm + Vc = 20+5= 25VVmin = Vm - Vc = 20-5= 15Vm= Vmax – Vmin / Vmax + Vmin =25-15 /25+15 = 0.25

36. An FM transmitter has a rest frequency fc =96MHz and a deviation sensitivity K1 = 4 KHz/V. Determine the frequency deviation for a modulating signal Vm(t) = 8sin(2π 2000t). Determine the modulation index.

Vm=8V, fm =2000Hz and K1 =4 kHz /VFrequency deviation = δ = K1Vm = 4 kHz/v * 8V = 32kHzModulation index = m = δ/ fm = 32 kHz/2000Hz = 16

37. For an FM receiver with an input frequency deviation ∆f=4 kHz and a transfer ratio K= 0.01 V/k Hz, determine Vout.

Vout = K * ∆f =0.01* 40 =0.4V

PART –B

1. For an AM DSBFC wave with unmodulated carrier voltage of 10Vp and load resistance of 10 ohms, and m=1determine (i) Power in carrier,(ii)Power in upper and lower sidebands(iii)Total transmitted power.

Refer: Wayne Tomasi, “Advanced Electronic Communication Systems” Fifth edition, Page No.151

2. Derive the expression for total power in an AM DSBFC and draw the power spectrum.


3. For an AM DSBFC transmitter with an unmodulated carrier power Pc=100w that is simultaneously modulated by 3 modulating signals with co-

IT 2202 6

efficient of modulation m1=0.2, m2=0.4 and m3=0.5, determine, Total coefficient of modulation, Upper and lower sideband power, Total sideband power, Total transmitted power and then, Draw the output spectrum.


4. One input to an AM DSBFC modulator is 500 kHz carrier with peak amplitude of 20Vp. The second input is a 10 kHz modulating signal whose amplitude is sufficient to produce a ± 7.5Vp change in the amplitude of the envelope. Determine the following (i) Upper and lower side frequency,(ii)Modulation co-efficient and percent modulation.,(iii)Maximum and minimum amplitude of the envelope,(iv)Draw the output envelope,(v)Draw the output frequency spectrum.


5. Derive Eusf and Elsf in terms of Vmax and Vmin .


6. Derive the output expression for an AM DSBFC and also draw the AM spectrum.


7. Explain in detail about the Bandwidth requirements of angle-modulated wave.


8. Compare PM and FM.Refer: Wayne Tomasi, “Advanced Electronic Communication Systems”

Fifth edition, Page No.2819. Determine the side band frequencies of an angle-modulated wave.


10. Derive the expression for average power of an angle-modulated wave.Refer: Wayne Tomasi, “Advanced Electronic Communication Systems”

Fifth edition, Page No.293.

UNIT II--Digital CommunicationPART-A

1. What is digital modulation?When the information signal is digital and any one of the parameters

(amplitude, phase or frequency) of the analog carrier is varied proportional to the information signal is called ad digital modulation.

2. What is information capacity?

IT 2202 7

It is the number of independent symbols that can be carried through a system in a given unit of time.

3. Give the expression for Shannon limit for information capacity.I= B log2 [1+ S/N]Where, I= information capacity (bps)

B= bandwidthS/N=signal to noise power ratio (unit less)

4. Give the Nyquist formulation for channel capacity.fb =2B log2 M

Where, fb –channel capacity (bps)B-minimum Nyquist bandwidth (Hz)M- number of discrete level or voltage levels

5. Compare QASK and QPSK.

QPSK QASK

1. Quadrature phase modulation

Quadrature phase and amplitude modulation

2.All signal points placed on circumference of circle

Signal points are replaced symmetrically about origin

3. Circuit is simple. Relatively complex

4.Noise immunity better then QASK

Poor than QPSK.

5. Error probability less then AQSK

Higher than QPSK

6. What are Antipodal signals? In BPSK, the two symbols are transmitted with the help of following

signals,Symbol ‘1’ => s1 (t) = √2P cos (2πf0 t)Symbol ‘0’ => s2 (t) = √2P cos (2πf0 t + π)

Here observe that above two signals differ only in a relative phase shift of 1800. Such signals are called antipodal signals.

7. Define minimum shift keying. Minimum shift keying uses two orthogonal signal to transmit ‘0’ and ‘1’

in such a way the difference between these two frequencies is minimum. Hence, there is no abrupt change in the amplitude and the modulated signal is continuous and smooth.

8. Give the difference between standard FSK and MSK.

FSK MSK

1. The two frequencies are Difference between two

IT 2202 8

integer multiple of base band frequency and at the same time orthogonal.

frequencies minimum and at the same time they are orthogonal.

2. Bandwidth (BW) = 4fb BW = fb/2

3. Has discontinuities when phase changes from 0 to 1 or 1 to 0.

Phase discontinuities are removed by smooth phase transition.

9. What are the advantages of M-ary signaling scheme?i. M-ary signaling schemes transmit bits at a time.

ii. Bandwidth requirement of M-ary signaling schemes is reduced.10. What does correlative coding mean?

Correlative coding allows the signaling rate of 2B0 in the channel of bandwidth B0. This is made physically possible by allowing ISI in the transmitted signal in controlled manner. The receiver knows this ISI. Hence effects of ISI are eliminated at the receiver. Correlative coding is implemented by duobinary signaling and modified duobinary signaling.

11. Differentiate coherent and noncoherent methods.Coherent (synchronous) detection: In coherent detection, the local

carrier generated at the receiver is phase locked with the carrier at the transmitter. The detection is done by correlating received noisy signal and locally generated carrier. The coherent detection is a synchronous detection.

Non- coherent (envelope) detection: This type of detection does not need receiver carrier to be phase locked with transmitter carrier. The advantage of such a system is that the system becomes simple, but the drawback is that error probability increases.

12. Define peak frequency deviation for FSK.Peak frequency deviation (Δf) is the half the difference between either the

mark and space frequency. (Δf)=|fm-fs|/2.13. 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).

14. Define Baud rate.The rate of change at the output of the modulator is called baud rate.Baud= 1/ts, where, ts- time of one signaling element (seconds).

15. Compare binary PSK with QPSK.

BPSK QPSK

1. One bit forms a symbol. Two bits form a symbol.

2. Two possible symbols Four possible symbols.

3. Minimum bandwidth is Minimum bandwidth is

IT 2202 9

twice of fb equal to fb.

4. Symbol duration = Tb. Symbol duration = 2Tb.16. 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.

17. What is a constellation diagram? It is also called as signal state-space diagram, similar to phasor diagram

where, the relative position of peaks of phasors is shown.18. Bring out the difference between DPSK and BPSK.

DPSK BPSK

1. It does not need a carrier at its receiver

It needs a carrier at receiver

2. Bandwidth reduced compared to BPSK

More bandwidth.

3.Probability of error or bit error rate more than BPSK

Comparatively low

4. Error propagation more, since it uses two bits for its reception

Comparatively low, since it uses only single bit

5. Noise interference more Comparatively low19. What is bandwidth efficiency?

It is also called as information density or spectral efficiency, is the ratio of the transmission bit rate to the minimum bandwidth required for particular modulation scheme.

20. What is an Offset QPSK?Offset QPSK (OQPSK) is a modified form of QPSK where the bit

waveforms on the I and Q channels are offset or shifted in phase from each other by one-half of a bit time.

21. Mention any four advantage of digital modulation over analog modulation. i. Maximum data rate

ii. Minimum probability of symbol erroriii. Minimum transmitted power.iv. Minimum channel bandwidth.v. Minimum circuit complexity

vi. Maximum resistance to interfering signals22. Define carrier recovery.

IT 2202 10

It is the process of extracting a phase-coherent reference carrier from a receiver signal. It is also called as phase referencing

23. What is DPSK?Differential phase-shift keying (DPSK) is an alternative form of digital

modulation where the binary input information is contained in the difference between successive signaling elements rather than the absolute phase. It is not necessary to recover phase-coherent carrier.

24.What do you mean by ASK?ASK (Amplitude Shift Keying) is a modulation technique

which converts digital data to analog signal. In ASK, the two binary values(0,1) are represented by two different amplitudes of the carrier signal.

S(t) = Acos2ft binary 1 0 binary 0

PART-B

1. Explain FSK bit rate, baud, bandwidth and modulation index.Refer: Wayne Tomasi, “Advanced Electronic Communication Systems”

Fifth edition, Page No.3692. Explain on-off keying (OOK) or ASK.


3. Explain QPSK transmitter and receiver. Refer: Wayne Tomasi, “Advanced Electronic Communication Systems”

Fifth edition, Page No.3814. Explain DPSK with an example.


5. Explain BPSK (transmitter and receiver) and also discuss about the bandwidth.


6. Discuss the operation of 16-QAM transmitters and receivers.Refer: Wayne Tomasi, “Advanced Electronic Communication Systems”

Fifth edition, Page No.3987. Explain in detail about carrier recovery.


UNIT III --Digital TransmissionPART-A

IT 2202 11

1. State the sampling theorem for band-limited signals of finite energy. If a finite energy signal g(t) contains no frequency higher than W Hz, it is

completely determined by specifying its ordinates at a sequence of points spaced 1/2W seconds apart.

2. What are the advantages of digital transmission?i. The advantage of digital transmission over analog transmission is noise

immunity. Digital pulses are less susceptible than analog signals to variations caused by noise.

ii. Digital signals are better suited to processing and multiplexing than analog signals.

iii. Digital transmission systems are more noise resistant than the analog transmission systems.

iv. Digital systems are better suited to evaluate error performance.3. What are the disadvantages of digital transmission?

i. The transmission of digitally encoded analog signals requires significantly more bandwidth than simply transmitting the original analog signal.

ii. 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.

4. 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.

5. 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.6. What is the Nyquist sampling rate?

Nyquist sampling rate states that, the minimum sampling rate is equal to twice the highest audio input frequency.

7. What is the principle of pulse modulation?Pulse modulation consists essentially of sampling analog information

signal and then converting those discrete pulses and transporting the pulses from a source to a destination over a physical transmission medium.

8. List the four predominant methods of pulse modulation.i. Pulse width modulation (PWM)

ii. Pulse position modulation (PPM)iii. Pulse amplitude modulation (PAM)iv. Pulse duration modulation (PDM)

9. What is codec?An integrated circuit that performs the PCM encoding and decoding

functions is called a Codec (coder/decoder).10. 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

IT 2202 12

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”.

11. Define overload distortion.If the magnitude of sample exceeds the highest quantization interval,

overload distortion occurs.12. Define quantization.

Quantization is a process of approximation or rounding off. Assigning PCM codes to absolute magnitudes is called quantizing.

13. Define dynamic range.Dynamic range is the ratio of the largest possible magnitude to the

smallest possible magnitude. Mathematically, dynamic range isDR= Vmax / Vmin

14. What is nonuniform or nonlinear encoding?With voice transmission, low-amplitude signals are more likely to occur

than large-amplitude signals. Therefore, if more codes are used for lower amplitude, it would increase accuracy and fewer codes are used for higher amplitudes, which would increase quantization error. This type of coding is called nonuniform or nonlinear encoding.

15. What is the advantage and disadvantage of midtread quantization?Advantage: less idle channel noise Disadvantage: largest possible magnitude for Qe

16. What is the necessity of companding?Companding is the process of compression and then expanding. Higher

amplitude signals are compressed prior to transmission and then expanded in the receiver. Companding is the means of improving dynamic range of communication systems.

17. What is idle channel noise?When there is no analog input signal, the only input to PAM sampler is

random, thermal noise. This noise is called idle channel noise.18. How is percentage error calculated?

12-bit encoded voltage-12-bit decoded voltage% Error = --------------------------------------------------------------X 100

12-bit decoded voltage19. Compare slope overload and granular noise.

Slope overload noise Granular noise

1. Slope of analog signal is greater than delta modulator can maintain

Original input signal has relatively constant amplitude and the reconstructed signal has variation the were not present in the original signal.

2. Caused when step- size is small.

2. Caused when step -size is large.

20. Give the concept of delta modulation PCM.

IT 2202 13

Rather than transmit a coded representation of the sample, only single bit is transmitted, which indicates whether the sample is larger or smaller then the previous sample.

21. What is ISI and give its causes.The ringing tails of several pulses have overlapped, thus interfering with

major pulse lobe. This interference is commonly called as intersymbol interference or ISI. The four primary causes of ISI are

i. Timing inaccuraciesii. Insufficient bandwidth

iii. Amplitude distortioniv. Phase distortion

22. What is an eye pattern?The performance of a digital transmission system can be measured by

displaying the received signal on an oscilloscope and triggering the time base at data rate. Thus, all waveform combinations are superimposed over adjacent signaling intervals. Such a display is called eye pattern or eye diagram.

23. List the significance of eye pattern.It used to determine the effects of degradations introduced into pulses as

they travel to the regenerator. It discloses any noise or errors in line equalization. It also gives the amount of ISI present.

24.What are the two fold effects of quantizing process?1. The peak-to-peak range of input sample values subdivided into a finite set of decision levels or decision thresholds2. The output is assigned a discrete value selected from a finite set of representation levels are reconstruction values that are aligned with the treads of the staircase.

25.Define quantization error?Quantization is the value of which equals the difference

between the output and input values of quantizer.26.What is nyquist rate?

The minimum sampling rate of 2W sample per second for a signal bandwidth of W hertz is called the nyquist rate.

27.What is PAM?PAM is the pulse amplitude modulation. In pulse amplitude

modulation, the amplitude of a carrier consisting of a periodic train of rectangular pulses is varied in proportion to sample values of a message signal.

28. What do you mean by slope overload distortion in delta modulation?Slope of analog signal is greater than delta modulator can maintain.

Caused when the step size is small.

IT 2202 14

PART-B

1. Explain PCM with a neat block diagram.Refer: Wayne Tomasi, “Advanced Electronic Communication Systems”

Fifth edition, Page No.4252. Explain PCM sampling with necessary diagrams and circuits. Write a note on

aliasing.Refer: Wayne Tomasi, “Advanced Electronic Communication Systems”

Fifth edition, Page No.4273. What is companding? Explain in detail Analog and digital companding.


4. With a neat block diagram explain Delta modulation. How slope over and granular noise can be minimized.


5. What is the advantage of DPCM? With a neat block diagram explain transmitter and receiver.


6. Write notes on ISI and eye pattern.Refer: Wayne Tomasi, “Advanced Electronic Communication Systems”

Fifth edition, Page No.460

UNIT IV -SPREAD SPECTRUM AND MULTIPLE ACCESS TECHNIQUES

PART A

1. Define spread spectrum.Spread spectrum is defined in two parts

i. Data of interest occupies a bandwidth in excess of the minimum bandwidth necessary to send the date.

ii. The spectrum spreading is accomplished before transmission through the use of code that is independent of data sequence. The same code is used in receiver to despread the signal so that the original data may be recovered.

2. What do you mean by direct sequence spread technique?

IT 2202 15

The data sequence directly modulates the pseudo noise sequence. Let the data signal be b (t) and pseudo-noise signal be c (t). Then the modulated signal is given as, m (t) = b (t) c(t).

3. What are the advantages of spread spectrum modulation?Spread spectrum modulation spreads the message signal over wide

bandwidth with the help of special code (key). It has following important advantages.

i) Unwanted interference is rejected.ii) Protection against antijamming signals is also provided.iii) Multipath interference rejection.

4. Define pseudo noise sequence.The pseudo noise sequence is a noise like high frequency signal. This

signal is binary in nature. It looks like pulses. The sequence is not completely random, but it is generated by a well-defined logic. The same logic is used at transmitter and receiver. Hence the sequence is rather ‘pseudo’ random. Hence it is called pseudo-random (or pseudo-noise) sequence. The pseudo noise sequence can be generated by a feedback shift register and combinational logic.

5. What is frequency hop spreading?In frequency hop spread spectrum, the carrier frequency randomly changes

among different slots. These frequency slots are called hops. The data is transmitted in these hops.

6. What are the applications of Spread spectrum modulation?i. The spread spectrum has the ability to resist the effect of intentional

jamming. Previously this antijam capability was used in military application. Some commercial applications also use spread spectrum because of its antijam capability.

ii. Low probability of intercept is an application of spread spectrum in military .In this case, the signal spectral density is kept small such that the presence of the signal is not detected easily.

iii. Spread spectrum is used in mobile communications. This is because the spread spectrum signal has the ability to resist the effects of multipath fading. Because of wide spectrum only small portion of the signals is in fade.

7. List the advantages of direct sequence systemsi. This system has best noise and antijam performance

ii. Unrecognized receivers find it most difficult to detect direct sequence signals

iii. It has best discrimination against multipath signals8. List the disadvantages of direct sequence systems

i. It requires wideband channel with small phase distortionii. It has long acquisition time

iii. The pseudo-noise generator should generate sequence at high ratesiv. This system is distance relative

9. List the advantages of frequency hopping systemsi. These systems bandwidth are very large

ii. They can be programmed to avoid some portions of the spectrum

IT 2202 16

iii. They have relatively short acquisition time.iv. The distance effect is less

10. List the disadvantages of frequency hopping systemsi. These systems need complex frequency synthesizers

ii. They are not useful for range and range –rate measurementiii. They need error correction.

11. Define slow frequency hoppingWhen several symbols are transmitted in one frequency hop (slot), then it

is called slow frequency hopping. This means the symbol rate is higher than hop rate.

12. Define fast frequency hoppingWhen several frequency hops take place to transmit one symbol, then it is

called fast frequency hopping. This means the symbol rate is less than hop rate.13. What is processing gain?

Processing gain is defined as the ratio of the bandwidth of spreaded signal to the bandwidth of the unspreaded signal

14. What are the properties of maximum length sequence?Balance property: The number of 1’s is always one more than the number of zeros in each period of a maximum length sequenceRun property: Run means subsequence of identical symbols i.e. 1’s or 0’s within one period of the sequence. The length of the run is equal to the length of the subsequence.Correlation property: The auto correlation function of maximum length sequence is periodic and it is binary valued.

15. Mention the classification of multiple access protocols.i. Time division multiple access (TDMA)

ii. Frequency division multiple access (FDMA)iii. Code division multiple access (CDMA).

16. Compare slow and fast frequency hopping.Slow Frequency Hopping Fast Frequency Hopping

1.More than one symbols are transmitted per frequency hop.

More than one frequency hops are required to transmit one symbol.

2.Chip rate is equal to symbol rate. Chip rate is equal to hop rate.3. Symbol rate higher than hop rate.

Hop rate higher than symbol rate.

17. What is TDMA?In time division multiple access, the time of the channel is shared by

multiple users. Complete bandwidth of the channel is available to the user in given time slot. TDMA is in satellite communication.

18. What does CEPT stand for?Conference of European Postal and Telecommunication Administrations

(CEPT) is a commonly used TDMA frame format for digital satellite systems.19. Give the expression for probability of gain.

The bandwidth of signal before and after encoding, using spread spectrum is related by processing gain (PG)

PG=Tb / Tc

IT 2202 17

Where, Tb – bit durationTc – chip duration

20. What are orthogonal codes?If half the bits within a code were made the same and half were exactly the

opposite, the resultant would be zero cross correlation between chip codes. Such a code is called orthogonal codes.

21. What are the two different techniques used in speech coding for wireless communication?

i. Multi-pulse excited Linear Predictive Coding (LPC).ii. Code-excited LPC

22.What are the two function of fast frequency hopping?1. Spread Jammer over the entire measure of the spectrum of transmitted signal.2. Retuning the Jamming signal over the frequency band of transmitted signal.

23.What are the features of code Division multiple Accesses?

1. It does not require external synchronization networks.2. CDMA offers gradual degradation in performance when the no. of users is increased But it is easy to add new user to the system.3. If offers an external interference rejection capability.

PART - B

1. Describe what a reference burst is for TDMA and explain the following terms: preamble, carrier recovery sequence, bit timing recovery, unique word and correlation spike.


2. Describe the operation of CEPT primary multiplex frame. Refer: Wayne Tomasi, “Advanced Electronic Communication Systems”

Fifth edition, Page No.11063. Explain in detail the notion of spread spectrum.

Refer: Simon Haykin, “Communication Systems” Fourth edition, Page No.488.

4. Explain CDMA and also give the orthogonal condition of the signals in CDMA.


5. Explain direct sequence-spread spectrum with appropriate waveforms and expressions.


6. Explain the generation of pseudo-noise sequence with an example.

IT 2202 18

Refer: Simon Haykin, “Communication Systems” Fourth edition, Page No.480

7. List the properties of maximum length sequence with examples.Refer: Simon Haykin, “Communication Systems” Fourth edition, Page

No.482.8. With the help of transmitter and receiver diagram explain frequency hopped

spread spectrum and discuss about slow –frequency hop and Fast-frequency hop.


UNIT-V --SATELLITE AND OPTICAL COMMUNICATION

PART-A1. Define orbit.

The satellite can be rotated around the earth through various paths. These paths are called orbits of the satellite. These orbits are used to cover the specific application areas.

2. What is a satellite system?A satellite system consists of one or more satellite space vehicles, a

ground – based station to control the operation of the system, and a user network of earth stations.

3. State the basic function of satellite transponder.The transmitter-receiver combination in the satellite is known as a

transponder. The basic function of a transponder is amplification and frequency translation.

4. List the satellite orbits.Satellite orbits about the earth are either circular or elliptical. The satellite

orbits are: • Inclined orbit• Polar orbit• Equatorial orbit

5. Define inclined orbit.Inclined orbits are virtually all orbits except those that travel directly

above the equator or directly over the north and south poles.6. Define polar orbit.

Satellite orbits with inclinations of 90°are called polar orbit. Polar orbits are used for special applications like navigational satellites.

7. Define geostationary orbit.The circular equatorial orbit is exactly in the plane of equator on earth. All

the points in this orbit are at equal distance from earth surface, and a satellite in this orbit appears to be stationary to the point of earth. Therefore this orbit is called geostationary orbit.

8. Define geosynchronous orbit.

IT 2202 19

When the inclination of the orbit is not zero and eccentricity is not zero, it is called as geosynchronous orbit. The period of geosynchronous orbit is equal to the period of revolution of earth with itself.

9. Define perigee and apogee.The point in the orbit where the satellite is closest to the earth is called the

perigee.The point in the orbit where the satellite is farthest from the earth is called

the apogee.10. Define angle of inclination and angle of elevation.

Angle of inclination: It is the angle between the earth’s equatorial plane and the orbital plane of a satellite measured counter clockwise at the point in the orbit where it crosses the equatorial plane traveling from south to north.

Angle of elevation: It 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.

11. State the laws of planetary motion.Kepler’s law may be simply stated as (1) the planets move in ellipses with

the sun at one focus,(2) the line joining the sun and a planet sweeps out equal areas in equal intervals of time, and (3) the square of the time of revolution of a planet divided by the cube of its mean distance form the sun gives a number that is the same for all planets.

12. How are satellites classified based on elevation?i. Low earth orbit (LEO): 1 GHz -2.5 GHz

ii. Medium Earth orbit (MEO): 1.2GHz-1.66 GHziii. Geosynchronous earth orbit (GEO): 2 GHz -18 GHz

13. Define Azimuth angle.It is defined as the horizontal pointing angle of an earth station antenna.

14. Define a transponder. What is its basic function?A satellite radio repeater is called a transponder. It is an RF-RF repeater.

15. What is a footprint?The geographical representation of a satellite antenna’s radiation pattern is

called a footprint or footprint map.

16. What is station keeping?The process of maneuvering a satellite within a pre assigned window is

called station keeping.17. State the uplink frequency and downlink frequency.

A typical uplink frequency is 6 GHz and a common downlink frequency is 4 GHz.

18. What are the techniques for increasing channel capacity?Two of the techniques for increasing channel capacity are:

• Frequency reuse• Spatial isolation

19. What are the major subsystems in a communication satellite?The major subsystems in a communication satellite are

IT 2202 20

• Communications subsystems• Power subsystems• Telemetry tracking and control (TTC) subsystems• Propulsion subsystems• Attitude stabilization subsystems• Antenna subsystems

20. What is the basic transponder configuration?There are three basic transponder configurations used in communication

systems. They are• Single conversion transponder• Double conversion transponder• Regenerative transponders

21. State the major subsystems in a satellite earth station.The major subsystems in a satellite earth station are:• Transmit subsystems• Receive subsystems• Power subsystems• Antenna subsystems• Telemetry tracking and control (TTC) subsystems• Ground control equipment (GCE) subsystems.

22. List the applications of a satelliteSome of the applications of a satellite are:

• Surveillance or observation• Navigation• TV broadcast• Satellite telephones

23. What is an optical communication system?It uses light as the carrier of information. It uses glass or plastic fiber

cables to contain the light waves and guide them 24. State the frequency range of optical fiber communication.

Light frequencies range - 1x1014 to 4x1014

25. List the optical sources used for optical fiber communication.The optical sources used for optical fiber communication are

• Light Emitting Diodes and• Solid-state lasers.

26. List the optical detectors.The optical detectors used for optical fiber communication are

• Photodiodes and• Avalanche photodiodes.

27. State the applications of optical fibers.a. Local and long distance telephone systemsb. TV studio-to-transmitter interconnection, eliminating microwave radio link.c. Closed circuit TV systems used in buildings for security.d. Computer networks, wide area and local area.f. Shipboard communications.

IT 2202 21

g. Aircraft communication and Aircraft controlsk. Nuclear plant instrumentation.

28. List the advantages of optical fiber systems.1. Wider bandwidth: they have higher information carrying capability.2. Lower loss: with fiber optic cables, there is less signal attenuation over long distances.3. Lightweight: glass or plastic cables are much lighter than copper cables and offer benefits in those areas where low weight is critical (i.e., aircraft).4. Small size: practical fiber optic cables are much smaller than electrical cables in diameter; therefore more can be contained in a smaller space.5. Strength: fiber optic cables are stronger than electrical cables and can support more weight.6. Security: fiber optic cables cannot be “tapped” as easily as electrical cables; they don’t radiate signals that can be picked up for eavesdropping purposes.7. Interference immunity: fiber optic cables do not radiate signals as some electrical cables do and cause interference to other cables. They are also immune to pickup of interference from other sources.8. Greater safety: fiber optic cables don’t carry electricity. Therefore, there is no shock hazard. They are also insulators so are not susceptible to lightning strikes as electrical cables are.

29. Define refractive index.Refractive index or index of refraction (n) is ratio of the velocity of

propagation of a light ray in free space to the velocity of propagation of a light in a given material.

n=c/vwhere, n-refractive index(unit less)

c-velocity of light in free space (3*10^8 meters per second)v- velocity of light in given material (meters per second)

30. Define critical angle.The angle of incidence at which the refracted angle becomes 90°to the

normal is said to be the critical angle.

31. Define total internal reflection.When the light ray strikes the interface between the air and the glass, at an

angle greater than the critical angle, the light ray does not pass through the interface into the glass. When this occurs the angle of reflection is equal to the angle of incidence as if a real mirror were used. This action is known as total internal reflection.

32. Describe the construction of optical fiber cable.The portion of a fiber optic cable that carries the light is made from either

glass (silica) or plastic. The fiber, which is called the core, is usually surrounded by a protective cladding. The cladding is also made of glass or plastic of low refractive index. This ensures that proper interface is achieved so that the light

IT 2202 22

waves remain within the core. In addition to protecting the fiber, the cladding adds strength. A plastic jacket is put over the cladding for insulation.

33. Define mode.Mode refers to the various paths that the light rays can take in passing

through the cable. There are two classifications: • Single mode and• Multimode.

34. What is meant by step index fiber and graded index fiber?Step index fiber refers to the fiber, in which there is a sharply defined step

in the index of refraction where the fiber core and cladding interface. It means the core has one constant refractive index N1 while the cladding has another constant refractive index N2.

Graded index fiber refers to the fiber, in which the refractive index of the core is not constant. Instead it varies smoothly and continuously over the diameter of the core, reaching a peak at the center and then declining as the outer edge of the core is reached.

35. What is single mode fiber and multimode fiber?A single mode fiber is one in which, the light follows a single path

through the core and a multimode fiber is one in which, the light takes many paths through the core.

36. What is meant by modal dispersion?The stretching of the pulse due to the attenuation of the light in the cable

and increase in duration of arrival times of various light rays is referred to as modal dispersion.

37. List the reasons for the losses in optical fiber cable.The main reasons for the losses in optical fiber cable are

• Light absorption• Scattering and• Dispersion.

38. Define absorption.Absorption refers to how the light waves are actually “soaked up” in the

core material.39. Define scattering.

Scattering refers to the light lost because of light waves entering at the wrong angle and being lost in the cladding due to refraction.

40. State the expression for attenuation in optical fiber cable.The attenuation of a fiber optic cable is expressed in decibels per unit of

length. The standard decibel formula used is

IT 2202 23

41. Write short notes on cable splicing and connectors.Connectors are special mechanical assemblies that allow fiber optic cables

to be connected to one another. Fiber optic connectors are the optical equivalent of electrical plugs and sockets. Fiber optic connectors can be spliced by gluing. Connectors are used at the repeater units and at the end of the cable applied to the light source or photo detector.

42. What are the advantages of lasers over LED’s?LED’s covers a narrow spectrum of frequencies, less intense and are good

only for short distances whereas, the Lasers are monochromatic, coherent, and highly intense and can be used over long distances.

43. What are the advantages of single mode step index fiber over multimode step index fiber?

Single mode step index fibers are of extremely small size and therefore difficult to make and very expensive. Whereas multimode step index fibers are the easiest to make and the least expensive.

44. Name the multiplexing scheme used for fiber optic communication.The multiplexing scheme used for fiber optic communication is WDM,

wavelength division multiplexing.

PART - B

1. Explain, in detail, about light wave propagation. Refer: Wayne Tomasi, “Advanced Electronic Communication Systems”

Fifth edition, Page No.5832. Classify and explain the Fiber optic cable based on index of refraction and

mode. Refer: Wayne Tomasi, “Advanced Electronic Communication Systems”

Fifth edition, Page No.5943. Explain in detail loses in optical fibers.


4. Draw the basic block diagram of a Fiber optic communication system and explain the function of each block.


5. Explain optical sources used in fiber optic communication. Refer: Wayne Tomasi, “Advanced Electronic Communication Systems”

Fifth edition, Page No.6076. Explain optical sources used in fiber optic communication.


7. Write notes on satellite elevation categories and satellite orbital patterns.

IT 2202 24


8. Explain the satellite system with a neat block diagram.Refer: Wayne Tomasi, “Advanced Electronic Communication Systems”

Fifth edition, Page No.10769. State and explain the laws of planetary motion.


10. What is a geosynchronous satellite? Discuss its advantages and disadvantages.


B.E./B. TECH. DEGREE EXAMNIATION, NOV/DEC 2005

Third Semester

Information Technology

IT1202-PRINCIPLES OF COMMUNICATION

Time: Three hours Maximum: 100 marksAnswer all Questions

IT 2202 25

PART A – (10 *2 =20marks)

1. If the rms value of the aerial current before modulation is 12.5A and during modulation is 14A.Calculate the percentage of modulation employed, assuming no distortion.

Itotal = Ic [1 + m2 / 2] ½ => Itotal/ Ic = [1 + m2 / 2] 1/2

14/12.5 = [1 + m2 / 2] ½ =1.2544 =1+m2/ 2 m=0..7133M= 71.33%

2. Draw the block diagram of a super heterodyne AM receiver and explain why the usual AM radio receiver uses a super heterodyne system.

AM uses a Super heterodyne receiver to mix two frequencies together in a nonlinear device or to translate one frequency to another, using nonlinear mixing.

3. Write down the comparison of frequency and amplitude modulation. Amplitude modulation Frequency modulation

1. Noise interference is more Noise interference is less2. 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.



But in FM the depth ofmodulation can be increased to any value by increasing the deviation.

4. Simple circuits used in transmitter and receiver.

Uses more complex circuits in transmitter and receiver.

5. Power varies in AM depending on depth of modulation.

The amplitude of FM is constant. Hence transmitter power remains constant in FM

4. An angle modulated signal is described by Xc(t)=10 cos[2π (106) t + 0.1 sin (103) π t] considering Xc(t) as PM Signal with Kp=10,find m(t).]

m(t)=Vc sin[ωc t + KpVm cos ωm t]

IT 2202 26

m(t)=10 cos[2π (106) t + 1 sin (103) π t].5. Sketch the waveform representation of ASK, FSK, PSK for NRZ coded

binary sequence and represent also each case mathematically.

(a)PSK

(b) ASK

( c) FSK6. Draw the signal constellation of QPSK and give comments on QPSK

QPSK is an M-ary encoding scheme with N=2 and M=4. With two bits there are four possible conditions and four possible output phases +450 , +1350, -450 and -1350.

IT 2202 27

7. Define sampling theorem.If a finite energy signal g(t) contains no frequency higher than W Hz, it is


8. Draw Eye pattern and interpret it properly.MN - Margin over noiseJT – distortion at zero crossingsST – time interval over which the wave can be sampled. DA – distortion at sampling time.

9. Mention the classification of multiple access protocols.i. Time Division Multiple access (TDMA)

ii. Code Division Multiple access (CDMA)iii. Frequency Division Multiple access (FDMA)

10. Sketch the structure of a CDMA system and give the orthogonal condition for the signals in CDMA.

CDMA Encoder

Data in Logic 1=+1 Logic 0 = -1X Orthogonal code 1 1 –1 1 –1 -1 -1 –1 1 –1 1 1------------------------- ---------------------------- ------------------------------Product code 1 1 -1 1 -1 -1 -1 –1 1 -1 1 1Recovered chip code 1 1 -1 –1 1 1 1 1 -1 -1 1 1--------------------------- ---------------------------- ------------------------------Correlation 1 1 1 -1 -1 -1 -1 -1 -1 1 1 1

PART B—(5*16=80 marks)

11. (i) Verify that the message signal m(t) is recovered from a modulated DSB signal by first multiplying it by a local sinusoidal carrier and then passing

IT 2202 28

BalancedModulator

PSKModulator

RFModulator

High-power amplifier & BPF

N-bit code word generator

IFCarrier

RFCarrier

the resultant signal through a low pass filter(1) in the time domain(2) in the frequency domain.(ii) Derive the efficiency η of ordinary AM and show that for a single tone AM, ηmax=33.3% at μ =1.

(i)

s(t) v(t) Vo(t)

(ii) Psb = ma 2Ec

2 / 4 Pt = Ec2/2[1+ma

2/2] η = Psb / Pt at μ =1 , ma = 1 substituting ,

ηmax = 33.3%12. (a) Consider an angle modulated signal Xc(t) = 10 cos (ωc(t) + 3 sin

ωm(t)).Assume FM and fm=1kHz.calculate the modulation index and find the bandwidth when (1)fm is doubled (2) fm is by one half

Carson’s rule is B=2(Δf +fm) Hz. m=3DR= Δf (max) / fm(max)

Δf= 3KHZ(1)fm is doubled , B=10 KHZ (2) fm is by one half, B= 7KHz

(b) (i)A block diagram of an indirect (Armstrong) FM transmitter is shown in fir.(1).Compute the maximum frequency deviation(∆f) of the output of the FM transmitter and the carrier frequency fc is f1=200 kHz, flo=10.8MHz, ∆f1=25Hz, n1=64 and n2=48

out m(t)

f1 X n1 f3 X n2

∆f1 f2=n1f1 ∆f3 fc

∆f2 ∆fflo

f2=n1f1 = 64*200 kHz=12.8 MHz.f3 = f2 - flo =2MHzfc =n2 f3 = 48*2 =96 MHz

∆f = n2∆f3 = 76.8kHz13. (a) With the help of a block diagram (transmitter and receiver) explain the

principle of a differential phase shift keying. Illustrate the generation of DPSK signal with an example sequence.


14. (b)Write short notes on intersymbol interference.Refer: Wayne Tomasi, “Advanced Electronic Communication Systems”

Fifth edition, Page No.460

IT 2202 29

~

Narrow band FM

Frequency multiplier X

Frequency multiplier

Product Modulator

LPF

LocalOscilltor

15. (a) With neat transmitter and receiver block diagram of Direct Sequence spread coherent phase shift keying, explain its principle of operation.

Refer: Simon Haykin, “Communication Systems” Fourth edition, Page No.490(b) (i) Describe what a reference burst is for TDMA and explain the following terms: preamble, carrier recovery sequence, bit timing recovery, unique word and correlation spike.

(ii) Briefly describe the operation of CEPT primary multiplex frame.Refer: Wayne Tomasi, “Advanced Electronic Communication

Systems” Fifth edition, Page No.1104

B.E./B. TECH. DEGREE EXAMNIATION, NOV/DEC 2006

Third Semester

Information Technology

IT1202-PRINCIPLES OF COMMUNICATIONTime: Three hours Maximum: 100 marks

Answer all Questions

PART A – (10 *2 =20marks)

1. A broadcast radio transmitter radiates 5 KW power when the modulation percentage is 60% .How much is the carrier power?

Pt = 5kW,m=0.6 or 60%Ptotal = Pc [1 + m2 / 2]

Pc = Ptotal / [1 + m2 / 2] = 5* 102 / [1+ (.6) 2 / 2] = 4.24 kW2. What are the two major limitations of the standard form of amplitude

modulation? i. AM less efficient because most of power is transmitted in carrier.ii. The effect of noise is more because of amplitude variations.

3. Illustrate the relationship between frequency modulation (FM) and phase modulation.

Phase modulation is the first integral of FM. Ө (t) = ∫ Ө’ (t) dt

where, Ө (t) – instantaneous phase deviation Ө’ (t) –instantaneous frequency deviation

4. A carrier is frequency modulated with a sinusoidal signal of 2 kHz resulting in a maximum frequency deviation of 5 kHz. Find the bandwidth of the modulated signal.

fm=2kHz∆f = 5 kHzBW= 2(∆f + fm ) =2(5 *103 + 2 * 103)=14 KHz

5. Compare QASK and QPSK. QPSK QASK

1. Quadrature phase modulation Quadrature phase and amplitude

IT 2202 30

modulation2.All signal points placed on circumference of circle


3. Circuit is simple. Relatively complex4.Noise immunity better then QASK.

Poor than QPSK.

5. Error probability less then AQSK Higher than QPSK6. What are Antipodal signals?

In BPSK, the two symbols are transmitted with the help of following signals,

Symbol ‘1’ => s1 (t) = √2P cos (2πf0 t)Symbol ‘0’ => s2 (t) = √2P cos (2πf0 t + π)

Here observe that above two signals differ only in a relative phase shift of 1800. Such signals are called antipodal signals.

7. State sampling theorem for band-limited signals of finite energy.If a finite energy signal g(t) contains no frequency higher than W Hz, it is


8. How is eye pattern used to measure intersymbol interference in a data transmission system?

i. The width of the eye opening defines the time interval over which the received wave can be sampled without error from intersymbol interference. It is apparent that the preferred time for sampling is the instant time at which the eye is open widest.

MN - Margin over noiseJT – distortion at zero crossingsST – time interval over which the wave can be sampled. DA – distortion at sampling time

ii. When the effect of intersymbol interference (ISI) is severe traces from the upper portion of the eye pattern cross traces from the lower portion with the result that the eye is completely closed.

9. Illustrate the two modes of an adaptive equalizer.a. Training modeb. Decision Directed mode

Decision device

IT 2202 31

x[n] y[n] a ^[n] a[n]

e[n]

10. List out the comparison between slow and fast frequency hopping.

Slow Frequency Hopping Fast Frequency Hopping1.More than one symbols are transmitted per frequency hop.




PART-B (5 X 16=80 MARKS)

11. (a) (i)Explain the process of demodulating DSBSC signal by using synchronous demodulator, with the help of neat block diagram and frequency spectrum.

Refer Nov/Dec 2005 –11 (i)12. (a)Determine the spectrum of single tone FM wave for an arbitrary value of

the modulation index β.Refer: Wayne Tomasi, “Advanced Electronic Communication Systems” Fifth edition, Page No.282.

(b)(i) A 20 MHz carrier is frequency modulated by a sinusoidal signal such that the maximum frequency deviation is 100 kHz. Determine the modulation index and the approximate bandwidth of the FM signal if the frequency of the modulating signal is 100 kHz.

m= ∆f/ fm =1 BW= 2(∆f + fm ) = 400kHz.

13. (a) (i) Explain BPSK transmitter and receiver with the help of block diagram.

Refer: Wayne Tomasi, “Advanced Electronic Communication Systems” Fifth edition, Page No.376.(b) (ii)The bit stream 1011100011 is to be transmitted using DPSK. Determine the encoded sequence and transmitted phase sequence

Input 1 0 1 1 1 0 0 0 1 1XNOR Output

0 0 1 1 1 1 0 1 0 0 0

Output 180 180 0 0 0 0 180 0 180 180 180

IT 2202 32

Adaptive equalizer[wk]

Trainingsequencegenerator

∑

phase 14. (a) (i) Draw the functional block diagram of a direct sequence spread

spectrum modulator and demodulator and explain.Refer: Simon Haykin, “Communication Systems” Fourth edition, Page

No.490.(ii) What are the applications of spread spectrum techniques?

Refer: Simon Haykin, “Communication Systems” Fourth edition, Page No.480.(b) With the help of transmitter and receiver block diagram, explain the principle of operation frequency hopped (FH) spread spectrum system


B.E./B. TECH. DEGREE EXAMNIATION, April/May 2008

Third SemesterInformation Technology

IT1202-PRINCIPLES OF COMMUNICATIONTime: Three hours Maximum: 100 marks

Answer all QuestionsPART A – (10 *2 =20marks)

1. Draw the spectrum of AM signal.

Vc

mVc/2 mVc/2

fm fm

fc-fmfc fc+fm frequency fLSB fUSB

2. Calculate the BW of FM signal whose frequency deviation is 75KHz and signal frequency is 2.5KHz.

fm=2.5kHz∆f = 75 kHzBW= 2(∆f + fm ) =2(75 *103 + 2.5 * 103)=155 KHz

3. Define phase modulation.Phase of a constant amplitude carrier is varied directly proportional to the

amplitude of the modulating signal at a rate equal to the frequency of the modulating signal.

4. Bring out the difference between DPSK and BPSK.DPSK BPSK

1.it does not need a carrier at It needs a carrier at receiver

IT 2202 33

its receiver2.Bandwidth reduced compared to BPSK

More bandwidth.

3.Probability of error or bit error rate more than BPSK

Comparatively low

4.error propagation more, since it uses two bits for its reception

Comparatively low, since it uses only single bit

5.noise interference more Comparatively low5. Mention any four advantages of Digital Modulation over analog modulation.

i. Maximum data rateii. Minimum probability of symbol error

iii. Minimum transmitted power.iv. Minimum channel bandwidth

6. What is PAM? The amplitude of constant width, constant-position pulse is varied

according to the amplitude of the sample of the analog signal.7. Define source-coding theorem.

The entropy represents a fundamental limit on the average number of bits pr source symbol necessary to represent a discrete memoryless source in that it can be made as small as, but smaller than entropy.

8. What are the properties of pseudo noise sequence?Balance property: The number of 1’s is always one more than the number of zeros in each period of a maximum length sequenceRun property: Run means subsequence of identical symbols i.e. 1’s or 0’s within one period of the sequence. The length of the run is equal to the length of the subsequence.Correlation property: The auto correlation function of maximum length sequence is periodic and it is binary valued.

9. Differentiate between Fast FH and Slow FH with respect to symbol rate and hop rate.

Slow Frequency Hopping Fast Frequency Hopping1.More than one symbols are transmitted per frequency hop.




10. Write the condition required to avoid the slope over load distortion in delta modulation.

a. Increasing the clock frequency reduces the probability of slope overload noise.

b. increase the magnitude of the minimum step size.

PART – B11. (a) (ii) Derive the waveform of power relation between carrier power and

total transmitter power of AM signal.

IT 2202 34

Refer: Wayne Tomasi, “Advanced Electronic Communication Systems” Fifth edition, Page No.149(b) (ii) Draw the waveform of AM signal for over modulation, under modulation and 100% modulation.

(a) Under modulation

(b) 100% modulation (c) over modulation

12. (b) (i) Distinguish between FM and AM.Refer: Wayne Tomasi, “Advanced Electronic Communication

Systems” Fifth edition, Page No.319. 13. (a) With block diagram explain QPSK transmitter and receiver. Draw the

QPSK waveform for a digital data 100110. Refer: Wayne Tomasi, “Advanced Electronic Communication Systems” Fifth edition, Page No.381.

(b)(ii)Define MSK and how is it different from FSK. Refer: Wayne Tomasi, “Advanced Electronic Communication Systems” Fifth edition, Page No.375

14. (a)(ii)What is meant by Intersymbol interference? (ISI). Suggest a suitable signal processing that has to be done to the baseband signal to overcome ISI.


15. (a)(i) What is spread spectrum modulation? Describe the following features of spread spectrum modulation:

(1) Antijamming(2) Ranging(3) Multiple accessing(4) Message security.

IT 2202 35

(ii) Draw a PN sequence generator to generate 7 bit sequence. Refer: Simon Haykin, “Communication Systems” Fourth edition, Page

No.479. (b) (i) Distinguish CMDA and TDMA.

Refer: Wayne Tomasi, “Advanced Electronic Communication Systems” Fifth edition, Page No.1104 & 1108

(ii) Draw DS spread spectrum modulation with block diagram. Derive an expression for processing gain PG.


2009 Anna University B.Tech Information Technology IT 2202-Principles Of Communication Question Paper Regulation 2008

Answer all questions PART-A

1. Define amplitude modulation.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.

2. What is modulation index?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 ism = Em/ Ec

Where m = Modulation coefficientEm = Peak change in the amplitude of the output waveform voltage.Ec = Peak amplitude of the unmodulated carrier voltage.

3. Why is ASK called as ON-OFF keying ?ASK (Amplitude Shift Keying) is a modulation technique which converts digital data to analog signal. In ASK, the two binary values(0,1) are represented by two different amplitudes of the carrier signal. S(t) = Acos2ft binary 1

0 binary 04. What is the difference between QASK and QPSK ?

QPSK QASK

1. Quadrature phase modulation

Quadrature phase and amplitude modulation

2.All signal points placed on circumference of circle


3. Circuit is simple. Relatively complex

IT 2202 36

4.Noise immunity better then QASK

Poor than QPSK.

5. Error probability less then AQSK

Higher than QPSK

5. Define sampling rate.sampling rate states that, the minimum sampling rate is equal to twice the highest audio input frequency.6. What is ISI ?The ringing tails of several pulses have overlapped, thus interfering with major pulse lobe. This interference is commonly called as intersymbol interference or ISI.

7. Define Pseudonoise sequence.The pseudo noise sequence is a noise like high frequency signal. This signal is binary in nature. It looks like pulses. The sequence is not completely random, but it is generated by a well-defined logic. The same logic is used at transmitter and receiver. Hence the sequence is rather ‘pseudo’ random. Hence it is called pseudo-random (or pseudo-noise) sequence. The pseudo noise sequence can be generated by a feedback shift register and combinational logic.

8. What are the different types of multiple access techniques ?iii. Time division multiple access (TDMA)iv. Frequency division multiple access (FDMA)v. Code division multiple access (CDMA).

9. Define Kepler's three laws of planetary motion.Kepler’s law may be simply stated as (1) the planets move in ellipses with the sun at one focus,(2) the line joining the sun and a planet sweeps out equal areas in equal intervals of time, and (3) the square of the time of revolution of a planet divided by the cube of its mean distance form the sun gives a number that is the same for all planets.10. What are the losses encountered by optical fiber ?Absorption lossMaterial or Rayleigh, scattering lossesChromatic (or) wavelength, dispersionRadiation lossesModel dispersionCoupling losses

PART-B

11. (a) (i)Write short notes on(1) AM voltage distribution. (4)(2) AM power distribution. (4)Refer: Wayne Tomasi, “Advanced Electronic Communication Systems” Fifth edition,

IT 2202 37

Page No.145-153

(ii) An audio frequency signal 10sin2p*500t is used to amplitude modulate a carrier of 50 sin2p*10^5t.Calculate.(1) Modulation index. (2)Vm/vc=10/50=0.2(2)Side band frequencies. (2)USB= fc+fm=500+100000=100500 HzLSB=fc-fm=90500 Hz(3)BW required. (2)2fm =500*2=1000Hz

(4)Total power delivered to the load of 600O. (2)Ptotal = Pc [1 + m2 / 2]

(OR)(b) (i)Compare FM and AM.(12)

Amplitude modulation Frequency modulation1. Noise interference is more Noise interference is less2. 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.



But in FM the depth of modulation can be increased to any value by increasing the deviation.

(ii) The phase deviation constant in a phase modulation system is K=0.01rad/v.Calculate the maximum phase deviation when the modulating signal of 10V is applied. (4)

12. (a) (i)Explain the principle of FSK transmitter and receiver (10)Refer: Wayne Tomasi, “Advanced Electronic Communication Systems” Fifth edition, Page No.364-375

(ii)Write short note on spectrum and bandwidth of FSK. (6)Refer: Wayne Tomasi, “Advanced Electronic Communication Systems” Fifth edition, Page No.369-372(OR)(b) (i)Compare the various types of digital modulation techniques. (8) Refer: Wayne Tomasi, “Advanced Electronic Communication Systems” Fifth edition,

IT 2202 38

(ii)Explain the eye pattern in base band digital transmission with neat diagram. (8)

Refer: Wayne Tomasi, “Advanced Electronic Communication Systems” Fifth edition, Page No.462-463

13. (a) (i)Explain the elements of PCM system with a neat block diagram. (12)Refer: Wayne Tomasi, “Advanced Electronic Communication Systems” Fifth edition, Page No.425-427(ii)What is companding ? (4)Companding is the process of compression and then expanding. Higher amplitude signals are compressed prior to transmission and then expanded in the receiver. Companding is the means of improving dynamic range of communication systems.(OR)(b) (i)Find the signal amplitude for minimum quantization error in a delta modulation system if step size is 1 volt having repetition period 1 ms. The information signal operates at 100Hz. (4)

(ii) Describe the operations of DPCM system with relevant diagram. (12)Refer: Wayne Tomasi, “Advanced Electronic Communication Systems” Fifth edition, Page No.455-456

14. (a) (i)Write short notes on Frequency hop spread spectrum. (10)Refer: Simon Haykin, “Communication Systems” Fourth edition, Page No.499.(ii)Explain the applications of spread spectrum techniques. (6) The spread spectrum has the ability to resist the effect of intentional jamming. Previously this antijam capability was used in military application. Some commercial applications also use spread spectrum because of its antijam capability.

iv. Low probability of intercept is an application of spread spectrum in military .In this case, the signal spectral density is kept small such that the presence of the signal is not detected easily.

v. Spread spectrum is used in mobile communications. This is because the spread spectrum signal has the ability to resist the effects of multipath fading. Because of wide spectrum only small portion of the signals is in fade.

IT 2202 39

(OR)(b) (i)Give a detailed account of multiple access techniques. (10) i.Time division multiple access (TDMA)

ii.Frequency division multiple access (FDMA) iii.Code division multiple access (CDMA).

(ii)Compare TDMA and CDMA. (6)Refer: Wayne Tomasi, “Advanced Electronic Communication Systems” Fifth edition, Page No.1104 & 1108 15. (a) (i)Discuss briefly the basic satellite communication system . (8)Refer: Wayne Tomasi, “Advanced Electronic Communication Systems” Fifth edition, Page No.1076(ii)Write short notes on LEO and GEO orbits. (8)Refer: Wayne Tomasi, “Advanced Electronic Communication Systems” Fifth edition, Page No.1062(OR)(b) Enumerate the elements of an optical fiber transmission link. (16) Refer: Wayne Tomasi, “Advanced Electronic Communication Systems” Fifth edition, Page No.580

IT 2202 40

it 2202 poc

Documents