UNIT 2 ADC.pdf

7/28/2019 UNIT 2 ADC.pdf

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VSA GROUP OF INSTITUTIONS, SALEM

DEPARTMENT OF ECE

ANALOG AND DIGITAL COMMUNICATION

(FOR III-SEM-CSE STUDENTS)

UNIT-II

DIGITAL COMMUNICATION

PREPARED BY

Ms.V.Ezhilya

HOD PRINCIPAL


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ANALOG AND DIGITAL COMMUNICATION

UNIT I FUNDAMENTALS OF ANALOG COMMUNICATION

Principles of amplitude modulation, AM envelope, Frequency spectrum andbandwidth, Modulation index and percent modulation, AM voltage distribution,

AM power distribution, Angle Modulation-FM and PM waveforms, phasedeviation and modulation index, Frequency deviation and percent modulation,Frequency analysis of angle modulated wave, Bandwidth requirements for anglemodulated waves.

UNIT II DIGITAL COMMUNICATION

Introduction, Shannon limit for information capacity, Digital amplitudemodulation, Frequency shift keying, FSK bit rate and baud, FSK transmitter, BWconsideration of FSK, FSK receiver, Phase shift keying- Binary phase shift keying QPSK, Quadrature amplitude modulation, Bandwidth efficiency, Carrierrecovery Squaring loop, Costas loop, DPSK.

UNIT III DIGITAL TRANSMISSION

Introduction, Pulse Modulation, PCM PCM sampling, Sampling rate, Signal toquantization noise rate, Companding Analog and Digital Percentage error,Delta modulation, Adaptive delta modulation, differential pulse codemodulation, Pulse transmission Intersymbol interference, eye patterns.

UNIT IV DATA COMMUNICATIONS

Introduction, History of Data Communications, Standards Organizations fordata communicaton, Data communication circuits, Data communication codes,

Error control, Error detection, Error correction, Data communication hardware,Serial and parallel interfaces, Data modems, Asynchronous modem,Synchronous modem, Low speed modem, Medium and high speed modem,Modem control.

UNIT V SPREAD SPECTRUM AND MULTIPLE ACCESS TECHNIQUES

Introduction, Pseudo-noise sequences, DS spread spectrum with coherent binaryPSK, Processing gain, FH spread spectrum, Multiple access techniques Wirelesscommunication, TDMA and CDMA in wireless communication systems, Soursecoding of speech for wireless communication.


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Unit II DIGITAL COMMUNICATION

S. NoTopic Reference

TeachingAids

No ofhours

UNIT II DIGITAL COMMUNICATION

1Introduction to DigitalCommunication

T 1, Ch 2 CB 1

2Shannon limit for information

capacityT 1, Ch - 2 CB 2

3 Digital amplitude modulation T 1, Ch - 2 CB 1

4Frequency shift keying, FSK bitrate and baud

T 1, Ch - 2 CB 1

5FSK transmitter and FSKreceiver, BW consideration ofFSK

T -1 , Ch - 2 CB 1

6Phase shift keying, Binaryphase shift keying

T -1 , Ch - 2 CB 1

7 QPSK, Quadrature amplitudemodulation T -1 , Ch - 2 CB 1

8Bandwidth efficiency, Carrierrecovery , Squaring loop

T -1 , Ch - 2 CB 2

9 Costas loop, DPSK T -1 , Ch - 2 CB 2

Total Hours= 12


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Block diagram of Basic digital communications system :

Shannons limit for Information Capacity and Shannon Hartley Theorem:

Shannons limit for Information Capacity

Transmission channel is a medium over which the electrical signals from atransmitter travel to the receiver. Two important characteristics of a transmission

channel are1. Signal to Noise ratio(SNR)2. Bandwidth

These two characteristics will ultimately decide the maximum capacity ofa channel to carry information. Nyquist and Shannon worked on finding themaximum channel capacity of a band limited channel.

Nyquist theorem

It states that if the bandwidth of a transmission channel is B which

carries a signal having L number of levels, then the maximum data rate R onthis channel is given byR = 2 B log2 L

The number of levels L can be two or more.


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Shannons Theorem :

If (S/N) is the signal to noise ratio then the maximum data rate is given byR = B log2 [1 + (S/N)]

Shannon extended Nyquist work. He included the effect of noise presenton the transmission channel. Shannons theorem puts a limit on the maximum

number of levels for a given (S/N) ratio and bandwidth.

Shannon Hartley Theorem :

This theorem is complementary to the Shannons theorem. It applies to achannel in which the noise is Gaussian.

Information Capacity :

The information capacity of a communication system represents the number

of independent symbols that can be carried through that system in a givenunit of time.

It is expressed in bits/sec.

Statement of the Theorem :

The information capacity of a white, band limited Gaussian channel isgiven by

I = B log2 [1 + (S/N)] bits/ secWhere, B = Channel Bandwidth

S = Signal Power

N = Noise within the Channel Bandwidth

Effect of S/N on I

If the communication channel is noiseless then N = 0. Therefore (S/N)tends to infinity and so I also will tend to infinity. Thus the noiseless channelwill have an infinite capacity.

Effect of Bandwidth on Information Capacity:

If the bandwidth approaches infinity the channel capacity does notbecome infinite since N = N0 B will also increase with the bandwidth B.This will reduce the value of (S/N) with increase in B, assuming the signal

power S to be constant. Thus we conclude that an ideal system with infinitebandwidth has a finite channel capacity. It is denoted by I (Infinity) and givenby

I (Infinity) = 1.44 (S/N0)


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Shannons Information rate

The maximum rate of transmission is equal to the information capacityRmax = Imax = 1.44 (S/N0)

It is very difficult to achieve it practically because to achieve this rate thechannel bandwidth needs to be equal to infinity, and practically it is extremelydifficult to have a transmission channel with an infinite bandwidth.

Digital Modulation:

It is the process of varying the angle of a wave in a carrier in order totransmit analog or digital data. For digital signals, phase modulation (PM) iswidely used in conjunction with amplitude modulation (AM). For example,quadrature amplitude modulation (QAM) uses both phase and amplitudemodulation to create different binary states for transmission (see QAM). Seemodulation and carrier.

Fig : Digital modulated wave

In PM modulation, the angle of the carrier wave is varied by the incoming
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n this example, the modulating wave implies an analog signal.

Fig:Digital Phase Shift Keying (PSK)

For digital signals, phase shift keying (PSK) uses two phases for 0 and 1 asin this example. See DPSK.

There are three major classes of digital modulation techniques used fortransmission of digitally represented data:

Amplitude-shift keying (ASK)Frequency-shift keying (FSK)Phase-shift keying (PSK)

All convey data by changing some aspect of a base signal, the carrier wave,(usually a sinusoid) in response to a data signal. In the case of PSK, the phase ischanged to represent the data signal. There are two fundamental ways ofutilizing the phase of a signal in this way:

By viewing the phase itself as conveying the information, in which casethe demodulator must have a reference signal to compare the receivedsignal's phase against; or
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By viewing the change in the phase as conveying information differentialschemes, some of which do not need a reference carrier (to a certainextent).

A convenient way to represent PSK schemes is on a constellation diagram.This shows the points in the Argand plane where, in this context, the real and

imaginary axes are termed the in-phase and quadrature axes respectively due totheir 90 separation. Such a representation on perpendicular axes lends itself tostraightforward implementation. The amplitude of each point along the in-phaseaxis is used to modulate a cosine (or sine) wave and the amplitude along thequadrature axis to modulate a sine (or cosine) wave.

In PSK, the constellation points chosen are usually positioned withuniform angular spacing around a circle. This gives maximum phase-separationbetween adjacent points and thus the best immunity to corruption. They arepositioned on a circle so that they can all be transmitted with the same energy. In

this way, the moduli of the complex numbers they represent will be the sameand thus so will the amplitudes needed for the cosine and sine waves. Twocommon examples are "binary phase-shift keying" (BPSK) which uses twophases, and "quadrature phase-shift keying" (QPSK) which uses four phases,although any number of phases may be used. Since the data to be conveyed areusually binary, the PSK scheme is usually designed with the number ofconstellation points being a power of 2.

PSK system:

Phase-shift keying (PSK) is a digital modulation scheme that conveys

data by changing, or modulating, the phase of a reference signal (the carrierwave).

Any digital modulation scheme uses a finite number of distinct signals torepresent digital data. PSK uses a finite number of phases, each assigned aunique pattern of binary bits. Usually, each phase encodes an equal number ofbits. Each pattern of bits forms the symbol that is represented by the particularphase. The demodulator, which is designed specifically for the symbol-set usedby the modulator, determines the phase of the received signal and maps it backto the symbol it represents, thus recovering the original data. This requires the

receiver to be able to compare the phase of the received signal to a referencesignal such a system is termed coherent.

Alternatively, instead of using the bit patterns to set the phase of the wave,it can instead be used to change it by a specified amount. The demodulator thendetermines the changes in the phase of the received signal rather than the phaseitself. Since this scheme depends on the difference between successive phases, itis termed differential phase-shift keying (DPSK). DPSK can be significantly
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simpler to implement than ordinary PSK since there is no need for thedemodulator to have a copy of the reference signal to determine the exact phaseof the received signal (it is a non-coherent scheme). In exchange, it producesmore erroneous demodulations. The exact requirements of the particularscenario under consideration determine which scheme is used.

Binary phase-shift keying (BPSK)

Fig : Constellation diagram for BPSK

BPSK is the simplest form of PSK. It uses two phases which are separatedby 180 and so can also be termed 2-PSK. It does not particularly matter exactlywhere the constellation points are positioned, and in this figure they are shownon the real axis, at 0 and 180. This modulation is the most robust of all the PSKssince it takes serious distortion to make the demodulator reach an incorrectdecision. It is, however, only able to modulate at 1 bit/symbol (as seen in the

figure) and so is unsuitable for high data-rate applications when bandwidth islimited.

The bit error rate (BER) of BPSK in AWGN can be calculated as:

Since there is only one bit per symbol, this is also the symbol error rate.

In the presence of an arbitrary phase-shift introduced by thecommunications channel, the demodulator is unable to tell which constellationpoint is which. As a result, the data is often differentially encoded prior tomodulation.
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Implementation

Binary data is often conveyed with the following signals:

for binary "0"

for binary "1"

wherefc is the frequency of the carrier-wave.

Hence, the signal-space can be represented by the single basis function

where 1 is represented by and 0 is represented by . Thisassignment is, of course, arbitrary.

The use of this basis function is shown at the end of the next section in asignal timing diagram. The topmost signal shows PSK modulating a cosinewave, and is the signal that the BPSK modulator would produce. The bit-streamthat causes this output is shown above the signal (the other parts of this figureare relevant only to QPSK).

QPSK system with the help of transmitter and receiver:

Quadrature phase-shift keying (QPSK)

Constellation diagram for QPSK with Gray coding
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Each adjacent symbol only differs by one bit. Sometimes known asquaternary or quadriphase PSK or 4-PSK, QPSK uses four points on theconstellation diagram, equispaced around a circle. With four phases, QPSK canencode two bits per symbol, shown in the diagram with Gray coding to minimizethe BER twice the rate of BPSK. Analysis shows that this may be used either to

double the data rate compared to a BPSK system while maintaining thebandwidth of the signal or to maintain the data-rate of BPSK but halve thebandwidth needed.

Although QPSK can be viewed as a quaternary modulation, it is easier tosee it as two independently modulated quadrature carriers. With thisinterpretation, the even (or odd) bits are used to modulate the in-phasecomponent of the carrier, while the odd (or even) bits are used to modulate thequadrature-phase component of the carrier. BPSK is used on both carriers andthey can be independently demodulated.

As a result, the probability of bit-error for QPSK is the same as for BPSK:

.

However, with two bits per symbol, the symbol error rate is increased:

.

If the signal-to-noise ratio is high (as is necessary for practical QPSKsystems) the probability of symbol error may be approximated:

As with BPSK, there are phase ambiguity problems at the receiver anddifferentially encoded QPSK is more normally used in practice.

Implementation

The implementation of QPSK is more general than that of BPSK and alsoindicates the implementation of higher-order PSK. Writing the symbols in the
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constellation diagram in terms of the sine and cosine waves used to transmitthem:

.

This yields the four phases / 4, 3 / 4, 5 / 4 and 7 / 4 as needed.

This results in a two-dimensional signal space with unit basis functions

The first basis function is used as the in-phase component of the signaland the second as the quadrature component of the signal.

Hence, the signal constellation consists of the signal-space 4 points

.

The factors of 1 / 2 indicate that the total power is split equally between

the two carriers.

Comparing these basis functions with that for BPSK shows clearly howQPSK can be viewed as two independent BPSK signals. Note that the signal-space points for BPSK do not need to split the symbol (bit) energy over the twocarriers in the scheme shown in the BPSK constellation diagram.

QPSK systems can be implemented in a number of ways. An illustrationof the major components of the transmitter and receiver structure are shownbelow.
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Fig : Conceptual transmitter structure for QPSK

The binary data stream is split into the in-phase and quadrature-phasecomponents. These are then separately modulated onto two orthogonal basisfunctions. In this implementation, two sinusoids are used. Afterwards, the two

signals are superimposed, and the resulting signal is the QPSK signal. Note theuse of polar non-return-to-zero encoding. These encoders can be placed beforefor binary data source, but have been placed after to illustrate the conceptualdifference between digital and analog signals involved with digital modulation.

Fig : Receiver structure for QPSK

The matched filters can be replaced with correlators. Each detectiondevice uses a reference threshold value to determine whether a 1 or 0 is detected.

QPSK signal in the time domain

The modulated signal is shown below for a short segment of a randombinary data-stream. The two carrier waves are a cosine wave and a sine wave, asindicated by the signal-space analysis above. Here, the odd-numbered bits havebeen assigned to the in-phase component and the even-numbered bits to thequadrature component (taking the first bit as number 1). The total signal thesum of the two components is shown at the bottom. Jumps in phase can beseen as the PSK changes the phase on each component at the start of each bit-period. The topmost waveform alone matches the description given for BPSKabove.


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Timing diagram for QPSK

The binary data stream is shown beneath the time axis. The two signal

components with their bit assignments are shown the top and the total,combined signal at the bottom. Note the abrupt changes in phase at some of thebit-period boundaries.

The binary data that is conveyed by this waveform is: 1 1 0 0 0 1 1 0.

The odd bits, highlighted here, contribute to the in-phase component: 1 1 00 0 1 1 0The even bits, highlighted here, contribute to the quadrature-phasecomponent: 1 1 0 0 0 1 1 0

Offset QPSK (OQPSK) :

Offset quadrature phase-shift keying (OQPSK) is a variant of phase-shift

keying modulation using 4 different values of the phase to transmit. It issometimes called Staggered quadrature phase-shift keying (SQPSK). Taking fourvalues of the phase (two bits) at a time to construct a QPSK symbol can allow thephase of the signal to jump by as much as 180 at a time. When the signal is low-pass filtered (as is typical in a transmitter), these phase-shifts result in largeamplitude fluctuations, an undesirable quality in communication systems.
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Difference of the phase between QPSK and OQPSK

By offsetting the timing of the odd and even bits by one bit-period, or halfa symbol-period, the in-phase and quadrature components will never change atthe same time. In the constellation diagram shown on the left, it can be seen thatthis will limit the phase-shift to no more than 90 at a time. This yields much

lower amplitude fluctuations than non-offset QPSK and is sometimes preferredin practice.

The picture on the right shows the difference in the behavior of the phasebetween ordinary QPSK and OQPSK. It can be seen that in the first plot thephase can change by 180 at once, while in OQPSK the changes are never greaterthan 90.

The modulated signal is shown below for a short segment of a randombinary data-stream. Note the half symbol-period offset between the two

component waves. The sudden phase-shifts occur about twice as often as forQPSK (since the signals no longer change together), but they are less severe. Inother words, the magnitude of jumps is smaller in OQPSK when compared toQPSK.


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Timing diagram for offset-QPSK

The binary data stream is shown beneath the time axis. The two signalcomponents with their bit assignments are shown the top and the total,combined signal at the bottom. Note the half-period offset between the twosignal components.

/ 4QPSK:

Fig : Dual constellation diagram for /4-QPSK

This shows the two separate constellations with identical Gray coding butrotated by 45 with respect to each other.

This final variant of QPSK uses two identical constellations which arerotated by 45 ( / 4 radians, hence the name) with respect to one another.Usually, either the even or odd data bits are used to select points from one of the

constellations and the other bits select points from the other constellation. Thisalso reduces the phase-shifts from a maximum of 180, but only to a maximum of135 and so the amplitude fluctuations of / 4QPSK are between OQPSK andnon-offset QPSK.

One property this modulation scheme possesses is that if the modulatedsignal is represented in the complex domain, it does not have any paths throughthe origin. In other words, the signal does not pass through the origin. Thislowers the dynamical range of fluctuations in the signal which is desirable whenengineering communications signals.

On the other hand, / 4QPSK lends itself to easy demodulation and hasbeen adopted for use in, for example, TDMA cellular telephone systems.
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Fig : Timing diagram for /4-QPSK

The modulated signal is shown below for a short segment of a randombinary data-stream. The construction is the same as above for ordinary QPSK.Successive symbols are taken from the two constellations shown in the diagram.Thus, the first symbol (1 1) is taken from the 'blue' constellation and the secondsymbol (0 0) is taken from the 'green' constellation. Note that magnitudes of thetwo component waves change as they switch between constellations, but thetotal signal's magnitude remains constant. The phase-shifts are between those ofthe two previous timing-diagrams.

The binary data stream is shown beneath the time axis. The two signalcomponents with their bit assignments are shown the top and the total,combined signal at the bottom. Note that successive symbols are taken

alternately from the two constellations, starting with the 'blue' one.

Higher-order PSK :

Fig : Constellation diagram for 8-PSK with Gray coding


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Any number of phases may be used to construct a PSK constellation but 8-PSK is usually the highest order PSK constellation deployed. With more than 8phases, the error-rate becomes too high and there are better, though morecomplex, modulations available such as quadrature amplitude modulation(QAM). Although any number of phases may be used, the fact that theconstellation must usually deal with binary data means that the number of

symbols is usually a power of 2 this allows an equal number of bits-per-symbol.

For the general M-PSK there is no simple expression for the symbol-errorprobability ifM> 4. Unfortunately, it can only be obtained from:

where

,

,

,

and are jointly-Gaussianrandom variables.

Fig : Bit-error rate curves for BPSK, QPSK, 8-PSK and 16-PSK
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This may be approximated for highMand high Eb / N0 by:

.

The bit-error probability for M-PSK can only be determined exactly once the bit-

mapping is known. However, when Gray coding is used, the most probable errorfrom one symbol to the next produces only a single bit-error and

.

The graph on the left compares the bit-error rates of BPSK, QPSK (whichare the same, as noted above), 8-PSK and 16-PSK. It is seen that higher-ordermodulations exhibit higher error-rates; in exchange however they deliver ahigher raw data-rate.

Bounds on the error rates of various digital modulation schemes can becomputed with application of the union bound to the signal constellation.

Differential phase-shift keying :

Differential encoding

There is an ambiguity of phase if the constellation is rotated by some effectin the communications channel the signal passes through. This problem can beovercome by using the data to change rather than set the phase.

For example, in differentially-encoded BPSK a binary '1' may betransmitted by adding 180 to the current phase and a binary '0' by adding 0 tothe current phase. In differentially-encoded QPSK, the phase-shifts are 0, 90,180, -90 corresponding to data '00', '01', '11', '10'. This kind of encoding may bedemodulated in the same way as for non-differential PSK but the phaseambiguities can be ignored. Thus, each received symbol is demodulated to one oftheMpoints in the constellation and a comparator then computes the differencein phase between this received signal and the preceding one. The differenceencodes the data as described above.

The modulated signal is shown below for both DBPSK and DQPSK asdescribed above. It is assumed that the signal starts with zero phase, and so there isa phase shift in both signals at t = 0.
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Fig : Timing diagram for DBPSK and DQPSK

The binary data stream is above the DBPSK signal. The individual bits ofthe DBPSK signal are grouped into pairs for the DQPSK signal, which onlychanges every Ts = 2Tb.

Analysis shows that differential encoding approximately doubles the errorrate compared to ordinary M-PSK but this may be overcome by only a smallincrease in Eb / N0. Furthermore, this analysis (and the graphical results below)are based on a system in which the only corruption is additive white Gaussiannoise. However, there will also be a physical channel between the transmitterand receiver in the communication system. This channel will, in general,introduce an unknown phase-shift to the PSK signal; in these cases thedifferential schemes can yield a bettererror-rate than the ordinary schemes whichrely on precise phase information.

Example: Differentially-encoded BPSK

Fig : Differential encoding/decoding system diagram.

At the kth time-slot call the bit to be modulated bk, the differentially-encoded bit ekand the resulting modulated signal mk(t). Assume that the constellation diagrampositions the symbols at 1 (which is BPSK). The differential encoder produces:

where indicates binary or modulo-2 addition.
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Fig : BER comparison between BPSK and differentially-encoded BPSK withgray-coding operating in white noise

So ek only changes state (from binary '0' to binary '1' or from binary '1' to

binary '0') if bk is a binary '1'. Otherwise it remains in its previous state. This is thedescription of differentially-encoded BPSK given above.

The received signal is demodulated to yield ek = 1 and then thedifferential decoder reverses the encoding procedure and produces:

since binary subtraction is the same as binary addition.

Therefore, bk = 1 if ek and ek - 1 differ and bk = 0 if they are the same. Hence,if both ek and ek - 1 are inverted, bk will still be decoded correctly. Thus, the 180

phase ambiguity does not matter.

Differential schemes for other PSK modulations may be devised alongsimilar lines. The waveforms for DPSK are the same as for differentially-encodedPSK given above since the only change between the two schemes is at thereceiver.

The BER curve for this example is compared to ordinary BPSK on theright. As mentioned above, whilst the error-rate is approximately doubled, theincrease needed in Eb / N0 to overcome this is small. The performancedegradation is a result of noncoherent transmission - in this case it refers to the

fact that tracking of the phase is completely ignored.


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Fig : BER comparison between DBPSK, DQPSK and their non-differentialforms using gray-coding and operating in white noise.

For a signal that has been differentially encoded, there is an obvious

alternative method of demodulation. Instead of demodulating as usual andignoring carrier-phase ambiguity, the phase between two successive receivedsymbols is compared and used to determine what the data must have been.When differential encoding is used in this manner, the scheme is known asdifferential phase-shift keying (DPSK). Note that this is subtly different to justdifferentially-encoded PSK since, upon reception, the received symbols are notdecoded one-by-one to constellation points but are instead compared directly toone another.

Call the received symbol in the kth timeslot rk and let it have phase k.

Assume without loss of generality that the phase of the carrier wave is zero.Denote the AWGN term as nk. Then

.

The decision variable for the k - 1th symbol and the kth symbol is the phasedifference between rk and rk - 1. That is, if rk is projected onto rk - 1, the decision istaken on the phase of the resultant complex number:

where superscript * denotes complex conjugation. In the absence of noise, thephase of this is k - k - 1, the phase-shift between the two received signals which

can be used to determine the data transmitted.

The probability of error for DPSK is difficult to calculate in general, but, inthe case of DBPSK it is:
http://www.answers.com/topic/complex-conjugatehttp://www.answers.com/topic/complex-conjugate


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,

which, when numerically evaluated, is only slightly worse than ordinary BPSK,particularly at higher Eb / N0 values.

Using DPSK avoids the need for possibly complex carrier-recoveryschemes to provide an accurate phase estimate and can be an attractivealternative to ordinary PSK.

In optical communications, the data can be modulated onto the phase of alaser in a differential way. The modulation is a laser which emits a continuouswave, and a Mach-Zehnder modulator which receives electrical binary data. Forthe case of BPSK for example, the laser transmits the field unchanged for binary'1', and with reverse polarity for '0'. The demodulator consists of a delay lineinterferometer which delays one bit, so two bits can be compared at one time. In

further processing, a photo diode is used to transform the optical field into anelectric current, so the information is changed back into its original state.

The bit-error rates of DBPSK and DQPSK are compared to their non-differential counterparts in the graph to the right. The loss for using DBPSK issmall enough compared to the complexity reduction that it is often used incommunications systems that would otherwise use BPSK. For DQPSK though,the loss in performance compared to ordinary QPSK is larger and the systemdesigner must balance this against the reduction in complexity.

FSK system with the help of transmitter and receiver :

Frequency-shift keying (FSK) is a modulation scheme in which digitalinformation is transmitted through discrete frequency changes of a carrier wave.The most common form of frequency shift keying is 2-FSK. As suggested by thename, 2-FSK uses two discrete frequencies to transmit binary (0's and 1's)information. With this scheme, the "1" is called the mark frequency and the "0" iscalled the space frequency. The time domain of an FSK modulated carrier isillustrated in the figures at right.
http://www.answers.com/topic/optical-communicationhttp://www.answers.com/topic/continuous-wavehttp://www.answers.com/topic/continuous-wavehttp://www.answers.com/topic/delay-line-interferometerhttp://www.answers.com/topic/delay-line-interferometerhttp://www.answers.com/topic/photodiodehttp://www.answers.com/topic/optical-fieldhttp://www.answers.com/topic/optical-fieldhttp://www.answers.com/topic/photodiodehttp://www.answers.com/topic/delay-line-interferometerhttp://www.answers.com/topic/delay-line-interferometerhttp://www.answers.com/topic/delay-line-interferometerhttp://www.answers.com/topic/continuous-wavehttp://www.answers.com/topic/continuous-wavehttp://www.answers.com/topic/continuous-wavehttp://www.answers.com/topic/optical-communication


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Fig : An example of binary FSK

Comparison of AM and ASK signals :

S. No Parameter AM ASK

1Variable characteristics ofthe carrier

Amplitude Amplitude

2Nature of the modulatingsignal

Analog Digital

3Variation in the carrieramplitude

Continuous variationin accordance with theamplitude ofmodulating signal

Carrier ON orOFF depending onwhether a 1 or 0 isto be transmitted

4Number of sidebandsproduced

Two Two

5 Bandwidth 2fm (1+r)R

6 Noise Immunity Poor Poor

7 Application Radio BroadcastingData Transmissionin low bit rate

8 Detection method Envelope Envelope


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Comparison of various digital modulation systems:

S. No Parameter Binary ASK Binary FSK Binary PSK

1 Variable characteristic Amplitude Frequency Phase

2 Bandwidth 2R f1 f0 + (1+r)R (1+r)R

3 Noise Immunity Low High High

4 Error probability High Low Low5 Performance in

presence of noisePoor Better than

ASKBetter thanFSK

6 Complexity Simple Moderatelycomplex

Very complex

7 Bit rate Suitable upto 100bits/sec

Suitable up toabout 1200bits/sec

Suitable forhigh bit rates

8 Detection method Envelope Envelope coherent

Comparison BPSK and DPSK, BPSK and QPSK and QPSK and QAM :

S. No Modulation Method Transmission rate MinimumBandwidth

1 Variable characteristic Phase Phase

2 Bandwidth fb Fb

3 Effect of noise Low Higher than BPSK

4 Error probability Low Higher than BPSK

5 Need of synchronous carrier Needed Not needed

6 Complexity Lower than DPSK Higher than BPSK

7 Detection method Synchronous Synchronous8 Bit determination at the

receiverBased on single bitinterval

Based on signalreceived in twosuccessive bitintervals

S.No

Parameters BPSK QPSK

1 Variable characteristic Phase Phase

2 Type of modulation Two Level(binary) Four Level

3 Type of representation A binary bit isrepresented by onephase state

A group of two binarybits is represented byone phase state

4 Bit rate / Baud rate Bit rate = Baud rate Bit rate = 2 Baud rate

5 Detection method Coherent Coherent

6 Complexity Complex Very Complex


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S.No Parameters QPSK QAM

1 Type of modulation Quadrature phasemodulation

Quadrature amplitudeand phase modulation

2 Location of signal points On the circumference

of a circle

Equally spaced and

placed symmetricallyabout origin

3 Distance between thesignal points

d = 2 (Eb)1/2 for N = 2 d = 2 (0.4 Eb)1/2 for N = 2or M= 16

4 Noise Immunity Better than QASK Poorer than QPSK

5 Probability of error Less than QASK More than QPSK

6 Type of demodulation Synchronous Synchronous

7 System complexity Less complex thanQASK

More complex thanQPSK

Comparison of bit transmission and symbol transmission.

S. No Bit Transmission Symbol Transmission

1 Each bit is treated as a separate symbol N Successive bits areclubbed together toform a symbol

2 Only two different symbols can betransmitted i.e. 0 or 1

M number of symbolscan be transmittedwhere M = 2N

3 Binary systems such as BPSK, BFSK are bittransmitting systems

QPSK , M ary PSK,QAM etc. Are symboltransmission systems.They are also called asmultilevel modulationsystem

4 The channel bandwidth requirement isdependent on the bit rate of transmission

The channel bandwidthrequirement isdependent on thesymbol rate of

transmission5 Large bandwidth is required For PSK systems,bandwidth reduces withincrease in M. But forthe FSK systems thebandwidth increaseswith increase in M


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Comparison of Orthogonal BFSK and Non orthogonal BFSK and CoherentBFSK reception and Non Coherent BFSK reception.

S. No Orthogonal BFSK Non - Orthogonal BFSK

1 The frequencies fH and fL are the integral

multiples of fb

The frequencies fH and fL

are not the integralmultiples of fb

2 The signals SH(t) and SL(t) lie on the axesu1(t) and u2(t) in the signal spacerepresentation

The signals SH(t) andSL(t) do not lie on theaxes u1(t) and u2(t) in thesignal spacerepresentation

3 The Euclidean distance d is more than thatfor non orthogonal BFSK

The Euclidean distanced is less than that fororthogonal BFSK

4 Error probability Pe is less than that for thenon orthogonal BFSK system Error probability Pe ishigher than that for the

orthogonal BFSK system

S. No Coherent BFSK reception Non Coherent BFSKreception

1 It uses a regenerated carrier, multiplierand integrator to recover the originalbinary signal

It uses two band passfilters, envelopedetectors, and

comparator to recoverthe original signal

2 The receiver is more complicated The receiver is lesscomplicated

3 Reception process will not work unlessthe regenerated carrier of exactly the samefrequency is generated at the receiver

Regenerated carrier is notrequired for the reception

4 Error Probability is low Error Probability is high

5 Less effect of noise Effect of noise is more


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Comparison of the digital CW systems:

S.No

Paramet

er

ASK

BPSK

QPSK

QAM

M

ary

PSK

BFSK

M-ary

FSK

MSK

1

Informatio

nis

transmitte

dby

Amplitude

Phase

Phase

Amplitude

andphase

Phase

Frequency

Frequency

Frequency

2

Expressionfor

transmittedsignal

(2P

s)1/2

coswctfor

sym

bol1,

0for

(2P

s)1/2coswct,

b(t)=1for1,

b(t)

=-1for0

(2P

s)1/2

cos[wct+

(2m

+1)/4

]

M=

0,

1,

2,

3

K1(0.2

Ps)

1/2coswc

t+

k2ii

3

Number

ofbits

per

symbol

N=1

N=1

N=2

N N N=1

N N=2

4

Number

of

possible

symbols

=

N

Two

Two

Four

M=

2N

M=

2N

Two

M=

2N

Four

5

Detection

method

Coherent

Coherent

Coherent

Coherent

Coherent

Non-

Coherent

Non-

Coherent

Coherent

6

Minimum

Euclidean

distance

(Eb)1/2

2(Eb)1/2

2(Eb)1/2

(0.4

Es)

1/2For

M=

16

2(E

s)1/2Sin

(/

M)

(2Eb)1/2

(2NEb)1/2

(2Eb)1/2

7

Minimum

bandwidth

2fb

2fb

fb

2fb/N

2fb/N

4fb

(2N+1

fb)/

N 1.

5fb

8

Symbol

duratio

n Tb

Tb

2Tb

NTb

NTb

Tb

NTb

2Tb


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TWO MARKS

1. What does Digital Communication mean?

The digital communication is type of communication in which the digitaldata is transmitted directly or modulates some carrier. It covers a broad area ofcommunications techniques, including digital transmission and digital radio.

2. What are the advantages of digital communication?

Digital transmission has better noise immunity than analog transmission.

The digital communication systems are more flexible and adaptable.

Multiplexing and processing becomes easier due to digitalcommunication.

Error correction and decoding becomes more effective in digitalcommunication.

3. What are the two types of digital communication?

Digital TransmissionThere is no modulation. Digital data is transmitted directly over a pair of wires,co-axial cable or fiber optic cable.Digital radio

The digital data modulates some high frequency carrier. And then the signalis transmitted in free space.

4. What do you mean by transmission channel and what are the two importantparameters of a transmission channel?

Transmission channel is a medium over which the electrical signals from atransmitter travel to the receiver. Two important characteristics of a transmission

channel are1.Signal to Noise ratio(SNR)2.Bandwidth

5. Define band pass transmission and base band transmission.

Band pass transmission uses modulated carrier along with its sideband. Baseband transmission, the data is transmitted without any modulation. There is nocarrier used.

6. Define Nyquist theorem.

It states that if the bandwidth of a transmission channel is B which carries asignal having L number of levels, then the maximum data rate R on thischannel is given byR = 2 B log2 LThe number of levels L can be two or more.


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7. State Shannon limit for information capacity (or) State Shannons Theorem.

If (S/N) is the signal to noise ratio then the maximum data rate is given byR = B log2 [1 + (S/N)]where B = bandwidth (Hz)

S/N = signal to noise ratio (unit less)Shannon extended Nyquist work. He included the effect of noise present on the

transmission channel. Shannons theorem puts a limit on the maximum numberof levels for a given (S/N) ratio and bandwidth.

8. Define Shannon Hartley Theorem.

The information capacity of a white, band limited Gaussian channel is given byI = B log2 [1 + (S/N)] bits/ secWhere, B = Channel BandwidthS = Signal PowerN = Noise within the Channel Bandwidth

9. What is Information Capacity?The information capacity of a communication system represents the numberof independent symbols that can be carried through that system in a givenunit of time.It is expressed in bits/sec.

10. What is the Effect of S/N on I?

If the communication channel is noiseless then N = 0. Therefore (S/N) tends toinfinity and so I also will tend to infinity. Thus the noiseless channel will havean infinite capacity.

11. What is the effect of Bandwidth on Information Capacity?If the bandwidth approaches infinity the channel capacity does not becomeinfinite since N = N0 B will also increase with the bandwidth B. This will reducethe value of (S/N) with increase in B, assuming the signal power S to beconstant. Thus we conclude that an ideal system with infinite bandwidth has afinite channel capacity. It is denoted by I (Infinity) and given byI (Infinity) = 1.44 (S/N0)

12. What is Shannons Information rate?

The maximum rate of transmission is equal to the information capacity

Rmax = Imax = 1.44 (S/N0)It is very difficult to achieve it practically because to achieve this rate the channelbandwidth needs to be equal to infinity, and practically it is extremely difficult tohave a transmission channel with an infinite bandwidth.


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13. What is meant by Digital Amplitude Modulation (DAM) or AmplitudeShift Keying (ASK)?

The digital amplitude modulation is simply double sideband, full carrieramplitude modulation where the input-modulating signal is a binary waveform.The ASK output will be present only when a binary 1 is to be transmitted. TheASK output corresponding to a binary 0 is zero.

14. What is the Baud rate in ASK?

In ASK we use 1 bit (0 or 1) to represent one symbol. So the rate of symboltransmission is the baud rate. Nb is same as bit rate RBaud rate = Bit rate

15. What is the Transmission bandwidth of the ASK signal?

The bandwidth of the ASK signal is dependant on the bit rate R.R = 1/ TbBandwidth (Max) = R HZ

The frequency spectrum shows that the spectrum consists of the carrier signal offrequency fc with upper and lower sidebands.

16. How will you restrict the transmission bandwidth?

The transmission bandwidth BW of the ASK signal can be restricted by using afilter. The restricted value of bandwidth is given as:BW = (1 + r) RWhere r is a factor related to the filter characteristics and its value lies between0 and 1. Fc is the carrier frequency i.e. frequency of the sine wave beingtransmitted.

17. Derive the bandwidth of ASK in terms of Baud Rate?The bandwidth in terms of bit rate is given byBW = fc +(R/2)-(fc-(R/2))Where R = 1/Tb = bit rate and Tb = one bit intervalSince bit rate = baud rate for ASK, the expression for bandwidth is given byBW = fc + Nb/2 (fc-Nb/2) = Nb/2 + Nb/2 = NbPractically the bandwidth requirement of ASK is given byBW = (1 + d)* Nb

18. What is the use of sample and hold circuit in Coherent reception of ASK?

The integrator output is sampled at a particular instant corresponding to themaximum possible value of output and the sampled value is held by the sampleand hold circuit.


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19. What is meant by Frequency Shift Keying (FSK)?

Frequency Shift Keying is the relatively simple, low performance type of digitalmodulation. Binary FSK is a form of constant amplitude angle modulationsimilar to conventional frequency modulation except that the modulating signalis a binary signal that varies between two discrete voltage levels rather than acontinuously changing analog waveform.

20. What are all the types of FSK systems & explain them?

There are two types of FSK system.1) Non coherent FSK2) Coherent FSK

With non-coherent FSK, the transmitter and receiver are not frequency or phasesynchronized. With coherent FSK, local receiver reference signals are infrequency and phase lock with the transmitted signals.

21. Derive the Bandwidth for FSK Signals.

The bandwidth of FSK signal is dependant on the pulse width Tb or bit rate R =1/ Tb and the separation between the frequencies f0 and f1. The maximumbandwidth is given as:Bmax= (f1 + R/2) (f0 R/2)= (f1 f0 + R)The separation between f1 and f0 is kept at least 2R/ 3. ThenBmax= 2R /3 + R= 5R/3The FSK requires larger bandwidth than ASK and PSK.

22. What is the Bandwidth of FSK Signals in terms of Baud rate?

In FSKBit rate = Baud rate = NbThe FSK spectrum can be considered to be as combination of two ASK spectrumscentered at frequencies fH and FL.The expression for bandwidth can be given asBW= Nb/2 + (f1-f0) +Nb/2= (f1 f0) + NbThe minimum bandwidth will correspond to the situation in which (f1 f0) = NbBW (min) = Nb + Nb = 2 Nb

23. How will you restrict the bandwidth requirement in FSK?The bandwidth can be restricted by using a band pass filter. The restrictedbandwidth is given asB = (f1 f0) + ((1 + r) R/2)


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24. Define FSK bite rate & baud.The rate of change at the input to the modulator is called the bit rate and has theunit of bits per second. The rate of change at the output of the modulator is calledbaud.

25. Compare Amplitude Shift Keying (ASK) & Frequency Shift Keying (FSK).

S. no ASK FSK

1Amplitude of the signal ismodulated as per digital data

Frequency of the signal ismodulated as per digital data

2 Minimum BW = 2b Minimum BW = 4b

3Transmitted power keeps onchanging

Transmitted power remainsconstant

26. Give the truth table of the diode ring modulator.

Binary Input Binary OutputLogic 0 0 Degree

Logic 1 180 Degree

27. What do you meant by M-ary encoding?

M-ary is a term derived from the word binary. M is simply a digit that representsthe number of conditions or combinations possible for a given number of binaryvariables.

28. Give the formula for the error distance of the PSK.The error distance of the PSK is given by,d = (2 sin 180/M)*D

Where, d error distance M number of phasesD Peak signal amplitude.

29. Derive the Bandwidth of BPSK.

Bandwidth= Highest frequency Lowest frequency= (fc + fb) (fc - fb)= 2fbWhere fb= 1/TbThus the minimum bandwidth of BPSK signal is equal to twice the highestfrequency contained in the base band signal.

30. What are the applications of BPSK?

1. Phase Shift Keying is the most efficient of the three modulation methodsand it is used for high bit rates even higher than 1800 bits/sec.


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2. Due to low bandwidth requirement the BPSK modems are preferred overthe FSK modems, at higher operating speeds.

31. What is the probability of error in BPSK?

1. The expression for error probability is as follows:Pe = erfc (E/N0)1/2

2. The expression indicates that the error probability is dependant only onthe energy contents of the signal i.e. E.3. As the energy increases, value of complementary error function erfc

decreases and the value of Pe will reduce

32. What is meant by DPSK?

Differential Phase Shift Keying (DPSK) is an alternative form of digitalmodulation where the binary input information is contained in the differencebetween two successive signaling elements rather than the absolute phase.

33. What are the basic operations of DPSK?There are two basic operations and they are1. The differential encoding2. Phase Shift Keying

34. Derive the bandwidth for DPSK signal.

The phase shift of DPSK signal is dependant on the existing bit and one previousbit. Thus in DPSK two bits form a symbol.

Symbol duration Ts= 2 TbThe bandwidth is given by

BW = 2 / Ts = 2/2Tb = 1/ Tb = fbThus the bandwidth is half the bandwidth of BPSK.

35. What is the probability of error in DPSK?

1. The symbol duration in DPSK is equal to two bit duration i.e. = T s =2 Tb.2. The energy associated with each symbol is 2 Eb.3. Hence the error probability is given byPe = e (Eb/N0)4. The error probability is same as that of the non coherent BFSK

36. What does QPSK mean?

Quaternary Phase Shift Keying (QPSK), or quadrature PSK as it is sometimescalled, is another form of angle modulated, constant amplitude digitalmodulation. QPSK is an M-ary encoding technique where M=4. IN QPSK, twosuccessive bits are taken together. Two bits are taken from four distinct symbols.When the symbol is changed to next symbol, the phase of the carrier is changed


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by 45 degree. Since QPSK transmits two bits at a time, its bandwidth requirementis reduced. The phase changes in QPSK are not as abrupt as BPSK.

37. What is meant by offset QPSK?

Offset QPSK is a modified form of QPSK where the bit waveforms on the I and Qchannels are offset or shifted in phase from each other by one half a bit time.

38. What is the symbol transmission rate in QPSK?

In QPSK two bits are grouped together to form a symbol. Therefore when thesymbols are transmitted, the signal changes occur at the symbol rate which ishalf the bit rate.The symbol time Ts = 2 Tb

39. Derive the Bandwidth of QPSK signal.

The Bandwidth of QPSK signal is one half of the Bandwidth of BPSK signal.BW = 2fb / 2 = fb

Thus the advantage of multilevel modulation is reduction in requiredbandwidth.

40. Why QPSK is better than PSK?

1. Due to multilevel modulation used in QPSK, it is possible to increase thebit rate to double the bit rate of PSK without increasing the bandwidth.

2. The noise immunity of QPSK is same as that of PSK system.3. Available channel bandwidth is utilized in a better way by the QPSK

system than PSK system.

41. What is a 8 PSK system?

If three bits are brought together to form a message or symbol, then therewilll be 23 = 8 messages.

A PSK system that uses eight different phase shifts to transmit 8 symbolsis known as 8 PSK system.

The baud rate is one third of bit rate for 8 PSK system.The messages and the

corresponding phases are given in the following table:

Symbol 000 001 010 011 100 101 110 111

Phase 0 45 090 135 180 225 270 315

42. What does QAM stands for?

Quadrature amplitude Modulation (QAM) is a form of digital modulation wherethe digital information is contained in both the amplitude and phase of thetransmitted carrier.


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43. What are the types of QAM?

Depending on the types of bits per message the QAM signals are classified asfollows:

Name Bits per Symbol Number of Symbols

4 QAM 2 4(22)

8 QAM 3 8(23)16 QAM 4 16(24)

32 QAM 5 32(25)

64 QAM 6 64(26)

44. What is the Bandwidth of QAM?

The minimum bandwidth required for QAM transmission is same as that forASK or PSK.

45. Define Bandwidth efficiency.

It is defined as the ratio of the transmission bit rate to the minimum bandwidthrequired for a particular modulation scheme.BW efficiency = transmission rate (bps) / minimum BW (Hz) bits/cycle.

46. Define carrier recovery & what are all the methods used for this?

Carrier recovery is the process of extracting a phase coherent reference carrierfrom a receiver signal. This is sometimes called phase referencing.There are two carrier recovery circuits.

1. Squaring loop2.

Costas loop

47. What is meant by Probability of error & Bit Error Rate?

Probability of error P (e) & Bit Error Rate (BER) are often used interchangeably,although in practice they do have slightly different meanings. P (e) is atheoretical expectation of the bit error rate for a given system. BER is anempirical record of a systems actual bit error performance.

48. What is meant by antipodal signaling?

The phase relationship between signaling elements for BPSK (i.e., 180 degree outof phase) is the optimum-signaling format, referred to as antipodal signaling,

and occurs only when two binary signal levels are allowed and when one signalis the exact negative of the other.

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