DIGITAL COMMUNICATION ANSWER WITH QUESTIONS

DIGITAL COMMUNICATION

Data transmission, digital transmission, or digital communications is the physical transfer

ofdata (a digital bit stream) over a point-to-point or point-to-multipoint communication channel.

Examples of such channels are copper wires, optical fibres, wireless communication channels,

and storage media. The data are represented as an electromagnetic signal, such as an electrical

voltage,radiowave, microwave, or infrared signal.

DIGITAL COMMUNICATION TECHNIQUE

1.what is communication ?

Ans-The way in which data transferring information from one place to another another place is known as communication.

2.What is sampling ?

Ans- The process by which a signal is divided into number of samples which carries the characteristics of the original signal is called sampling.

3.what is white noise ? where is it encountered ?

Ans- White noise:

*White noise is noise whose power spectral density is uniform over the entire frequency range of interest.

*It occurs due to the superposition of all visible light spectral components.

*Gn (f)=ή/2

Where ή= constant.

*It is encountered in the communication channel.

*This noise is always additive in nature. That why the channel is called “AWGN channel” “Additive white Guassian Noise“.

4.Difine “code efficiency”?

Ans-code efficiency , n = Lmin/L

As per source coding theorem :

L ≥ H (x)

Lmin = H (x)

N = H(x)/L

5. Define “code redundancy” ?

Ans- code redundancy = 1 – n = 1 – code efficiency.

Y = 1 – n.

6. What is scrambling ?

Ans-*It is a process by which data is randomized.

*in general , a scrambler  tends to make the data more random by removing long strings of 1’s or 0’s. Scrambling can be helpful  in timing extraction by removing long strings of 0’s in binary data.The digital network must be able to cope with these long zero strings using zero suppression techniques.

7.What is QPSK ?

Ans- QPSK is known as Quadrature Phase Shift Keying.

*The channel bandwith depends upon the bit rate or signaling rate fb .In digital bandpass transmission , a carrier is used for transmission . This carrier is transmitted over a channel.

*If two or more bits are combined in some symbols, then the signaling rate will be reduced.

8.What is Carrier Recovery ?

Ans- *Carrer recovery is the process  of extracting a phase coherent reference carrier from a received signal. So it is also called as phase referencing.

*There are two loop in carrier recovery.

i. Squaring loop.

ii. Costas loop.

9. Difference between BPSK and DPSK.

Ans-

 BPSK

i. Low error of probability.

ii. complexity is lower.

iii. Low interface of Noise.

iv. Synchronous carrier is needed.

v. Bit determination at the receiver based on signal bit interval.

  DPSK

i. error of probability higher than BPSK.

ii.Complexity higher than BPSK.

iii. Interference of Noise is high.

iv. Synchronous carrier is not needed.

v. Bit determination at the receiver based on signal received in two successive bit intervals.

10. What is need of signaling in a PCM system?

Ans- i. Increase in transmission bandwith.

ii. Increase the power efficency.

iii. Better Transparency.

iv. Increase the erorr detection capability.

11..What is Shannon’s Theorem? What is the mean of Capacity of a Gaussian Channel?

*It is the most fundamental theorem of communication, is concerned with the rate of transmission of information over a communication channel.

*It based on the principles that a communication system will transmit information with an arbitary small probability of erorr provided that the information rate R is less than or equal to a rate c called the “channel capacity” ; this technical approach is called “coding” .

*Defination: This states that , A source M equally likely messages, with M >>1, which is generating information at a rate R .Given by a channel capacity C, IFR<=C, then there exist a coding techniques such that the output of the source may be transmitted over the channel with a probability of error in the received message which may be made arbitrarily small.’

capacity of Gaussian channel

A theorem which is complementary to shannon’s theorem and applies to a channel in which the noise  is gaussian is known as the “shannon’s – Hartly theorem”.

Theorem: The channel capacity of a white , band limited Gaussian is

C = Blog2(1+S/N)bits / sec

B = band width of the channel

where

S=signal power

N= Total noise within the channel bandwith =ήB

ή/2=double sided power spectral density.

*firstly , we find that channels are encountered in physical systems generally,are at least approximately,Gaussian.

* second , it turns out that the results obtained for a Gaussian often provide a lower bound on the performance of a system operating over a non Gaussian channel.

12.What is pulse Width Modulation(PWM) and describe it ?

Defination:

PWM (pulse width modulation) is a modulation system in which the width of the pulses is varied according to modulating signal where as Amplitude and position of pulses remains unchanged.

*The other type of a analog modulation is the pulse width modulation (pwm).

*In PWM , the width of the modulated pulses varies in proportion with the amplitude of modulating signal .

*As seen from the waveforms , the amplitude and the frequency of the PWM wave  remains constant.

*Only the width changes .That is why the “information” is contained in the width variation.

*This is similar to FM.As the noise is normally “additive”noise,it changes the amplitude of the PWM signal.

*At the receiver , it is possible  to remove these unwanted amplitude variations very easily by the means of a limiter circuit.

Generation of PWM:

* As the information is contained width variation .It is unaffected by the amplitude variations introduced by the noise.Thus the PWM system is more immune to noise than the PAM signal.

* A sawtooth generates a  sawtooth signal of frequency fs, therefore the sawtooth signal in this case is a sampling signal .It is applied to the inverting terminal of a comparator.

* The modulating signal x(t) is applied to the non inverting terminal of the same comparator .

*The comparator output will remain high as long as the instantaneous amplitude of x(t) is higher than that of the ramp signal.

* Note that the leading edges of the PWM wave from coincide with the falling edges of the ramp signal.

Detection Of PWM Signal:

*The pwm signal received at the input of the  detection circuit is contaminated with noise.This signal is applied to pulse generator circuit which regenerates the pwm signal.Thus some of the noise is recovered and the pulse are squared up.

*The generated pulse are applied to a reference pulse generator .It produces a train of constant amplitude , constant width pulses. These pulses are synchronized to the leading edges of the regenerated PWM pulses but delayed by a fixed interval.

Advantages of PWM:

i. Less effect of noise.

ii. Very good noise immunity.

iii. Synchronization between the transmitter and receiver is not essential.

Disadvantages of PWM:

I. Due to the variable pulse width, the pulses have variable power contents. So the transmission must be powerful enough to handle the maximum width, pulse, though the average power transmitted can be as low as 50% of this maximum power.

13. Write short notes:

(a). Multiplexer:

* Multiplexing may be defined as , ” a technique which allows many users to share a common communication channel simultaneously.”

i. There are two major types of multiplexing techniques ,they are

a. Time Division Multiplexing (TDM)

B. Frequency Division Multiplexing (FDM)

ii. The digital  multiplexing of digital  signals can be accomplished by using a bit by bit interleaving procedure with a selector switch that sequentially takes a bit from each incoming line and applies it to the high speed common the output of this common line separated out into it ‘s low speed individual components and then delivered to their respective destination.

iii. Digital multiplexer is of 2 types.

a. Low speed multiplexer

b. High speed multiplexer

iv. Low speed multiplexer is designed to combine relatively low speed digital signals upto a maximum rate of 4800bit/ sec into a high speed  multiplexed signal at a rate of 9600bit/ sec.

*It is used in voice grade channel and modern.

14. Regenerative Repeater:

*It is used at regularly spaced interval along a digital transmission line to detect the incoming digital signal and regenerate new clean pulses for further transmission.

* Normally it performs three major functions

a. Reshaping incoming pulses by using equalizer.

b. The extraction of information required to sample incoming pulse at optimum level.

c.The last part is decision making which is based on the pulses samples.

Operation:

i. The input pulses is attenuated and distorted by the transmitting system.So this distortion results dispersion which is caused due to high frequency components are attenuated.

ii. To restrict it equalizer is used which frequency characteristics is the inverse of transmitting medium.Actually equalizer restore higher frequency components and eliminates the dispersion.

15. short notes

(a). Linear equalizer:

* Channel equalizer approaches to compensate for the ISI means to minimize the ISI.

* It is used to regain the pulse shape by filtering the signal by a filter, the characteristics which is close approximationof inverse of the channel.

*This is usually done through transverse filtering which uses weighted tap delay lines.The weights on the tap are adjusted in a such manner that the pulse shape is best and isi is minimized.

* The filter structure has a computation complexity which is a linear function of the channel dispersion length L.

* If the received pulse side lobes dont go through zero at all sample time.

* since we are interested in sampling the equalized waveform at any predefined sample times,so we need a equalizing circuit called transversal filter a each sample point of the main lobe

* the main contribution is from the central tap with “Co” coefficient and other taps contributing echoes of main signal at symbol intervals on either side of main signal.

16.Timing Extraction:

* The received signal need to be sampled at precise instants, which requires a clock signal at the receiver in synchronisim with the clock signal at the transmitter. This process is known as bit or symbol synchronization ,for this purpose different methods are followed.

* Master slave concept is suitable for large number of data and high speed communication systems.

* Transmitting a separate synchronizing signal i.e , pilot clock.

* Self synchronization is very efficient method because the timing is derived from the digital signal it self.

17.Timing Jitter:

* Small random deviations of the incoming pulses from their ideal solution known as timing jitters which are always present even in the sophisticated system.

* Variations of pulse position or sampling instant cause timing jitters.

* This results from the several reasons as some of which are dependent on the pulse pattern being transmitted where as others are not .

* The random forms of jitters are caused by noise, interference and mistuning   of the clock circuit  for timing extraction must be large enough to provide an adequate suppression of timing jitters.

* The pattern  jitters results from clock mistuning , amplitude to phase conversion in the clock circuit and ISI which alters the position of the picks of the input signal according to pattern .

* The rms value of jitter over a long chain of ‘N’ repeaters is proportional to N.

* Jitter accumulation over a digital link can be reduced by buffering the link with an elastic store and clocking out the digit stream under the control of highly stable PLL.

SHORT ANSWERED QUESTIONS

UNIT : PULSE MODULATION

18. Define Nyquist rate.

Let the signal be bandlimited to „W‟ Hz. Then Nyquist rate is given as,

Nyquist rate = 2W samples/sec

Aliasing will not take place if sampling rate is greater than Nyquist rate.

19.. What is meant by aliasing effect?

Aliasing effect takes place when sampling frequency is less than Nyquist rate.

Under such condition, the spectrum of the sampled signal overlaps with itself.

Hence higher frequencies take the form of lower frequencies. This interference of

the frequency components is called as aliasing effect.

20. Define PWM.

PWM is basically pulse width modulation. Width of the pulse changes according to

amplitude of the modulating signal. It also referred as pulse duration modulation or

PDM.

21. State Sampling theorem.

A bandlimited signal of finite energy, which has no frequency components higher

than W Hz, may be completely recovered from the knowledge of its samples taken

at the rate of 2W samples per second.

22. Mention the merits of DPCM.

1. Bandwidth requirement of DPCM is less compared to PCM.

2. Quantization error is reduced because of prediction filter

3. Numbers of bits used to represent one sample value are also reduced compared to PCM.

23. What is the main difference in DPCM and DM?

DM encodes the input sample by one bit. It sends the information about + δ or -δ, ie step rise or fall.

DPCM can have more than one bit of encoding the sample. It sends the information about difference

between actual sample value and the predicted sample value.

24.How the message can be recovered from PAM?

The message can be recovered from PAM by passing the PAM signal through reconstruction filter

integrates amplitude of PAM pulses. Amplitude reconstruction signal is done to remove amplitude

discontinuities due to pulses.

25.Write an expression for bandwidth of binary PCM with N messages each with a maximum frequency

of fm Hz.

If „v‟ number of bits are used to code each input sample, then bandwidth of PCM isgiven as,

BT ≥ N.v.fm

Here v. fm is the bandwidth required by one message.

26. How is PDM wave converted into PPM message?

The PDM is signal is clock signal to monostable multivibrator. The multivibraor triggers on falling edge.

Hence a PPM pulse of fixed width is produced after falling edge of PDM pulse. PDM represents the input

signal amplitude in the form of width of the pulse. A PPM pulse is produced after the width of PDM

pulse. In other words, the position of the PPM pulse depends upon input signal amplitude.

27. Mention the use of adaptive quantizer in adaptive digital waveform coding schemes.

Adaptive quantizer changes its step size according variance of the input signal. Hence quantization error

is significantly reduced due to the adaptive quantization.

ADPCM uses adaptive quantization. The bit rate of such schemes is reduced due to adaptive quantization.

28.. What do u understand from adaptive coding?

In adaptive coding, the quantization step size and prediction filter coefficients are changed as per

properties of input signal. This reduces the quantization error and number of bits to represent the sample

value. Adaptive coding is used for speech coding at low bits rates.

29. What is meant by quantization?

While converting the signal value from analog to digital, quantization is performed.The  analog value is

assigned to nearest digital value. This is called quantization.

The quantized value is then converted into equivalent binary value. The quantization levels are fixed

depending upon the number of bits. Quantization is performed in every Analog to Digital Conversion.

30. The signal to quantization noise ratio in a PCM system depends on what criteria?

The signal to quantisation noise ratio in PCM is given as,

(S/N)db ≤(4.8+6v)dB

Here  v is the number of bits used to represent samples in PCM. Hence signal to quantization noise ratio

in PCM depends upon the number of bits or quantization levels.

31. What is meant by adaptive delta modulation?

In adaptive delta modulation, the step size is adjusted as per the slope of the input signal. Step size is

made high if slope of the input signal is high. This avoids slope overload distortion.

32.. What is the advantage of delta modulation over pulse modulation schemes?

Delta modulation encodes one bit per samples. Hence signalling rate is reduced in DM.

33. What should be the minimum bandwidth required to transmit a PCM channel?

The minimum transmission bandwidth in PCM is given as,

BT = vW

Here v is the number of bits used to represent one pulse.

W is the maximum signal frequency.

34. What is the advantage of delta modulation over PCM?

Delta modulation uses one bit to encode on sample. Hence bit rate of delta modulation is low compared to

PCM.

35. What are the two limitations of delta modulation?

1. Slope of overload distortion.

2. Granular noise.

36. How does Granular noise occurs?

It occurs due to large step size and very small amplitude variation in the input signal.

37. What are the advantages of the Delta modulation?

1. Delta modulation transmits only one bit for one sample. Thus the signaling rate and

transmission channel bandwidth is quite small for delta modulation.

2. The transmitter and receiver implementation is very much simple for delta modulation. There

is no analog to digital converter involved in delta modulation.

UNIT : BASEBAND PULSE TRANSMISSION

38. What is intersymbol interference in baseband binary PAM systems?

In baseband binary PAM, symbols are transmitted one after another. These symbols are separated by

sufficient time durations. The transmitter, channel and receiver acts as a filter to this baseband data.

Because of the filtering characteristics, transmitted PAM pulses are spread in time.

39. What is correlative coding?

Correlative level coding is used to transmit a baseband signal with the signaling rate of 2Bo over the

channel of bandwidth Bo. This is made physically possible by allowing ISI in the transmitted in

controlled manner. This ISI is known to receiver. The correlative coding is implemented by duo-binary

signaling and

modified duo-binary signaling.

40. Define Duo-binary baseband PAM system.

Duo-binary encoding reduces the maximum frequency of the baseband signal. The word „duo‟ means to

double the transmission capacity of the binary system.

Let the PAM signal ak represents kth bit. Then the encoder the new waveform as

Ck =ak + ak-1

Thus two successive bits are added to get encoded value of the kth bit. Hence Ck becomes a correlated

signal even though ak is not correlated. This introduces intersymbol interference in the controlled manner

to reduce the bandwidth.

41. What are eye pattern?

Eye pattern is used to study the effect of ISI in baseband transmission.

1.) Width of eye opening defines the interval over which the receivedwave can be sampled

without error from ISI.

2.) The sensitivity of the system to timing error is determined by the rate of closure of the eye as

the sampling time is varied.

3.) Height of the eye opening at sampling time is called margin over noise.

42. How is eye pattern obtained on the CRO?

Eye pattern can be obtained on CRO by applying the signal to one of the input channels and given an

external trigger of 1/Tb Hz. This makes one sweep of beam equal to Tb seconds.

43. Why do you need adaptive equalization in a switched telephone network.

In switched telephone network the distortion depends upon

1) Transmission characteristics of individual links.

2) Number of links in connection.

Hence fixed pair of transmit and receive filters will not serve the equalization problem. The transmission

characteristics keep on changing. Therefore adaptive equalization is used.

44.What are the necessity of adaptive equalization?

Ans. Most of the channels are made up of individual links in switched telephone

Network the distortion induced depends upon

1) transmission characteristics of individual links

2) number of links in connection

45. Define the principle of adaptive equalization?

Ans. The filters adapt themselves to the dispersive effects of the channel that is the coefficients of the

filters are changed continuously according to the received data. The filter coefficients are changed in such

a way that thedistortion in the data is reduced

46. Define the term ISI?

Ans. The presence of outputs due to other bits interference with the output of required bit . this effect is

called inter symbol interference (ISI)

47. Write the performance of data transmission system using eye

Pattern technique?

Ans. The width of the eye opening defines  the interval over which the received wave can can be  sampled

without error from inter symbol interference . The sensitivity of the system to timing error is determined

by the rate of closure of the eye as the sampling time is varied

48. What is the necessity  of equalization?

Ans. When the signal is passed through the channel distortion is introduced in terms of 1) amplitude 2)

delay this distortion creates problem of ISI. The detection of the signal also become difficult this

distraction can be compensated with the help of equalizer.

49. What is raised cosine spectrum?

In the raised cosine spectrum, the frequency response P(f) decreases towards zero gradually That is there

is no abrupt transition).

50. What is nyquist Bandwidth?

The B0 is called nyquist bandwidth.  The nyquist bandwidth is the minimum transmission bandwidth for

zero ISI.

UNIT : PASSBAND DELTA TRANSMISSION

51. Mention the need of optimum transmitting and receiving filter in baseband data transmission.

When binary data is transmitted over the baseband channel, noise interfaces with it. Because of this noise

interference, errors are introduced in signal detection. Optimum filter performs two functions while

receiving the noisy

signal:

1) Optimum filter integrates the signal during the bit interval and checks the output at the  time

instant where signal to noise ratio is maximum

2) Transfer function of the optimum filter is selected so as to maximize signal to noise ratio.

3) Optimum filter minimizes the probability of error.

52. Define ASK.

In ASK, carrier is switched on when binary 1 is to be transmitted and it is switched off when binary D is

to be transmitted ASK is also called on-off keying.

53. What is meant by DPSK?

In DPSK, the input sequence is modified. Let input sequence be d(t) and output sequence be b(t).

Sequence b(t) changes level at the beginning of each interval in which d(t)=1 and it does not changes

level when d(t)=0.When b(t) changes level, phase

of the carrier is changed. And as stated above, b(t) changes t=its level only when d(t) =1. This means

phase of the carrier is changed only if d(t)=1. Hence the technique is called Differential PSK.

54. Explain coherent 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.

55. What is the difference between PSK and FSK?

In PSK, phase of the carrier is switched according to input bit sequence. In FSK frequency of the carrier

is switched according to input bit sequence. FSK needs double of the bandwidth of PSK.

56. What is meant by coherent ASK?

In coherent ASK, correlation receiver is used to detect the signal. Locally generated carrier is correlated

with incoming ASK signal. The locally generated carrier is in exact phase with the transmitted carrier.

Coherent ASK is also called as synchronous ASK.

57. What is the major advantage of coherent PSK over coherent ASK?

ASK is on-off signaling, where as the modulated carrier is continuously transmitted in PSK. Hence peak

power requirement is more ASK, whereas it is reduced in case of PSK.

58. Explain the model of band pass digital data transmission system?

The band pass digital data transmission system consists of source, encoder and modulator in the

transmitter. Similarly receiver, decoder and destination form the transmitter.

59. What is baseband signal receiver?

A baseband signal receiver increases the signal to noise  ratio at the instant of sampling. This reduces the

probability of error. The baseband signal receiver is also called optimum receiver.

60. What is matched filter?

The matched filter is a baseband signal receiver, which works in presence of white Gaussian  noise. The

impulse response of the matched response of the matched filter is matched to the shape pf the input

signal.

61. What is the value of maximum signal to noise ratio of the matchedfilter? When it becomes maximum?

Maximum signal to noise ratio is the ratio of energy to psd of white noise. i.e.,

ρmax = E/ (N0/2)

This maximum value occurs at the end of bit duration i.e. Tb.

62. What is correlator ?

Correlator is the coherent receiver. It correlates the received noisy signal  f(t) with the locally generated

replica of the unknown signal x(t). It‟s output is denoted as r(t).

63. On what factor, the error probability of matched filter depends.

Error probability is given as

Pe = 1/2erfc√E/No

This equation shows that error probability depends only  on energy but not on shape of the signal.

64. Bring out the difference between coherent & non coherent binary modulation scheme.

a. Coherent detection:

In this method the local carrier generated at the receiver is phase locked with the carrier at the transmitter.

Hence it is called synchronous detection

b. Non coherent detection:

In this method, the receiver carrier need not be phase locked with transmitter carrier. Hence it is called

envelope detection.

65. Write the expression for bit error rate for coherent binary FSK.

Bit error rate for coherent binary FSK is given as,

Pe = 1/2erfc√0.6E/No

66. Highlight the major difference between a QPSK & MSK signal.

MSK signal have continuous phase in all the cases, where as QPSK has phase shift of π/2 or π.

67. What is the error probability of MSK & DPSK?

Error probability of MSK: Pe = 1/2erfc√E/No

Error probability of DPSK: Pe = 1/2e-Eb/No

68. In minimum shift keying what is the relation between the signal frequencies & bit rate.

Let the bit rate be fb and the frequency of carrier be f0. The higher and lower MSK signal frequencies are

given as,

fH = f0 + fb/4

fL = f0 - fb/4

69. List the advantages of Pass band transmission.

a. Long distance.

b. Analog channels can be used for transmission.

c. Multiplexing techniques can be used for bandwidth conservation.

d. Transmission can be done by using wireless channel also.

70. List the requirements of Passband transmission.

a. Maximum data transmission rate.

b. Minimum probability of symbol error.

c. Minimum transmitted power.

UNIT : INFORMATION THEORY & CODING

71. Define code efficiency.

The code efficiency is the ratio of message bits in a block to the transmitted bits for

that block by the encoder i.e.,

Code efficiency= (k/n)

k=message bits

n=transmitted bits.

72. What is meant by systematic and non-systematic codes?

In a Systematic block code, message bits appear first and then check bits. In the non-systematic code,

message and check bits cannot be identified in the code vector.

73. What is meant by linear code?

A code is linear if modulo-2 sum of any two code vectors produces another code vector. This means any

code vector can be expressed as linear combination of other code vectors.74. Define entropy.

Entropy is the measure of the average information content per second. It is givenby the expression

H(X)=_I P(xi)log2P(xi) bits/sample.75.Define mutual information.

Mutual information I(X,Y) of a channel is defined byI(X,Y)=H(X)-H(X/Y) bits/symbolH(X)- entropy of the source

H(X/Y)- conditional entropy of Y.76.State the properties of mutual information.

1. I(X,Y)=I(Y,X)2. I(X,Y)>=03. I(X,Y)=H(Y)-H(Y/X)4. I(X,Y)=H(X)+H(Y)-H(X,Y).

77.Give the relation between the different entropies.H(X,Y)=H(X)+H(Y/X)

=H(Y)+H(X/Y)H(X)- entropy of the source,H(Y/X),H(X/Y)-conditional entropyH(Y)-entropy of destinationH(X,Y)- Joint entropy of the source and destination78.Define information rate.If the time rate at which source X emits symbols is r symbols per second. Theinformation rate R of the source is given byR=r H(X) bits/second H(X)- entropy of the source

79.State the property of entropy.1.0< H(X) < log2K , is the radix of the alphabet X of the source.

80.What is differential entropy?The average amount of information per sample value of x(t) is measured by_H(X)= - INTEGRATION OF ( _ fx(x)log fx(x))dx bit/sampleH(X) –differential entropy of X.81 .What is the channel capacity of a discrete signal?The channel capacity of a discrete signal C= max I(X,Y)

P(xi)I(X,Y)-mutual information.82. What is source coding and entropy coding?A conversion of the output of a DMS into a sequence of binary symbols is calledsource coding. he design of a variable length code such that its average cod word lengthapproaches the entropy of the DMS is often referred to as entropy coding.83.State Shannon Hartley theorem.The capacity ‘C’ of a additive Gaussian noise channel is C=B log2 (1+S/N)B= channel bandwidth ,S/N=signal to noise ratio.84.What is the entropy of a binary memory-less source?The entropy of a binary memory-less source H(X)=-p0 log2p0-(1-p0)log2(1-p0)p0-probability of symbol ‘0’,p1=(1- p0 ) =probability of transmittingsymbol ‘1’85.How is the efficiency of the coding technique measured?Efficiency of the code =H(X) /L

L=SUMMATION OF p(xi)li average code word length .li=length of the code word.86.What happens when the number of coding alphabet increases?When the number of coding alphabet increases the efficiency of the codingtechnique decreases.87.What is channel diagram and channel matrix?The transition probability diagram of the channel is called the channel diagramand its matrix representation is called the channel matrix.88.What is information theory?Information theory deals with the mathematical modeling and analysis of acommunication system rather than with physical sources and physical channels89 .What is the channel capacity of a BSC and BEC?For BSC the channel capacity C=1+plog2 p +(1-p)log2(1-p).

For BEC the channel capacity C=(1-p)90. Explain the different types of channel.Loss less ChannelH (X/ Y) = 0 , I(X,Y) = H (X)Channel capacity = Max (I(X,Y)) = H (X)Channel diagram - ExplanationDeterministic channelH (Y/ X) = 0I(X,Y) = H (Y)Channel capacity = Max (I(X,Y)) = H (Y) , Channel diagram - ExplanationNoise less ChannelH (X/ Y) = 0H (Y/ X) = 0Channel capacity = Max (I(X,Y)) = H (Y) = H (X)

Binary Symmetric ChannelChannel capacity = Max (I(X,Y)) = H (Y) = Plog2 P + (1- P) log2 (1-P)

92.What is multiplexing?Multiplexing is the transmission of information from one or more sourceto one or more destination over the same transmission medium.93. What are the advantages of digital transmission?• The advantage of digital transmission over analog transmission is noiseimmunity. Digital pulses are less susceptible than analog signals to variations caused by noise.• Digital signals are better suited to processing and multiplexing than analog signals.• Digital transmission systems are more noise resistant than the analog transmission systems.• Digital systems are better suited to evaluate error performance.94.What are the disadvantages of digital transmission?_ The transmission of digitally encoded analog signals requires significantly more bandwidth than simply transmitting the original analog signal._ Analog signal must be converted to digital codes prior to transmission and converted back to analog form at the receiver, thus necessitating additional encoding anddecoding circuitry.

95.What is the purpose of the sample and hold circuit?The sample and hold circuit periodically samples the analog input signal andconverts those samples to a multilevel PAM signal.

96.Define and state the causes of fold over distortion.The minimum sampling rate(fs) is equal to twice the highest audio inputfrequency(fa).If fs is less than two times fa ,distortion will result. The distortion isCalled aliasing or fold over distortion.The side frequencies from one harmonic fold over into the sideband of anotherharmonic. The frequency that folds over is an alias of the input signal hence ,thenames “aliasing” or “fold over distortion” .

97.Define dynamic range.Dynamic range is the ratio of the largest possible magnitude to the smallestpossible magnitude. Mathematically, dynamic range isDR= Vmax/Vmin98.Define bit rate.In digital modulation, the rate of change at the input to the modulator is called thebit rate (fb) and has the unit of bits per second (bps).

99.Define Baud rate.The rate of change at the output of the modulator is called baud.100.Define QAM.Quadrature amplitude modulation is a form of digital modulation where thedigital information is contained in both the amplitude and phase of the transmittedcarrier.101.Write the relationship between the minimum bandwidth required for an FSKsystem and the bit rate.The minimum bandwidth can be approximated asB=2_f +2fbWhere B=minimum bandwidth (hertz)_f=minimum peak frequency deviation (hertz)Fb=bitrate102. Compare binary PSK with QPSK.SI.No BPSK QPSK1. One bit forms a symbol. Two bits form a symbol.2. Two possible symbols. Four possible symbols.3. Minimum bandwidth is twice of fb. Minimum bandwidth is equal to fb.4. Symbol duration = Tb. Symbol duration = 2Tb.103. What are the advantages of M-ary signaling scheme?1. M-ary signaling schemes transmit bits at a time.2. Bandwidth requirement of M-ary signaling schemes is reduced.104: What happens to the probability of error in M-ary FSK as the value of Mincrease?As the value of ‘M’ increases, the Euclidean distance between the symbolsreduces. Hence the symbols come closer to each other. This increases the probability of error in M-ary systems.

Prepared by : giridhari muduli