The Question bank download for CS2060 High Speed Networks are found below. You can view and download it from here.
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Thursday, 6 December 2012
IT 2202 Principles of Communication question bank download
1. Derive an expression for the Amplitude modulated signal and draw its frequency spectrum
2. Derive an expression for the AM power calculation, current calculation & modulation by a complex information signal
3. Compare FM & AM.
4. Describe the frequency analysis of Angle modulated waves. Explain their Bandwidth requirements
5. Draw FSK Transmitter and explain. Describe its Bandwidth Considerations
6. Define FSK and give the symbols of BFSK and explain the modulation mechanism with relevant waveforms and expressions
7. Explain the QPSK system with appropriate expressions and diagrams
8. Compare the various types of digital modulation techniques
9. Explain the modulation and demodulation process of PCM in detail Discuss on companding
10.With a block diagram, explain the adaptive Delta Modulation technique
11. Describe the operation of DPCM system with neat block diagram
12.Determine 12-bit linear code, 8-bit compressed code, the decoded 12- bit code, the c quantization error and compression error for a resolution of 0.01 voltage for analog sample voltages of a) 0.083V b) -0.398Vc) +9.234V d) -20.124 V
13. Explain TDMA system with its frame structure, features, efficiency
14.Explain CDMA System with its features. Explain the various problems available in CDMA system. Compare TDMA & FDMA
15. Give a detail account of the different types of multiple access techniques
16.Explain the principle and operation of direct sequence spread spectrum system in detail with an appropriate example
17. Explain the principles and operation of LEO, MEO and GEO satellites
18.Describe in detail the operations of various types of optical detectors
19.Enumerate the elements of an optical fiber transmission link
20.Draw the neat diagram of satellite system link model. Explain its each part in detail
Tuesday, 4 December 2012
Important question with answers in digital communication engineering
UNIT 1: PULSE MODULATION
1. 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.
2. 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.
3. 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.
4. 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.
5. 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.
6. 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.
7. 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.
8. 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 is given as, BT ≥ N.v.fm
Here v. fm is the bandwidth required by one message.
9. 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.
10. 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.
11. 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.
12. 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.
13. 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.
14. 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.
15. 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.
16. 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.
17. 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.
18. What are the two limitations of delta modulation?
1. Slope of overload distortion.
2. Granular noise.
19. How does Granular noise occurs?
It occurs due to large step size and very small amplitude variation in the input signal.
20. What are the advantages of the Delta modulation?
1. Delta modulation transmits only one bit for one sample. Thus the signalling 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 II: BASEBAND PULSE TRANSMISSION
1. 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.
2. What is correlative coding?
Correlative level coding is used to transmit a baseband signal with the signalling 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 duobinary signalling and modified duobinary signalling.
3. Define Duobinary baseband PAM system.
Duobinary 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.
4. 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 received wave 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.
5. 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.
6. 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.
7.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
8. Define the principle of adaptive equalization?
Ans. The filters adapt themselves to the dispersive effects of the channel that is the cofficients of the filters are changed contineously according to the received data. The filter cofficients are changed in such a way that the distortion in the data is reduced
9. Define duobinary encoding?
Ans. Duobinary encoding reduces the maximum frequency of the base band signal the “word duo” means to the double transmission capacity of the binary system
10. Write a note on correlative level coding?
Correlative level coding allows the signal scaling rate of 2Bo in the channel of bandwidth Bo this is made physically possible by allowing ISI in the transmitted signal in controlled manner this ISI IS KNOWN TO THE RECEIVER
11. 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)
12. 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
13. 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.
14. 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).
15. What is nyquist Bandwidth?
The B0 is called nyquist bandwidth. The nyquist bandwidth is the minimum transmission bandwidth for zero ISI.
UNIT III: PASSBAND DELTA TRANSMISSION
1. 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 maximise signal to noise ratio.
3) Optimum filter minimizes the probability of error.
2. 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.
3. 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.
4. 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.
5. 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.
6. 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.
7. What is the major advantage of coherent PSK over coherent ASK?
ASK is on-off signalling, 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.
8. Explain the model of bandpass digital data transmission system?
The bandpass digital data transmission system consists of source, encoder and modulator in the transmitter. Similarly receiver, decoder and destination form the transmitter.
9. 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.
10. 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.
11. What is the value of maximum signal to noise ratio of the matched filter? 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.
12. 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).
13. 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.
14. 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.
15. 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
16. 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.
17. 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
18. 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
19. List the advantages of Passband 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.
20. List the requirements of Passband transmission.
a. Maximum data transmission rate.
b. Minimum probability of symbol error.
c. Minimum transmitted power.
UNIT IV: ERROR CONTROL CODING
1.What is hamming distance?
The hamming distance between two code vectors is equal to the number of elements in which they differ. For example, let the two code words be, X = (101) and Y= (110) These two code words differ in second and third bits. Therefore the hamming distance between X and Y is two.
2. 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.
3. 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.
4. 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.
5. What are the error detection and correction capabilities of hamming codes?
The minimum distance (dmin) of hamming codes is „3‟. Hence it can be used to detect double errors or correct single errors. Hamming codes are basically linear block codes with dmin =3.
6. What is meant by cyclic codes?
Cyclic codes are the subclasses of linear block codes. They have the property that a cyclic shift of one codeword produces another code word.
7. How syndrome is calculated in Hamming codes and cyclic codes?
In hamming codes the syndrome is calculated as, S=YHT Here Y is the received and HT is the transpose of parity check matrix.
8. What is BCH code?
BCH codes are most extensive and powerful error correcting cyclic codes. The decoding of BCH codes is comparatively simpler. For any positive integer „m‟ and „t‟ (where t<2 m-1 )there exists a BCH code with following parameters: Block length: n= 2m-1 Number of parity check bits : n-k<=mt Minimum distance: dmin>=2t+1
9. What is RS code?
These are non binary BCH codes. The encoder for RS code operates on multiple bits simultaneously. The (n, k) RS code takes the groups of m- bit symbols of incoming binary data stream. It takes such „k‟ number of symbols in one block. Then the encoder acts (n – k) redundant symbols to form the code word of „n‟ symbols
RS code has: Block Length : n=2 m-1 symbols Message size: K symbols Parity check size: n-k= 2t symbols Minimum distance: dmin=2t+a symbols
10. What is difference between block codes and convolutional codes?
Block codes takes‟k‟ number of bits simultaneously form „n‟-bit code vector. This code vector is also called block. Convolutional code takes one message bits at a time and generates two or more encoded bits. Thus convolutional codes generate a string of encoded bits for input message string.
11. Define constraint length in convolutional code?
Constraint length is the number of shift over which the single message bit can influence the encoder output. It is expressed in terms of message bits.
12. Define free distance and coding gain.
Free distance is the minimum distance between code vectors. It is also equal to minimum weight of the code vectors. Coding gain is used as a basis of comparison for different coding methods. To achieve the same bit error rate the coding gain is defined as,
A= (Eb/No) encoded
(Eb/No) coded For convolutional coding, the coding gain is given as,
A = rdf /2 Here „r‟ is the code rate And „df is the free distance.
13. What is convolution code?
Fixed number of input bits is stored in the shift register & they are combined with the help of mod 2 adders. This operation is equivalent to binary convolution coding.
14. What is meant by syndrome of linear block code?
The non zero output of the produce YHT is called syndrome & it is used to detect errors in y. Syndrome is denoted by S & given as, S=YHT
15. What are the advantages of convolutional codes?
Advantages:
1. The decoding delay is small in convolutional codes since they operate o smaller blocks of data.
2. The storage hardware required by convolutional decoder is less since the block sizes are smaller.
Disadvantages:
1. Convolutional codes are difficult to analyze since their analysis is complex.
2. Convolutional codes are not developed much as compared to block codes.
16. Define sates of encoder?
The constraint length of the given convolutional encoder is K=2. Its rate is ½ means for single message bit input, two bits x1 and x2 are encoded at the output. „S1‟ represents the input message bit and S2 stores the previous message bit. Since only one previous message bit is stored, this encoder can have states depending upon this stored message bit. Let‟s represent, S2 = 0 state „a‟ and S2 = 1 state „b‟
17. Compare between code tree and trellis diagram?
Code tree Trellis diagram
1Code tree indicates flow of the coded signal along the nodes of the tree.Trellis diagram indicates transitions from current to next states.
2.Code tree is lengthy way of representing coding process. Code trellis diagram is shorter or compact way of representing coding process.
18. Write the futures of BCH Codes?
BCH codes are most extensive and powerful error correcting cyclic codes. The decoding of BCH codes is comparatively simpler. The decoding schemes of BCH codes can be implemented on digital computer. Because of software implementation of decoding schemes they are quite flexible compared to hardware implementation of other schemes.
19. What is Golay codes?
Golay code is the (23,12) cyclic code whose generating polynomial is,
G(p) = p11+p9+p7+p6+p5+p+1
This code has minimum distance of dmin = 7. This code can correct upto 3 errors. But Golay code cannot be generalized to other combinations of n and k.
20. Define constraint length in convolutional codes?
Constraint length is the number of shifts over which the single message bit can influence the encoder output. This expressed in terms of message bits.
UNIT – 5: SPREAD SPECTRUM MODULATION
1. Define spread spectrum communication
Ans: in spread spectrum communication the transmitted data sequence occupies much more band width than the minimum required bandwidth. Special code is used to aspread the bandwidth of the message signal. This special code is known only to authorized receiver. Hence the transmitted signal is received only by authorized receiver. Unwanted receivers cannot receive the signal. Thus spread spectrum communication provides secure transmission of data.
2. What is pseudo noise sequence ?
Ans. Pseudo noise sequence is a noise like high frequency signal. The sequence is not completely random, but it is generated by well defined logic. Hence it is called „pseudo‟ noise sequence. Pseudo noise sequences are used in spread spectrum communication for spreading message signals.
3. What is direct sequence spread spectrum modulation
Ans. In direct sequence spread spectrum modulation, the pseudo-noise sequence is directly modulated with data sequence. Thus pseudo-noise sequence acts as high frequency carrier and data sequence acts as low frequency modulating signal. The pseudo-noise sequence and data sequence are applied to a product modulator. The output of product modulator can be used directly or it can further generate BPSK signal for long distance communication.
4. What is frequency hap spread spectrum modulation?
Ans. In frequency hop spread spectrum, the data is transmitted in different frequency slots. These frequency slots are selected with the help of pseudo-noise sequence. Selection of frequency slots is called frequency hopping. The bandwidth of frequency hop spread spectrum is very much large compared to direct sequences spread spectrum. Frequency hop spread spectrum is of two types: I) slow frequency hopping and ii) fast frequency hopping.
5. What is processing gain?
Ans: processing gain is given as,
Bandwidth of spreaded signal
Processing gain (PG) = Bandwidth of unspreaded signal
For DS-SS, processing gain is given as Tb PG = Tb Here Tb is bit period of data sequence and Tc is bit period of PN sequence.
And for FH-SS, processing gain is given as, PG = 2t Here ‘t’ is the number of bits in PN
6. What is jamming margin ?
Ans. Jamming margin is the ratio of average power of interference (J) to average
power of data signal (Ps) i.e., J
Jamming margin = —–Ps
It is also given mathematically as, J Eb
— PG db ——–
Ps No dB
7. State four applications of spread spectrum.
Ans.
I) Spread spectrum has the ability to resist the effect of intentional jamming
ii) Spread spectrum is used in mobile communications. This is because the spread
spectrum signal has the ability to resist the effects of multipath fading.
iii) Spread spectrum communication are used in distance measurement.
iv) Spread spectrum communications are secure. This secrecy capability of spread spectrum is used in military as well as in many commercial applications.
8. When is the PN sequence called as maximal length sequence?
Ans. When the PN sequence has the length of 2m – 1, it is called maximal length sequence.
9. What is meant by processing gain of DS spread spectrum system?
Ans. The processing gain of DS-SS is given as the ratio of bit period to chip
period. Tb
P.G. = —- Tc
10. What is the period of the maximal length sequence generated using 3
bit shift register.
Ans. Here m=3, The period is given as,
N = 2n- 1 = 2(3)– 1 = 7 bits.
11. What are the application of spread spectrum modulation.
Ans. Application
(i) Multipath access capability.
(ii) Multipath protection in mobile communication
(iii) Low probability intercept.
(iv) Interference rejection.
(v) To provide antijam capability.
(vi) Distance measurements.
(vii) Selective calling.
12. Define frequency hopping.
Ans. Definition : The frequency of the carrier is changed (hopped) according
to bits of PN sequence.
Types: I) Slow frequency hopping
II) Fast frequency hopping
13. What are the Advantages of DS-SS system.
Ans.
1. This system has best noise and antijam performance.
2. Unrecognized receivers find it most difficult to detect direct sequence
signals.
3. It has best discrimination against multipath signals.
14.What are the Disadvantages of DS-SS system.
Ans.
1. It required wideband channel with small phase distortion.
2. It has long acquisition time.
3. The pseudo –noise generator should generate sequence at high rates.
4. This system is distance relative.
15. What are the Advantages of FH-SS System
Ans.
1. These systems bandwidth (spreads) are very large.
2. They can be programmed to avoid some portions of the spectrum.
3. They have relatively short acquisition time.
4. The distance effect is less.
16. What are the Disadvantages of FH-SS System
Ans.
1. Those systems need complex frequency synthesizers.
2. They are not useful for range and range-rate measurement.
3. They need error correction.
17. Define synchronization in Spread Spectrum Systems
Spread spectrum systems are essentially synchronous. The pseudo noise sequences generated at the receiver and the transmitted must be same and locked t each other so that the transmitted signal can be extracted. The synchronization of the spread spectrum systems can be considered in two parts : Acquisition and tracking.
18. Comparison between DS-SS and FH-SS
Parameter Direct sequence spread spectrum
Frequency hop spread spectrum
1 Definition PN sequence of large bandwidth is mulitplied with narrow band data signal . Data bits are transmitted in different frequency slots which are changed by PN sequence.
2 Effect of distance
This system is distance relative Effect of distance is less in this system
3 Acquisition time Acquisition time is long Acquisition time is short.
19. What are the Application of Direct Sequence Spread Spectrum Signals
Anti jamming with the help of direct sequence spread spectrum signals. Low delectability signal transmission or low probability intercept. Code division multiple access with direct sequence SS ( SSMA)
1. 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.
2. 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.
3. 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.
4. 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.
5. 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.
6. 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.
7. 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.
8. 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 is given as, BT ≥ N.v.fm
Here v. fm is the bandwidth required by one message.
9. 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.
10. 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.
11. 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.
12. 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.
13. 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.
14. 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.
15. 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.
16. 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.
17. 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.
18. What are the two limitations of delta modulation?
1. Slope of overload distortion.
2. Granular noise.
19. How does Granular noise occurs?
It occurs due to large step size and very small amplitude variation in the input signal.
20. What are the advantages of the Delta modulation?
1. Delta modulation transmits only one bit for one sample. Thus the signalling 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 II: BASEBAND PULSE TRANSMISSION
1. 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.
2. What is correlative coding?
Correlative level coding is used to transmit a baseband signal with the signalling 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 duobinary signalling and modified duobinary signalling.
3. Define Duobinary baseband PAM system.
Duobinary 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.
4. 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 received wave 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.
5. 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.
6. 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.
7.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
8. Define the principle of adaptive equalization?
Ans. The filters adapt themselves to the dispersive effects of the channel that is the cofficients of the filters are changed contineously according to the received data. The filter cofficients are changed in such a way that the distortion in the data is reduced
9. Define duobinary encoding?
Ans. Duobinary encoding reduces the maximum frequency of the base band signal the “word duo” means to the double transmission capacity of the binary system
10. Write a note on correlative level coding?
Correlative level coding allows the signal scaling rate of 2Bo in the channel of bandwidth Bo this is made physically possible by allowing ISI in the transmitted signal in controlled manner this ISI IS KNOWN TO THE RECEIVER
11. 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)
12. 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
13. 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.
14. 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).
15. What is nyquist Bandwidth?
The B0 is called nyquist bandwidth. The nyquist bandwidth is the minimum transmission bandwidth for zero ISI.
UNIT III: PASSBAND DELTA TRANSMISSION
1. 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 maximise signal to noise ratio.
3) Optimum filter minimizes the probability of error.
2. 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.
3. 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.
4. 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.
5. 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.
6. 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.
7. What is the major advantage of coherent PSK over coherent ASK?
ASK is on-off signalling, 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.
8. Explain the model of bandpass digital data transmission system?
The bandpass digital data transmission system consists of source, encoder and modulator in the transmitter. Similarly receiver, decoder and destination form the transmitter.
9. 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.
10. 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.
11. What is the value of maximum signal to noise ratio of the matched filter? 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.
12. 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).
13. 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.
14. 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.
15. 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
16. 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.
17. 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
18. 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
19. List the advantages of Passband 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.
20. List the requirements of Passband transmission.
a. Maximum data transmission rate.
b. Minimum probability of symbol error.
c. Minimum transmitted power.
UNIT IV: ERROR CONTROL CODING
1.What is hamming distance?
The hamming distance between two code vectors is equal to the number of elements in which they differ. For example, let the two code words be, X = (101) and Y= (110) These two code words differ in second and third bits. Therefore the hamming distance between X and Y is two.
2. 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.
3. 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.
4. 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.
5. What are the error detection and correction capabilities of hamming codes?
The minimum distance (dmin) of hamming codes is „3‟. Hence it can be used to detect double errors or correct single errors. Hamming codes are basically linear block codes with dmin =3.
6. What is meant by cyclic codes?
Cyclic codes are the subclasses of linear block codes. They have the property that a cyclic shift of one codeword produces another code word.
7. How syndrome is calculated in Hamming codes and cyclic codes?
In hamming codes the syndrome is calculated as, S=YHT Here Y is the received and HT is the transpose of parity check matrix.
8. What is BCH code?
BCH codes are most extensive and powerful error correcting cyclic codes. The decoding of BCH codes is comparatively simpler. For any positive integer „m‟ and „t‟ (where t<2 m-1 )there exists a BCH code with following parameters: Block length: n= 2m-1 Number of parity check bits : n-k<=mt Minimum distance: dmin>=2t+1
9. What is RS code?
These are non binary BCH codes. The encoder for RS code operates on multiple bits simultaneously. The (n, k) RS code takes the groups of m- bit symbols of incoming binary data stream. It takes such „k‟ number of symbols in one block. Then the encoder acts (n – k) redundant symbols to form the code word of „n‟ symbols
RS code has: Block Length : n=2 m-1 symbols Message size: K symbols Parity check size: n-k= 2t symbols Minimum distance: dmin=2t+a symbols
10. What is difference between block codes and convolutional codes?
Block codes takes‟k‟ number of bits simultaneously form „n‟-bit code vector. This code vector is also called block. Convolutional code takes one message bits at a time and generates two or more encoded bits. Thus convolutional codes generate a string of encoded bits for input message string.
11. Define constraint length in convolutional code?
Constraint length is the number of shift over which the single message bit can influence the encoder output. It is expressed in terms of message bits.
12. Define free distance and coding gain.
Free distance is the minimum distance between code vectors. It is also equal to minimum weight of the code vectors. Coding gain is used as a basis of comparison for different coding methods. To achieve the same bit error rate the coding gain is defined as,
A= (Eb/No) encoded
(Eb/No) coded For convolutional coding, the coding gain is given as,
A = rdf /2 Here „r‟ is the code rate And „df is the free distance.
13. What is convolution code?
Fixed number of input bits is stored in the shift register & they are combined with the help of mod 2 adders. This operation is equivalent to binary convolution coding.
14. What is meant by syndrome of linear block code?
The non zero output of the produce YHT is called syndrome & it is used to detect errors in y. Syndrome is denoted by S & given as, S=YHT
15. What are the advantages of convolutional codes?
Advantages:
1. The decoding delay is small in convolutional codes since they operate o smaller blocks of data.
2. The storage hardware required by convolutional decoder is less since the block sizes are smaller.
Disadvantages:
1. Convolutional codes are difficult to analyze since their analysis is complex.
2. Convolutional codes are not developed much as compared to block codes.
16. Define sates of encoder?
The constraint length of the given convolutional encoder is K=2. Its rate is ½ means for single message bit input, two bits x1 and x2 are encoded at the output. „S1‟ represents the input message bit and S2 stores the previous message bit. Since only one previous message bit is stored, this encoder can have states depending upon this stored message bit. Let‟s represent, S2 = 0 state „a‟ and S2 = 1 state „b‟
17. Compare between code tree and trellis diagram?
Code tree Trellis diagram
1Code tree indicates flow of the coded signal along the nodes of the tree.Trellis diagram indicates transitions from current to next states.
2.Code tree is lengthy way of representing coding process. Code trellis diagram is shorter or compact way of representing coding process.
18. Write the futures of BCH Codes?
BCH codes are most extensive and powerful error correcting cyclic codes. The decoding of BCH codes is comparatively simpler. The decoding schemes of BCH codes can be implemented on digital computer. Because of software implementation of decoding schemes they are quite flexible compared to hardware implementation of other schemes.
19. What is Golay codes?
Golay code is the (23,12) cyclic code whose generating polynomial is,
G(p) = p11+p9+p7+p6+p5+p+1
This code has minimum distance of dmin = 7. This code can correct upto 3 errors. But Golay code cannot be generalized to other combinations of n and k.
20. Define constraint length in convolutional codes?
Constraint length is the number of shifts over which the single message bit can influence the encoder output. This expressed in terms of message bits.
UNIT – 5: SPREAD SPECTRUM MODULATION
1. Define spread spectrum communication
Ans: in spread spectrum communication the transmitted data sequence occupies much more band width than the minimum required bandwidth. Special code is used to aspread the bandwidth of the message signal. This special code is known only to authorized receiver. Hence the transmitted signal is received only by authorized receiver. Unwanted receivers cannot receive the signal. Thus spread spectrum communication provides secure transmission of data.
2. What is pseudo noise sequence ?
Ans. Pseudo noise sequence is a noise like high frequency signal. The sequence is not completely random, but it is generated by well defined logic. Hence it is called „pseudo‟ noise sequence. Pseudo noise sequences are used in spread spectrum communication for spreading message signals.
3. What is direct sequence spread spectrum modulation
Ans. In direct sequence spread spectrum modulation, the pseudo-noise sequence is directly modulated with data sequence. Thus pseudo-noise sequence acts as high frequency carrier and data sequence acts as low frequency modulating signal. The pseudo-noise sequence and data sequence are applied to a product modulator. The output of product modulator can be used directly or it can further generate BPSK signal for long distance communication.
4. What is frequency hap spread spectrum modulation?
Ans. In frequency hop spread spectrum, the data is transmitted in different frequency slots. These frequency slots are selected with the help of pseudo-noise sequence. Selection of frequency slots is called frequency hopping. The bandwidth of frequency hop spread spectrum is very much large compared to direct sequences spread spectrum. Frequency hop spread spectrum is of two types: I) slow frequency hopping and ii) fast frequency hopping.
5. What is processing gain?
Ans: processing gain is given as,
Bandwidth of spreaded signal
Processing gain (PG) = Bandwidth of unspreaded signal
For DS-SS, processing gain is given as Tb PG = Tb Here Tb is bit period of data sequence and Tc is bit period of PN sequence.
And for FH-SS, processing gain is given as, PG = 2t Here ‘t’ is the number of bits in PN
6. What is jamming margin ?
Ans. Jamming margin is the ratio of average power of interference (J) to average
power of data signal (Ps) i.e., J
Jamming margin = —–Ps
It is also given mathematically as, J Eb
— PG db ——–
Ps No dB
7. State four applications of spread spectrum.
Ans.
I) Spread spectrum has the ability to resist the effect of intentional jamming
ii) Spread spectrum is used in mobile communications. This is because the spread
spectrum signal has the ability to resist the effects of multipath fading.
iii) Spread spectrum communication are used in distance measurement.
iv) Spread spectrum communications are secure. This secrecy capability of spread spectrum is used in military as well as in many commercial applications.
8. When is the PN sequence called as maximal length sequence?
Ans. When the PN sequence has the length of 2m – 1, it is called maximal length sequence.
9. What is meant by processing gain of DS spread spectrum system?
Ans. The processing gain of DS-SS is given as the ratio of bit period to chip
period. Tb
P.G. = —- Tc
10. What is the period of the maximal length sequence generated using 3
bit shift register.
Ans. Here m=3, The period is given as,
N = 2n- 1 = 2(3)– 1 = 7 bits.
11. What are the application of spread spectrum modulation.
Ans. Application
(i) Multipath access capability.
(ii) Multipath protection in mobile communication
(iii) Low probability intercept.
(iv) Interference rejection.
(v) To provide antijam capability.
(vi) Distance measurements.
(vii) Selective calling.
12. Define frequency hopping.
Ans. Definition : The frequency of the carrier is changed (hopped) according
to bits of PN sequence.
Types: I) Slow frequency hopping
II) Fast frequency hopping
13. What are the Advantages of DS-SS system.
Ans.
1. This system has best noise and antijam performance.
2. Unrecognized receivers find it most difficult to detect direct sequence
signals.
3. It has best discrimination against multipath signals.
14.What are the Disadvantages of DS-SS system.
Ans.
1. It required wideband channel with small phase distortion.
2. It has long acquisition time.
3. The pseudo –noise generator should generate sequence at high rates.
4. This system is distance relative.
15. What are the Advantages of FH-SS System
Ans.
1. These systems bandwidth (spreads) are very large.
2. They can be programmed to avoid some portions of the spectrum.
3. They have relatively short acquisition time.
4. The distance effect is less.
16. What are the Disadvantages of FH-SS System
Ans.
1. Those systems need complex frequency synthesizers.
2. They are not useful for range and range-rate measurement.
3. They need error correction.
17. Define synchronization in Spread Spectrum Systems
Spread spectrum systems are essentially synchronous. The pseudo noise sequences generated at the receiver and the transmitted must be same and locked t each other so that the transmitted signal can be extracted. The synchronization of the spread spectrum systems can be considered in two parts : Acquisition and tracking.
18. Comparison between DS-SS and FH-SS
Parameter Direct sequence spread spectrum
Frequency hop spread spectrum
1 Definition PN sequence of large bandwidth is mulitplied with narrow band data signal . Data bits are transmitted in different frequency slots which are changed by PN sequence.
2 Effect of distance
This system is distance relative Effect of distance is less in this system
3 Acquisition time Acquisition time is long Acquisition time is short.
19. What are the Application of Direct Sequence Spread Spectrum Signals
Anti jamming with the help of direct sequence spread spectrum signals. Low delectability signal transmission or low probability intercept. Code division multiple access with direct sequence SS ( SSMA)
CS2204 Analog and Digital Communication question paper download
B.E/B.Tech. DEGREE EXAMINATION, APRIL/MAY 2010
Third Semester
Computer Science and Engineering
CS2204 - ANALOG AND DIGITAL COMMUNICATION
(Regulation 2008)
Time: Three hours Maximum: 100 Marks
Answer ALL Questions
PART A- (10 x 2 = 20 Marks)
1. In a Amplitude modulation system, the carrier frequency is Fc= 100KHz. The maximum frequency of the signal is 5 KHz. Determine the lower and upper side bonds and the bond width of AM signal.
2. The maximum frequency deviation in an FM is 10 KHz and signal frequency is 10 KHz. Find out the bandwidth using Carson's rule and the modulation index.
3. Draw the ASK and FSK signals for the binary signal s(t)= 1011001.
4. What are the advantages of QPSK?
5. Define Nyquist sampling theorem.
6. For the signal m(t)= 3 cos 500^t + 4 sin 1000^t, Determine the Nyquist sampling rate.
7. What is meant by ASCII code?
8. Which error detection technique is simple and which one is more reliable?
9. What are the applications of spread spectrum modulation?
10. Design processing gain in spread spectrum modulation.
PART B- (5 x 16 = 80 Marks)
11. (a) (i) Distinguish between FM and PM by giving its mathematical analysis. (8 Marks)
(ii) Derive the relationship between the voltage amplitudes of the side band frequencies and the carrier and draw the frequency spectrum. (8 Marks)
(Or)
(b) (i) Discuss about the sets of side bands produced when a carrier is frequency modulated by a single frequency sinusoid. (8 Marks)
(ii) In an AM modulator, 500 KHz carrier of amplitude 20 V is modulated by 10 KHz modulating signal which causes a change in the output wave of +_ 7.5 V. Determine:
(1) Upper and lower side band frequencies
(2) Modulation Index
(3) Peak amplitude of upper and lower side frequency
(4) Maximum and minimum amplitudes of envelope. (8 Marks)
12. (a) What is known as Binary phase shift keying? Discuss in detail the BPSK transmitter and Receiver and also obtain the minimum double sided Nyquist bandwidth. (16 Marks)
(Or)
(b) (i) Illustrate the concept of 8 QAM transmitter with the truth table. (8 Marks)
(ii) What is the need for carrier Recovery? Explain the Costas loop method of carrier recovery. (8 Marks)
13. (a) (i) What is called companding? Briefly discuss the Analog companding. (8 Marks)
(ii) Discuss about the causes of ISI. (8 Marks)
(Or)
(b) (i) Explain in detail the Delta modulation transmitter and Receiver. (10 Marks)
(ii) Discuss the draw backs of delta modulation and explain the significance of adaptive delta modulator. (6 Marks)
14. (a) (i) Describe the most common error detection techniques. (12 Marks)
(ii) Discuss the function of a data modem. (4 Marks)
(Or)
(b) (i) Explain in detail the characteristics of IEEE 488 Bus. (10 Marks)
(ii) Briefly explain the three methods of error connection. (6 Marks)
15. (a) (i) What is a Pseudo noise sequence? What are the properties of Pseudo noise sequence? (8 Marks)
(ii) Describe the application of CDMA in wireless communication system. (8 Marks)
(Or)
(b) (i) With a block diagram explain, DS spread spectrum with coherent binary PSK. (10 Marks)
(ii) Explain the near- far problem in spread spectrum modulation? (6 Marks)
Third Semester
Computer Science and Engineering
CS2204 - ANALOG AND DIGITAL COMMUNICATION
(Regulation 2008)
Time: Three hours Maximum: 100 Marks
Answer ALL Questions
PART A- (10 x 2 = 20 Marks)
1. In a Amplitude modulation system, the carrier frequency is Fc= 100KHz. The maximum frequency of the signal is 5 KHz. Determine the lower and upper side bonds and the bond width of AM signal.
2. The maximum frequency deviation in an FM is 10 KHz and signal frequency is 10 KHz. Find out the bandwidth using Carson's rule and the modulation index.
3. Draw the ASK and FSK signals for the binary signal s(t)= 1011001.
4. What are the advantages of QPSK?
5. Define Nyquist sampling theorem.
6. For the signal m(t)= 3 cos 500^t + 4 sin 1000^t, Determine the Nyquist sampling rate.
7. What is meant by ASCII code?
8. Which error detection technique is simple and which one is more reliable?
9. What are the applications of spread spectrum modulation?
10. Design processing gain in spread spectrum modulation.
PART B- (5 x 16 = 80 Marks)
11. (a) (i) Distinguish between FM and PM by giving its mathematical analysis. (8 Marks)
(ii) Derive the relationship between the voltage amplitudes of the side band frequencies and the carrier and draw the frequency spectrum. (8 Marks)
(Or)
(b) (i) Discuss about the sets of side bands produced when a carrier is frequency modulated by a single frequency sinusoid. (8 Marks)
(ii) In an AM modulator, 500 KHz carrier of amplitude 20 V is modulated by 10 KHz modulating signal which causes a change in the output wave of +_ 7.5 V. Determine:
(1) Upper and lower side band frequencies
(2) Modulation Index
(3) Peak amplitude of upper and lower side frequency
(4) Maximum and minimum amplitudes of envelope. (8 Marks)
12. (a) What is known as Binary phase shift keying? Discuss in detail the BPSK transmitter and Receiver and also obtain the minimum double sided Nyquist bandwidth. (16 Marks)
(Or)
(b) (i) Illustrate the concept of 8 QAM transmitter with the truth table. (8 Marks)
(ii) What is the need for carrier Recovery? Explain the Costas loop method of carrier recovery. (8 Marks)
13. (a) (i) What is called companding? Briefly discuss the Analog companding. (8 Marks)
(ii) Discuss about the causes of ISI. (8 Marks)
(Or)
(b) (i) Explain in detail the Delta modulation transmitter and Receiver. (10 Marks)
(ii) Discuss the draw backs of delta modulation and explain the significance of adaptive delta modulator. (6 Marks)
14. (a) (i) Describe the most common error detection techniques. (12 Marks)
(ii) Discuss the function of a data modem. (4 Marks)
(Or)
(b) (i) Explain in detail the characteristics of IEEE 488 Bus. (10 Marks)
(ii) Briefly explain the three methods of error connection. (6 Marks)
15. (a) (i) What is a Pseudo noise sequence? What are the properties of Pseudo noise sequence? (8 Marks)
(ii) Describe the application of CDMA in wireless communication system. (8 Marks)
(Or)
(b) (i) With a block diagram explain, DS spread spectrum with coherent binary PSK. (10 Marks)
(ii) Explain the near- far problem in spread spectrum modulation? (6 Marks)
Tuesday, 28 August 2012
GE2111 Engineering Graphics Question paper download
GE2111 ENGINEERING GRAPHICS
Time: Three Hours Maximum: 100 Marks
Answer ALL Questions
(5 x 20 = 100 Marks)
1. (a) Draw a hyperbola when the distance between its focus and directrix is 50 mm
and eccentricity is 3/2. Also draw the tangent and normal at a point 25 mm
from the directrix. (20)
OR
1. (b) Make free hand sketches of front, top and right side views of the 3D object
shown below : (20)
Fig. 1.(b)
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2. (a) A line PQ has its end P, 10 mm above the HP and 20 mm in front of the VP.
The end Q is 35 mm in front of the VP. The front view of the line measures 75
mm. The distance between the end projectors is 50 mm. Draw the projections
of the line and ¯nd its true length and its true inclinations with the VP and
HP. (20)
OR
2. (b) Draw the projections of a circle of 70 mm diameter resting on the H.P. on a
point A of the circumference. The plane is inclined to the H.P. such that the
top view of it i s an ellipse of minor axis 40 mm. The top view of the diameter,
through the point A is making an angle of 45± with the V.P. Determine the
inclination of the plane with the H.P. (20)
3. (a) An equilateral triangular prism 20 mm side of base and 50 mm long rests with
one of its shorter edges on HP such that the rectangular face containing the
edge on which the prism rests is inclined at 30± to H.P. The shorter edge resting
on HP is perpendicular to VP. (20)
OR
3. (b) A square pyramid of base 40 mm and axis 70 mm long has one of its triangular
faces on VP and the edge of base contained by that face perpendicular to HP.
Draw its projections. (20)
4. (a) A hexagonal prism of side of base 35 mm and axis length 55 mm rests with its
base on H.P such that two of the vertical surfaces are perpendicular to V.P. It
is cut by a plane inclined at 50± to H.P and perpendicular to V.P and passing
through a point on the axis at a distance 15 mm from the top. Draw its front
view, sectional top view and true shape of section. (20)
OR
4. (b) Draw the development of the lateral surface of the lower portion of a cylinder
of diameter 50 mm and axis 70 mm. The solid is cut by a section plane inclined
at 40± to HP and perpendicular to VP and passing through the midpoint of
the axis. (20)
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5. (a) Draw the isometric projection of the object from the views shown in Figure
5(a). (20)
OR
5. (b) Draw the perspective projection of a cube of 25 mm edge, lying on a face on
the ground plane, with an edge touching the picture plane and all vertical faces
equally inclined to the picture plane. The station point is 50 mm in front of
the picture plane, 35 mm above the ground plane and lies in a central plane
which is 10 mm to the left of the center of the cube. (20)
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Time: Three Hours Maximum: 100 Marks
Answer ALL Questions
(5 x 20 = 100 Marks)
1. (a) Draw a hyperbola when the distance between its focus and directrix is 50 mm
and eccentricity is 3/2. Also draw the tangent and normal at a point 25 mm
from the directrix. (20)
OR
1. (b) Make free hand sketches of front, top and right side views of the 3D object
shown below : (20)
Fig. 1.(b)
Nitro PDF Software
100 Portable Document Lane
Wonderland
2. (a) A line PQ has its end P, 10 mm above the HP and 20 mm in front of the VP.
The end Q is 35 mm in front of the VP. The front view of the line measures 75
mm. The distance between the end projectors is 50 mm. Draw the projections
of the line and ¯nd its true length and its true inclinations with the VP and
HP. (20)
OR
2. (b) Draw the projections of a circle of 70 mm diameter resting on the H.P. on a
point A of the circumference. The plane is inclined to the H.P. such that the
top view of it i s an ellipse of minor axis 40 mm. The top view of the diameter,
through the point A is making an angle of 45± with the V.P. Determine the
inclination of the plane with the H.P. (20)
3. (a) An equilateral triangular prism 20 mm side of base and 50 mm long rests with
one of its shorter edges on HP such that the rectangular face containing the
edge on which the prism rests is inclined at 30± to H.P. The shorter edge resting
on HP is perpendicular to VP. (20)
OR
3. (b) A square pyramid of base 40 mm and axis 70 mm long has one of its triangular
faces on VP and the edge of base contained by that face perpendicular to HP.
Draw its projections. (20)
4. (a) A hexagonal prism of side of base 35 mm and axis length 55 mm rests with its
base on H.P such that two of the vertical surfaces are perpendicular to V.P. It
is cut by a plane inclined at 50± to H.P and perpendicular to V.P and passing
through a point on the axis at a distance 15 mm from the top. Draw its front
view, sectional top view and true shape of section. (20)
OR
4. (b) Draw the development of the lateral surface of the lower portion of a cylinder
of diameter 50 mm and axis 70 mm. The solid is cut by a section plane inclined
at 40± to HP and perpendicular to VP and passing through the midpoint of
the axis. (20)
2 Q2901
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5. (a) Draw the isometric projection of the object from the views shown in Figure
5(a). (20)
OR
5. (b) Draw the perspective projection of a cube of 25 mm edge, lying on a face on
the ground plane, with an edge touching the picture plane and all vertical faces
equally inclined to the picture plane. The station point is 50 mm in front of
the picture plane, 35 mm above the ground plane and lies in a central plane
which is 10 mm to the left of the center of the cube. (20)
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Thursday, 23 August 2012
GE2111 Engineering Graphics Question paper download
GE2111 ENGINEERING GRAPHICS
Time: Three Hours Maximum: 100 Marks
Answer ALL Questions
(5 x 20 = 100 Marks)
1. (a) Draw a hyperbola when the distance between its focus and directrix is 50 mm
and eccentricity is 3/2. Also draw the tangent and normal at a point 25 mm
from the directrix. (20)
OR
1. (b) Make free hand sketches of front, top and right side views of the 3D object
shown below : (20)
Fig. 1.(b)
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2. (a) A line PQ has its end P, 10 mm above the HP and 20 mm in front of the VP.
The end Q is 35 mm in front of the VP. The front view of the line measures 75
mm. The distance between the end projectors is 50 mm. Draw the projections
of the line and ¯nd its true length and its true inclinations with the VP and
HP. (20)
OR
2. (b) Draw the projections of a circle of 70 mm diameter resting on the H.P. on a
point A of the circumference. The plane is inclined to the H.P. such that the
top view of it i s an ellipse of minor axis 40 mm. The top view of the diameter,
through the point A is making an angle of 45± with the V.P. Determine the
inclination of the plane with the H.P. (20)
3. (a) An equilateral triangular prism 20 mm side of base and 50 mm long rests with
one of its shorter edges on HP such that the rectangular face containing the
edge on which the prism rests is inclined at 30± to H.P. The shorter edge resting
on HP is perpendicular to VP. (20)
OR
3. (b) A square pyramid of base 40 mm and axis 70 mm long has one of its triangular
faces on VP and the edge of base contained by that face perpendicular to HP.
Draw its projections. (20)
4. (a) A hexagonal prism of side of base 35 mm and axis length 55 mm rests with its
base on H.P such that two of the vertical surfaces are perpendicular to V.P. It
is cut by a plane inclined at 50± to H.P and perpendicular to V.P and passing
through a point on the axis at a distance 15 mm from the top. Draw its front
view, sectional top view and true shape of section. (20)
OR
4. (b) Draw the development of the lateral surface of the lower portion of a cylinder
of diameter 50 mm and axis 70 mm. The solid is cut by a section plane inclined
at 40± to HP and perpendicular to VP and passing through the midpoint of
the axis. (20)
2 Q2901
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100 Portable Document Lane
Wonderland
5. (a) Draw the isometric projection of the object from the views shown in Figure
5(a). (20)
OR
5. (b) Draw the perspective projection of a cube of 25 mm edge, lying on a face on
the ground plane, with an edge touching the picture plane and all vertical faces
equally inclined to the picture plane. The station point is 50 mm in front of
the picture plane, 35 mm above the ground plane and lies in a central plane
which is 10 mm to the left of the center of the cube. (20)
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Time: Three Hours Maximum: 100 Marks
Answer ALL Questions
(5 x 20 = 100 Marks)
1. (a) Draw a hyperbola when the distance between its focus and directrix is 50 mm
and eccentricity is 3/2. Also draw the tangent and normal at a point 25 mm
from the directrix. (20)
OR
1. (b) Make free hand sketches of front, top and right side views of the 3D object
shown below : (20)
Fig. 1.(b)
Nitro PDF Software
100 Portable Document Lane
Wonderland
2. (a) A line PQ has its end P, 10 mm above the HP and 20 mm in front of the VP.
The end Q is 35 mm in front of the VP. The front view of the line measures 75
mm. The distance between the end projectors is 50 mm. Draw the projections
of the line and ¯nd its true length and its true inclinations with the VP and
HP. (20)
OR
2. (b) Draw the projections of a circle of 70 mm diameter resting on the H.P. on a
point A of the circumference. The plane is inclined to the H.P. such that the
top view of it i s an ellipse of minor axis 40 mm. The top view of the diameter,
through the point A is making an angle of 45± with the V.P. Determine the
inclination of the plane with the H.P. (20)
3. (a) An equilateral triangular prism 20 mm side of base and 50 mm long rests with
one of its shorter edges on HP such that the rectangular face containing the
edge on which the prism rests is inclined at 30± to H.P. The shorter edge resting
on HP is perpendicular to VP. (20)
OR
3. (b) A square pyramid of base 40 mm and axis 70 mm long has one of its triangular
faces on VP and the edge of base contained by that face perpendicular to HP.
Draw its projections. (20)
4. (a) A hexagonal prism of side of base 35 mm and axis length 55 mm rests with its
base on H.P such that two of the vertical surfaces are perpendicular to V.P. It
is cut by a plane inclined at 50± to H.P and perpendicular to V.P and passing
through a point on the axis at a distance 15 mm from the top. Draw its front
view, sectional top view and true shape of section. (20)
OR
4. (b) Draw the development of the lateral surface of the lower portion of a cylinder
of diameter 50 mm and axis 70 mm. The solid is cut by a section plane inclined
at 40± to HP and perpendicular to VP and passing through the midpoint of
the axis. (20)
2 Q2901
Nitro PDF Software
100 Portable Document Lane
Wonderland
5. (a) Draw the isometric projection of the object from the views shown in Figure
5(a). (20)
OR
5. (b) Draw the perspective projection of a cube of 25 mm edge, lying on a face on
the ground plane, with an edge touching the picture plane and all vertical faces
equally inclined to the picture plane. The station point is 50 mm in front of
the picture plane, 35 mm above the ground plane and lies in a central plane
which is 10 mm to the left of the center of the cube. (20)
For more information contact us via Studentstrainer@gmail.com. Caring is sharing, show your caring to us by sharing our post in social sites and keep like us,,,