Multiple Choice Questions and Answers on Digital Communication

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1)   In uniform quantization process

DRex Electronics

a. The step size remains same
b. Step size varies according to the values of the input signal
c. The quantizer has linear characteristics
d. Both a and c are correct

ANSWER: (d) Both a and c are correct

2)   The process of converting the analog sample into discrete form is called

a. Modulation
b. Multiplexing
c. Quantization
d. Sampling

ANSWER:(c) Quantization

3)   The characteristics of compressor in μ-law companding are

a. Continuous in nature
b. Logarithmic in nature
c. Linear in nature
d. Discrete in nature

ANSWER: (a) Continuous in nature

4)   The modulation techniques used to convert analog signal into digital signal are

a. Pulse code modulation
b. Delta modulation
c. Adaptive delta modulation
d. All of the above

ANSWER: (d) All of the above

5)   The sequence of operations in which PCM is done is

a. Sampling, quantizing, encoding
b. Quantizing, encoding, sampling
c. Quantizing, sampling, encoding
d. None of the above

ANSWER:(a) Sampling, quantizing, encoding

6)   In PCM, the parameter varied in accordance with the amplitude of the modulating signal is

a. Amplitude
b. Frequency
c. Phase
d. None of the above

ANSWER: (d) None of the above

7)   One of the disadvantages of PCM is

a. It requires large bandwidth
b. Very high noise
c. Cannot be decoded easily
d. All of the above

ANSWER: (a) It requires large bandwidth

8)   The expression for bandwidth BW of a PCM system, where v is the number of bits per sample and fm is the modulating frequency, is given by

a. BW ≥ vfm
b. BW ≤ vfm
c. BW ≥ 2 vfm
d. BW ≥ 1/2 vfm

ANSWER: (a) BW ≥ vfm

9)   The error probability of a PCM is

a. Calculated using noise and inter symbol interference
b. Gaussian noise + error component due to inter symbol interference
c. Calculated using power spectral density
d. All of the above

ANSWER: (d) All of the above

10)   In Delta modulation,

a. One bit per sample is transmitted
b. All the coded bits used for sampling are transmitted
c. The step size is fixed
d. Both a and c are correct

ANSWER: (d) Both a and c are correct

11)   In digital transmission, the modulation technique that requires minimum bandwidth is

a. Delta modulation
b. PCM
c. DPCM
d. PAM

ANSWER: (a) Delta modulation

12)   In Delta Modulation, the bit rate is

a. N times the sampling frequency
b. N times the modulating frequency
c. N times the nyquist criteria
d. None of the above

ANSWER: (a) N times the sampling frequency

13)   In Differential Pulse Code Modulation techniques, the decoding is performed by

a. Accumulator
b. Sampler
c. PLL
d. Quantizer

ANSWER: (a) Accumulator

14)   DPCM is a technique

a. To convert analog signal into digital signal
b. Where difference between successive samples of the analog signals are encoded into n-bit data streams
c. Where digital codes are the quantized values of the predicted value
d. All of the above

ANSWER: (d) All of the above

15)   DPCM suffers from

a. Slope over load distortion
b. Quantization noise
c. Both a & b
d. None of the above

ANSWER:(c)  Both a & b

16)   The noise that affects PCM

a. Transmission noise
b. Quantizing noise
c. Transit noise
d. Both a and b are correct

ANSWER: (d) Both a and b are correct

17)   The factors that cause quantizing error in delta modulation are

a. Slope overload distortion
b. Granular noise
c. White noise
d. Both a and b are correct

ANSWER:(d) Both a and b are correct

18)   Granular noise occurs when

a. Step size is too small
b. Step size is too large
c. There is interference from the adjacent channel
d. Bandwidth is too large

ANSWER: (b) Step size is too large

19)   The crest factor of a waveform is given as –

a. 2Peak value/ rms value
b. rms value / Peak value
c. Peak value/ rms value
d. Peak value/ 2rms value

ANSWER: (c) Peak value/ rms value

20)   The digital modulation technique in which the step size is varied according to the variation in the slope of the input is called

a. Delta modulation
b. PCM
c. Adaptive delta modulation
d. PAM

ANSWER: (c) Adaptive delta modulation

21)   The digital modulation scheme in which the step size is not fixed is

a. Delta modulation
b. Adaptive delta modulation
c. DPCM
d. PCM

ANSWER:(b) Adaptive delta modulation

22)   In Adaptive Delta Modulation, the slope error reduces and

a. Quantization error decreases
b. Quantization error increases
c. Quantization error remains same
d. None of the above

ANSWER: (b) Quantization error increases

23)   The number of voice channels that can be accommodated for transmission in T1 carrier system is

a. 24
b. 32
c. 56
d. 64

ANSWER: (a) 24

24)   The maximum data transmission rate in T1 carrier system is

a. 2.6 megabits per second
b. 1000 megabits per second
c. 1.544 megabits per second
d. 5.6 megabits per second

ANSWER: (c) 1.544 megabits per second

25)   T1 carrier system is used

a. For PCM voice transmission
b. For delta modulation
c. For frequency modulated signals
d. None of the above

ANSWER: (a) For PCM voice transmission

26)   Matched filter may be optimally used only for

a. Gaussian noise
b. Transit time noise
c. Flicker
d. All of the above

ANSWER:(a) Gaussian noise

27)   Characteristics of Matched filter are

a. Matched filter is used to maximize Signal to noise ratio even for non Gaussian noise
b. It gives the output as signal energy in the absence of noise
c. They are used for signal detection
d. All of the above

ANSWER: (d) All of the above

28)   Matched filters may be used

a. To estimate the frequency of the received signal
b. In parameter estimation problems
c. To estimate the distance of the object
d. All of the above

ANSWER: (d) All of the above

29)   The process of coding multiplexer output into electrical pulses or waveforms for transmission is called

a. Line coding
b. Amplitude modulation
c. FSK
d. Filtering

ANSWER:(a)  Line coding

30)   For a line code, the transmission bandwidth must be

a. Maximum possible
b. As small as possible
c. Depends on the signal
d. None of the above

ANSWER: (b) As small as possible

31)   Regenerative repeaters are used for

a. Eliminating noise
b. Reconstruction of signals
c. Transmission over long distances
d. All of the above

ANSWER:(d) All of the above

32)   Scrambling of data is

a. Removing long strings of 1’s and 0’s
b. Exchanging of data
c. Transmission of digital data
d. All of the above

ANSWER: (a) Removing long strings of 1’s and 0’s

33)   In polar RZ format for coding, symbol ‘0’ is represented by

a. Zero voltage
b. Negative voltage
c. Pulse is transmitted for half the duration
d. Both b and c are correct

ANSWER: (d) Both b and c are correct

34)   In a uni-polar RZ format,

a. The waveform has zero value for symbol ‘0’
b. The waveform has A volts for symbol ‘1’
c. The waveform has positive and negative values for ‘1’ and ‘0’ symbol respectively
d. Both a and b are correct

ANSWER: (d) Both a and b are correct

35)   Polar coding is a technique in which

a. 1 is transmitted by a positive pulse and 0 is transmitted by negative pulse
b. 1 is transmitted by a positive pulse and 0 is transmitted by zero volts
c. Both a & b
d. None of the above

ANSWER: (a) 1 is transmitted by a positive pulse and 0 is transmitted by negative pulse

36)   The polarities in NRZ format use

a. Complete pulse duration
b. Half duration
c. Both positive as well as negative value
d. Each pulse is used for twice the duration

ANSWER: (a) Complete pulse duration

37)   The format in which the positive half interval pulse is followed by a negative half interval pulse for transmission of ‘1’ is

a. Polar NRZ format
b. Bipolar NRZ format
c. Manchester format
d. None of the above

ANSWER: (c) Manchester format

38)   The maximum synchronizing capability in coding techniques is present in

a. Manchester format
b. Polar NRZ
c. Polar RZ
d. Polar quaternary NRZ

ANSWER: (a) Manchester format

39)   The advantage of using Manchester format of coding is

a. Power saving
b. Polarity sense at the receiver
c. Noise immunity
d. None of the above

ANSWER: (a) Power saving

40)   Alternate Mark Inversion (AMI) is also known as

a. Pseudo ternary coding
b. Manchester coding
c. Polar NRZ format
d. None of the above

ANSWER:(a) Pseudo ternary coding

41)   In DPSK technique, the technique used to encode bits is

a. AMI
b. Differential code
c. Uni polar RZ format
d. Manchester format

ANSWER: (b)Differential code

42)   The channel capacity according to Shannon’s equation is

a. Maximum error free communication
b. Defined for optimum system
c. Information transmitted
d. All of the above

ANSWER: (d) All of the above

43)   For a binary symmetric channel, the random bits are given as

a. Logic 1 given by probability P and logic 0 by (1-P)
b. Logic 1 given by probability 1-P and logic 0 by P
c. Logic 1 given by probability P2 and logic 0 by 1-P
d. Logic 1 given by probability P and logic 0 by (1-P)2

ANSWER: (a) Logic 1 given by probability P and logic 0 by (1-P)

44)   The technique that may be used to increase average information per bit is

a. Shannon-Fano algorithm
b. ASK
c. FSK
d. Digital modulation techniques

ANSWER: (a) Shannon-Fano algorithm

45)   Code rate r, k information bits and n as total bits, is defined as

a. r = k/n
b. k = n/r
c. r = k * n
d. n = r * k

ANSWER: (a) r = k/n

46)   The information rate R for given average information H= 2.0 for analog signal band limited to B Hz is

a. 8 B bits/sec
b. 4 B bits/sec
c. 2 B bits/sec
d. 16 B bits/sec

ANSWER:(b) 4 B bits/sec

47)   Information rate is defined as

a. Information per unit time
b. Average number of bits of information per second
c. rH
d. All of the above

ANSWER: (d) All of the above

48)   The mutual information

a. Is symmetric
b. Always non negative
c. Both a and b are correct
d. None of the above

ANSWER: (c) Both a and b are correct

49)   The relation between entropy and mutual information is

a. I(X;Y) = H(X) – H(X/Y)
b. I(X;Y) = H(X/Y) – H(Y/X)
c. I(X;Y) = H(X) – H(Y)
d. I(X;Y) = H(Y) – H(X)

ANSWER:(a)  I(X;Y) = H(X) – H(X/Y)

50)   Entropy is

a. Average information per message
b. Information in a signal
c. Amplitude of signal
d. All of the above

ANSWER: (a) Average information per message

51)   The memory less source refers to

a. No previous information
b. No message storage
c. Emitted message is independent of previous message
d. None of the above

ANSWER: (c) Emitted message is independent of previous message

52)   The information I contained in a message with probability of occurrence is given by (k is constant)

a. I = k log21/P
b. I = k log2P
c. I = k log21/2P
d. I = k log21/P2

ANSWER:(a) I = k log21/P

53)   The expected information contained in a message is called

a. Entropy
b. Efficiency
c. Coded signal
d. None of the above

ANSWER: (a) Entropy

54)   Overhead bits are

a. Framing and synchronizing bits
b. Data due to noise
c. Encoded bits
d. None of the above

ANSWER: (a) Framing and synchronizing bits

55)   ISI may be removed by using

a. Differential coding
b. Manchester coding
c. Polar NRZ
d. None of the above

ANSWER: (a) Differential coding

56)   Timing jitter is

a. Change in amplitude
b. Change in frequency
c. Deviation in location of the pulses
d. All of the above

ANSWER: (c) Deviation in location of the pulses

57)   Probability density function defines

a. Amplitudes of random noise
b. Density of signal
c. Probability of error
d. All of the above

ANSWER: (a) Amplitudes of random noise

58)   Impulse noise is caused due to

a. Switching transients
b. Lightening strikes
c. Power line load switching
d. All of the above

ANSWER: (d) All of the above

59)   In coherent detection of signals,

a. Local carrier is generated
b. Carrier of frequency and phase as same as transmitted carrier is generated
c. The carrier is in synchronization with modulated carrier
d. All of the above

ANSWER: (d) All of the above

60)   Synchronization of signals is done using

a. Pilot clock
b. Extracting timing information from the received signal
c. Transmitter and receiver connected to master timing source
d. All of the above

ANSWER:(d) All of the above

61)   Graphical representation of linear block code is known as

a. Pi graph
b. Matrix
c. Tanner graph
d. None of the above

ANSWER: (c) Tanner graph

62)   A linear code

a. Sum of code words is also a code word
b. All-zero code word is a code word
c. Minimum hamming distance between two code words is equal to weight of any non zero code word
d. All of the above

ANSWER: (d) All of the above

63)   For decoding in convolution coding, in a code tree,

a. Diverge upward when a bit is 0 and diverge downward when the bit is 1
b. Diverge downward when a bit is 0 and diverge upward when the bit is 1
c. Diverge left when a bit is 0 and diverge right when the bit is 1
d. Diverge right when a bit is 0 and diverge left when the bit is 1

ANSWER: (a)Diverge upward when a bit is 0 and diverge downward when the bit is 1

64)   The code in convolution coding is generated using

a. EX-OR logic
b. AND logic
c. OR logic
d. None of the above

ANSWER: (a) EX-OR logic

65)   Interleaving process permits a burst of B bits, with l as consecutive code bits and t errors when

a. B ≤ 2tl
b. B ≥ tl
c. B ≤ tl/2
d. B ≤ tl

ANSWER: (d) B ≤ tl

66)   For a (7, 4) block code, 7 is the total number of bits and 4 is the number of

a. Information bits
b. Redundant bits
c. Total bits- information bits
d. None of the above

ANSWER: (a) Information bits

67)   Parity bit coding may not be used for

a. Error in more than single bit
b. Which bit is in error
c. Both a & b
d. None of the above

ANSWER: (c) Both a & b

68)   Parity check bit coding is used for

a. Error correction
b. Error detection
c. Error correction and detection
d. None of the above

ANSWER: (b) Error detection

69)   For hamming distance dmin and t errors in the received word, the condition to be able to correct the errors is

a. 2t + 1 ≤ dmin
b. 2t + 2 ≤ dmin
c. 2t + 1 ≤ 2dmin
d. Both a and b

ANSWER: (d) Both a and b

70)   For hamming distance dmin and number of errors D, the condition for receiving invalid codeword is

a. D ≤ dmin + 1
b. D ≤ dmin – 1
c. D ≤ 1 – dmin
d. D ≤ dmin

ANSWER:(b)  D ≤ dmin – 1

71)   Run Length Encoding is used for

a. Reducing the repeated string of characters
b. Bit error correction
c. Correction of error in multiple bits
d. All of the above

ANSWER: (a) Reducing the repeated string of characters

72)   The prefix code is also known as

a. Instantaneous code
b. Block code
c. Convolutional code
d. Parity bit

ANSWER: (a) Instantaneous code

73)   The minimum distance for unextended Golay code is

a. 8
b. 9
c. 7
d. 6

ANSWER: (c) 7

74)   The Golay code (23,12) is a codeword of length 23 which may correct

a. 2 errors
b. 3 errors
c. 5 errors
d. 8 errors

ANSWER: (b) 3 errors

75)   Orthogonality of two codes means

a. The integrated product of two different code words is zero
b. The integrated product of two different code words is one
c. The integrated product of two same code words is zero
d. None of the above

ANSWER: (a) The integrated product of two different code words is zero

76)   The probability density function of a Markov process is

a. p(x1,x2,x3…….xn) = p(x1)p(x2/x1)p(x3/x2)…….p(xn/xn-1)
b. p(x1,x2,x3…….xn) = p(x1)p(x1/x2)p(x2/x3)…….p(xn-1/xn)
c. p(x1,x2,x3……xn) = p(x1)p(x2)p(x3)…….p(xn)
d. p(x1,x2,x3……xn) = p(x1)p(x2 * x1)p(x3 * x2)……..p(xn * xn-1)

ANSWER:(a) p(x1,x2,x3…….xn) = p(x1)p(x2/x1)p(x3/x2)…….p(xn/xn-1)

77)   The capacity of Gaussian channel is

a. C = 2B(1+S/N) bits/s
b. C = B2(1+S/N) bits/s
c. C = B(1+S/N) bits/s
d. C = B(1+S/N)2 bits/s

ANSWER: (c) C = B(1+S/N) bits/s

78)   For M equally likely messages, the average amount of information H is

a. H = log10M
b. H = log2M
c. H = log10M2
d. H = 2log10M

ANSWER:(b)  H = log2M

79)   The channel capacity is

a. The maximum information transmitted by one symbol over the channel
b. Information contained in a signal
c. The amplitude of the modulated signal
d. All of the above

ANSWER: (a) The maximum information transmitted by one symbol over the channel

80)   The capacity of a binary symmetric channel, given H(P) is binary entropy function is

a. 1 – H(P)
b. H(P) – 1
c. 1 – H(P)2
d. H(P)2 – 1

ANSWER:(a)  1 – H(P)

81)   According to Shannon Hartley theorem,

a. The channel capacity becomes infinite with infinite bandwidth
b. The channel capacity does not become infinite with infinite bandwidth
c. Has a tradeoff between bandwidth and Signal to noise ratio
d. Both b and c are correct

ANSWER: (d) Both b and c are correct

82)   The negative statement for Shannon’s theorem states that

a. If R > C, the error probability increases towards Unity
b. If R < C, the error probability is very small
c. Both a & b
d. None of the above

ANSWER: (a) If R > C, the error probability increases towards Unity

83)   For M equally likely messages, M>>1, if the rate of information R ≤ C, the probability of error is

a. Arbitrarily small
b. Close to unity
c. Not predictable
d. Unknown

ANSWER: (a) Arbitrarily small

84)   For M equally likely messages, M>>1, if the rate of information R > C, the probability of error is

a. Arbitrarily small
b. Close to unity
c. Not predictable
d. Unknown

ANSWER: (b) Close to unity

85)   In Alternate Mark Inversion (AMI) is

a. 0 is encoded as positive pulse and 1 is encoded as negative pulse
b. 0 is encoded as no pulse and 1 is encoded as negative pulse
c. 0 is encoded as negative pulse and 1 is encoded as positive pulse
d. 0 is encoded as no pulse and 1 is encoded as positive or negative pulse

ANSWER: (b) 0 is encoded as no pulse and 1 is encoded as positive or negative pulse

86)   Advantages of using AMI

a. Needs least power as due to opposite polarity
b. Prevents build-up of DC
c. May be used for longer distance
d. All of the above

ANSWER: (d)All of the above

87)   The interference caused by the adjacent pulses in digital transmission is called

a. Inter symbol interference
b. White noise
c. Image frequency interference
d. Transit time noise

ANSWER: (a) Inter symbol interference

88)   Eye pattern is

a. Is used to study ISI
b. May be seen on CRO
c. Resembles the shape of human eye
d. All of the above

ANSWER: (d) All of the above

89)   The time interval over which the received signal may be sampled without error may be explained by

a. Width of eye opening of eye pattern
b. Rate of closure of eye of eye pattern
c. Height of the eye opening of eye pattern
d. All of the above

ANSWER:(a)  Width of eye opening of eye pattern

90)   For a noise to be white Gaussian noise, the optimum filter is known as

a. Low pass filter
b. Base band filter
c. Matched filter
d. Bessel filter

ANSWER:(c) Matched filter

91)   Matched filters are used

a. For maximizing signal to noise ratio
b. For signal detection
c. In radar
d. All of the above

ANSWER: (d) All of the above

92)   The number of bits of data transmitted per second is called

a. Data signaling rate
b. Modulation rate
c. Coding
d. None of the above

ANSWER: (a) Data signaling rate

93)   Pulse shaping is done

a. to control Inter Symbol Interference
b. by limiting the bandwidth of transmission
c. after line coding and modulation of signal
d. All of the above

ANSWER: (d) All of the above

94)   The criterion used for pulse shaping to avoid ISI is

a. Nyquist criterion
b. Quantization
c. Sample and hold
d. PLL

ANSWER: (a) Nyquist criterion

95)   The filter used for pulse shaping is

a. Raised – cosine filter
b. Sinc shaped filter
c. Gaussian filter
d. All of the above

ANSWER: (d) All of the above

96)   Roll – off factor is defined as

a. The bandwidth occupied beyond the Nyquist Bandwidth of the filter
b. The performance of the filter or device
c. Aliasing effect
d. None of the above

ANSWER: (a) The bandwidth occupied beyond the Nyquist Bandwidth of the filter

97)   Nyquist criterion helps in

a. Transmitting the signal without ISI
b. Reduction in transmission bandwidth
c. Increase in transmission bandwidth
d. Both a and b

ANSWER: (d) Both a and b

98)   The Nyquist theorem is

a. Relates the conditions in time domain and frequency domain
b. Helps in quantization
c. Limits the bandwidth requirement
d. Both a and c

ANSWER: (d) Both a and c

99)   The difficulty in achieving the Nyquist criterion for system design is

a. There are abrupt transitions obtained at edges of the bands
b. Bandwidth criterion is not easily achieved
c. Filters are not available
d. None of the above

ANSWER: (a) There are abrupt transitions obtained at edges of the bands

100)   Equalization in digital communication

a. Reduces inter symbol interference
b. Removes distortion caused due to channel
c. Is done using linear filters
d. All of the above

ANSWER:  (d) All of the above

101)   Zero forced equalizers are used for

a. Reducing ISI to zero
b. Sampling
c. Quantization
d. None of the abov

ANSWER:  (a)Reducing ISI to zero

102)   The transmission bandwidth of the raised cosine spectrum is given by

a. Bt = 2w(1 + α)
b. Bt = w(1 + α)
c. Bt = 2w(1 + 2α)
d. Bt = 2w(2 + α)

ANSWER: (a) Bt = 2w(1 + α)

103)   The preferred orthogonalization process for its numerical stability is

a. Gram- Schmidt process
b. House holder transformation
c. Optimization
d. All of the above

ANSWER:  (b) House holder transformation

104)   For two vectors to be orthonormal, the vectors are also said to be orthogonal. The reverse of the same

a. Is true
b. Is not true
c. Is not predictable
d. None of the above

ANSWER:  (b) Is not true

105)   Orthonormal set is a set of all vectors that are

a. Mutually orthonormal and are of unit length
b. Mutually orthonormal and of null length
c. Both a & b
d. None of the above

ANSWER:  (a) Mutually orthonormal and are of unit length

106)   In On-Off keying, the carrier signal is transmitted with signal value ‘1’ and ‘0’ indicates

a. No carrier
b. Half the carrier amplitude
c. Amplitude of modulating signal
d. None of the above

ANSWER: (a)  No carrier

107)   ASK modulated signal has the bandwidth

a. Same as the bandwidth of baseband signal
b. Half the bandwidth of baseband signal
c. Double the bandwidth of baseband signal
d. None of the above

ANSWER:  (a) Same as the bandwidth of baseband signal

108)   Coherent detection of binary ASK signal requires

a. Phase synchronization
b. Timing synchronization
c. Amplitude synchronization
d. Both a and b

ANSWER:  (d) Both a and b

109)   The probability of error of DPSK is ______________ than that of BPSK.

a. Higher
b. Lower
c. Same
d. Not predictable

ANSWER:  (a) Higher

110)   In Binary Phase Shift Keying system, the binary symbols 1 and 0 are represented by carrier with phase shift of

a. Π/2
b. Π
c.
d. 0

ANSWER: (b)  Π

111)   BPSK system modulates at the rate of

a. 1 bit/ symbol
b. 2 bit/ symbol
c. 4 bit/ symbol
d. None of the above

ANSWER: (a) 1 bit/ symbol

112)   The BPSK signal has +V volts and -V volts respectively to represent

a. 1 and 0 logic levels
b. 11 and 00 logic levels
c. 10 and 01 logic levels
d. 00 and 11 logic levels

ANSWER:  (a) 1 and 0 logic levels

113)   The binary waveform used to generate BPSK signal is encoded in

a. Bipolar NRZ format
b. Manchester coding
c. Differential coding
d. None of the above

ANSWER:  (a) Bipolar NRZ format

114)   The bandwidth of BFSK is ______________ than BPSK.

a. Lower
b. Same
c. Higher
d. Not predictable

ANSWER:  (c) Higher

115)   In Binary FSK, mark and space respectively represent

a. 1 and 0
b. 0 and 1
c. 11 and 00
d. 00 and 11

ANSWER:  (a) 1 and 0

116)   The frequency shifts in the BFSK usually lies in the range

a. 50 to 1000 Hz
b. 100 to 2000 Hz
c. 200 to 500 Hz
d. 500 to 10 Hz

ANSWER: (a) 50 to 1000 Hz

117)   The spectrum of BFSK may be viewed as the sum of

a. Two ASK spectra
b. Two PSK spectra
c. Two FSK spectra
d. None of the above

ANSWER:  (a) Two ASK spectra

118)   The maximum bandwidth is occupied by

a. ASK
b. BPSK
c. FSK
d. None of the above

ANSWER: (c) FSK

119)   QPSK is a modulation scheme where each symbol consists of

a. 4 bits
b. 2 bits
c. 1 bits
d. M number of bits, depending upon the requireme

ANSWER: (b)  2 bits

120)   The data rate of QPSK is ___________ of BPSK.

a. Thrice
b. Four times
c. Twice
d. Same

ANSWER:  (c) Twice

121)   QPSK system uses a phase shift of

a. Π
b. Π/2
c. Π/4
d.

ANSWER:  (b) Π/2

122)   Minimum shift keying is similar to

a. Continuous phase frequency shift keying
b. Binary phase shift keying
c. Binary frequency shift keying
d. QPSK

ANSWER: (a) Continuous phase frequency shift keying

123)   In MSK, the difference between the higher and lower frequency is

a. Same as the bit rate
b. Half of the bit rate
c. Twice of the bit rate
d. Four time the bit rate

ANSWER:  (b) Half of the bit rate

124)   The technique that may be used to reduce the side band power is

a. MSK
b. BPSK
c. Gaussian minimum shift keying
d. BFSK

ANSWER:  (c) Gaussian minimum shift keying