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Communications MCQ Set 1

1. Average energy per bit is given by
a) average energy symbol/log2 M
b) average energy symbol * log2 M
c) log2 M/ Average energy symbol
d) none of the mentioned

View Answer

Answer: a [Reason:] Average energy per bit is given by average energy symbol divided by log2 M.

2. Which FSK has no phase discontinuity?
a) Continuous FSK
b) Discrete FSK
c) Uniform FSK
d) None of the mentioned

View Answer

Answer: a [Reason:] Continuous frequency shift keying has no phase discontinuity between symbols.

3. FSK reception is
a) Phase Coherent
b) Phase non coherent
c) Phase Coherent & non coherent
d) None of the mentioned

View Answer

Answer: c [Reason:] Reception of FSK can be either phase coherent or phase non coherent.

4. FSK reception uses
a) Correlation receiver
b) PLL
c) Correlation receiver & PLL
d) None of the mentioned

View Answer

Answer: c [Reason:] Frequency shift keying uses correlation receiver and phase locked loop.

5. In non coherent reception _____ is measured.
a) Phase
b) Energy
c) Power
d) None of the mentioned

View Answer

Answer: b [Reason:] In non coherent reception of FSK, energy in each frequency is measured.

6. Every frequency has ____ orthogonal functions.
a) One
b) Two
c) Four
d) Six

View Answer

Answer: b [Reason:] Every frequency has two orthogonal functions – sine and cosine.

7. If we correlate the received signal with any one of the two orthogonal function, the obtained inner product will be
a) In phase
b) Quadrature
c) Zero
d) Cannot be determined

View Answer

Answer: c [Reason:] If we correlate the received signal with only one of the orthogonal function for example sine, the inner product obtained will be zero.

8. If we correlate the received signal with both orthogonal function the inner product will be
a) In phase
b) Quadrature
c) Both of the mentioned
d) Unity

View Answer

Answer: c [Reason:] If we correlate the received signal with both the orthogonal function, the two inner products obtained will be in phase and quadrature.

9. Simulation is used to determine
a) Bit error rate
b) Symbol error rate
c) Bit error
d) Symbol error

View Answer

Answer: a [Reason:] A simulation of digital communication system is used to estimate bit error rate.

10. Matched filter is a _____ technique.
a) Modulation
b) Demodulation
c) Modulation & Demodulation
d) None of the mentioned

View Answer

Answer: b [Reason:] Matched filter is a demodulation technique with LTI.

11. Which is called as on-off keying?
a) Amplitude shift keying
b) Uni-polar PAM
c) Both of the mentioned
d) None of the mentioned

View Answer

Answer: c [Reason:] Amplitude shift keying and uni-polar PAM both schemes are called as on off keying.

12. QAM uses ______ as the dimensions.
a) In phase
b) Quadrature
c) In phase & Quadrature
d) None of the mentioned

View Answer

Answer: c [Reason:] QAM uses in phase and quadrature which is cosine and sine respectively as the dimensions.

13. Which has same probability of error?
a) BPSK and QPSK
b) BPSK and ASK
c) BPSK and PAM
d) BPSK and QAM

View Answer

Answer: c [Reason:] BPSK is similar to bipolar PAM and both has same probability of error.

14. Which system uses QAM?
a) Digital microwave relay
b) Dial up modem
c) Digital microwave relay & Dial up modem
d) None of the mentioned

View Answer

Answer: c [Reason:] Digital microwave relay, dial up modem and etc uses QAM modulation technique.

Communications MCQ Set 2

1. A local area network can be used to interconnect
a) Computers
b) Terminals
c) Printers
d) All of the mentioned

View Answer

Answer: d [Reason:] A local area network can be used to interconnect computers, terminals, printers and so on located within a building or a small set of buildings.

2. LAN uses _________ cables.
a) Low bandwidth
b) High bandwidth
c) Low & High bandwidth
d) None of the mentioned

View Answer

Answer: b [Reason:] Local area network uses high bandwidth cables for designing.

3. The maximum packet size of Ethernet is
a) 1426 bytes
b) 1526 bytes
c) 1256 bytes
d) 1626 bytes

View Answer

Answer: b [Reason:] The maximum packet size of Ethernet is 1526 bytes where a byte is 8 bits.

4. The minimum packet size of Ethernet is
a) 81 bytes
b) 75 bytes
c) 72 bytes
d) 64 bytes

View Answer

Answer: c [Reason:] The minimum packet size is 72 bytes, consisting of 8 byte preamble, 14 byte header, 46 byte data, 4 byte parity.

5. The interface between the ring and the stations are passive.
a) True
b) False

View Answer

Answer: b [Reason:] The interface between the rings and the stations are active rather than passive.

6. Bit stuffing is done by inserting _____, after ____ consecutive ones.
a) Zero, seven
b) Zero, five
c) One, seven
d) One, five

View Answer

Answer: a [Reason:] Bit stuffing is done by inserting a zero in the data stream after seven consecutive ones.

7. As channel capacity increases, channel throughput
a) Increases
b) Decreases
c) Remains the same
d) None of the mentioned

View Answer

Answer: a [Reason:] As channel capacity R increases, the channel throughput should also increase.

8. In Aloha technique, the re-transmission takes place immediately.
a) True
b) False

View Answer

Answer: b [Reason:] In Aloha technique if negative acknowledgement is received, the re-transmission is done after a random delay so that it doesn’t collide again.

9. In which technique, packet size are constant?
a) Pure Aloha
b) Slotted Aloha
c) Pure & Slotted Aloha
d) None of the mentioned

View Answer

Answer: a [Reason:] The Pure Aloha technique has constant packet lengths.

10. For a throughput of value less than 0.20, which has less average delay?
a) S-ALOHA
b) Pure ALOHA
c) R-ALOHA
d) None of the mentioned

View Answer

Answer: a [Reason:] For a throughput value of less than 0.20, S-ALOHA has less average delay than R-ALOHA.

11. For a throughput value of 0.20 to 0.67, which has less average delay?
a) S-ALOHA
b) Pure ALOHA
c) R-ALOHA
d) None of the mentioned

View Answer

Answer: c [Reason:] For throughput value of 0.20 to 0.67, it is clear that R-ALOHA is superior and has less average delay than S-ALOHA.

12. Binary tree search is used for
a) Multiple users
b) Single users
c) Single & Multiple users
d) None of the mentioned

View Answer

Answer: a [Reason:] Binary tree search is used for rapid polling of user population and it is used for multiple users.

13. Which method is better suited for large population?
a) Straight polling
b) Binary tree search
c) Both of the mentioned
d) None of the mentioned

View Answer

Answer: b [Reason:] Binary tree search is more efficient method than straight polling in case of large population.

14. For a higher throughput value which method works efficiently?
a) S-ALOHA
b) Pure ALOHA
c) R-ALOHA
d) None of the mentioned

View Answer

Answer: c [Reason:] At lower throughput values, S-ALOHA works better and for higher throughput values R-ALOHA works better.

Communications MCQ Set 3

1. SQNR can be improved by _______ sampling rate.
a) Increasing
b) Decreasing
c) Does not depend
d) None of the mentioned

View Answer

Answer: a [Reason:] SQNR can be improved by increasing the sampling rate.

2. Which offers better SQNR?
a) Delta modulation
b) PCM
c) Delta modulation & PCM
d) None of the mentioned

View Answer

Answer: b [Reason:] PCM offers better SQNR than delta modulation.

3. Two functions are called as orthogonal if on intergrating the product we get
a) One
b) Zero
c) Infinity
d) None of the mentioned

View Answer

Answer: b [Reason:] Two functions are called as orthogonal if the integral value of integrating the product is zero.

4. Eigen functions are not orthogonal.
a) True
b) False

View Answer

Answer: b [Reason:] An important property of eigen function is that they are orthogonal.

5. Shot noise is produced by
a) Electrons
b) Photons
c) Electrons & Photons
d) None of the mentioned

View Answer

Answer: c [Reason:] Shot noise occurs due to particle like behaviour of electron and photon.

6. Shot noise is avoidable when current is
a) Zero
b) One
c) Infinity
d) None of the mentioned

View Answer

Answer: a [Reason:] Shot noise is avoidable only when current is zero.

7. Shot noise is
a) Stochastic process
b) Poisson process
c) Stochastic & Poisson process
d) None of the mentioned

View Answer

Answer: b [Reason:] Shot noise is a Poisson process.

8. Which occurs due to equilibrium fluctuations?
a) Thermal noise
b) Johnson noise
c) Thermal & Johnson noise
d) None of the mentioned

View Answer

Answer: c [Reason:] Thermal noise occurs due to equilibrium fluctuations. This noise is also called as Johnson noise.

9. Flicker noise is called as
a) White noise
b) Pink noise
c) Brown noise
d) None of the mentioned

View Answer

Answer: b [Reason:] Flicker noise is called as pink noise.

10. Which has same power spectral density?
a) White noise
b) Brown noise
c) White & Brown noise
d) None of the mentioned

View Answer

Answer: a [Reason:] White noise has same power spectral density where as it decreases in case of brown noise.

Communications MCQ Set 4

1. Noise figure measures the
a) Power degradation
b) Noise degradation
c) SNR degradation
d) None of the mentioned

View Answer

Answer: c [Reason:] Noise figure relates the SNR of the input to SNR of the output. It measures the SNR degradation caused by the network.

2. Noise figure is a parameter that represents a ______ of the system.
a) Noisiness
b) Efficiency
c) Maximum output
d) Maximum power handling capacity

View Answer

Answer: a [Reason:] Noise figure is a parameter that reperesents the noisiness of a two port network or device such as an amplifier, compared with a reference noise.

3. When two networks are connected in series, its composite noise figure can be given as
a) F1+(F2-1)/G1
b) F1-(F2-1)/G1
c) F2+(F1-1)/G1
d) F1G1+(F2-1)

View Answer

Answer: a [Reason:] When two networks are connected in series , the composite noise figure can be written as F1+(F2-1)/G1.

4. The space window region ranges from
a) 1GHz to 10 GHz
b) 10GHz to 100GHZ
c) 5GHz to 50GHz
d) 1MHz to 10MHz

View Answer

Answer: a [Reason:] There is a region between 1GHz and 10GHz where the temperature will be lowest. This region is called as microwave window or space window.

5. Atmospheric window region is used for
a) Satellite communication
b) Deep-space communication
c) Satellite & Deep-space communication
d) None of the mentioned

View Answer

Answer: c [Reason:] The region between 1GHz and 10GHz is called as space window or atmospheric window. Here the temperature will be lowest. This region is used for satellite communication and deep-space communication.

6. Which parameter is referred to as receiver sensitivity?
a) S/N ratio
b) G/T ratio
c) EIRP
d) PR/N0

View Answer

Answer: b [Reason:] The figure of parameter G/T0 is called as receiver sensitivity.

7. What are the functions of satellite repeaters?
a) Re transmits
b) Regenerates
c) Reconstitutes
d) All of the mentioned

View Answer

Answer: d [Reason:] Satellite repeaters re transmit the message received. It regenerates, demodulates and reconstitutes the digital information embedded in the received waveform.

8. What are the functions of non-regenerative repeaters?
a) Amplifies
b) Re transmits
c) Amplifies & Re transmits
d) None of the mentioned

View Answer

Answer: c [Reason:] A non regenerative repeaters can be used with many different modulation formats. It only amplifies and re transmits the message received.

9. In which repeaters uplink is decoupled from the downlink?
a) Regenerative
b) Non regenerative
c) Regenerative & Non regenerative
d) None of the mentioned

View Answer

Answer: a [Reason:] The principal advantage of regenerative repeaters over non regenerative repeaters is that the uplink is decoupled from the downlink so that the uplink noise is not transmitted on the downlink.

10. AM-to-AM conversion brings about
a) Amplitude variation
b) Phase variation
c) Frequency variation
d) None of the mentioned

View Answer

Answer: a Exp lanation: AM-to-AM conversion is a phenomenon common to non linear devices and it undergoes a non linear transformation and results in amplitude variation.

11. AM-to-PM conversion brings about
a) Amplitude variation
b) Phase variation
c) Frequency variation
d) None of the mentioned

View Answer

Answer: b [Reason:] AM-to-PM conversion is another common phenomenon in non linear devices. It brings about phase variation that can effect the error performance.

Communications MCQ Set 5

1. Pretty good privacy (PGP) security system uses
a) Public key cryptosystem
b) Private key cryptosystem
c) Public & Private key cryptosystem
d) None of the mentioned

View Answer

Answer: c [Reason:] PGP uses many encryption techniques such as private key cryptosystem and also public key cryptosystem.

2. Public key cryptosystem is used for the encryption of
a) Messages
b) Session key
c) Session key & Messages
d) None of the mentioned

View Answer

Answer: b [Reason:] Public key cryptosystem is used for encryption of messages and private key cryptosystem is used for encryption of messages.

3. For digital signatures private key cryptosystem is used.
a) True
b) False

View Answer

Answer: b [Reason:] Public key cryptosystem is used for the creation of digital signatures.

4. Data compression includes
a) Removal of redundant character
b) Uniform distribution of characters
c) Both of the mentioned
d) None of the mentioned

View Answer

Answer: c [Reason:] Data compression removes redundant character strings in a file and produces a more uniform distribution of characters.

5. PGP offers _____ block ciphers for message encryption.
a) Triple-DES
b) CAST
c) IDEA
d) All of the mentioned

View Answer

Answer: d [Reason:] Pretty good privacy security system offers three block ciphers for message encryption – Triple-DES, IDEA and CAST.

6. Which block cipher has key length of 128 bits?
a) IDEA
b) CAST
c) IDEA & CAST
d) None of the mentioned

View Answer

Answer: c [Reason:] The block ciphers IDEA and CAST has key length of 128 bits where as Triple-DES has key size of 168 bits.

7. These three ciphers can operate of ______ of plaintext and cipher text.
a) 128 bit blocks
b) 64 bit blocks
c) 256 bit blocks
d) 156 bit blocks

View Answer

Answer: b [Reason:] The three ciphers – Triple-DES, IDEA and CAST can operate on 64 bit blocks of plain text and cipher text.

8. The key size of DES is
a) 56 bits
b) 64 bits
c) 128 bits
d) 168 bits

View Answer

Answer: a [Reason:] The key size of DES algorithm is 56 bits and with Triple-DES, the message to be encrypted is run on DES algorithm 3 times.

9. Which operation is used in encryption using IDEA?
a) Addition modulo 216
b) Bit wise XOR
c) Addition modulo 216 & Bit wise XOR
d) None of the mentioned

View Answer

Answer: c [Reason:] The arithmetic operations used in IDEA are addition modulo 216, multiplication modulo 216+1, and bit wise XOR.

10. What is the key size allowed in PGP?
a) 1024-1056
b) 1024-4056
c) 1024-4096
d) 1024-2048

View Answer

Answer: c [Reason:] Pretty good privacy security system allows 1024 to 4096 bits of key size.

11. Which algorithm is used for public key encryption?
a) RSA
b) Diffie-Hellman
c) RSA & Diffie-Hellman
d) None of the mentioned

View Answer

Answer; c [Reason:] The two algorithms used for public key encryption in PGP are RSA and Diffie-Hellman algorithms.

12. Which should be used first to decrypt?
a) Public key
b) Private key
c) Public & Private key
d) None of the mentioned

View Answer

Answer: b [Reason:] The recipient should first decrypt the message using private key and then with the public key.

13. The digital signature provides authentication to the
a) Sender
b) Message
c) Sender & Message
d) None of the mentioned

View Answer

Answer: c [Reason:] The digital signature provides authentication to both the sender and the message.

14. In hash function output length is fixed.
a) True
b) False

View Answer

Answer: a [Reason:] In hash function the output length is fixed and it is easy to compute.

15. The hash function
a) Is collision free
b) Has manageable collision
c) Has high unmanageable level of collision
d) None of the mentioned

View Answer

Answer: a [Reason:] Hash function is collision free and it is unfeasible that two different messages will create the same hash value.

16. DSA protocol is based on
a) Discrete logarithm problem
b) Continuous logarithm problem
c) Discrete & Continuous logarithm problem
d) None of the mentioned

View Answer

Answer: a [Reason:] The DSA algorithm uses and is based on discrete logarithm problem.