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Data Communication

Topic in Computer Networks & Cloud Computing

210 total MCQsShowing 30 with explanations10 Easy10 Medium10 Hard

About This Topic

Data communication is the exchange of digital data between two devices over a transmission medium such as copper wire, optical fibre or radio. Questions here usually start with the five components of a communication system and the simplex, half-duplex and full-duplex modes, then move to signal theory: bandwidth versus throughput, bit rate versus baud rate, and converting a signal-to-noise ratio into decibels. Expect Nyquist and Shannon capacity calculations, baseband and broadband transmission, synchronous and asynchronous framing, modulation schemes like ASK, FSK and PSK, multiplexing with FDM, TDM and WDM, and impairments such as attenuation, noise and intersymbol interference.

Below are 30 practice questions from a pool of 210 Data Communication MCQs, one of 10 topics in Computer Networks & Cloud Computing. Each shows the correct answer with an explanation; when you are ready, take a timed quiz to test recall under exam conditions.

Practice Questions

Each question below shows the correct answer with a full explanation. Use these to build conceptual understanding before attempting a timed quiz.

Data CommunicationEasy

Q1. What is the primary purpose of data communication?

  1. A.To exchange data between two devices✓ Correct
  2. B.To design hardware circuit boards
  3. C.To store data permanently on disks
  4. D.To compile programs into binaries

Explanation

Data communication is the exchange of data between two devices via some form of transmission medium such as a wire cable or wireless.

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Data CommunicationEasy

Q2. Which of the following is NOT a component of a data communication system?

  1. A.Compiler✓ Correct
  2. B.Receiver
  3. C.Sender
  4. D.Protocol

Explanation

The five components of a data communication system are sender, receiver, message, transmission medium, and protocol. A compiler is not a component of data communication.

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Data CommunicationEasy

Q3. In simplex mode of data transmission, data flows in:

  1. A.One direction only✓ Correct
  2. B.Both directions at once
  3. C.Both ways but alternating
  4. D.Multiple directions

Explanation

In simplex mode, the communication is unidirectional. Only one device can send and the other can only receive. Example: keyboard to CPU.

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Data CommunicationEasy

Q4. Which transmission mode allows data to flow in both directions simultaneously?

  1. A.Half-duplex
  2. B.Multiplex
  3. C.Simplex
  4. D.Full-duplex✓ Correct

Explanation

Full-duplex mode allows data to travel in both directions at the same time. Example: telephone communication.

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Data CommunicationEasy

Q5. What does bandwidth refer to in data communication?

  1. A.The count of devices on the network
  2. B.The physical width of the cable used
  3. C.The range of frequencies a channel passes✓ Correct
  4. D.The length of the transmission medium

Explanation

Bandwidth is the range of frequencies that a communication channel can pass, typically measured in Hertz (Hz) or bits per second (bps).

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Data CommunicationEasy

Q6. Walkie-talkie is an example of which transmission mode?

  1. A.Full-duplex
  2. B.Half-duplex✓ Correct
  3. C.Simplex
  4. D.Automatic

Explanation

A walkie-talkie uses half-duplex mode where both parties can communicate but not at the same time. One must finish before the other can respond.

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Data CommunicationEasy

Q7. Which of the following is a guided transmission medium?

  1. A.Infrared
  2. B.Microwaves
  3. C.Radio waves
  4. D.Twisted pair✓ Correct

Explanation

Twisted pair cable is a guided (wired) transmission medium. Radio waves, microwaves, and infrared are unguided (wireless) media.

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Data CommunicationEasy

Q8. What is the unit of data transmission speed?

  1. A.Amperes (A)
  2. B.Hertz (Hz)
  3. C.Watts (W)
  4. D.Bits per second (bps)✓ Correct

Explanation

Data transmission speed is measured in bits per second (bps). Higher-order units include Kbps, Mbps, and Gbps.

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Data CommunicationEasy

Q9. In data communication, what is a protocol?

  1. A.A hardware device on a NIC
  2. B.A software application suite
  3. C.A set of rules for data exchange✓ Correct
  4. D.A type of transmission cable

Explanation

A protocol is a set of rules that governs data communication. It defines what is communicated, how it is communicated, and when it is communicated.

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Data CommunicationEasy

Q10. Which signal type has a continuous varying pattern?

  1. A.Analog signal✓ Correct
  2. B.Digital signal
  3. C.Binary signal
  4. D.Discrete signal

Explanation

An analog signal has a continuous varying pattern that changes smoothly over time. Digital signals have discrete values (typically 0 and 1).

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Data CommunicationMedium

Q11. According to the Nyquist theorem, the maximum data rate of a noiseless channel with bandwidth B and L signal levels is:

  1. A.2B log2(L)✓ Correct
  2. B.2B / log2(L)
  3. C.B log2(L)
  4. D.B^2 log2(L)

Explanation

The Nyquist theorem states that the maximum data rate = 2B log2(L), where B is bandwidth in Hz and L is the number of signal levels.

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Data CommunicationMedium

Q12. What is the Shannon capacity formula for a noisy channel?

  1. A.C = B * SNR
  2. B.C = B log2(1 + SNR)✓ Correct
  3. C.C = 2B log2(SNR)
  4. D.C = B / log2(SNR)

Explanation

Shannon's capacity formula is C = B log2(1 + SNR), where C is capacity in bps, B is bandwidth in Hz, and SNR is the signal-to-noise ratio.

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Data CommunicationMedium

Q13. Which multiplexing technique divides the bandwidth of a link into frequency bands?

  1. A.CDM
  2. B.FDM✓ Correct
  3. C.TDM
  4. D.WDM

Explanation

Frequency Division Multiplexing (FDM) divides the bandwidth into multiple frequency bands, each carrying a separate signal.

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Data CommunicationMedium

Q14. In PCM (Pulse Code Modulation), what are the three steps in order?

  1. A.Encoding, Sampling, Quantization
  2. B.Quantization, Sampling, Encoding
  3. C.Sampling, Quantization, Encoding✓ Correct
  4. D.Sampling, Encoding, Quantization

Explanation

PCM involves three steps: Sampling (measuring the analog signal at regular intervals), Quantization (rounding to nearest level), and Encoding (converting to binary).

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Data CommunicationMedium

Q15. What type of encoding uses the transition at the middle of a bit interval to represent data?

  1. A.Unipolar encoding
  2. B.Manchester encoding✓ Correct
  3. C.NRZ-I encoding
  4. D.NRZ-L encoding

Explanation

Manchester encoding uses a transition at the middle of each bit interval. A low-to-high transition represents 1, and high-to-low represents 0 (or vice versa in differential Manchester).

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Data CommunicationMedium

Q16. If a signal-to-noise ratio (SNR) is 1000, what is the SNR in decibels?

  1. A.20 dB
  2. B.30 dB✓ Correct
  3. C.10 dB
  4. D.40 dB

Explanation

SNRdB = 10 log10(SNR) = 10 log10(1000) = 10 * 3 = 30 dB.

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Data CommunicationMedium

Q17. Which type of transmission sends bits one at a time over a single channel?

  1. A.Serial transmission✓ Correct
  2. B.Parallel transmission
  3. C.Synchronous transmission
  4. D.Isochronous transmission

Explanation

Serial transmission sends data one bit at a time over a single channel. It is used for long-distance communication.

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Data CommunicationMedium

Q18. In asynchronous serial transmission, what is the purpose of start and stop bits?

  1. A.Compressing the data being sent
  2. B.Detecting transmission errors in data
  3. C.Encrypting the transmitted payload
  4. D.Synchronizing sender and receiver clocks✓ Correct

Explanation

Start and stop bits in asynchronous transmission provide synchronization at the character level, alerting the receiver to the beginning and end of each character.

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Data CommunicationMedium

Q19. Wavelength Division Multiplexing (WDM) is conceptually similar to which multiplexing technique?

  1. A.TDM
  2. B.FDM✓ Correct
  3. C.CDM
  4. D.STDM

Explanation

WDM is conceptually the same as FDM but applied to optical fiber channels. It combines multiple light signals at different wavelengths.

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Data CommunicationMedium

Q20. What is throughput in data communication?

  1. A.The delay in delivering packets
  2. B.Number of bits in each signal
  3. C.Maximum capacity of a channel
  4. D.Actual data rate through a point✓ Correct

Explanation

Throughput is the actual rate of data transfer, which may be less than the bandwidth due to congestion, protocol overhead, and other factors.

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Data CommunicationHard

Q21. A channel has a bandwidth of 4 KHz and SNR of 63. What is the maximum achievable data rate according to Shannon's theorem?

  1. A.24 Kbps✓ Correct
  2. B.12 Kbps
  3. C.48 Kbps
  4. D.36 Kbps

Explanation

C = B log2(1 + SNR) = 4000 * log2(64) = 4000 * 6 = 24,000 bps = 24 Kbps.

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Data CommunicationHard

Q22. In CDMA (Code Division Multiple Access), if a station's chip sequence is (-1, -1, -1, +1, +1, -1, +1, +1) and the received data is (-1, -1, -3, +1, -1, -3, +1, +1), what is the station's data bit?

  1. A.Indeterminate
  2. B.Bit is -1
  3. C.Bit is +1✓ Correct
  4. D.Bit is 0

Explanation

The inner product is computed: (1+1+3+1-1+3+1+1)/8 = 10/8 > 0. Since the normalized result is positive, the data bit is +1.

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Data CommunicationHard

Q23. What is the main disadvantage of Differential Manchester encoding compared to standard Manchester encoding?

  1. A.It is more complex to implement✓ Correct
  2. B.It requires more bandwidth
  3. C.It lacks self-clocking ability
  4. D.It has higher error rates

Explanation

Differential Manchester encoding is more complex to implement because the encoding depends on the previous bit's value, requiring state tracking. Both use twice the bandwidth of NRZ.

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Data CommunicationHard

Q24. In a QAM-16 constellation diagram, how many bits does each signal element represent?

  1. A.2 bits
  2. B.3 bits
  3. C.8 bits
  4. D.4 bits✓ Correct

Explanation

QAM-16 has 16 different signal combinations, so each element represents log2(16) = 4 bits.

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Data CommunicationHard

Q25. A signal has passed through 3 cascaded amplifiers, each with a gain of 10 dB. If the original signal power is 5 mW, what is the output power?

  1. A.5000 mW✓ Correct
  2. B.50000 mW
  3. C.500 mW
  4. D.50 mW

Explanation

Total gain = 10 + 10 + 10 = 30 dB. 30 dB means 10^3 = 1000 times amplification. Output = 5 * 1000 = 5000 mW = 5 W.

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Data CommunicationHard

Q26. In Statistical Time Division Multiplexing (STDM), what is the key advantage over synchronous TDM?

  1. A.It allocates time slots based on demand✓ Correct
  2. B.It guarantees lower latency always
  3. C.It offers simpler implementation
  4. D.It provides higher total bandwidth

Explanation

STDM allocates time slots dynamically based on demand rather than pre-assigning them. This avoids wasting capacity when some sources are idle.

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Data CommunicationHard

Q27. A digitized voice channel is made by sampling at 8000 samples/sec with 8 bits per sample. What is the required data rate?

  1. A.32 Kbps
  2. B.128 Kbps
  3. C.64 Kbps✓ Correct
  4. D.48 Kbps

Explanation

Data rate = sampling rate * bits per sample = 8000 * 8 = 64,000 bps = 64 Kbps. This is the standard rate for a single PCM voice channel.

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Data CommunicationHard

Q28. Which scrambling technique replaces 8 consecutive zeros with a specific violation pattern to maintain synchronization?

  1. A.AMI
  2. B.MLT-3
  3. C.B8ZS✓ Correct
  4. D.HDB3

Explanation

B8ZS (Bipolar with 8-Zero Substitution) replaces 8 consecutive zeros with a pattern containing intentional violations to ensure transitions for clock recovery.

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Data CommunicationHard

Q29. What is the relationship between bit rate and baud rate when using 8-PSK modulation?

  1. A.Bit rate = 8 * baud rate
  2. B.Bit rate = 3 * baud rate✓ Correct
  3. C.Bit rate = 2 * baud rate
  4. D.Bit rate = baud rate

Explanation

In 8-PSK, each signal element represents log2(8) = 3 bits, so the bit rate is 3 times the baud rate.

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Data CommunicationHard

Q30. In spread spectrum communication, what happens to the signal bandwidth compared to the original data bandwidth?

  1. A.It is halved exactly
  2. B.It decreases significantly
  3. C.It remains the same
  4. D.It increases significantly✓ Correct

Explanation

Spread spectrum deliberately spreads the signal over a much wider bandwidth than required. This makes it resistant to interference and eavesdropping.

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