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?
- A.To exchange data between two devices✓ Correct
- B.To design hardware circuit boards
- C.To store data permanently on disks
- 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?
- A.Compiler✓ Correct
- B.Receiver
- C.Sender
- 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:
- A.One direction only✓ Correct
- B.Both directions at once
- C.Both ways but alternating
- 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?
- A.Half-duplex
- B.Multiplex
- C.Simplex
- 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?
- A.The count of devices on the network
- B.The physical width of the cable used
- C.The range of frequencies a channel passes✓ Correct
- 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?
- A.Full-duplex
- B.Half-duplex✓ Correct
- C.Simplex
- 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?
- A.Infrared
- B.Microwaves
- C.Radio waves
- 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?
- A.Amperes (A)
- B.Hertz (Hz)
- C.Watts (W)
- 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?
- A.A hardware device on a NIC
- B.A software application suite
- C.A set of rules for data exchange✓ Correct
- 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?
- A.Analog signal✓ Correct
- B.Digital signal
- C.Binary signal
- 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:
- A.2B log2(L)✓ Correct
- B.2B / log2(L)
- C.B log2(L)
- 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?
- A.C = B * SNR
- B.C = B log2(1 + SNR)✓ Correct
- C.C = 2B log2(SNR)
- 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?
- A.CDM
- B.FDM✓ Correct
- C.TDM
- 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?
- A.Encoding, Sampling, Quantization
- B.Quantization, Sampling, Encoding
- C.Sampling, Quantization, Encoding✓ Correct
- 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?
- A.Unipolar encoding
- B.Manchester encoding✓ Correct
- C.NRZ-I encoding
- 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?
- A.20 dB
- B.30 dB✓ Correct
- C.10 dB
- 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?
- A.Serial transmission✓ Correct
- B.Parallel transmission
- C.Synchronous transmission
- 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?
- A.Compressing the data being sent
- B.Detecting transmission errors in data
- C.Encrypting the transmitted payload
- 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?
- A.TDM
- B.FDM✓ Correct
- C.CDM
- 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?
- A.The delay in delivering packets
- B.Number of bits in each signal
- C.Maximum capacity of a channel
- 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?
- A.24 Kbps✓ Correct
- B.12 Kbps
- C.48 Kbps
- 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?
- A.Indeterminate
- B.Bit is -1
- C.Bit is +1✓ Correct
- 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?
- A.It is more complex to implement✓ Correct
- B.It requires more bandwidth
- C.It lacks self-clocking ability
- 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?
- A.2 bits
- B.3 bits
- C.8 bits
- 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?
- A.5000 mW✓ Correct
- B.50000 mW
- C.500 mW
- 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?
- A.It allocates time slots based on demand✓ Correct
- B.It guarantees lower latency always
- C.It offers simpler implementation
- 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?
- A.32 Kbps
- B.128 Kbps
- C.64 Kbps✓ Correct
- 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?
- A.AMI
- B.MLT-3
- C.B8ZS✓ Correct
- 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?
- A.Bit rate = 8 * baud rate
- B.Bit rate = 3 * baud rate✓ Correct
- C.Bit rate = 2 * baud rate
- 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?
- A.It is halved exactly
- B.It decreases significantly
- C.It remains the same
- 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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