200+ Communication Systems MCQs with Answers & Explanations | Power Sector Job & BCS Exam Preparation Guide

Looking for the most important MCQs on Communication Systems, Data Communication, and Networking for competitive exams? This complete guide brings you 200+ frequently asked multiple-choice questions with answers and clear explanations, carefully selected for Power Sector Job Exams (BREB, BPDB, DESCO, PGCB), BUET/MIST admission and recruitment tests, BCS preliminary exam, and GATE/ECE exams.

Covering every key topic — from Analog & Digital Communication, Modulation (AM/FM/PM), PCM, OFDM, Noise & Information Theory, Multiplexing, Fading & Propagation, GSM/LTE/5G Standards, OSI Layers, TCP/IP Protocols, Routing, Switching, Ethernet, IPv4/IPv6, DNS, DHCP, ARP, Firewalls, VPNs, NAT, and Optical Fiber Communication — this MCQ bank is your one-stop solution for quick revision and exam practice.

Whether you’re preparing for a government job, power sector exam, or engineering entrance, these MCQs with step-by-step explanations will strengthen your concepts, boost confidence, and help you crack competitive tests with ease.

A. Fundamentals & Noise (1–10)

1. Nyquist Sampling Theorem
The minimum sampling rate for a signal of bandwidth $B$ is:
A) $B$ B) $2B$ C) $B/2$ D) $1/B$
Answer: B) $2B$
Explanation: Nyquist criterion states sampling must be ≥ twice the highest frequency to avoid aliasing.

2. Shannon Channel Capacity
For an AWGN channel with bandwidth $B$ and SNR $S/N$:
A) $B\log_2(1+S/N)$ B) $2B\log_2(1+S/N)$ C) $B\log_{10}(1+S/N)$ D) ${\log }_{2}B\mathrm{<br>}$ 

Answer: A
Explanation: Shannon–Hartley theorem gives maximum data rate.

3. Thermal Noise Power
Over bandwidth $B$:
A) $kTB$ B) $2kTB$ C) $kT/B$ D) $1/kTB$
Answer: A
Explanation: Johnson noise = $kTB$

4. Noise Figure (F)
Defined as:
A) $SNR_{out}/SNR_{in}$ B) $SNR_{in}/SNR_{out}$ C) $N_{out}/N_{in}$ D) $G/N$
Answer: B
Explanation: Ratio of input to output SNR.

5. White Noise PSD is:
A) Flat B) Exponential C) Gaussian D) Impulse
Answer: A
Explanation: Constant power density across frequencies.

6. RMS Thermal Noise Voltage ∝
A) $\sqrt{T}$ B) $T$ C) $1/T$ D) Constant
Answer: A
Explanation: $V_n=\sqrt{4kTBR}$.

7. If signal power doubles, SNR (dB) increases by:
A) 1 dB B) 2 dB C) 3 dB D) 6 dB
Answer: C
Explanation: Doubling = +3 dB.

8. Channel with $B=3kHz$, SNR=30 dB → Capacity ≈
A) 3 kbps B) 10 kbps C) 30 kbps D) 30 Mbps
Answer: C
Explanation: $C=B{\log }_2(1+SNR)$

9. Quantization noise distribution is:
A) Uniform B) Gaussian C) Exponential D) Rayleigh
Answer: A
Explanation: Uniform error between steps.

10. Companding in PCM improves:
A) Bandwidth B) Dynamic range C) Timing D) Carrier power
Answer: B
Explanation: Protects weak signals against noise.

B. Amplitude Modulation (11–20)

11. Modulation index $m$ from envelope:
A) $(V_{max}+V_{min})/(V_{max}-V_{min})$
B) $(V_{max}-V_{min})/(V_{max}+V_{min})$
C) $V_{max}/V_{min}$
D) $V_{min}/V_{max}$
Answer: B
Explanation: Derived from envelope peaks.

12. Bandwidth of AM (DSB-FC):
A) $W$ B) $2W$ C) $W/2$ D) $4W$
Answer: B
Explanation: Two sidebands.

13. Power in AM carrier =
A) $V_c^2/(2R)$ B) $V_c^2/R$ C) $m^2V_c^2/(2R)$ D) None
Answer: A
Explanation: Power = $V_{rms}^2/R$

14. Efficiency of SSB vs AM:
A) Lower B) Higher C) Same D) Zero
Answer: B
Explanation: Only one sideband → more efficient.

15. Envelope detector is used for:
A) DSB-SC B) AM with carrier C) SSB D) PM
Answer: B
Explanation: Simple detection of AM.

16. DSB-SC efficiency:
A) 33% B) 50% C) 100% D) 25%
Answer: C
Explanation: All transmitted power is useful sidebands.

17. VSB is used in:
A) AM radio B) FM radio C) TV video D) Radar
Answer: C
Explanation: Reduces TV video bandwidth.

18. Overmodulation in AM leads to:
A) Distortion B) Noise reduction C) More bandwidth D) Higher SNR
Answer: A
Explanation: Envelope crosses zero → distortion.

19. DSB-SC detection requires:
A) Envelope detector B) Coherent detector C) PLL D) Filter
Answer: B
Explanation: Needs carrier reference.

20. AM waveform consists of:
A) One component B) Carrier + sidebands C) Two carriers D) Noise only
Answer: B

C. Frequency & Phase Modulation (21–30)

21. FM bandwidth (Carson’s rule):
A) $2(\Delta f+f_m)$ B) $2f_m$ C) $\Delta f+f_m$ D) $2\Delta f$
Answer: A
Explanation: Includes deviation + message freq.

22. Pre-emphasis in FM boosts:
A) Low frequencies B) High frequencies C) Carrier D) Sidebands
Answer: B
Explanation: Compensates high-frequency noise.

23. De-emphasis in FM:
A) Boosts low freq. B) Restores spectrum C) Removes carrier D) Adds noise
Answer: B

24. Narrowband FM bandwidth 
A) $2f_m$ B) $f_m$ C) $2\Delta f$ D) $f_m+\Delta f$
Answer: A

25. PM modulation index =
A) Δf/fm B) Phase deviation / mod amplitude C) Power/Noise D) SNR
Answer: B

26. FM better than AM because:
A) Lower power B) Ignored amplitude noise C) Simpler transmitter D) Higher distortion
Answer: B

27. FM discriminator is used for:
A) AM detection B) FM demodulation C) Phase detection D) Noise filtering
Answer: B

28. Capture effect occurs in:
A) AM B) FM C) PM D) SSB
Answer: B

29. Increasing deviation ratio improves:
A) Noise immunity B) ISI C) Bandwidth reduction D) Carrier stability
Answer: A

30. PM and FM relation:
A) PM is integral of FM B) FM is derivative of PM C) Related by modulating signal derivative D) No relation
Answer: C

D. Pulse Modulation (31–40)

31. PCM bit rate:
A) $n f_s$ B) $f_s/n$ C) $2nf_s$ D) $n/f_s$
Answer: A

32. SQNR of PCM:
A) $6.02n+1.76$ dB B) $3n$ dB C) $10n$ dB D) $20n$ dB
Answer: A

33. Delta modulation slope overload occurs when:
A) Step too small B) Step too large C) High sampling D) Low SNR
Answer: A

34. Adaptive delta modulation solves:
A) Slope overload B) Aliasing C) Noise D) ISI
Answer: A

35. DPCM reduces:
A) Power B) Redundancy C) Bandwidth only D) Carrier
Answer: B

36. Aliasing occurs if:
A) $f_s<2f_{max}$ B) $f_s=2f_{max}$ C) $f_s>2f_{max}$ D) $f_s=\infty$
Answer: A

37. PAM represents info by:
A) Pulse amplitude B) Width C) Position D) Phase
Answer: A

38. PWM varies:
A) Amplitude B) Width C) Position D) Phase
Answer: B

39. PPM varies:
A) Amplitude B) Width C) Position D) Phase
Answer: C

40. TDM separates channels by:
A) Frequency B) Time slots C) Power D) Phase
Answer: B

E. Digital Modulation (41–50)

41. ASK, FSK, PSK are:
A) Analog B) Digital passband C) Analog passband D) Line codes
Answer: B

42. BPSK BER vs BFSK:
A) Lower B) Higher C) Same D) Zero
Answer: A

43. QPSK advantage:
A) More bandwidth B) Doubles bit rate with same bandwidth as BPSK C) Higher power D) Lower SNR
Answer: B

44. Bits per M-QAM symbol:
A) $\log_2 M$ B) $M$ C) $2M$ D) $1/M$
Answer: A

45. Gray coding reduces:
A) Bandwidth B) Errors per symbol error C) Latency D) Noise
Answer: B

46. Coherent detection requires:
A) Carrier phase reference B) Random phase C) PLL only D) Envelope detector
Answer: A

47. DPSK avoids:
A) Bandwidth B) Coherent detection C) Power D) ISI
Answer: B

48. Bandwidth efficiency improves with:
A) Lower M B) Higher M (QAM, PSK) C) Low SNR D) More noise
Answer: B

49. BER of BPSK in AWGN =
A) $Q(\sqrt{2E_b/N_0})$ B) $Q(\sqrt{E_b/N_0})$ C) $1/E_b$ D) $N_0/E_b$
Answer: A

50. FSK tones separated by:
A) 1/T B) 2T C) 1/2T D) Random
Answer: A

F. Multiplexing & OFDM (51–60)

51. Processing gain in CDMA =
A) Data rate / Chip rate
B) Chip rate / Data rate
C) $E_b/N_0$
D) Symbol rate × bandwidth
Answer: B
Explanation: Spreading a signal increases chip rate. The ratio (chip rate / data rate) is the processing gain, which indicates resistance to interference.

52. Orthogonality in OFDM requires subcarrier spacing =
A) $1/T$
B) $T$
C) $2/T$
D) $1/2T$
Answer: A
Explanation: Orthogonality is achieved when spacing is $1/T$, where $T$ is symbol duration. This ensures no inter-carrier interference.

53. Cyclic prefix in OFDM combats:
A) AWGN
B) ISI due to multipath
C) Frequency offset
D) Clock drift
Answer: B
Explanation: The cyclic prefix absorbs multipath delay, preventing inter-symbol interference in multipath channels.

54. Guard band in FDM prevents:
A) Noise
B) Adjacent-channel interference
C) Multipath fading
D) Doppler spread
Answer: B
Explanation: Guard bands separate adjacent channels to reduce overlap and interference.

55. Framing bits in TDM are used for:
A) Error correction
B) Identifying slot boundaries
C) Synchronization only
D) Noise removal
Answer: B
Explanation: Framing bits allow receivers to know which timeslot belongs to which user.

56. OFDM uses IFFT at the transmitter to:
A) Create orthogonal subcarriers
B) Filter noise
C) Demodulate PSK
D) Add redundancy
Answer: A
Explanation: IFFT converts frequency-domain symbols into orthogonal time-domain subcarriers for transmission.

57. Raised-cosine filter is used to:
A) Reduce ISI
B) Increase power
C) Remove noise
D) Reduce carrier
Answer: A
Explanation: Raised cosine pulses satisfy Nyquist criterion, reducing intersymbol interference.

58. Nyquist criterion ensures:
A) Zero ISI
B) Zero noise
C) Infinite bandwidth
D) High power
Answer: A
Explanation: Proper pulse shaping per Nyquist ensures symbols do not overlap at sampling instants.

59. OFDM spectral efficiency is high because:
A) Non-orthogonal carriers
B) Overlapped subcarriers without interference
C) Guard bands are large
D) Carriers are random
Answer: B
Explanation: OFDM subcarriers overlap but remain orthogonal, saving bandwidth.

60. A drawback of OFDM is:
A) Low efficiency
B) High PAPR (Peak-to-Average Power Ratio)
C) Large guard intervals
D) More ISI
Answer: B
Explanation: OFDM signals have large amplitude variations, requiring linear high-power amplifiers.

G. Error Control Coding (61–70)

61. Hamming code with distance 3 can:
A) Detect 1 error only
B) Detect 2 errors, correct 1
C) Correct 2 errors
D) Detect 3 errors
Answer: B
Explanation: With distance 3, minimum 1-bit error correction and 2-bit error detection are possible.

62. Viterbi algorithm is used for:
A) Source coding
B) Convolutional code decoding
C) Modulation
D) Equalization
Answer: B
Explanation: Viterbi is an efficient maximum-likelihood decoder for convolutional codes.

63. CRC is mainly used for:
A) Error correction
B) Error detection
C) Synchronization
D) Bandwidth saving
Answer: B
Explanation: Cyclic redundancy check detects burst errors using polynomial division.

64. Parity check detects:
A) 1 error always
B) All odd number of errors
C) All even number of errors
D) Burst errors
Answer: B
Explanation: Parity ensures total bits are even/odd; odd errors are detected, even errors may go unnoticed.

65. Forward error correction (FEC) adds:
A) Redundancy
B) Guard bands
C) Synchronization bits
D) Filters
Answer: A
Explanation: Extra redundant bits are added to correct errors without retransmission.

66. Interleaving helps combat:
A) AWGN
B) Burst errors
C) Bandwidth expansion
D) Co-channel interference
Answer: B
Explanation: Interleaving spreads burst errors into separate codewords, improving error correction.

67. Reed–Solomon codes are widely used in:
A) FM
B) CDs/DVDs
C) GSM
D) WiFi
Answer: B
Explanation: RS codes correct burst errors, ideal for storage and optical media.

68. Turbo codes are important because:
A) Easy to decode
B) Approach Shannon capacity limit
C) Zero redundancy
D) Used in AM radio
Answer: B
Explanation: Turbo codes provide near-optimal error correction performance.

69. LDPC codes are used in:
A) Old analog systems
B) Modern wireless standards (WiFi, 5G)
C) FM broadcasting
D) AM radio
Answer: B
Explanation: LDPC codes provide excellent BER performance close to Shannon limit.

70. ARQ works by:
A) Error correction
B) Retransmission on error detection
C) Preventing fading
D) Encoding source
Answer: B
Explanation: Automatic Repeat Request uses acknowledgments and retransmissions for reliability.

H. Channel & Propagation (71–80)

71. Free-space path loss (FSPL) ∝
A) $d/\lambda$
B) $(d/\lambda)^2$
C) $\lambda/d$
D) $1/d^2$
Answer: B
Explanation: Path loss increases with square of distance and frequency.

72. Doppler shift =
A) $v\lambda$
B) $v/\lambda$
C) $\lambda/v$
D) $2v/\lambda$
Answer: B
Explanation: Doppler shift = relative velocity / wavelength.

73. Coherence time ≈
A) $1/B$
B) $1/(2f_D)$
C) $1/f_c$
D) $1/\sqrt{SNR}$
Answer: B
Explanation: Inverse of Doppler spread, indicates how long channel response is stable.

74. Coherence bandwidth ≈
A) $1/\tau_{rms}$
B) $1/f_c$
C) $\tau_{rms}$
D) $\sqrt{\tau_{rms}}$
Answer: A
Explanation: Inverse relation to RMS delay spread.

75. Multipath propagation causes:
A) Noise only
B) Fading and delay spread
C) More bandwidth
D) Higher power
Answer: B
Explanation: Reflections cause constructive/destructive interference → fading.

76. Rayleigh fading occurs when:
A) LOS present
B) No LOS, many scatterers
C) Only free space
D) Only indoors
Answer: B
Explanation: Typical for dense urban environments without direct LOS.

77. Rician fading differs because:
A) No fading
B) Strong LOS + scatterers
C) Weak LOS only
D) Random noise
Answer: B
Explanation: Rician includes both direct path and scattered paths.

78. Shadowing causes:
A) Fast fading
B) Slow fading
C) Doppler shift
D) Diversity gain
Answer: B
Explanation: Obstacles cause slow variations in received signal strength.

79. Diversity combining reduces:
A) Noise
B) Fading effects
C) Bandwidth
D) Power
Answer: B
Explanation: Multiple antennas/paths improve reliability under fading.

80. MIMO improves:
A) Capacity and reliability
B) Noise only
C) Path loss
D) Delay spread
Answer: A
Explanation: Exploits multiple antennas for spatial multiplexing and diversity.

I. Receivers & Filters (81–90)

81. Superheterodyne receiver converts RF to:
A) Baseband
B) Intermediate Frequency (IF)
C) Image frequency
D) LO frequency
Answer: B
Explanation: IF simplifies filtering and amplification.

82. Image rejection improved by:
A) IF filter
B) High-Q RF filter
C) AGC
D) Audio filter
Answer: B
Explanation: Pre-selection at RF blocks image frequencies.

83. Automatic Gain Control (AGC) keeps:
A) Bandwidth constant
B) Output amplitude constant
C) Noise constant
D) Frequency stable
Answer: B
Explanation: Compensates fading, maintains output level.

84. Matched filter maximizes:
A) Bit rate
B) SNR at sampling instant
C) Bandwidth efficiency
D) Power
Answer: B
Explanation: Provides optimal detection in noise.

85. Zero-forcing equalizer drawback:
A) Complexity
B) Amplifies noise
C) Cannot remove ISI
D) Nonlinear
Answer: B
Explanation: Cancels ISI but boosts noise at spectral nulls.

86. MMSE equalizer minimizes:
A) ISI only
B) Noise only
C) Mean squared error (ISI + noise)
D) Delay
Answer: C
Explanation: Balances ISI cancellation with noise amplification.

87. Eye diagram closing vertically indicates:
A) Timing jitter
B) Noise/ISI
C) No error
D) Synchronization only
Answer: B
Explanation: Vertical eye closure → high noise/ISI.

88. PLL can be used for:
A) AM detection
B) FM/FSK demodulation and carrier recovery
C) Sampling
D) Synchronization only
Answer: B
Explanation: PLL locks onto carrier phase/frequency.

89. Scrambler is used to:
A) Compress data
B) Randomize sequences (avoid long runs)
C) Increase power
D) Reduce bandwidth
Answer: B
Explanation: Prevents long sequences of 0s/1s for synchronization.

90. Manchester coding provides:
A) Zero DC component and self-clocking
B) High noise
C) Lower bandwidth
D) Higher power
Answer: A
Explanation: Each bit has a transition → easy clock recovery.

J. Modern Standards (91–100)

91. GSM channel bandwidth =
A) 25 kHz
B) 100 kHz
C) 200 kHz
D) 1.25 MHz
Answer: C
Explanation: GSM allocates 200 kHz per carrier.

92. GSM modulation =
A) QPSK
B) GMSK
C) 16-QAM
D) OFDM
Answer: B
Explanation: Gaussian Minimum Shift Keying reduces spectrum.

93. LTE downlink multiple access =
A) SC-FDMA
B) OFDMA
C) TDMA
D) CDMA
Answer: B
Explanation: OFDMA supports multiuser and high throughput.

94. LTE uplink multiple access =
A) OFDMA
B) SC-FDMA
C) TDMA
D) CDMA
Answer: B
Explanation: SC-FDMA reduces PAPR for mobile devices.

95. 5G NR subcarrier spacing =
A) 15 kHz fixed
B) $15\times 2^\mu$ kHz
C) 7.5 kHz only
D) 240 kHz only
Answer: B
Explanation: Flexible numerology adapts to scenarios.

96. Massive MIMO in 5G improves:
A) Bandwidth
B) Capacity and spectral efficiency
C) Noise
D) Delay
Answer: B
Explanation: Uses large antenna arrays.

97. CDMA uses:
A) Time slots
B) Spreading codes
C) Frequency channels
D) Guard bands
Answer: B
Explanation: Orthogonal codes separate users.

98. OFDM combats:
A) Doppler shift
B) Multipath fading and ISI
C) Path loss
D) Noise figure
Answer: B
Explanation: Subcarrier orthogonality and CP handle multipath.

99. WiFi (IEEE 802.11) uses:
A) AM
B) FM
C) OFDM
D) TDMA only
Answer: C
Explanation: Modern WLAN standards use OFDM.

100. Satellite links suffer mainly from:
A) Doppler shift
B) Rain attenuation
C) Thermal noise only
D) Fading only
Answer: B
Explanation: Rain absorption at high frequencies is major limiting factor.

K. Analog Communication (101–110)

101. If carrier frequency is 1 MHz and modulating signal is 5 kHz, AM signal bandwidth =
A) 5 kHz
B) 10 kHz
C) 1 MHz
D) 2 MHz
Answer: B) 10 kHz
Explanation: AM bandwidth = $2f_m = 2×5kHz = 10kHz$, independent of carrier frequency.

102. In AM, maximum transmitted power occurs when modulation index =
A) 0
B) 0.5
C) 1
D) >1
Answer: C) 1
Explanation: At 100% modulation, sideband power is maximized without distortion.

103. The modulation index of FM is defined as:
A) $Δf/f_c$
B) $Δf/f_m$
C) $f_m/f_c$
D) $P_c/P_m$
Answer: B
Explanation: FM index $β = Δf/f_m$, ratio of frequency deviation to modulating frequency.

104. Narrowband FM (NBFM) is valid when modulation index is:
A) <1
B) =1
C) >1
D) >>1
Answer: A
Explanation: NBFM approximates AM+PM when β << 1.

105. An advantage of SSB over AM is:
A) Higher distortion
B) Wider bandwidth
C) Lower power and bandwidth
D) More noise
Answer: C
Explanation: SSB transmits only one sideband, saving bandwidth and power.

106. A synchronous detector requires:
A) Envelope detector
B) Local carrier oscillator
C) Mixer with diode
D) Comparator
Answer: B
Explanation: Needs carrier reference to demodulate suppressed carrier signals.

107. In FM, noise at high frequencies is reduced using:
A) Pre-emphasis & De-emphasis
B) Companding
C) SSB
D) PLL
Answer: A
Explanation: Pre-emphasis boosts highs at transmitter; de-emphasis attenuates highs at receiver, restoring spectrum while reducing noise.

108. Phase deviation in PM depends on:
A) Amplitude of modulating signal
B) Frequency of modulating signal
C) Carrier amplitude
D) Noise
Answer: A
Explanation: PM = deviation in carrier phase proportional to modulating signal amplitude.

109. In AM, efficiency at modulation index 1 =
A) 16.67%
B) 25%
C) 33.3%
D) 50%
Answer: C
Explanation: Efficiency = $m^2/(2+m^2)$. For m=1, η=1/3=33.3%.

110. A balanced modulator suppresses:
A) Sidebands
B) Carrier
C) Noise
D) Both sidebands
Answer: B
Explanation: Produces DSB-SC by suppressing carrier, leaving both sidebands.

L. Digital Communication (111–120)

111. In PCM, increasing quantization levels improves:
A) Bandwidth
B) Quantization SNR
C) Noise figure
D) Interference
Answer: B
Explanation: More levels = finer resolution = higher SQNR.

112. If 8-bit PCM uses fs = 8 kHz, bit rate =
A) 8 kbps
B) 64 kbps
C) 16 kbps
D) 32 kbps
Answer: B
Explanation: Bit rate = n × fs = 8×8k = 64 kbps.

113. DPCM differs from PCM because it transmits:
A) Actual samples
B) Difference between samples
C) Analog pulses
D) Noise only
Answer: B
Explanation: DPCM predicts current sample from previous and encodes difference.

114. In Delta Modulation, step size too large causes:
A) Slope overload
B) Granular noise
C) Distortion removal
D) Zero noise
Answer: B
Explanation: Large step size causes small variations to be overestimated → granular noise.

115. BER of coherent BPSK in AWGN is:
A) Q(√Eb/N0)
B) Q(√2Eb/N0)
C) 1/Eb
D) N0/Eb
Answer: B
Explanation: BER formula = Q(√(2Eb/N0)).

116. QPSK transmits how many bits per symbol?
A) 1
B) 2
C) 3
D) 4
Answer: B
Explanation: 4 constellation points = log2(4)=2 bits per symbol.

117. 16-QAM transmits how many bits per symbol?
A) 2
B) 3
C) 4
D) 8
Answer: C
Explanation: log2(16) = 4 bits per symbol.

118. DPSK avoids:
A) Bandwidth expansion
B) Need for coherent carrier reference
C) Frequency errors
D) ISI
Answer: B
Explanation: Differential encoding eliminates need for carrier phase recovery.

119. Gray coding in QAM is used to:
A) Increase power
B) Reduce bandwidth
C) Reduce bit errors per symbol error
D) Eliminate ISI
Answer: C
Explanation: Adjacent symbols differ by one bit, reducing BER.

120. Spectral efficiency (bps/Hz) improves by using:
A) High-order modulation
B) Low-order modulation
C) AM
D) NRZ coding
Answer: A
Explanation: Higher-order QAM/PSK transmits more bits per Hz.

M. Noise & Information Theory (121–130)

121. The unit of information is:
A) Joule
B) Watt
C) Bit
D) dB
Answer: C
Explanation: Information is measured in bits.

122. Entropy measures:
A) Redundancy
B) Average information content
C) Power
D) Noise
Answer: B
Explanation: Higher entropy → more uncertainty → more information.

123. Channel capacity increases with:
A) Decreasing bandwidth
B) Increasing SNR
C) Decreasing SNR
D) Higher noise
Answer: B
Explanation: Capacity ∝ log2(1+SNR).

124. Hartley law relates information with:
A) Bandwidth and time
B) Noise and power
C) Amplitude and phase
D) Delay and spread
Answer: A
Explanation: H = 2B·T log2 M.

125. White noise autocorrelation is:
A) Delta function
B) Gaussian
C) Exponential
D) Cosine
Answer: A
Explanation: White noise is uncorrelated except at zero lag.

126. SNR (dB) =
A) 10 log(Ps/Pn)
B) 20 log(Ps/Pn)
C) Ps/Pn
D) log(Ps)
Answer: A
Explanation: Ratio in power → 10 log formula.

127. If Ps=10 mW, Pn=1 mW, SNR(dB)=
A) 0
B) 10
C) 20
D) 100
Answer: B
Explanation: 10 log(10/1)=10 dB.

128. Noise figure F =
A) SNRin/SNRout
B) SNRout/SNRin
C) Ps/Pn
D) kTB
Answer: A
Explanation: NF quantifies degradation of SNR.

129. A channel with infinite bandwidth and finite power has capacity:
A) Infinite
B) Zero
C) Finite
D) Undefined
Answer: C
Explanation: As B→∞, SNR→0, capacity finite.

130. Shannon limit defines:
A) Minimum power for reliable transmission
B) Maximum entropy
C) Maximum bandwidth
D) ISI limit
Answer: A
Explanation: Sets theoretical minimum Eb/N0 for error-free communication.

N. Multiplexing & Switching (131–140)

131. FDM separates users by:
A) Time
B) Frequency
C) Power
D) Phase
Answer: B

132. TDM separates users by:
A) Frequency
B) Time slots
C) Code
D) Power
Answer: B

133. CDMA separates users by:
A) Time slots
B) Frequency slots
C) Codes
D) Guard bands
Answer: C

134. OFDM combats:
A) ISI in multipath channels
B) Noise figure
C) Path loss
D) Shadowing only
Answer: A

135. Guard interval in OFDM prevents:
A) Frequency offset
B) ISI
C) Multiplexing
D) Power loss
Answer: B

136. Statistical TDM improves efficiency by:
A) Fixed slots
B) Dynamic slot allocation
C) Guard bands
D) Pre-emphasis
Answer: B

137. Packet switching advantage:
A) Constant delay
B) Bandwidth sharing
C) Low noise
D) Carrier efficiency
Answer: B

138. Circuit switching is used in:
A) Internet
B) PSTN (old telephone)
C) LTE
D) WiFi
Answer: B

139. Burst error correction is improved by:
A) Interleaving
B) Equalization
C) FDM
D) Sampling
Answer: A

140. GSM uses which multiplexing?
A) TDMA + FDMA
B) CDMA
C) OFDM
D) SDMA
Answer: A

O. Propagation & Antennas (141–150)

141. Free-space path loss increases with:
A) Higher frequency and distance
B) Lower frequency
C) Shorter distance
D) Antenna gain
Answer: A

142. Doppler effect causes:
A) Frequency shift
B) Bandwidth increase
C) Delay
D) Noise
Answer: A

143. Delay spread causes:
A) Fading
B) ISI
C) Bandwidth expansion
D) Better SNR
Answer: B

144. Rayleigh fading occurs:
A) With LOS
B) Without LOS
C) With optical fiber
D) With high power
Answer: B

145. Rician fading includes:
A) Only scattered paths
B) LOS + scattered paths
C) Thermal noise only
D) Shadowing only
Answer: B

146. Slow fading is mainly due to:
A) Shadowing
B) Doppler
C) Noise
D) ISI
Answer: A

147. Fast fading occurs due to:
A) Shadowing
B) Rapid motion + multipath
C) Path loss
D) Guard bands
Answer: B

148. Coherence time is inverse of:
A) Doppler spread
B) Delay spread
C) Bandwidth
D) Noise
Answer: A

149. Coherence bandwidth is inverse of:
A) RMS delay spread
B) Doppler shift
C) SNR
D) Bit rate
Answer: A

150. Diversity combining improves:
A) Bandwidth
B) Reliability against fading
C) Power
D) Delay
Answer: B

P. Modern Standards & Applications (151–160)

151. GSM channel spacing =
A) 25 kHz
B) 100 kHz
C) 200 kHz
D) 1.25 MHz
Answer: C

152. GSM multiple access technique:
A) TDMA/FDMA
B) CDMA
C) OFDMA
D) SDMA
Answer: A

153. LTE downlink uses:
A) OFDMA
B) SC-FDMA
C) TDMA
D) FDMA
Answer: A

154. LTE uplink uses:
A) OFDMA
B) SC-FDMA
C) CDMA
D) FDMA
Answer: B

155. WiFi (802.11) uses:
A) OFDM
B) CDMA
C) AM
D) FM
Answer: A

156. Bluetooth uses:
A) FHSS (Frequency hopping spread spectrum)
B) DSSS
C) OFDM
D) QAM
Answer: A

157. 5G supports subcarrier spacing:
A) 15 kHz only
B) 15×2^μ kHz
C) 7.5 kHz
D) 200 kHz
Answer: B

158. Massive MIMO in 5G improves:
A) Spectral efficiency and capacity
B) Path loss
C) Noise figure
D) ISI only
Answer: A

159. Satellite communication major impairment:
A) Rain attenuation
B) Doppler spread
C) Shadowing
D) ISI
Answer: A

160. Optical fiber communication is immune to:
A) Electromagnetic interference
B) Attenuation
C) Dispersion
D) Nonlinearity
Answer: A

Q. Basics of Networking (161–170)

161. OSI model has how many layers?
A) 5 B) 6 C) 7 D) 8
Answer: C) 7
Explanation: The OSI model defines 7 layers: Physical, Data Link, Network, Transport, Session, Presentation, and Application.

162. Which layer is responsible for routing?
A) Data Link B) Network C) Transport D) Application
Answer: B) Network
Explanation: The network layer handles logical addressing and routing of packets.

163. TCP and UDP belong to which OSI layer?
A) Application B) Network C) Transport D) Data Link
Answer: C) Transport
Explanation: TCP/UDP provide end-to-end communication at the transport layer.

164. Which device operates at the Data Link layer?
A) Hub B) Switch C) Router D) Gateway
Answer: B) Switch
Explanation: Switches use MAC addresses to forward frames at the data link layer.

165. The main function of ARP is to:
A) Map IP to MAC address
B) Map MAC to IP address
C) Encrypt data packets
D) Assign IP addresses
Answer: A
Explanation: ARP resolves an IP address into its physical MAC address.

166. IP address 192.168.1.1 belongs to:
A) Class A B) Class B C) Class C D) Class D
Answer: C) Class C
Explanation: Class C range = 192–223; often used in private LANs.

167. Default subnet mask of Class B is:
A) 255.0.0.0
B) 255.255.0.0
C) 255.255.255.0
D) 255.255.255.255
Answer: B
Explanation: Class B supports 16-bit network and 16-bit host fields.

168. Which protocol is connection-oriented?
A) UDP B) TCP C) IP D) ICMP
Answer: B) TCP
Explanation: TCP establishes a connection before data transfer.

169. Which protocol is used for email retrieval?
A) SMTP B) POP3 C) SNMP D) FTP
Answer: B) POP3
Explanation: SMTP sends emails, POP3 retrieves emails.

170. Which addressing is used in the Data Link layer?
A) IP address
B) Port number
C) MAC address
D) Domain name
Answer: C) MAC address
Explanation: Layer 2 uses hardware addresses.

R. Protocols & Transmission (171–180)

171. Which protocol is used for secure web browsing?
A) FTP B) HTTP C) HTTPS D) SNMP
Answer: C) HTTPS
Explanation: HTTPS uses SSL/TLS for encryption.

172. DNS works on which layer?
A) Transport B) Application C) Network D) Data Link
Answer: B
Explanation: DNS translates domain names to IP at the application layer.

173. In Ethernet, CSMA/CD is used to:
A) Avoid collision
B) Detect and handle collision
C) Encrypt data
D) Route packets
Answer: B
Explanation: Carrier Sense Multiple Access with Collision Detection manages Ethernet access.

174. Maximum length of IPv4 address is:
A) 16 bits B) 32 bits C) 64 bits D) 128 bits
Answer: B
Explanation: IPv4 uses 32-bit addressing.

175. Maximum length of IPv6 address is:
A) 64 bits B) 128 bits C) 256 bits D) 32 bits
Answer: B
Explanation: IPv6 uses 128-bit addresses.

176. Which protocol is used for file transfer?
A) SNMP B) FTP C) SMTP D) ARP
Answer: B
Explanation: FTP is a standard file transfer protocol.

177. Which protocol is used to dynamically assign IP addresses?
A) ARP B) RARP C) DHCP D) ICMP
Answer: C
Explanation: DHCP automates IP configuration.

178. ICMP is used for:
A) Routing
B) Error and diagnostic messages
C) File transfer
D) Encryption
Answer: B
Explanation: ICMP provides ping, traceroute, and error reporting.

179. Bandwidth of a standard Ethernet (Fast Ethernet) is:
A) 10 Mbps B) 100 Mbps C) 1 Gbps D) 10 Gbps
Answer: B
Explanation: Fast Ethernet provides 100 Mbps.

180. Bandwidth of a standard Gigabit Ethernet is:
A) 100 Mbps B) 1 Gbps C) 10 Gbps D) 40 Gbps
Answer: B
Explanation: Gigabit Ethernet runs at 1 Gbps.

S. Advanced Topics (181–190)

181. In star topology, central device is usually a:
A) Switch/Hub
B) Router
C) Gateway
D) Repeater
Answer: A
Explanation: Star topology connects devices to a central hub/switch.

182. Which topology provides highest reliability?
A) Bus B) Star C) Ring D) Mesh
Answer: D) Mesh
Explanation: Each node connected to every other node.

183. The PDU (Protocol Data Unit) of Transport layer is:
A) Frame B) Packet C) Segment D) Bit
Answer: C
Explanation: Transport layer data unit = Segment.

184. A router works at which OSI layer?
A) Data Link B) Network C) Transport D) Application
Answer: B
Explanation: Router operates using IP addresses.

185. SMTP operates over port:
A) 21 B) 25 C) 110 D) 443
Answer: B
Explanation: SMTP default port is 25.

186. POP3 operates over port:
A) 80 B) 21 C) 110 D) 143
Answer: C
Explanation: POP3 default port is 110.

187. IMAP operates over port:
A) 25 B) 110 C) 143 D) 53
Answer: C
Explanation: IMAP uses port 143.

188. DNS operates over port:
A) 25 B) 110 C) 143 D) 53
Answer: D
Explanation: DNS standard port = 53.

189. Which protocol is used for remote login?
A) FTP B) SSH C) SNMP D) DHCP
Answer: B
Explanation: SSH provides secure remote login.

190. In networking, throughput is defined as:
A) Maximum bandwidth of channel
B) Actual data rate achieved
C) Signal-to-noise ratio
D) Packet error rate
Answer: B
Explanation: Throughput is the actual achieved transfer rate, less than theoretical bandwidth.

T. Security & Miscellaneous (191–190)

191. Which protocol is used for network management?
A) SMTP B) SNMP C) FTP D) POP3
Answer: B
Explanation: Simple Network Management Protocol monitors and manages devices.

192. SSL/TLS mainly provides:
A) Encryption and authentication
B) Multiplexing
C) Routing
D) Error correction
Answer: A
Explanation: SSL/TLS secures communication by encrypting traffic and verifying identity.

193. Firewall works at which OSI layer?
A) Application only
B) Transport only
C) Network and above
D) Physical only
Answer: C
Explanation: Firewalls can filter traffic at network, transport, and application layers.

194. Ping command uses:
A) TCP
B) UDP
C) ICMP
D) FTP
Answer: C
Explanation: Ping sends ICMP Echo Request packets.

195. Traceroute command uses:
A) ICMP & TTL
B) FTP
C) SMTP
D) DHCP
Answer: A
Explanation: Traceroute uses ICMP and TTL to find route path.

196. VPN (Virtual Private Network) is used for:
A) High speed
B) Secure remote access
C) Error control
D) Frequency division
Answer: B
Explanation: VPN provides encrypted tunnels for secure communication over public internet.

197. NAT (Network Address Translation) is used for:
A) Mapping private IP to public IP
B) Error correction
C) Routing only
D) Encryption
Answer: A
Explanation: NAT enables multiple private hosts to share a public IP.

198. Which cable has maximum bandwidth?
A) Coaxial
B) Twisted pair
C) Optical fiber
D) Wireless
Answer: C
Explanation: Fiber optic cable offers very high bandwidth.

199. Maximum transmission unit (MTU) for Ethernet =
A) 512 bytes
B) 1024 bytes
C) 1500 bytes
D) 2048 bytes
Answer: C
Explanation: Ethernet MTU is 1500 bytes.

200. Which is NOT a wireless standard?
A) 802.11
B) 802.15
C) 802.16
D) 802.3
Answer: D
Explanation: 802.3 is Ethernet (wired LAN).