Digital Communication — Part 1 MCQs with Explanations
Practice high-yield Electronics & Communications Engineering MCQs on Digital Communication — Part 1 with worked academic solutions for FPSC, PPSC, PTA, PAEC, NTS.
Questions
40
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Exam-Level
Target Exams
FPSC, PPSC, PTA...
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48 mins
Exam-Level FPSC PPSC PTA PAEC NTS
Verified for Commission Syllabus • Oct 14, 2026
40
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Question 1 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
A 4 GHz carrier is DSB-SC modulated by a low pass message signal with maximum frequency of 2 MHz. The resultant signal is to be ideally sampled. The minimum frequency of the sampling impulse train should be
Correct Answer: Option B — 8 MHz
Detailed Academic Explanation:
8 MHz is the correct answer to “A 4 GHz carrier is DSB-SC modulated by a low pass message signal with maximum frequency of 2 MHz. The resultant signal is to be ideally sampled. The minimum frequency of the sampling impulse train should be”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 2 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
A video transmission system transmits 625 picture frames per second. Each frame consists of a 400 × 400 pixel grid with 64 intensity levels per pixel. The data rate of the system is
Correct Answer: Option D — 6.4 Gbps
Detailed Academic Explanation:
6.4 Gbps is the correct answer to “A video transmission system transmits 625 picture frames per second. Each frame consists of a 400 × 400 pixel grid with 64 intensity levels per pixel. The data rate of the system is”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 3 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
The raised cosine pulse p(t) is used for zero ISI in digital communications. The expression for p(t) with unity roll-off factor is given by $$p\left( t \right) = \frac{{\sin 4\pi \omega t}}{{4\pi \omega t\left( {1 - 16{\omega ^2}{t^2}} \right)}}.$$ The value of p(t) at $$t = \frac{1}{{4\omega }}$$ is
Correct Answer: Option C — 0.5
Detailed Academic Explanation:
0.5 is the correct answer to “The raised cosine pulse p(t) is used for zero ISI in digital communications. The expression for p(t) with unity roll-off factor is given by $$p\left( t \right) = \frac{{\sin 4\pi \omega t}}{{4\pi \omega t\left( {1 - 16{\omega ^2}{t^2}} \right)}}.$$ The value of p(t) at $$t = \frac{1}{{4\omega }}$$ is”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 4 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
A sinusoidal signal of 2 kHz frequency is applied to a delta modulator. The sampling rate and step-size Δ of the delta modulator are 20,000 samples per second and 0.1 V, respectively. To prevent slope overload, the maximum amplitude of the sinusoidal signal (in Volts) is
Correct Answer: Option A — $$\frac{1}{{2\pi }}$$
Detailed Academic Explanation:
$$\frac{1}{{2\pi }}$$ is the correct answer to “A sinusoidal signal of 2 kHz frequency is applied to a delta modulator. The sampling rate and step-size Δ of the delta modulator are 20,000 samples per second and 0.1 V, respectively. To prevent slope overload, the maximum amplitude of the sinusoidal signal (in Volts) is”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 5 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
Coherent orthogonal binary FSK modulation is used to transmit two equiprobable symbol waveforms s 1 (t) = αcos2πf 1 t & s 2 (t) = αcos2πf 2 t 1 where α = 4 mV. Assume an AWGN channel with two-sided noise power spectral density $$\frac{{{N_0}}}{2} = 0.5 \times {10^{ - 12}}W/Hz.$$ Using an optimal receiver and the relation $$Q\left( v \right) = \frac{1}{{\sqrt {2\pi } }}\int\limits_v^\infty {{\theta ^{\frac{{{u^2}}}{2}}}du,} $$ the bit error probability for a data rate of 500 kbps is
Correct Answer: Option C — Q(4)
Detailed Academic Explanation:
Q(4) is the correct answer to “Coherent orthogonal binary FSK modulation is used to transmit two equiprobable symbol waveforms s 1 (t) = αcos2πf 1 t & s 2 (t) = αcos2πf 2 t 1 where α = 4 mV. Assume an AWGN channel with two-sided noise power spectral density $$\frac{{{N_0}}}{2} = 0.5 \times {10^{ - 12}}W/Hz.$$ Using an optimal receiver and the relation $$Q\left( v \right) = \frac{1}{{\sqrt {2\pi } }}\int\limits_v^\infty {{\theta ^{\frac{{{u^2}}}{2}}}du,} $$ the bit error probability for a data rate of 500 kbps is”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 6 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
During transmission over a certain binary communication channel, bit errors occur independently with probability p. The probability of AT MOST one bit in error in a block of n bits is given by
np (1 - p)n - 1 + (1 - p)n is the correct answer to “During transmission over a certain binary communication channel, bit errors occur independently with probability p. The probability of AT MOST one bit in error in a block of n bits is given by”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 7 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
The message bit sequence to a DPSK modulator is 1, 1, 0, 0, 1, 1. The carrier phase during the reception of the first two message bits is π, π. The carrier phase for the remaining four message bits is
Correct Answer: Option C — 0, π, π, π
Detailed Academic Explanation:
0, π, π, π is the correct answer to “The message bit sequence to a DPSK modulator is 1, 1, 0, 0, 1, 1. The carrier phase during the reception of the first two message bits is π, π. The carrier phase for the remaining four message bits is”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 8 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
A deterministic signal has the power spectrum given in figure. The minimum sampling rate needed to completely represent signal is
Correct Answer: Option C — 3 kHz
Detailed Academic Explanation:
3 kHz is the correct answer to “A deterministic signal has the power spectrum given in figure. The minimum sampling rate needed to completely represent signal is”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 9 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
During transmission over a communication channel, bit errors occur independently with probability p. If a block of n bits is transmitted, the probability of at most one bit error is equal to
(1 - p)n + np(1 - p)n - 1 is the correct answer to “During transmission over a communication channel, bit errors occur independently with probability p. If a block of n bits is transmitted, the probability of at most one bit error is equal to”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 10 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
A signal having uniformly distributed amplitude in the interval (-V to + V), is to be encoded using PCM with uniform quantization. The signal to quantizing noise ratio is determined by the
Correct Answer: Option C — Number of quantizing levels
Detailed Academic Explanation:
Number of quantizing levels is the correct answer to “A signal having uniformly distributed amplitude in the interval (-V to + V), is to be encoded using PCM with uniform quantization. The signal to quantizing noise ratio is determined by the”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 11 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
Companding in PCM systems lead to improved signal to quantization noise ratio. This improvement is for
Correct Answer: Option C — Lower amplitudes only
Detailed Academic Explanation:
Lower amplitudes only is the correct answer to “Companding in PCM systems lead to improved signal to quantization noise ratio. This improvement is for”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 12 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
Consider a binary digital communication system with equally likely 0's and 1's. When binary 0 is transmitted the detector input can lie between the levels -0.25V and +0.25V with equal probability: when binary 1 is transmitted, the voltage at the detector can have any value between 0 and 1 V with equal probability. If the detector has a threshold of 0.2 V (i.e., if the received signal is greater than 0.2 V, the bit is taken as 1), the average bit error probability is
Correct Answer: Option A — 0.15
Detailed Academic Explanation:
0.15 is the correct answer to “Consider a binary digital communication system with equally likely 0's and 1's. When binary 0 is transmitted the detector input can lie between the levels -0.25V and +0.25V with equal probability: when binary 1 is transmitted, the voltage at the detector can have any value between 0 and 1 V with equal probability. If the detector has a threshold of 0.2 V (i.e., if the received signal is greater than 0.2 V, the bit is taken as 1), the average bit error probability is”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 13 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
A source produces binary data at the rate of 10 kbps. The binary symbols are represented as shown in the figure. The source output is transmitted using two modulation schemes, namely Binary PSK (BPSK) and Quadrature PSK (QPSK). Let and B2 be the bandwidth requirements of BPSK and QPSK respectively. Assuming that the bandwidth of the above rectangular pulses is 10 kHz, B 1 and B 2 are
B1 = 20 kHz, B2 = 10 kHz is the correct answer to “A source produces binary data at the rate of 10 kbps. The binary symbols are represented as shown in the figure. The source output is transmitted using two modulation schemes, namely Binary PSK (BPSK) and Quadrature PSK (QPSK). Let and B2 be the bandwidth requirements of BPSK and QPSK respectively. Assuming that the bandwidth of the above rectangular pulses is 10 kHz, B 1 and B 2 are”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 14 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
The modulation scheme commonly used for transmission from GSM mobile terminals is
Correct Answer: Option D — Gaussian Minimum Shift Keying (GMSK)
Detailed Academic Explanation:
Gaussian Minimum Shift Keying (GMSK) is the correct answer to “The modulation scheme commonly used for transmission from GSM mobile terminals is”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 15 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
Consider a Binary Symmetric Channel (BSC) with probability of error being p. To transmit a bit, say 1, we transmit a sequence of three 1 s. The receiver will interpret the received sequence to represent 1 if at least two bits are 1. The probability that the transmitted bit will be received in error is
Correct Answer: Option A — p3 + 3p2 (1- p)
Detailed Academic Explanation:
p3 + 3p2 (1- p) is the correct answer to “Consider a Binary Symmetric Channel (BSC) with probability of error being p. To transmit a bit, say 1, we transmit a sequence of three 1 s. The receiver will interpret the received sequence to represent 1 if at least two bits are 1. The probability that the transmitted bit will be received in error is”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 16 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
The signal to quantization noise ratio in an n-bit PCM system
Correct Answer: Option C — Increasing with increasing value of 'n'
Detailed Academic Explanation:
Increasing with increasing value of 'n' is the correct answer to “The signal to quantization noise ratio in an n-bit PCM system”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 17 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
Which one of the following statements about differential pulse code modulation (DPCM) is true?
Correct Answer: Option D — The difference of message signal sample with its prediction is quantized
Detailed Academic Explanation:
The difference of message signal sample with its prediction is quantized is the correct answer to “Which one of the following statements about differential pulse code modulation (DPCM) is true”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 18 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
In delta modulation, the slope overload distortion can be reduced by
Correct Answer: Option D — Increasing the step size
Detailed Academic Explanation:
Increasing the step size is the correct answer to “In delta modulation, the slope overload distortion can be reduced by”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 19 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
Let x(t) = 2cos(800πt) + cos(1400πt).x(t) is sampled with the rectangular pulse train shown in the figure. The only spectral components (in kHz) present in the sampled signal in the frequency range 2.5 kHz to 3.5 kHz are
Correct Answer: Option D — 2.7, 3.3
Detailed Academic Explanation:
2.7, 3.3 is the correct answer to “Let x(t) = 2cos(800πt) + cos(1400πt).x(t) is sampled with the rectangular pulse train shown in the figure. The only spectral components (in kHz) present in the sampled signal in the frequency range 2.5 kHz to 3.5 kHz are”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 20 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
If E b , the energy per bit of a binary digital signal, is 10 -5 Watt-sec and the one-sided power spectral density of the white noise, N 0 = 10 -6 W/Hz, then the output SNR of the matched filter is
Correct Answer: Option B — 10 dB
Detailed Academic Explanation:
10 dB is the correct answer to “If E b , the energy per bit of a binary digital signal, is 10 -5 Watt-sec and the one-sided power spectral density of the white noise, N 0 = 10 -6 W/Hz, then the output SNR of the matched filter is”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 21 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
In a digital communication system employing Frequency Shift Keying (FSK), the 0 and 1 bit are represented by sine waves of 10 kHz and 25 kHz respectively. These waveforms will be orthogonal for a bit interval of
Correct Answer: Option B — 200 μsec
Detailed Academic Explanation:
200 μsec is the correct answer to “In a digital communication system employing Frequency Shift Keying (FSK), the 0 and 1 bit are represented by sine waves of 10 kHz and 25 kHz respectively. These waveforms will be orthogonal for a bit interval of”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 22 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
If the number of bits per sample in a PCM system is increased from a n to n + 1, the improvement in signal to quantization nose ratio will be
Correct Answer: Option C — 2n dB
Detailed Academic Explanation:
2n dB is the correct answer to “If the number of bits per sample in a PCM system is increased from a n to n + 1, the improvement in signal to quantization nose ratio will be”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 23 of 40
★ FPSC
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★ FPSC
★ PPSC
Flat top sampling of low pass signals
Correct Answer: Option A — Gives rise to aperture effect
Detailed Academic Explanation:
Gives rise to aperture effect is the correct answer to “Flat top sampling of low pass signals”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 24 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
The input to a matched filter is given by \[s\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {10\sin \left( {2\pi \times {{10}^6}} \right)}&{0 < t < {{10}^{ - 4}}\sec } \\ 0&{{\text{Otherwise}}} \end{array}} \right.\] The peak amplitude of the filter output is
Correct Answer: Option A — 10 Volts
Detailed Academic Explanation:
10 Volts is the correct answer to “The input to a matched filter is given by \[s\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {10\sin \left( {2\pi \times {{10}^6}} \right)}&{0 < t < {{10}^{ - 4}}\sec } \\ 0&{{\text{Otherwise}}} \end{array}} \right.\] The peak amplitude of the filter output is”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 25 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
At a given probability of error, binary coherent FSK is inferior to binary coherent PSK by
Correct Answer: Option B — 3 dB
Detailed Academic Explanation:
3 dB is the correct answer to “At a given probability of error, binary coherent FSK is inferior to binary coherent PSK by”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 26 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
For a bit-rate of 8 Kbps, the best possible values of the transmitted frequencies in a coherent binary FSK system are
Correct Answer: Option D — 32 kHz and 40 kHz
Detailed Academic Explanation:
32 kHz and 40 kHz is the correct answer to “For a bit-rate of 8 Kbps, the best possible values of the transmitted frequencies in a coherent binary FSK system are”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 27 of 40
★ FPSC
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★ FPSC
★ PPSC
In the following figure the minimum value of the constant "C", which is to be added to y 1 (t) such that y 1 (t) and y 2 (t) are different, is
Correct Answer: Option B — $$\frac{\Delta }{2}$$
Detailed Academic Explanation:
$$\frac{\Delta }{2}$$ is the correct answer to “In the following figure the minimum value of the constant "C", which is to be added to y 1 (t) such that y 1 (t) and y 2 (t) are different, is”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 28 of 40
★ FPSC
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★ FPSC
★ PPSC
A signal x(t) = 100cos(24π × 10 3 )t is ideally sampled with a sampling period of 50 psec and then passed through an ideal lowpass filter with cutoff frequency of 15 kHz. Which of the following frequencies is/are present at the filter output?
Correct Answer: Option D — 12 kHz and 8 kHz
Detailed Academic Explanation:
12 kHz and 8 kHz is the correct answer to “A signal x(t) = 100cos(24π × 10 3 )t is ideally sampled with a sampling period of 50 psec and then passed through an ideal lowpass filter with cutoff frequency of 15 kHz. Which of the following frequencies is/are present at the filter output”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 29 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
In the output of a DM speech encoder, the consecutive pulses are of opposite polarity during time interval t 1 ≤ t ≤ t 2 . This indicates that during this interval
Correct Answer: Option A — The input to the modulator is essentially constant
Detailed Academic Explanation:
The input to the modulator is essentially constant is the correct answer to “In the output of a DM speech encoder, the consecutive pulses are of opposite polarity during time interval t 1 ≤ t ≤ t 2 . This indicates that during this interval”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 30 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
A BPSK scheme operating over an AWGN channel with noise power spectral density of $$\frac{{{N_0}}}{2},$$ uses equiprobable signals $${s_1}\left( t \right) = \sqrt {\frac{{2E}}{T}} \sin \left( {{\omega _c}t} \right)$$ and $${s_2}\left( t \right) = \sqrt {\frac{{2E}}{T}} \sin \left( {{\omega _c}t} \right)$$ over the symbol internal (0, T). If the local oscillator in a coherent receiver is ahead in phase by 45° with respect to the received signal, the probability of error in the resulting system is
Correct Answer: Option B — $$Q\left( {\sqrt {\frac{E}{{{N_0}}}} } \right)$$
Detailed Academic Explanation:
$$Q\left( {\sqrt {\frac{E}{{{N_0}}}} } \right)$$ is the correct answer to “A BPSK scheme operating over an AWGN channel with noise power spectral density of $$\frac{{{N_0}}}{2},$$ uses equiprobable signals $${s_1}\left( t \right) = \sqrt {\frac{{2E}}{T}} \sin \left( {{\omega _c}t} \right)$$ and $${s_2}\left( t \right) = \sqrt {\frac{{2E}}{T}} \sin \left( {{\omega _c}t} \right)$$ over the symbol internal (0, T). If the local oscillator in a coherent receiver is ahead in phase by 45° with respect to the received signal, the probability of error in the resulting system is”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 31 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
Consider a sample signal \[y\left( t \right) = 5 \times {10^{ - 6}}x\left( t \right)\sum\nolimits_{ - \infty }^{ + \infty } {\delta \left( {t - n{T_s}} \right)} \] where x(t) = 10cos(8π × 10 3 )t and T s =100 μsec. When y(t) is passed through an ideal lowpass filter with a cutoff frequency of 5 kHz, the output of the filter is
5 × 10-1 cos(8π × 103)t is the correct answer to “Consider a sample signal \[y\left( t \right) = 5 \times {10^{ - 6}}x\left( t \right)\sum\nolimits_{ - \infty }^{ + \infty } {\delta \left( {t - n{T_s}} \right)} \] where x(t) = 10cos(8π × 10 3 )t and T s =100 μsec. When y(t) is passed through an ideal lowpass filter with a cutoff frequency of 5 kHz, the output of the filter is”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 32 of 40
★ FPSC
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★ PPSC
The Nyquist sampling frequency (in Hz) of a signal given by 6 × 10 4 sinC 3 (400t) ∗ 10 6 sinC 3 (100t) is
Correct Answer: Option D — 1000
Detailed Academic Explanation:
1000 is the correct answer to “The Nyquist sampling frequency (in Hz) of a signal given by 6 × 10 4 sinC 3 (400t) ∗ 10 6 sinC 3 (100t) is”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 33 of 40
★ FPSC
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★ FPSC
★ PPSC
In a PCM system, if the code word length is increased from 6 to 8 bits, the signal to quantization noise ratio improves by the factor
Correct Answer: Option B — 12
Detailed Academic Explanation:
12 is the correct answer to “In a PCM system, if the code word length is increased from 6 to 8 bits, the signal to quantization noise ratio improves by the factor”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 34 of 40
★ FPSC
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★ PPSC
Three analog signals, having bandwidths 1200 Hz, 600 Hz and 600 Hz, are sampled at their respective Nyquist rates, encoded with 12-bit words, and time division multiplexed. The bit rate for the multiplexed signal is
Correct Answer: Option C — 57.6 kpbs
Detailed Academic Explanation:
57.6 kpbs is the correct answer to “Three analog signals, having bandwidths 1200 Hz, 600 Hz and 600 Hz, are sampled at their respective Nyquist rates, encoded with 12-bit words, and time division multiplexed. The bit rate for the multiplexed signal is”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 35 of 40
★ FPSC
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★ PPSC
The peak-to-peak input to an 8-bit PCM coder is 2 Volts. The signal power-to-quantization noise power ratio (in dB) for an input of 0.5cos(ω m t) is
Correct Answer: Option C — 95.6
Detailed Academic Explanation:
95.6 is the correct answer to “The peak-to-peak input to an 8-bit PCM coder is 2 Volts. The signal power-to-quantization noise power ratio (in dB) for an input of 0.5cos(ω m t) is”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 36 of 40
★ FPSC
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The Nyquist sampling rate for the signal $$s\left( t \right) = \frac{{\sin \left( {500\pi t} \right)}}{{\pi t}} \times \frac{{\sin \left( {700\pi t} \right)}}{{\pi t}}$$ is given by
Correct Answer: Option C — 1200 Hz
Detailed Academic Explanation:
1200 Hz is the correct answer to “The Nyquist sampling rate for the signal $$s\left( t \right) = \frac{{\sin \left( {500\pi t} \right)}}{{\pi t}} \times \frac{{\sin \left( {700\pi t} \right)}}{{\pi t}}$$ is given by”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 37 of 40
★ FPSC
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★ PPSC
The bit rate of digital communication system is R kbits/s. The modulation used is 32-QAM. The minimum bandwidth required for |S| free transmission is
Correct Answer: Option B — $$\frac{R}{{10}}kHz$$
Detailed Academic Explanation:
$$\frac{R}{{10}}kHz$$ is the correct answer to “The bit rate of digital communication system is R kbits/s. The modulation used is 32-QAM. The minimum bandwidth required for |S| free transmission is”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 38 of 40
★ FPSC
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★ PPSC
A binary symmetric channel (BSC) has a transition probability of $$\frac{1}{8}.$$ If the binary transmit symbol X is such that $$P\left( {X = 0} \right) = \frac{9}{{10}},$$ then the probability of error for an optimum receiver will be
Correct Answer: Option A — $$\frac{7}{{80}}$$
Detailed Academic Explanation:
$$\frac{7}{{80}}$$ is the correct answer to “A binary symmetric channel (BSC) has a transition probability of $$\frac{1}{8}.$$ If the binary transmit symbol X is such that $$P\left( {X = 0} \right) = \frac{9}{{10}},$$ then the probability of error for an optimum receiver will be”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 39 of 40
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★ PPSC
Choose the correct one from among the alternatives a, b, c, dafter matching an item from Group 1 with the most appropriate item in Group 2 . Group 1 Group 2 1. FM P. Slope overload 2. DM Q. H-law 3. PSK R. Envelope detector 4. PCM S. Capture effect T. Hilbert transform U. Matched filter
Correct Answer: Option C — 1-S, 2-P, 3-U, 4-Q
Detailed Academic Explanation:
1-S, 2-P, 3-U, 4-Q is the correct answer to “Choose the correct one from among the alternatives a, b, c, dafter matching an item from Group 1 with the most appropriate item in Group 2 . Group 1 Group 2 1. FM P. Slope overload 2. DM Q. H-law 3. PSK R. Envelope detector 4. PCM S. Capture effect T. Hilbert transform U. Matched filter”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
Question 40 of 40
★ FPSC
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A source emits bit 0 with probability $$\frac{1}{3}$$ and bit 1 with probability $$\frac{2}{3}.$$ The emitted bits are communicated to the receiver. The receiver decides for either 0 or 1 based on the received value R. It is given that the conditional density functions of R as \[\begin{gathered} {f_{R|0}}\left( r \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{4},}&{ - 3 \leqslant r \leqslant 1} \\ {0,}&{{\text{otherwise;}}} \end{array}} \right. \hfill \\ {f_{R|1}}\left( r \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{6},}&{ - 1 \leqslant r \leqslant 5} \\ {0,}&{{\text{otherwise;}}} \end{array}} \right. \hfill \\ \end{gathered} \] The minimum decision error probability is
Correct Answer: Option D — $$\frac{1}{6}$$
Detailed Academic Explanation:
$$\frac{1}{6}$$ is the correct answer to “A source emits bit 0 with probability $$\frac{1}{3}$$ and bit 1 with probability $$\frac{2}{3}.$$ The emitted bits are communicated to the receiver. The receiver decides for either 0 or 1 based on the received value R. It is given that the conditional density functions of R as \[\begin{gathered} {f_{R|0}}\left( r \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{4},}&{ - 3 \leqslant r \leqslant 1} \\ {0,}&{{\text{otherwise;}}} \end{array}} \right. \hfill \\ {f_{R|1}}\left( r \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{6},}&{ - 1 \leqslant r \leqslant 5} \\ {0,}&{{\text{otherwise;}}} \end{array}} \right. \hfill \\ \end{gathered} \] The minimum decision error probability is”. This question is standard in competitive examination syllabi for Digital Communication.
💡Key Conceptual Anchor:Digital Communication
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Analog Electronics — Part 1
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💡Examiner's Advisory for Electronics & Communications Engineering
In competitive commission exams (FPSC, PPSC, KPPSC, SPSC, CSS, PMS), questions are frequently phrased with distractor options designed to catch rote learners. Whenever you answer an MCQ, read the Detailed Academic Explanation and Conceptual Anchor to understand why the remaining options are incorrect.