Twelve-bit PCM samples at 20 ksamples/s produce what raw rate?
A 20 kb/s
B 120 kb/s
C 240 kb/s
D 480 kb/s
Answer: C. 240 kb/s. The rate is 12 bits/sample × 20,000 samples/s.
At 15 kbaud, what information rate results from rate-2/3 coding with 64-QAM?
A 45 kb/s
B 90 kb/s
C 30 kb/s
D 60 kb/s
Answer: D. 60 kb/s. Six bits/symbol × 15 kbaud × 2/3 gives 60 kb/s.
Set B=2.5 kHz and S/N=7; calculate Shannon capacity.
A 5 kb/s
B 10 kb/s
C 2.5 kb/s
D 7.5 kb/s
Answer: D. 7.5 kb/s. C = 2.5 kHz × log₂(1+7) = 7.5 kb/s.
Forty-eight QPSK data tones send one OFDM symbol in 4 μs. Find raw rate.
A 12 Mb/s
B 96 Mb/s
C 24 Mb/s
D 48 Mb/s
Answer: C. 24 Mb/s. Each symbol carries 48 × 2 = 96 bits; 96 bits per 4 μs is 24 Mb/s.
Useful OFDM time is 8 μs with a 2 μs prefix; what is efficiency?
A 90%
B 80%
C 20%
D 75%
Answer: B. 80%. Useful time is 8 μs out of a 10 μs total symbol interval.
Two PCM voice channels each use 8 ksamples/s and 8 bits/sample. Find aggregate rate.
A 64 kb/s
B 256 kb/s
C 128 kb/s
D 96 kb/s
Answer: C. 128 kb/s. Each channel carries 64 kb/s, so two channels carry 128 kb/s.
Using 16 equal intervals across an 8 V full-scale range, what is the quantizer step?
A 0.25 V
B 2 V
C 1 V
D 0.5 V
Answer: D. 0.5 V. An 8 V full-scale range divided into 16 equal intervals gives 0.5 V per interval.
Delta modulation has maximum slope 2000 V/s and step 0.25 V. What minimum sample rate avoids slope overload?
A 8 kHz
B 16 kHz
C 4 kHz
D 2 kHz
Answer: A. 8 kHz. The slope limit is step × sample rate; 2000/0.25 requires 8,000 samples/s.
Probabilities 1/2, 1/4, 1/8, and 1/8 give what source entropy?
A 1 bit
B 1.75 bits
C 1.5 bits
D 2 bits
Answer: B. 1.75 bits. H = Σp log₂(1/p) = 0.5 + 0.5 + 0.75 = 1.75 bits.
What is the entropy of an equiprobable 64-symbol source?
A 5 bits per symbol
B 6 bits per symbol
C 8 bits per symbol
D 4 bits per symbol
Answer: B. 6 bits per symbol. Six bits distinguish 64 equally likely symbols because 2⁶ = 64.
How many parity bits appear in a (7,4) block code?
A 2 bits
B 4 bits
C 1 bit
D 3 bits
Answer: D. 3 bits. A seven-bit codeword contains four data bits and three redundant bits.
What does a nonzero syndrome locate after a single-bit error?
A Error position
B Sample rate
C Source entropy
D Carrier phase
Answer: A. Error position. A syndrome points to the error pattern; for a single-bit error it identifies the bit position.
When signal likelihoods are equal but priors differ, which symbol does MAP favor?
A Higher-prior symbol
B Lower-prior symbol
C Lower-frequency symbol
D Higher-energy symbol
Answer: A. Higher-prior symbol. With equal likelihoods, the posterior probability is larger for the symbol with the larger prior.
Uncoded QPSK has symbol energy 2 mJ. What is energy per bit?
A 4 mJ
B 2 mJ
C 0.5 mJ
D 1 mJ
Answer: D. 1 mJ. QPSK carries two bits per symbol, so each bit has half the symbol energy.
Raised-cosine parameters Rs=80 kbaud and α=0.25 yield what one-sided bandwidth?
A 50 kHz
B 20 kHz
C 40 kHz
D 80 kHz
Answer: A. 50 kHz. B = 1.25 × 80/2 = 50 kHz.
Orthogonal OFDM subcarrier spacing for Tu=4 μs equals what?
A 62.5 kHz
B 125 kHz
C 500 kHz
D 250 kHz
Answer: D. 250 kHz. Orthogonal tone spacing is 1/Tu = 1/(4 μs) = 250 kHz.
Given dmin=6, how many bit errors are guaranteed correctable?
A 1 error
B 2 errors
C 5 errors
D 3 errors
Answer: B. 2 errors. Guaranteed correction is floor((dmin−1)/2), which equals two.
For coherent BPSK in AWGN, which expression gives bit-error probability?
A Q(√(2Eb/N0))
B Q(√(Eb/N0))
C Q(2Eb/N0)
D 1−Q(√(2Eb/N0))
Answer: A. Q(√(2Eb/N0)). The standard coherent BPSK BER is Q(√(2Eb/N0)).
Which code family uses encoder memory across input bits?
A Gray code
B Manchester code
C Convolutional code
D Single parity code
Answer: C. Convolutional code. A convolutional encoder’s output depends on current and previous input bits.
Which decoder finds a best path through a convolutional-code trellis?
A Huffman decoder
B Lempel-Ziv decoder
C Viterbi decoder
D Manchester decoder
Answer: C. Viterbi decoder. The Viterbi algorithm finds the minimum-metric trellis path.
What does Huffman coding minimize for a known source distribution?
A Average code length
B Carrier frequency
C Sample interval
D Peak power
Answer: A. Average code length. Huffman coding constructs a prefix code with minimum average length.
Huffman coding assigns shorter codewords to which symbols?
A Longer messages
B More probable symbols
C Higher-frequency carriers
D Less probable symbols
Answer: B. More probable symbols. Frequent symbols receive shorter codewords to reduce average length.
What does Shannon capacity represent for a noisy channel?
A Guaranteed practical throughput
B Minimum symbol energy
C Reliable-rate upper bound
D Maximum code length
Answer: C. Reliable-rate upper bound. Capacity is the theoretical upper bound for reliable information transmission.
If the information rate is below channel capacity, what does Shannon’s theorem permit?
A Zero bandwidth
B Arbitrarily low error
C Infinite symbol rate
D No coding
Answer: B. Arbitrarily low error. With suitable coding, error probability can be made arbitrarily small below capacity.
What does passband QAM vary to represent symbols?
A Frequency alone
B Pulse width alone
C Amplitude and phase
D Sampling rate
Answer: C. Amplitude and phase. Quadrature amplitude modulation combines amplitude and phase variations.