Nepal Engineering Council · Electronics & Communication Engineering · Chapter 7
Communication System
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180 questions in 6 syllabus topics.
7.1 Communications system
30 questions · AExE0701
1. In the basic block diagram of a communication system, which block lies between the transmitter and the receiver?
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Think of the medium the signal travels through.
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Answer: A. Channel
The channel is the physical medium (wire, fibre, free space) that carries the transmitted signal from transmitter to receiver.
2. The main purpose of the source encoder in a digital communication system is to:
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Compare source coding with channel coding.
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Answer: D. Remove redundancy so the message is represented with fewer bits
Source coding (compression) removes redundancy to represent the source efficiently; adding controlled redundancy is the job of the channel encoder.
3. The channel encoder in a digital communication system deliberately:
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It fights the effects of the channel, not the source.
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Answer: B. Adds controlled redundancy to detect or correct errors
Channel coding inserts extra (parity) bits in a controlled way so the receiver can detect or correct transmission errors.
4. Which block of a communication system converts a non-electrical message (such as sound) into an electrical signal?
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A microphone is an example.
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Answer: C. Input transducer
A transducer such as a microphone converts the physical message into an electrical signal before transmission.
5. A major advantage of digital communication over analog communication is that:
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Think of what a repeater can do with a 0/1 pulse.
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Answer: D. Regenerative repeaters can remove accumulated noise along the link
Digital pulses can be detected and regenerated afresh at each repeater, so noise does not accumulate as it does with analog amplifiers.
6. A common disadvantage of digital transmission of a voice signal compared with analog transmission is that it:
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Compare 4 kHz voice with 64 kbps PCM.
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Answer: B. Generally requires a larger transmission bandwidth
A 4 kHz voice channel becomes a 64 kbps PCM stream, which needs much more bandwidth than the original analog signal.
7. The two primary communication resources that a designer must trade off against each other are:
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Shannon's formula contains both of them.
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Answer: D. Transmitted power and channel bandwidth
Power and bandwidth are the basic resources; for a given capacity, one can be traded against the other (Shannon-Hartley).
8. A walkie-talkie, in which both ends can transmit but only one at a time, is an example of:
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Two-way, but taking turns.
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Answer: B. Half-duplex communication
Half duplex allows two-way communication but not simultaneously; push-to-talk radios are the classic example.
9. One important reason for using modulation in radio communication is to:
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Antenna length scales with wavelength.
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Answer: C. Make the antenna size practical by using a high carrier frequency
Antenna length is of the order of a quarter wavelength; shifting the message to a high carrier frequency makes the antenna practically small.
10. A quarter-wave antenna is to radiate a 3 kHz baseband signal directly. What length would it need? (c = 3×10⁸ m/s)
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Find λ first, then take a quarter.
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Answer: A. 25 km
λ = c/f = 3×10⁸/3000 = 100 km, so λ/4 = 25 km, which is impractical; hence modulation is needed.
11. Noise generated by the random thermal motion of electrons in a resistor is called:
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It depends on temperature.
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Answer: D. Thermal (Johnson) noise
Thermal or Johnson-Nyquist noise arises from random thermal agitation of charge carriers in any conductor above 0 K.
12. Shot noise in electronic devices is mainly due to:
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It exists only when DC current flows through a device.
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Answer: B. The random arrival of discrete charge carriers across a junction
Shot noise arises because current consists of discrete charges crossing a junction or gap at random instants; its mean-square current is 2qIB.
13. Which type of noise has a power spectral density roughly proportional to 1/f and dominates at low frequencies?
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Its other name is 1/f noise.
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Answer: B. Flicker noise
Flicker (1/f or pink) noise increases as frequency falls and is significant below about a few kHz.
14. Transit-time noise in a transistor becomes significant:
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Compare transit time with the signal period.
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Answer: D. At very high frequencies, when the carrier transit time is comparable to the signal period
When the period of the signal approaches the time carriers take to cross the device, the device becomes noisy; this sets a high-frequency limit.
15. Which of the following is an example of external (not device-generated) noise?
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Where does it originate: inside or outside the receiver?
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Answer: B. Atmospheric noise from lightning discharges
Atmospheric, extraterrestrial (solar, cosmic) and man-made noise originate outside the receiver; the others are internal noise sources.
16. Noise that is simply added to the transmitted signal in the channel, independent of the signal, is modelled as:
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Recall the 'A' in AWGN.
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Answer: C. Additive noise
The AWGN channel model is r(t) = s(t) + n(t); the noise is added and does not depend on the signal.
17. A signal power of 10 mW is received along with a noise power of 1 µW. The signal-to-noise ratio is:
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Use 10 log for a power ratio.
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Answer: D. 40 dB
SNR = 10 log₁₀(10×10⁻³ / 1×10⁻⁶) = 10 log₁₀(10⁴) = 40 dB.
18. A received signal has an RMS voltage of 2 V and the noise across the same load has an RMS voltage of 50 mV. The SNR is about:
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Voltage ratios use 20 log.
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Answer: B. 32 dB
For voltages across the same resistance, SNR = 20 log₁₀(2/0.05) = 20 log₁₀(40) ≈ 32 dB.
19. An SNR of 30 dB corresponds to a signal-to-noise power ratio of:
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Divide the dB value by 10 and take the antilog.
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Answer: A. 1000
Power ratio = 10^(30/10) = 10³ = 1000. (31.6 would be the voltage ratio.)
20. The noise figure of an amplifier is defined as:
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A noisy amplifier makes the SNR worse, so the ratio is at least 1.
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Answer: C. The ratio of input SNR to output SNR, expressed in dB
Noise factor F = (S/N)in / (S/N)out ≥ 1; noise figure NF = 10 log₁₀ F dB. A noiseless amplifier has NF = 0 dB.
21. The SNR at the input of an amplifier is 40 dB and at its output is 34 dB. The noise figure of the amplifier is:
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In dB, a ratio becomes a difference.
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Answer: C. 6 dB
NF (dB) = SNRin (dB) − SNRout (dB) = 40 − 34 = 6 dB.
22. An amplifier has a noise factor F = 2. Its noise figure and equivalent noise temperature (T₀ = 290 K) are:
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Te = (F − 1)T₀.
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Answer: A. 3 dB and 290 K
NF = 10 log₁₀ 2 ≈ 3 dB; Te = (F − 1)T₀ = (2 − 1)×290 = 290 K.
23. Two amplifiers are cascaded. Stage 1 has noise factor 2 and power gain 10; stage 2 has noise factor 5. The overall noise factor is:
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Use the Friis formula for cascaded stages.
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Answer: C. 2.4
Friis formula: F = F₁ + (F₂ − 1)/G₁ = 2 + (5 − 1)/10 = 2.4.
24. According to the Friis formula, to keep the overall noise figure of a receiver low, the first stage should have:
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Later-stage noise is divided by earlier gains.
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Answer: A. Low noise figure and high gain
Noise of later stages is divided by the gain of the earlier stages, so a low-noise, high-gain first stage (LNA) dominates the overall noise figure.
25. A 10 W transmitter feeds a channel with a total loss of 20 dB. The power reaching the receiver is:
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Convert 20 dB to a power ratio.
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Answer: A. 0.1 W
20 dB loss is a factor of 100, so P = 10/100 = 0.1 W.
26. A transmitter output of 2 W expressed in dBm is about:
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dBm is referred to 1 mW.
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Answer: A. 33 dBm
P(dBm) = 10 log₁₀(2000 mW / 1 mW) = 10 log₁₀ 2000 ≈ 33 dBm.
27. A diode carries a DC current of 1 mA. The RMS shot-noise current in a 1 MHz bandwidth is about (q = 1.6×10⁻¹⁹ C):
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Mean-square shot noise current is 2qIB.
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Answer: B. 17.9 nA
i_n = √(2qIB) = √(2×1.6×10⁻¹⁹×10⁻³×10⁶) = √(3.2×10⁻¹⁶) ≈ 1.79×10⁻⁸ A = 17.9 nA.
28. Which of these is NOT normally classed as a channel impairment?
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Which item is done on purpose by the transmitter?
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Answer: D. Source coding
Attenuation, distortion, interference and noise degrade a signal in the channel; source coding is a deliberate processing step at the transmitter.
29. Interference differs from random noise in that interference:
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Think of a neighbouring radio station.
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Answer: A. Comes from other man-made signals, often at known frequencies
Interference is contamination by other signals (e.g. adjacent transmitters), which is often predictable and can be reduced by filtering or planning; noise is random.
30. In a digital communication system, the overall performance is usually measured by the:
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What goes wrong when a 0 becomes a 1?
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Answer: C. Bit error rate (probability of bit error)
The fidelity of a digital link is expressed by the probability of bit error (BER) for a given Eb/N0; analog links use output SNR.
7.2 Representation of signals and systems in communication
30 questions · AExE0702
31. A low-pass (baseband) signal is one whose spectrum:
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Where on the frequency axis is its energy?
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Answer: B. Is concentrated around zero frequency, from 0 up to some maximum frequency
A low-pass signal has significant spectral content from DC up to a highest frequency W; speech and video baseband signals are examples.
32. A band-pass signal has its spectral content:
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Modulated signals are typical examples.
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Answer: C. Confined to a band around a centre (carrier) frequency fc, away from zero
A band-pass signal occupies fc − B/2 to fc + B/2, with no significant content near DC; modulated signals are band-pass.
33. A band-pass signal is called narrowband when:
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Compare B with fc.
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Answer: B. Its bandwidth is much smaller than its centre frequency
The narrowband condition B << fc allows the signal to be written as a slowly varying envelope and phase on a carrier.
34. A band-pass signal can be written as x(t) = xI(t) cos 2πfct − xQ(t) sin 2πfct. Here xI(t) and xQ(t) are:
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These components vary slowly compared with the carrier.
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Answer: A. Low-pass in-phase and quadrature components
Any band-pass signal can be represented by two low-pass components, the in-phase xI(t) and quadrature xQ(t), which modulate quadrature carriers.
35. The complex envelope of the band-pass signal x(t) = xI(t) cos 2πfct − xQ(t) sin 2πfct is:
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Expand Re{x̃(t) e^(j2πfct)} and match terms.
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Answer: C. xI(t) + j xQ(t)
x(t) = Re{[xI(t) + j xQ(t)] e^(j2πfct)}, so the complex envelope is x̃(t) = xI(t) + j xQ(t).
36. At some instant, the in-phase and quadrature components of a band-pass signal are xI = 3 V and xQ = 4 V. The envelope at that instant is:
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Magnitude of the complex envelope.
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Answer: C. 5 V
Envelope a(t) = √(xI² + xQ²) = √(9 + 16) = 5 V.
37. A band-pass signal occupies the band 9.9 MHz to 10.1 MHz. Its bandwidth and centre frequency are:
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Bandwidth is highest minus lowest frequency.
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Answer: A. 200 kHz and 10 MHz
B = 10.1 − 9.9 = 0.2 MHz = 200 kHz; fc = (9.9 + 10.1)/2 = 10 MHz.
38. The 3-dB bandwidth of a system is the frequency range over which:
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−3 dB in power is a factor of about 0.5.
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Answer: B. The power gain is at least half of its maximum value
At the 3-dB points the power falls to 1/2 (voltage to 1/√2 ≈ 0.707) of its peak value.
39. An RC low-pass filter has R = 1 kΩ and C = 0.1 µF. Its 3-dB bandwidth is about:
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f = 1/(2πRC).
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Answer: D. 1.59 kHz
f3dB = 1/(2πRC) = 1/(2π × 10³ × 10⁻⁷) ≈ 1591 Hz ≈ 1.59 kHz.
40. The noise-equivalent bandwidth of a simple RC low-pass filter with R = 1 kΩ and C = 0.1 µF is:
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For an RC filter B_N = 1/(4RC), slightly larger than the 3-dB bandwidth.
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Answer: A. 2.5 kHz
For a first-order RC filter, B_N = (π/2) f3dB = 1/(4RC) = 1/(4 × 10³ × 10⁻⁷) = 2500 Hz.
41. A rectangular pulse of width 1 µs is transmitted. The first null of its amplitude spectrum occurs at:
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The spectrum is a sinc function with nulls at multiples of 1/T.
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Answer: D. 1 MHz
The spectrum of a rectangular pulse of width T is T sinc(fT), with its first zero at f = 1/T = 1/(1 µs) = 1 MHz.
42. Using the rule of thumb tr ≈ 0.35/B, an amplifier with 10 MHz bandwidth has a rise time of about:
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Rise time is inversely proportional to bandwidth.
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Answer: A. 35 ns
tr ≈ 0.35/B = 0.35/(10 × 10⁶) = 35 × 10⁻⁹ s = 35 ns.
43. A tuned band-pass circuit has a centre frequency of 10 MHz and a 3-dB bandwidth of 100 kHz. Its quality factor Q is:
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Q is centre frequency divided by bandwidth.
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Answer: A. 100
Q = f0/B = 10 MHz / 100 kHz = 100.
44. For distortionless transmission through a linear system, the transfer function must have:
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The output should be a scaled, delayed copy of the input.
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Answer: C. Constant magnitude and phase varying linearly with frequency over the signal band
H(f) = K e^(−j2πft₀) gives y(t) = K x(t − t₀): only scaling and delay, with no change of waveform shape.
45. Which of the following systems provides distortionless transmission?
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Look for only a gain and a delay.
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Answer: B. y(t) = 3 x(t − 2)
Only a constant gain and pure delay keep the waveform shape; squaring is non-linear, an echo causes amplitude ripple, and differentiation weights frequencies unequally.
46. A system has a constant magnitude response but a phase response that is not linear with frequency. The output will suffer from:
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Different components arrive at different times.
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Answer: A. Phase (delay) distortion
If phase is not linear, different frequency components are delayed by different amounts, which distorts the waveform even though their amplitudes are unchanged.
47. The group delay of a system with phase response θ(f) is given by:
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It involves the slope of the phase curve.
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Answer: D. τg = −(1/2π) dθ/df
Group delay is the negative derivative of phase with respect to angular frequency, τg = −dθ/dω = −(1/2π) dθ/df.
48. A distortionless channel has a linear phase response that reaches −π/2 rad at 1 kHz. The time delay introduced by the channel is:
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Linear phase θ = −2πft₀.
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Answer: C. 0.25 ms
θ(f) = −2πft₀, so t₀ = (π/2)/(2π × 1000) = 1/4000 s = 0.25 ms.
49. Human hearing is relatively insensitive to which kind of distortion, so that voice channels tolerate it well?
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Video and data are much more sensitive to this.
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Answer: A. Phase (delay) distortion
The ear responds mainly to the magnitude spectrum, so moderate phase distortion is tolerable in speech; video and data are far more sensitive to it.
50. Two tones at 100 MHz and 101 MHz pass through a non-linear amplifier. The third-order intermodulation products that fall near the wanted band are:
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Use 2f₁ − f₂ and 2f₂ − f₁.
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Answer: B. 99 MHz and 102 MHz
Third-order products 2f₁ − f₂ = 200 − 101 = 99 MHz and 2f₂ − f₁ = 202 − 100 = 102 MHz lie close to the wanted signals.
51. The signal x(t) = cos(2π·2000t) + cos(2π·7000t) is applied to an ideal low-pass filter with a 5 kHz cut-off. The output is:
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Which tone is below the cut-off?
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Answer: A. cos(2π·2000t) only
An ideal LPF passes components below 5 kHz unchanged and rejects those above, so only the 2 kHz tone remains.
52. An ideal (brick-wall) low-pass filter cannot be physically built because its impulse response:
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Look at the sinc response for t < 0.
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Answer: D. Is non-zero for negative time, so it is non-causal
The impulse response of an ideal LPF is a sinc function extending from −∞ to +∞, so it would respond before the input arrives.
53. The Hilbert transformer is a linear filter whose transfer function is:
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It is a pure ±90° phase shifter.
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Answer: B. H(f) = −j sgn(f)
The Hilbert transformer shifts positive frequencies by −90° and negative frequencies by +90° with unit magnitude: H(f) = −j sgn(f).
54. The impulse response of the Hilbert transformer is:
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It is not causal and decays as 1/t.
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Answer: C. h(t) = 1/(πt)
The Hilbert transform is the convolution x̂(t) = x(t) * 1/(πt), so the impulse response is 1/(πt), which is non-causal.
55. The Hilbert transform of 3 cos(100t) is:
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Delay the phase of each component by 90°.
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Answer: D. 3 sin(100t)
The Hilbert transform shifts a cosine by −90°: cos(ωt − 90°) = sin(ωt), so 3 cos(100t) → 3 sin(100t).
56. If the Hilbert transform is applied twice to a signal x(t) (with no DC), the result is:
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Two 90° shifts in a row.
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Answer: B. −x(t)
Two successive −90° shifts give −180°, i.e. multiplication by (−j sgn f)² = −1, so the result is −x(t).
57. Which statement about a signal x(t) and its Hilbert transform x̂(t) is correct?
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The Hilbert transformer has unit magnitude response.
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Answer: D. They have the same amplitude spectrum and are orthogonal
|H(f)| = 1 so the amplitude spectrum and energy are unchanged, and the integral of x(t)x̂(t) over all time is zero (orthogonal).
58. The analytic signal (pre-envelope) x₊(t) = x(t) + j x̂(t) has a spectrum that is:
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1 + sgn(f) takes values 2 and 0.
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Answer: D. Zero for negative frequencies and twice X(f) for positive frequencies
X₊(f) = X(f)[1 + sgn(f)], which equals 2X(f) for f > 0, X(0) at f = 0 and 0 for f < 0.
59. A key application of the Hilbert transform in communication is:
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One sideband is cancelled using a 90°-shifted copy of the message.
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Answer: B. Generating SSB signals by the phase-shift method
The phase-shift SSB modulator combines m(t)cos ωct and m̂(t)sin ωct; the 90° shifted message m̂(t) is the Hilbert transform of m(t).
60. For a band-pass signal x(t), the natural envelope can be obtained from the Hilbert transform as:
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Take the magnitude of the analytic signal.
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Answer: C. |x(t) + j x̂(t)|
The magnitude of the pre-envelope x(t) + j x̂(t) gives the envelope √(x² + x̂²) of the band-pass signal.
7.3 Modulation
31 questions · AExE0703
61. The signal s(t) = Ac[1 + ka m(t)] cos 2πfct represents:
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Expand the bracket: is a pure carrier term present?
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Answer: B. Conventional AM (DSB with full carrier)
The carrier term Ac cos 2πfct is present together with sidebands from Ac ka m(t) cos 2πfct, so this is standard DSB-FC AM.
62. An AM wave displayed on an oscilloscope has a maximum envelope amplitude of 15 V and a minimum of 5 V. The modulation index is:
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μ = (Vmax − Vmin)/(Vmax + Vmin).
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Answer: C. 0.5
μ = (Vmax − Vmin)/(Vmax + Vmin) = (15 − 5)/(15 + 5) = 0.5.
63. An AM transmitter has a carrier power of 100 W and is modulated to a depth of 80% by a sinusoid. The total transmitted power is:
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Pt = Pc(1 + μ²/2).
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Answer: C. 132 W
Pt = Pc(1 + μ²/2) = 100(1 + 0.64/2) = 132 W.
64. A 1 kW carrier is amplitude modulated by a sinusoid with μ = 0.5. The total power in both sidebands together is:
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Each sideband carries Pc μ²/4.
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Answer: A. 125 W
P_SB = Pc μ²/2 = 1000 × 0.25/2 = 125 W (62.5 W in each sideband).
65. The maximum power efficiency (sideband power / total power) of single-tone conventional AM, reached at μ = 1, is:
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η = μ²/(2 + μ²).
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Answer: D. 33.3%
η = μ²/(2 + μ²) = 1/3 ≈ 33.3% at μ = 1; two-thirds of the power is wasted in the carrier.
66. The unmodulated antenna current of an AM transmitter is 8 A. When modulated with μ = 0.6, the antenna current becomes about:
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Current goes as the square root of power.
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Answer: D. 8.69 A
It = Ic √(1 + μ²/2) = 8 √(1 + 0.18) ≈ 8.69 A.
67. A 1 MHz carrier is amplitude modulated (DSB-FC) by a 5 kHz tone. The spectrum contains components at:
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Carrier plus upper and lower side frequencies.
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Answer: B. 995 kHz, 1000 kHz and 1005 kHz
Standard AM contains the carrier fc and the two side frequencies fc ± fm = 995 kHz and 1005 kHz; bandwidth 2fm = 10 kHz.
68. The transmission bandwidth of an SSB signal for a message of maximum frequency fm is:
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Only one of the two sidebands is sent.
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Answer: D. fm
SSB transmits only one sideband, so it needs half the bandwidth of DSB, i.e. fm.
69. Vestigial sideband (VSB) modulation was used for the video signal in analog television mainly because:
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Think of the spectrum of a video signal near 0 Hz.
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Answer: B. Video has significant low-frequency content, which makes pure SSB filtering impractical
Video extends almost to DC, so a sharp SSB filter is impossible; VSB keeps one sideband plus a vestige of the other, saving bandwidth compared with DSB.
70. Which AM variant cannot be demodulated by a simple envelope detector and requires coherent (synchronous) detection?
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Which one has no transmitted carrier?
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Answer: D. DSB-SC
In DSB-SC the envelope is |m(t)|, not m(t), so phase reversals are lost; a locally generated synchronous carrier is required.
71. A balanced modulator is used to generate:
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Its output has the carrier suppressed.
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Answer: B. DSB-SC signals
A balanced modulator cancels the carrier and outputs only the product m(t) cos ωct, i.e. DSB-SC.
72. For an envelope detector to follow an AM envelope correctly, the RC time constant should satisfy:
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It must smooth the carrier but track the message.
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Answer: B. 1/fc << RC << 1/fm
RC must be long compared with the carrier period (to smooth the RF ripple) but short compared with the message period (to follow the envelope).
73. When conventional AM is over-modulated (μ > 1), the result is:
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What happens to the envelope when it would go negative?
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Answer: A. Envelope distortion and generation of extra sideband components (splatter)
With μ > 1 the envelope crosses zero and is clipped, so an envelope detector output is distorted and spurious frequencies spread outside the channel.
74. An AM superheterodyne receiver with an IF of 455 kHz is tuned to 1000 kHz (local oscillator above the signal). The image frequency is:
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The image is 2 × IF away from the wanted signal.
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Answer: A. 1910 kHz
f_image = fs + 2 IF = 1000 + 910 = 1910 kHz.
75. An FM broadcast signal has a peak frequency deviation of 75 kHz and the highest audio frequency is 15 kHz. The modulation index is:
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β = Δf / fm.
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Answer: B. 5
β = Δf/fm = 75/15 = 5.
76. Using Carson's rule, the bandwidth of an FM signal with 75 kHz peak deviation and 15 kHz maximum modulating frequency is:
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Carson: B = 2(Δf + fm).
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Answer: A. 180 kHz
B = 2(Δf + fm) = 2(75 + 15) = 180 kHz.
77. The FM signal s(t) = 10 cos(2π×10⁶t + 5 sin 2π×10³t) has a peak frequency deviation of:
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Δf = β × fm.
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Answer: C. 5 kHz
The phase term is β sin 2πfmt with β = 5 and fm = 1 kHz, so Δf = βfm = 5 kHz.
78. An FM modulator has a frequency sensitivity kf = 10 kHz/V. A 2 V peak, 5 kHz sinusoid is applied. The peak deviation and modulation index are:
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Deviation is proportional to the message amplitude.
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Answer: C. 20 kHz and 4
Δf = kf Am = 10 × 2 = 20 kHz; β = Δf/fm = 20/5 = 4.
79. An FM signal can be generated with a phase modulator by:
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Frequency is the derivative of phase.
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Answer: A. Integrating the message before applying it to the phase modulator
In FM the phase is proportional to the integral of m(t); so integrating m(t) and then phase-modulating gives FM (the basis of the Armstrong method).
80. Which statement about the total power of a sinusoidally modulated FM signal is correct?
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Look at the envelope of an FM wave.
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Answer: A. It remains equal to the unmodulated carrier power for any modulation index
FM has a constant envelope Ac, so total power is Ac²/2 regardless of β; modulation only redistributes power among the carrier and sidebands.
81. In FM with sinusoidal modulation, the carrier component disappears completely when the modulation index is about:
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Find the first zero of J₀(β).
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Answer: D. 2.405
The carrier amplitude is proportional to J₀(β), whose first zero is at β ≈ 2.405; this is used to calibrate deviation.
82. Pre-emphasis in an FM transmitter:
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Noise in FM is worst at high audio frequencies.
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Answer: C. Boosts the higher audio frequencies before modulation to improve SNR at the receiver
FM output noise rises with frequency, so high audio frequencies are boosted before modulation and restored by de-emphasis at the receiver, improving SNR.
83. Which of the following is commonly used to demodulate an FM signal?
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The detector must respond to frequency changes.
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Answer: C. Phase-locked loop (PLL)
A PLL tracks the instantaneous frequency of the input, and its control voltage is the demodulated message; discriminators and ratio detectors are other FM detectors.
84. In binary amplitude shift keying in its on-off (OOK) form, the two binary symbols are represented by:
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On-off.
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Answer: B. Presence and absence of the carrier
OOK transmits the carrier for a 1 and nothing for a 0, which is the simplest form of ASK.
85. Which binary digital modulation scheme cannot be demodulated non-coherently in its basic form?
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Which scheme carries information only in the phase?
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Answer: C. BPSK
Information in BPSK is in the absolute carrier phase, so a phase reference is required; ASK and FSK can use envelope detection. DPSK avoids this by differential encoding.
86. For the same Eb/N0 in an AWGN channel, which coherent binary scheme gives the lowest bit error probability?
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Antipodal signals are farthest apart.
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Answer: B. BPSK
BPSK uses antipodal signals with the largest distance 2√Eb, giving Pe = Q(√(2Eb/N0)), about 3 dB better than orthogonal FSK.
87. A BPSK signal carries data at 10 kbps using rectangular pulses. The null-to-null bandwidth of the transmitted signal is:
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The main lobe spans ±Rb around the carrier.
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Answer: C. 20 kHz
The main lobe of BPSK extends from fc − Rb to fc + Rb, so the null-to-null bandwidth is 2Rb = 20 kHz.
88. A BFSK system sends 1 kbps using tones of 10 kHz and 12 kHz. Taking B ≈ |f₂ − f₁| + 2Rb, the required bandwidth is about:
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Add the tone separation to twice the bit rate.
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Answer: D. 4 kHz
B ≈ (12 − 10) kHz + 2 × 1 kHz = 4 kHz.
89. A modem transmits 9600 bps using 16-QAM. The symbol (baud) rate is:
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Bits per symbol = log₂ M.
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Answer: A. 2400 baud
16-QAM carries log₂16 = 4 bits per symbol, so the symbol rate is 9600/4 = 2400 baud.
90. In M-ary PSK, as M increases for a fixed bit rate:
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Points on the circle get closer together.
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Answer: D. Bandwidth efficiency improves but a higher Eb/N0 is needed for the same error rate
More bits per symbol reduces symbol rate and bandwidth, but the constellation points come closer together, so more power is needed for the same BER.
91. Minimum shift keying (MSK) can be described as:
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The 'minimum' refers to the frequency separation.
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Answer: A. Continuous-phase FSK with modulation index h = 0.5
MSK is CPFSK with frequency separation equal to half the bit rate (h = 0.5), the minimum spacing for orthogonal coherent tones; it has a constant envelope.
7.4 Digital communication systems
30 questions · AExE0704
92. The correct order of operations in converting an analog message into a PCM signal is:
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Time is discretised before amplitude.
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Answer: B. Sampling, quantising, encoding
The analog signal is first sampled in time, the sample values are then quantised to discrete levels, and each level is encoded into a binary code word.
93. According to the sampling theorem, a signal band-limited to W Hz can be recovered exactly from its samples if the sampling rate is:
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Twice the highest frequency.
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Answer: A. At least 2W samples per second
A band-limited signal is uniquely determined by samples taken at fs ≥ 2W; the minimum rate 2W is the Nyquist rate.
94. The Nyquist rate for x(t) = 5 cos(200πt) + 2 cos(600πt) is:
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Convert ω to f, then double the highest.
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Answer: B. 600 Hz
The components are at 100 Hz and 300 Hz; the highest is 300 Hz, so the Nyquist rate is 2 × 300 = 600 Hz.
95. A 7 kHz sinusoid is sampled at 10 kHz and reconstructed with an ideal 5 kHz low-pass filter. The output is a tone at:
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Look for fs − f.
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Answer: A. 3 kHz
Sampling below the Nyquist rate causes aliasing: the 7 kHz tone appears at |10 − 7| = 3 kHz inside the 0–5 kHz band.
96. The anti-aliasing filter in an A/D conversion chain is a:
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It must act before the damage is done.
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Answer: A. Low-pass filter placed before the sampler
An LPF before sampling removes components above fs/2 so they cannot fold back into the baseband as aliases.
97. In flat-top (sample-and-hold) sampling, the high-frequency attenuation of the reconstructed signal caused by the finite pulse width is called:
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It depends on the width (aperture) of the pulse.
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Answer: B. Aperture effect
Holding each sample for time T multiplies the spectrum by a sinc(fT) shape, attenuating higher frequencies; an equaliser can correct this aperture effect.
98. In which analog pulse modulation does the position (timing) of constant-width pulses vary with the message?
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The name says what varies.
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Answer: B. PPM
Pulse position modulation shifts the time of each pulse in proportion to the sample value; PAM varies amplitude and PWM varies width.
99. Among PAM, PWM and PPM, the one whose transmitted pulses vary in amplitude and is therefore most affected by additive noise is:
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Noise adds directly to the amplitude.
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Answer: B. PAM
In PAM the information is in the amplitude, which noise directly corrupts; PWM and PPM keep constant amplitude so limiters/slicers can remove much amplitude noise.
100. A telephone voice signal is sampled at 8 kHz and each sample is coded with 8 bits. The PCM bit rate is:
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Bits per sample × samples per second.
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Answer: A. 64 kbps
Bit rate = n fs = 8 × 8000 = 64 000 bps.
101. What is the minimum (Nyquist) transmission bandwidth needed for a binary PCM signal of 8 bits per sample sampled at 8 kHz?
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Minimum bandwidth is half the bit rate.
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Answer: D. 32 kHz
Rb = 64 kbps and the minimum baseband bandwidth for binary signalling is Rb/2 = 32 kHz.
102. A quantiser must provide at least 200 levels. The minimum number of bits per sample is:
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Find the smallest n with 2ⁿ ≥ 200.
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Answer: D. 8
2⁷ = 128 < 200 ≤ 256 = 2⁸, so 8 bits are needed.
103. A signal ranging from −1 V to +1 V is uniformly quantised with 8 bits. The step size is about:
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Divide the full range by the number of levels.
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Answer: A. 7.8 mV
Δ = (Vmax − Vmin)/2ⁿ = 2/256 ≈ 7.81 mV.
104. For uniform quantisation with step size Δ, the mean-square quantisation noise (error uniformly distributed) is:
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Variance of a uniform distribution of width Δ.
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Answer: C. Δ²/12
The error is uniform over −Δ/2 to Δ/2, whose variance is Δ²/12.
105. For a full-scale sinusoid, the signal-to-quantisation-noise ratio of an 8-bit uniform PCM system is about:
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Use 6.02n + 1.76 dB.
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Answer: C. 49.9 dB
SQNR ≈ 6.02n + 1.76 dB = 6.02 × 8 + 1.76 ≈ 49.9 dB.
106. In uniform PCM, increasing the number of bits per sample by one improves the SQNR by about:
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Halving Δ reduces Δ²/12 by a factor of four.
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Answer: D. 6 dB
Each extra bit doubles the number of levels and halves Δ, reducing noise power by 4, i.e. about 6 dB.
107. The main purpose of non-uniform quantisation (companding) in speech PCM is to:
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Speech has many low-amplitude samples.
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Answer: A. Give a more uniform SQNR for weak and strong signals
Companding uses small steps for small amplitudes and larger steps for large ones, so weak speech signals get adequate SQNR without more bits.
108. The μ-law companding standard with μ = 255 is used mainly in:
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Its counterpart is A-law.
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Answer: B. North America and Japan
μ-law (μ = 255) is used in North American and Japanese systems; A-law (A = 87.6) is used in Europe and most other countries.
109. A mid-tread uniform quantiser is one that:
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Think of a staircase with a flat tread at the origin.
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Answer: C. Has zero as one of its output levels
In a mid-tread quantiser zero is an output (reconstruction) level; in a mid-rise quantiser zero is a decision threshold, so there is no zero output level.
110. Delta modulation (DM) transmits:
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It is the simplest form of differential coding.
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Answer: C. One bit per sample indicating whether the signal is above or below its staircase approximation
DM is 1-bit DPCM: each bit tells the receiver to step the staircase up or down by Δ.
111. A 1 kHz sinusoid of amplitude 1 V is delta modulated at 64 kHz. To avoid slope overload, the step size must be at least about:
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Compare Δ·fs with the maximum slope 2πfmAm.
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Answer: A. 98 mV
No slope overload requires Δ fs ≥ 2πfmAm, so Δ ≥ 2π × 1000 × 1 / 64 000 ≈ 0.098 V.
112. Granular noise in a delta modulator is most noticeable when:
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It is the opposite case to slope overload.
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Answer: D. The step size is too large compared with a slowly varying input
With a large Δ, the staircase hunts up and down around a nearly constant input, producing granular noise; too small a Δ causes slope overload.
113. Differential PCM (DPCM) reduces the bit rate compared with PCM because it:
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Neighbouring samples are similar.
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Answer: C. Quantises the difference between the sample and its predicted value, which has a smaller range
Adjacent speech samples are highly correlated, so the prediction error has a smaller variance and needs fewer bits for the same quality.
114. In Huffman source coding, symbols with higher probability are assigned:
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Morse code uses the same idea.
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Answer: B. Shorter code words
Huffman coding is a variable-length prefix code that gives short code words to frequent symbols, minimising the average code length.
115. The Shannon-Hartley theorem gives the channel capacity of an AWGN channel of bandwidth B and signal-to-noise ratio S/N as:
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The logarithm is to base 2.
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Answer: B. C = B log₂(1 + S/N)
Shannon-Hartley: C = B log₂(1 + S/N) bits per second, with S/N as a power ratio (not in dB).
116. A telephone channel has a bandwidth of 3 kHz and an SNR of 30 dB. Its Shannon capacity is about:
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Convert 30 dB to a ratio first.
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Answer: C. 29.9 kbps
S/N = 10³; C = 3000 log₂(1001) ≈ 3000 × 9.97 ≈ 29.9 kbps.
117. The Shannon limit states that reliable communication over an AWGN channel is impossible, however large the bandwidth, if Eb/N0 is below about:
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It equals ln 2 as a ratio.
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Answer: C. −1.6 dB
As B → ∞, the minimum Eb/N0 for error-free transmission tends to ln 2 = 0.693, i.e. −1.59 dB.
118. A T1 frame carries 24 voice channels of 8 bits each plus 1 framing bit, at 8000 frames per second. The T1 line rate is:
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Count the bits in one frame first.
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Answer: D. 1.544 Mbps
(24 × 8 + 1) bits × 8000 frames/s = 193 × 8000 = 1 544 000 bps.
119. The European E1 TDM system has a line rate of 2.048 Mbps and carries:
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Two slots are reserved for overhead.
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Answer: A. 32 time slots, of which 30 are used for voice
E1 has 32 × 8-bit slots per 125 µs frame (2.048 Mbps); slot 0 is for framing/synchronisation and slot 16 usually for signalling, leaving 30 voice channels.
120. Guard bands are a feature of which multiplexing technique?
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Gaps between adjacent channel spectra.
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Answer: D. Frequency-division multiplexing
In FDM, unused frequency gaps (guard bands) are left between adjacent channels so practical filters can separate them without crosstalk.
121. Wavelength-division multiplexing (WDM) is used mainly in:
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Wavelength suggests light.
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Answer: D. Optical fibre systems
WDM sends several optical carriers of different wavelengths over one fibre; it is essentially FDM at optical frequencies.
7.5 Baseband and band pass data communication systems
30 questions · AExE0705
122. The amount of information carried by a message is:
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A surprising message tells you more.
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Answer: B. Larger for less probable messages
Self-information I = log₂(1/p) increases as probability p decreases; a certain event (p = 1) carries no information.
123. A symbol occurs with probability 1/8. The information it conveys is:
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I = log₂(1/p).
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Answer: A. 3 bits
I = log₂(1/p) = log₂ 8 = 3 bits.
124. When information is measured using the natural logarithm (base e), the unit is the:
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Natural log.
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Answer: A. Nat
log₂ gives bits, ln gives nats and log₁₀ gives hartleys (decits); 1 nat ≈ 1.443 bits.
125. A source emits four symbols with probabilities 1/2, 1/4, 1/8 and 1/8. Its entropy is:
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H = Σ p log₂(1/p).
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Answer: A. 1.75 bits/symbol
H = ½(1) + ¼(2) + ⅛(3) + ⅛(3) = 0.5 + 0.5 + 0.375 + 0.375 = 1.75 bits/symbol.
126. The source with symbol probabilities 1/2, 1/4, 1/8, 1/8 emits 1000 symbols per second. Its average information rate is:
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Rate = symbol rate × entropy.
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Answer: C. 1750 bits/s
R = r H = 1000 × 1.75 = 1750 bits/s.
127. The entropy of a binary memoryless source is maximum, equal to 1 bit/symbol, when the probability of a 1 is:
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Maximum uncertainty.
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Answer: A. 0.5
H(p) = −p log₂p − (1−p) log₂(1−p) peaks at p = 0.5 with H = 1 bit; it is zero when p = 0 or 1.
128. For a discrete source with M symbols, the entropy is maximum when:
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Maximum uncertainty again.
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Answer: A. All symbols are equally likely, giving H = log₂ M
Entropy is bounded by 0 ≤ H ≤ log₂ M, the upper bound reached for equiprobable symbols.
129. The source with probabilities 1/2, 1/4, 1/8, 1/8 is encoded with a fixed 2-bit code. The coding efficiency is:
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Efficiency = entropy ÷ average code length.
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Answer: C. 87.5%
η = H / L̄ = 1.75 / 2 = 0.875 = 87.5%. (A Huffman code would reach 100% here.)
130. A binary symmetric channel has a crossover (bit error) probability of 0.1. Its capacity is about:
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C = 1 − H(p) for a BSC.
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Answer: D. 0.53 bit per channel use
C = 1 − H(0.1) = 1 − 0.469 ≈ 0.531 bit per use.
131. Mutual information I(X;Y) between the channel input X and output Y is equal to:
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Uncertainty before minus uncertainty after.
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Answer: B. H(X) − H(X|Y)
Mutual information is the reduction in uncertainty about X after observing Y: I(X;Y) = H(X) − H(X|Y); channel capacity is its maximum over input distributions.
132. A prefix (instantaneous) code is one in which:
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Think of decoding without waiting for the next bits.
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Answer: B. No code word is the beginning of any other code word
In a prefix code no code word is a prefix of another, so each code word can be decoded as soon as it ends; Huffman codes are prefix codes.
133. A disadvantage of unipolar NRZ line coding is that it:
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What is the average voltage of a 0/+V signal?
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Answer: B. Has a significant DC component and no guaranteed transitions for timing
Unipolar NRZ uses 0 and +V only, so the average is non-zero (DC), and long runs of 1s or 0s give no transitions for clock recovery.
134. In bipolar AMI line coding, binary 1s are represented by:
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'Alternate mark inversion' says it all.
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Answer: D. Pulses of alternating polarity, with 0s as zero voltage
AMI sends 0 as zero volts and successive 1s as +V, −V, +V …, which removes the DC component and allows detection of single errors as bipolar violations.
135. The data 1 0 1 1 is coded in bipolar AMI, with the first 1 sent as a positive pulse. The transmitted levels are:
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Only the 1s alternate.
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Answer: D. +V, 0, −V, +V
Zeros are sent as 0 V and each 1 alternates polarity: first 1 → +V, second 1 → −V, third 1 → +V.
136. Manchester line coding is widely used because:
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Look at the middle of each bit.
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Answer: A. Every bit has a mid-bit transition, so it is self-clocking with no DC component
The guaranteed mid-bit transition carries timing and makes the average zero, at the cost of about twice the bandwidth of NRZ.
137. HDB3 line coding improves on plain AMI by:
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The '3' is the maximum run of zeros allowed.
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Answer: A. Replacing strings of four consecutive zeros with a code that contains deliberate bipolar violations
Long zero runs in AMI cause loss of timing; HDB3 substitutes 000V or B00V patterns for every four zeros, keeping enough transitions while staying DC-free.
138. Inter-symbol interference (ISI) in a baseband system is caused mainly by:
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Neighbouring pulses overlap.
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Answer: C. Spreading of pulses beyond their symbol interval due to limited channel bandwidth
A band-limited channel spreads each pulse in time, so the tails of neighbouring pulses add to the sample of the current pulse.
139. According to the Nyquist criterion, the maximum symbol rate for zero-ISI transmission through an ideal channel of bandwidth 4 kHz is:
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Rs,max = 2B.
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Answer: C. 8000 symbols/s
An ideal low-pass channel of bandwidth B supports up to 2B = 8000 symbols/s without ISI (sinc pulses).
140. Binary data at 10 kbps is sent with raised-cosine pulse shaping of roll-off factor α = 0.5. The required baseband bandwidth is:
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B = (R/2)(1 + α).
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Answer: D. 7.5 kHz
B = (Rb/2)(1 + α) = 5 kHz × 1.5 = 7.5 kHz.
141. Raised-cosine pulses with roll-off α > 0 are preferred over ideal sinc pulses (α = 0) because:
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Compare how fast the pulse tails die away.
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Answer: D. Their tails decay faster, so timing errors cause less ISI and the filter is realisable
The ideal Nyquist filter is unrealisable and its sinc tails decay slowly (1/t); raised-cosine tails decay as 1/t³, trading extra bandwidth for robustness.
142. In an eye diagram, the vertical opening of the eye indicates the:
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A closing eye means more errors.
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Answer: D. Noise margin at the sampling instant
The height of the eye opening shows how much noise can be tolerated; its width shows the time interval over which sampling can be done without ISI errors.
143. A matched filter at the receiver is designed to:
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It is matched to the known pulse shape in white noise.
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Answer: C. Maximise the output signal-to-noise ratio at the sampling instant
The matched filter, with impulse response h(t) = s(T − t), maximises the peak SNR at t = T in white noise, giving the best detection performance.
144. The Hamming distance between the code words 1011010 and 1001011 is:
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Count positions where the bits differ.
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Answer: B. 2
They differ in the 3rd and 7th positions, so the Hamming distance is 2.
145. A block code has a minimum Hamming distance of 5. The number of bit errors per code word it can always correct is:
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dmin ≥ 2t + 1.
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Answer: A. 2
Correction of t errors needs dmin ≥ 2t + 1, so t = ⌊(5 − 1)/2⌋ = 2.
146. To guarantee detection of up to 3 errors in a code word, the minimum Hamming distance must be at least:
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Detection needs dmin ≥ s + 1.
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Answer: D. 4
Detecting s errors requires dmin ≥ s + 1 = 4.
147. For a Hamming code carrying 11 data bits, the number of parity bits needed for single-error correction is:
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Use 2^r ≥ k + r + 1.
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Answer: B. 4
The condition 2^r ≥ k + r + 1 gives r = 4 since 16 ≥ 11 + 4 + 1 = 16, while r = 3 gives 8 < 15. This is the (15, 11) Hamming code.
148. The code rate of the (7, 4) Hamming code and the number of bit errors it can correct per code word are:
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Rate = k/n; dmin of Hamming codes is 3.
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Answer: B. About 0.57 and one
Code rate = k/n = 4/7 ≈ 0.57; with dmin = 3 it corrects one error per code word.
149. The 7-bit data word 1101001 is to be sent with an even parity bit. The parity bit is:
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Count the 1s.
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Answer: B. 0
The word already has four 1s (an even number), so the even parity bit is 0.
150. Which statement correctly distinguishes ARQ from FEC?
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Which one needs a feedback channel?
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Answer: C. ARQ detects errors and requests retransmission; FEC corrects errors at the receiver without a return channel
Automatic repeat request needs a feedback channel and retransmits erroneous frames; forward error correction adds enough redundancy to correct errors directly.
151. The Viterbi algorithm is used for decoding:
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It works on a trellis diagram.
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Answer: C. Convolutional codes
The Viterbi algorithm performs maximum-likelihood decoding of convolutional codes by searching the trellis for the most likely path.
7.6 Random signals and noise in communication system
29 questions · AExE0706
152. A random process is best described as:
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Fix the time and you get a random variable.
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Answer: D. An ensemble (collection) of time functions, one of which occurs in each trial
A random process X(t, s) assigns a sample function of time to each outcome s; at any fixed time it is a random variable.
153. A random process X(t) is wide-sense stationary (WSS) if:
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Only the mean and autocorrelation are checked.
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Answer: A. Its mean is constant and its autocorrelation depends only on the time difference τ
WSS needs only first- and second-order conditions: E[X(t)] = constant and R(t, t+τ) = R(τ). Full shift-invariance of all distributions is strict-sense stationarity.
154. A random process is ergodic in the mean when:
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Can one long record represent the whole ensemble?
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Answer: B. The time average of a single sample function equals the ensemble mean
Ergodicity lets ensemble averages be estimated from time averages of one long sample function, which is what a measuring instrument does.
155. For a WSS random process, the autocorrelation at zero lag, R(0), equals:
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Set τ = 0 in E[X(t)X(t + τ)].
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Answer: B. The mean-square value (total average power)
R(0) = E[X²(t)], which is the average power of the process; it equals the variance plus the square of the mean.
156. Which of the following is NOT a property of the autocorrelation function R(τ) of a real WSS process?
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Where is the autocorrelation largest?
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Answer: C. R(τ) can exceed R(0) for some τ ≠ 0
R(τ) is even and its magnitude never exceeds R(0); it is the inverse Fourier transform of the PSD (Wiener-Khinchin).
157. The Wiener-Khinchin theorem states that the power spectral density of a WSS process is:
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Transform from the τ domain to the f domain.
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Answer: A. The Fourier transform of its autocorrelation function
S(f) = ∫ R(τ) e^(−j2πfτ) dτ; the PSD and autocorrelation form a Fourier transform pair.
158. The power spectral density of a real WSS random process is always:
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It represents power per hertz.
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Answer: C. Real, non-negative and an even function of frequency
Since R(τ) is real and even, S(f) is real and even; it is also non-negative because it represents power per unit bandwidth.
159. The process X(t) = 2 cos(ω₀t + Θ), with Θ uniformly distributed over (0, 2π), is WSS. Its average power is:
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Average power of a sinusoid is A²/2.
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Answer: D. 2 W
R(τ) = (A²/2) cos ω₀τ, so the power R(0) = A²/2 = 4/2 = 2 W (normalised to 1 Ω).
160. A WSS process with PSD Sx(f) is passed through an LTI system with frequency response H(f). The output PSD is:
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Power scales with the square of the gain.
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Answer: A. |H(f)|² Sx(f)
For an LTI system, Sy(f) = |H(f)|² Sx(f); the phase of H(f) does not affect the output PSD.
161. If a Gaussian random process is applied to a stable LTI filter, the output process is:
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Linear combinations of Gaussians.
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Answer: A. Also Gaussian
Linear operations (weighted sums/integrals) on jointly Gaussian variables give Gaussian variables, so the output stays Gaussian.
162. Two random variables that are statistically independent are always:
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The implication works in one direction only.
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Answer: C. Uncorrelated, but uncorrelated variables need not be independent
Independence implies E[XY] = E[X]E[Y] (uncorrelated), but the converse holds in general only for jointly Gaussian variables.
163. White noise is defined as noise whose power spectral density:
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By analogy with white light.
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Answer: D. Is constant (flat) over all frequencies
White noise has a flat two-sided PSD of N0/2 W/Hz at all frequencies, by analogy with white light containing all colours equally.
164. The autocorrelation function of white noise with two-sided PSD N0/2 is:
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Inverse Fourier transform of a constant.
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Answer: A. (N0/2) δ(τ)
The inverse Fourier transform of a constant N0/2 is the impulse (N0/2)δ(τ).
165. Since the autocorrelation of white noise is an impulse at τ = 0, any two samples of white noise taken at different instants are:
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What is R(τ) for τ ≠ 0?
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Answer: C. Uncorrelated
R(τ) = 0 for every τ ≠ 0, so samples at distinct times are uncorrelated (and independent if the noise is Gaussian).
166. Ideal white noise cannot exist physically because it would have:
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Integrate a constant over all frequencies.
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Answer: B. Infinite average power
Integrating a constant PSD over all frequencies gives infinite power; real noise is white only over the bandwidth of interest.
167. White noise with two-sided PSD N0/2 = 2×10⁻¹² W/Hz passes through an ideal low-pass filter of bandwidth 1 MHz. The output noise power is:
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Integrate the two-sided PSD over −B to +B.
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Answer: C. 4 µW
P = ∫ from −B to B of N0/2 df = 2B(N0/2) = 2×10⁶ × 2×10⁻¹² = 4×10⁻⁶ W.
168. Band-limited white noise with PSD N0/2 for |f| < B has an autocorrelation function of the form:
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The transform of a rectangle is a sinc.
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Answer: B. N0B sinc(2Bτ)
The inverse Fourier transform of a rectangular PSD of height N0/2 and width 2B is N0B sinc(2Bτ), with sinc(x) = sin(πx)/(πx).
169. For band-limited white noise of bandwidth B = 1 MHz, the autocorrelation first becomes zero at a lag of:
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Zeros of sinc(2Bτ) occur at τ = k/(2B).
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Answer: D. 0.5 µs
R(τ) = N0B sinc(2Bτ) is zero at τ = k/(2B); the first zero is at 1/(2 × 10⁶) = 0.5 µs.
170. Samples of band-limited white noise (bandwidth B) taken at the Nyquist rate 2B are:
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Where do the zeros of the sinc fall?
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Answer: D. Uncorrelated with one another
Samples spaced 1/(2B) apart fall on the zeros of N0B sinc(2Bτ), so they are uncorrelated.
171. White noise of two-sided PSD N0/2 is applied to an RC low-pass filter. The average output noise power is:
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Integrate |H(f)|² N0/2 over all frequencies.
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Answer: A. N0/(4RC)
Sy(f) = (N0/2)/(1 + (2πfRC)²); integrating over all f gives (N0/2)(1/(2RC)) = N0/(4RC), which is finite because the filter limits the bandwidth.
172. The mean-square open-circuit thermal noise voltage of a resistor R at absolute temperature T over bandwidth B is:
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Don't confuse it with available power.
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Answer: A. 4kTRB
Johnson-Nyquist: v̄² = 4kTRB; kTB is the available noise power and 2qIB is shot noise.
173. The RMS thermal noise voltage of a 10 kΩ resistor at 300 K over a 10 kHz bandwidth is about (k = 1.38×10⁻²³ J/K):
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Take the square root of 4kTRB.
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Answer: D. 1.29 µV
v = √(4kTRB) = √(4 × 1.38×10⁻²³ × 300 × 10⁴ × 10⁴) = √(1.66×10⁻¹²) ≈ 1.29 µV.
174. The maximum thermal noise power that a resistor can deliver to a matched load is kTB. This power:
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Divide 4kTRB by 4R.
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Answer: B. Does not depend on the value of the resistance
Available power = v̄²/(4R) = 4kTRB/(4R) = kTB, so R cancels out.
175. The available thermal noise power at T = 290 K in a 1 MHz bandwidth is about:
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Start from −174 dBm/Hz and add 10 log B.
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Answer: B. −114 dBm
kTB = 1.38×10⁻²³ × 290 × 10⁶ ≈ 4.0×10⁻¹⁵ W ≈ −114 dBm (−174 dBm/Hz + 60 dB).
176. If the absolute temperature of a resistor is increased four times (bandwidth unchanged), its RMS thermal noise voltage:
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The RMS voltage goes as the square root of T.
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Answer: B. Doubles
v_rms ∝ √T, so a fourfold increase in T doubles the RMS noise voltage.
177. Two resistors R₁ and R₂ at the same temperature are connected in series. The total mean-square thermal noise voltage equals that of:
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Uncorrelated noise adds in power, not in voltage.
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Answer: C. A single resistor of value R₁ + R₂
Independent noise sources add in mean-square: 4kTB R₁ + 4kTB R₂ = 4kTB(R₁ + R₂).
178. Thermal noise is usually modelled as white Gaussian noise because:
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Think of the central limit theorem.
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Answer: A. Its PSD is practically flat up to very high (THz-range) frequencies and it results from many independent electron motions
The PSD of thermal noise is flat up to about 10¹² Hz at room temperature, and by the central limit theorem its amplitude is Gaussian.
179. Noise whose PSD is not flat (for example, falling as 1/f) is called:
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The opposite of white.
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Answer: C. Coloured noise
Any noise with a frequency-dependent PSD is coloured; 1/f noise is often called pink noise.
180. The envelope of narrowband Gaussian noise (zero mean) follows a:
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The magnitude of two independent zero-mean Gaussians.
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Answer: D. Rayleigh distribution
Narrowband noise n(t) = nI cos ωct − nQ sin ωct has independent Gaussian nI and nQ, so the envelope √(nI² + nQ²) is Rayleigh and the phase is uniform.