Communications MCQ Set 1
1. Which of the following is not an outdoor propagation model?
a) Longley-Rice model
b) Ericson Multiple Breakpoint Model
c) Hata model
d) Okumura model
Answer: b [Reason:] Ericson multiple breakpoint model is an indoor propagation model. Longley-Rice, Hata and Okumura model are outdoor propagation models. Most of these models are based on a systematic interpretation of measurement data obtained in the service area.
2. Longley –Rice model is applicable to _________
a) Point to point communication
b) All to all communication
c) Point to multipoint communication
d) Multipoint microwave distribution sstem
Answer: a [Reason:] The Longley-Rice model is applicable to point-to-point communication systems in the frequency range from 40 MHz to 100 GHz. They are applicable for different kinds of terrain. Terrain profile may vary from a simple curved Earth profile to a highly mountainous profile.
3. Longley-Rice prediction model is also referred as _________
a) Okumura model
b) Hata model
c) ITS irregular terrain model
d) Bertoni model
Answer: c [Reason:] The Longley Rice prediction model is also referred to as ITS irregular terrain model. The model is based on electromagnetic theory and on statistical analyses of both terrain features and radio measurements. It predicts the median attenuation of a radio signal as a function of distance and the variability of the signal in time and in space.
4. The extra term for additional attenuation due to urban clutter near the receiving antenna is called _____
a) Power factor
b) Urban gain
c) Clutter factor
d) Urban factor
Answer: d [Reason:] The urban factor (UF) is derived by comparing the predictions by the original Longley –Rice model with those obtained by Okumura. It deals with radio propagation in urban areas and is relevant to mobile radio.
5. Longley Rice model’s merit is to provide corrections due to environmental factors. State whether True or False.
Answer: b [Reason:] One shortcoming of the Longley –Rice model is that it does not provide a way of determining corrections due to environmental factors. It does not consider correction factors to account for the effects of buildings and foliage. Multipath is also not considered.
6. Which method is used by Edwards and Durkin algorithm to calculate the loss associated with diffraction edges?
a) Epstein and Peterson method
b) Interpolation method
c) Knife edge diffraction method
d) Fresnel- Kirchoff method
Answer: a [Reason:] The Edwards and Durkin algorithm uses Epstein and Peterson method to calculate the loss associated with two diffraction edges. It is the sum of two attenuations. First is loss at second diffraction edge caused by first diffraction edge. And second is the loss at receiver caused by second diffraction edge.
7. Durkin’s model can read digital elevation map. State whether True or False.
Answer: a [Reason:] Durkin’s model is very attractive because it can read in a digital elevation map and perform a site specific propagation computation on the elevation data. It can produce a signal strength contour that is reported to be good within a few dB.
8. Which of the most widely used model for signal prediction in urban areas?
a) Ericsson Multiple Breakpoint Model
b) Log distance path loss model
c) Okumura model
d) Attenuation factor model
Answer: c [Reason:] Okumura’s model is one of the most widely used models for signal prediction in urban areas. This model is applicable for frequencies in the range 150 MHz to 1920 MHz (Extrapolated upto 3000 MHz).
9. Okumura model is applicable for distances of _________
a) 1 m to 10 m
b) 1 km to 100 km
c) 100 km to 1000 km
d) 10 km to 10000 km
Answer: b [Reason:] Okumura’s model is applicable for distances of 1 km to 100 km. It can be used for base station antenna heights ranging from 30 m to 1000 m. Okumura developed a set of curves giving the median attenuation relative to free space in an urban area.
10. Okumura model is considered to be complex in predicting path loss. State whether True or False.
Answer: b [Reason:] Okumura’s model is considered to be among the simplest and best in terms of accuracy in path loss prediction for mature cellular and land mobile radio system. It is very practical and has become a standard for system planning in modern land mobile system in Japan.
11. Which of the following is the major disadvantage of Okumura model?
c) Not practical
d) Slow response to rapid change in terrain
Answer: d [Reason:] The major disadvantage with the model is its slow response to rapid changes in terrain. Therefore the model is fairly good in urban and suburban areas, but not as good in rural areas. Common standard deviations between predicted and measured path loss values are 10 dB to 14 dB.
12. The Hata model is empirical formulation of which model?
a) Okumura model
b) Longley- Rice model
c) Durkin’s model
d) Walfisch and Bertoni model
Answer: a [Reason:] The Hata model is an empirical formulation of the graphical path loss data provided by Okumura. It is valid from 150 MHz to 1500 MHz. Hata presented the urban area propagation loss as a standard formulation. It supplied correct Equations for application to other situations.
13. Hata model is well suited for _________
a) Personal communication system
b) Large cell mobile radio system
c) Small cell mobile radio system
d) Every mobile radio system
Answer: b [Reason:] Hata model is well suited for large cell mobile radio systems. But it is not well suited for personal communication system (PCS) which have cells on the order of 1 km radius. Hata model does not have any path specific corrections which are available in Okumura model.
14. Which of the following considers the impact of rooftops and building?
a) Okumura model
b) Hata model
c) Walfisch and Bertoni model
d) Longley- Rice model
Answer: c [Reason:] The impact of rooftops and building height is considered by Walfisch and Bertoni model. It uses diffraction to predict average signal strength at street level. It considers path loss to be a product of three factors.
Communications MCQ Set 2
1. In ______ technique, many subscribers attempt to access a single channel in an uncoordinated manner.
a) Packet radio
b) Multiple access
d) Spread spectrum
Answer: a [Reason:] In packet radio (PR) access techniques, many subscribers attempt to access a single channel in an uncoordinated (or minimally coordinated) manner. Transmission is done by using bursts of data.
2. Packet radio uses ____ and _____ signals for perfect feedback.
a) Forward, reverse
b) ACK, NACK
c) Active, passive
d) Linear, non linear
Answer: b [Reason:] Packet radio uses ACK and NACK signals for perfect feedback. The ACK signal indicates an acknowledgment of a received burst and NACK indicates that the previous burst was not received correctly by base station.
3. Packet radio has high spectral efficiency.
Answer: b [Reason:] Packet radio multiple access is very easy to implement, but has low spectral efficiency and may induce delays. The subscriber uses a contention technique to transmit on a common channel.
4. ______ is defined as average number of messages successfully transmitted per unit time in packet radio multiple access technique.
a) Average delay
b) Figure of merit
Answer: c [Reason:] The performance of contention technique can be evaluated by the throughput (T), which is defined as the average number of messages successfully transmitted per unit time, and the average delay experienced by a typical message burst.
5. What is the time period during which the packets are susceptible to collisions with transmissions from other users?
a) Delay time
b) Latency period
c) Average delay time
d) Vulnerable period
Answer: d [Reason:] Vulnerable period is the time period during which the packets are susceptible to collisions with transmissions from other users. It is used in order to determine the throughput.
6. For a PR multiple access technique, packet transmissions occur with _____ distribution.
Answer: a [Reason:] For a PR multiple access technique, packet transmissions occur with Poisson distribution. To study packet radio protocols, it is assumed that all packets sent by all users have a constant packet length and fixed channel rate.
7. If λ is mean arrival rate and τ is the packet duration, then traffic occupancy of a packet radio network is equal to ____
Answer: a [Reason:] If λ is mean arrival rate and τ is the packet duration, then traffic occupancy or throughput of a packet radio network is equal to λτ. The unit of R is in Erlangs.
8. Pure ALOHA is _____
a) Modulation technique
b) Multiple access technique
c) Random access technique
d) Spread spectrum technique
Answer: c [Reason:] The pure ALOHA protocol is a random access protocol used for data transfer. A user accesses a channel as soon as a message is ready to be transmitted.
9. For pure ALOHA protocol, the vulnerable period is ______ the packet duration.
c) Ten times
d) Not equal
Answer: b [Reason:] For the pure ALOHA protocol, the vulnerable period is double the packet duration. After a transmission, the user waits for an acknowledgement on either the same channel or a separate feedback channel.
10. The vulnerable period of slotted ALOHA is ____ packet duration.
Answer: c [Reason:] The vulnerable period of slotted ALOHA is one packet duration. Since, partial collisions are prevented through synchronization.
11. In slotted ALOHA, time is divided into different length of time slots. State whether True or False.
Answer: b [Reason:] In slotted ALOHA, time is divided into equal time slots of length greater than the packet duration. The subscribers each have synchronized clocks and transmit a message only at the beginning of a new time slot.
12. CSMA stands for _______
a) Carrier sense multiple access
b) Code sense multiple access
c) Carrier sense modulation access
d) Carry sense multiple access
Answer: a [Reason:] CSMA stands for carrier sense multiple access. CSMA protocol is based on the fact that each terminal on the network is able to monitor the status of the channel before transmitting information.
13. _______ is the time required for a terminal to sense whether or not the channel is idle.
a) Propagation delay
b) Average delay
c) Detection delay
d) Time delay
Answer: c [Reason:] In CSMA protocols, detection delay and propagation delay are two important parameters. Detection delay is a function of the receiver hardware and is the time required for a terminal to sense whether or not the channel is idle.
14. Which of the following is based on time division multiplexing?
a) Slotted ALOHA
b) Pure ALOHA
d) Reservation ALOHA
Answer: d [Reason:] Reservation ALOHA is a packet access scheme based on time division multiplexing. In this protocol, certain packet slots are assigned with priority, and it is possible for users to reserve slots for the transmission of packets.
15. Discrete packet time technique, PRMA stands for _________
a) Packet reservation multiple access
b) Photo reflector multiple access
c) Proton reflector modulating access
d) Packet reflection multiple access
Answer: a [Reason:] PRMA stands for Packet Reservation Multiple Access. PRMA uses a discrete packet time technique similar to reservation ALOHA and combines the cyclical frame structure of TDMA in a manner that allows each TDMA time slot to carry either voice or data, where voice is the priority.
Communications MCQ Set 3
1. Intersymbol interference (ISI) leads to ________ probability of the receiver for making an error in detecting the symbols.
Answer: a [Reason:] ISI leads to increased probability of the receiver making an error in detecting a symbol. When rectangular pulses are passed through a bandlimited channel, the pulses will spread in time, and the pulse for each symbol will smear into the time intervals of succeeding symbols.
2. ISI is ________ by increasing channel bandwidth.
Answer: b [Reason:] Increasing channel bandwidth is one of the method to minimize intersymbol interference. But mobile communication systems use minimal bandwidth, thus other methods to reduce ISI are desirable.
3. Why is pulse shaping technique used?
a) To increase ISI
b) To increase spectral width of modulated signal
c) To reduce ISI
d) To reduce power spectral density
Answer: c [Reason:] Pulse shaping techniques reduces the intersymbol interference. They are also used to reduce the spectral width of the modulated digital signal.
4. Who was the first to solve the problem of ISI?
c) Graham Bell
Answer: d [Reason:] Nyquist was the first to solve the problem of ISI. He overcome the problem of ISI while keeping the transmission bandwidth low. He observed that ISI can be completely nullified if at every instant, the response due to all symbols except the current symbol is equal to zero.
5. According to Nyquist, the impulse response of the overall communication system should have ______ decay with _______ magnitude for sample values not equal to zero.
a) Fast, small
b) Slow, small
c) Slow, large
d) Fast. Large
Answer: a [Reason:] According to Nyquist, the impulse response of the overall communication system should have fast decay with small magnitude for sample values not equal to zero. If the channel is ideal then it should be possible to realize approximate shaping filters at both transmitter and receiver.
6. Raised cosine filter does not satisfy Nyquist criteria. State whether True or False.
Answer: b [Reason:] Raised cosine filter is the most popular pulse shaping filter used in mobile communication. It belongs to the class of filters that satisfy Nyquist criterion.
7. As the roll off factor in raised cosine rolloff filter ________, the occupied bandwidth ______.
a) Increases, decreases
b) Decreases, constant
c) Increases, increases
d) Decreases, increases
Answer: c [Reason:] As the rolloff factor increases, the bandwidth of the filter also increases and the time sidelobe levels decrease in adjacent symbol slots. Thus, it implies that increasing rolloff factor decreases the sensitivity to timing jitter but increases the occupied bandwidth.
8. Gaussian pulse shaping filter follows Nyquist criterion. State whether True or False.
Answer: b [Reason:] Gaussian pulse shaping filter uses non Nyquist technique. It is effective when used in conjunction with minimum shift keying (MSK) modulation, or other modulation which is well suited for power efficient nonlinear amplifiers.
9. Gaussian filter has zero crossings at adjacent symbol peaks. State whether True or False.
Answer: b [Reason:] Nyquist filters have zero crossings at adjacent symbol peaks and a truncated transfer function. Gaussian filter does not follow Nyquist criterion and has a smooth transfer function with no zero crossings.
10. Which of the following is true for a Gaussian filter?
a) Large bandwidth
b) Minimum ISI
c) High overshoot
d) Sharp cut off
Answer: d [Reason:] The Gaussian filter has a narrow absolute bandwidth, and has a sharp cut off, low overshoot and pulse area preservation properties. This makes it attractive for use in mobile communication that uses nonlinear RF amplifiers.
11. Gaussian pulse shaping filter reduces the spectral occupancy and ISI. State whether True or False.
Answer: b [Reason:] Gaussian pulse shaping does not satisfy Nyquist criterion for ISI cancellation. Thus, it reduces the spectral occupancy but there is degradation in the performance due to increased ISI.
12. Gaussian pulses are used when cost and power efficiency are major factors. State whether True or False.
Answer: a [Reason:] Gaussian pulses are used when cost and power efficiency are major factors. But the bit error rates due to ISI are deemed to be lower than what is nominally required. Thus, there is a trade-off between desired RF bandwidth and irreducible error due to ISI.
Communications MCQ Set 4
1. Quantization is a process of mapping a ________ range of amplitude of a signal into a finite set of _____ amplitudes.
a) Continuous, discrete
b) Discrete, continuous
c) Discrete, discrete
d) Continuous, continuous
Answer: a [Reason:] Quantization is a process of mapping a continuous range of amplitude of a signal into a finite set of discrete amplitudes. Quantizers are the devices that remove the irrelevancies in the signal.
2. For a n bit quantizer, number of levels is equal to _____
Answer: b [Reason:] A quantizer that uses n bits can have M = 2n discrete amplitude levels. Amplitude quantization is an important step in any speech coding process, and it determines to a great extent the overall distortion as well as bit rate necessary to represent each waveform.
3. Distortion produced by quantization is directly proportion to ________ and inversely proportional to ______
a) Number of levels, step size
b) Amplitude of the signal, step size
c) Square of step size, number of levels
d) Amplitude of the signal, number of levels
Answer: c [Reason:] The distortion introduced by any quantization operation is directly proportional to the square of the step size. It is inversely proportional to the number of levels for a given amplitude range.
4. Signal to quantization noise ratio of a PCM encoder is given by _________
a) 6.02n + α
b) 3n + 6.02α
c) n + 6.02α
d) 6.02n + α
Answer: d [Reason:] Signal to quantization noise (SQNR) ratio of a PCM encoder is related to the number of bits, n used for encoding. It follows the following relation, SQNR= 6.02n + α. Here, α = 4.77 dB for peak SQNR and α = 0 dB for average SQNR.
5. Companding technique used in US is called ______
a) μ law
b) A law
c) Hybrid companding
d) Direct companding
Answer: a [Reason:] Companding technique known as μ law is used in US. In Europe, A law companding technique is used.
6. An adaptive quantizer varies its _____ in accordance to input speech signal power.
b) Step size
Answer: b [Reason:] An adaptive quantizer varies its step size in accordance to the input speech signal power. Its characteristics shrink and expand in time like an accordion.
7. Shannon predicted that better performance can be achieved by coding one sample at a time. State whether True or False.
Answer: b [Reason:] Shannon predicted that better performance can be achieved by coding many samples at a time instead of one sample at a time.
8. ________ is a delayed decision coding technique.
a) Adaptive quantization
b) Uniform quantization
c) Vector quantization
d) Non-uniform quantization
Answer: c [Reason:] Vector (VQ) quantization is a delayed decision coding technique. It maps a group of input samples called a vector to a code book index. A code book is set up consisting of a finite set of vectors covering the entire anticipated range of values.
9. The characteristics of compressor in μ-law companding are ______
a) Continuous in nature
b) Logarithmic in nature
c) Linear in nature
d) Discrete in nature
Answer: a [Reason:] In the μ law companding, the compressor characteristic is continuous. It approximates the linear dependence on x for low input signals and a logarithmic one for high input signals.
10. What is the sequence of operations in PCM?
a) Sampling, quantizing, encoding
b) Quantizing, encoding, sampling
c) Quantizing, sampling, encoding
d) None of the above
Answer: a [Reason:] Sequence of operation in PCM is sampling, quantization and encoding. Sampling and quantizing operations transform an analogue signal to a digital signal. The quantizing and encoding operations are usually performed in the same circuit at the transmitter, which is called an Analogue to Digital Converter (ADC). At the receiver end the decoding operation converts the pulse back into an analogue voltage in a Digital to Analogue Converter (DAC).
11. In Adaptive Delta Modulation, the slope error reduces and _____
a) Quantization error decreases
b) Quantization error increases
c) Quantization error remains same
d) None of the above
Answer: b [Reason:] ADM reduces slope error, at the expense of increasing quantizing error. This error can be reduced by using a low-pass filter.
Communications MCQ Set 5
1. Which of the following is a combined linear and constant envelope technique?
Answer: a [Reason:] M-ary phase shift keying (MPSK) is a combined linear and constant envelope technique. It is a part of M-ary modulation techniques. These modern modulation techniques exploit the fact that digital baseband data may be sent by varying both the envelope and phase of a RF carrier.
2. In an M-ary signalling scheme two or more bits are grouped together to form a _______
Answer: b [Reason:] In an M-ary signalling scheme two or more bits are grouped together to form symbols. And one of the M possible signals is transmitted during each symbol period of duration Ts.
3. The number of possible signal in M-ary signalling is given by M and M=_____, where n is an integer.
Answer: c [Reason:] Two or more bits are grouped to form a symbol in M-ary modulation. And the number of possible symbols should be equal to 2n, where n is an integer.
4. M-ary signalling techniques are not sensitive to timing jitters. State whether True or False.
Answer: b [Reason:] Timing errors increase when smaller distances between signals in the constellation diagram are used. M-ary signalling techniques are attractive for use in bandlimited channel, but are limited in their applications due to sensitivity in timing jitter.
5. M-ary modulation schemes have very good power efficiency. State whether True or False.
Answer: b [Reason:] M-ary modulation schemes have poor power efficiency, but they have a better bandwidth efficiency. An 8-PSK system requires a bandwidth that is 3 times smaller than a BPSK system, whereas its BER performance is very worse since signals are packed more closely in the signal constellation.
6. In M-ary PSK, the carrier ______ takes one of M possible values.
Answer: d [Reason:] In an M-ary PSK, the carrier phase takes one of the M possible values. The possible values of phase are θi=2(i-1)/M, where i=1,2,…….,M.
7. The constellation of M-ary PSK is _______ dimensional.
b) Does not exist
Answer: c [Reason:] The constellation of M-ary PSK is two dimensional. It is due to the presence of two basis signals. And the M-ary message points are equally spaced on a circle.
8. What is the radius of the circle in M-ary PSK on which message points are equaly spaced?
Answer: a [Reason:] The M-ary message points are equally spaced on a circle of radius √Es centred at the origin. Here Es is energy per symbol. Thus, MPSK is a constant envelope signal when no pulse shaping is used.
9. As the value of M _________ the bandwidth efficiency ________
a) Increases, same.
b) Increases, decreases
c) Increases, increases
d) Decreases, same
Answer: c [Reason:] The first null bandwidth of M-ary PSK signals decrease as M increases while Rb is held constant. Therefore, as the value of M increases, the bandwidth efficiency also increases.
10. The power efficiency of the M ary PSK decreases because of the _____
a) Freely packed constellation
b) Increment of bandwidth efficiency
c) Fixed null bandwidth
d) Densely packed constellation
Answer: d [Reason:] Bandwidth efficiency increases as the value of M increases. But at the same time, increasing M implies that the constellation is more densely packed. Hence the power efficiency or noise tolerance is decreased.
11. In QAM, the amplitude is _______ and phase is _______
a) Varied, constant
b) Varied, varied
c) Constant, varied
d) Constant, constant
Answer: b [Reason:] Quadrature amplitude modulation (QAM) is obtained by allowing the amplitude to also vary with the phase. Thus, the constellation consists of square lattice of signal points.
12. M-ary QAM signal have constant energy per symbol. State whether True or False.
Answer: b [Reason:] M-ary QAM does not have constant energy per symbol. It also does not have constant distance between possible symbol states. It reasons that particular values of M-ary QAM signal will be detected with higher probability than others.
13. In comparison to M-ary PSK, M-ary QAM bandwidth efficiency is _____ and power efficiency is ________
a) Identical, superior
b) Less, superior
c) Identical, identical
d) Superior, superior
Answer: a [Reason:] The power spectrum and bandwidth efficiency of QAM modulator is identical to M-ary PSK modulation. But, in terms of power efficiency QAM is superior to M-ary PSK.
14. The bandwidth efficiency of an M-ary FSK signal ________ with ________ in M.
a) Constant, increase
b) Increases, increase
c) Decreases, increase
d) Decreases, decrease
Answer: c [Reason:] The bandwidth efficiency of an M-ary FSK signal decreases with increase in M. Therefore, unlike M-PSK signals, M-FSK signals are bandwidth inefficient.
15. Power efficiency of M-ary FSK increases, since _________
a) Constellation is densely packed
b) M signals are non-orthogonal
c) Fixed null bandwidth
d) M-signals are orthogonal
Answer: d [Reason:] In M-ary FSK , all the M signals are orthogonal and there is no crowding in the signal space. Hence, power efficiency of M-ary FSK increases with M.