True-False Questions

Question 1

A fiber link has a larger bandwidth than twisted-pair cable. On the contrary, a twisted-pair cable has a lower attenuation coefficient per km.

Answer 1

False, twisted-pair cable has a higher attenuation coefficient per kilometer.

Question 2

BPSK and QPSK use two orthogonal basis functions, but 8-PSK uses three orthogonal basis functions.

Answer 2

False, 8-PSK also uses two orthogonal basis functions.

Question 3

OFDM is less sensitive to inter-symbol interference than single carrier communication.

Answer 3

True, the symbol duration Ts is large and ISI can be completely eliminated using a cyclic prefix of length greater than maximum delay spread.

Question 4

Assuming perfect synchronization, the symbol error probability of a 16-PSK constellation increases if the binary labeling is changed to a non-Gray labeling.

Answer 4

False, The symbol error probability does not change if the binary labeling is changed, the bit error probability does.

Question 5

Code-division multiple access (CDMA) is used in the GSM standard as a multiple access technique.

Answer 5

False, GSM standard uses FDMA and TDMA as multiple access techniques.

Question 6

You are given two communication systems, one using BPSK and the other using QPSK. If the minimum distance between the constellation points is d, both constellations have the same average symbol energy.

Answer 6

False, for BPSK it is d^2/4 and for QPSK it is d^2/2

Question 7

Constant-envelope modulations are good because they do not introduce inter-symbol interference.

Answer 7

False, they are good because such signals allow to use amplifiers in a nonlinear regime, i.e. with high efficiency.

Question 8

In a QPSK communication system with no noise, in which the phase synchronization unit stopped working, the receiver can tolerate a phase drift |theta| < 40 deg. without making any errors.

Answer 8

True, you can tolerate up to |theta| < 45 deg.

Question 9

In the case of using correlator receivers, the transmitted signal must satisfy the Nyquist criterion to avoid ISI.

Answer 9

False, to avoid ISI with correlator receivers, the signal must be T-orthogonal.

Question 10

The convolution of a T-orthogonal pulse with its matched version creates a Nyquist pulse.

Answer 10

True, The matched filter receiver works without ISI based on this.

Question 11

Two constellations with the same minimum distance have always the same symbol error probability at high SNR.

Answer 11

False, Average energy of the constellation is also important. Thus, it’s important to compare the normalized minimum distances of the constellations. Also, the number of pairs of points at minimum distance have an effect.

Question 12

For a lowpass bandwidth of B = 4 Mhz, ISI free transmission is not possible with the symbol rate Rs=1/Ts=10 Msymbol/s.

Answer 12

True, In order to have ISI free transmission the symbol rate must be at most Rs=1/TS=8Msymbol/s.

Question 13

In a QPSK modulation, doubling the bit energy increases the symbol energy by a factor of four.

Answer 13

False, The relation between symbol and bit energy is linear. Symbol energy also increase with a factor of two.

Question 14

In the Shannon’s communication model, the block channel encoder compresses the data to perform a more efficient transmission.

Answer 14

False, Channel encoder adds redundancy to the data, which helps in detecting and correcting erroneous bits.

Question 15

If the sampling rate in a communication system is doubled, the symbol rate would be doubled accordingly.

Answer 15

False, the symbol rate is set by the transmitter, whereas the sampling rate is a property of the receiver, and changes in one doesn’t lead to changes in the other.

Question 16

A digital communication system uses a phase shift keying constellation with eight points (8-PSK) over the additive white gaussian noise channel. The matched filter receiver implementation requires at least 8 filters matched to the signals {sm(t), 1≤m≤8}

Answer 16

False, the minimum number of filters needed is the dimension of the signal space, which is N=2 for 8-PSK.

Question 17

The symbol error rate for an envelope detector in an AWGN channel with equiprobable M-PSK signaling is M-1/M. (The envelope detector decides only on the amplitude/magnitude of the received signal.)

Answer 17

True, because all signals will have the same amplitude, which means that the detector can’t differentiate symbols. Hence the probability of the detector being correct is 1/M –> symbol error rate is 1-1/M which is equal to (M-1)/M.

Question 18

In binary pulse-amplitude modulation (PAM) using a rectangular pulse, the length if the pulse is reduced from 1 second to 0.25 second. To maintain the same probability of error, the amplitude of the pulse should be increased by a factor of four.

Answer 18

False, the amplitude needs to increase by a factor of two, so that the symbol energy is constant, and the same performance maintained.

Question 19

One of the challenges for the frequency shift keying (FSK) signals is phase discontinuities resulting in large spectral lobes. To avoid this, continuous phase modulation (CPM) schemes are preferred.

Answer 19

True, there are no phase discontinuities in CPM.

Question 20

In the class of orthogonal signal constellations, for a fixed average bit energy, Eb, the minimum distance in the constellation dmin decreases with increasing M.

Answer 20

False, in orthogonal signal constellations dmin=sqrt(2Eblog2(M)). Which increases with increasing M.

Question 21

Minimum-shift keying (MSK) signaling is an orthogonal signaling scheme exploiting minimum frequency separation required for both orthogonality of transmitted signals, and providing minimum probability of error in the coherent receiver.

Answer 21

False, MSK uses minimum frequency separation, delta f =1/2T for orthogonality. However, a larger frequency separation delta f = 0.715/T = 1.43/2T is required to minimize the BER in the coherent receiver.

Question 22

The nearest neighbor (or min distance) detector is the optimal detector provided that the signals are both equiprobable and have equal energy.

Answer 22

False, Nearest-neighbor detector is optimal in an AWGN channel provided that the messages are equiprobable a priori (wuuuut?) Having equal energy is not necessary.

Question 23

In CDMA, different users are allotted small portions of the available bandwidth and each user transmits only in the allotted bandwidth, without interfering with other user’s bandwidth.

Answer 23

False, in CDMA, all users transmit simultaneously over the entire available spectrum.

Question 24

For one-sided bandwidth B=5 MHz, ISI-free transmission with a symbol rate of Rs=1/Ts=10 Msymbol/s is theoretically possible using RRC pulse with alpha > 0, together with a matched filter receiver.

Answer 24

False, ISI-free transmission is theoretically possible with the maximum symbol rate Rs=1/Ts=10 Msymbol/s, but using sinc.

Question 25

In Shannon’s communication model, the block modulator serves as the interface to the communication channel and its primary purpose is to map the binary information sequence into signal waveforms.

Answer 25

True, the process of mapping a digital/binary sequence to signals for transmission over the communication channel is done in the modulator block.

Question 26

For a communication system with a fixed bit rate, two modulation formats can be chosen: 64 QAM, 8-PSK. If the objective is to lower the bandwidth, 64-QAM should be chosen.

Answer 26

True. m = 6 for 64 QAM and m=3 for 8-PSK. Since 64-QAM fit more data into each symbol less symbol have to be transmitted.

Question 27

In modern wireless communication systems, OFDM is often preferable to standard linear modulation because it occupies less bandwidth.

Answer 27

False, OFDM is preferable in modern communication since it mitigates ISI because of higher symbol time Ts. The prefix is also chosen longer as the maximum delay spread of the channel.

Question 28

Two constellations with the same number of symbols and minimum distance have same symbol error rate.

Answer 28

False, we should examine the normalized minimum distance since the average symbol energy plays a part in SER. Also K, pair at minimum distance has an effect.

Question 29

In CDMA, different users are allotted small portions of the available bandwidth and each user transmits only in the allotted bandwidth without interfering in other users bandwidth.

Answer 29

False, In CDMA each user is allotted the entire bandwidth without interfering with eachother due to coding that separates users from eachother.

Question 30

In M-ary PSK modulation schemes with gray labeling, the most likely symbol detection error results in declaration of a message which is different from the transmitted message in one bit.

Answer 30

True, since with gray labeling the neighboring symbols only differ in one bit.

Question 31

For a given constellation with equal energy signals, the higher the prior probability is, the larger the corresponding decision region will be.

Answer 31

False, only true for a MAP.

Question 32

In Gram Schmidt procedure, we need to first normalize the signals and then apply the procedure to them.

Answer 32

False, they are linearized at the end. (Phi_n = Theta_n/sqrt(E_n)).

Question 33

Rectangular QAM signal constellations have a distinct advantage of being easily generated as two PAM signals, one on the in-phase and one on the quadrature carrier.

Answer 33

True, since they are both deciding on amplitude.

Question 34

Although the frequency is switched from one to another in FSK signaling, fine-tuned oscillators may prevent the relatively large spectral side lobes .

Answer 34

False, we can use CPFSK to prevent spectral spread.

Question 35

Consider a digital communication system using BPSK modulation scheme. The thermal noise at the receiver increases because of an inappropriate heat sink. When the variance of noise goes higher than d/2, where d is the distance between symbols, the BER reaches to 0.5.

Answer 35

False, since BER = Q(d/sigma)=Q(sqrt(Dmin^2/2N0)). Dmin will be smaller than d/2 in this case, but the BER is still

Question 36

A linear modulation scheme is defined as a signaling scheme for which the sum of two lowpass equivalent signals is another lowpass signal for that specific scheme. With this definition, PSK and QAM signaling methods are not linear.

Answer 36

False, QAM signaling methods are linear.

Question 37

Suppose that we want to send data over an unreliable link which introduces both additive noise n(t) and multiplicative gain alpha which can be either 0 or 1, so that the received signal r(t) can be written as r(t)=alpha*s(t)+n(t), where s(t) is the transmitted signal. The channel state information (knowledge of alpha) is available at the transmitter before each signaling interval, but the receiver does not know alpha. The most power efficient signaling scheme is an adaptive scheme which does not send anything in the signaling intervals when alpha=0.

Answer 37

True, since when alpha = o the receiver won’t get any usefull information. Only noise.

Question 38

In a digital communication system with 8-PSK signaling, without any phase synchronization unit, the receiver can still tolerate phase drift, theta with |theta| < 45 deg. without making any errors.

Answer 38

False, 8-PSK can tolerate phase shift theta with |theta| < 22.5 deg. without making any errors, in noiseless case.

Question 39

RC. For a given Rs, ISI free transmission is only possible for…..

Answer 39

W ≥ 1+alpha/2*Ts. In addition, Symbol Rate = 1/Ts

Question 40

If a zero-mean one-dimensional constellation is shifted to the right so that the new constellation is not zero-mean anymore, for a given energy constraint, the performance of the new constellation will be identical to the first one.

Answer 40

False, since the shift increases average symbol energy, the new constellation will perform worse under the energy constraint.

Question 41

QPSK modulation with root-raised cosine pulse shaping is a constant envelope modulation.

Answer 41

False, it uses non-constant amplitude pulses for communication.

Question 42

A communication system using electromagnetic waves as a carrier is designed to transmit a receive data between earth and a planet. The planet is at a distance of 300 x 10^9m from earth. Is it possible to transmit and then receive a packet with 400 bits data in less than 15s.

Answer 42

False, the time needed for the EM-wave to travel the distance back and forth is approximatly 2000s

Question 43

Continuous phase FSK (CPFSK) signals occupy less bandwidth compared to FSK.

Answer 43

True, the sudden changes in the phase require more higher frequencies.

Question 44

Assume S to be an N-dimensional space. Any N orthogonal vectors in S form a basis for S.

Answer 44

True, they span the N-dimensional space as long as they are orthogonal and N many.

Question 45

The ML detection criterion is used to draw decision boundaries for a given channel when the symbols are equally likely. Irrespective of the channel given, the perpendicular bisector of the line joining any two symbols forms the ML decision boundary for the detection of those symbols.

Answer 45

True, a line in the middle of the two symbols make up the decision boundary in the ML detection scheme for AWGN.

Question 46

It is absolutely necessary to implement a gain control mechanism to detect symbols at a receiver when a 16-PSK modulation is used in an AWGN channel.

Answer 46

False, since the information lies in the phase, an amplitude correction isn’t absolutely needed.

Question 47

A passband signal is obtained by passing its baseband version through a passband filter.

Answer 47

False, you add a carrier by multiplying for example a sine or cosine with the wanted passband frequency.

Question 48

Root raised cosine (RRC) is a Nyquist pulse

Answer 48

False, the impulse response is not zero for all -+n*Ts. However, the combined transmit and receive filters form a raised-cosine filter which does have zero at the intervals of ±Ts (is a Nyquist pulse).

Question 49

Raised cosine is a Nyquist pulse.

Answer 49

True, a RC-pulse is zero for all +-n*Ts