McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Computer Science Engineering (CSE) .
| 151. |
In an M-ary signalling scheme two or more bits are grouped together to form a |
| A. | chip |
| B. | symbol |
| C. | byte |
| D. | pattern |
| Answer» B. symbol | |
| Explanation: 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. | |
| 152. |
M-ary signalling techniques are not sensitive to timing jitters. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: 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. | |
| 153. |
M-ary modulation schemes have very good power efficiency. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: 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. | |
| 154. |
What is the radius of the circle in M-ary PSK on which message points are equaly spaced? |
| A. | √es |
| B. | √eb |
| C. | eb |
| D. | es |
| Answer» A. √es | |
| Explanation: 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. | |
| 155. |
As the value of M the bandwidth efficiency |
| A. | increases, same. |
| B. | increases, decreases |
| C. | increases, increases |
| D. | decreases, same |
| Answer» C. increases, increases | |
| Explanation: 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. | |
| 156. |
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. densely packed constellation | |
| Explanation: 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. | |
| 157. |
In QAM, the amplitude is and phase is |
| A. | varied, constant |
| B. | varied, varied |
| C. | constant, varied |
| D. | constant, constant |
| Answer» B. varied, varied | |
| Explanation: 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. | |
| 158. |
M-ary QAM signal have constant energy per symbol. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: 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. | |
| 159. |
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. identical, superior | |
| Explanation: 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. | |
| 160. |
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. decreases, increase | |
| Explanation: the bandwidth efficiency of an m-ary fsk signal decreases with increase in | |
| 161. |
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. m-signals are orthogonal | |
| Explanation: 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. | |
| 162. |
The method in which the tail of one pulse smears into adjacent symbol interval is called as |
| A. | intersymbol interference |
| B. | interbit interference |
| C. | interchannel interference |
| D. | none of the mentioned |
| Answer» A. intersymbol interference | |
| Explanation: due to the effect of system filtering the received pulse can overlap on one and another. the tail of one pulse smears | |
| 163. |
If each pulse of the sequence to be detected is in shape, the pulse can be detected without ISI. |
| A. | sine |
| B. | cosine |
| C. | sinc |
| D. | none of the mentioned |
| Answer» C. sinc | |
| Explanation: the sinc shaped pulse is the ideal nyquist pulse. if each pulse in the sequence to be detected is in sinc shape the pulses can be detected without isi. | |
| 164. |
What is symbol rate packing? |
| A. | maximum possible symbol transmission rate |
| B. | maximum possible symbol receiving rate |
| C. | maximum bandwidth |
| D. | maximum isi value allowed |
| Answer» A. maximum possible symbol transmission rate | |
| Explanation: a system with bandwidth rs/2 can support a maximum transmission rate of rs without isi. thus for ideal nyquist filtering the maximum possible symbol transmission rate is called as symbol rate packing and it is equal to 2 symbols/s/hz. | |
| 165. |
A nyquist pulse is the one which can be represented by shaped pulse multiplied by another time function. |
| A. | sine |
| B. | cosine |
| C. | sinc |
| D. | none of the mentioned |
| Answer» C. sinc | |
| Explanation: a nyquist filter is one whose frequency transfer function can be represented by a rectangular function convolved with any real even symmetric frequency function and a nyquist pulse is one | |
| 166. |
Examples of nyquist filters are |
| A. | root raised cosine filter |
| B. | raised cosine filter |
| C. | root raised & raised cosine filter |
| D. | none of the mentioned |
| Answer» C. root raised & raised cosine filter | |
| Explanation: the most popular among the class of nyquist filters are raised cosine and root raised cosine filter. | |
| 167. |
The minimum nyquist bandwidth for the rectangular spectrum in raised cosine filter is |
| A. | 2t |
| B. | 1/2t |
| C. | t2 |
| D. | 2/t |
| Answer» B. 1/2t | |
| Explanation: for raised cosine spectrum the minimum nyquist bandwidth is equal to 1/2t. | |
| 168. |
Roll off factor is the fraction of |
| A. | excess bandwidth and absolute bandwidth |
| B. | excess bandwidth and minimum nyquist bandwidth |
| C. | absolute bandwidth and minimum nyquist bandwidth |
| D. | none of the mentioned |
| Answer» B. excess bandwidth and minimum nyquist bandwidth | |
| Explanation: the roll off factor is defined by a fraction of excess bandwidth and the | |
| 169. |
A pulse shaping filter should satisfy two requirements. They are |
| A. | should be realizable |
| B. | should have proper roll off factor |
| C. | should be realizable & have proper roll off factor |
| D. | none of the mentioned |
| Answer» C. should be realizable & have proper roll off factor | |
| Explanation: a pulse shaping filter should provide the desired roll off and should be realizable, that is the impulse response needs to be truncated to a finite length. | |
| 170. |
Examples of double side band signals are |
| A. | ask |
| B. | psk |
| C. | ask & psk |
| D. | none of the mentioned |
| Answer» C. ask & psk | |
| Explanation: ask and psk needs twice the transmission bandwidth of equivalent baseband signals. thus these are called as double side band signals. | |
| 171. |
ISI is by increasing channel bandwidth. |
| A. | maximized |
| B. | minimized |
| C. | zero |
| D. | infinite |
| Answer» B. minimized | |
| Explanation: 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. | |
| 172. |
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. to reduce isi | |
| Explanation: pulse shaping techniques reduces the intersymbol interference. they are also used to reduce the spectral width of the modulated digital signal. | |
| 173. |
Raised cosine filter does not satisfy Nyquist criteria. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: 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. | |
| 174. |
As the roll off factor in raised cosine rolloff filter the occupied bandwidth |
| A. | manchester |
| B. | faraday |
| C. | graham bell |
| D. | nyquist |
| Answer» D. nyquist | |
| Explanation: 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. | |
| 175. |
Gaussian pulse shaping filter follows Nyquist criterion. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: 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. | |
| 176. |
Gaussian filter has zero crossings at adjacent symbol peaks. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: 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. | |
| 177. |
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. sharp cut off | |
| Explanation: 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. | |
| 178. |
Gaussian pulse shaping filter reduces the spectral occupancy and ISI. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: gaussian pulse shaping does | |
| 179. |
Gaussian pulses are used when cost and power efficiency are major factors. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: 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. | |
| 180. |
The method in which small amount of controlled ISI is introduced into the data stream rather than trying to eliminate it completely is called as |
| A. | correlative coding |
| B. | duobinary signalling |
| C. | partial response signalling |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: the interference at the detector can be cancelled out using these methods in which some controlled amount of isi is introduced into the data stream. | |
| 181. |
From digital filter we will get the output pulse as the of the current and the previous pulse. |
| A. | summation |
| B. | difference |
| C. | product |
| D. | ratio |
| Answer» A. summation | |
| Explanation: the digital filter incorporates | |
| 182. |
In duobinary signalling method, for M-ary transmission, the number of output obtained is |
| A. | 2m |
| B. | 2m+1 |
| C. | 2m-1 |
| D. | m2 |
| Answer» C. 2m-1 | |
| Explanation: in duobinary coding, the number of output obtained for m-ary transmission is 2m-1. | |
| 183. |
The method using which the error propagation in dubinary signalling can be avoided is |
| A. | filtering |
| B. | precoding |
| C. | postcoding |
| D. | none of the mentioned |
| Answer» B. precoding | |
| Explanation: in duobinary signalling method if one error occurs it repeats everywhere through out the next steps. to avoid this precoding method can be used. | |
| 184. |
In precoding technique, the binary sequence is with the previous precoded bit. |
| A. | and-ed |
| B. | or-ed |
| C. | exor-ed |
| D. | added |
| Answer» C. exor-ed | |
| Explanation: to avoid error propogation precoding method is used. in this each bit is encoded individually without having any effect due to its prior bit or decisions. | |
| 185. |
The duobinary filter, He (f) is called as |
| A. | sine filter |
| B. | cosine filter |
| C. | raised cosine filter |
| D. | none of the mentioned |
| Answer» B. cosine filter | |
| Explanation: the transfer function is 2t cos(πft) which is called as cosine filter. | |
| 186. |
The method which has greater bandwidth efficiency is called as |
| A. | duobinary signalling |
| B. | polybinary signalling |
| C. | correlative coding |
| D. | all of the mentioned |
| Answer» B. polybinary signalling | |
| Explanation: if more than three levels are introduced in duobinary signalling technique the bandwidth efficiency increases this method is called as polybinary signalling. | |
| 187. |
In polybinary signalling method the present bit of binary sequence is algebraically added with number of previous bits. |
| A. | j |
| B. | 2j |
| C. | j+2 |
| D. | j-2 |
| Answer» D. j-2 | |
| Explanation: in polybinary signalling method the present binary digit of the sequence is formed from the modulo-2 addition of the j-2 preceding digits of the sequence and the present digit. | |
| 188. |
The primary advantage of this method is |
| A. | redistribution of spectral density |
| B. | to favor low frequencies |
| C. | redistribution of spectral density & to favor low frequencies |
| D. | none of the mentioned |
| Answer» C. redistribution of spectral density & to favor low frequencies | |
| Explanation: each bit can be independently detected in-spite of strong correlation and this provides redistribution of spectral density and also favors low frequencies. | |
| 189. |
Source encoding procedure does |
| A. | sampling |
| B. | quantization |
| C. | compression |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: source encoding includes a sampling of continuous time signals, quantization of continuous valued signals and compression of those sources. | |
| 190. |
The range of amplitude difference gives the value of |
| A. | width |
| B. | distortion |
| C. | timing jitter |
| D. | noise margin |
| Answer» B. distortion | |
| Explanation: in the eye pattern, the amplitude difference gives the value of distortion caused by isi. | |
| 191. |
As the eye opens, ISI |
| A. | increases |
| B. | decreases |
| C. | remains the same |
| D. | none of the mentioned |
| Answer» B. decreases | |
| Explanation: as the eye closes, isi increases and as the eye opens isi decreases. | |
| 192. |
The index value n, in transversal filter can be used as. |
| A. | time offset |
| B. | filter coefficient identifier |
| C. | time offset & filter coefficient identifier |
| D. | none of the mentioned |
| Answer» C. time offset & filter coefficient identifier | |
| Explanation: the index n can be used as both time offset and the filter coefficient identifier, which is the address in the filter. | |
| 193. |
The over-determined set of equations can be solved using |
| A. | zero forcing |
| B. | minimum mean square error |
| C. | zero forcing & minimum mean square error |
| D. | none of the mentioned |
| Answer» C. zero forcing & minimum mean square error | |
| Explanation: the matrix x in transversal equalizer if non square with dimensions 4n+1 and 2n+1. such equations are called as over-determined set. this can be solved by two methods called as zero forcing method and minimum mean square error method. | |
| 194. |
If the filter’s tap weight remains fixed during transmission of data, then the equalization is called as |
| A. | preset equalization |
| B. | adaptive equalization |
| C. | fixed equalization |
| D. | none of the mentioned |
| Answer» A. preset equalization | |
| Explanation: if the weight remains fixed during transmission of data then the equalization is called as preset equalization. it is a simple method which consists of setting the tap weight according to some average knowledge of the channel. | |
| 195. |
Equalization method which is done by tracking a slowly time varying channel response is |
| A. | preset equalization |
| B. | adaptive equalization |
| C. | variable equalization |
| D. | none of the mentioned |
| Answer» B. adaptive equalization | |
| Explanation: this method is implemented to perform tap weight adjustment periodically or continually. equalization is done by tracking a slowly varying channel response. | |
| 196. |
Preamble is used for |
| A. | detect start of transmission |
| B. | to set automatic gain control |
| C. | to align internal clocks |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: the receiver uses preamble for detecting the start of transmission, to set automatic gain control, and to align internal clocks and local oscillator with the received signal. | |
| 197. |
The disadvantage of preset equalizer is that |
| A. | it doesnot requires initial training pulse |
| B. | time varying channel degrades the performance of the system |
| C. | all of the mentioned |
| D. | none of the mentioned |
| Answer» B. time varying channel degrades the performance of the system | |
| Explanation: the disadvantage of preset equalization is that it requires an initial training period that must be invoked at the start of any new transmission. also time varying channel can degrade system performance due to isi, since the tap weights are fixed. | |
| 198. |
For AWGN, the noise variance is |
| A. | n0 |
| B. | n0/2 |
| C. | 2n0 |
| D. | n0/4 |
| Answer» B. n0/2 | |
| Explanation: the noise variance out of the correlator for awgn is n0/2. | |
| 199. |
A Gaussian distribution into the non linear envelope detector yields |
| A. | rayleigh distribution |
| B. | normal distribution |
| C. | poisson distribution |
| D. | binary distribution |
| Answer» A. rayleigh distribution | |
| Explanation: the two output signals of gaussian distribution yields rayleigh and rician distribution. | |
| 200. |
The DPSK needs Eb/N0 than BPSK. |
| A. | 1db more |
| B. | 1db less |
| C. | 3db more |
| D. | 3db less |
| Answer» A. 1db more | |
| Explanation: the dpsk system is easier to implement than psk and it needs 1db more eb/n0 than bpsk. | |
Done Studing? Take A Test.
Great job completing your study session! Now it's time to put your knowledge to the test. Challenge yourself, see how much you've learned, and identify areas for improvement. Don’t worry, this is all part of the journey to mastery. Ready for the next step? Take a quiz to solidify what you've just studied.