McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Computer Science Engineering (CSE) .
| 251. |
For a error free channel, conditional probability should be |
| A. | zero |
| B. | one |
| C. | equal to joint probability |
| D. | equal to individual probability |
| Answer» A. zero | |
| Explanation: for a error free channel | |
| 252. |
Average effective information is obtained by |
| A. | subtracting equivocation from entropy |
| B. | adding equivocation with entropy |
| C. | ratio of number of error bits by total number of bits |
| D. | none of the mentioned |
| Answer» A. subtracting equivocation from entropy | |
| Explanation: according to shannon the average effective information is obtained by subtracting the equivocation from the entropy of the source. | |
| 253. |
Turbo codes are |
| A. | forward error correction codes |
| B. | backward error correction codes |
| C. | error detection codes |
| D. | none of the mentioned |
| Answer» A. forward error correction codes | |
| Explanation: turbo codes are a class of high performance forward error correction codes. | |
| 254. |
Components used for generation of turbo codes are |
| A. | inter leavers |
| B. | punching pattern |
| C. | inter leavers & punching pattern |
| D. | none of the mentioned |
| Answer» C. inter leavers & punching pattern | |
| Explanation: there are many instances of turbo codes, using different component encoders, input/output ratios, inter leavers, punching patterns. | |
| 255. |
Decoders are connected in series. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: two elementary decoders are | |
| 256. |
The inter leaver connected between the two decoders is used to |
| A. | remove error bursts |
| B. | scatter error bursts |
| C. | add error bursts |
| D. | none of the mentioned |
| Answer» B. scatter error bursts | |
| Explanation: an inter leaver installed between the two decoders connected in series is used to scatter error bursts. | |
| 257. |
In soft decision approach what does -127 mean? |
| A. | certainly one |
| B. | certainly zero |
| C. | very likely zero |
| D. | very likely one |
| Answer» B. certainly zero | |
| Explanation: the decoder front end produces an integer for each bit in the data stream. this integer is the measure of how likely it is that the bit 0 or 1 and is called as soft bit. it ranges from -127 to 127. here -127 represents certainly zero. | |
| 258. |
In soft decision approach 100 means? |
| A. | certainly one |
| B. | certainly zero |
| C. | very likely zero |
| D. | very likely one |
| Answer» D. very likely one | |
| Explanation: the decoder front end produces an integer for each bit in the data stream. this integer is the measure of how likely it is that the bit 0 or 1 and is called as soft bit. it ranges from -127 to 127. here 100 represents very likely one. | |
| 259. |
In soft decision approach 0 represents |
| A. | certainly one |
| B. | certainly zero |
| C. | very likely zero |
| D. | could be either zero or one |
| Answer» D. could be either zero or one | |
| Explanation: the decoder front end produces an integer for each bit in the data stream. this integer is the measure of how likely it is that the bit 0 or 1 and is called as soft bit. it ranges from -127 to 127. here 0 represents ‘could be either zero or one’. | |
| 260. |
In layering, n layers provide service to |
| A. | n layer |
| B. | n-1 layer |
| C. | n+1 layer |
| D. | none of the mentioned |
| Answer» C. n+1 layer | |
| Explanation: in layering n layer provides service to n+1 layer and use the service provided by n-1 layer. | |
| 261. |
Which can be used as an intermediate device in between transmitter entity and receiver entity? |
| A. | ip router |
| B. | microwave router |
| C. | telephone switch |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: ip router, microwave router and telephone switch can be used as an intermediate device between communication of two entities. | |
| 262. |
Which has comparatively high frequency component? |
| A. | sine wave |
| B. | cosine wave |
| C. | square wave |
| D. | none of the mentioned |
| Answer» C. square wave | |
| Explanation: square wave has comparatively high frequency component in them. | |
| 263. |
Which has continuous transmission? |
| A. | asynchronous |
| B. | synchronous |
| C. | asynchronous & synchronous |
| D. | none of the mentioned |
| Answer» B. synchronous | |
| Explanation: synchronous has continuous transmission where as asynchronous have sporadic transmission. | |
| 264. |
Which requires bit transitions? |
| A. | asynchronous |
| B. | synchronous |
| C. | asynchronous & synchronous |
| D. | none of the mentioned |
| Answer» B. synchronous | |
| Explanation: synchronous transmission needs bit transition. | |
| 265. |
In synchronous transmission, receiver must stay synchronous for |
| A. | 4 bits |
| B. | 8 bits |
| C. | 9 bits |
| D. | 16 bits |
| Answer» C. 9 bits | |
| Explanation: in synchronous transmission, receiver must stay synchronous for 9 bits. | |
| 266. |
How error detection and correction is done? |
| A. | by passing it through equalizer |
| B. | by passing it through filter |
| C. | by amplifying it |
| D. | by adding redundancy bits |
| Answer» D. by adding redundancy bits | |
| Explanation: error can be detected and corrected by adding additional information that is by adding redundancy bits. | |
| 267. |
Which is more efficient? |
| A. | parity check |
| B. | cyclic redundancy check |
| C. | parity & cyclic redundancy check |
| D. | none of the mentioned |
| Answer» B. cyclic redundancy check | |
| Explanation: cyclic redundancy check is more efficient than parity check. | |
| 268. |
Which can detect two bit errors? |
| A. | parity check |
| B. | cyclic redundancy check |
| C. | parity & cyclic redundancy check |
| D. | none of the mentioned |
| Answer» B. cyclic redundancy check | |
| Explanation: crc is more powerful and it can detect various kind of errors like 2 bit errors. | |
| 269. |
CRC uses |
| A. | multiplication |
| B. | binary division |
| C. | multiplication & binary division |
| D. | none of the mentioned |
| Answer» C. multiplication & binary division | |
| Explanation: crc uses more math like multiplication and binary division. | |
| 270. |
Block codes can achieve a larger coding gain than convolution coding. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: convolution code can achieve a larger coding gain that can be achieved using a block coding with the same complexity. | |
| 271. |
Which of the following indicates the number of input bits that the current output is dependent upon? |
| A. | constraint length |
| B. | code length |
| C. | search window |
| D. | information rate |
| Answer» A. constraint length | |
| Explanation: constraint length determines the number of input data bits that the current output is dependent upon. the constraint length determines how powerful and complex the code is. | |
| 272. |
Which of the following is not a way to represent convolution code? |
| A. | state diagram |
| B. | trellis diagram |
| C. | tree diagram |
| D. | linear matrix |
| Answer» D. linear matrix | |
| Explanation: linear matrix is not a way to represent convolution code. various ways of representing convolution codes are generator matrix, generator polynomial, logic tables, state diagram, tree diagram and trellis diagram. | |
| 273. |
Which of the following is not an algorithm for decoding convolution codes? |
| A. | viterbi algorithm |
| B. | stack algorithm |
| C. | fano’s sequential coding |
| D. | ant colony optimization |
| Answer» D. ant colony optimization | |
| Explanation: there are a number of techniques for decoding convolution codes. the most important of these methods is viterbi algorithm. other decoding algorithms for convolutional codes are fano’s sequential coding, stack algorithm and feedback coding. | |
| 274. |
Fano’s algorithm searches all the paths of trellis diagram at same time to find the most probable path. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: fano’s algorithm searches for the most probable path through the trellis diagram by examining one path at a time. the error rate performance of fano’s algorithm is comparable to viterbi’s algorithm. | |
| 275. |
Which of the following is not an advantage of Fano’s algorithm in comparison to Viterbi’s algorithm? |
| A. | less storage |
| B. | large constraint length |
| C. | error rate |
| D. | small delays |
| Answer» D. small delays | |
| Explanation: in comparison to viterbi decoding, sequential decoding has a significantly larger delay. in advantage over viterbi decoding is that it requires less storage, and thus codes with larger constraint lengths can be employed. | |
| 276. |
In comparison to stack algorithm, Fano’s algorithm is simpler. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: in comparison to fano’s | |
| 277. |
Which of the following is not an error correction and detection code? |
| A. | block code |
| B. | convolutional codes |
| C. | passive codes |
| D. | turbo codes |
| Answer» C. passive codes | |
| Explanation: there are three basic types of error correction and detection codes. they are block codes, convolutional codes and turbo codes. a channel coder operates on digital message data by encoding the source information into a code sequence. | |
| 278. |
Which decoding method involves the evaluation by means of Fano’s algorithm? |
| A. | maximum likelihood decoding |
| B. | sequential decoding |
| C. | maximum a priori |
| D. | minimum mean square |
| Answer» B. sequential decoding | |
| Explanation: fano’s algorithm involves sequential decoding. it searches for the most probable path through the trellis by examining one path at a time. | |
| 279. |
In Viterbi’s algorithm, the selected paths are regarded as |
| A. | survivors |
| B. | defenders |
| C. | destroyers |
| D. | carriers |
| Answer» A. survivors | |
| Explanation: in viterbi’s algorithm, the selected paths are regarded as survivors. the path thus defined is unique and corresponds to the decoded output. | |
| 280. |
Pseudorandom signal predicted. |
| A. | can be |
| B. | cannot be |
| C. | maybe |
| D. | none of the mentioned |
| Answer» A. can be | |
| Explanation: random signals cannot be predicted whereas pseudorandom sequence can be predicted. | |
| 281. |
The properties used for pseudorandom sequence are |
| A. | balance |
| B. | run |
| C. | correlation |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: the three basic properties that can be applied for pseudorandom sequence are balance, run and correlation properties. | |
| 282. |
The shift register needs to be controlled by clock pulses. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: the shift register operation is controlled by clock pulses. | |
| 283. |
A linear feedback shift register consists of |
| A. | feedback path |
| B. | modulo 2 adder |
| C. | four stage register |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: a linear feedback shift register | |
| 284. |
If the initial pulse of 1000 is fed to shift register, after how many clock pulses does the sequence repeat? |
| A. | 15 |
| B. | 16 |
| C. | 14 |
| D. | 17 |
| Answer» A. 15 | |
| Explanation: if the initial pulse 1000 is given to shift register, the foregoing sequence repeats after 15 clock pulses. | |
| 285. |
The sequences produced by shift register depends on |
| A. | number of stages |
| B. | feedback tap connections |
| C. | initial conditions |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: the sequences produced by shift register depends on the number of stages, the feedback tap connections and initial conditions. | |
| 286. |
For maximal length sequence, the sequence repetition clock pulses p is given by |
| A. | 2n + 1 |
| B. | 2n -1 |
| C. | 2n |
| D. | none of the mentioned |
| Answer» B. 2n -1 | |
| Explanation: for maximal length sequence, produced by n stage linear feedback shift register the sequence repetition clock pulses p is given by 2n -1 . | |
| 287. |
For any cyclic shift, the auto-correlation function is equal to |
| A. | 1/p |
| B. | -1/p |
| C. | –p |
| D. | p |
| Answer» B. -1/p | |
| Explanation: for any cyclic shift the auto- correlation function is equal to -1/p. | |
| 288. |
Which method is better? |
| A. | to share same bandwidth |
| B. | to share different bandwidth |
| C. | to share same & different bandwidth |
| D. | none of the mentioned |
| Answer» B. to share different bandwidth | |
| Explanation: if the jammer noise shares the same bandwidth, the result could be destructive. | |
| 289. |
Pulse jammer consists of |
| A. | pulse modulated excess band noise |
| B. | pulse modulated band-limited noise |
| C. | pulse width modulated excess band noise |
| D. | pulse width modulated band-limited noise |
| Answer» B. pulse modulated band-limited noise | |
| Explanation: pulse jammer consists of pulse modulated band-limited noise. | |
| 290. |
Which are the design options for anti jam communicator? |
| A. | time diversity |
| B. | frequency diversity |
| C. | special discrimination |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: the design options for anti-jam communicator are time diversity, frequency diversity and special discrimination. | |
| 291. |
The ratio (J/S)reqd gives the measure of |
| A. | vulnerability to interference |
| B. | invulnerability to interference |
| C. | all of the mentioned |
| D. | none of the mentioned |
| Answer» B. invulnerability to interference | |
| Explanation: the ratio (j/s)reqd gives the measure of how invulnerable the system is to interference. | |
| 292. |
The system should have |
| A. | larger (j/s)reqd |
| B. | greater system’s noise rejection capability |
| C. | larger (j/s)reqd & greater system’s noise rejection capability |
| D. | none of the mentioned |
| Answer» C. larger (j/s)reqd & greater system’s noise rejection capability | |
| Explanation: the system will be efficient if it has greater (j/s)reqd and larger system’s noise rejection capability. | |
| 293. |
The broadband jammer jams the entire |
| A. | w |
| B. | wss |
| C. | w & wss |
| D. | none of the mentioned |
| Answer» B. wss | |
| Explanation: the broadband jammer or wide-band jammer is the one which jams the entire wss with its fixed power. | |
| 294. |
To increase error probability, the processing gain should be |
| A. | increased |
| B. | decreased |
| C. | exponentially increased |
| D. | exponentially decreased |
| Answer» A. increased | |
| Explanation: in a system, to increase the error probability the processing gain should be increased. | |
| 295. |
Which jamming method produces greater degradation? |
| A. | broadband jamming |
| B. | partial jamming |
| C. | broadband & partial jamming |
| D. | none of the mentioned |
| Answer» B. partial jamming | |
| Explanation: greater degradation is possible more with partial jamming than broadband jamming. | |
| 296. |
The jammer which monitors a communicator’s signal is known as |
| A. | frequency follower jammers |
| B. | repeat back jammers |
| C. | frequency follower & repeat back jammers |
| D. | none of the mentioned |
| Answer» C. frequency follower & repeat back jammers | |
| Explanation: the smart jammers which monitor a communicator’s signals is known as frequency follower or repeat back jammers. | |
| 297. |
DS/BPSK includes |
| A. | despreading |
| B. | demodulation |
| C. | despreading & demodulation |
| D. | none of the mentioned |
| Answer» C. despreading & demodulation | |
| Explanation: ds/bpsk is a two step precess which includes despreading and demodulation. | |
| 298. |
In direct sequence process which step is performed first? |
| A. | de-spreading |
| B. | demodulation |
| C. | despreading & demodulation |
| D. | none of the mentioned |
| Answer» A. de-spreading | |
| Explanation: in direct sequence process, de- spreading correlator is followed by a modulator. | |
| 299. |
Chip is defined as |
| A. | shortest uninterrupted waveform |
| B. | largest uninterrupted waveform |
| C. | shortest diversion |
| D. | none of the mentioned |
| Answer» A. shortest uninterrupted waveform | |
| Explanation: a chip is defined as the shortest uninterrupted waveform in the system. | |
| 300. |
Processing gain is given as |
| A. | wss/r |
| B. | rch/r |
| C. | wss/r & rch/r |
| D. | none of the mentioned |
| Answer» C. wss/r & rch/r | |
| Explanation: processing gain is given as both as the ratio of the minimum bandwidth of the data to data rate and also the by the ratio of code chip rate and data rate as minimum bandwidth is approximately equal to code chip rate. | |
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