McqMate
| 201. |
Coherent PSK and non coherent orthogonal FSK have a difference of in PB. |
| A. | 1db |
| B. | 3db |
| C. | 4db |
| D. | 6db |
| Answer» C. 4db | |
| Explanation: the difference of pb is approximately 4db for the best ( coherent psk ) and the worst (non coherent orthogonal fsk). | |
| 202. |
Which is easier to implement and is preferred? |
| A. | coherent system |
| B. | non coherent system |
| C. | coherent & non coherent system |
| D. | none of the mentioned |
| Answer» B. non coherent system | |
| Explanation: a non coherent system is desirable because there may be difficulty is establishing and maintaining a coherent reference. | |
| 203. |
Which is the main system consideration? |
| A. | probability of error |
| B. | system complexity |
| C. | random fading channel |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: the major system considerations are error probability, complexity and random fading channel. considering all this non coherent system is more desirable than coherent. | |
| 204. |
Self information should be |
| A. | positive |
| B. | negative |
| C. | positive & negative |
| D. | none of the mentioned |
| Answer» A. positive | |
| Explanation: self information is always non negative. | |
| 205. |
The unit of average mutual information is |
| A. | bits |
| B. | bytes |
| C. | bits per symbol |
| D. | bytes per symbol |
| Answer» A. bits | |
| Explanation: the unit of average mutual information is bits. | |
| 206. |
When probability of error during transmission is 0.5, it indicates that |
| A. | channel is very noisy |
| B. | no information is received |
| C. | channel is very noisy & no information is received |
| D. | none of the mentioned |
| Answer» C. channel is very noisy & no information is received | |
| Explanation: when probability of error during transmission is 0.5 then the channel is very noisy and thus no information is received. | |
| 207. |
The event with minimum probability has least number of bits. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: in binary huffman coding the event with maximum probability has least number of bits. | |
| 208. |
The method of converting a word to stream of bits is called as |
| A. | binary coding |
| B. | source coding |
| C. | bit coding |
| D. | cipher coding |
| Answer» B. source coding | |
| Explanation: source coding is the method of converting a word to stream of bits that is 0’s and 1’s. | |
| 209. |
When the base of the logarithm is 2, then the unit of measure of information is |
| A. | bits |
| B. | bytes |
| C. | nats |
| D. | none of the mentioned |
| Answer» A. bits | |
| Explanation: when the base of the logarithm is 2 then the unit of measure of information is bits. | |
| 210. |
When X and Y are statistically independent, then I (x,y) is |
| A. | 1 |
| B. | 0 |
| C. | ln 2 |
| D. | cannot be determined |
| Answer» B. 0 | |
| Explanation: when x and y are statistically independent the measure of information i (x,y) is 0. | |
| 211. |
The self information of random variable is |
| A. | 0 |
| B. | 1 |
| C. | infinite |
| D. | cannot be determined |
| Answer» C. infinite | |
| Explanation: the self information of a random variable is infinity. | |
| 212. |
Entropy of a random variable is |
| A. | 0 |
| B. | 1 |
| C. | infinite |
| D. | cannot be determined |
| Answer» C. infinite | |
| Explanation: entropy of a random variable is also infinity. | |
| 213. |
Which is more efficient method? |
| A. | encoding each symbol of a block |
| B. | encoding block of symbols |
| C. | encoding each symbol of a block & encoding block of symbols |
| D. | none of the mentioned |
| Answer» B. encoding block of symbols | |
| Explanation: encoding block of symbols is more efficient than encoding each symbol of a block. | |
| 214. |
Lempel-Ziv algorithm is |
| A. | variable to fixed length algorithm |
| B. | fixed to variable length algorithm |
| C. | fixed to fixed length algorithm |
| D. | variable to variable length algorithm |
| Answer» A. variable to fixed length algorithm | |
| Explanation: lempel-ziv algorithm is a variable to fixed length algorithm. | |
| 215. |
While recovering signal, which gets attenuated more? |
| A. | low frequency component |
| B. | high frequency component |
| C. | low & high frequency component |
| D. | none of the mentioned |
| Answer» B. high frequency component | |
| Explanation: high frequency components are attenuated more when compared to low frequency components while recovering the signals. | |
| 216. |
Mutual information should be |
| A. | positive |
| B. | negative |
| C. | positive & negative |
| D. | none of the mentioned |
| Answer» C. positive & negative | |
| Explanation: mutual information can also be negative. | |
| 217. |
ASCII code is a |
| A. | fixed length code |
| B. | variable length code |
| C. | fixed & variable length code |
| D. | none of the mentioned |
| Answer» A. fixed length code | |
| Explanation: ascii code is a fixed length code. it has a fixed length of 7 bits. | |
| 218. |
Which reduces the size of the data? |
| A. | source coding |
| B. | channel coding |
| C. | source & channel coding |
| D. | none of the mentioned |
| Answer» A. source coding | |
| Explanation: source coding reduces the size of the data and channel coding increases the size of the data. | |
| 219. |
In digital image coding which image must be smaller in size? |
| A. | input image |
| B. | output image |
| C. | input & output image |
| D. | none of the mentioned |
| Answer» B. output image | |
| Explanation: in digital image coding, output image must be smaller than the input image. | |
| 220. |
Which coding method uses entropy coding? |
| A. | lossless coding |
| B. | lossy coding |
| C. | lossless & lossy coding |
| D. | none of the mentioned |
| Answer» B. lossy coding | |
| Explanation: lossy source coding uses entropy coding. | |
| 221. |
Which achieves greater compression? |
| A. | lossless coding |
| B. | lossy coding |
| C. | lossless & lossy coding |
| D. | none of the mentioned |
| Answer» B. lossy coding | |
| Explanation: lossy coding achieves greater compression where as lossless coding achieves only moderate compression. | |
| 222. |
Which are uniquely decodable codes? |
| A. | fixed length codes |
| B. | variable length codes |
| C. | fixed & variable length codes |
| D. | none of the mentioned |
| Answer» A. fixed length codes | |
| Explanation: fixed length codes are uniquely decodable codes where as variable length codes may or may not be uniquely decodable. | |
| 223. |
A rate distortion function is a |
| A. | concave function |
| B. | convex function |
| C. | increasing function |
| D. | none of the mentioned |
| Answer» B. convex function | |
| Explanation: a rate distortion function is a monotone decreasing function and also a convex function. | |
| 224. |
Which of the following algorithms is the best approach for solving Huffman codes? |
| A. | exhaustive search |
| B. | greedy algorithm |
| C. | brute force algorithm |
| D. | divide and conquer algorithm |
| Answer» B. greedy algorithm | |
| Explanation: greedy algorithm is the best approach for solving the huffman codes problem since it greedily searches for an optimal solution. | |
| 225. |
How many printable characters does the ASCII character set consists of? |
| A. | 120 |
| B. | 128 |
| C. | 100 |
| D. | 98 |
| Answer» C. 100 | |
| Explanation: out of 128 characters in an ascii set, roughly, only 100 characters are printable while the rest are non-printable. | |
| 226. |
Which bit is reserved as a parity bit in an ASCII set? |
| A. | first |
| B. | seventh |
| C. | eighth |
| D. | tenth |
| Answer» C. eighth | |
| Explanation: in an ascii character set, seven bits are reserved for character representation while the eighth bit is a parity bit. | |
| 227. |
How many bits are needed for standard encoding if the size of the character set is X? |
| A. | log x |
| B. | x+1 |
| C. | 2x |
| D. | x2 |
| Answer» A. log x | |
| Explanation: if the size of the character set is x, then [log x] bits are needed for representation in a standard encoding. | |
| 228. |
The code length does not depend on the frequency of occurrence of characters. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: the code length depends on the frequency of occurrence of characters. the more frequent the character occurs, the less is the length of the code. | |
| 229. |
From the following given tree, what is the code word for the character ‘a’? |
| A. | 011 |
| B. | 010 |
| C. | 100 |
| D. | 101 |
| Answer» A. 011 | |
| Explanation: by recording the path of the node from root to leaf, the code word for character ‘a’ is found to be 011. | |
| 230. |
From the following given tree, what is the computed codeword for ‘c’? |
| A. | 111 |
| B. | 101 |
| C. | 110 |
| D. | 011 |
| Answer» C. 110 | |
| Explanation: by recording the path of the node from root to leaf, assigning left branch as 0 and right branch as 1, the codeword for c is 110. | |
| 231. |
What will be the cost of the code if character ci is at depth di and occurs at frequency fi? |
| A. | cifi |
| B. | ∫cifi |
| C. | ∑fidi |
| D. | fidi |
| Answer» C. ∑fidi | |
| Explanation: if character ci is at depth di and occurs at frequency fi, the cost of the codeword obtained is ∑fidi. | |
| 232. |
An optimal code will always be present in a full tree. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: an optimal tree will always | |
| 233. |
The type of encoding where no character code is the prefix of another character code is called? |
| A. | optimal encoding |
| B. | prefix encoding |
| C. | frequency encoding |
| D. | trie encoding |
| Answer» B. prefix encoding | |
| Explanation: even if the character codes are of different lengths, the encoding where no character code is the prefix of another character code is called prefix encoding. | |
| 234. |
What is the running time of the Huffman encoding algorithm? |
| A. | o(c) |
| B. | o(log c) |
| C. | o(c log c) |
| D. | o( n log c) |
| Answer» C. o(c log c) | |
| Explanation: if we maintain the trees in a priority queue, ordered by weight, then the running time is given by o(c log c). | |
| 235. |
What is the running time of the Huffman algorithm, if its implementation of the priority queue is done using linked lists? |
| A. | o(c) |
| B. | o(log c) |
| C. | o(c log c) |
| D. | o(c2) |
| Answer» D. o(c2) | |
| Explanation: if the implementation of the priority queue is done using linked lists, the running time of huffman algorithm is o(c2). | |
| 236. |
Notch is a |
| A. | high pass filter |
| B. | low pass filter |
| C. | band stop filter |
| D. | band pass filter |
| Answer» C. band stop filter | |
| Explanation: notch filter is a band stop filter that allows most frequencies to pass through it, except frequencies in a specific range. it is just opposite of a band-pass filter. high pass filter allows higher frequencies to pass while low pass filter allows lower frequencies to pass through it. | |
| 237. |
Sin wave is |
| A. | aperiodic signal |
| B. | periodic signal |
| C. | random signal |
| D. | deterministic signal |
| Answer» B. periodic signal | |
| Explanation: periodic signal is that which repeats itself after a regular interval. sin wave is a periodic function since it’s value can be determined at any point of time, as it repeats itself at a regular interval. aperiodic signal does not repeat itself at regular interval of time. random signals are the signals which have uncertain values at any time. while deterministic signals are the signals which are constant over a period of time. | |
| 238. |
What is the role of channel in communication system? |
| A. | acts as a medium to send message signals from transmitter to receiver |
| B. | converts one form of signal to other |
| C. | allows mixing of signals |
| D. | helps to extract original signal from incoming signal |
| Answer» A. acts as a medium to send message signals from transmitter to receiver | |
| Explanation: channel acts as a medium to transmit message signal from source transmitter to the destination receiver. | |
| 239. |
Sum of a periodic and aperiodic signal always be an aperiodic signal. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: periodic signal is a signal which repeats itself after a regular interval. while aperiodic signal does not repeat itself at regular interval of time. | |
| 240. |
Noise is added to a signal |
| A. | in the channel |
| B. | at receiving antenna |
| C. | at transmitting antenna |
| D. | during regeneration of information |
| Answer» A. in the channel | |
| Explanation: noise is an unwanted signal that gets mixed with the transmitted signal while passing through the channel. the noise interferes with the signal and provides distortion in received signal. the transmitting antenna transmits modulated message signal while the receiving antenna receives the transmitted signal. regeneration of information refers to demodulating the received signal to produce the original message signal. | |
| 241. |
Agreement between communication devices are called |
| A. | transmission medium |
| B. | channel |
| C. | protocol |
| D. | modem |
| Answer» C. protocol | |
| Explanation: protocol is a set of rules that looks after data communication, by acting as an agreement between communication devices. channel is the transmission medium or the path through which information travels. modem is a device that modulates and demodulates data. | |
| 242. |
What is the advantage of superheterodyning? |
| A. | high selectivity and sensitivity |
| B. | low bandwidth |
| C. | low adjacent channel rejection |
| D. | low fidelity |
| Answer» A. high selectivity and sensitivity | |
| Explanation: the main advantage of superheterodyning is that it provides high selectivity and sensitivity. it’s bandwidth remains same. it has high adjacent channel rejection and high fidelity. | |
| 243. |
Low frequency noise is |
| A. | flicker noise |
| B. | shot noise |
| C. | thermal noise |
| D. | partition noise |
| Answer» A. flicker noise | |
| Explanation: flicker noise is a type of electronic noise which is generated due to fluctuations in the density of carrier. it’s also known as 1/f as it’s power spectral density increases with a decrease in frequency or increase in offset from a signal. | |
| 244. |
Relationship between amplitude and frequency is represented by |
| A. | time-domain plot |
| B. | phase-domain plot |
| C. | frequency-domain plot |
| D. | amplitude-domain plot |
| Answer» C. frequency-domain plot | |
| Explanation: relationship between amplitude and frequency is represented by a frequency-domain plot. also, it represents the | |
| 245. |
A function f(x) is even, when? |
| A. | f(x) = -f(x) |
| B. | f(x) = f(-x) |
| C. | f(x) = -f(x)f(-x) |
| D. | f(x) = f(x)f(-x) |
| Answer» B. f(x) = f(-x) | |
| Explanation: geometrically a function f(x) is even, if plot of the function is symmetric over y-axis. algebraically, for any function f(x) to be even, f(x) = f(-x). | |
| 246. |
The minimum nyquist bandwidth needed for baseband transmission of Rs symbols per second is |
| A. | rs |
| B. | 2rs |
| C. | rs/2 |
| D. | rs2 |
| Answer» C. rs/2 | |
| Explanation: theoretical minimum nyquist bandwidth needed for the baseband transmission of rs symbols per second without isi is rs/2. | |
| 247. |
The capacity relationship is given by |
| A. | c = w log2 ( 1+s/n ) |
| B. | c = 2w log2 ( 1+s/n ) |
| C. | c = w log2 ( 1-s/n ) |
| D. | c = w log10 ( 1+s/n ) |
| Answer» A. c = w log2 ( 1+s/n ) | |
| Explanation: the capacity relationship from shannon-hartley capacity theorem is given by c = w log2 ( 1+s/n ). | |
| 248. |
Which parameter is called as Shannon limit? |
| A. | pb/n0 |
| B. | eb/n0 |
| C. | ebn0 |
| D. | none of the mentioned |
| Answer» B. eb/n0 | |
| Explanation: there exists a limiting value for eb/n0 below which they can be no error free communication at any information rate. this eb/n0 is called as shannon limit. | |
| 249. |
Entropy is the measure of |
| A. | amount of information at the output |
| B. | amount of information that can be transmitted |
| C. | number of error bits from total number of bits |
| D. | none of the mentioned |
| Answer» A. amount of information at the output | |
| Explanation: entropy is defined as the average amount of information per source output. | |
| 250. |
Equivocation is the |
| A. | conditional entropy |
| B. | joint entropy |
| C. | individual entropy |
| D. | none of the mentioned |
| Answer» A. conditional entropy | |
| Explanation: shannon uses a correction factor called equivocation to account for uncertainty in the received signal. it is defined as the conditional entropy of the message x given y. | |
| 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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