MCQOPTIONS
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MCQOPTIONS
This section includes 375 Mcqs, each offering curated multiple-choice questions to sharpen your Computer Science Engineering (CSE) knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
are used to resolve and combine multipath components. |
| A. | equalizer |
| B. | registers |
| C. | rake receiver |
| D. | frequency divider |
| Answer» D. frequency divider | |
| 2. |
IS-95 is specified for reverse link operation in band. |
| A. | 869-894 mhz |
| B. | 849-894 mhz |
| C. | 849-869 mhz |
| D. | 824-849 mhz |
| Answer» E. | |
| 3. |
of TDMA system is a measure of the percentage of transmitted data that contains information as opposed to providing overhead for the access scheme. |
| A. | efficiency |
| B. | figure of merit |
| C. | signal to noise ratio |
| D. | mean |
| Answer» B. figure of merit | |
| 4. |
synchronization overhead is required in TDMA due to transmission. |
| A. | high, burst |
| B. | high, continuous |
| C. | low, burst |
| D. | no, burst |
| Answer» B. high, continuous | |
| 5. |
Because of transmissions in TDMA, the handoff process in |
| A. | continuous, complex |
| B. | continuous, simple |
| C. | discontinuous, complex |
| D. | discontinuous, simple |
| Answer» E. | |
| 6. |
are utilized to allow synchronization of the receivers between different slots and frames. |
| A. | preamble |
| B. | data |
| C. | guard bits |
| D. | trail bits |
| Answer» D. trail bits | |
| 7. |
is based on FDMA/FDD. |
| A. | gsm |
| B. | w-cdma |
| C. | cordless telephone |
| D. | amps |
| Answer» E. | |
| 8. |
is undesired RF radiation. |
| A. | intermodulation frequency |
| B. | intermediate frequency |
| C. | instantaneous frequency |
| D. | instrumental frequency |
| Answer» B. intermediate frequency | |
| 9. |
Frequency division multiple access (FDMA) assigns channels to users. |
| A. | individual, individual |
| B. | many, individual |
| C. | individual, many |
| D. | many, many |
| Answer» B. many, individual | |
| 10. |
The processing gain is given as modulation scheme. |
| A. | wss/r |
| B. | r/wss |
| C. | wss/2r |
| D. | r/2wss |
| Answer» B. r/wss | |
| 11. |
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» D. none of the mentioned | |
| 12. |
In Viterbi’s algorithm, the selected paths are regarded as |
| A. | survivors |
| B. | defenders |
| C. | destroyers |
| D. | carriers |
| Answer» B. defenders | |
| 13. |
Pseudorandom signal predicted. |
| A. | can be |
| B. | cannot be |
| C. | maybe |
| D. | none of the mentioned |
| Answer» B. cannot be | |
| 14. |
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» C. maximum a priori | |
| 15. |
In comparison to stack algorithm, Fano’s algorithm is simpler. |
| A. | true |
| B. | false |
| Answer» C. | |
| 16. |
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» E. | |
| 17. |
Fano’s algorithm searches all the paths of trellis diagram at same time to find the most probable path. |
| A. | true |
| B. | false |
| Answer» C. | |
| 18. |
From the following given tree, what is the computed codeword for ‘c’? |
| A. | 111 |
| B. | 101 |
| C. | 110 |
| D. | 011 |
| Answer» D. 011 | |
| 19. |
From the following given tree, what is the code word for the character ‘a’? |
| A. | 011 |
| B. | 010 |
| C. | 100 |
| D. | 101 |
| Answer» B. 010 | |
| 20. |
The DPSK needs Eb/N0 than BPSK. |
| A. | 1db more |
| B. | 1db less |
| C. | 3db more |
| D. | 3db less |
| Answer» B. 1db less | |
| 21. |
Coherent PSK and non coherent orthogonal FSK have a difference of in PB. |
| A. | 1db |
| B. | 3db |
| C. | 4db |
| D. | 6db |
| Answer» D. 6db | |
| 22. |
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» E. | |
| 23. |
In precoding technique, the binary sequence is with the previous precoded bit. |
| A. | and-ed |
| B. | or-ed |
| C. | exor-ed |
| D. | added |
| Answer» D. added | |
| 24. |
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» B. difference | |
| 25. |
As the roll off factor in raised cosine rolloff filter the occupied bandwidth |
| A. | manchester |
| B. | faraday |
| C. | graham bell |
| D. | nyquist |
| Answer» E. | |
| 26. |
ISI is by increasing channel bandwidth. |
| A. | maximized |
| B. | minimized |
| C. | zero |
| D. | infinite |
| Answer» C. zero | |
| 27. |
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» D. none of the mentioned | |
| 28. |
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» D. none of the mentioned | |
| 29. |
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» D. decreases, decrease | |
| 30. |
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» B. less, superior | |
| 31. |
In QAM, the amplitude is and phase is |
| A. | varied, constant |
| B. | varied, varied |
| C. | constant, varied |
| D. | constant, constant |
| Answer» C. constant, varied | |
| 32. |
As the value of M the bandwidth efficiency |
| A. | increases, same. |
| B. | increases, decreases |
| C. | increases, increases |
| D. | decreases, same |
| Answer» D. decreases, same | |
| 33. |
Which of the following is not a detection technique used for detection of π/4 QPSK signals? |
| A. | baseband differential detection |
| B. | if differential detection |
| C. | fm discriminator detection |
| D. | envelope detection |
| Answer» E. | |
| 34. |
The bandwidth of OQPSK is to QPSK. |
| A. | identical |
| B. | twice |
| C. | half |
| D. | four times |
| Answer» B. twice | |
| 35. |
QPSK provides twice the bandwidth efficiency and energy efficiency as compared to BPSK. |
| A. | twice |
| B. | half |
| C. | same |
| D. | four times |
| Answer» D. four times | |
| 36. |
QPSK has the bandwidth efficiency of BPSK. |
| A. | twice |
| B. | same |
| C. | half |
| D. | four times |
| Answer» B. same | |
| 37. |
In DPSK system, input signal is differentially encoded and then modulated using a modulator. |
| A. | amplitude |
| B. | frequency |
| C. | bpsk |
| D. | qpsk |
| Answer» D. qpsk | |
| 38. |
In BPSK, the of constant amplitude carrier signal is switched between two values according to the two possible values. |
| A. | amplitude |
| B. | phase |
| C. | frequency |
| D. | angle |
| Answer» C. frequency | |
| 39. |
The real part of an antenna’s input impedance is due to |
| A. | swr |
| B. | radiated signal |
| C. | reflected signal |
| D. | refracted signal |
| Answer» C. reflected signal | |
| 40. |
is often called the formant of the speech signal. |
| A. | pitch frequency |
| B. | voice pitch |
| C. | pole frequency |
| D. | central frequency |
| Answer» D. central frequency | |
| 41. |
Vocoders the voice at the receiver. |
| A. | analyse |
| B. | synthesize |
| C. | modulate |
| D. | evaluate |
| Answer» C. modulate | |
| 42. |
Sample resolution for LPCM bits per sample. |
| A. | 8 |
| B. | 16 |
| C. | 24 |
| D. | all of the mentioned |
| Answer» E. | |
| 43. |
Delta modulation uses bits per sample. |
| A. | one |
| B. | two |
| C. | four |
| D. | eight |
| Answer» B. two | |
| 44. |
If the channel is noiseless information conveyed is and if it is useless channel information conveyed is |
| A. | 0,0 |
| B. | 1,1 |
| C. | 0,1 |
| D. | 1,0 |
| Answer» E. | |
| 45. |
Quantization is a process. |
| A. | few to few mapping |
| B. | few to many mapping |
| C. | many to few mapping |
| D. | many to many mapping |
| Answer» D. many to many mapping | |
| 46. |
What is the frequency of the stereo sub carrier signal in FM broadcasting? |
| A. | 19 khz |
| B. | 45 khz |
| C. | 55 khz |
| D. | 38 khz |
| Answer» E. | |
| 47. |
In angle modulation, signal to noise ratio before detection is a function of |
| A. | modulation index |
| B. | input signal to noise ratio |
| C. | maximum frequency of the message |
| D. | if filter bandwidth |
| Answer» E. | |
| 48. |
PLL in FM detection stands for |
| A. | phase locked loop |
| B. | programmable logic loop |
| C. | phase locked logic |
| D. | programmable locked loop |
| Answer» B. programmable logic loop | |
| 49. |
Frequency demodulator is a frequency to amplitude converter circuit. |
| A. | true |
| B. | false |
| Answer» B. false | |
| 50. |
Frequency modulation index defines the relationship between the and bandwidth of transmitted signal. |
| A. | frequency of message signal |
| B. | amplitude of message signal |
| C. | amplitude of carrier signal |
| D. | frequency of carrier signal |
| Answer» C. amplitude of carrier signal | |