McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Computer Science Engineering (CSE) .
| 301. |
Which modulation scheme is preferred for direct sequence spread spectrum process? |
| A. | bpsk |
| B. | qpsk |
| C. | bpsk & qpsk |
| D. | none of the mentioned |
| Answer» C. bpsk & qpsk | |
| Explanation: both the modulation scheme bpsk and qpsk can be used for direct sequence spread spectrum process. | |
| 302. |
The processing gain is given as modulation scheme. |
| A. | wss/r |
| B. | r/wss |
| C. | wss/2r |
| D. | r/2wss |
| Answer» A. wss/r | |
| Explanation: the processing gain is given by | |
| 303. |
The minimum spacing between consecutive hop positions gives the |
| A. | minimum number of chips necessary |
| B. | maximum number of chips necessary |
| C. | chip rate |
| D. | none of the mentioned |
| Answer» A. minimum number of chips necessary | |
| Explanation: the minimum spacing between consecutive hop positions given the minimum number of chips necessary in the frequency word. | |
| 304. |
Which system allows larger processing gain? |
| A. | direct sequence |
| B. | frequency hopping |
| C. | direct sequence & frequency hopping |
| D. | none of the mentioned |
| Answer» B. frequency hopping | |
| Explanation: frequency hopping spread spectrum system allows greater processing gain than direct sequence spread spectrum technique. | |
| 305. |
In which technique is phase coherence hard to maintain? |
| A. | direct sequence |
| B. | frequency hopping |
| C. | direct sequence & frequency hopping |
| D. | none of the mentioned |
| Answer» B. frequency hopping | |
| Explanation: in frequency hopping spread spectrum phase coherence is hard to maintain from hop to hop. | |
| 306. |
Which type of demodulator is used in the frequency hopping technique? |
| A. | coherent |
| B. | non coherent |
| C. | coherent & non coherent |
| D. | none of the mentioned |
| Answer» B. non coherent | |
| Explanation: as it is difficult to maintain phase coherence, non coherent demodulator is used. | |
| 307. |
Robustness gives the inability of a signal to withstand the impairments. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: robustness gives the ability of a signal to withstand the impairments such as noise, jamming etc. | |
| 308. |
Chips are the |
| A. | repeated symbols |
| B. | non repeated symbols |
| C. | smallest length symbols |
| D. | none of the mentioned |
| Answer» A. repeated symbols | |
| Explanation: the repeated symbols are called as chips. | |
| 309. |
Slow frequency hopping is |
| A. | several hops per modulation |
| B. | several modulations per hop |
| C. | several symbols per modulation |
| D. | none of the mentioned |
| Answer» B. several modulations per hop | |
| Explanation: slow frequency hopping is several modulation per frequency hop. | |
| 310. |
Fast frequency hopping is |
| A. | several modulations per hop |
| B. | several modulations per symbol |
| C. | several symbols per modulation |
| D. | none of the mentioned |
| Answer» D. none of the mentioned | |
| Explanation: fast frequency hopping is several frequency hops per modulation. | |
| 311. |
Which duration is shorter? |
| A. | hop duration |
| B. | symbol duration |
| C. | chip duration |
| D. | none of the mentioned |
| Answer» A. hop duration | |
| Explanation: in frequency hopping technique hop duration is shorter than the symbol duration. | |
| 312. |
Frequency division multiple access (FDMA) assigns channels to users. |
| A. | individual, individual |
| B. | many, individual |
| C. | individual, many |
| D. | many, many |
| Answer» A. individual, individual | |
| Explanation: frequency division multiple access (fdma) assigns individual channels to individual users. each user is allocated a unique frequency band or channel. these channels are assigned on demand to users who request service. | |
| 313. |
If the FDMA channel is not in use, it can be used by other users. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: if an fdma channel is not in use, it sits idle and cannot be used by other users to increase or share capacity. it is essentially a wasted resource. | |
| 314. |
The bandwidth of FDMA channel is |
| A. | wide |
| B. | narrow |
| C. | large |
| D. | zero |
| Answer» B. narrow | |
| Explanation: the bandwidth of fdma channels is relatively narrow as each channel supports only one circuit per carrier. that is, fdma is usually implemented in narrow band systems. | |
| 315. |
Which of the following is not true for FDMA systems as compared to TDMA systems? |
| A. | low complexity |
| B. | lower cell site system cost |
| C. | tight rf filtering |
| D. | narrow bandwidth |
| Answer» B. lower cell site system cost | |
| Explanation: fdma systems have higher cell site system costs as compared to tdma systems. it is due to single channel per carrier design, and the need to use costly bandpass filters to eliminate spurious radiation at the base station. | |
| 316. |
is undesired RF radiation. |
| A. | intermodulation frequency |
| B. | intermediate frequency |
| C. | instantaneous frequency |
| D. | instrumental frequency |
| Answer» A. intermodulation frequency | |
| Explanation: intermodulation (im) frequency is undesired rf radiation which can interfere with other channels in the fdma systems. the nonlinearities cause signal spreading in the frequency domain and generate im frequency. | |
| 317. |
is based on FDMA/FDD. |
| A. | gsm |
| B. | w-cdma |
| C. | cordless telephone |
| D. | amps |
| Answer» D. amps | |
| Explanation: the first us analog cellular system, the advanced mobile phone system (amps) is based on fdma/fdd. a single user occupies a single channel while the call is in progress. | |
| 318. |
In US AMPS, 416 channels are allocated to various operators with 10 kHz guard band and channel between them is 30 kHz. What is the spectrum allocation given to each operator? |
| A. | 12.5 khz |
| B. | 30 khz |
| C. | 12.5 mhz |
| D. | 30 mhz |
| Answer» C. 12.5 mhz | |
| Explanation: spectrum allocated to each cellular operator is 12.5 mhz. as bt = nbc + 2bguard; which is equal to 416*30*103+2(10*103) = 12.5 mhz. | |
| 319. |
TDMA systems transmit in a continuous way. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: tdma systems transmit data in a buffer and burst method. thus, the transmission for any user is not continuous. | |
| 320. |
are utilized to allow synchronization of the receivers between different slots and frames. |
| A. | preamble |
| B. | data |
| C. | guard bits |
| D. | trail bits |
| Answer» C. guard bits | |
| Explanation: guard times are utilized to allow synchronization of the receivers between different slots and frames. | |
| 321. |
Which of the following is not true for TDMA? |
| A. | single carrier frequency for single user |
| B. | discontinuous data transmission |
| C. | no requirement of duplexers |
| D. | high transmission rates |
| Answer» A. single carrier frequency for single user | |
| Explanation: tdma share a single carrier frequency with several users, where each user makes use of non-overlapping time slots. the number of time slots per frame depends on several factors, such as modulation technique, available bandwidth etc. | |
| 322. |
Because of transmissions in TDMA, the handoff process in |
| A. | continuous, complex |
| B. | continuous, simple |
| C. | discontinuous, complex |
| D. | discontinuous, simple |
| Answer» D. discontinuous, simple | |
| Explanation: because of discontinuous transmissions in tdma, the handoff process is much simpler for a subscriber unit, since it is able to listen for other base stations during idle time slots. | |
| 323. |
synchronization overhead is required in TDMA due to transmission. |
| A. | high, burst |
| B. | high, continuous |
| C. | low, burst |
| D. | no, burst |
| Answer» A. high, burst | |
| Explanation: high synchronization overhead is required in tdma systems because of burst transmissions. tdma transmissions are slotted, and this requires the receivers to be synchronized for each data burst. | |
| 324. |
TDMA allocates a single time per frame to different users. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: tdma has an advantage that it can allocate different numbers of time slots per frame to different users. thus, bandwidth can be supplied on demand to different users by concatenating or reassigning time slots based on priority. | |
| 325. |
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» A. efficiency | |
| Explanation: efficiency of tdma system is a measure of the percentage of transmitted data that contains information as opposed to providing overhead for the access scheme. | |
| 326. |
What is the time duration of a bit if data is transmitted at 270.833 kbps in the channel? |
| A. | 270.833 s |
| B. | 3 μs |
| C. | 3.692 μs |
| D. | 3.692 s |
| Answer» C. 3.692 μs | |
| Explanation: if data is transmitted at 270.833 kbps in the channel, the time duration of a bit will be 3.692 μs, as tb = (1/270.833 kbps) = 3.692 μs. | |
| 327. |
US digital cellular system based on CDMA was standardized as |
| A. | is-54 |
| B. | is-136 |
| C. | is-95 |
| D. | is-76 |
| Answer» C. is-95 | |
| Explanation: a us digital cellular system based on cdma was standardized as interim standard 95 (is-95). it was standardized by us telecommunication industry association (tia) and promised increased capacity. | |
| 328. |
IS-95 was not compatible with existing AMPS frequency band. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: like is-136, is-95 system was designed to be compatible with the existing us analog cellular system (amps) frequency band. hence, mobile and base stations can be economically produced for dual mode operation. | |
| 329. |
Which of the following is used by IS-95? |
| A. | dsss |
| B. | fhss |
| C. | thss |
| D. | hybrid |
| Answer» A. dsss | |
| Explanation: is-95 uses a direct sequence spread spectrum cdma system. it allows each user within a cell to use the same radio channel, and users in adjacent cell also use the same radio channel. | |
| 330. |
IS-95 uses same modulation technique for forward and reverse channel. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: is-95 uses different modulation and spreading technique for the forward and reverse links. on the forward link, the base station simultaneously transmits the user data for all mobiles in the cell by using different spreading sequence for each mobile. | |
| 331. |
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» D. 824-849 mhz | |
| Explanation: is-95 is specified for reverse link operation in the 824-849 mhz band and 869-894 mhz for the forward link. the pcs version of is-95 has also been designed for international use in the 1800-2000 mhz bands. | |
| 332. |
User data in IS-95 is spread to a channel chip rate of |
| A. | 1.2288 mchip/s |
| B. | 9.6 mchip/s |
| C. | 12.288 mchip/s |
| D. | 0.96 mchip/s |
| Answer» A. 1.2288 mchip/s | |
| Explanation: user data is spread to a channel chip rate of 1.2288 mchip/s (a total spreading | |
| 333. |
are used to resolve and combine multipath components. |
| A. | equalizer |
| B. | registers |
| C. | rake receiver |
| D. | frequency divider |
| Answer» C. rake receiver | |
| Explanation: at both the base station and the subscriber, rake receivers are used to resolve and combine multipath components, thereby reducing the degree of fading. a rake receiver exploits the multipath time delays in a channel and combines the delayed replicas of transmitted signal. | |
| 334. |
CT2 was the first generation of cordless telephones. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: ct2 was the second generation of cordless telephones introduced in great britain in 1989. it is used to provide telepoint services which allow a subscriber to use ct2 handsets at a public telepoint. | |
| 335. |
CT2 is analog version of first generation cordless telephones. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: ct2 is a digital version of the first generation, analog, cordless telephones. when compared with analog cordless phones, ct2 offers good speech quality and is more resistant to interference. | |
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