McqMate
| 101. |
In which waveform logic 1 is represented by equal amplitude alternating pulses? |
| A. | unipolar rz |
| B. | bipolar rz |
| C. | rz-ami |
| D. | manchester coding |
| Answer» C. rz-ami | |
| Explanation: in rz-ami logic 1 is represented by equal amplitude alternating pulses and logic 0 is represented by the absence of a pulse. | |
| 102. |
The signals which are obtained by encoding each quantized signal into a digital word is called as |
| A. | pam signal |
| B. | pcm signal |
| C. | fm signal |
| D. | sampling and quantization |
| Answer» B. pcm signal | |
| Explanation: pulse code modulation is the name for the class of signals which are obtained by encoding the quantized signals into a digital word. | |
| 103. |
The length of the code-word obtained by encoding quantized sample is equal to |
| A. | l=log(to the base 2)l |
| B. | l=log(to the base 10)l |
| C. | l=2log(to the base 2)l |
| D. | l=log(to the base 2)l/2 |
| Answer» A. l=log(to the base 2)l | |
| Explanation: the quantized sample which are digitally encoded into l bit value code- word. the length l can be calculated as l=log(to the base 2)l. | |
| 104. |
In PCM encoding, quantization level varies as a function of |
| A. | frequency |
| B. | amplitude |
| C. | square of frequency |
| D. | square of amplitude |
| Answer» B. amplitude | |
| Explanation: in linear pcm the quantization levels are uniform. but in normal pcm encoding the quantization level vary according to the amplitude, based of a-law of myu-law. | |
| 105. |
What is bit depth? |
| A. | number of quantization level |
| B. | interval between two quantization levels |
| C. | number of possible digital values to represent each sample |
| D. | none of the mentioned |
| Answer» C. number of possible digital values to represent each sample | |
| Explanation: one of the properties of pcm signal which determines its stream fidelity is bit depth which is the number of possible digital values that can be used to represent each sample. | |
| 106. |
Choosing a discrete value that is near but not exactly at the analog signal level leads to |
| A. | pcm error |
| B. | quantization error |
| C. | pam error |
| D. | sampling error |
| Answer» B. quantization error | |
| Explanation: one of the limitations of pcm is quantization error which occurs when we choose a discrete value at some near by value and not at the analog signal level. | |
| 107. |
In PCM the samples are dependent on |
| A. | time |
| B. | frequency |
| C. | quanization leavel |
| D. | interval between quantization level |
| Answer» A. time | |
| Explanation: the samples depend on time,an accurate clock is required for accurate reproduction. | |
| 108. |
DPCM encodes the PCM values based on |
| A. | quantization level |
| B. | difference between the current and predicted value |
| C. | interval between levels |
| D. | none of the mentioned |
| Answer» B. difference between the current and predicted value | |
| Explanation: differential pcm encodes the pcm value based on the difference between the previous sample and the present sample value. | |
| 109. |
Delta modulation uses bits per sample. |
| A. | one |
| B. | two |
| C. | four |
| D. | eight |
| Answer» A. one | |
| Explanation: delta modulation is used for analog to digital conversion and vice versa. it is a simple form of dpcm. its uses 1 bit per sample. it also depends on the difference between the current and previous sample values. | |
| 110. |
Sample resolution for LPCM bits per sample. |
| A. | 8 |
| B. | 16 |
| C. | 24 |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: common sampling resolution for lpcm are 8, 16, 20, 24 bits per sample. | |
| 111. |
Adaptive DPCM is used to |
| A. | increase bandwidth |
| B. | decrease bandwidth |
| C. | increase snr |
| D. | none of the mentioned |
| Answer» B. decrease bandwidth | |
| Explanation: adaptive dpcm is used to decrease required bandwidth for the given snr. | |
| 112. |
Vocoders analyse the speech signals at |
| A. | transmitter |
| B. | receiver |
| C. | channel |
| D. | if filter |
| Answer» A. transmitter | |
| Explanation: vocoders are a class of speech coding systems. they analyse the speech signal at the transmitter. and then transmit the parameters derived from the analysis. | |
| 113. |
Vocoders the voice at the receiver. |
| A. | analyse |
| B. | synthesize |
| C. | modulate |
| D. | evaluate |
| Answer» B. synthesize | |
| Explanation: vocoders synthesize the voice at the receiver. all vocoder systems attempt to model the speech generation process as a dynamic system and try to quantify certain physical constraints of the system. | |
| 114. |
6 CHANNEL VOCODER |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: vocoders are much more complex than the waveform coders. they can achieve very high economy in transmission bit rate but are less robust. | |
| 115. |
Which of the following is not a vocoding system? |
| A. | linear predictive coder |
| B. | channel vocoder |
| C. | waveform coder |
| D. | formant vocoder |
| Answer» C. waveform coder | |
| Explanation: waveform coder is not a vocoding system. lpc (linear predictive coding) is the most popular vocoding system. other vocoding systems are channel vocoder, formant vocoder, cepstrum vocoder etc. | |
| 116. |
Which of the following pronunciations lead to voiced sound? |
| A. | ‘f’ |
| B. | ‘s’ |
| C. | ‘sh’ |
| D. | ‘m’ |
| Answer» D. ‘m’ | |
| Explanation: voiced sounds are ‘m’, ‘n’ and ‘v’ pronounciations. they are a result of quasiperiodic vibrations of the vocal chord. | |
| 117. |
Channel vocoders are the time domain vocoders. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: channel vocoders are frequency domain vocoders. they determine the envelope of the speech signal for a number of frequency bands and then sample, encode and multiplex these samples with the encoded outputs of the other filters. | |
| 118. |
is often called the formant of the speech signal. |
| A. | pitch frequency |
| B. | voice pitch |
| C. | pole frequency |
| D. | central frequency |
| Answer» C. pole frequency | |
| Explanation: the pole frequencies correspond to the resonant frequencies of the vocal tract. they are often called the formants of the speech signal. for adult speakers, the formants are centered around 500 hz, 1500 hz, 2500 hz and 3500 hz. | |
| 119. |
Formant vocoders use large number of control signals. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: formant vocoders use fewer control signals. therefore, formant vocoders can operate at lower bit rates than the channel vocoder. instead of transmitting the power spectrum envelope, formant vocoders attempt to transmit the position of peaks of spectral envelope. | |
| 120. |
Cepstrum vocoder uses |
| A. | wavelet transform |
| B. | inverse wavelet transform |
| C. | cosine transform |
| D. | inverse fourier transform |
| Answer» D. inverse fourier transform | |
| Explanation: cepstrum vocoders use inverse fourier transform. it separates the excitation and vocal tract spectrum by fourier transforming spectrum to produce the cepstrum of the signal. | |
| 121. |
The real part of an antenna’s input impedance is due to |
| A. | swr |
| B. | radiated signal |
| C. | reflected signal |
| D. | refracted signal |
| Answer» B. radiated signal | |
| Explanation: in antenna impedance, impedance related the voltage and current at the input of the antenna. the real part of antenna impedance represents power that is either radiated away or absorbed within the antenna and the imaginary part of antenna impedance represents power that is stored in the near field of antenna. | |
| 122. |
What is the other name for half-wave dipole antenna? |
| A. | helical antenna |
| B. | isotropic antenna |
| C. | hertz antenna |
| D. | maxwell antenna |
| Answer» C. hertz antenna | |
| Explanation: the hertz antenna is also known as half wave dipole antenna. it consists of two straight collinear conductors of equal length separated by a small feeding gap. | |
| 123. |
Measured on the ground, the field strength of a horizontally polarized half wave dipole antenna is strongest |
| A. | in one direction |
| B. | in two directions |
| C. | depends on the number of elements |
| D. | depends on the shape of antenna |
| Answer» B. in two directions | |
| Explanation: as the name suggests, half wave dipole is half wavelength long. this antenna has the shortest resonant length that can be used for a resonant dipole. the field strength of a horizontally polarized half wave dipole antenna is strongest in two directions. | |
| 124. |
When an antenna radiates more energy in one direction than in other directions, it is called |
| A. | selectivity |
| B. | directivity |
| C. | active antenna |
| D. | resonance |
| Answer» B. directivity | |
| Explanation: when an antenna radiates more energy in one direction than in other directions is called directivity. an antenna that radiates equally in all directions has effectively zero directionality, and the directivity of this type of antenna should be 1 (or 0db). | |
| 125. |
What is the full form of ERP? |
| A. | effective radiated power |
| B. | effective reflected power |
| C. | equivalent radiated power |
| D. | equivalent reflected power |
| Answer» A. effective radiated power | |
| Explanation: erp stands for effective radiated power. effective radiated power (erp) is always given with respect to a certain direction. | |
| 126. |
The polarization of plane waves received from satellite is changed by |
| A. | faraday rotation |
| B. | gamma rays |
| C. | helical rotation |
| D. | distance travelled |
| Answer» A. faraday rotation | |
| Explanation: generally for satellite communication circular polarization is required. the polarization received by waves from satellite is changed by faraday rotation. | |
| 127. |
What is the input impedance to a lossless antenna, at resonance? |
| A. | infinite |
| B. | 0 |
| C. | resistive |
| D. | capacitive |
| Answer» C. resistive | |
| Explanation: in antenna impedance, impedance related the voltage and current at the input of the antenna. the real part of antenna impedance represents power that is either radiated away or absorbed within the antenna and the imaginary part of antenna impedance represents power that is stored in the near field of antenna. the input impedance of a lossless antenna is purely resistive. | |
| 128. |
TDMA stands for |
| A. | time division multiple access |
| B. | time domain multiple access |
| C. | time division mutual access |
| D. | time domain mutual access |
| Answer» A. time division multiple access | |
| Explanation: tdma stands for time division multiple access. it can be seen as a channel access method for shared-medium networks. | |
| 129. |
Which term is used when signals move from one line to another? |
| A. | path switching |
| B. | space switching |
| C. | line switching |
| D. | cross-point switching |
| Answer» B. space switching | |
| Explanation: space switching is the used term for signals moving from one line to another. | |
| 130. |
PSK stands for Pulse Shift Keying. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: psk stands for phase shift keying. it is a modulation scheme that conveys information by changing the phase of carrier. | |
| 131. |
Which term is used for moving PCM samples from one time slot to another? |
| A. | time switching |
| B. | space switching |
| C. | phase switching |
| D. | frequency switching |
| Answer» A. time switching | |
| Explanation: time switching is the used term for moving pcm samples moving from one time slot to another. | |
| 132. |
Power can be coupled into or out of a waveguide with a magnetic field probe. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: a waveguide is a line through which electromagnetic waves are passed for | |
| 133. |
What is the full form of LOS? |
| A. | level of signal |
| B. | line of sight |
| C. | loss of signal |
| D. | level of sight |
| Answer» B. line of sight | |
| Explanation: line of sight is a line between two points. it is a straight path between a transmitting antenna and a receiving antenna. | |
| 134. |
How we can define the satisfactory performance of an analog microwave system? |
| A. | carrier to noise ratio that exceeds a given value |
| B. | carrier to noise ratio that is below a given value |
| C. | an erp value that exceeds a given value |
| D. | an erp value that is below a given value |
| Answer» A. carrier to noise ratio that exceeds a given value | |
| Explanation: we can measure performance of an analog microwave system by calculating the carrier to noise ratio that exceeds a given value. it gives the signal to noise ratio. | |
| 135. |
RGB stands for |
| A. | red green brown |
| B. | red green black |
| C. | red gold blue |
| D. | red green blue |
| Answer» D. red green blue | |
| Explanation: rgb stands for red green blue. it is an additive color model in which red, green and blue light intensity and different shades are added together in various ways to reproduce a broad variety of colors. | |
| 136. |
How many lines are there in an NTSC signal? |
| A. | 1024 |
| B. | 1856 |
| C. | 625 |
| D. | 525 |
| Answer» D. 525 | |
| Explanation: ntsc stands for national television system committee. in ntsc, it is standardized fixed that it has total 525 lines. | |
| 137. |
Luminance refers to |
| A. | contrast |
| B. | diffusion |
| C. | brightness |
| D. | aperture |
| Answer» C. brightness | |
| Explanation: luminance refers to brightness. it is a photometric measure of luminous intensity per unit area of light travelling in a given direction. | |
| 138. |
Linear modulation techniques are not bandwidth efficient. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: linear modulation techniques are bandwidth efficient. they are used in wireless communication systems when there is an increasing demand to accommodate more and more users within a limited spectrum. | |
| 139. |
Which of the following is not a linear modulation technique? |
| A. | oqpsk |
| B. | π/4 qpsk |
| C. | fsk |
| D. | bpsk |
| Answer» C. fsk | |
| Explanation: oqpsk, π/4 qpsk and bpsk are the most popular linear modulation techniques. they have very good spectral efficiency. however, fsk is an non-linear modulation technique. | |
| 140. |
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» B. phase | |
| Explanation: in binary phase shift keying (bpsk), the phase of a constant amplitude carrier signal is switched between two possible values m1 and m2. these two values corresponds to binary 1 and 0 respectively. | |
| 141. |
BPSK uses non-coherent demodulator. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: bpsk uses coherent or synchronous demodulation. it requires the information about the phase and frequency of the carrier be available at the receiver. | |
| 142. |
DPSK uses coherent form of PSK. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: differential phase shift keying uses noncoherent form of phase shift keying. noncoherent form avoids the need for a coherent reference signal at the receiver. | |
| 143. |
In DPSK system, input signal is differentially encoded and then modulated using a modulator. |
| A. | amplitude |
| B. | frequency |
| C. | bpsk |
| D. | qpsk |
| Answer» C. bpsk | |
| Explanation: in dpsk system, input binary sequence is first differentially encoded and then modulated using a bpsk modulator. the differentially encoded sequence is generated | |
| 144. |
QPSK has the bandwidth efficiency of BPSK. |
| A. | twice |
| B. | same |
| C. | half |
| D. | four times |
| Answer» A. twice | |
| Explanation: quadrature phase shift keying (qpsk) has twice the bandwidth of bpsk. it is because two bits are transmitted in a single modulation symbol. the phase of the carrier takes on one of the four equally spaced values, where each value of phase corresponds to a unique pair of message bit. | |
| 145. |
QPSK provides twice the bandwidth efficiency and energy efficiency as compared to BPSK. |
| A. | twice |
| B. | half |
| C. | same |
| D. | four times |
| Answer» C. same | |
| Explanation: the bit error probability of qpsk is identical to bpsk but twice as much data can be sent in the same bandwidth. thus, when compared to bpsk, qpsk provides twice the spectral efficiency with exactly the same efficiency. | |
| 146. |
What is the full form of OQPSK? |
| A. | optical quadrature phase shift keying |
| B. | orthogonal quadrature pulse shift keying |
| C. | orthogonal quadrature phase shift keying |
| D. | offset quadrature phase shift keying |
| Answer» D. offset quadrature phase shift keying | |
| Explanation: oqpsk stands for offset quadrature phase shift keying. it is a modified form of qpsk which is less susceptible to deleterious effects and supports more efficient amplification. oqpsk is sometimes also called staggered qpsk. | |
| 147. |
The bandwidth of OQPSK is to QPSK. |
| A. | identical |
| B. | twice |
| C. | half |
| D. | four times |
| Answer» A. identical | |
| Explanation: the spectrum of an oqpsk signal is identical to that of qpsk signal. hence, both signals occupy the same bandwidth. the staggered alignment of the even and odd bit streams in oqpsk signal does not change the nature of spectrum. | |
| 148. |
QPSK signals perform better than OQPSK in the presence of phase jitter. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: oqpsk signal perform better than qpsk in the presence of phase jitter. it is due to the presence of noisy reference signal at the receiver. | |
| 149. |
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» D. envelope detection | |
| Explanation: there are various types of detection techniques used for the detection of π/4 qpsk signals. they include baseband differential detection, if differential detection and fm discriminator detection. | |
| 150. |
Which of the following is a combined linear and constant envelope technique? |
| A. | mpsk |
| B. | psk |
| C. | bpsk |
| D. | qpsk |
| Answer» A. mpsk | |
| Explanation: m-ary phase shift keying (mpsk) is a combined linear and constant envelope technique. it is a part of m-ary modulation techniques. these modern modulation techniques exploit the fact that digital baseband data may be sent by varying both the envelope and phase of an rf carrier. | |
| 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. | |
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