The theory of radio waves was originated by:
DIGITAL COMMUNICATIONSA measure of how much information can be
propagated through a communications system and is a function of
bandwidth and transmission time.system capacitybandwidthC.
information density D. information capacityThe number of symbols
occurring per second in a data transmission is called thebit
ratebaudsymbol ratetransmission speedWhat is the channel capacity
for a signal power of 200W, noise power of 10 W, and a bandwidth of
2 kHz of a digital system?8.784 kbps5.170 kbps7.510 kbps2.644 kbpsA
highly theoretical study of the efficient use of bandwidth to
propagate information through electronic communications systemsA.
information capacity B. bandwidth efficiencyC. information theoryD.
Nyquist criterionInformation capacity is a function of bandwidth
andA. transmission timeB. modulation technique usedC. bit rateD.
none of the choicesHartleys law is a relationship of information
capacity to bandwidth andA. transmission timeB. modulation
technique usedC. bit rateD. none of the choicesRefers to the rate
of change of a signal on the transmission medium after encoding and
modulation have occurredbit ratebaudinformation ratenone of the
choicesFor a standard telephone circuit with a signal power of 3000
W, a noise power of 3 W, and a bandwidth of 2.7 kHz, the Shannon
limit for information capacity is26.9 kbps46 kbps52.3 kbpsnone of
the choicesModulation used in low-performance, low-cost,
asynchronous data modemsASKFSKPSKnone of the choicesThe circuit
used to produce BPSKXOR gateXNOR gatebalanced modulatorshift
registerThe circuit used to demodulate a BPSK signalXOR gateXNOR
gatebalanced modulatorshift registerIn digital communication using
FSK, the rate of change of frequency at the input to the modulator
is called thebaudbit rateband ratenone of the choicesFor quaternary
phase shift keying (QPSK) modulation, data with carrier frequency
of 70 MHz, and an input bit rate of 10 Mbps, determine the minimum
Nyquist bandwidth.5 MHz7 MHz9 MHz3 MHzThe modulator transmits
symbols, each of which has 16 different possible states, 5000 times
per second. How much is the bit rate?5 kbps30 kbps20 kbps16 kbpsA
constellation diagram is sometimes called asignal-state space
diagramphasor diagramphase-coordinate diagrambit-space phasor
diagramBPSK is sometimes calledphase reversal keyingbiphase
modulationdibit phase modulationanswers A and BA VCO FSK modulator
is operated in the sweep mode. The peak binary modulating voltage
is 0.005 V and the deviation sensitivity is 20 kHz/V. What is the
distance between the mark and space frequencies?100 Hz200Hz50
Hznone of the choicesA GMSK transmitter has a bit rate of 20 kbps.
What is the bandwidth required?20 kHz40 kHz50 kHznone of the
choicesGAUSSIAN MINIMUM-SHIFT KEYING(GMSK)
Special case of FSK where the mark and space frequencies are
separated by half the bit rate
Uses less bandwidth than conventional FSK
A GMSK transmitter has a bit rate of 20 kbps. What is the
bandwidth required?20 kHz40 kHz50 kHznone of the choicesA system
using GMSK has a bit rate of 20 kHz. If the mark frequency is 40
kHz, what is the space frequency?30 kHz60 kHz20 kHznone of the
choicesFor an FSK modulator with a peak frequency deviation of 1
kHz and a bit rate of 42 kbps, the necessary bandwidth is23 kHz45
kHz86 kHznone of the choicesAmplitude shift keying is sometimes
calleddigital amplitude modulationbinary amplitude modulationbit
amplitude modulationanswers A and CDetermine the baud of a
transmission using amplitude shift keying with a bit rate of 10
kbps10,00020,0005,000none of the choicesA modulation technique
which varies both the amplitude and the phase of the
signal.QPSKQAMDPSKnone of the choicesAn QPSK signal can undergo how
many degrees of phase shift and still retain its integrity+/
11.25+/ 45+/ 22.5+/ 908-QAM usesdibittribitquad bitsfive bitsA
balanced modulator is a/anOP-Amplinear summerproduct modulatorall
of the choices128-QAM has a minimum bandwidth equal tofb / 4fb / 5
fb / 6fb / 7A carrier recovery circuit is not needed with ____ PSK
methodminimumquaternarydifferentialnone of the choicesThe key
circuit used in a DPSK modulator isXOR gateXNOR gatebalanced
modulatorshift registerThe key circuit used in a DPSK demodulator
isXOR gateXNOR gatebalanced modulatorshift registerDetermine the
bandwidth efficiency of QPSK modulation scheme at a transmission
rate of 10 Mbps2 bits/cycle3 bits/cycle4 bits/cycle5 bits/cycleThe
scheme used in transmission of two separate information signals
using two amplitude-modulated carriers at the same frequency but
differing in phase by 90 degrees is known asVSBQAMQPSKanswers B and
CGray code is sometimes calledminimal error codeminimum distance
codemaximum distance codeadjacent phasor codeThe ratio of the
transmission bit rate to the minimum bandwidth required for a
particular modulation scheme.bandwidth efficiencyinformation
densityspectral widthanswers A and BA system has a P(e) of 10-5.
How many errors can you expect in every 100,000 bits
transmitted?110100None of the choicesFor an 8-PSK system, operating
with an information bit rate of 24 kbps, determine the bandwidth
efficiency.2345An alternative form of digital modulation where the
binary input information is contained in the difference between two
successive signaling elementsdifferential FSKdifferential
PSKdifferential ASKnone of the choicesIf the signals are the same,
a logic 1 is generated. If they are different, a logic 0 is
generated. What does this function? XOR gateXNOR gatebalanced
modulatorshift registerIf the signals are the same, a logic 1 is
generated. If they are different, a logic 0 is generated. What does
this function? XOR gateXNOR gatebalanced modulatorshift registerThe
process of extracting a phase-coherent reference carrier from a
receiver signal.Carrier recoveryPhase recoveryCarrier
referencingAnswers A and CThe circuit used in clock recovery
circuitsXORXNORShift registerPLLA carrier recovery method which
uses two parallel tracking loops simultaneously to derive the
product of the I and Q components of the signal that drives the
VCO.squaring loopCostas loopremodulatorparallel tracking loopA
carrier recovery method which produces a loop error voltage that is
proportional to twice the phase error between the incoming signal
and the VCO signal.squaring loopCostas loopremodulatorparallel
tracking loopFor a QPSK system with carrier power = 10-12 W and a
bit rate = 60 kbps, calculate the energy per bit in dBJ-132.5-167.8
263.8None of the choicesThe process whereby the binary data are
encoded as a precise phase of the transmitted carrierphase
referencingabsolute phase encodingphase correspondencenone of the
choicesInvented Trellis Code ModulationWilliam TrellisDr.
UngerboekAlex ReevesBob WildarSometimes thought of as a magical
method of increasing transmission bit rates over communications
systems using QAM or PSK with fixed bandwidths.PCMTCMCCMQCMAn
empirical (historical) record of a systems actual bit error
performance.BERP(e)answers A and Bnone of the choicesEncoding
technique which made possible data transmission in excess of 56
kbps over a standard telephone circuit.PCMTCMCCMQCMThe parameter
used to compare two or more digital modulation systems that use
different transmission rates (bitrates), modulation schemes
(FSK,PSK, QAM), or encoding techniques (M-ary).bit error
rateC/Nprobability of errorEb/N0 ratioTCM involves the use
ofcorrelationconvolutioncompandingcomparisonThe error in delta
modulation which can be reduced by reducing the minimum step
sizeslope overloadgranular noiseoverload distortionAnswers A and
CAn encoding scheme where the step size increases with the
amplitude of the input signal.linearnonlinearmidrisemidtreadWhich
coder is most suitable for high-speed
applicationslevel-at-a-timedigit-at-a-timeword-at-a-timenone of the
choicesThe companding technique used in the United States and
JapanA-law-lawJ-law-lawA single-channel PCM system with a sample
rate of 8000 Hz and an eight-bit compressed PCM code has a line
speed of 92 kbps32 kbps8 kbps64 kbpsThe spectral power of most
speech energy concentrates at three or four peak frequencies
calledchannelsformantstrenchesdrainsDecreasing the step size is the
solution to which noise?granular noiseslope overloadoverload
distortionnone of the choicesA problem in delta PCM which occurs
when the slope of the analog signal is greater than the delta
modulator can maintain.granular noiseslope overloadoverload
distortionnone of the choicesA modulation technique in which only
the difference in the amplitude of two successive samples is
transmitted rather than the actual sample.delta PCMdifferential
PCMadaptive delta PCMnone of the choicesWhat vocoder did Homer
Dudley invent in 1928?formant vocoderchannel vocoderlinear
predictive codernone of the choicesThe noise which is inputted in
the PAM sampler in the absence of a signal.random noisethermal
noiseidle channel noisequantization noiseA method to reduce noise
in the PAM sampler in the absence of a signal.midrise quantizingB.
midtread quantizingC. nonlinear quantizingD. linear quantizingA
larger possible magnitude for quantization error in the lowest
quantization interval is a disadvantage ofmidrise
quantizingmidtread quantizingnonlinear quantizinglinear quantizingA
vocoder which extracts the most significant portions of speech
information directly from the time waveform rather than from the
frequency spectrum.formant vocoderchannel vocoderlinear predictive
coderThe line speed of a T1 carrier system1.544 Mbps3.152 Mbps6.312
Mbps44.736 MbpsThe line speed of a T2 carrier system1.544 Mbps3.152
Mbps6.312 Mbps44.736 MbpsIf a bit is active only 20% of the time,
what is its duty cycle10%20%30%50%The error that is most prevalent
in natural sampling.Aperture errorOverload
distortionAliasingAnswers A and BStorage time is also calledA/D
Conversion timeaperture timeacquisition timeanswers B and CFor a
PCM system with a maximum audio input frequency of 5 kHz, what is
the minimum sample rate?5 kHz7 kHz10 kHz15 kHzThis occurs if the
magnitude of the sample exceeds the highest quantization
intervalOverload distortionSlope overloadAperture errorAnswers A
and BThe acquisition time in the sample-and-hold circuit must
beVery longVery shortEnough for analog-to-digital conversionNone of
the choicesThe circuit which converts PAM signals to parallel PCM
codes.ADCDACSample and Hold None of the choicesFor a resolution of
0.5 V, the worst Qe is0.5 V0.25 V0.4 Vnone of the
choicesMathematically, the worst-case SQR is0.512none of the
choices
