Roy Blake Electronics Communica
Chapter 1: Introduction to Communication SystemsMULTIPLE
CHOICE1.The theory of radio waves was originated by:
a. Marconic. Maxwell
b. Bell
ANS: Cd. Hertz
2. The person who sent the first radio signal across the
Atlantic ocean was:
a. Marconi c. Maxwell b. Bell d. Hertz
ANS: A
3. The transmission of radio waves was first done by:
a. Marconi c. Maxwell b. Bell d. Hertz
ANS: D
4. A complete communication system must include:
a. a transmitter and receiver
b. a transmitter, a receiver, and a channel
c. a transmitter, a receiver, and a spectrum analyzer d. a
multiplexer, a demultiplexer, and a channel
ANS: B
5. Radians per second is equal to:
a. 2 f c. the phase angleb. f 2 d. none of the aboveANS: A
6. The bandwidth required for a modulated carrier depends
on:
a. the carrier frequency c. the signal-plus-noise to noise ratio
b. the signal-to-noise ratio d. the baseband frequency range
ANS: D
7. When two or more signals share a common channel, it is
called:
a. sub-channeling
b. signal switchingc.
d.SINAD
multiplexing
ANS: D
8.TDM stands for:
a. Time-Division Multiplexingc.Time Domain Measurement
b. Two-level Digital Modulationd.none of the above
9.ANS: A
FDM stands for:
a. Fast Digital Modulation
b. Frequency Domain Measurementc. Frequency-Division
Multiplexing
d. none of the above
ANS: C
10.The wavelength of a radio signal is:
a. equal to f cb. equal to c c. the distance a wave travels in
one period
d. how far the signal can travel without distortion
ANS: C
11. Distortion is caused by:
a. creation of harmonics of baseband frequencies b. baseband
frequencies "mixing" with each other
c. shift in phase relationships between baseband frequencies
d. all of the above
ANS: D
12. The collection of sinusoidal frequencies present in a
modulated carrier is called its:
a. frequency-domain representation c. spectrum
b. Fourier series d. all of the above
ANS: D
13. The baseband bandwidth for a voice-grade (telephone) signal
is:
a. approximately 3 kHz c. at least 5 kHz
b. 20 Hz to 15,000 Hz d. none of the above
ANS: A
14. Noise in a communication system originates in:
a. the sender c. the channel
b. the receiver d. all of the above
ANS: D
15. "Man-made" noise can come from:
a. equipment that sparks c. static
b. temperature d. all of the above
ANS: A
16. Thermal noise is generated in:
a. transistors and diodes c. copper wire
b. resistors d. all of the above
ANS: D
17.Shot noise is generated in: a. transistors and diodes b.
resistorsc. d.copper wire
none of the above
ANS: A
18.The power density of "flicker" noise is:
a. the same at all frequenciesc.greater at low frequencies
b. greater at high frequenciesd.the same as "white" noise
ANS: C
19.So called "1/f" noise is also called:
a. random noisec.white noise
b. pink noised.partition noise
ANS: B
20."Pink" noise has:
a. equal power per Hertzc.constant power
b. equal power per octaved.none of the above
ANS: B
21. When two noise voltages, V1 and V2, are combined, the total
voltage VT is:
a. VT = sqrt(V1 V1 + V2 V2)c.VT = sqrt(V1 V2)
b. VT = (V1 + V2)/2d.VT = V1 + V2
ANS: A
22.Signal-to-Noise ratio is calculated as:
a. signal voltage divided by noise voltage
b. signal power divided by noise power
c. first add the signal power to the noise power, then divide by
noise power d. none of the above
ANS: B
23. SINAD is calculated as:
a. signal voltage divided by noise voltage b. signal power
divided by noise power
c. first add the signal power to the noise power, then divide by
noise power
d. none of the above
ANS: D
24. Noise Figure is a measure of:
a. how much noise is in a communications system b. how much
noise is in the channel
c. how much noise an amplifier adds to a signal
d. signal-to-noise ratio in dB ANS: C
25. The part, or parts, of a sinusoidal carrier that can be
modulated are:
a. its amplitude c. its amplitude, frequency, and direction
b. its amplitude and frequency d. its amplitude, frequency, and
phase angle
ANS: D
COMPLETION1. The telephone was invented in the year .
ANS: 1863
2. Radio signals first were sent across the Atlantic in the year
.
ANS: 1901
3. The frequency band used to modulate the carrier is called the
band.
ANS: base
4. The job of the carrier is to get the information through the
.
ANS: channel
5. The bandwidth of an unmodulated carrier is .
ANS: zero
6. The 'B' in Hartley's Law stands for .
ANS: bandwidth
7. The more information per second you send, the the bandwidth
required.
ANS: greater larger wider
8. In , you split the bandwidth of a channel into sub-channels
to carry multiple signals.
ANS: FDM
9. In , multiple signal streams take turns using the
channel.
ANS: TDM
10. VHF stands for the frequency band.
ANS: very high
11. The VHF band starts at MHz.
ANS: 30
12. The UHF band starts at MHz.
ANS: 300
13. A radio signal's is the distance it travels in one cycle of
the carrier.
ANS: wavelength
14. In free space, radio signals travel at approximately meters
per second.
ANS: 300 million
15. The equipment used to show signals in the frequency domain
is the .
ANS: spectrum analyzer
16. Mathematically, a spectrum is represented by a series.
ANS: Fourier
17. Disabling a receiver during a burst of atmospheric noise is
called .
ANS:
noise blanking blanking
18. For satellite communications, noise can be a serious
problem.
ANS: solar
19. Thermal noise is caused by the random motions of in a
conductor.
ANS: electrons
SHORT ANSWER1. Name the five elements in a block diagram of a
communications system.
ANS:
Source, Transmitter, Channel, Receiver, Destination
2. Name five types of internal noise.
ANS:
Thermal, Shot, Partition, 1/f, transit-time
3. Why is thermal noise called "white noise"?
ANS:
White light is composed of equal amounts of light at all visible
frequencies. Likewise, thermal noise has equal power density over a
wide range of frequencies.
4. What is "pink noise"?
ANS:
Light is pink when it contains more red than it does other
colors, and red is at the low end of the visible spectrum.
Likewise, pink noise has higher power density at lower
frequencies.
5. Suppose there is 30 V from one noise source that is combined
with 40 V from another noise source.
Calculate the total noise voltage.
ANS:
50 V
6. If you have 100 mV of signal and 10 mV of noise, both across
the same 100-ohm load, what is the signal- to-noise ratio in
dB?
ANS:
20 dB
7. The input to an amplifier has a signal-to-noise ratio of 100
dB and an output signal-to-noise ratio of 80 dB. Find NF, both in
dB and as a ratio.
ANS:
20 dB, NF = 100
8. A microwave receiver has a noise temperature of 145 K. Find
its noise figure.
ANS:
1.5
9. Two cascaded amplifiers each have a noise figure of 5 and a
gain of 10. Find the total NF for the pair.
ANS:
5.4
10. Explain why you could use a diode as a noise source with a
spectrum close to that of pure thermal noise.
How would you control the amount of noise generated?
ANS:
When current flows through a diode, it generates shot noise that
can be represented as a current source,
the output of which is a noise current. The equation for the
noise current is very similar to the equation for thermal noise
voltage. Since the power in the shot noise is proportional to the
diode current, controlling
the diode current controls the noise power.
Chapter 2: Radio-Frequency CircuitsMULTIPLE CHOICE1. The time it
takes a charge carrier to cross from the emitter to the collector
is called:
a. base time
b. transit timec.
d.charge time
Miller time
ANS: B
2.A real capacitor actually contains:
a. capacitance and resistance onlyc.capacitance, inductance, and
resistance
b. capacitance and inductance onlyd.reactance only
ANS: C
3.Bypass capacitors are used to:
a. remove RF from non-RF circuitsc.neutralize amplifiers
b. couple RF around an amplifierd.reduce the Miller effect
ANS: A
4.A resonant circuit is:
a. a simple form of bandpass filterc.both a and b
b. used in narrowband RF amplifiersd.none of the above
ANS: C
5.Loading down a tuned-circuit amplifier will:
a. raise the Q of the tuned circuitc."multiply" the Q
b. lower the Q of the tuned circuitd.have no effect on Q
ANS: B
6.The "Miller Effect" can:
a. cause an amplifier to oscillatec.reduce the bandwidth of an
amplifier
b. cause an amplifier to lose gaind.all of the above
ANS: D
7.The Miller Effect can be avoided by:
a. using a common-emitter amplifierc.increasing the Q of the
tuned circuit
b. using a common-base amplifierd.it cannot be avoided
ANS: B
8.In a BJT, the Miller Effect is due to:
a. inductance of collector leadc.base-to-emitter capacitance
b. collector-to-emitter capacitanced.base-to-collector
capacitance
ANS: D
9. In RF amplifiers, impedance matching is usually done
with:
a. RC coupling c. direct coupling
b. transformer coupling d. lumped reactance
ANS: B
10. Neutralization cancels unwanted feedback by:
a. adding feedback out of phase with the unwanted feedback b.
bypassing the feedback to the "neutral" or ground plane
c. decoupling it
d. none of the above
ANS: A
11. For a "frequency multiplier" to work, it requires:
a. a nonlinear circuit b. a linear amplifier
c. a signal containing harmonics
d. an input signal that is an integer multiple of the desired
frequency
ANS: A
12. A sinusoidal oscillation from an amplifier requires:
a. loop gain equal to unity
b. phase shift around loop equal to 0 degrees c. both a and b,
but at just one frequency
d. none of the above
ANS: C
13. The conditions for sinusoidal oscillation from an amplifier
are called:
a. the loop-gain criteria b. the Hartley criteriac. d.the Bode
criteria
the Barkhausen criteria
ANS: D
14.The Hartley oscillator uses:
a. a tapped inductorc.an RC time constant
b. a two-capacitor dividerd.a piezoelectric crystal
ANS: A
15.The Colpitts VFO uses:
a. a tapped inductorc.an RC time constant
b. a two-capacitor dividerd.a piezoelectric crystal
ANS: B
16.The Clapp oscillator is:
a. a modified Hartley oscillatorc.a type of crystal-controlled
oscillator
b. a modified Colpitts oscillatord.only built with FETs
ANS: B
17.A varactor is:
a. a voltage-controlled capacitorc. used in tuner circuits
b. a diode
ANS: Dd. all of the above
18.Crystal-Controlled oscillators are:
a. used for a precise frequency
b. used for very low frequency drift (parts per million)
c. made by grinding quartz to exact dimensions d. all of the
above
ANS: D
19. If two signals, Va = sin(at) and Vb = sin(bt), are fed to a
mixer, the output:
a. will contain 1 = a + b and 2 = a bb. will contain 1 = a / b
and 2 = b / ac. will contain = (a + b ) / 2 d. none of the
above
ANS: A
20. In a balanced mixer, the output:
a. contains equal (balanced) amounts of all input frequencies b.
contains the input frequencies
c. does not contain the input frequencies
d. is a linear mixture of the input signals
21.ANS: C
"VFO" stands for:
a. Voltage-Fed Oscillator
b. Variable-Frequency Oscillatorc. Varactor-Frequency
Oscillator
d. Voltage-Feedback Oscillator
ANS: B
22.A "frequency synthesizer" is:
a. a VCO phase-locked to a reference frequency
b. a VFO with selectable crystals to change frequency c. a
fixed-frequency RF generator
d. same as a mixer
ANS: A
COMPLETION1. Generally, conductor lengths in RF circuits should
be .
ANS: short
2. At UHF frequencies and above, elements must be considered as
instead of as being "lumped".
ANS: distributed
3. When one side of a double-sided pc board is used for ground,
it is called a .
ANS: ground-plane
4. Interactions between parts of an RF circuit can be reduced by
using between them.
ANS: shielding
5. In high-frequency RF circuits, the placement of wires and can
be critical.
ANS: components
6. A circuit is used to remove RF from the DC voltage bus.
ANS: decoupling
7. A capacitor is used to short unwanted RF to ground.
ANS: bypass
8. The bandwidth of a tuned-circuit amplifier depends on the of
the tuned circuit.
ANS: Q
9. A value of or more for Q is required for the approximate
tuned circuit equations to be valid.
ANS: 10
10. In a class C RF amplifier, the extracts one frequency from
all the harmonics contained in the device current (e.g. collector
current).
ANS: tuned circuit
11. Using additional feedback to compensate for "stray" feedback
is called .
ANS: neutralization
12. A Colpitts oscillator uses a voltage divider to provide
feedback.
ANS: capacitive
13. Electrically, a piezoelectric crystal has both a and a
resonant frequency.
ANS: series, parallel
14. To produce sum and difference frequencies, a mixer must be a
non- circuit.
ANS: linear
15. At some bias point, a diode or a transistor can act as a
-law mixer.
ANS: square
SHORT ANSWER1. What inductance would you use with a 47-pF
capacitor to make a tuned circuit for 10 MHz?
ANS:
5.4 H
2. What value of Q is required for a 10-MHz tuned circuit to
have a bandwidth of 100 kHz?
ANS:
100
3. A tuned-circuit amplifier with a gain of 10 is being used to
make an oscillator. What should be the value of the feedback ratio
to satisfy the Barkhausen criteria?
ANS:
0.1
4. What is the advantage of a Clapp oscillator compared to a
Colpitts oscillator?
ANS:
It is more stable because it "swamps" the device capacitance
with large value capacitors in the feedback divider.
5. If a varactor has a capacitance of 90 pF at zero volts, what
will be the capacitance at 4 volts?
ANS:
30 pF
6. An oscillator has a frequency of 100 MHz at 20C, and a tempco
of +10 ppm per degree Celsius. What will be the shift in frequency
at 70C? What percentage is that?
ANS:
50 kHz, 0.05%
7. Two sinusoidal signals, V1 and V2, are fed into an ideal
balanced mixer. V1 is a 20-MHz signal; V2 is a 5- MHz signal. What
frequencies would you expect at the output of the mixer?
ANS:
15 MHz and 25 MHz
8. Suppose the phase-locked-loop frequency synthesizer of Figure
2.39 has a reference frequency of 1 MHz and a fixed-modulus divider
of 10. What should be the value of the programmable divider to get
an output frequency of 120 MHz?
ANS:
12
Chapter 3: Amplitude ModulationMULTIPLE CHOICE:
ANS: C
7. If Va sin(at) amplitude modulates the carrier Vc sin(ct), it
will produce the frequencies:
a. c + a and c a c.b. (c + a)/2 and (c a)/2 d.c + a and 2c +
2anone of the above
ANS: A
8.At 100% modulation, the total sideband power is:
a. equal to the carrier power c.half the carrier power
b. twice the carrier power d.1.414 carrier power
ANS: C
9. If a 5-kHz signal modulates a 1-MHz carrier, the bandwidth of
the AM signal will be:
a. 5 kHz c. 1.005 MHz
b. 10 kHz d. none of the above
ANS: B
10. If an AM radio station increases its modulation index, you
would expect:
a. the audio to get louder at the receiver c. the
signal-to-noise ratio to increase b. the received RF signal to
increase d. all of the above
ANS: D
11. The modulation index can be derived from:
a. the time-domain signal c. both a and b
b. the frequency-domain signal d. none of the above
ANS: C
12. The main problem in using quadrature AM would be:
a. requires too much bandwidth c. incompatibility with ordinary
AM radios b. requires too much power d. all of the above
ANS: C
13. As compared to plain AM, SSB AM:
a. is more efficient
b. requires a more complex demodulator circuit c. requires less
bandwidth
d. all of the above
14.ANS: D
The SC in SSB SC stands for:
a. single-carrier
b. suppressed-carrierc.
d.sideband-carrier
none of the above
ANS: B
15.PEP stands for:
a. Peak Envelope Powerc.Peak Envelope Product
b. Peak Efficiency Powerd.none of the above
ANS: A
16. If an SSB transmitter radiates 1000 watts at peak
modulation, what will it radiate with no modulation?
a. 1000 watts c. 250 watts b. 500 watts d. 0 watts
ANS: D
17. Music on AM radio stations is "low-fidelity" because:
a. AM is susceptible to noise
b. commercial AM stations use low power
c. commercial AM stations have a narrow bandwidth d. all of the
above
ANS: C
18. The type of information that can be sent using AM is:
a. audio c. digital data
b. video d. all of the above
ANS: D
19. Two tones modulate an AM carrier. One tone causes a
modulation index of m1 and the other tone causes a modulation index
of m2. The total modulation index is:
a. m1 + m2 c. sqrt(m1 m2 + m2 m1)
b. (m1 + m2) / 2 d. sqrt(m1 m1 + m2 m2)ANS: D
20. To demodulate a USB SSB signal, the receiver must:
a. be set to USB mode c. both a and b
b. reinsert the carrier d. none of the above
ANS: C
COMPLETION1. An advantage of AM is that the receiver can be very
.
ANS: simple
2. A disadvantage of AM is its use of power.
ANS: inefficient
3. The of an AM signal resembles the shape of the baseband
signal.
ANS: envelope
4. In AM, modulating with a single audio tone produces
sidebands.
ANS: two
5. Compared to the USB, the information in the LSB is .
ANS: the same
6. Compared to the USB, the power in the LSB is .
ANS: the same
7. In AM, total sideband power is always than the carrier
power.
ANS: less
8. In AM, as the modulation index increases, the carrier power
.
ANS: remains constant
9. The power in an AM signal is maximum when the modulation
index is .
ANS: one
10. In AM, a voice-band signal of 300 Hz to 3000 Hz will require
a bandwidth of .
ANS: 6000 Hz
11. With a 1-MHz carrier, if the LSB extends down to 990 kHz,
then the USB will extend up to
.
ANS: 1010 kHz
12. If an AM transmitter puts out 100 watts with no modulation,
it will put out watts with 100% modulation.
ANS: 150
SHORT ANSWER1. An AM transmitter generates 100 watts with 0%
modulation. How much power will it generate with 20%
modulation?
ANS:
102 watts
2. If the carrier power is 1000 watts, what is the power in the
USB at 70.7% modulation?
ANS:
125 watts
3. A carrier is modulated by three audio tones. If the
modulation indexes for the tones are 0.3, 0.4, and 0.5, then what
is the total modulation index?
ANS:
0.707
4. You look at an AM signal with an oscilloscope and see that
the maximum Vpp is 100 volts and the minimum Vpp is 25 volts. What
is the modulation index?
ANS:
0.6
5. A SSB transmitter is connected to a 50-ohm antenna. If the
peak output voltage of the transmitter is 20 volts, what is the
PEP?
ANS:
4 watts
Chapter 4: Angle ModulationMULTIPLE CHOICE1. The FM modulation
index:
a. increases with both deviation and modulation frequency
b. increases with deviation and decreases with modulation
frequency c. decreases with deviation and increases with modulation
frequency d. is equal to twice the deviation
ANS: B
2. One way to derive FM from PM is:
a. integrate the modulating signal before applying to the PM
oscillator b. integrate the signal out of the PM oscillator
c. differentiate the modulating signal before applying to the PM
oscillator d. differentiate the signal out of the PM oscillator
ANS: A
3. The bandwidth of an FM signal is considered to be limited
because:
a. there can only be a finite number of sidebands b. it is equal
to the frequency deviation
c. it is band-limited at the receiver
d. the power in the outer sidebands is negligible
ANS: D
4. Mathematically, the calculation of FM bandwidth requires the
use of:
a. ordinary trigonometry and algebra
b. Bessel functionsc.
d.Taylor series
fractals
ANS: B
5.FM bandwidth can be approximated by:
a. Armstrong's Rulec.Carson's Rule
b. Bessel's Ruled.none of the above
ANS: C
6.NBFM stands for:
a. National Broadcast FMc.Near Band FM
b. Non-Broadcast FMd.Narrowband FM
ANS: D
7. When FM reception deteriorates abruptly due to noise, it is
called: a. the capture effect c. the noise effect b. the threshold
effect d. the limit effect
ANS: B
8. An FM receiver switching suddenly between two stations on
nearby frequencies is called:
a. the capture effect c. the "two-station" effect b. the
threshold effect d. none of the above
ANS: A
9. Pre-emphasis is used to:
a. increase the signal to noise ratio for higher audio
frequencies b. increase the signal to noise ratio for lower audio
frequencies
c. increase the signal to noise ratio for all audio
frequencies
d. allow stereo audio to be carried by FM stations
ANS: A
10. A pre-emphasis of 75 s refers to:
a. the time it takes for the circuit to work
b. the "dead time" before de-emphasis occurs
c. the time delay between the L and R channels d. the
time-constant of the filter circuits used
11.ANS: D
FM stereo:
a. uses DSBSC AM modulation
b. is implemented using an SCA signalc.
d.has a higher S/N than mono FM
is not compatible with mono FM
ANS: A
12.An SCA signal:
a. can use amplitude modulationc.is monaural
b. can use FM modulationd.all of the above
ANS: D
13. The modulation index of an FM signal can be determined
readily:
a. using measurements at points where J0 equals one b. using
measurements at points where J0 equals zero
c. using measurements at points where the deviation equals
zero
d. only by using Bessel functions
ANS: B
COMPLETION1. FM and PM are two forms of modulation.
ANS: angle
2. PM is extensively used in communication.
ANS: data
3. Compared to AM, the signal-to-noise ratio of FM is usually
.
ANS: better
4. Compared to AM, the bandwidth of FM is usually .
ANS: wider greater
5. FM transmitters can use Class amplifiers since amplitude
linearity is not important.
ANS: C
6. Both the power and amplitude of an FM signal as modulation is
applied.
ANS: stay constant
7. In FM, the frequency deviation is proportional to the
instantaneous of the modulating signal.
ANS: amplitude
8. The frequency deviation of an FM signal occurs at a rate
equal to the of the modulating signal.
ANS: frequency
9. Mathematically, the number of sidebands in an FM signal is
.
ANS: infinite
10. As FM sidebands get farther from the center frequency, their
power .
ANS: decreases
11. Mathematically, the value of an FM modulation index can be
as high as .
ANS: any number
12. In FM, as the modulating frequency decreases, the modulation
index .
ANS: increases
13. In FM, as the frequency deviation decreases, the modulation
index .
ANS: decreases
14. As the FM modulation index increases, the number of
significant sidebands .
ANS: increases
15. For certain values of mf, such as 2.4, the amplitude of the
carrier frequency .
ANS: disappears goes to zero
16. The bandwidth of an FM signal can be approximated using
rule.
ANS: Carson's
17. FM bandwidth can be calculated precisely using
functions.
ANS: Bessel
18. The effect is characteristic of FM reception in a noisy
environment.
ANS: threshold
19. The effect is seen when an FM receiver is exposed to two FM
signals that are close to each other in frequency.
ANS: capture
20. Rest frequency is another name for an FM frequency.
ANS: carrier
SHORT ANSWER1. If a 2-volt instantaneous value of modulating
signal amplitude causes a 10-kHz deviation in carrier frequency,
what is the deviation sensitivity of the modulator?
ANS:
5 kHz / volt
2. If a 2-kHz audio tone causes a frequency deviation of 4 kHz,
what is the modulation index?
ANS:
2
3. What will be the deviation caused by a 3-kHz tone if the
modulation index is 3?
ANS:
9 kHz
4. If the deviation sensitivity of an FM modulator is 2 kHz /V,
what will be the modulation index caused by a 1-volt, 1-kHz audio
signal?
ANS:
2
5. At a modulation index of 2, how much power is in the carrier
of a 1000-watt FM transmitter?
ANS:
48.4 watts
6. At a modulation index of 2, how much power is in the first
pair of sidebands of a 1000-watt FM
transmitter?
ANS:
673 watts
7. At a modulation index of 2, how much power is in the fifth
pair of sidebands of a 1000-watt FM
transmitter?
ANS:
200 mW (0.2 watt)
8. Using Carson's rule, what is the approximate bandwidth of an
FM signal with a modulation index of 2 being modulated by a 5-kHz
signal?
ANS:
30 kHz
9. Using the Bessel chart of Figure 4.1, what is the bandwidth
of an FM signal with a modulation index of 2 being modulated by a
5-kHz signal if we ignore sidebands containing less than 1% of the
total power?
ANS:
30 kHz
10. How would you use the fact that J0 is zero for certain known
values of mf (2.4, 5.5, etc) to measure the frequency deviation of
an FM modulator?
ANS:
Use an audio frequency generator to modulate the FM carrier.
Using a spectrum analyzer, adjust the audio frequency until the
carrier amplitude vanishes. Record the audio frequency. Then do the
calculation: =
fm mf where mf will have one of the known values. For example,
if fm is measured to be 2 kHz when mf is
5.5, then is 11 kHz.
Chapter 5: TransmittersMULTIPLE CHOICE1. The ability to change
operating frequency rapidly without a lot of retuning is
called:
a. agility c. VFO
b. expansion d. spread-spectrum
ANS: A
2. The difference between the DC power into a transmitter and
the RF power coming out:
a. is a measure of efficiency c. may require water cooling b.
heats the transmitter d. all of the above
ANS: D
3. Baseband compression produces:
a. a smaller range of frequencies from low to high b. a smaller
range of amplitude from soft to loud
c. a smaller number of signals d. none of the above
4.ANS: B
ALC stands for:
a. Amplitude Level Control
b. Automatic Level Controlc.
d.Accurate Level Control
none of the above
ANS: B
5.In an AM transmitter, ALC is used to:
a. keep the modulation close to 100%c.maximize transmitted
power
b. keep the modulation below 100%d.all of the above
ANS: D
6.With high-level AM:
a. all RF amplifiers can be nonlinearc.minimum RF power is
required
b. minimum modulation power is requiredd.all of the above
ANS: A
7.With high-level AM:
a. the RF amplifiers are typically Class Ac.the RF amplifiers
are typically Class C
b. the RF amplifiers are typically Class Bd.the RF amplifiers
are typically Class AB
ANS: C
8.With low-level AM:
a. the RF amplifiers must be Class Ac.the RF amplifiers must be
linear
b. the RF amplifiers must be Class Bd.the RF amplifiers must be
low-power
ANS: C
9. Power amplifiers must be linear for any signal that:
a. is complex c. has variable frequency b. has variable
amplitude d. all of the above
ANS: B
10. In high-level AM, "high-level" refers to:
a. the power level of the carrier c. the power level of the
final RF amplifier b. the power level of the modulation d. none of
the above
ANS: D
11. In high-level AM, the power in the sidebands comes from:
a. the modulating amplifier c. the driver stage b. the RF
amplifier d. the carrier
ANS: A
12. In an AM transmitter with 100% modulation, the voltage of
the final RF stage will be:
a. approximately half the DC supply voltage
b. approximately twice the DC supply voltage
c. approximately four times the DC supply voltage d. none of the
above
ANS: C
13. Practical transmitters are usually designed to drive a load
impedance of: a. 50 ohms resistive c. 300 ohms resistive b. 75 ohms
resistive d. 600 ohms resistive
ANS: A
14. Which of the following can be used for impedance
matching?
a. pi network c. both a and b
b. T network d. a bridge circuit
ANS: C
15. When a transmitter is connected to a resistor instead of an
antenna, the resistor is called:
a. a heavy load c. a temporary load b. a dummy load d. a test
load
ANS: B
16. When a transmitter is connected to a resistor instead of an
antenna, the resistor must be:
a. wire-wound
b. noninductivec.
d.1% tolerance or better
all of the above
ANS: B
17.A Class D amplifier is:
a. very efficientc.essentially pulse-duration modulation
b. essentially pulse-width modulation
ANS: Dd.all of the above
18.To generate a SSB signal:
a. start with full-carrier AMc.start with a quadrature
signal
b. start with DSBSCd.all of the above
ANS: B
19.The carrier is suppressed in:
a. a balanced modulatorc.a frequency multiplier
b. a mixerd.none of the above
ANS: A
20. To remove one AM sideband and leave the other you could
use:
a. a mechanical filter
b. a crystal filterc.
d.both a and b
none of the above
ANS: C
21.A direct FM modulator:
a. varies the frequency of the carrier oscillator
b. integrates the modulating signal c. both a and b
d. none of the above
ANS: A
22. An indirect FM modulator:
a. requires a varactor in the carrier oscillator b. varies the
phase of the carrier oscillator
c. both a and b
d. none of the above
23.ANS: B
AFC stands for:
a. Amplitude to Frequency Conversion
b. Automatic Frequency Centeringc.
d.Automatic Frequency Control
Audio Frequency Control
ANS: C
24.Frequency multipliers are:
a. essentially balanced modulatorsc.essentially mixers
b. essentially Class C amplifiersd.none of the above
ANS: B
25.With mixing:
a. the carrier frequency can be raised
b. the carrier frequency can be lowered
c. the carrier frequency can be changed to any required
value
d. the deviation is altered
ANS: C
COMPLETION1. The accuracy and stability of a transmitter
frequency is fixed by the oscillator.
ANS: carrier
2. In the USA, the sets requirements for accuracy and stability
of a transmitter's frequency.
ANS: FCC
3. In Canada, sets requirements for accuracy and stability of a
transmitter's frequency.
ANS: Industry Canada
4. Frequency is the ability of a transmitter to change frequency
without a lot of retuning.
ANS: agility
5. Power output of SSB transmitters is rated by .
ANS: PEP
6. Reducing the dynamic range of a modulating signal is called
.
ANS: compression
7. The opposite of compression is called .
ANS: expansion
8. ALC is a form of .
ANS: compression
9. High-level modulation allows the RF amplifiers to operate
more .
ANS: efficiently
10. Low-level modulation requires the RF amplifiers to be .
ANS: linear
11. To isolate the oscillator from load changes, a stage is
used.
ANS: buffer
12. The peak collector voltage in a Class C RF amplifier is than
the DC supply voltage.
ANS: higher
13. Most practical transmitters are designed to operate into a
-ohm load.
ANS: 50
14. Transmitters built with transistor RF amplifiers often use a
network for impedance matching.
ANS: T
15. Matching networks also act as filters to help reduce
levels.
ANS: harmonic
16. Severe impedance can destroy a transmitter's output
stage.
ANS: mismatch
17. Transceivers combine a transmitter and a into one "box".
ANS: receiver
18. To allow a high modulation percentage, it is common to
modulate the as well as the power amplifier in transistor
modulators.
ANS: driver
19. Pulse-width modulation is the same as pulse- modulation.
ANS: duration
20. Switching amplifiers are sometimes called Class
amplifiers.
ANS: D
21. Because the sideband filter in a SSB transmitter is fixed,
is used to operate at more than one frequency.
ANS: mixing
22. To generate a SSB signal, it is common to start with a
signal.
ANS: DSBSC
23. Indirect FM is derived from modulation.
ANS: phase
24. Using a varactor to generate FM is an example of a
modulator.
ANS: reactance
25. The modern way to make a stable VFO is to make it part of a
loop.
ANS: phase-locked
SHORT ANSWER1. If a 50-MHz oscillator is accurate to within
0.001%, what is the range of possible frequencies?
ANS:
50 MHz 500 hertz
2. What is the efficiency of a 100-watt mobile transmitter if it
draws 11 amps from a 12-volt car battery?
ANS:
75.8%
3. The power amplifier of an AM transmitter draws 100 watts from
the power supply with no modulation.
Assuming high-level modulation, how much power does the
modulation amplifier deliver for 100%
modulation?
ANS:
50 watts
4. If the final RF amplifier of an AM transmitter is powered by
100 volts DC, what is the maximum collector voltage at 100%
modulation?
ANS:
400 volts
5. Suppose the output of a balanced modulator has a center
frequency of 10 MHz. The audio modulation frequency range is 1 kHz
to 10 kHz. To pass the USB, what should be the center frequency of
an ideal crystal filter?
ANS:
10.005 MHz
6. Suppose you have generated a USB SSB signal with a nominal
carrier frequency of 10 MHz. What is the minimum frequency the SSB
signal can be mixed with so that the output signal has a nominal
carrier frequency of 50 MHz?
ANS:
40 MHz
7. Suppose you have an FM modulator that puts out 1 MHz carrier
with a 100-hertz deviation. If frequency multiplication is used to
increase the deviation to 400 hertz, what will be the new carrier
frequency?
ANS:
4 MHz
8. Suppose you had an FM signal with a carrier of 10 MHz and a
deviation of 10 kHz. Explain how you could use it to get an FM
signal at 100 MHz with a deviation of 20 kHz.
ANS:
First, put the signal through a frequency doubler to get a
20-MHz carrier with a 20-kHz deviation. Then mix that signal with
an 80-MHz carrier to generate a 100-MHz carrier with 20-kHz
deviation.
Chapter 6: ReceiversMULTIPLE CHOICE1. The two basic
specifications for a receiver are:
a. the sensitivity and the selectivity
b. the number of converters and the number of IFs c. the
spurious response and the tracking
d. the signal and the noise
ANS: A
2. The superheterodyne receiver was invented by:
a. Foster c. Armstrong b. Seeley d. Hertz
ANS: C
3. Trimmers and padders are:
a. two types of adjusting tools c. small adjustable inductors b.
small adjustable resistors d. small adjustable capacitors
ANS: D
4. "Skin effect" refers to:
a. the way radio signals travel across a flat surface b. the
tissue-burning effect of a strong RF signal
c. the increase of wire resistance with frequency d. none of the
above
5.ANS: C
The "front end" of a receiver can include:
a. the tuner
b. the RF amplifierc.
d.the mixer
all of the above
ANS: D
6."IF" stands for:
a. intermediate frequencyc.indeterminate frequency
b. intermodulation frequencyd.image frequency
ANS: A
7.AGC stands for:
a. Audio Gain Controlc.Active Gain Control
b. Automatic Gain Controld.Active Gain Conversion
ANS: B
8.The frequency of the local oscillator:
a. is above the RF frequency
b. is below the RF frequency
c. can be either above of below the RF frequency d. is fixed,
typically at 455 kHz.
ANS: C
9. The local oscillator and mixer are combined in one device
because:
a. it gives a greater reduction of spurious responses b. it
increases sensitivity
c. it increases selectivity
d. it is cheaper
ANS: D
10. Basically, sensitivity measures:
a. the weakest signal that can be usefully received
b. the highest-frequency signal that can be usefully received c.
the dynamic range of the audio amplifier
d. none of the above
ANS: A
11. Basically, selectivity measures:
a. the range of frequencies that the receiver can select
b. with two signals close in frequency, the ability to receive
one and reject the other c. how well adjacent frequencies are
separated by the demodulator
d. how well the adjacent frequencies are separated in the
mixer
12.ANS: B
When comparing values for shape factor:
a. a value of 1.414 dB is ideal
b. a value of 0.707 is idealc.
d.a value of 1.0 is ideal
there is no ideal value
ANS: C
13.When comparing values for shape factor:
a. a value of 2 is better than a value of 4c.both values are
basically equivalent
b. a value of 4 is better than a value of 2d.none of the
above
ANS: A
14.Distortion in a receiver can occur in:
a. the mixerc.the IF amplifiers
b. the detectord.all of the above
ANS: D
15.Phase distortion is important in:
a. voice communications systemsc.monochrome video receivers
b. color video receiversd.all of the above
ANS: B
16. The response of a receiver to weak signals is usually
limited by:
a. the AGC c. the dynamic range of the receiver
b. noise generated in the receiver d. the type of detector
circuit being used
ANS: B
17. Image frequencies occur when two signals:
a. are transmitted on the same frequency
b. enter the mixer, with one being a reflected signal equal to
the IF frequency
c. enter the mixer, one below and one above the local oscillator
by a difference equal to the
IF
d. enter the mixer, and the difference between the two signals
is equal to twice the IF
ANS: C
18.An image must be rejected:
a. prior to mixingc. prior to detection
b. prior to IF amplificationd. images cannot be rejected
ANS: A
19. Image frequency problems would be reduced by:
a. having an IF amplifier with the proper shape factor b. having
a wideband RF amplifier after the mixer
c. having a narrowband RF amplifier before the mixer
d. none of the above
20.ANS: C
A common AM detector is the:
a. PLL
b. envelope detectorc.
d.ratio detector
all of the above
ANS: B
21.An FM detector is the:
a. PLLc.quadrature detector
b. ratio detectord.all of the above
ANS: D
22. Germanium diodes are used in AM detectors because:
a. they are faster than silicon diodes
b. they are cheaper than silicon diodes
c. they minimize distortion from nonlinearity d. all of the
above
ANS: C
23.A common SSB detector is:
a. a PLLc. a BFO
b. a dioded. a product detector
ANS: D
24.BFO stands for:
a. Beat Frequency Oscillatorc. Bipolar Frequency Oscillator
b. Barrier Frequency Oscillator
ANS: Ad. Bistable Frequency Oscillator
25. To demodulate both SSB and DSBSC, you need to:
a. use a Foster-Seeley discriminator b. reinject the carrier
c. use double conversion
d. use one diode for SSB and two diodes for DSBSC
26.ANS: B
Which would be best for DSBSC:
a. carrier detection
b. coherent detectionc. envelope detection
d. ratio detection
ANS: B
27. An FM detector that is not sensitive to amplitude variations
is:
a. Foster-Seeley detector
b. a quadrature detectorc.
d.a PLL detector
all of the above
ANS: C
28.The function of a limiter is:
a. to remove amplitude variationsc.to limit dynamic range
b. to limit spurious responsesd.to limit noise response
ANS: A
29. Suppressing the audio when no signal is present is
called:
a. AGC
b. squelchc.
d.AFC
limiting
ANS: B
30.LNA stands for:
a. Limited-Noise Amplifierc.Low-Noise Audio
b. Low-Noise Amplifierd.Logarithmic Noise Amplification
ANS: B
31.AFC stands for:
a. Audio Frequency Compensatorc.Automatic Frequency Control
b. Autodyne Frequency Compensationd.Autonomous Frequency
Control
ANS: C
32. The function of AFC is:
a. maintain a constant IF frequency
b. match the local oscillator to the received signal
c. lock the discriminator to the IF frequency d. none of the
above
ANS: B
33.SAW stands for:
a. Symmetrical Audio Wavec.Silicon-Activated Wafer
b. Surface Acoustic Waved.Software-Activated Wave
ANS: B
34.The important property of a SAW is:
a. it stabilizes the audio in a receiverc.it is a stable
bandpass filter
b. it allows software radios to be builtd.none of the above
ANS: C
35.The main function of the AGC is to:
a. keep the gain of the receiver constant
b. keep the gain of the IF amplifiers constant
c. keep the input to the detector at a constant amplitude d. all
of the above
36.ANS: C
DSP stands for:
a. Dynamic Signal Properties
b. Direct Signal Phasec. Distorted Signal Packet
d. Digital Signal Processor
ANS: D
37.SINAD stands for:
a. Sinusoidal Amplitude Distortion
b. Signal and Noise Amplitude Distortion c.
Signal-plus-Noise-to-Noise Ratio
d. Signal-plus-Noise and Distortion-to-Noise and Distortion
Ratio
ANS: D
38.TRF stands for:
a. Tuned Radio Frequencyc. Transmitted Radio Frequency
b. Tracking Radio Frequencyd. Tuned Receiver Function
ANS: A
COMPLETION1. Almost all modern receivers use the principle.
ANS: superheterodyne
2. The first radio receiver of any kind was built in the year
.
ANS: 1887
3. When two tuned circuits each other, it means that when the
frequency of one is adjusted, the other changes with it.
ANS: track
4. The effect causes the resistance of wire to increase with
frequency.
ANS: skin
5. The superhet was invented in the year .
ANS: 1918
6. In a receiver, the refers to the input filter and RF
stage.
ANS: front end
7. In a superhet, the output of the goes to the IF
amplifiers.
ANS: mixer
8. In a superhet, the frequency is the difference between the
local oscillator frequency and the received signal frequency.
ANS: intermediate IF
9. The circuit adjusts the gain of the IF amplifiers in response
to signal strength.
ANS: AGC
10. An converter uses the same transistor for both the local
oscillator and the mixer.
ANS: autodyne
11. In low-side injection, the local oscillator is than the
received signal frequency.
ANS: lower
12. is the ability of a receiver to separate two signals that
are close to each other in frequency.
ANS: Selectivity
13. is the ability of a receiver to receive and successfully
demodulate a very weak signal.
ANS: Sensitivity
14. A receiver with two different IF frequencies is called a
double- receiver.
ANS: conversion
15. A multiple-conversion receiver will have better rejection of
frequencies.
ANS: image
16. A demodulator is also called a .
ANS: detector
17. An detector uses a diode to half-wave rectify an AM
signal.
ANS: envelope
18. A detector is used for SSB signals.
ANS: product
19. A BFO produces a locally generated .
ANS: carrier
20. A DSBSC signal requires a detection circuit.
ANS: coherent
21. FM detectors have a characteristic -shaped curve.
ANS: S
22. While still commonly found, the Foster-Seeley and ratio
detectors are .
ANS: obsolescent
23. Unlike the PLL detector, the quadrature detector is
sensitive to changes in of the input signal.
ANS: amplitude
24. A dual- MOSFET is useful for AGC.
ANS: gate
25. Diode mixers are too to be practical in most
applications.
ANS: noisy
26. The IF amplifiers in an AM receiver must be Class .
ANS: A
27. A double-tuned IF transformer is usually coupled for the
response to have a flat top and steep sides.
ANS: over
28. Multiple IF stages can be -tuned to increase the
bandwidth.
ANS: stagger
29. Compared to tuned circuits, ceramic and crystal IF filters
do not require .
ANS: adjustment
30. Up-conversion is when the output of the mixer is a frequency
than the incoming signal.
ANS: higher
31. In a block converter, the frequency of the first local
oscillator is .
ANS: fixed constant
32. Typically, AGC reduces the gain of the amplifiers.
ANS: IF
33. An -meter is designed to indicate signal strength in many
communications receivers.
ANS: S
34. The effectiveness of FM is measured by a receivers quieting
sensitivity.
ANS: limiting
35. A refers to any kind of FM or PM detector.
ANS: discriminator
SHORT ANSWER1. Suppose the bandwidth of a tuned circuit is 10
kHz at 1 MHz. Approximately what bandwidth would you expect it to
have at 4 MHz?
ANS:
20 kHz
2. Using high-side injection for a 1-MHz IF, what is the
frequency of the local oscillator when the receiver is tuned to 5
MHz?
ANS:
6 MHz
3. An IF filter has a 60 dB bandwidth of 25 kHz and a 6 dB
bandwidth of 20 kHz. What is the shape factor value?
ANS:
1.25
4. Suppose a receiver uses a 5-MHz IF frequency. Assuming
high-side injection, what would be the image frequency if the
receiver was tuned to 50 MHz?
ANS:
60 MHz
5. Suppose a SSB receiver requires an injected frequency of 1.5
MHz. What would be the acceptable frequency range of the BFO if the
maximum acceptable baseband shift is 100 hertz?
ANS:
1.5 MHz 100 hertz
6. The transformer of a double-tuned IF amplifier has a Q of 25
for both primary and secondary. What value of kc do you need to
achieve optimal coupling?
ANS:
0.06
7. What value of transformer coupling would a double-tuned
10-MHz IF amplifier with optimal coupling need to get a bandwidth
of 100 kHz?
ANS:
0.01
Chapter 7: Digital CommunicationsMULTIPLE CHOICE1.The first
digital code was the:
a. ASCII code b. Baudot codec. Morse code
d. none of the above
ANS: C
2. In digital transmission, signal degradation can be removed
using:
a. an amplifierc.a regenerative repeater
b. a filterd.all of the above
ANS: C
3.TDM stands for:
a. Time-Division Multiplexingc.Ten-Digital Manchester
b. Time-Domain Multiplexingd.Ten Dual-Manchester
ANS: A
4.Hartley's Law is:
a. I = ktBb. C = 2B log2Mc.
d.C = B log2(1 + S/N)SR = 2fmax
ANS: A
5.The Shannon-Hartley theorem is:
a. I = ktBb. C = 2B log2Mc.
d.C = B log2(1 + S/N)SR = 2fmax
ANS: B
6.The Shannon Limit is given by:
a. I = ktBb. C = 2B log2Mc.
d.C = B log2(1 + S/N)SR = 2fmax
ANS: C
7.The Nyquist Rate can be expressed as:
a. I = ktBb. C = 2B log2Mc.
d.C = B log2(1 + S/N)SR = 2fmax
ANS: D
8.Natural Sampling does not use:
a. a sample-and-hold circuitc.a fixed sample rate
b. true binary numbersd.an analog-to-digital converter
ANS: A
9. Which is true about aliasing and foldover distortion?
a. They are two types of sampling error.
b. You can have one or the other, but not both.
c. Aliasing is a technique to prevent foldover distortion. d.
They are the same thing.
10.ANS: D
Foldover distortion is caused by:
a. noise
b. too many samples per secondc.
d.too few samples per second
all of the above
ANS: C
11.The immediate result of sampling is:
a. a sample aliasc.PCM
b. PAMd.PDM
ANS: B
12. Which of these is not a pulse-modulation technique:
a. PDM
b. PWMc. PPM
d. PPS
ANS: D
13.Quantizing noise (quantization noise):
a. decreases as the sample rate increases
b. decreases as the sample rate decreases
c. decreases as the bits per sample increases d. decreases as
the bits per sample decreases
ANS: C
14. The dynamic range of a system is the ratio of:
a. the strongest transmittable signal to the weakest discernible
signal
b. the maximum rate of conversion to the minimum rate of
conversion c. the maximum bits per sample to the minimum bits per
sample
d. none of the above
ANS: A
15. Companding is used to:
a. compress the range of base-band frequencies b. reduce dynamic
range at higher bit-rates
c. preserve dynamic range while keeping bit-rate low
d. maximize the useable bandwidth in digital transmission
ANS: C
16. In North America, companding uses:
a. the Logarithmic Law c. the Law (alpha law)b. the A Law d. the
Law (mu law)17.ANS: D
In Europe, companding uses:
a. the Logarithmic Law
b. the A Lawc.
d.the Law (alpha law)
the Law (mu law)
ANS: B
18.Codec stands for:
a. Coder-Decoderc.Code-Compression
b. Coded-Carrierd.none of the above
ANS: A
19. A typical codec in a telephone system sends and
receives:
a. 4-bit numbers
b. 8-bit numbersc.
d.12-bit numbers
16-bit numbers
ANS: B
20.Compared to PCM, delta modulation:
a. transmits fewer bits per samplec.can suffer slope
overload
b. requires a much higher sampling rated.all of the above
ANS: D
21. In delta modulation, "granular noise" is produced when:
a. the signal changes too rapidly c. the bit rate is too high b.
the signal does not change d. the sample is too large
ANS: B
22. Compared to PCM, adaptive delta modulation can transmit
voice:
a. with a lower bit rate but reduced quality c. only over
shorter distances
b. with a lower bit rate but the same quality d. only if the
voice is band-limited
ANS: B
23. Which coding scheme requires DC continuity:
a. AMI c. unipolar NRZ
b. Manchester d. bipolar RZ
ANS: C
24. Manchester coding:
a. is a biphase code
b. has a level transition in the middle of every bit period c.
provides strong timing information
d. all of the above
ANS: D
25. The number of framing bits in DS-1 is:
a. 1c.4
b. 2d.8
ANS: A
26.Framing bits in DS-1 are used to:
a. detect errorsc.synchronize the transmitter and receiver
b. carry signalingd.all of the above
ANS: C
27.So-called "stolen" bits in DS-1 are used to:
a. detect errorsc.synchronize the transmitter and receiver
b. carry signalingd.all of the above
ANS: B
28.The number of bits per sample in DS-1 is:
a. 1c.4
b. 2d.8
ANS: D
29. The number of samples per second in DS-1 is:
a. 8 k c. 64 k
b. 56 k d. 1.544 106ANS: A
30. The bit rate for each channel in DS-1 is:
a. 1.544 Mb/s c. 56 kb/s b. 64 kb/s d. 8 kb/s
ANS: B
31. In DS-1, bits are transmitted over a T-1 cable at:
a. 1.544 MB/s c. 56 kb/s b. 64 kb/s d. 8 kb/s
ANS: A
32. A T-1 cable uses:
a. Manchester coding c. NRZ coding
b. bipolar RZ AMI coding d. pulse-width coding
ANS: B
33. The number of frames in a superframe is:
a. 6 c. 24 b. 12 d. 48
ANS: B
34. A typical T-1 line uses:
a. twisted-pair wire b. coaxial cablec. d.fiber-optic cable
microwave
ANS: A
35."Signaling" is used to indicate:
a. on-hook/off-hook conditionc.ringing
b. busy signald.all of the above
ANS: D
36.A vocoder implements compression by:
a. constructing a model of the transmission medium
b. constructing a model of the human vocal system c. finding
redundancies in the digitized data
d. using lossless techniques
ANS: B
37. Compared to standard PCM systems, the quality of the output
of a vocoder is:
a. much better c. about the same b. somewhat better d. not as
good
ANS: D
COMPLETION1. Digitizing a signal often results in transmission
quality.
ANS: improved better
2. To send it over an analog channel, a digital signal must be
onto a carrier.
ANS: modulated
3. To send it over a digital channel, an analog signal must
first be .
ANS: digitized
4. In analog channels, the signal-to-noise ratio of an analog
signal gradually as the length of the channel increases.
ANS: decreases gets worse
5. The value of a pulse is the only information it carries on a
digital channel.
ANS: binary
6. A repeater is used to restore the shape of pulses on a
digital cable.
ANS: regenerative
7. There are techniques to detect and some errors in digital
transmission.
ANS: correct
8. Converting an analog signal to digital form is another source
of in digital transmission systems.
ANS: error noise
9. -division multiplexing is easily done in digital
transmission.
ANS: Time
10. All practical communications channels are band- .
ANS: limited
11. Law gives the relationship between time, information
capacity, and bandwidth.
ANS: Hartley's
12. Ignoring noise, the theorem gives the maximum rate of data
transmission for a given bandwidth.
ANS: Shannon-Hartley
13. The limit gives the maximum rate of data transmission for a
given bandwidth and a given signal-to-noise ratio.
ANS: Shannon
14. sampling is done without a sample-and-hold circuit.
ANS: Natural
15. The Rate is the minimum sampling rate for converting analog
signals to digital format.
ANS: Nyquist
16. distortion occurs when an analog signal is sampled at too
slow a rate.
ANS: Foldover
17. means that higher frequency baseband signals from the
transmitter "assume the identity" of low-frequency baseband signals
at the receiver when sent digitally.
ANS: Aliasing
18. The output of a sample-and-hold circuit is a pulse-
modulated signal.
ANS: amplitude
19. modulation is the most commonly used digital modulation
scheme.
ANS: Pulse-code
20. noise results from the process of converting an analog
signal into digital format.
ANS: Quantizing
21. is used to preserve dynamic range using a reasonable
bandwidth.
ANS: Companding
22. In North America, compression is done using the -law
equation.
ANS:
mu
23. In Europe, compression is done using the -law equation.
ANS: A
24. A is an IC that converts a voice signal to PCM and vice
versa.
ANS: codec
25. In a PCM system, the samples of the analog signal are first
converted to bits before being compressed to 8 bits.
ANS: 12
26. The number of bits per sample transmitted in delta
modulation is .
ANS:
1 one
27. Delta modulation requires a sampling rate than PCM for the
same quality of reproduction.
ANS: higher
28. noise is produced by a delta modulator if the analog signal
doesn't change.
ANS: Granular
29. In delta modulation, overload can occur if the analog signal
changes too fast.
ANS: slope
30. The size varies in adaptive delta modulation.
ANS: step
31. Adaptive delta modulation can transmit PCM-quality voice at
about the bit rate of PCM.
ANS: half
32. Unipolar NRZ is not practical because most channels do not
have continuity.
ANS: DC
33. In AMI, binary ones are represented by a voltage that
alternates in .
ANS: polarity
34. Long strings of should be avoided in AMI.
ANS: zeros
35. Manchester code has a level in the center of each bit
period.
ANS: transition
36. Manchester coding provides information regardless of the
pattern of ones and zeros.
ANS: timing
37. There are channels in a DS-1 frame.
ANS: 24
38. DS-1 uses a bit to synchronize the transmitter and
receiver.
ANS: framing
39. In DS-1, each channel is sampled times per second.
ANS: 8000
40. Data is carried over a T-1 line at a rate of bits per
second.
ANS: 1.544 10641. A group of 12 DS-1 frames is called a .
ANS: superframe
42. From a group of twelve frames, signaling bits are "stolen"
from every frame.
ANS: sixth
43. compression transmits all the data in the original signal
but uses fewer bits to do it.
ANS: Lossless
SHORT ANSWER1. Use Hartley's Law to find how much time it would
take to send 100,000 bits over a channel with a bandwidth of 2,000
hertz and a channel constant of k = 10.
ANS:
5 seconds
2. Use the Shannon-Hartley theorem to find the bandwidth
required to send 12,000 bits per second if the number of levels
transmitted is 8.
ANS:
2000 hertz
3. What is the Shannon Limit of a channel that has a bandwidth
of 4000 hertz and a signal-to-noise ratio of
15?
ANS:
16 kbps
4. What is the minimum required number of samples per second to
digitize an analog signal with frequency components ranging from
300 hertz to 3300 hertz?
ANS:
6600 samples/second
5. What is the approximate dynamic range, in dB, of a linear PCM
system that uses 12 bits per sample?
ANS:
74 dB
6. What is the approximate data rate for a system using 8 bits
per sample and running at 8000 samples per second?
ANS:
64 kbps
7. If bits were "stolen" from every DS-1 frame, what would the
useable data-rate be for each channel in the frame?
ANS:
56 kbps
8. Assuming maximum input and output voltages of 1 volt, what is
the output voltage of a -law compressor if the input voltage is
0.388 volt?
ANS:
0.833 volt
Chapter 8: The Telephone SystemMULTIPLE CHOICE1.DTMF stands
for:
a. Digital Telephony Multiple Frequency b. Dial Tone Master
Frequencyc. d.Dual-Tone Multifrequency
Digital Trunk Master Frequency
ANS: C
2.PSTN stands for:
a. Public Switched Telephone Networkc.Primary Service Telephone
Network
b. Private Switched Telephone Networkd.Primary Service Telephone
Numbers
ANS: A
3.POTS stands for:
a. Private Office Telephone Systemc.Primary Operational Test
System
b. Primary Office Telephone Serviced.Plain Old Telephone
Service
ANS: D
4.LATA stands for:
a. Local Access and Transport Areac.Local Area Telephone
Access
b. Local Access Telephone Aread.Local Area Transport Access
ANS: A
5.A LATA is a:
a. a local calling areac.a way of accessing a tandem office
b. a type of digital local networkd.a way of accessing a central
office
ANS: A
6.Central offices are connected by:
a. local loopsc.both a and b
b. trunk linesd.none of the above
ANS: B
7.Local loops terminate at:
a. a tandem officec.a central office
b. a toll stationd.an interexchange office
ANS: C
8.Call blocking:
a. cannot occur in the public telephone network
b. occurs on the local loop when there is an electrical power
failure c. occurs only on long-distance cables
d. occurs when the central office capacity is exceeded
9.ANS: D
In telephony, POP stands for:
a. Post Office Protocol
b. Point Of Presencec.
d.Power-On Protocol
none of the above
ANS: B
10.The cable used for local loops is mainly:
a. twisted-pair copper wirec.coaxial cable
b. shielded twisted-pair copper wired.fiber-optic
ANS: A
11.FITL stands for:
a. Framing Information for Toll Loopsc.Framing In The Loop
b. Fiber In the Toll Loopd.Fiber-In-The-Loop
ANS: D
12.Loading coils were used to:
a. increase the speed of the local loop for digital data
b. reduce the attenuation of voice signals c. reduce
crosstalk
d. provide C-type conditioning to a local loop
13.ANS: B
DC current flows through a telephone:
a. when it is on hook
b. when it is off hookc. as long as it is attached to a local
loop
d. only when it is ringing
ANS: B
14. The range of DC current that flows through a telephone
is:
a.
b.20 A to 80 A
200 A to 800 Ac. 2 mA to 8 mA
d. 20 mA to 80 mA
ANS: D
15. The separation of control functions from signal switching is
known as:
a. step-by-step switching control c. common control b. crossbar
control d. ESS
ANS: C
16. The typical voltage across a telephone when on-hook is:
a. 48 volts DC c. 90 volts DC
b. 48 volts, 20 hertz AC d. 90 volts, 20 hertz AC
ANS: A
17. The typical voltage needed to "ring" a telephone is:
a. 48 volts DC c. 90 volts DC
b. 48 volts, 20 hertz AC d. 90 volts, 20 hertz AC
ANS: D
18. The bandwidth of voice-grade signals on a telephone system
is restricted in order to: a. allow lines to be "conditioned" c.
allow signals to be multiplexed b. prevent "singing" d. all of the
above
ANS: C
19. VNL stands for:
a. voltage net loss c. via net loss
b. volume net loss d. voice noise level
ANS: C
20. Signal loss is designed into a telephone system to:
a. eliminate reflections c. improve signal-to-noise ratio b.
prevent oscillation d. reduce power consumption
ANS: B
21. The reference noise level for telephony is:
a. 1 mW c. 1 pW
b. 0 dBm d. 0 dBr
ANS: C
22. The number of voice channels in a basic FDM group is:
a. 6c.24
b. 12d.60
ANS: B
23.Basic FDM groups can be combined into:
a. supergroupsc.jumbogroups
b. mastergroupsd.all of the above
ANS: D
24. In telephone system FDM, voice is put on a carrier using: a.
SSB c. PDM b. DSBSC d. PCM
ANS: A
25. PABX stands for:
a. Power Amplification Before Transmission b. Private Automatic
Branch Exchange
c. Public Automated Branch Exchange
d. Public Access Branch Exchange
ANS: B
26.SLIC stands for:
a. Single-Line Interface Circuit b. Standard Line Interface
Cardc. d.Subscriber Line Interface Card
Standard Local Interface Circuit
ANS: C
27.In DS-1, bits are "robbed" in order to:
a. provide synchronizationc.cancel echoes
b. carry signalingd.check for errors
ANS: B
28."Bit-stuffing" is more formally called:
a. compensationc.justification
b. rectificationd.frame alignment
ANS: C
29.ISDN stands for:
a. Integrated Services Digital Networkc.Integrated Services Data
Network
b. Information Services Digital Networkd.Information Systems
Digital Network
ANS: A
30. Basic ISDN has not been widely adopted because:
a. it took to long to develop b. it is too slow
c. it has been surpassed by newer technologies d. all of the
above
ANS: D
31. ADSL stands for:
a. All-Digital Subscriber Line c. Allocated Digital Service Line
b. Asymmetrical Digital Subscriber Line d. Access to Data Services
Line
ANS: B
32. Compared to ISDN, internet access using ADSL is
typically:
a. much faster c. much more expensive b. about the same speed d.
none of the above
ANS: A
COMPLETION1. A is a local calling area.
ANS: LATA
2. Central offices are connected together by lines.
ANS: trunk
3. One central office can be connected to another through a
office.
ANS: tandem
4. With 7-digit phone numbers, thousand telephones can connect
to a central office.
ANS: ten
5. Call is when it becomes impossible for a subscriber to place
a call due to an overload of lines being used.
ANS: blocking
6. New switching equipment uses TDM to combine signals.
ANS: digital
7. Most local loops still use copper wire.
ANS: twisted-pair
8. As compared to a hierarchical network, a network never needs
more than one intermediate switch.
ANS: flat
9. coils were used to reduce the attenuation of voice
frequencies.
ANS: Loading
10. In a twisted-pair telephone cable, the red wire is called
.
ANS: ring
11. In a twisted-pair telephone cable, the green wire is called
.
ANS: tip
12. Of the red and green 'phone wires, the wire is positive with
respect to the other.
ANS: green
13. A telephone is said to have the line when the central office
sends it dial tone.
ANS: seized
14. The functions are provided by a SLIC.
ANS: BORSCHT
15. A coil prevents loss of signal energy within a telephone
while allowing full- duplex operation over a single pair of
wires.
ANS: hybrid
16. In a crosspoint switch, not all can be in use at the same
time.
ANS: lines
17. The old carbon transmitters generated a relatively signal
voltage.
ANS: large
18. The generic term for Touch-Tone signaling is .
ANS: DTMF
19. A line provides more bandwidth than a standard line.
ANS: conditioned
20. In the telephone system, amplifiers are called .
ANS: repeaters
21. An echo converts a long-distance line from full-duplex to
half-duplex operation.
ANS: suppressor
22. weighting is an attempt to adjust the noise or signal level
to the response of a typical telephone receiver.
ANS: C-message
23. In FDM telephony, the modulation is usually .
ANS: SSB SSBSC
24. In FDM telephony, bands separate the channels in a
group.
ANS: guard
25. Because of "bit robbing", a channel in a DS-1 frame allows
only kbps when used to send digital data.
ANS: 56
26. A is a group of 12 DS-1 frames with signaling information in
the sixth and twelfth frames.
ANS: superframe
27. In DS-1C, bits are used to compensate for differences
between clock rates.
ANS: stuff
28. Busy and dial tone are referred to as signals because they
use the same pair of wires as the voice signal.
ANS: in-channel
29. SS7 is the current version of signaling.
ANS: common-channel
30. SS7 is a -switched data network.
ANS: packet
31. In ISDN, the channel is used for common-channel
signaling.
ANS: D
32. In ISDN, the channels are used for voice or data.
ANS: B
33. Terminal equipment especially designed for ISDN is
designated equipment.
ANS: TE1
34. The A in ADSL stands for .
ANS: asymmetrical
35. In ADSL, the speed from the network to the subscriber is
than the speed in the opposite direction.
ANS: greater faster
SHORT ANSWER1. For a certain telephone, the DC loop voltage is
48 V on hook and 8 V off hook. If the loop current is 40 mA, what
is the DC resistance of the local loop?
ANS:
1000 ohms
2. For a certain telephone, the DC loop voltage is 48 V on hook
and 8 V off hook. If the loop current is 40 mA, what is the DC
resistance of the telephone?
ANS:
200 ohms
3. Which two DTMF tones correspond to the digit "1"? (Use the
table in the text.) ANS:
697 Hz and 1209 Hz
4. Calculate the dB of VNL required for a channel with a 3 ms
delay.
ANS:
1 dB
5. If a telephone voice signal has a level of 0 dBm, what is its
level in dBrn?
ANS:
90 dBrn
6. A telephone test-tone has a level of 80 dBrn at a point where
the level is +5dB TLP. If C-weighting produces a 10-dB loss, what
would the signal level be in dBrnc0?
ANS:
65 dBrnc TLP
Chapter 9: Data TransmissionMULTIPLE CHOICE1. In practical
terms, parallel data transmission is sent:
a. over short distances only c. over any distance
b. usually over long distances d. usually over a coaxial
cable
ANS: A
2. The five-level teletype code was invented by:
a. the Morkum Company c. Western Union b. the Teletype Company
d. Emile Baudot
ANS: D
3. Data codes are also called:
a. character codes c. they do not have any other name b.
character sets d. both a and b
ANS: C
4. Digital data that is not being used to carry characters is
called:
a. FIGS data c. numerical data b. binary data d. all of the
above
ANS: B
5. Character codes include:
a. alphanumeric characters c. graphic control characters b. data
link control characters d. all of the above
ANS: D
6. ASCII stands for:
a. American Standard Character-set 2
b. American Standard Code for Information Interchange c.
American Standard Code 2
d. Alphanumeric Standard Code for Information Interchange
7.ANS: B
BS, FF, and CR are examples of:
a. nonstandard character codes
b. escape charactersc.
d.control characters
none of the above
ANS: C
8.LF stands for:
a. Line Feedc.Line Forward
b. Link Feed d. Link Forward
ANS: A
9. UART stands for:
a. Universal Asynchronous Receiver-Transmitter
b. Unidirectional Asynchronous Receiver-Transmitter c. Unaltered
Received Text
d. Universal Automatic Receiver for Text
ANS: A
10. In asynchronous transmission, the transmitter and receiver
are:
a. frame-by-frame synchronized using the data bits
b. frame-by-frame synchronized using a common clock
c. frame-by-frame synchronized using the start and stop bits d.
not synchronized at all, hence the name "asynchronous"
ANS: C
11. In asynchronous transmission, the time between consecutive
frames is:
a. equal to zero c. equal to the start and stop bit-times b.
equal to one bit-time d. not a set length
ANS: D
12. In synchronous transmission, the frames are:
a. about the same length as ten asynchronous frames b. much
longer than asynchronous frames
c. 128 bytes long d. 1024 bytes long
ANS: B
13. Synchronous transmission is used because:
a. no start and stop bits means higher efficiency
b. it is cheaper than asynchronous since no UARTS are required
c. it is easier to implement than asynchronous
d. all of the above
ANS: A
14. In synchronous transmission, the receiver "syncs-up" with
the transmitter by using:
a. the clock bits c. the CRC bits
b. the data bits d. a separate clock line
ANS: B
15. To maintain synchronization in synchronous transmission:
a. long strings of 1s and 0s must not be allowed
b. transmission must stop periodically for resynchronization c.
the clock circuits must be precisely adjusted
d. the channel must be noise-free
16.ANS: A
BISYNC:
a. is an IBM product
b. is a character-oriented protocolc.
d.requires the use of DLE
all of the above
ANS: D
17.HDLC:
a. is an IBM productc.is