blake edited reviewer

8/3/2019 Blake Edited Reviewer

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Chapter 1: Introduction to Communication

SystemsMULTIPLE CHOICE

1. The theory of radio waves was originated by:a. Marconi c. Maxwell

b. Bell d. Hertz ANS: C2. The person who sent the first radio signal across the

Atlantic ocean was:a. Marconi c. Maxwell

b. Bell d. Hertz ANS: A3. The transmission of radio waves was first done by:

a. Marconi c. Maxwell

b. Bell d. Hertz ANS: D4. A complete communication system must include:

a. a transmitter and receiverb. a transmitter, a receiver, and a channelc. 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. 2Tvfc. the phase angle

b.fz2Td. none of the above ANS: A6. The bandwidth required for a modulated carrierdepends on:

a. the carrier frequency c. the signal-plus-noise to noise

ratiob. the signal-to-noise ratio d. the baseband frequencyrange ANS: D

7. When two or more signals share a common channel, itis called:

a. sub-channeling c. SINAD

b. signal switching d. multiplexing ANS: D8. TDM stands for:a. Time-Division Multiplexing c. Time Domain

Measurement

b. Two-level Digital Modulation d. none of the aboveANS: A9. FDM stands for:

a. Fast Digital Modulation c. Frequency-Division

Multiplexingb. Frequency Domain Measurement d. none of the above

ANS: C10. The wavelength of a radio signal is:

a. equal tofzcb. equal to c zP

c. the distance a wave travels in one periodd. 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

frequenciesd. all of the above ANS: D

12. The collection of sinusoidal frequencies present in amodulated 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: D15. "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: D17. Shot noise is generated in:

a. transistors and diodes c. copper wire b. resistors d. none of the above ANS: A

18. The power density of "flicker" noise is:

a. the same at all frequencies c. greater at lowfrequencies

b. greater at high frequencies d. the same as "white"

noise ANS: C

19. So called "1/f" noise is also called:a. random noise c. white noise

b. pink noise d. partition noise ANS: B

20. "Pink" noise has:a. equal power per Hertz c. constant power

b. equal power per octave d. none of the above ANS: B

21. When two noise voltages, V1 and V2, are combined,

the total voltage VTis:

a. VT= sqrt(V1vV1 + V2vV2) c. VT= sqrt(V1 vV2)b. VT= (V1 + V2)/2 d. VT= V1 + V2 ANS: A

22. Signal-to-Noise ratio is calculated as:

a. signal voltage divided by noise voltageb. signal power divided by noise power

c. first add the signal power to the noise power, thendivide by noise power

d. none of the above ANS: B23. SINAD is calculated as:

a. signal voltage divided by noise voltageb. signal power divided by noise power

c. first add the signal power to the noise power, thendivide 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 signald. signal-to-noise ratio in dB ANS: C

25. The part, or parts, of a sinusoidal carrier that can be

modulated are:


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a. its amplitude c. its amplitude, frequency, and directionb. its amplitude and frequency d. its amplitude,frequency, and phase angle ANS: DCOMPLETION

1. The telephone was invented in the year

____________________. ANS: 18632. Radio signals first were sent across the Atlantic in theyear ____________________. ANS: 1901

3. The frequency band used to modulate the carrier iscalled the ____________________ band.

ANS: base4. The job of the carrier is to get the information throughthe ____________________. ANS: channel

5. The bandwidth of an unmodulated carrier is

____________________. ANS: zero

6. The 'B' in Hartley's Law stands for ____________________. ANS: bandwidth7. 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 multiplesignals. ANS: FDM9. In ____________________, multiple signal streams

take turns using the channel. ANS: TDM

10. VHF stands for the ____________________frequency band. ANS: very high11. The VHF band starts at ____________________

MHz. ANS: 3012. 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 million15. 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: Fourier17. Disabling a receiver during a burst of atmosphericnoise is called ____________________.

ANS: noise blanking/blanking18. For satellite communications,

____________________ noise can be a serious problem.

ANS: solar19. Thermal noise is caused by the random motions of

____________________ in a conductor.

ANS: electronsSHORT ANSWER

1. Name the five elements in a block diagram of acommunications system.

ANS:Source, Transmitter, Channel, Receiver, Destination2. 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 hasequal 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 QV from one noise source that is

combined with 40 QV from another noise source.Calculate the total noise voltage.

ANS: 50 QV6. If you have 100 mV of signal and 10 mV of noise,

both across the same 100-ohm load, what is the signalto-

noise ratio in dB?

ANS: 20 dB7. The input to an amplifier has a signal-to-noise ratio of100 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: 1.59. Two cascaded amplifiers each have a noise figure of 5

and a gain of 10. Find the total NF for the pair.

ANS: 5.410. Explain why you could use a diode as a noise sourcewith 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 forthe 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, controllingthe diode current controls the noise power.

Chapter 2: Radio-Frequency CircuitsMULTIPLE CHOICE

1. The time it takes a charge carrier to cross from theemitter to the collector is called:a. base time c. charge time

b. transit time d. Miller time ANS: B

2. A real capacitor actually contains:


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a. capacitance and resistance only c. capacitance,inductance, and resistance

b. capacitance and inductance only d. reactance only

ANS: C

3. Bypass capacitors are used to:

a. remove RF from non-RF circuits c. neutralizeamplifiers

b. couple RF around an amplifier d. reduce the Miller

effect ANS: A4. A resonant circuit is:

a. a simple form of bandpass filter c. both a and bb. used in narrowband RF amplifiers d. none of theabove ANS: C

5. Loading down a tuned-circuit amplifier will:

a. raise the Q of the tuned circuit c. "multiply" the Q

b. lower the Q of the tuned circuit d. have no effect on QANS: B6. The "Miller Effect" can:

a. cause an amplifier to oscillate c. reduce the bandwidthof an amplifier

b. cause an amplifier to lose gain d. all of the above

ANS: D7. The Miller Effect can be avoided by:

a. using a common-emitter amplifier c. increasing the Q

of the tuned circuit

b. using a common-base amplifier d. it cannot beavoided ANS: B8. In a BJT, the Miller Effect is due to:

a. inductance of collector lead c. base-to-emittercapacitance

b. collector-to-emitter capacitance d. base-to-collector

capacitance ANS: D

9. In RF amplifiers, impedance matching is usually donewith:

a. RC coupling c. direct coupling

b. transformer coupling d. lumped reactanceANS: B10. Neutralization cancels unwanted feedback by:

a. adding feedback out of phase with the unwantedfeedback

b. bypassing the feedback to the "neutral" or ground

planec. decoupling itd. 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 harmonicsd. an input signal that is an integer multiple of thedesired frequency ANS: A

12. A sinusoidal oscillation from an amplifier requires:

a. loop gain equal to unityb. phase shift around loop equal to 0 degreesc. both a and b, but at just one frequency

d. none of the above ANS: C13. The conditions for sinusoidal oscillation from anamplifier are called:

a. the loop-gain criteria c. the Bode criteria

b. the Hartley criteria d. the Barkhausen criteria

ANS: D14. The Hartley oscillator uses:a. a tapped inductor c. an RC time constant

b. a two-capacitor divider d. a piezoelectric crystalANS: A

15. The Colpitts VFO uses:a. a tapped inductor c. an RC time constant

b. a two-capacitor divider d. a piezoelectric crystal

ANS: B

16. The Clapp oscillator is:

a. a modified Hartley oscillator c. a type of crystal-controlled oscillator

b. a modified Colpitts oscillator d. only built with FETs

ANS: B17. A varactor is:

a. a voltage-controlled capacitor c. used in tuner circuits

b. a diode d. all of the above ANS: D18. 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 dimensionsd. all of the above ANS: D

19. If two signals, Va = sin([at) and Vb= sin([bt), are fedto a mixer, the output:

a. will contain [1 = [a + [b and [2 = [a[b

b. will contain[1 = [a / [b and [2 = [b / [a

c. will contain [= ([a + [b ) / 2d. none of the above ANS: A

20. In a balanced mixer, the output:

a. contains equal (balanced) amounts of all inputfrequencies

b. contains the input frequenciesc. does not contain the input frequencies

d. is a linear mixture of the input signals ANS: C21. "VFO" stands for:

a. Voltage-Fed Oscillator c. Varactor-FrequencyOscillator

b. Variable-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 frequencyc. a fixed-frequency RF generator

d. same as a mixer ANS: ACOMPLETION

1. Generally, conductor lengths in RF circuits should be ____________________. ANS: short2. At UHF frequencies and above, elements must be

considered as ____________________ instead of as


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being "lumped". ANS: distributed3. When one side of a double-sided pc board is used forground, it is called a ____________________.

ANS: ground-plane

4. Interactions between parts of an RF circuit can be

reduced by using ____________________ betweenthem. ANS: shielding5. In high-frequency RF circuits, the placement of wires

and ____________________ can be critical.ANS: components

6. A ____________________ circuit is used to removeRF from the DC voltage bus. ANS: decoupling7. 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: Q9. A value of ____________________ or more for Q is

required for the approximate tuned circuit equationsto be valid. ANS: 10

10. In a class C RF amplifier, the

____________________ extracts one frequency from allthe 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: neutralization12. 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, parallel14. To produce sum and difference frequencies, a mixer

must be a non-____________________ circuit.

ANS: linear15. At some bias point, a diode or a transistor can act asa ____________________-law mixer.

ANS: squareSHORT ANSWER

1. What inductance would you use with a 47-pF

capacitor to make a tuned circuit for 10 MHz?

ANS:5.4 QH2. 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 valueof the feedback ratio to satisfy the Barkhausen criteria?ANS:0.1

4. What is the advantage of a Clapp oscillator comparedto a Colpitts oscillator?

ANS:

It is more stable because it "swamps" the devicecapacitance with large value capacitors in the feedbackdivider.

5. If a varactor has a capacitance of 90 pF at zero volts,

what will be the capacitance at 4 volts?

ANS:30 pF6. 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 percentageis 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 MHz8. Suppose the phase-locked-loop frequency synthesizerof Figure 2.39 has a reference frequency of 1 MHz

and a fixed-modulus divider of 10. What should be thevalue of the programmable divider to get an output

frequency of 120 MHz?

ANS:12

Chapter 3: Amplitude ModulationMULTIPLE CHOICE

1. AM stands for:

a. Audio Modulation c. Angle Modulationb. Amplitude Modulation d. Antenna Modulation

ANS: B

2. The "envelope" of an AM signal is due to:a. the baseband signal c. the amplitude signal

b. the carrier signal d. none of the above ANS: A

3. If the audio Va sin([at) modulates the carrierVc

sin([ct), then the modulation index, m, is:

a. m= [a / [c c. m= (Va / Vc)2

b. m= Va / Vc d. m= Va / [a ANS: B4. The equation for full-carrier AM is:

a. v(t) = (Ec +Em) vsin([ct) c. v(t) = (EcvEm) vsin([mt)

vsin([ct)

b. v(t) = (Ec +Em) vsin([mt) + sin([ct) d. v(t) = (Ec +Em

sin([mt)) vsin([ct) ANS: D5. Overmodulation causes:

a. distortion c. both a and b

b. splatter d. none of the above ANS: C6. The peak voltage of an AM signal goes fromEmax to

Emin. The modulation index, m, is:a. m=Emin/Emax c. m= (EmaxEmin) / (Emax +Emin)

b. m=Emax /Emind. m= (Emax +Emin) / (EmaxEmin)

ANS: C

7. IfVa sin([at) amplitude modulates the carrierVc

sin([ct), it will produce the frequencies:

a. [c + [a and [c[a c. [c + [a and 2[c + 2[a

b. ([c + [a)/2 and ([c[a)/2 d. none of the above


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ANS: A8. 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 vcarrier powerANS: 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 modulationindex, you would expect:

a. the audio to get louder at the receiver c. the signal-to-

noise ratio to increaseb. the received RF signal to increase d. all of the aboveANS: D

11. The modulation index can be derived from:

a. the time-domain signal c. both a and bb. the frequency-domain signal d. none of the aboveANS: C

12. The main problem in using quadrature AM would

be:a. requires too much bandwidth c. incompatibility with

ordinary AM radiosb. requires too much power d. all of the aboveANS: C

13. As compared to plain AM, SSB AM:

a. is more efficientb. requires a more complex demodulator circuitc. requires less bandwidth

d. all of the above ANS: D

14. The SC in SSB SC stands for:a. single-carrier c. sideband-carrier

b. suppressed-carrier d. none of the aboveANS: B15. PEP stands for:

a. Peak Envelope Power c. Peak Envelope Product

b. Peak Efficiency Power d. none of the aboveANS: A

16. If an SSB transmitter radiates 1000 watts at peakmodulation, 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 noiseb. commercial AM stations use low power

c. commercial AM stations have a narrow bandwidth

d. all of the above ANS: C18. The type of information that can be sent using AMis:

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 ofm1 and the other tone causes a

modulation index ofm2. The total modulation index is:

a. m1 + m2 c. sqrt(m1vm2 + m2 vm1)

b. (m1 + m2) / 2 d. sqrt(m1vm1 + m2vm2) ANS: D20. 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: CCOMPLETION

1. An advantage of AM is that the receiver can be very ____________________. ANS: simpl2. A disadvantage of AM is its ____________________

use of power. ANS: inefficient

3. The ____________________ of an AM signalresembles the shape of the baseband signal.

ANS: envelope4. 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 sam6. Compared to the USB, the power in the LSB is

____________________. ANS: the sam

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 constant9. 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 Hz11. With a 1-MHz carrier, if the LSB extends down to

990 kHz, then the USB will extend up to ____________________. ANS: 1010 kH

12. If an AM transmitter puts out 100 watts with nomodulation, it will put out ____________________

watts with 100% modulation. ANS: 150SHORT ANSWER

1. An AM transmitter generates 100 watts with 0%

modulation. How much power will it generate with 20%modulation? ANS:102 watts2. If the carrier power is 1000 watts, what is the power in

the USB at 70.7% modulation? ANS:125 watts3. 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: 0.7074. You look at an AM signal with an oscilloscope andsee that the maximum Vpp is 100 volts and the

minimum Vpp is 25 volts. What is the modulation index?

ANS: 0.65. 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


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Chapter 4: Angle ModulationMULTIPLE CHOICE

1. The FM modulation index:

a. increases with both deviation and modulationfrequency

b. increases with deviation and decreases withmodulation frequency

c. decreases with deviation and increases withmodulation frequency

d. is equal to twice the deviation ANS: B2. 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 oscillatorc. differentiate the modulating signal before applying to

the PM oscillatord. differentiate the signal out of the PM oscillator

ANS: A3. 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 negligibleANS: D

4. Mathematically, the calculation of FM bandwidthrequires the use of:

a. ordinary trigonometry and algebra c. Taylor series b. Bessel functions d. fractals ANS: B

5. FM bandwidth can be approximated by:

a. Armstrong's Rule c. Carson's Rule b. Bessel's Rule d. none of the above ANS: C6. NBFM stands for:

a. National Broadcast FM c. Near Band FM

b. Non-Broadcast FM d. Narrowband FMANS: D7. 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: B8. An FM receiver switching suddenly between twostations on nearby frequencies is called:

a. the capture effect c. the "two-station" effect

b. the threshold effect d. none of the aboveANS: A

9. Pre-emphasis is used to:a. increase the signal to noise ratio for higher audiofrequencies

b. increase the signal to noise ratio for lower audio

frequenciesc. increase the signal to noise ratio for all audiofrequencies

d. allow stereo audio to be carried by FM stations

ANS: A

10. A pre-emphasis of 75 Qs refers to:a. the time it takes for the circuit to work

b. the "dead time" before de-emphasis occursc. the time delay between the L and R channels

d. the time-constant of the filter circuits used

ANS: D11. FM stereo:a. uses DSBSC AM modulation c. has a higher S/N than

mono FMb. is implemented using an SCA signal d. is not

compatible with mono FM ANS: A

12. An SCA signal:a. can use amplitude modulation c. is monaural

b. can use FM modulation d. all of the above ANS: D

13. The modulation index of an FM signal can be

determined readily:a. using measurements at points whereJ0 equals one

b. using measurements at points whereJ0 equals zero

c. using measurements at points where the deviation

equals zerod. only by using Bessel functions ANS: BCOMPLETION

1. FM and PM are two forms of ____________________ modulation. ANS: angle

2. PM is extensively used in ____________________

communication. ANS: data3. Compared to AM, the signal-to-noise ratio of FM isusually ____________________. ANS: better

4. Compared to AM, the bandwidth of FM is usually

____________________. ANS:wider/greater5. FM transmitters can use Class

____________________ amplifiers since amplitudelinearity is not important. ANS: C6. Both the power and amplitude of an FM signal

____________________ as modulation is applied.

ANS: stay constant7. In FM, the frequency deviation is proportional to the

instantaneous ____________________ of themodulating signal. ANS: amplitude

8. The frequency deviation of an FM signal occurs at arate equal to the ____________________ of the

modulating signal. ANS: frequency9. Mathematically, the number of sidebands in an FM

signal is ____________________. ANS: infinite10. As FM sidebands get farther from the center

frequency, their power _________. ANS: decreases

11. Mathematically, the value of an FM modulationindex can be as high as ________. ANS: any number12. In FM, as the modulating frequency decreases, the

modulation index ___________. ANS: increases13. In FM, as the frequency deviation decreases, the

modulation index _________. ANS: decreases


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14. As the FM modulation index increases, the numberof significant sidebands ______. ANS: increases15. For certain values ofmf, such as 2.4, the amplitude of

the carrier frequency ____________________.

ANS: disappears/goes to zero

16. The bandwidth of an FM signal can be approximatedusing __________ rule. ANS: Carson's17. FM bandwidth can be calculated precisely using

________ functions. ANS: Bessel18. The _________ effect is characteristic of FM

reception in a noisy environment. ANS: threshold19. The ____________________ effect is seen when anFM 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: carrierSHORT ANSWER

1. If a 2-volt instantaneous value of modulating signal

amplitude causes a 10-kHz deviation in carrierfrequency, what is the deviation sensitivity of the

modulator? ANS: 5 kHz / volt

2. If a 2-kHz audio tone causes a frequency deviation of4 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 2kHz /V, what will be the modulation index caused bya 1-volt, 1-kHz audio signal? ANS:2

5. At a modulation index of 2, how much power is in thecarrier 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 watts7. 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 kHz9. 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 ignoresidebands containing less than 1% of the total power?ANS:30 kHz

10. How would you use the fact that J0 is zero for certainknown values ofmf(2.4, 5.5, etc) to measure thefrequency deviation of an FM modulator?

ANS:Use an audio frequency generator to modulate the FMcarrier. Using a spectrum analyzer, adjust the audio

frequency until the carrier amplitude vanishes. Record

the audio frequency. Then do the calculation: H=

fmvmfwhere mfwill have one of the known values. Forexample, iffmis measured to be 2 kHz when mfis

5.5, then His 11 kHz.

Chapter 5: TransmittersMULTIPLE CHOICE

1. The ability to change operating frequency rapidly

without a lot of retuning is called:a. agility c. VFO

b. expansion d. spread-spectrum ANS: A2. The difference between the DC power into atransmitter and the RF power coming out:

a. is a measure of efficiency c. may require watercooling

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 loudc. a smaller number of signalsd. none of the above ANS: B

4. ALC stands for:

a. Amplitude Level Control c. Accurate Level Controlb. Automatic Level Control d. none of the above

ANS: B5. 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 aboveANS: D6. With high-level AM:

a. all RF amplifiers can be nonlinear c. minimum RFpower is required

b. minimum modulation power is required d. all of the

above ANS: A7. With high-level AM:a. the RF amplifiers are typically Class A c. the RF

amplifiers are typically Class C

b. the RF amplifiers are typically Class B d. the RFamplifiers are typically Class AB ANS: C8. With low-level AM:

a. the RF amplifiers must be Class A c. the RFamplifiers must be linear

b. the RF amplifiers must be Class B d. the RF

amplifiers must be low-power ANS: C9. 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 amplifierb. the power level of the modulation d. none of the

above ANS: D11. 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


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12. In an AM transmitter with 100% modulation, thevoltage 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: C13. Practical transmitters are usually designed to drive aload 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 impedancematching?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 insteadof an antenna, the resistor must be:

a. wire-wound c. 1% tolerance or better

b. noninductive d. all of the above ANS: B17. A Class D amplifier is:

a. very efficient c. essentially pulse-duration modulation

b. essentially pulse-width modulation d. all of the above

ANS: D18. To generate a SSB signal:a. start with full-carrier AM c. start with a quadrature

signal b. start with DSBSC d. all of the above ANS: B

19. The carrier is suppressed in:

a. a balanced modulator c. a frequency multiplier

b. a mixer d. none of the above ANS: A20. To remove one AM sideband and leave the other you

could use:

a. a mechanical filter c. both a and b b. a crystal filter d. none of the above ANS: C21. A direct FM modulator:

a. varies the frequency of the carrier oscillatorb. integrates the modulating signal

c. both a and b

d. none of the above ANS: A22. An indirect FM modulator:a. requires a varactor in the carrier oscillator

b. varies the phase of the carrier oscillatorc. both a and bd. none of the above ANS: B

23. AFC stands for:a. Amplitude to Frequency Conversion c. AutomaticFrequency Control

b. Automatic Frequency Centering d. Audio Frequency

Control ANS: C24. Frequency multipliers are:a. essentially balanced modulators c. essentially mixers

b. essentially Class C amplifiers d. none of the aboveANS: B25. 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 requiredvalued. the deviation is altered ANS: CCOMPLETION

1. The accuracy and stability of a transmitter frequency

is fixed by the __________ oscillator. ANS: carrier2. In the USA, the ____________________ setsrequirements for accuracy and stability of a transmitter's

frequency. ANS: FCC

3. In Canada, _________________________ sets

requirements for accuracy and stability of a transmitter'sfrequency. ANS: Industry Canada4. Frequency ____________________ is the ability of a

transmitter to change frequency without a lot ofretuning. ANS: agility

5. Power output of SSB transmitters is rated by

____________________. ANS: PEP6. Reducing the dynamic range of a modulating signal is

called _____________. ANS: compression

7. The opposite of compression is called

____________________. ANS: expansion8. ALC is a form of ______. ANS: compression9. High-level modulation allows the RF amplifiers to

operate more ___________. ANS: efficiently10. Low-level modulation requires the RF amplifiers to

be ____________________. ANS: linear

11. To isolate the oscillator from load changes, a

___________ stage is used. ANS: buffer12. The peak collector voltage in a Class C RF amplifier

is ____________________ than the DC supply

voltage. ANS: higher13. Most practical transmitters are designed to operateinto a ___________-ohm load. ANS: 50

14. Transmitters built with transistor RF amplifiers oftenuse a ____________________ network for impedance

matching. ANS: T

15. Matching networks also act as filters to help reduce ____________________ levels. ANS: harmonic16. Severe impedance ____________________ can

destroy a transmitter's output stage. ANS: mismatch17. Transceivers combine a transmitter and a _________into one "box". ANS: receiver

18. To allow a high modulation percentage, it iscommon to modulate the ____________________ aswell as the power amplifier in transistor modulators.

ANS: driver

19. Pulse-width modulation is the same as pulse- _________ modulation. ANS: duratio


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20. Switching amplifiers are sometimes called Class _____________ amplifiers. ANS: D21. 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 witha _______________ signal. ANS: DSBSC23. Indirect FM is derived from _______________

modulation. ANS: phase24. Using a varactor to generate FM is an example of a

_____________ modulator. ANS: reactance25. The modern way to make a stable VFO is to make it

part of a ____________________ loop.

ANS: phase-lockedSHORT ANSWER

1. If a 50-MHz oscillator is accurate to within 0.001%,what is the range of possible frequencies?

ANS:50 MHz s500 hertz2. 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 doesthe modulation amplifier deliver for 100%

modulation? ANS:50 watts

4. If the final RF amplifier of an AM transmitter ispowered by 100 volts DC, what is the maximumcollector voltage at 100% modulation?

ANS:400 volts

5. Suppose the output of a balanced modulator has acenter 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 idealcrystal filter? ANS:10.005 MHz

6. Suppose you have generated a USB SSB signal with a

nominal carrier frequency of 10 MHz. What is theminimum frequency the SSB signal can be mixed with

so that the output signal has a nominal carrierfrequency of 50 MHz? ANS:40 MHz

7. Suppose you have an FM modulator that puts out 1MHz carrier with a 100-hertz deviation. If frequency

multiplication is used to increase the deviation to 400hertz, what will be the new carrier frequency?

ANS:4 MHz8. 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 adeviation of 20 kHz.ANS:

First, put the signal through a frequency doubler to get a20-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 CHOICE

1. The two basic specifications for a receiver are:

a. the sensitivity and the selectivityb. the number of converters and the number of IFsc. 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 ANS3. Trimmers and padders are:a. two types of adjusting tools c. small adjustable

inductors

b. small adjustable resistors d. small adjustablecapacitors 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 signalc. the increase of wire resistance with frequency

d. none of the above ANS: C5. The "front end" of a receiver can include:a. the tuner c. the mixer

b. the RF amplifier d. all of the above ANS: D

6. "IF" stands for:a. intermediate frequency c. indeterminate frequency

b. intermodulation frequency d. image frequencyANS: A

7. AGC stands for:a. Audio Gain Control c. Active Gain Control

b. Automatic Gain Control d. Active Gain Conversion

ANS: B8. 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 frequencyd. is fixed, typically at 455 kHz. ANS: C9. 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 selectivityd. it is cheaper ANS: D10. Basically, sensitivity measures:

a. the weakest signal that can be usefully received

b. the highest-frequency signal that can be usefullyreceivedc. the dynamic range of the audio amplifier

d. none of the above ANS: A11. Basically, selectivity measures:

a. the range of frequencies that the receiver can select

b. with two signals close in frequency, the ability toreceive one and reject the otherc. how well adjacent frequencies are separated by the

demodulator


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d. how well the adjacent frequencies are separated in themixer ANS: B12. When comparing values for shape factor:

a. a value of 1.414 dB is ideal c. a value of 1.0 is ideal

b. a value of 0.707 is ideal d. there is no ideal value

ANS: C13. When comparing values for shape factor:a. a value of 2 is better than a value of 4 c. both values

are basically equivalentb. a value of 4 is better than a value of 2 d. none of the

above ANS: A14. Distortion in a receiver can occur in:a. the mixer c. the IF amplifiers

b. the detector d. all of the above ANS: D

15. Phase distortion is important in:

a. voice communications systems c. monochrome videoreceivers

b. color video receivers d. all of the above ANS: B

16. The response of a receiver to weak signals is usuallylimited by:

a. the AGC c. the dynamic range of the receiver

b. noise generated in the receiver d. the type of detectorcircuit being used ANS: B17. Image frequencies occur when two signals:

a. are transmitted on the same frequency

b. enter the mixer, with one being a reflected signalequal to the IF frequencyc. enter the mixer, one below and one above the local

oscillator by a difference equal to theIF

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 mixing c. prior to detection

b. prior to IF amplification d. images cannot be rejected

ANS: A19. 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 mixerc. having a narrowband RF amplifier before the mixer

d. none of the above ANS: C

20. A common AM detector is the:a. PLL c. ratio detector

b. envelope detector d. all of the above ANS: B

21. An FM detector is the:a. PLL c. quadrature detector

b. ratio detector d. all of the above ANS: D

22. Germanium diodes are used in AM detectorsbecause:a. they are faster than silicon diodes

b. they are cheaper than silicon diodes

c. they minimize distortion from nonlinearityd. all of the above ANS: C23. A common SSB detector is:

a. a PLL c. a BFO b. a diode d. a product detector ANS: D24. BFO stands for:

a. Beat Frequency Oscillator c. Bipolar Frequency

Oscillator

b. Barrier Frequency Oscillator d. Bistable FrequencyOscillator ANS: A25. To demodulate both SSB and DSBSC, you need to:

a. use a Foster-Seeley discriminatorb. reinject the carrier

c. use double conversiond. use one diode for SSB and two diodes for DSBSCANS: B

26. Which would be best for DSBSC:

a. carrier detection c. envelope detection

b. coherent detection d. ratio detection ANS: B27. An FM detector that is not sensitive to amplitudevariations is:

a. Foster-Seeley detector c. a PLL detector b. a quadrature detector d. all of the above ANS: C

28. The function of a limiter is:

a. to remove amplitude variations c. to limit dynamicrange

b. to limit spurious responses d. to limit noise response

ANS: A

29. Suppressing the audio when no signal is present iscalled:a. AGC c. AFC

b. squelch d. limiting ANS:30. LNA stands for:

a. Limited-Noise Amplifier c. Low-Noise Audio

b. Low-Noise Amplifier d. Logarithmic Noise

Amplification ANS: B31. AFC stands for:

a. Audio Frequency Compensator c. Automatic

Frequency Controlb. Autodyne Frequency Compensation d. AutonomousFrequency 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 frequencyd. none of the above ANS: B33. SAW stands for:

a. Symmetrical Audio Wave c. Silicon-Activated Waferb. Surface Acoustic Wave d. Software-Activated WaveANS: B

34. The important property of a SAW is:a. it stabilizes the audio in a receiver c. it is a stable

bandpass filter

b. it allows software radios to be built d. none of the

above ANS: C35. The main function of the AGC is to:a. keep the gain of the receiver constant


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b. keep the gain of the IF amplifiers constantc. keep the input to the detector at a constant amplituded. all of the above ANS: C

36. DSP stands for:

a. Dynamic Signal Properties c. Distorted Signal Packet

b. Direct Signal Phase d. Digital Signal ProcessorANS: D37. SINAD stands for:

a. Sinusoidal Amplitude Distortionb. Signal and Noise Amplitude Distortion

c. Signal-plus-Noise-to-Noise Ratiod. Signal-plus-Noise and Distortion-to-Noise andDistortion Ratio ANS: D

38. TRF stands for:

a. Tuned Radio Frequency c. Transmitted Radio

Frequencyb. Tracking Radio Frequency d. Tuned ReceiverFunction ANS: ACOMPLETION

1. Almost all modern receivers use the

_________________________ principle.

ANS: superheterodyne2. 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 isadjusted, the other changes with it. ANS: track4. The __________effect causes the resistance of wire to

increase with frequency. ANS: skin5. The superhet was invented in the year

______________. ANS: 1918

6. In a receiver, the _______________ refers to the input

filter and RF stage. ANS: front end7. In a superhet, the output of the ___________ goes to

the IF amplifiers. ANS: mixer

8. In a superhet, the __________ frequency is thedifference between the local oscillator frequency and thereceived signal frequency. ANS: intermediate IF

9. The ______ circuit adjusts the gain of the IFamplifiers in response to signal strength. ANS: AGC

10. An ____________________ converter uses the same

transistor for both the local oscillator and the mixer.ANS: autodyne11. In low-side injection, the local oscillator is

____________________ than the received signalfrequency. ANS: lower12. ____________________ is the ability of a receiver

to separate two signals that are close to each other infrequency. ANS: Selectivity13. ____________________ is the ability of a receiver

to receive and successfully demodulate a very weak

signal. ANS: Sensitivity14. A receiver with two different IF frequencies is calleda double-________ receiver. ANS: conversion

15. A multiple-conversion receiver will have betterrejection of _______ frequencies. ANS: image16. A demodulator is also called a ________________.

ANS: detector

17. An ______________ detector uses a diode to half-

wave rectify an AM signal. ANS: envelope18. A _______________ detector is used for SSBsignals. ANS: product

19. A BFO produces a locally generated ____________________. ANS: carrie

20. A DSBSC signal requires a ____________________detection circuit. ANS: coherent21. FM detectors have a characteristic

__________________-shaped curve. ANS: S

22. While still commonly found, the Foster-Seeley and

ratio detectors are ______. ANS: obsolescent23. Unlike the PLL detector, the quadrature detector issensitive to changes in ____________________ of the

input signal. ANS: amplitude24. 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 tohave 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 localoscillator is __________. ANS: fixed constant32. Typically, AGC reduces the gain of the __________

amplifiers. ANS: IF33. An ____________________-meter is designed to

indicate signal strength in many communications

receivers. ANS: S34. The effectiveness of FM ____________ is measured

by a receivers quieting sensitivity. ANS: limiting

35. A ____________________ refers to any kind of FMor PM detector. ANS: discriminatorSHORT ANSWER

1. Suppose the bandwidth of a tuned circuit is 10 kHz at1 MHz. Approximately what bandwidth would youexpect 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 receiveris tuned to 5 MHz? ANS:6 MHz


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3. An IF filter has a60 dB bandwidth of 25 kHz and a 6 dB bandwidth of 20 kHz. What is the shapefactor 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 MHz5. Suppose a SSB receiver requires an injected frequency

of 1.5 MHz. What would be the acceptablefrequency range of the BFO if the maximum acceptable

baseband shift is 100 hertz?ANS:1.5 MHz s100 hertz6. The transformer of a double-tuned IF amplifier has aQof 25 for both primary and secondary. What valueofkc 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 couplingneed to get a bandwidth of 100 kHz? ANS:0.01

Chapter 7: Digital CommunicationsMULTIPLE CHOICE1. The first digital code was the:a. ASCII code c. Morse code

b. Baudot code d. none of the above ANS: C

2. In digital transmission, signal degradation can be

removed using:a. an amplifier c. a regenerative repeater

b. a filter d. all of the above ANS: C

3. TDM stands for:a. Time-Division Multiplexing c. Ten-Digital

Manchester

b. Time-Domain Multiplexing d. Ten Dual-ManchesterANS: A4. Hartley's Law is:

a. I= ktB c. C=B log2(1 + S/N)

b. C= 2B log2Md. SR = 2fmax ANS: A5. The Shannon-Hartley theorem is:a. I= ktB c. C=B log2(1 + S/N)

b. C= 2B log2Md. SR = 2fmax ANS: B6. The Shannon Limit is given by:

a. I= ktB c. C=B log2(1 + S/N)

b. C= 2B log2Md. SR = 2fmax ANS: C7. The Nyquist Rate can be expressed as:a. I= ktB c. C=B log2(1 + S/N)

b. C= 2B log2Md. SR = 2fmax ANS: D

8. Natural Sampling does not use:a. a sample-and-hold circuit c. a fixed sample rate

b. true binary numbers d. an analog-to-digital converterANS: 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. ANS: D

10. Foldover distortion is caused by:a. noise c. too few samples per second

b. too many samples per second d. all of the above

ANS: C

11. The immediate result of sampling is:

a. a sample alias c. PCM b. PAM d. PDM ANS12. Which of these is not a pulse-modulation technique:

a. PDM c. PPM b. PWM d. PPS ANS

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 weakestdiscernible signal

b. the maximum rate of conversion to the minimum rateof conversion

c. the maximum bits per sample to the minimum bits per

sampled. 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-ratesc. preserve dynamic range while keeping bit-rate lowd. maximize the useable bandwidth in digital

transmission ANS: C16. In North America, companding uses:

a. the Logarithmic Law c. the ELaw (alpha law)

b. the A Law d. the QLaw (mu law) ANS: D17. In Europe, companding uses:

a. the Logarithmic Law c. the ELaw (alpha law)

b. the A Law d. the QLaw (mu law) ANS: B18. Codec stands for:a. Coder-Decoder c. Code-Compression

b. Coded-Carrier d. none of the above ANS: A

19. A typical codec in a telephone system sends andreceives:

a. 4-bit numbers c. 12-bit numbers b. 8-bit numbers d. 16-bit numbers ANS: B20. Compared to PCM, delta modulation:

a. transmits fewer bits per sample c. can suffer slope

overloadb. requires a much higher sampling rate d. all of the

above ANS: D21. 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 largeANS: B22. Compared to PCM, adaptive delta modulation can

transmit voice:


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a. with a lower bit rate but reduced quality c. only overshorter 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: C24. Manchester coding:

a. is a biphase codeb. has a level transition in the middle of every bit period

c. provides strong timing informationd. all of the above ANS: D25. The number of framing bits in DS-1 is:

a. 1 c. 4

b. 2 d. 8 ANS: A

26. Framing bits in DS-1 are used to:a. detect errors c. synchronize the transmitter andreceiver

b. carry signaling d. all of the above ANS: C27. So-called "stolen" bits in DS-1 are used to:

a. detect errors c. synchronize the transmitter and

receiver b. carry signaling d. all of the above ANS: B

28. The number of bits per sample in DS-1 is:

a. 1 c. 4

b. 2 d. 8 ANS: D29. The number of samples per second in DS-1 is:a. 8 k c. 64 k

b. 56 k d. 1.544 v106 ANS: A30. 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 codingb. bipolar RZ AMI coding d. pulse-width coding

ANS: B33. 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 c. fiber-optic cable

b. coaxial cable d. microwave ANS: A35. "Signaling" is used to indicate:

a. on-hook/off-hook condition c. ringing

b. busy signal d. all of the above ANS: D36. A vocoder implements compression by:a. constructing a model of the transmission medium

b. constructing a model of the human vocal systemc. finding redundancies in the digitized data

d. using lossless techniques ANS: B

37. Compared to standard PCM systems, the quality ofthe output of a vocoder is:a. much better c. about the same

b. somewhat better d. not as good ANS: DCOMPLETION

1. Digitizing a signal often results in ________________transmission quality. ANS: improved better2. To send it over an analog channel, a digital signal

must be _______ onto a carrier. ANS: modulated3. To send it over a digital channel, an analog signal

must first be ____________________. ANS: digitized4. In analog channels, the signal-to-noise ratio of ananalog 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: binary6. 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 anothersource of ____________________ in digital

transmission systems. ANS: error/noise

9. ____________________-division multiplexing is

easily done in digital transmission. ANS: Time10. All practical communications channels are band-

____________________. ANS: limite

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-Hartley13. The ________________ limit gives the maximum

rate of data transmission for a given bandwidth and a

given signal-to-noise ratio. ANS: Shannon14. ____________________ sampling is done without asample-and-hold circuit. ANS: Natural

15. The ____________________ Rate is the minimumsampling rate for converting analog signals to digital

format. ANS: Nyquist

16. _______ distortion occurs when an analog signal issampled at too slow a rate. ANS: Foldover17. ____________________ means that higher

frequency baseband signals from the transmitter "assumethe identity" of low-frequency baseband signals at thereceiver when sent digitally. ANS: Aliasing

18. The output of a sample-and-hold circuit is a pulse- ______ modulated signal. ANS: amplitude19. ________ modulation is the most commonly used

digital modulation scheme. ANS: Pulse-code

20. ____________________ noise results from theprocess of converting an analog signal into digitalformat. ANS: Quantizing


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21. _________ is used to preserve dynamic range usinga reasonable bandwidth. ANS: Companding22. In North America, compression is done using the

_______-law equation. ANS:Qmu

23. In Europe, compression is done using the

_________-law equation. ANS: A24. A ____________________ is an IC that converts avoice signal to PCM and vice versa. ANS: codec

25. In a PCM system, the samples of the analog signalare first converted to ________________ bits

before being compressed to 8 bits. ANS: 12

26. The number of bits per sample transmitted in deltamodulation is ____________________. ANS:1/one27. Delta modulation requires a ____________________

sampling rate than PCM for the same quality of

reproduction. ANS: higher28. _______ noise is produced by a delta modulator ifthe analog signal doesn't change. ANS: Granular

29. In delta modulation, _________ overload can occur

if the analog signal changes too fast. ANS: slope30. The ____________________ size varies in adaptive

delta modulation. ANS: step31. Adaptive delta modulation can transmit PCM-qualityvoice at about ____________________ the bit rate

of PCM. ANS: half

32. Unipolar NRZ is not practical because most channelsdo not have ______ continuity. ANS: DC33. In AMI, binary ones are represented by a voltage that

alternates in ________. ANS: polarity

34. Long strings of ____________________ should beavoided in AMI. ANS: zeros

35. Manchester code has a level ______________ in thecenter of each bit period. ANS: transition36. Manchester coding provides ________________

information regardless of the pattern of ones and

zeros. ANS: timing37. There are ____________________ channels in a DS-

1 frame. ANS: 2438. DS-1 uses a _______________ bit to synchronize

the transmitter and receiver. ANS: framing39. In DS-1, each channel is sampled ______________

times per second. ANS: 800040. Data is carried over a T-1 line at a rate of _________

bits per second. ANS: 1.544v10641. 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: sixth43. ____________________ compression transmits allthe data in the original signal but uses fewer bits to do

it. ANS: LosslessSHORT ANSWER

1. Use Hartley's Law to find how much time it wouldtake to send 100,000 bits over a channel with a

bandwidth of 2,000 hertz and a channel constant ofk=

10. ANS:5 seconds

2. Use the Shannon-Hartley theorem to find the

bandwidth required to send 12,000 bits per second if thenumber of levels transmitted is 8. ANS:2000 hertz3. What is the Shannon Limit of a channel that has a

bandwidth of 4000 hertz and a signal-to-noise ratio of15? ANS:16 kbps

4. What is the minimum required number of samples persecond to digitize an analog signal with frequencycomponents 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 dB6. What is the approximate data rate for a system using 8

bits per sample and running at 8000 samples persecond? ANS:64 kbps

7. If bits were "stolen" from every DS-1 frame, what

would the useable data-rate be for each channel in theframe? ANS:56 kbps

8. Assuming maximum input and output voltages of 1

volt, what is the output voltage of a Q-law compressorif the input voltage is 0.388 volt? ANS:0.833 volt

Chapter 8: The Telephone SystemMULTIPLE CHOICE

1. DTMF stands for:

a. Digital Telephony Multiple Frequency c. Dual-Tone

Multifrequencyb. Dial Tone Master Frequency d. Digital Trunk MasterFrequency ANS: C

2. PSTN stands for:

a. Public Switched Telephone Network c. PrimaryService Telephone Network

b. Private Switched Telephone Network d. Primary

Service Telephone Numbers ANS: A3. POTS stands for:

a. Private Office Telephone System c. Primary

Operational Test Systemb. Primary Office Telephone Service d. Plain OldTelephone Service ANS: D

4. LATA stands for:

a. Local Access and Transport Area c. Local AreaTelephone Access

b. Local Access Telephone Area d. Local Area TransportAccess ANS: A

5. A LATA is a:a. a local calling area c. a way of accessing a tandem

officeb. a type of digital local network d. a way of accessing acentral office ANS: A


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6. Central offices are connected by:a. local loops c. both a and b

b. trunk lines d. none of the above ANS: B

7. Local loops terminate at:

a. a tandem office c. a central office

b. a toll station d. an interexchange office ANS: C8. Call blocking:a. cannot occur in the public telephone network

b. occurs on the local loop when there is an electricalpower failure

c. occurs only on long-distance cablesd. occurs when the central office capacity is exceededANS: D

9. In telephony, POP stands for:

a. Post Office Protocol c. Power-On Protocol

b. Point Of Presence d. none of the above ANS: B10. The cable used for local loops is mainly:a. twisted-pair copper wire c. coaxial cable

b. shielded twisted-pair copper wire d. fiber-opticANS: A

11. FITL stands for:

a. Framing Information for Toll Loops c. Framing InThe Loop

b. Fiber In the Toll Loop d. 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 crosstalkd. provide C-type conditioning to a local loop

ANS: B

13. DC current flows through a telephone:

a. when it is on hook c. as long as it is attached to a localloop

b. when it is off hook d. only when it is ringing

ANS: B14. The range of DC current that flows through atelephone is:

a. 20 QA to 80 QA c. 2 mA to 8 mA

b. 200 QA to 800 QA d. 20 mA to 80 mA ANS: D15. 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-hookis:a. 48 volts DC c. 90 volts DC

b. 48 volts, 20 hertz AC d. 90 volts, 20 hertz AC

ANS: A17. 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 ACANS: D

18. The bandwidth of voice-grade signals on a telephonesystem 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 consumptionANS: B21. 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 groupis:a. 6 c. 24

b. 12 d. 60 AN23. Basic FDM groups can be combined into:

a. supergroups c. jumbogroups

b. mastergroups d. all of the above ANS: D24. In telephone system FDM, voice is put on a carrier

using:

a. SSB c. PDM

b. DSBSC d. PCM ANS25. PABX stands for:a. Power Amplification Before Transmission

b. Private Automatic Branch Exchangec. Public Automated Branch Exchange

d. Public Access Branch Exchange ANS: B

26. SLIC stands for:

a. Single-Line Interface Circuit c. Subscriber LineInterface Card

b. Standard Line Interface Card d. Standard Local

Interface Circuit ANS: C27. In DS-1, bits are "robbed" in order to:a. provide synchronization c. cancel echoes

b. carry signaling d. check for errors ANS: B28. "Bit-stuffing" is more formally called:

a. compensation c. justification

b. rectification d. frame alignment ANS: C29. ISDN stands for:a. Integrated Services Digital Network c. Integrated

Services Data Networkb. Information Services Digital Network d. InformationSystems 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: D31. ADSL stands for:


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a. All-Digital Subscriber Line c. Allocated DigitalService 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: ACOMPLETION

1. A ____________________ is a local calling area.ANS: LATA2. Central offices are connected together by _______

lines. ANS: trunk

3. One central office can be connected to another

through a _________ office. ANS: tandem4. With 7-digit phone numbers, _______________thousand telephones can connect to a central

office. ANS: ten5. Call ____________________ is when it becomes

impossible for a subscriber to place a call due to an

overload of lines being used. ANS: blocking6. New ____________________ switching equipment

uses TDM to combine signals. ANS: digital

7. Most local loops still use ____________________

copper wire. ANS: twisted-pair8. As compared to a hierarchical network, a

____________________ network never needs more than

one intermediate switch. ANS: flat9. ____________________ 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: ring11. 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 tothe other. ANS: green

13. A telephone is said to have __________ the linewhen the central office sends it dial tone. ANS: seized

14. The ____________________ functions are provided

by a SLIC. ANS: BORSCHT15. A ____________________ coil prevents loss ofsignal energy within a telephone while allowing

fullduplex operation over a single pair of wires.ANS: hybrid16. In a crosspoint switch, not all _______________ can

be in use at the same time. ANS: lines17. The old carbon transmitters generated a relatively

_________ signal voltage. ANS: large

18. The generic term for Touch-Tone signaling is

____________________. ANS: DTMF19. A ______________ line provides more bandwidththan a standard line. ANS: conditioned

20. In the telephone system, amplifiers are called ____________________. ANS: repea21. 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 atypical telephone receiver. ANS: C-message23. In FDM telephony, the modulation is usually

____________________. ANS:SSB/SSBSC24. In FDM telephony, ____________________ bands

separate the channels in a group. ANS: guard25. Because of "bit robbing", a channel in a DS-1 frameallows 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 andtwelfth frames. ANS: superframe27. In DS-1C, ___________ bits are used to compensate

for differences between clock rates. ANS: stuff28. Busy and dial tone are referred to as ____________

signals because they use the same pair of wires as the

voice signal. ANS: in-channel29. SS7 is the current version of __________________

signaling. ANS: common-channel

30. SS7 is a ____________________-switched data

network. ANS: packet31. In ISDN, the ____________________ channel isused for common-channel signaling. ANS: D

32. In ISDN, the ____________________ channels areused 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 inthe opposite direction. ANS: greater/fasterSHORT ANSWER

1. For a certain telephone, the DC loop voltage is 48 Von hook and 8 V off hook. If the loop current is 40

mA, what is the DC resistance of the local loop?

ANS:1000 ohms2. For a certain telephone, the DC loop voltage is 48 Von hook and 8 V off hook. If the loop current is 40

mA, what is the DC resistance of the telephone?ANS:200 ohms3. Which two DTMF tones correspond to the digit "1"?

(Use the table in the text.) ANS:697 Hz and 1209 Hz4. Calculate the dB of VNL required for a channel with a3 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 dBrn6. A telephone test-tone has a level of 80 dBrn at a pointwhere the level is +5dB TLP. If C-weighting


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produces a 10-dB loss, what would the signal level be indBrnc0? ANS:65 dBrnc TLP

Chapter 9: Data TransmissionMULTIPLE CHOICE

1. 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 coaxialcable 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 charactersis 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 Interchangec. American Standard Code 2

d. Alphanumeric Standard Code for InformationInterchange ANS: B

7. BS, FF, and CR are examples of:

a. nonstandard character codes c. control characters b. escape characters d. none of the above ANS: C8. LF stands for:

a. Line Feed c. Line Forward

b. Link Feed d. Link Forward ANS: A9. UART stands for:a. Universal Asynchronous Receiver-Transmitter

b. Unidirectional Asynchronous Receiver-Transmitterc. Unaltered Received Text

d. Universal Automatic Receiver for Text ANS: A

10. In asynchronous transmission, the transmitter andreceiver 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 stopbits

d. not synchronized at all, hence the name"asynchronous" ANS: C11. 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: D12. In synchronous transmission, the frames are:

a. about the same length as ten asynchronous frames

b. much longer than asynchronous framesc. 128 bytes longd. 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 arerequiredc. it is easier to implement than asynchronous

d. all of the above ANS: A14. 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 allowedb. transmission must stop periodically forresynchronization

c. the clock circuits must be precisely adjustedd. the channel must be noise-free ANS: A

16. BISYNC:

a. is an IBM product c. requires the use of DLEb. is a character-oriented protocol d. all of the aboveANS: D

17. HDLC:

a. is an IBM product c. is identical to SDLCb. is a bit-oriented protocol d. all of the above ANS: B18. The use of flags in SDLC requires:

a. "bit-stuffing" c. FECb. different flags at either end of a frame d. ARQ

ANS: A

19. The initials ARQ are used to designate:

a. automatic request for resynchronization c. automaticreceiver queue

b. automatic request for retransmission d. automatic

request for queue ANS: B20. ARQ is used to:a. correct bit errors c. put data into a temporary buffer

b. correct synchronization problems d. none of the aboveANS: A

21. FEC stands for:

a. Fixed Error Control c. Forward Error Correctionb. Forward Error Control d. False Error ConditionANS: C

22. VRC is another name for:a. FEC c. LRC

b. ARQ d. parity ANS

23. CRC stands for:a. Control Receiver Code c. Cyclic Redundancy Check

b. Correct Received Character d. Cycle Repeat Character

ANS: C

24. Huffman codes:a. allow errors to be detected but not corrected

b. allow errors to be detected and corrected


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c. allow alphanumeric data to be correctedd. allow alphanumeric data to be compressedANS: D

25. Run-length encoding is used to:

a. encrypt data c. correct data

b. compress data d. none of the above ANS: B26. Public-key encryption:a. allows the use of digital signatures c. avoids the

"password problem"b. is used to convey symmetric keys d. all of the above

ANS: D27. SDLC stands for:a. Synchronous Data Link Control c. Synchronous Data

Link Character

b. Synchronous Data Line Control d. Synchronous Data

Line Character ANS: A28. HDLC is:a. a bit-oriented protocol c. an ISO standard

b. based on SDLC d. all of the above ANS: DCOMPLETION

1. Parallel transmission can be used only for _________

distances. ANS: short2. The term "baud" was named after Emil

_______________. ANS: Baudot

3. Data codes are also called ____________________

codes. ANS: character4. The ____________________ code is a 7-bit codecommonly used in communication between personal

computers. ANS: ASCII5. The two letters ____________________ designate the

code character used to advance a printer to the next

page. ANS: FF

6. An asynchronous frame begins with the ____________________ bit. ANS: start

7. An asynchronous frame ends with the

____________________ bit. ANS: stop8. At the end of an asynchronous frame, the line will beat the _______ level. ANS: mark/binary 1

9. An integrated circuit called a ___________ is used inan asynchronous communication system to convert

between parallel and serial data. ANS: UART

10. When receiving digital data, __________ are used tohold data until they can be read. ANS: buffers11. Synchronous communication is more ____________

than asynchronous since there are fewer "overhead" bits.ANS: efficient12. There must be sufficient 1-to-0 ______________ to

maintain synchronization in synchronous transmission.ANS: transitions13. Clock sync is derived from the stream of

____________________ bits in synchronous

transmission. ANS: data14. In the _______ protocol, each frame begins with atleast two SYN characters. ANS: BISYNC

15. In HDLC, each frame starts with an 8-bit ____________________. ANS: flag16. The first eight bits of an SDLC frame are

____________________. ANS: 01111

17. BCC stands for ____________________ check

character. ANS: block18. DLE stands for data link ____________________.ANS: escape

19. HDLC uses bit-____________________ to preventaccidental flags. ANS: stuffing

20. ____________________ errors cause manyconsecutive bits to be bad. ANS: Burst21. FEC stands for ____________________ error

correction. ANS: forward

22. An __________ scheme corrects errors by requiring

the retransmission of bad blocks. ANS: ARQ23. Parity fails when an ____________________number of bits are in error. ANS: even

24. CRC codes are particularly good at detecting ____________________ errors. ANS: burst

25. Huffman coding and run-length encoding are

examples of data _________. ANS: compression26. A ____________________ is an encoding scheme

that is not public in order to protect data. ANS: cipher

27. A ________ is often used to generate an encryption

key because it is easier to remember. ANS: password28. If the key is ____________ enough, private-keyencryption can be quite secure. ANS: long

29. Messages cannot be ____________________ using a public key. ANS: dec

30. Because it is ___________-intensive, public-key

encryption can be slow. ANS: computation

SHORT ANSWER1. How many different characters could be encoded

using a six-bit code?

ANS:642. What is the numerical difference between ASCII 'a'

and ASCII 'A' if you treat them as hexadecimal (hex)numbers? ANS:20 hex (32 decimal)

3. The ASCII codes for the characters '0' through '9' are

what hex numbers? ANS:30H to 39H4. If an asynchronous frame is used to send ASCIIcharacters in the form of bytes (8 bits), what is the

shortest time it could take to send 1000 characters ifeach bit in a frame is 1 msec long? ANS:10 sec5. Suppose an asynchronous frame holds 8 bits of data, a

parity bit, and two stop bits (it could happen). Calculatethe efficiency of the communication system. ANS:66.7%6. Suppose a synchronous frame has 16 bits of non-data

in the front and a 16-bit BCC at the end. The frame

carries 1024 bytes of actual data. Calculate the efficiencyof the communication system. ANS:97.0%


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Chapter 10: Local Area NetworksMULTIPLE CHOICE

1. CSMA stands for:

a. Client-Server Multi-Access c. Carrier Server MasterApplication

b. Carrier Sense Multiple Access d. none of the aboveANS: B

2. The CD in CSMA/CD stands for:a. Carrier Detection c. Collision Detection

b. Carrier Delay d. Collision Delay ANS: C3. The Internet is:a. a network of networks c. a very large CSMA/CD

network

b. a very large client-server network d. not really anetwork at all ANS: A

4. Most LANs:a. are based on Ethernet c. use UTP cable

b. use CSMA/CD d. all of the above ANS: D5. Dumb terminals are still used:

a. in token-passing networksb. in networks requiring central monitoringc. in networks that cannot provide central monitoring

d. none of the above ANS: B

6. In a circuit-switched network:a. communication is half-duplex only

b. each channel carries only one data streamc. connection is usually done using a bus topology

d. all of the above ANS: B7. Each computer on a network is called a:

a. hub c. node

b. token d. circuit ANS: C8. Compared to CSMA/CD systems, token-passing ringsare:

a. slower c. not as widely used

b. more expensive d. all of the above ANS: D9. The key feature of a star network is that individualworkstations are connected to:

a. a central ring c. a node b. a central bus d. none of the above ANS: D

10. On networks, long messages are divided into

"chunks" called:a. packets c. carriers

b. nodes d. tokens ANS: A

11. When two or more PCs try to access a baseband

network cable at the same time, it is called:a. a collision c. excess traffic

b. contention d. multiple access ANS: B12. When two PCs send data over a baseband networkcable at the same time, it is called:

a. a collision c. excess traffic

b. contention d. multiple access ANS: A13. One type of network that never has a collision is:a. CSMA c. token-passing

b. Ethernet d. all networks have collisions

ANS: C14. In an Ethernet-based network, a switch can be usedto reduce the number of:

a. nodes c. packets

b. users d. collisions ANS

15. The effect of too many collisions is:a. the network goes down c. the cable overheats

b. the network slows down d. data is lost ANS: B

16. MAU stands for:a. Multistation Access Unit c. Multiple Auxiliary Units

b. Multiple Access Unit d. none of the above ANS: A17. The standard that describes Ethernet-type networksis:

a. EIA 232 c. IEEE 802.3

b. IEEE 488.1 d. CCITT ITU-E ANS: C

18. Ethernet was invented by:a. IBM c. Xerox

b. INTEL d. Digital Equipment Corporation ANS: C

19. An Ethernet running at 10 Mbits / second uses:a. Manchester encoding c. NRZ encoding

b. Three-Level encoding d. AMI encoding ANS: A

20. A 100BaseT cable uses:a. fiber-optic cable c. RG-58U coaxial cable

b. twisted-pair copper wires d. 50-ohm coaxial cable

ANS: B

21. The word "Base" in 10BaseT means:a. the cable carries baseband signals

b. the cable has a base speed of 10 Mbps

c. it can be used as the base for a backbone cable systemd. none of the above ANS: A

22. The reason a CSMA/CD network has a minimum

length for packets is:

a. to increase the data rateb. to prevent packets from reaching all other nodes

during transmission

c. to make sure all other nodes hear a collision inprogressd. all of the above ANS: C

23. The reason a CSMA/CD network has a maximumlength for cables is:

a. to increase the data rate

b. to prevent packets from reaching all other nodesduring transmissionc. to make sure all other nodes hear a collision in

progressd. all of the above ANS: C24. NIC stands for:

a. Network Interface Card c. Network Interface Codeb. Network Interface Cable d. Network Internal CodeANS: A

25. 10BaseT cable typically uses:

a. a BNC connector c. an RJ45 connector b. a T connector d. an RS11 connector ANS: C26. UTP stands for:


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a. Untwisted-Pair copper wire c. UninterruptibleTerminal Packet

b. Unshielded Twisted-Pair copper wire d. Unicode Text

Packet ANS: B

27. Compared to twisted-pair telephone cables, CAT-5

cables:a. are cheaper c. allow faster bit rates

b. are easier to crimp connectors onto d. all of the above

ANS: C28. A hub:

a. sends incoming packets out to all other terminalsconnected to it

b. sends incoming packets out to specific ports

c. cannot be used in an Ethernet-type network

d. are more common in token-passing networks

ANS: A29. A switch:a. sends incoming packets out to all other terminals

connected to itb. sends incoming packets out to specific ports

c. cannot be used in an Ethernet-type network

d. are more common in token-passing networksANS: B

30. An advantage of using a switch instead of a hub is:

a. it is cheaper when used in large networks

b. it is faster when used in large networksc. it reduces the number of collisions in large networksd. all of the above ANS: C

31. Broadband LANs:a. modulate the data onto a carrier

b. use coaxial cables

c. are provided by cable TV companies for Internet

accessd. all of the above ANS: D

32. Using one node in the network to hold all the

application software is done in:a. peer-to-peer networks c. both a and b

b. client-server networks d. none of the above

ANS: B33. Record locking is used to:

a. store records securely on a server

b. prevent multiple users from looking at a documentsimultaneouslyc. prevent one user from reading a record that another

user is writing tod. none of the above ANS: C34. The software that runs a client-server network must

be:a. UNIX-based c. multitasking

b. WINDOWS-based d. Novell certified ANS: C

35. A "thin" client is:

a. basically, a PC with no disk drives c. same as a"dumb" terminal

b. a node that rarely sends data d. all of the above

ANS: ACOMPLETION

1. A LAN is a ________________ Area Network.

ANS: Local

2. The Internet is a network of ____________________.

ANS: networks3. In a ____________________ network, all nodes areconnected to a central computer. ANS: star

4. In a ____________-switched network, users have adedicated channel for the duration of communications.

ANS: circuit5. The __________ of a network describes how it is

physically connected together. ANS: topology

6. Ring networks often use _______________-passing.

ANS: token

7. A ____________________ is a short section of amessage in digital form. ANS: packet8. _______________ is when two nodes try to seize the

same cable at the same time. ANS: Contention9. A __________ occurs when two nodes transmit

simultaneously on the same baseband cable.

ANS: collision10. In CSMA/CD networks, all collisions must be

____________________. ANS: detec

11. Carrier-Sense means that a node "listens" for the

cable to be _________ before using it.ANS: quiet/free/unused/available12. A "____________________" cable links clusters of

computers together. ANS: backbone13. 100BaseT cables can reliably carry up to

____________ bits per second. ANS: 100 mega

14. In CSMA/CD, packets must have a

____________________ length to ensure that collisionsare detected. ANS: minimum

15. In CSMA/CD, the ___________ of a cable is limited

to ensure that collisions are detected. ANS: length16. A unique numerical address is provided to a node byits ____________________. ANS: NIC

17. A 100BaseTX cable is a ____________________cable. ANS: fiber-optic

18. Hubs can be ____________________ to form, in

effect, one big hub. ANS: stacked19. A switch looks at the ____________________ ofeach incoming packet. ANS: address

20. The effect of a switch is to greatly reduce ____________________. ANS: contentionSHORT ANSWER

1. Explain how a network can be a physical bus but alogical ring.ANS:

A token-passing network sends the token from node to

node in a prescribed order. So it doesn't matterhow the physical connection is made. It still works like atoken-passing ring.


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2. What is the key difference between a hub and aswitch?ANS:

A hub sends incoming packets out to all other ports on

the hub. A switch sends a packet to a specific port

based on the address in the packet.3. What is the advantage of a CSMA/CD network over a

basic star network?

ANS:If the central computer in a star network fails, the entire

network is inoperative. If a node fails in aCSMA/CD network, it can be disconnected and thenetwork still

blake edited reviewer

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