Amateur Radio Extra Class (Element 4) [#0001]

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  • 1/47 Questions

    What is meant by automatic control of a station?

    • The use of devices and procedures for control so that the control operator does not have to be present at a control point
    • A station operating with its output power controlled automatically
    • Remotely controlling a station's antenna pattern through a directional control link
    • The use of a control link between a control point and a locally controlled station
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About This Quiz


Practice Examination for a FCC issued Amateur Radio Extra Class license (2008-2012 question pool). Class members are required to enter their callsign in the box labeled "Your Name" (appearing below) for homework credit.

Amateur Radio Extra Class (Element 4) [#0001] - Quiz

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  • 2. 

    What is the only emission type permitted to be transmitted on the 60 meter band by an amateur station?

    • CW

    • RTTY Frequency shift keying

    • Single sideband, upper sideband only

    • Single sideband, lower sideband only

    Correct Answer
    A. Single sideband, upper sideband only
    Explanation
    The only emission type permitted to be transmitted on the 60 meter band by an amateur station is single sideband, upper sideband only. This means that amateur operators are not allowed to use other emission types such as CW (Continuous Wave) or RTTY (Radio Teletype) frequency shift keying on this band.

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  • 3. 

    What is the minimum number of qualified VEs required to administer an Element 4 amateur operator license examination?

    • 5

    • 2

    • 4

    • 3

    Correct Answer
    A. 3
    Explanation
    The minimum number of qualified VEs required to administer an Element 4 amateur operator license examination is 3. This is because the Federal Communications Commission (FCC) requires a minimum of three Volunteer Examiners (VEs) to be present during an examination. These VEs are responsible for administering and grading the examination to ensure that the applicant meets the necessary requirements to obtain an amateur radio license.

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  • 4. 

    What is the amateur-satellite service?

    • A radio navigation service using satellites for the purpose of self-training, intercommunication and technical studies carried out by amateurs

    • A spacecraft launching service for amateur-built satellites

    • A radio communications service using amateur stations on satellites

    • A radio communications service using stations on Earth satellites for weather information gathering

    Correct Answer
    A. A radio communications service using amateur stations on satellites
    Explanation
    The amateur-satellite service refers to a radio communications service that utilizes amateur stations on satellites. This means that amateur radio operators can use satellites to communicate with each other and engage in radio communications. This service allows amateurs to experiment, learn, and communicate with other operators using satellite technology.

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  • 5. 

    What is the Radio Amateur Civil Emergency Service (RACES)?

    • A radio service using amateur service frequencies on a regular basis for communications that can reasonably be furnished through other radio services

    • A radio service of amateur stations for civil defense communications during periods of local, regional, or national civil emergencies

    • A radio service using amateur service frequencies for broadcasting to the public during periods of local, regional or national civil emergencies

    • A radio service using local government frequencies by Amateur Radio operators for civil emergency communications

    Correct Answer
    A. A radio service of amateur stations for civil defense communications during periods of local, regional, or national civil emergencies
    Explanation
    RACES stands for Radio Amateur Civil Emergency Service, which is a radio service that utilizes amateur stations for civil defense communications during times of civil emergencies at the local, regional, or national level. This service allows amateur radio operators to provide crucial communication support during emergencies, helping to ensure the safety and well-being of communities in times of crisis.

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  • 6. 

    Who may be the control operator of an auxiliary station?

    • Any licensed amateur operator

    • Only Technician, General, Advanced or Amateur Extra Class operators

    • Only General, Advanced or Amateur Extra Class operators

    • Only Amateur Extra Class operators

    Correct Answer
    A. Only Technician, General, Advanced or Amateur Extra Class operators
    Explanation
    The control operator of an auxiliary station may only be a Technician, General, Advanced, or Amateur Extra Class operator. This means that any licensed amateur operator cannot be the control operator of an auxiliary station.

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  • 7. 

    How should you generally sign your call when attempting to contact a DX station working a "pileup" or in a contest?

    • Send your full call sign once or twice

    • Send only the last two letters of your call sign until you make contact

    • Send your full call sign and grid square

    • Send the call sign of the DX station three times, the words "this is", then your call sign three times

    Correct Answer
    A. Send your full call sign once or twice
    Explanation
    When attempting to contact a DX station working a "pileup" or in a contest, it is generally recommended to send your full call sign once or twice. This allows the DX station to clearly identify your call sign and make contact with you. By sending your full call sign, you provide the necessary information for the DX station to acknowledge and respond to your call. This approach is efficient and ensures clear communication in a busy or competitive radio environment.

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  • 8. 

    On which of the following frequencies is one likely to find FMTV transmissions?

    • 14.230 MHz

    • 29.6 MHz

    • 52.525 MHz

    • 1255 MHz

    Correct Answer
    A. 1255 MHz
  • 9. 

    Which of the following HF digital modes uses variable-length coding for bandwidth efficiency?

    • RTTY

    • PACTOR

    • MT63

    • PSK31

    Correct Answer
    A. PSK31
    Explanation
    PSK31 is the correct answer because it uses variable-length coding, which allows for efficient use of bandwidth. Variable-length coding means that different characters in PSK31 can be represented by different numbers of bits, depending on their frequency of occurrence. This allows for more frequently used characters to be represented by shorter codes, resulting in higher bandwidth efficiency. RTTY, PACTOR, and MT63 do not use variable-length coding, so they are not the correct answers.

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  • 10. 

    What is the orbital period of a satellite?

    • The point of maximum height of a satellite's orbit

    • The point of minimum height of a satellite's orbit

    • The time it takes for a satellite to complete one revolution around the Earth

    • The time it takes for a satellite to travel from perigee to apogee

    Correct Answer
    A. The time it takes for a satellite to complete one revolution around the Earth
    Explanation
    The orbital period of a satellite refers to the time it takes for the satellite to complete one revolution around the Earth. This means that the satellite will travel in a circular or elliptical path and return to its starting point after a certain amount of time. The orbital period is determined by the satellite's altitude and the gravitational pull of the Earth.

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  • 11. 

    When scheduling EME contacts, which of these conditions will generally result in the least path loss?

    • When the moon is at perigee

    • When the moon is full

    • When the moon is at apogee

    • When the MUF is above 30 MHz

    Correct Answer
    A. When the moon is at perigee
    Explanation
    When the moon is at perigee, it is at its closest point to the Earth in its orbit. This means that the distance between the Earth and the moon is shorter, resulting in less path loss. Path loss is the reduction in power density (attenuation) of an electromagnetic wave as it propagates through space. Therefore, when the moon is at perigee, the signal strength will be stronger and there will be less loss of signal during EME (Earth-Moon-Earth) contacts.

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  • 12. 

    From the contiguous 48 states, in which approximate direction should an antenna be pointed to take maximum advantage of auroral propagation?

    • South

    • North

    • East

    • West

    Correct Answer
    A. North
    Explanation
    Auroral propagation occurs when radio waves bounce off the ionized particles in the Earth's atmosphere during an aurora borealis (northern lights) event. To take maximum advantage of this propagation, an antenna should be pointed north. This is because the aurora borealis phenomenon is predominantly visible in the northern hemisphere, specifically near the North Pole. By pointing the antenna north, it can capture and utilize the radio waves that are reflected off the ionized particles during an aurora borealis event, maximizing the signal strength and reception.

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  • 13. 

    What is the cause of gray-line propagation?

    • At midday, the sun, being directly overhead, superheats the ionosphere causing increased refraction of radio waves

    • At twilight, solar absorption drops greatly, while atmospheric ionization is not weakened enough to reduce the MUF

    • At darkness, solar absorption drops greatly, while atmospheric ionization remains steady

    • At mid afternoon, the sun heats the ionosphere, increasing radio wave refraction and the MUF

    Correct Answer
    A. At twilight, solar absorption drops greatly, while atmospheric ionization is not weakened enough to reduce the MUF
    Explanation
    Gray-line propagation refers to a phenomenon where radio waves can travel over long distances during the transition periods between daylight and darkness. This occurs because at twilight, solar absorption drops significantly, allowing radio waves to penetrate the ionosphere more easily. However, atmospheric ionization remains steady, which means that the Maximum Usable Frequency (MUF) is not weakened enough to hinder the propagation of radio waves. Therefore, the cause of gray-line propagation is the combination of reduced solar absorption and maintained atmospheric ionization during twilight.

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  • 14. 

    Which of the following procedures is an important precaution to follow when connecting a spectrum analyzer to a transmitter output?

    • Use high quality double shielded coaxial cables to reduce signal losses

    • Attenuate the transmitter output going to the spectrum analyzer

    • Match the antenna to the load

    • All of these choices are correct

    Correct Answer
    A. Attenuate the transmitter output going to the spectrum analyzer
    Explanation
    Attenuating the transmitter output going to the spectrum analyzer is an important precaution when connecting a spectrum analyzer to a transmitter output. This is necessary to prevent damage to the spectrum analyzer due to excessive power levels. Attenuation helps to reduce the signal strength and ensure that the spectrum analyzer receives a safe and manageable signal. By attenuating the transmitter output, the spectrum analyzer can accurately measure and analyze the signal without any risk of overload or damage.

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  • 15. 

    What does the MDS of a receiver represent?

    • The meter display sensitivity

    • The minimum discernible signal

    • The multiplex distortion stability

    • The maximum detectable spectrum

    Correct Answer
    A. The minimum discernible signal
    Explanation
    The MDS of a receiver represents the minimum discernible signal. This means that it is the smallest signal that can be detected by the receiver. The MDS is an important parameter as it determines the sensitivity of the receiver and its ability to detect weak signals. A lower MDS indicates a more sensitive receiver that can detect smaller signals.

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  • 16. 

    What is the significance of voltmeter sensitivity expressed in ohms per volt?

    • The full scale reading of the voltmeter multiplied by its ohms per volt rating will provide the input impedance of the voltmeter

    • When used as a galvanometer, the reading in volts multiplied by the ohms/volt will determine the power drawn by the device under test

    • When used as an ohmmeter, the reading in ohms divided by the ohms/volt will determine the voltage applied to the circuit

    • When used as an ammeter, the full scale reading in amps divided by ohms/volt will determine the size of shunt needed

    Correct Answer
    A. The full scale reading of the voltmeter multiplied by its ohms per volt rating will provide the input impedance of the voltmeter
    Explanation
    The significance of voltmeter sensitivity expressed in ohms per volt is that it allows us to determine the input impedance of the voltmeter. By multiplying the full scale reading of the voltmeter by its ohms per volt rating, we can calculate the input impedance. This is important because the input impedance of a voltmeter affects the accuracy of its measurements. A higher input impedance means that the voltmeter will draw less current from the circuit being measured, minimizing the impact on the circuit's voltage levels.

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  • 17. 

    How can intermodulation interference between two repeaters occur?

    • When the repeaters are in close proximity and the signals cause feedback in one or both transmitter final amplifiers

    • When the repeaters are in close proximity and the signals mix in one or both transmitter final amplifiers

    • When the signals from the transmitters are reflected out of phase from airplanes passing overhead

    • When the signals from the transmitters are reflected in phase from airplanes passing overhead

    Correct Answer
    A. When the repeaters are in close proximity and the signals mix in one or both transmitter final amplifiers
    Explanation
    Intermodulation interference between two repeaters can occur when the repeaters are in close proximity and the signals mix in one or both transmitter final amplifiers. This mixing of signals can result in the creation of new frequencies that were not originally present, causing interference and degradation of the desired signals.

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  • 18. 

    How many watts are consumed in a circuit having a power factor of 0.6 if the input is 200V AC at 5 amperes?

    • 200 watts

    • 1000 watts

    • 1600 watts

    • 600 watts

    Correct Answer
    A. 600 watts
    Explanation
    The power consumed in a circuit can be calculated using the formula P = V * I * cos(θ), where P is power, V is voltage, I is current, and θ is the power factor angle. In this case, the power factor is given as 0.6, the voltage is 200V, and the current is 5 amperes. Plugging these values into the formula, we get P = 200 * 5 * 0.6 = 600 watts. Therefore, the correct answer is 600 watts.

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  • 19. 

    How long does it take for an initial charge of 20 V DC to decrease to 7.36 V DC in a 0.01-microfarad capacitor when a 2-megohm resistor is connected across it?

    • 0.02 seconds

    • 0.04 seconds

    • 20 seconds

    • 40 seconds

    Correct Answer
    A. 0.02 seconds
    Explanation
    When a resistor is connected across a capacitor, the capacitor discharges over time. The time it takes for the voltage across the capacitor to decrease is determined by the RC time constant, where R is the resistance and C is the capacitance. In this case, the resistance is 2 megohms and the capacitance is 0.01 microfarads. The RC time constant is calculated by multiplying the resistance and capacitance together, giving us 20 milliseconds. Since the time constant is the time it takes for the voltage to decrease to approximately 36.8% of its initial value, we can calculate the time it takes for the voltage to decrease to 7.36 V DC using the formula: t = RC ln(Vi/Vf), where Vi is the initial voltage and Vf is the final voltage. Plugging in the values, we get t = (2 x 10^6 ohms) x (0.01 x 10^-6 farads) x ln(20/7.36) ≈ 0.02 seconds. Therefore, the correct answer is 0.02 seconds.

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  • 20. 

    What is the phase relationship between the current through and the voltage across a parallel resonant circuit?

    • The voltage leads the current by 90 degrees

    • The current leads the voltage by 90 degrees

    • The voltage and current are in phase

    • The voltage and current are 180 degrees out of phase

    Correct Answer
    A. The voltage and current are in phase
    Explanation
    In a parallel resonant circuit, the voltage and current are in phase. This means that they reach their maximum and minimum values at the same time. The voltage and current waveforms have the same shape and are synchronized. This is because at resonance, the reactive components cancel each other out, resulting in a purely resistive circuit. Therefore, there is no phase shift between the voltage and current in a parallel resonant circuit.

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  • 21. 

    In polar coordinates, what is the impedance of a network consisting of a 300-ohm-reactance capacitor, a 600-ohm-reactance inductor, and a 400-ohm resistor, all connected in series?

    • 500 ohms at an angle of 37 degrees

    • 900 ohms at an angle of 53 degrees

    • 400 ohms at an angle of 0 degrees

    • 1300 ohms at an angle of 180 degrees

    Correct Answer
    A. 500 ohms at an angle of 37 degrees
    Explanation
    The impedance of a network consisting of components connected in series can be found by summing the individual impedances. In polar coordinates, the impedance is represented by a magnitude (in ohms) and an angle (in degrees).

    To find the impedance, we need to calculate the total reactance and resistance. The reactance of the capacitor is -300 ohms (since it is a reactance, we use a negative sign) and the reactance of the inductor is +600 ohms. The resistance is 400 ohms.

    The total impedance can be found using the formula Z = R + jX, where R is the resistance and X is the reactance. In this case, Z = 400 + j(-300 + 600) = 400 + j300.

    To find the magnitude and angle of the impedance, we can use the Pythagorean theorem and the arctangent function. The magnitude of the impedance is sqrt(400^2 + 300^2) ≈ 500 ohms. The angle of the impedance is arctan(300/400) ≈ 37 degrees.

    Therefore, the correct answer is 500 ohms at an angle of 37 degrees.

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  • 22. 

    What type of semiconductor device varies its internal capacitance as the voltage applied to its terminals varies?

    • Varactor diode

    • Tunnel diode

    • Silicon-controlled rectifier

    • Zener diode

    Correct Answer
    A. Varactor diode
    Explanation
    A varactor diode is a type of semiconductor device that can vary its internal capacitance as the voltage applied to its terminals changes. This property makes it useful for applications such as voltage-controlled oscillators, frequency multipliers, and voltage-controlled filters.

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  • 23. 

    Which of these filter bandwidths would be a good choice for use with standard double-sideband AM transmissions?

    • 1 kHz at -6 dB

    • 500 Hz at -6 dB

    • 6 kHz at -6 dB

    • 15 kHz at -6 dB

    Correct Answer
    A. 6 kHz at -6 dB
    Explanation
    A filter bandwidth of 6 kHz at -6 dB would be a good choice for use with standard double-sideband AM transmissions because it allows for a wide range of frequencies to pass through while attenuating frequencies outside of this range. Double-sideband AM transmissions typically have a bandwidth of around 5 kHz, so a 6 kHz filter would provide enough room for the transmission without cutting off any essential frequencies. Additionally, the -6 dB cutoff indicates that the filter's attenuation starts gradually, which is desirable for maintaining the integrity of the transmitted signal.

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  • 24. 

    What is a liquid-crystal display (LCD)?

    • A modern replacement for a quartz crystal oscillator which displays its fundamental frequency

    • A display that uses a crystalline liquid to change the way light is refracted

    • A frequency-determining unit for a transmitter or receiver

    • A display that uses a glowing liquid to remain brightly lit in dim light

    Correct Answer
    A. A display that uses a crystalline liquid to change the way light is refracted
    Explanation
    An LCD is a display that uses a crystalline liquid to change the way light is refracted. The liquid crystal molecules in an LCD can be manipulated to allow or block the passage of light, creating the desired image or text on the screen. By applying an electric current to the liquid crystal cells, the display can control the amount of light passing through and create different colors and shades. This technology is widely used in televisions, computer monitors, and other electronic devices for its high resolution and energy efficiency.

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  • 25. 

    What happens to the conductivity of a photoconductive material when light shines on it?

    • It increases

    • It decreases

    • It stays the same

    • It becomes unstable

    Correct Answer
    A. It increases
    Explanation
    When light shines on a photoconductive material, it increases its conductivity. This is because the photons from the light excite the electrons in the material, causing them to move more freely and conduct electricity better. This phenomenon is known as the photoconductivity effect. As a result, the material becomes more conductive when exposed to light.

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  • 26. 
    In Figure E6-5, what is the schematic symbol for a NAND gate? - ProProfs

    In Figure E6-5, what is the schematic symbol for a NAND gate?

    • 1

    • 2

    • 3

    • 4

    Correct Answer
    A. 2
    Explanation
    The schematic symbol for a NAND gate is represented by option 2.

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  • 27. 

    What type of semiconductor material contains more free electrons than pure germanium or silicon crystals? 

    • N-type

    • P-type

    • Bipolar

    • Insulated gate

    Correct Answer
    A. N-type
    Explanation
    N-type semiconductor material contains more free electrons than pure germanium or silicon crystals. In N-type semiconductors, impurities are added to the crystal lattice, which introduces extra electrons into the material. These impurities are called donor impurities, and they provide additional free electrons that can participate in electrical conduction. This makes N-type semiconductors conductive and allows for the flow of current.

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  • 28. 

    What is a Gunn diode oscillator?

    • An oscillator based on the negative resistance properties of properly-doped semiconductors

    • An oscillator based on the argon gas diode

    • A highly stable reference oscillator based on the tee-notch principle

    • A highly stable reference oscillator based on the hot-carrier effect

    Correct Answer
    A. An oscillator based on the negative resistance properties of properly-doped semiconductors
    Explanation
    A Gunn diode oscillator is an oscillator that operates based on the negative resistance properties of properly-doped semiconductors. These semiconductors are specifically designed to exhibit negative resistance, which means that as the current through the diode increases, the voltage across it decreases. This negative resistance characteristic allows the diode to generate microwave signals and function as an oscillator.

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  • 29. 

    What alternate method of determining frequency, other than by directly counting input pulses, is used by some frequency counters?

    • GPS averaging

    • Period measurement

    • Prescaling

    • D/A conversion

    Correct Answer
    A. Period measurement
    Explanation
    Some frequency counters use period measurement as an alternate method of determining frequency. Instead of directly counting input pulses, these counters measure the time period of the signal and then calculate the frequency based on the reciprocal of the period. This method is useful when the frequency of the signal is too high to be accurately counted or when the signal is irregular or intermittent. By measuring the period, the frequency counter can provide an accurate frequency measurement without relying on pulse counting.

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  • 30. 

    Which of the following is a circuit that continuously alternates between two unstable states without an external clock?

    • Monostable multivibrator

    • J-K Flip-Flop

    • T Flip-Flop

    • Astable Multivibrator

    Correct Answer
    A. Astable Multivibrator
    Explanation
    An astable multivibrator is a circuit that continuously alternates between two unstable states without the need for an external clock signal. It is commonly used in applications such as oscillators and timers. Unlike the other options listed, which require an external clock signal to change their states, the astable multivibrator is able to oscillate between its unstable states on its own. Therefore, it is the correct answer to the question.

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  • 31. 

    What is a Pi-L network, as used when matching a vacuum-tube final amplifier to a 50-ohm unbalanced output?

    • A Phase Inverter Load network

    • A network consisting of two series inductors and two shunt capacitors

    • A network with only three discrete parts

    • A matching network in which all components are isolated from ground

    Correct Answer
    A. A network consisting of two series inductors and two shunt capacitors
    Explanation
    A Pi-L network is a matching network used to match a vacuum-tube final amplifier to a 50-ohm unbalanced output. It consists of two series inductors and two shunt capacitors. This configuration helps in achieving impedance matching between the amplifier and the output, ensuring maximum power transfer and minimizing reflections. The inductors and capacitors are strategically placed to create the desired impedance transformation and provide a balanced load for the amplifier. This configuration is commonly used in RF and audio applications.

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  • 32. 

    Which of the following describes how the loading and tuning capacitors are to be adjusted when tuning a vacuum tube RF power amplifier that employs a pi-network output circuit?

    • The loading capacitor is set to maximum capacitance and the tuning capacitor is adjusted for minimum allowable plate current

    • The tuning capacitor is set to maximum capacitance and the loading capacitor is adjusted for minimum plate permissible current

    • The loading capacitor is adjusted to minimum plate current while alternately adjusting the tuning capacitor for maximum allowable plate current

    • The tuning capacitor is adjusted for minimum plate current, while the loading capacitor is adjusted for maximum permissible plate current

    Correct Answer
    A. The tuning capacitor is adjusted for minimum plate current, while the loading capacitor is adjusted for maximum permissible plate current
    Explanation
    The correct answer explains the correct procedure for adjusting the loading and tuning capacitors in a vacuum tube RF power amplifier with a pi-network output circuit. It states that the tuning capacitor should be adjusted for minimum plate current, while the loading capacitor should be adjusted for maximum permissible plate current. This ensures that the amplifier is properly tuned for maximum power transfer and efficiency.

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  • 33. 
    What is the purpose of C3 in the circuit shown in Figure E7-3? - ProProfs

    What is the purpose of C3 in the circuit shown in Figure E7-3?

    • It prevents self-oscillation

    • It provides brute force filtering of the output

    • It provides fixed bias for Q1

    • It clips the peaks of the ripple

    Correct Answer
    A. It prevents self-oscillation
    Explanation
    C3 is used in the circuit to prevent self-oscillation. Self-oscillation can occur when there is positive feedback in the circuit, causing the output to continuously oscillate or ring. By including C3 in the circuit, it helps to stabilize the circuit and prevent any unwanted oscillations from occurring.

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  • 34. 
    What voltage gain can be expected from the circuit in Figure E7-4 when R1 is 10 ohms and RF is 470 ohms? - ProProfs

    What voltage gain can be expected from the circuit in Figure E7-4 when R1 is 10 ohms and RF is 470 ohms?

    • 0.21

    • 94

    • 47

    • 24

    Correct Answer
    A. 47
    Explanation
    The voltage gain of an amplifier circuit is determined by the ratio of the feedback resistor (RF) to the input resistor (R1). In this case, when R1 is 10 ohms and RF is 470 ohms, the voltage gain can be calculated as RF/R1 = 470/10 = 47. Therefore, the expected voltage gain from the circuit is 47.

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  • 35. 

    What is the PEP output of a transmitter that has a maximum peak of 30 volts to a 50-ohm load as observed on an oscilloscope?

    • 4.5 watts

    • 9 watts

    • 16 watts

    • 18 watts

    Correct Answer
    A. 9 watts
    Explanation
    The PEP (Peak Envelope Power) output of a transmitter can be calculated using the formula PEP = (Vp^2)/(2*R), where Vp is the peak voltage and R is the load resistance. In this case, the peak voltage is given as 30 volts and the load resistance is 50 ohms. Plugging these values into the formula, we get PEP = (30^2)/(2*50) = 900/100 = 9 watts. Therefore, the correct answer is 9 watts.

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  • 36. 

    What is the necessary bandwidth of a 4800-Hz frequency shift, 9600-baud ASCII FM transmission?

    • 15.36 kHz

    • 9.6 kHz

    • 4.8 kHz

    • 5.76 kHz

    Correct Answer
    A. 15.36 kHz
    Explanation
    The necessary bandwidth of a frequency shift, baud ASCII FM transmission is determined by the formula: bandwidth = (baud rate) + (frequency shift). In this case, the baud rate is 9600 and the frequency shift is 4800 Hz. Plugging these values into the formula, we get: bandwidth = 9600 + 4800 = 14400 Hz = 14.4 kHz. Therefore, the correct answer is 15.36 kHz.

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  • 37. 

    What type of wave has a rise time significantly faster than its fall time (or vice versa)?

    • A cosine wave

    • A square wave

    • A sawtooth wave

    • A sine wave

    Correct Answer
    A. A sawtooth wave
    Explanation
    A sawtooth wave has a rise time significantly faster than its fall time. This means that the wave rises rapidly from its lowest point to its highest point, but falls back to its lowest point more gradually. The shape of a sawtooth wave resembles the teeth of a saw, hence the name. This type of wave is commonly used in music synthesis and as a test signal in electronics.

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  • 38. 

    What is the deviation ratio of an FM-phone signal having a maximum frequency swing of plus-or-minus 5 kHz and accepting a maximum modulation rate of 3 kHz?

    • 60

    • 0.167

    • 0.6

    • 1.67

    Correct Answer
    A. 1.67
    Explanation
    The deviation ratio is calculated by dividing the maximum frequency swing by the maximum modulation rate. In this case, the maximum frequency swing is plus-or-minus 5 kHz and the maximum modulation rate is 3 kHz. Therefore, the deviation ratio is 5 kHz / 3 kHz = 1.67.

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  • 39. 

    How does the total amount of radiation emitted by a directional (gain) antenna compare with the total amount of radiation emitted from an isotropic antenna, assuming each is driven by the same amount of power?

    • The total amount of radiation from the directional antenna is increased by the gain of the antenna

    • The total amount of radiation from the directional antenna is stronger by its front to back ratio

    • There is no difference between the two antennas

    • The radiation from the isotropic antenna is 2.15 dB stronger than that from the directional antenna

    Correct Answer
    A. There is no difference between the two antennas
    Explanation
    The total amount of radiation emitted by a directional (gain) antenna is the same as the total amount of radiation emitted from an isotropic antenna, assuming each is driven by the same amount of power.

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  • 40. 

    Which of the following is often determined using a Smith chart?

    • Beam headings and radiation patterns

    • Satellite azimuth and elevation bearings

    • Impedance and SWR values in transmission lines

    • Trigonometric functions

    Correct Answer
    A. Impedance and SWR values in transmission lines
    Explanation
    The Smith chart is a graphical tool used in electrical engineering to analyze and design radio frequency circuits. It is commonly used to determine impedance and standing wave ratio (SWR) values in transmission lines. The chart provides a visual representation of complex impedance, making it easier to calculate and analyze the characteristics of transmission lines. By using the Smith chart, engineers can easily determine the impedance and SWR values at any point along the transmission line, allowing for efficient design and troubleshooting of RF circuits.

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  • 41. 

    Why should an HF mobile antenna loading coil have a high ratio of reactance to resistance?

    • To swamp out harmonics

    • To maximize losses

    • To minimize losses

    • To minimize the Q

    Correct Answer
    A. To minimize losses
    Explanation
    An HF mobile antenna loading coil should have a high ratio of reactance to resistance in order to minimize losses. This is because a high reactance helps to match the impedance of the antenna to the impedance of the transmission line, resulting in efficient power transfer. By minimizing losses, the antenna can effectively radiate the signal and reduce the amount of power that is wasted as heat.

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  • 42. 

    What is the radiation pattern of two 1/4-wavelength vertical antennas spaced 1/2-wavelength apart and fed 180 degrees out of phase?

    • A cardioid

    • Omnidirectional

    • A figure-8 broadside to the axis of the array

    • A figure-8 oriented along the axis of the array

    Correct Answer
    A. A figure-8 oriented along the axis of the array
    Explanation
    When two 1/4-wavelength vertical antennas are spaced 1/2-wavelength apart and fed 180 degrees out of phase, they create a figure-8 radiation pattern oriented along the axis of the array. This means that the antennas radiate most of their energy in two opposite directions, forming a pattern that resembles the number 8. The radiation is strongest along the axis of the array and weakest in the perpendicular direction. This configuration is often used to create directional antennas with nulls in specific directions, which can be useful in certain applications.

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  • 43. 

    What is the triangulation method of direction finding?

    • The geometric angle of sky waves from the source are used to determine its position

    • A fixed receiving station plots three headings from the signal source on a map

    • Antenna headings from several different receiving stations are used to locate the signal source

    • A fixed receiving station uses three different antennas to plot the location of the signal source

    Correct Answer
    A. Antenna headings from several different receiving stations are used to locate the signal source
    Explanation
    The triangulation method of direction finding involves using antenna headings from several different receiving stations to locate the signal source. By measuring the angles at which the signal is received by multiple antennas, the position of the signal source can be determined. This method relies on the principle that the signal will arrive at each antenna from a slightly different angle, allowing for the calculation of the source's location based on the intersection of these angles.

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  • 44. 

    Which of the following describes an isotropic antenna?

    • A grounded antenna used to measure earth conductivity

    • A horizontal antenna used to compare Yagi antennas

    • A theoretical antenna used as a reference for antenna gain

    • A spacecraft antenna used to direct signals toward the earth

    Correct Answer
    A. A theoretical antenna used as a reference for antenna gain
    Explanation
    An isotropic antenna is a theoretical antenna used as a reference for antenna gain. It is an imaginary antenna that radiates power uniformly in all directions. It is used as a standard to compare the performance of other antennas, as it provides a baseline for measuring their gain. An isotropic antenna is often used in antenna testing and design to evaluate the effectiveness and efficiency of different antenna systems.

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  • 45. 

    What is an effective way of matching a feed-line to a VHF or UHF antenna when the impedances of both the antenna and feed-line are unknown?

    • Use a 50-ohm 1:1 balun between the antenna and feed-line

    • Use the "universal stub" matching technique

    • Connect a series-resonant LC network across the antenna feed terminals

    • Connect a parallel-resonant LC network across the antenna feed terminals

    Correct Answer
    A. Use the "universal stub" matching technique
    Explanation
    The "universal stub" matching technique is an effective way of matching a feed-line to a VHF or UHF antenna when the impedances of both the antenna and feed-line are unknown. This technique involves using a length of transmission line, referred to as a stub, connected in parallel to the feed-line. By adjusting the length and position of the stub, the impedance can be matched to achieve maximum power transfer between the feed-line and antenna. This technique is versatile and can be used for a wide range of impedance matching scenarios, making it an effective solution in this situation.

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  • 46. 

    What kind of impedance does a 1/4-wavelength transmission line present to a generator when the line is open at the far end?

    • A very high impedance

    • A very low impedance

    • The same as the characteristic impedance of the line

    • The same as the input impedance to the final generator stage

    Correct Answer
    A. A very low impedance
    Explanation
    When a 1/4-wavelength transmission line is open at the far end, it acts as a short circuit. This means that it presents a very low impedance to the generator. The open end reflects the signal back towards the generator, causing constructive interference and resulting in a low impedance.

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  • 47. 

    What, if any, are the differences between the radiation produced by radioactive materials and the electromagnetic energy radiated by an antenna?

    • There is no significant difference between the two types of radiation

    • Only radiation produced by radioactivity can injure human beings

    • RF radiation does not have sufficient energy to break apart atoms and molecules; radiation from radioactive sources does

    • Radiation from an antenna will damage unexposed photographic film, ordinary radioactive materials do not cause this problem

    Correct Answer
    A. RF radiation does not have sufficient energy to break apart atoms and molecules; radiation from radioactive sources does
    Explanation
    RF radiation, which is the electromagnetic energy radiated by an antenna, does not have enough energy to break apart atoms and molecules. On the other hand, radiation from radioactive sources, which is produced by radioactivity, does have enough energy to break apart atoms and molecules. This is the main difference between the two types of radiation.

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Quiz Review Timeline (Updated): Feb 26, 2024 +

Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.

  • Current Version
  • Feb 26, 2024
    Quiz Edited by
    ProProfs Editorial Team
  • Aug 04, 2009
    Quiz Created by
    K0rar

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