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Subelement L11
Test Equipment and measurements.
Section L11
What is the easiest amplitude dimension to measure by viewing a pure sine wave on an oscilloscope?
  • Average voltage
  • Correct Answer
    Peak-to-peak voltage
  • Peak voltage
  • RMS voltage

Key word: EASIEST. Reading the peak-to-peak amplitude of a waveform is easier than attempting to precisely center the waveform to determine the peak value. RMS and Average values can only be computed from the peak value or peak-to-peak values.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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What is the RMS value of a 340 volt peak-to-peak pure sine wave?
  • 300 volts
  • Correct Answer
    120 volts
  • 170 volts
  • 240 volts

Peak is half the peak-to-peak. RMS (or effective) is 0.707 times the peak value. 340 volts divided by 2 times 0.707 = 120 volts RMS.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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What is the equivalent to the RMS value of an AC voltage?
  • The AC voltage found by taking the square root of the average AC value
  • Correct Answer
    The AC voltage causing the same heating of a given resistor as a DC voltage of the same value
  • The AC voltage found by taking the square root of the peak AC voltage
  • The DC voltage causing the same heating of a given resistor as the peak AC voltage

A kettle would bring the same quantity of water to a boil within the same time whether it runs on 120 volts DC or 120 volts RMS AC.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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If the peak value of a 100 Hz sinusoidal waveform is 20 volts, the RMS value is:
  • 7.07 volts
  • 16.38 volts
  • Correct Answer
    14.14 volts
  • 28.28 volts

Peak is half the peak-to-peak. RMS (or effective) is 0.707 times the peak value. 20 volts times 0.707 = 14.14 volts RMS.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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In applying Ohm's law to AC circuits, current and voltage values are:
  • average values
  • average values times 1.414
  • none of the proposed answers
  • Correct Answer
    peak values times 0.707

Peak is half the peak-to-peak. RMS (or effective) is 0.707 times the peak value.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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The effective value of a sine wave of voltage or current is:
  • 50% of the maximum value
  • 100% of the maximum value
  • 63.6% of the maximum value
  • Correct Answer
    70.7% of the maximum value

Peak is half the peak-to-peak. RMS (or effective) is 0.707 times the peak value.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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AC voltmeter scales are usually calibrated to read:
  • peak voltage
  • instantaneous voltage
  • average voltage
  • Correct Answer
    RMS voltage

Voltmeters normally read the RMS (or effective) value; i.e., 0.707 times the peak value.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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An AC voltmeter is calibrated to read the:
  • peak value
  • Correct Answer
    effective value
  • peak-to-peak value
  • average value

Voltmeters normally read the RMS (or effective) value; i.e., 0.707 times the peak value.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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Which AC voltage value will produce the same amount of heat as a DC voltage, when applied to the same resistance?
  • The average value
  • The peak value
  • The peak-to-peak value
  • Correct Answer
    The RMS value

A kettle would bring the same quantity of water to a boil within the same time whether it runs on 120 volts DC or 120 volts RMS AC.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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What is the peak-to-peak voltage of a sine wave that has an RMS voltage of 120 volts?
  • 84.8 volts
  • 169.7 volts
  • 204.8 volts
  • Correct Answer
    339.5 volts

Knowing that RMS is peak times 0.707, peak must be RMS divided by 0.707 . Peak-to-peak is twice the peak value. 120 volts RMS divided by 0.707 = 170 volts peak. 170 volts peak times 2 = 340 volts peak-to-peak.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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A sine wave of 17 volts peak is equivalent to how many volts RMS?
  • 24 volts
  • 34 volts
  • 8.5 volts
  • Correct Answer
    12 volts

Peak is half the peak-to-peak. RMS (or effective) is 0.707 times the peak value. 17 volts times 0.707 = 12 volts RMS.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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The power supplied to the antenna transmission line by a transmitter during an RF cycle at the highest crest of the modulation envelope is known as:
  • mean power
  • carrier power
  • full power
  • Correct Answer
    peak-envelope power

Peak Envelope Power (PEP) in SSB is defined as the power during one cycle on a modulation peak. Knowing that P = E squared divided by R, that peak voltage is half the peak-to-peak value and that RMS voltage is peak times 0.707 . PEP can be computed as ((peak voltage times 0.707) squared ) divided by R. When given peak-to-peak, first divide by 2 to obtain peak voltage.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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To compute one of the following, multiply the peak-envelope voltage by 0.707 to obtain the RMS value, square the result and divide by the load resistance. Which is the correct answer?
  • power factor
  • Correct Answer
    PEP
  • PIV
  • ERP

PIV = Peak Inverse Voltage (diode rating). ERP = Effective Radiated Power (transmit power minus line losses plus antenna gain). Peak Envelope Power (PEP) in SSB is defined as the power during one cycle on a modulation peak. Knowing that P = E squared divided by R, that peak voltage is half the peak-to-peak value and that RMS voltage is peak times 0.707 . PEP can be computed as ((peak voltage times 0.707) squared ) divided by R. When given peak-to-peak, first divide by 2 to obtain peak voltage.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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Peak-Envelope Power (PEP) for SSB transmission is:
  • a hypothetical measurement
  • Correct Answer
    Peak-Envelope Voltage (PEV) multiplied by 0.707, squared and divided by the load resistance
  • peak-voltage multiplied by peak current
  • equal to the RMS power

Peak Envelope Power (PEP) in SSB is defined as the power during one cycle on a modulation peak. Knowing that P = E squared divided by R, that peak voltage is half the peak-to-peak value and that RMS voltage is peak times 0.707 . PEP can be computed as ((peak voltage times 0.707) squared ) divided by R. When given peak-to-peak, first divide by 2 to obtain peak voltage.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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The formula to be used to calculate the power output of a transmitter into a resistor load using a voltmeter is:
  • P = EI/R
  • P = EI cos 0
  • P = IR
  • Correct Answer
    P = E2 /R

Peak Envelope Power (PEP) in SSB is defined as the power during one cycle on a modulation peak. Knowing that P = E squared divided by R, that peak voltage is half the peak-to-peak value and that RMS voltage is peak times 0.707 . PEP can be computed as ((peak voltage times 0.707) squared ) divided by R. When given peak-to-peak, first divide by 2 to obtain peak voltage.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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How is the output Peak-Envelope Power of a transmitter calculated if an oscilloscope is used to measure the Peak-Envelope Voltage across a dummy resistive load (where PEP = Peak-Envelope Power, PEV = Peak-Envelope Voltage, Vp = peak-voltage, RL = load resistance)?
  • PEP = [(1.414 PEV)(1.414 PEV)] / RL
  • Correct Answer
    PEP = [(0.707 PEV)(0.707 PEV)] / RL
  • PEP = [(Vp)(Vp)] / (RL)
  • PEP = (Vp)(Vp)(RL)

Peak Envelope Power (PEP) in SSB is defined as the power during one cycle on a modulation peak. Knowing that P = E squared divided by R, that peak voltage is half the peak-to-peak value and that RMS voltage is peak times 0.707 . PEP can be computed as ((peak voltage times 0.707) squared ) divided by R. When given peak-to-peak, first divide by 2 to obtain peak voltage.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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What is the output PEP from a transmitter if an oscilloscope measures 200 volts peak-to-peak across a 50-ohm dummy load connected to the transmitter output?
  • 400 watts
  • 1000 watts
  • 200 watts
  • Correct Answer
    100 watts

Peak Envelope Power (PEP) in SSB is defined as the power during one cycle on a modulation peak. Knowing that P = E squared divided by R, that peak voltage is half the peak-to-peak value and that RMS voltage is peak times 0.707 . PEP can be computed as ((peak voltage times 0.707) squared ) divided by R. When given peak-to-peak, first divide by 2 to obtain peak voltage. In this example, 200 divided by 2 = 100 volts peak ; (100 times 0.707) squared = 5000 ; 5000 divided by 50 = 100 watts PEP.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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What is the output PEP from a transmitter if an oscilloscope measures 500 volts peak-to-peak across a 50-ohm dummy load connected to the transmitter output?
  • 2500 watts
  • 500 watts
  • Correct Answer
    625 watts
  • 1250 watts

Peak Envelope Power (PEP) in SSB is defined as the power during one cycle on a modulation peak. Knowing that P = E squared divided by R, that peak voltage is half the peak-to-peak value and that RMS voltage is peak times 0.707 . PEP can be computed as ((peak voltage times 0.707) squared ) divided by R. When given peak-to-peak, first divide by 2 to obtain peak voltage. In this example, 500 divided by 2 = 250 volts peak ; (250 times 0.707) squared = 31 240 ; 31 240 divided by 50 = 625 watts PEP.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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What is the output PEP of an unmodulated carrier transmitter if a wattmeter connected to the transmitter output indicates an average reading of 1060 watts?
  • 1500 watts
  • 530 watts
  • Correct Answer
    1060 watts
  • 2120 watts

Key word: UNMODULATED. This is a catch. There would be no difference between output power and PEP in the absence of modulation. [ With modulation, the answer would have been different. Some wattmeters are calibrated to read peak power. ]

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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What is the output PEP from a transmitter, if an oscilloscope measures 400 volts peak-to-peak across a 50 ohm dummy load connected to the transmitter output?
  • 600 watts
  • 1000 watts
  • Correct Answer
    400 watts
  • 200 watts

Peak Envelope Power (PEP) in SSB is defined as the power during one cycle on a modulation peak. Knowing that P = E squared divided by R, that peak voltage is half the peak-to-peak value and that RMS voltage is peak times 0.707 . PEP can be computed as ((peak voltage times 0.707) squared ) divided by R. When given peak-to-peak, first divide by 2 to obtain peak voltage. In this example, 400 divided by 2 = 200 volts peak ; (200 times 0.707) squared = 20 000 ; 20 000 divided by 50 = 400 watts PEP.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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What is the output PEP from a transmitter, if an oscilloscope measures 800 volts peak-to-peak across a 50 ohm dummy load connected to the transmitter output?
  • 6400 watts
  • 3200 watts
  • Correct Answer
    1600 watts
  • 800 watts

Peak Envelope Power (PEP) in SSB is defined as the power during one cycle on a modulation peak. Knowing that P = E squared divided by R, that peak voltage is half the peak-to-peak value and that RMS voltage is peak times 0.707 . PEP can be computed as ((peak voltage times 0.707) squared ) divided by R. When given peak-to-peak, first divide by 2 to obtain peak voltage. In this example, 800 divided by 2 = 400 volts peak ; (400 times 0.707) squared = 80 000 ; 80 000 divided by 50 = 1600 watts PEP.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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An oscilloscope measures 500 volts peak-to-peak across a 50 ohm dummy load connected to the transmitter output during unmodulated carrier conditions. What would an average-reading power meter indicate under the same transmitter conditions?
  • Correct Answer
    625 watts
  • 427.5 watts
  • 884 watts
  • 442 watts

Peak Envelope Power (PEP) in SSB is defined as the power during one cycle on a modulation peak. Knowing that P = E squared divided by R, that peak voltage is half the peak-to-peak value and that RMS voltage is peak times 0.707 . PEP can be computed as ((peak voltage times 0.707) squared ) divided by R. When given peak-to-peak, first divide by 2 to obtain peak voltage. In this example, 500 divided by 2 = 250 volts peak ; (250 times 0.707) squared = 31 240 ; 31 240 divided by 50 = 625 watts PEP.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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What is a dip meter?
  • A marker generator
  • A field-strength meter
  • Correct Answer
    A variable frequency oscillator with metered feedback current
  • An SWR meter

A 'Dip Meter' is an instrument incorporating a Variable Frequency Oscillator whose resonant circuit can be placed near circuits to be tested. The oscillator's activity is monitored with a built-in meter. When the external circuit's resonant frequency matches the oscillator's frequency, energy is absorbed and a dip is observed on the meter. It works best on parallel-tuned circuits. It can be used to test antenna traps and tuned circuits in static conditions (i.e., not operating). Overcoupling, hand capacity and stray capacity all introduce errors.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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What does a dip meter do?
  • Correct Answer
    It gives an indication of the resonant frequency of a circuit
  • It measures transmitter output power accurately
  • It measures field strength accurately
  • It measures frequency accurately

A 'Dip Meter' is an instrument incorporating a Variable Frequency Oscillator whose resonant circuit can be placed near circuits to be tested. The oscillator's activity is monitored with a built-in meter. When the external circuit's resonant frequency matches the oscillator's frequency, energy is absorbed and a dip is observed on the meter. It works best on parallel-tuned circuits. It can be used to test antenna traps and tuned circuits in static conditions (i.e., not operating). Overcoupling, hand capacity and stray capacity all introduce errors.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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What two ways could a dip meter be used in an amateur station?
  • To measure antenna resonance and percentage modulation
  • To measure resonant frequency of antenna traps and percentage modulation
  • Correct Answer
    To measure resonant frequencies of antenna traps and to measure a tuned circuit resonant frequency
  • To measure antenna resonance and impedance

A 'Dip Meter' is an instrument incorporating a Variable Frequency Oscillator whose resonant circuit can be placed near circuits to be tested. The oscillator's activity is monitored with a built-in meter. When the external circuit's resonant frequency matches the oscillator's frequency, energy is absorbed and a dip is observed on the meter. It works best on parallel-tuned circuits. It can be used to test antenna traps and tuned circuits in static conditions (i.e., not operating). Overcoupling, hand capacity and stray capacity all introduce errors.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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A dip meter supplies the radio frequency energy which enables you to check:
  • the adjustment of an inductor
  • Correct Answer
    the resonant frequency of a circuit
  • the calibration of an absorption-type wavemeter
  • the impedance mismatch in a circuit

A 'Dip Meter' is an instrument incorporating a Variable Frequency Oscillator whose resonant circuit can be placed near circuits to be tested. The oscillator's activity is monitored with a built-in meter. When the external circuit's resonant frequency matches the oscillator's frequency, energy is absorbed and a dip is observed on the meter. It works best on parallel-tuned circuits. It can be used to test antenna traps and tuned circuits in static conditions (i.e., not operating). Overcoupling, hand capacity and stray capacity all introduce errors.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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A dip meter may not be used directly to:
  • align transmitter-tuned circuits
  • determine the frequency of oscillations
  • align receiver-tuned circuits
  • Correct Answer
    measure the value of capacitance or inductance

Key word: DIRECTLY. A 'Dip Meter' cannot measure inductance or capacitance DIRECTLY but if an unknown component is first attached to a known inductance or capacitance, the unknown value can be computed once the resonant frequency of the combination is obtained. The oscillator in a dip meter can usually be disabled to turn the instrument into an absorption wavemeter.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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The dial calibration on the output attenuator of a signal generator:
  • reads half the true output when the attenuator is properly terminated
  • Correct Answer
    reads accurately only when the attenuator is properly terminated
  • always reads the true output of the signal generator
  • reads twice the true output when the attenuator is properly terminated

A 'Signal Generator' is an instrument capable of producing any of a wide range of frequencies (RF frequencies for radio work). Signal generators sometimes include a calibrated output attenuator so a given amplitude can be produced (e.g., a certain number of microvolts). Calibration in that case is only accurate if the generator feeds a circuit with the expected impedance. An instrument which "generates reference signals at exact frequency intervals" is a marker generator or crystal calibrator.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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What is a signal generator?
  • A low-stability oscillator used to inject a signal into a circuit under test
  • A high-stability oscillator which generates reference signals at exact frequency intervals
  • Correct Answer
    A high-stability oscillator which can produce a wide range of frequencies and amplitudes
  • A low-stability oscillator which sweeps through a range of frequencies

A 'Signal Generator' is an instrument capable of producing any of a wide range of frequencies (RF frequencies for radio work). Signal generators sometimes include a calibrated output attenuator so a given amplitude can be produced (e.g., a certain number of microvolts). Calibration in that case is only accurate if the generator feeds a circuit with the expected impedance. An instrument which "generates reference signals at exact frequency intervals" is a marker generator or crystal calibrator.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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A dip meter:
  • may be used only with series tuned circuits
  • accurately measures frequencies
  • Correct Answer
    should be loosely coupled to the circuit under test
  • should be tightly coupled to the circuit under test

A 'Dip Meter' is an instrument incorporating a Variable Frequency Oscillator whose resonant circuit can be placed near circuits to be tested. The oscillator's activity is monitored with a built-in meter. When the external circuit's resonant frequency matches the oscillator's frequency, energy is absorbed and a dip is observed on the meter. It works best on parallel-tuned circuits. It can be used to test antenna traps and tuned circuits in static conditions (i.e., not operating). Overcoupling, hand capacity and stray capacity all introduce errors.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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Which two instruments are needed to measure FM receiver sensitivity for a 12 dB SINAD ratio (signal + noise + distortion over noise + distortion)?
  • Oscilloscope and spectrum analyzer
  • Receiver noise bridge and total harmonic distortion analyser
  • Correct Answer
    Calibrated RF signal generator with FM tone modulation and total harmonic distortion (THD) analyzer
  • RF signal generator with FM tone modulation and a deviation meter

The SINAD (signal + noise + distortion over noise + distortion) ratio takes the SNR (signal + noise over noise ratio) one step further by including distortion. A 12 dB SINAD ratio ensures that speech remains intelligible. Sensitivity expressed in those terms is the lowest RF level that will produce a usable message. The RF signal generator must be calibrated so the number of microvolts is precisely determined. Total Harmonic Distortion compares unwanted harmonic components added to the desired fundamental frequency, an audio tone in this instance.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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The dip meter is most directly applicable to:
  • digital logic circuits
  • series tuned circuits
  • Correct Answer
    parallel tuned circuits
  • operational amplifier circuits

A 'Dip Meter' is an instrument incorporating a Variable Frequency Oscillator whose resonant circuit can be placed near circuits to be tested. The oscillator's activity is monitored with a built-in meter. When the external circuit's resonant frequency matches the oscillator's frequency, energy is absorbed and a dip is observed on the meter. It works best on parallel-tuned circuits. It can be used to test antenna traps and tuned circuits in static conditions (i.e., not operating). Overcoupling, hand capacity and stray capacity all introduce errors.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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Which of the following is not a factor affecting the frequency accuracy of a dip meter?
  • Over coupling
  • Correct Answer
    Transmitter power output
  • Hand capacity
  • Stray capacity

A 'Dip Meter' is an instrument incorporating a Variable Frequency Oscillator whose resonant circuit can be placed near circuits to be tested. The oscillator's activity is monitored with a built-in meter. When the external circuit's resonant frequency matches the oscillator's frequency, energy is absorbed and a dip is observed on the meter. It works best on parallel-tuned circuits. It can be used to test antenna traps and tuned circuits in static conditions (i.e., not operating). Overcoupling, hand capacity and stray capacity all introduce errors.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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What does a frequency counter do?
  • It generates broad-band white noise for calibration
  • It produces a reference frequency
  • Correct Answer
    It makes frequency measurements
  • It measures frequency deviation

Two methods exist for determining frequency: counting the number of cycles in a fixed time interval or counting the time interval for some integer number of input cycles. In both cases, the instrument relies on an internal time base ( a crystal reference oscillator ). The accuracy of the measurement depends on the accuracy and stability (short and long-term) of the time base. Speed limitations in the logic circuitry limit the frequency range for a given instrument.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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What factors limit the accuracy, frequency response and stability of a frequency counter?
  • Correct Answer
    Time base accuracy, speed of the logic, and time base stability
  • Time base accuracy, temperature coefficient of the logic and time base stability
  • Number of digits in the readout, speed of the logic, and time base stability
  • Number of digits in the readout, external frequency reference and temperature coefficient of the logic

Two methods exist for determining frequency: counting the number of cycles in a fixed time interval or counting the time interval for some integer number of input cycles. In both cases, the instrument relies on an internal time base ( a crystal reference oscillator ). The accuracy of the measurement depends on the accuracy and stability (short and long-term) of the time base. Speed limitations in the logic circuitry limit the frequency range for a given instrument.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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How can the accuracy of a frequency counter be improved?
  • By using faster digital logic
  • By improving the accuracy of the frequency response
  • Correct Answer
    By increasing the accuracy of the time base
  • By using slower digital logic

Two methods exist for determining frequency: counting the number of cycles in a fixed time interval or counting the time interval for some integer number of input cycles. In both cases, the instrument relies on an internal time base ( a crystal reference oscillator ). The accuracy of the measurement depends on the accuracy and stability (short and long-term) of the time base. Speed limitations in the logic circuitry limit the frequency range for a given instrument.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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If a frequency counter with a time base accuracy of +/- 0.1 PPM (parts per million) reads 146 520 000 Hz, what is the most that the actual frequency being measured could differ from that reading?
  • 0.1 MHz
  • 1.4652 Hz
  • 1.4652 kHz
  • Correct Answer
    14.652 Hz

The reading can be off by as much as the time base accuracy in Parts Per Million times the measured frequency in megahertz: 1 part per million is one hertz per megahertz. [ In reality, an added error of plus or minus 1 count exists as the last digit displayed may have been rounded up or down. ]

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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If a frequency counter, with a time base accuracy of 10 PPM (parts per million) reads 146 520 000 Hz, what is the most the actual frequency being measured could differ from that reading?
  • 1465.2 kHz
  • Correct Answer
    1465.2 Hz
  • 146.52 Hz
  • 146.52 kHz

The reading can be off by as much as the time base accuracy in Parts Per Million times the measured frequency in megahertz: 1 part per million is one hertz per megahertz. [ In reality, an added error of plus or minus 1 count exists as the last digit displayed may have been rounded up or down. ]

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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The clock in a frequency counter normally uses a:
  • Correct Answer
    crystal oscillator
  • self-oscillating Hartley oscillator
  • mechanical tuning fork
  • free-running multivibrator

Two methods exist for determining frequency: counting the number of cycles in a fixed time interval or counting the time interval for some integer number of input cycles. In both cases, the instrument relies on an internal time base ( a crystal reference oscillator ). The accuracy of the measurement depends on the accuracy and stability (short and long-term) of the time base. Speed limitations in the logic circuitry limit the frequency range for a given instrument.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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The frequency accuracy of a frequency counter is determined by:
  • the size of the frequency counter
  • type of display used in the counter
  • the number of digits displayed
  • Correct Answer
    the characteristics of the internal time-base generator

Two methods exist for determining frequency: counting the number of cycles in a fixed time interval or counting the time interval for some integer number of input cycles. In both cases, the instrument relies on an internal time base ( a crystal reference oscillator ). The accuracy of the measurement depends on the accuracy and stability (short and long-term) of the time base. Speed limitations in the logic circuitry limit the frequency range for a given instrument.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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Which device relies on a stable low-frequency oscillator, with harmonic output, to facilitate the frequency calibration of receiver dial settings?
  • Signal generator
  • Harmonic calibrator
  • Frequency counter
  • Correct Answer
    Frequency-marker generator

A frequency marker generator (or crystal calibrator) is a relatively stable and precise oscillator rich in harmonics. It is typically used to calibrate analog station receivers by providing reference signals at known intervals on the dial. For an HF receiver, harmonics may be generated every 25, 50 or 100 kHz throughout the HF spectrum. For microwave work, harmonics of 144 MHz are useful for 432 MHz (3x), 1296 (9x), 2304 (16x), 3456 (24x), 5760 (40x), 10368 (72x) and 24192 MHz (168x). The "harmonic calibrator" is a bogus answer.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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What is the traditional way of verifying the accuracy of a crystal calibrator?
  • Compare the oscillator with your transmitter
  • Use a dip-meter to determine the oscillator's fundamental frequency
  • Compare the oscillator with your receiver
  • Correct Answer
    Zero-beat the crystal oscillator against a standard frequency station such as WWV

One way to calibrate a crystal calibrator (frequency marker generator) is to turn it on so it is heard while receiving a standard frequency station on shortwave. "Zero beating", which can be done audibly, is bringing, through adjustment, the difference between two frequencies down to zero. When two RF signals (calibrator and reference station) are nearly at the same frequency, they produce a heterodyne (beat) which falls in the audio range; the lower the beat note, the closer the match.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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Out of the following oscillators, one is NOT, by itself, considered a high-stability reference:
  • oven-controlled crystal oscillator (OCXO)
  • GPS disciplined oscillator (GPSDO)
  • Correct Answer
    voltage-controlled crystal oscillator (VCXO)
  • temperature compensated crystal oscillator (TCXO)

Key word: NOT. The voltage-controlled crystal oscillator (VCXO) is always part of a larger ensemble, its stability is not controlled, but can be controlled by an external circuit. The temperature compensated crystal oscillator (TCXO) relies on a temperature sensor and some circuitry (analog or digital) to correct (compensate) the frequency. The oven-controlled crystal oscillator (OCXO) maintains the crystal at a precise and constant temperature well above ambient temperature. The GPS disciplined oscillator (GPSDO) uses a GPS (Global Positioning System) receiver to synchronize an oscillator with the highly accurate reference signals from the satellites.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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You want to calibrate your station frequency reference to the WWV signal on your receiver. The resulting beat tone must be:
  • a combined frequency above both
  • the mathematical mean of both frequencies
  • at the highest audio frequency possible
  • Correct Answer
    of a frequency as low as possible and with a period as long as possible

One way to calibrate a crystal calibrator (frequency marker generator) is to turn it on so it is heard while receiving a standard frequency station on shortwave. "Zero beating", which can be done audibly, is bringing, through adjustment, the difference between two frequencies down to zero. When two RF signals (calibrator and reference station) are nearly at the same frequency, they produce a heterodyne (beat) which falls in the audio range; the lower the beat note, the closer the match. [ Station WWV broadcasts time and frequency information from Fort Collins, Colorado; it is maintained by the Physical Measurement Laboratory (PML) of the National Institute of Standards and Technology (NIST). ]

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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If a 100 Hz signal is fed to the horizontal input of an oscilloscope and a 150 Hz signal is fed to the vertical input, what type of pattern should be displayed on the screen?
  • Correct Answer
    A looping pattern with 3 horizontal loops, and 2 vertical loops
  • A rectangular pattern 100 mm wide and 150 mm high
  • An oval pattern 100 mm wide and 150 mm high
  • A looping pattern with 100 horizontal loops and 150 vertical loops

Using the horizontal and vertical inputs on an oscilloscope, it is possible to compare the frequency of two sine waves or the phase relationship of two signals of equal frequency. The resulting pattern is called a "Lissajous Figure". Apply a known signal to the horizontal input. The ratio of the number of loops on the horizontal edge to the number on the vertical edge equals the ratio of the vertical frequency Y divided by horizontal frequency X: the number of cycles the Y frequency covers during one horizontal cycle. [French mathematician Jules-Antoine Lissajous]

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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What factors limit the accuracy, frequency response and stability of an oscilloscope?
  • Tube face voltage increments and deflection amplifier voltages
  • Correct Answer
    Accuracy of the time base and the linearity and bandwidth of the deflection amplifiers
  • Deflection amplifier output impedance and tube face frequency increments
  • Accuracy and linearity of the time base and tube face voltage increments

Similarly to the frequency counter, the accuracy and stability of the time base responsible for sweeping on the horizontal axis are paramount. The top frequency is limited by the highest horizontal sweep rate and bandwidth of the horizontal and vertical deflection amplifiers.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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How can the frequency response of an oscilloscope be improved?
  • By using triggered sweep and a crystal oscillator for the timebase
  • Correct Answer
    By increasing the horizontal sweep rate and the vertical amplifier frequency response
  • By using a crystal oscillator as the time base and increasing the vertical sweep rate
  • By increasing the vertical sweep rate and the horizontal amplifier frequency response

Similarly to the frequency counter, the accuracy and stability of the time base responsible for sweeping on the horizontal axis are paramount. The top frequency is limited by the highest horizontal sweep rate and bandwidth of the horizontal and vertical deflection amplifiers.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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You can use an oscilloscope to display the input and output of a circuit at the same time by:
  • measuring the input on the X axis and the output on the Y axis
  • measuring the input on the X axis and the output on the Z axis
  • measuring the input on the Y axis and the output on the X axis
  • Correct Answer
    utilizing a dual trace oscilloscope

A dual-trace oscilloscope has two distinct vertical channels. They can be used to take simultaneous measurements at different points in a circuit.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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An oscilloscope cannot be used to:
  • measure DC voltage
  • determine the amplitude of complex voltage wave forms
  • Correct Answer
    determine FM carrier deviation directly
  • measure frequency

Measuring frequency is a matter of using the horizontal sweep rate and number of divisions to determine the period of one cycle of the waveform. Measuring deviation requires a deviation meter.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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The bandwidth of an oscilloscope is:
  • a function of the time-base accuracy
  • Correct Answer
    the highest frequency signal the scope can display
  • directly related to gain compression
  • indirectly related to screen persistence

Similarly to the frequency counter, the accuracy and stability of the time base responsible for sweeping on the horizontal axis are paramount. The top frequency is limited by the highest horizontal sweep rate and bandwidth of the horizontal and vertical deflection amplifiers.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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When using Lissajous figures to determine phase differences, an indication of zero or 180 degrees is represented on the screen of an oscilloscope by:
  • Correct Answer
    a diagonal straight line
  • a horizontal straight line
  • an ellipse
  • a circle

Two in-phase signals of the same frequency applied to the X and Y amplifiers will grow exactly at the same rate: equal deflection on the X and Y create a trace at a 45 degree angle. Ponder these X and Y values: 1 and 1, 2 and 2, 3 and 3, etc. A precise 180 degree phase difference has a similar effect except the trace appears in the other two quadrants. Ponder these X and Y values: 1 and -1, 2 and -2, 3 and -3, etc. [other angles correspond to the sine function of the ratio of Y as the ellipse crosses the vertical centre axis to the maximum Y value.]

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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A 100-kHz signal is applied to the horizontal channel of an oscilloscope. A signal of unknown frequency is applied to the vertical channel. The resultant wave form has 5 loops displayed vertically and 2 loops horizontally. The unknown frequency is:
  • 20 kHz
  • 50 kHz
  • 30 kHz
  • Correct Answer
    40 kHz

Using the horizontal and vertical inputs on an oscilloscope, it is possible to compare the frequency of two sine waves or the phase relationship of two signals of equal frequency. The resulting pattern is called a "Lissajous Figure". Apply a known signal to the horizontal input. The ratio of the number of loops on the horizontal edge to the number on the vertical edge equals the ratio of the vertical frequency Y divided by horizontal frequency X: the number of cycles the Y frequency covers during one horizontal cycle. In this example, the unknown frequency is two fifths the known frequency. [French mathematician Jules-Antoine Lissajous]

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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An oscilloscope probe must be compensated:
  • Correct Answer
    every time the probe is used with a different oscilloscope
  • when measuring a sine wave
  • through the addition of a high-value series resistor
  • when measuring a signal whose frequency varies

"Probe compensation is the process of adjusting the probe's RC network divider so that the probe maintains its attenuation ratio over the probe's rated bandwidth. Most scopes have a square wave reference signal available on the front panel to use for compensating the probe. You can attach the probe tip to the probe compensation terminal and connect the probe to an input of the scope. Viewing the square wave reference signal, make the proper adjustments on the probe using a small screw driver so that the square waves on the scope screen look square" (Jae-yong Chang, Agilent Technologies). Given that the input circuitry cannot be identical between instruments, adjustment must be made every time a high-impedance probe is used with a different scope.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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What is the best instrument to use to check the signal quality of a CW or single-sideband phone transmitter?
  • A signal tracer and an audio amplifier
  • A field-strength meter
  • Correct Answer
    An oscilloscope
  • A sidetone monitor

An oscilloscope of sufficient bandwidth permits visualizing the transmitter's output. The sidetone monitor merely produces a tone when a CW transmitter is keyed. The field-strength meter reports relative field strength in proximity to antennas. The signal tracer permits troubleshooting audio circuitry.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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What is the best signal source to connect to the vertical input of an oscilloscope for checking the quality of a transmitted signal?
  • The IF output of a monitoring receiver
  • The audio input of the transmitter
  • Correct Answer
    The RF output of the transmitter through a sampling device
  • The RF signals of a nearby receiving antenna

The most accurate representation of a transmitted signal can be viewed at the RF output of a transmitter. The correct method supposes the use of a "sampler" in the transmission line. That way, proper impedance match is maintained, instruments are protected from large voltages and extraneous signals are minimized.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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A meter has a full-scale deflection of 40 microamperes and an internal resistance of 96 ohms. You want it to read 0 to 1 mA. The value of the shunt to be used is:
  • 40 ohms
  • Correct Answer
    4 ohms
  • 24 ohms
  • 16 ohms

Extending the range of an ammeter supposes placing a shunt resistor across the meter movement to divert part of the current: the shunt resistor = internal resistance divided by the multiplication factor from which we will first have deducted a quantity of one. For example, making a 40 microamperes movement, with a 96 ohms internal resistance, read 1000 microamperes at full scale is a factor of 25: the shunt = 96 ohms divided by 24 (i.e., 25 minus 1) = 4 ohms. Method B: you could have computed voltage across the meter as 40 times 96 = 3840 microvolts ; and then, computed the shunt resistor as 3840 microvolts divided by 960 microamperes = 4 ohms.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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A moving-coil milliammeter having a full-scale deflection of 1 mA and an internal resistance of 0.5 ohms is to be converted to a voltmeter of 20 volts full-scale deflection. It would be necessary to insert a:
  • shunt resistance of 19 999.5 ohms
  • shunt resistance of 19.5 ohms
  • Correct Answer
    series resistance of 19 999.5 ohms
  • series resistance of 1 999.5 ohms

Turning an ammeter into a voltmeter supposes inserting a suitable resistor in series with the instrument. For a current of 1 milliampere to flow under 20 volts, Ohm's Law tells us that a total resistance of 20 000 ohms is required. Subtract the internal resistance from that number and the actual series resistance needed is 19 999.5 ohms.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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A voltmeter having a range of 150 volts and an internal resistance of 150 000 ohms is to be extended to read 750 volts. The required multiplier resistor would have a value of:
  • 1 500 ohms
  • 750 000 ohms
  • 1 200 000 ohms
  • Correct Answer
    600 000 ohms

Extending the range of a voltmeter supposes placing a suitable multiplier resistor in series with the instrument: the multiplier resistance = total internal resistance times the multiplication factor from which we will first have deducted a quantity of one. For example, to turn a 150 volts meter, with an internal resistance of 150 000 ohms, to read 750 volts full scale is an increase of 5 times: the multiplier resistor = 150 000 times 4 (i.e., 5 minus 1) = 600 000 ohms. Method B: current for full-scale reading = 150 volts divided by 150 000 ohms = 1 milliampere. The total resistance that would allow 1 milliampere under 750 volts = 750 000 ohms. Deduct the internal resistance from this value to determine the multiplier resistor.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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The sensitivity of an ammeter is an expression of:
  • Correct Answer
    the amount of current causing full-scale deflection
  • the resistance of the meter
  • the loading effect the meter will have on a circuit
  • the value of the shunt resistor

Ammeter sensitivity is the current needed for a full-scale deflection.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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Voltmeter sensitivity is usually expressed in ohms per volt. This means that a voltmeter with a sensitivity of 20 kilohms per volt would be a:
  • Correct Answer
    50 microampere meter
  • 1 milliampere meter
  • 50 milliampere meter
  • 100 milliampere meter

Voltmeter sensitivity in ohms per volt: i.e., the full-scale reading times the sensitivity yields total instrument resistance. Given total resistance, sensitivity can be computed as resistance divided by full-scale voltage reading: e.g., a total of 150 000 ohms for a voltmeter reading 150 volts is a sensitivity of 1000 ohms per volt. Given sensitivity, the current needed for full-scale voltmeter reading follows Ohm's Law: e.g., a voltmeter with a sensitivity of 20 000 ohms per volt uses a 50 microampere movement ( I = 1 volt divided by 20 000 = 50 microamperes ).

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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The sensitivity of a voltmeter, whose resistance is 150 000 ohms on the 150-volt range, is:
  • 150 ohms per volt
  • Correct Answer
    1000 ohms per volt
  • 100 000 ohms per volt
  • 10 000 ohms per volt

Voltmeter sensitivity in ohms per volt: i.e., the full-scale reading times the sensitivity yields total instrument resistance. Given total resistance, sensitivity can be computed as resistance divided by full-scale voltage reading: e.g., a total of 150 000 ohms for a voltmeter reading 150 volts is a sensitivity of 1000 ohms per volt. Given sensitivity, the current needed for full-scale voltmeter reading follows Ohm's Law: e.g., a voltmeter with a sensitivity of 20 000 ohms per volt uses a 50 microampere movement ( I = 1 volt divided by 20 000 = 50 microamperes ).

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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The range of a DC ammeter can easily be extended by:
  • connecting an external resistance in series with the internal resistance
  • changing the internal inductance of the meter
  • changing the internal capacitance of the meter to resonance
  • Correct Answer
    connecting an external resistance in parallel with the internal resistance

Extending the range of an ammeter supposes placing a shunt resistor across the meter movement to divert part of the current: the shunt resistor = internal resistance divided by the multiplication factor from which we will first have deducted a quantity of one.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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What happens inside a multimeter when you switch it from a lower to a higher voltage range?
  • Resistance is added in parallel with the meter
  • Correct Answer
    Resistance is added in series with the meter
  • Resistance is reduced in series with the meter
  • Resistance is reduced in parallel with the meter

Extending the range of a voltmeter supposes placing a suitable multiplier resistor in series with the instrument; the higher the series resistance, the higher the voltage needed to obtain full-scale deflection on the meter.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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How can the range of an ammeter be increased?
  • By adding resistance in series with the meter
  • Correct Answer
    By adding resistance in parallel with the meter
  • By adding resistance in series with the circuit under test
  • By adding resistance in parallel with the circuit under test

Extending the range of an ammeter supposes placing a shunt resistor across the meter movement to divert part of the current: the shunt resistor = internal resistance divided by the multiplication factor from which we will first have deducted a quantity of one.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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Where should an RF wattmeter be connected for the most accurate readings of transmitter output power?
  • One-half wavelength from the transmitter output
  • One-half wavelength from the antenna feed point
  • At the antenna feed point
  • Correct Answer
    At the transmitter output connector

Measuring right at the transmitter output connector removes any line losses from the measurement.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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At what line impedance do most RF wattmeters usually operate?
  • 25 ohms
  • 100 ohms
  • 300 ohms
  • Correct Answer
    50 ohms

In-line RF wattmeters are designed to work with the most common Characteristic Impedance in radio work: 50 ohms.

Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.

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