Free Ham Radio Extra Class Questions and Answers — Questions and Answers
Question 1: Which ionospheric layer is the nearest to Earth's surface?
- The F1 layer
- The F2 layer
- The D layer (Correct answer)
- The E layer
Correct answer: The D layer
The D layer is the lowest region of the Earth's ionosphere, typically found at altitudes between 50 and 90 kilometers during daylight hours. It is primarily responsible for absorbing lower frequency radio waves, especially during the day, which can limit long-distance communication on those bands. The E and F layers are located at progressively higher altitudes above the D layer.
Question 2: What benefit does employing a ferrite core toroidal inductor provide?
- Most of the magnetic field is contained in the core
- Large values of inductance may be obtained
- The magnetic properties of the core may be optimized for a specific range of frequencies
- All of the above (Correct answer)
Correct answer: All of the above
Ferrite core toroidal inductors offer several significant advantages in electronic circuits. Their toroidal shape effectively contains most of the magnetic field within the core, minimizing external interference and improving efficiency. The high permeability of ferrite material allows for the creation of large inductance values in a compact size, and different ferrite compositions can be chosen to optimize the core's magnetic properties for specific frequency ranges, making them highly versatile.
Question 3: In a 146.52 MHz FM phone transmitter with a 5 kHz dispersion, what is the frequency deviation for a 12.21 MHz reactance modulated oscillator?
- 5 kHz
- 60 kHz
- 101.75 HZ
- 416.7 Hz (Correct answer)
Correct answer: 416.7 Hz
In an FM transmitter, the frequency deviation produced by the reactance modulated oscillator is multiplied to reach the final output frequency and deviation. First, calculate the frequency multiplication factor by dividing the final output frequency (146.52 MHz) by the oscillator frequency (12.21 MHz), which is 12. Then, divide the desired final frequency dispersion (5 kHz) by this multiplication factor (12) to find the oscillator's deviation, which is approximately 416.7 Hz.
Question 4: If an oscilloscope records 200 volts peak-to-peak across a 50 ohm dummy load attached to the transmitter output, what is the output PEP from the transmitter?
- 100 watts (Correct answer)
- 353.5 watts
- 400 watts
- 1.4 watts
Correct answer: 100 watts
To calculate the Peak Envelope Power (PEP) from a peak-to-peak voltage measurement across a resistive load, first determine the peak voltage. A 200-volt peak-to-peak signal means the peak voltage is 100 volts. Then, use the formula P = Vp^2 / (2 * R), where Vp is the peak voltage and R is the resistance. Plugging in the values, (100V)^2 / (2 * 50 ohms) equals 10000 / 100, resulting in an output PEP of 100 watts.
Question 5: What is a 22,000 picofarad (pF) capacitor's value in nanofarads (nF)?
- 220
- 22 (Correct answer)
- 2.2
- 0.22
Correct answer: 22
To convert picofarads (pF) to nanofarads (nF), you need to remember that 1 nanofarad is equal to 1000 picofarads. Therefore, to convert 22,000 picofarads to nanofarads, you divide 22,000 by 1000. This calculation yields a value of 22 nanofarads.
Question 6: How do the gains for the same antenna expressed in dBi and dBd compare?
- dBi gain figures are the reciprocal of dBd gain figures +2.15 dB
- dBi gain figures are 2.15 dB lower than dBd gain figures
- dBi gain figures are 2.15 dB higher than dBd gain figures (Correct answer)
- dBi gain figures are the same as the square root of dBd gain figures multiplied by 2.15
Correct answer: dBi gain figures are 2.15 dB higher than dBd gain figures
Antenna gain expressed in dBi (decibels relative to an isotropic radiator) uses a theoretical isotropic antenna as the reference, which radiates equally in all directions. In contrast, dBd (decibels relative to a dipole) uses a half-wave dipole antenna as the reference. A half-wave dipole inherently has a gain of 2.15 dB compared to an isotropic radiator, meaning dBi gain figures will always be 2.15 dB higher than dBd gain figures for the same antenna.
Question 7: What benefit does an oscilloscope have over a digital voltmeter, specifically?
- Complex waveforms can be measured (Correct answer)
- Input impedance is much lower
- An oscilloscope uses less power
- Complex impedances can be easily measured
Correct answer: Complex waveforms can be measured
An oscilloscope provides a visual representation of a signal's voltage over time, allowing for the analysis of complex waveforms, including their shape, frequency, and phase. While a digital voltmeter (DVM) can accurately measure RMS or DC voltage, it only provides a numerical value and cannot display the dynamic characteristics or intricate details of a non-sinusoidal or varying signal. This makes the oscilloscope indispensable for troubleshooting and analyzing complex electronic circuits.
Free Ham Radio Extra Class Questions and Answers
Which ionospheric layer is the nearest to Earth's surface?