Direct to Full Sample Questions

Three questions from each section of the Direct to Full syllabus. Pick an answer to see whether you're right, how everyone else did, and why.

Direct to Full exam guide

1 Licensing conditions and station identification

The EMF restrictions in the amateur licence are designed to protect:

  1. Domestic electronic equipment from overload by strong RF fields
  2. The licensee from excessive exposure to electromagnetic fields while operating
  3. Other radio services from interference
  4. People from excessive exposure to electromagnetic fields

Answer: D, People from excessive exposure to electromagnetic fields. Condition 9 requires EMF exposure to stay within ICNIRP limits for the general public. Its purpose is to protect people, meaning anyone who could be in an area exposed to your transmissions (syllabus 1G1).

A remotely controlled station must ensure that transmissions from the Radio Equipment:

  1. Are automatically logged by the remote software
  2. Are limited to 5 Watts ERP
  3. Are identified by the Licence Number instead of the call sign
  4. Can be terminated promptly

Answer: D, Can be terminated promptly. When operating by remote control, the licensee must ensure transmissions can be terminated promptly. This is a remote control requirement in licence Condition 6 (paragraph 10c) and in syllabus 1E1.

A UK amateur licence authorises operation aboard ships or aircraft registered in:

  1. The UK, Channel Islands or Isle of Man, in international water or airspace
  2. Any country in the Commonwealth, in international water or airspace
  3. Any country that is a CEPT member, in international waters or airspace
  4. Any EU or EEA member state, in international water or airspace

Answer: A, The UK, Channel Islands or Isle of Man, in international water or airspace. Licence Condition 4 authorises use aboard any ship or aircraft registered in the UK, the Channel Islands or the Isle of Man, but only in international waters or airspace (syllabus 1A1).

2 Operating practices and procedures

Which callsign prefix denotes the Republic of Ireland?

  1. IR
  2. RI
  3. GI
  4. EI

Answer: D, EI. EI is the callsign prefix for the Republic of Ireland. It is one of the common international prefixes candidates are expected to recall under Direct-to-Full syllabus 2A3.

The three ITU Regions divide the world for frequency allocation purposes. Which region includes Europe, Africa and the Middle East?

  1. Region 4
  2. Region 3
  3. Region 2
  4. Region 1

Answer: D, Region 1. The ITU divides the world into three regions for spectrum allocation. Region 1 covers Europe, Africa, the Middle East and the former Soviet Union. Region 2 covers the Americas; Region 3 covers Asia-Pacific.

Most amateur satellites are LEO (Low Earth Orbit), meaning:

  1. They orbit at less than 250 km above the Earth
  2. They cross the sky, in view for short periods
  3. They stay fixed above one point on the Earth
  4. They remain in view for several hours at a time

Answer: B, They cross the sky, in view for short periods. A LEO satellite is moving relative to the Earth and is above the horizon only at certain times, so it crosses the sky and is in view for short passes (syllabus 2F1).

3 Technical aspects

At the resonant frequency of a series LC circuit, the impedance is:

  1. Purely capacitive
  2. Purely inductive
  3. At its minimum value
  4. At its maximum value

Answer: C, At its minimum value. In a series LC circuit at resonance the inductive and capacitive reactances are equal and cancel, leaving only the small resistance. Impedance is therefore at its minimum and current at its maximum (syllabus 3F1).

A series circuit consists of a 50Ω resistor, an inductor with 100Ω reactance, and a capacitor with 100Ω reactance. What is the impedance?

  1. Infinity
  2. 0 Ω
  3. 50 Ω
  4. 250 Ω

Answer: C, 50 Ω. At series resonance XL = XC = 100 Ω, so they cancel exactly. Z = sqrt(R² + (XL − XC)²) = sqrt(50² + 0²) = 50 Ω per D2F syllabus 3F. Impedance is purely resistive and at its minimum.

In a common emitter amplifier using an NPN transistor, the input signal is applied to the:

  1. Ground connection
  2. Emitter
  3. Collector
  4. Base

Answer: D, Base. Per D2F syllabus 3G2, in a common emitter NPN amplifier the input signal is applied to the base. Small base-current variations cause large collector-current changes (gain = hFE/beta). The emitter is the common (reference) terminal; the collector is the output terminal.

4 Transmitters and receivers

You are operating a superheterodyne receiver with a fixed intermediate frequency (IF) of 9 MHz and using high-side injection. If the receiver is tuned to 14 MHz, what is the frequency of the image signal that could cause interference?

  1. 32 MHz
  2. 23 MHz
  3. 18 MHz
  4. 5 MHz

Answer: A, 32 MHz. With high-side injection the local oscillator is 14 + 9 = 23 MHz. The image lies the IF on the other side of the LO: 23 + 9 = 32 MHz, or the signal plus twice the IF (syllabus 4I3).

Over-modulation of an AM signal causes:

  1. Unwanted mixer products from the local oscillator
  2. Distorted audio with harmonics that widen the bandwidth
  3. Chirp, as the oscillator is pulled by a changing load
  4. A narrower bandwidth as the sidebands are suppressed

Answer: B, Distorted audio with harmonics that widen the bandwidth. Over-modulation distorts the modulating signal, producing audio harmonics that cause excessive transmitted bandwidth. This is in Direct-to-Full syllabus 4G1.

A Transverter mixes:

  1. The received antenna signal with a BFO to produce an audio tone
  2. A microphone audio signal with an RF carrier to give VHF/UHF
  3. A low-level drive signal with a local oscillator to give VHF/UHF
  4. An HF transmit signal with its own harmonics to produce VHF/UHF

Answer: C, A low-level drive signal with a local oscillator to give VHF/UHF. A transverter mixes a low-level drive signal from the main transceiver with a local oscillator to translate it to another band such as VHF or UHF. It generally needs low power drive. See syllabus 4N2.

5 Feeders and antennas

You measure 25W forward power and 6.25W reflected power. What is the SWR?

  1. 4.0:1
  2. 2.0:1
  3. 3.0:1
  4. 1.5:1

Answer: C, 3.0:1. Γ = √(P_ref / P_fwd) = √(6.25 / 25) = √0.25 = 0.5. VSWR = (1 + 0.5) / (1 − 0.5) = 1.5 / 0.5 = 3:1. Per D2F syllabus 5E2, VSWR is derived from the voltage reflection coefficient.

Which of the following describes a T-match antenna matching network?

  1. A single series inductor followed by a shunt capacitor
  2. Two shunt capacitors with a series inductor between them
  3. A series capacitor, shunt inductor, series capacitor arrangement
  4. A shunt inductor followed by a series capacitor

Answer: C, A series capacitor, shunt inductor, series capacitor arrangement. Per D2F syllabus 5F1, a T-match AMU consists of a series capacitor, shunt inductor, series capacitor — two series capacitors with a shunt inductor between them — giving a T topology.

The outer braid of a coaxial cable acts as:

  1. A support
  2. A screen to keep RF in and interference out
  3. A balun to balance the feeder currents
  4. A low-pass filter to remove harmonics

Answer: B, A screen to keep RF in and interference out. The braid is a screen. In a correctly connected coax the RF field stays inside the cable and is unaffected by outside objects, so RF is kept in and interference is kept out (syllabus 5A1).

6 Propagation

Which of the following weather conditions is most likely to cause significant signal attenuation at UHF frequencies?

  1. Strong anticyclonic high pressure
  2. Heavy rain
  3. Thick fog
  4. A temperature inversion

Answer: B, Heavy rain. Heavy rain, like falling snow and ice, can attenuate signals at UHF and above, because the water droplets absorb and scatter the energy (Direct-to-Full syllabus 6C2).

The power flux density of an electromagnetic wave is given by:

  1. The product of E and H fields
  2. The square of the wave frequency
  3. The sum of the E and H fields
  4. The ratio of the E to H fields

Answer: A, The product of E and H fields. Power flux density, in watts per square metre, is the product of the E and H fields (S = E × H). This is syllabus item 6A3, where E and H are in phase and at right angles to each other and to the direction of travel.

The sunspot cycle has an approximate period of:

  1. 7 years
  2. 11 years
  3. 22 years
  4. 5 years

Answer: B, 11 years. Syllabus 6B1 states the sunspot cycle is approximately 11 years. At solar maximum, increased ionisation enhances HF propagation, raising the MUF and opening higher frequency bands.

7 Electromagnetic compatibility

A Dummy Load is constructed from:

  1. A mains light bulb
  2. A variable capacitor
  3. A wire-wound inductor
  4. A non-reactive resistor

Answer: D, A non-reactive resistor. A dummy load uses a non-reactive resistor of suitable power rating, screened, so it absorbs the transmitter's power as heat without radiating it. This is useful for testing and fault finding (syllabus 7D4).

To minimise EMC, station cables should be:

  1. Unscreened, to avoid RF currents flowing on the screen
  2. Run in parallel with the house mains wiring
  3. Cut longer than needed so they can reach anywhere in the shack
  4. Kept short and neat

Answer: D, Kept short and neat. Keeping station cables short and neat reduces their ability to pick up and radiate RF, which helps EMC. This follows the good station design principles in syllabus 7E1.

Which of the following is a common source of non-radio interference?

  1. Thermostats and motors
  2. A length of coaxial cable
  3. A ferrite ring on a power lead
  4. A steadily lit filament lamp

Answer: A, Thermostats and motors. Syllabus 7B4 lists arcing thermostats and electric motors as non-radio sources of interference. They cause bursts of buzzing or clicking on analogue receivers.

8 Safety

High voltages carry a risk of:

  1. Injury from a slipping drill bit
  2. Hydrogen gas explosion
  3. Trips and falls from trailing cables
  4. Electrocution

Answer: D, Electrocution. Syllabus 8A1 says lethal voltages can exist in equipment, so the risk from high voltages is electrocution. Live circuits may be exposed once the case is removed.

When operating mobile, the radio equipment should be:

  1. Stowed loose on the rear parcel shelf out of sight
  2. Mounted in front of the passenger airbag for easy reach
  3. Held in place by its power and antenna leads
  4. Securely mounted so it cannot fly loose in a crash

Answer: D, Securely mounted so it cannot fly loose in a crash. When operating mobile, the equipment must be secure and cables protected, because loose equipment is a hazard in a crash or sudden braking. See the mobile precautions in Direct-to-Full syllabus 8F5.

Thunderstorms carry heavy static charges. The static charge from thunderclouds can ionise the air to form:

  1. A strong static magnetic field
  2. A low resistance path to ground
  3. A short circuit in the transmitter
  4. A region of reduced air pressure

Answer: B, A low resistance path to ground. Static charge from thunderclouds can ionise the air into a low resistance path to ground, allowing a very high current to flow as a lightning strike. This is syllabus item 8E1.

9 Measurements and construction

Connecting a test meter to a circuit can affect the measurement result because:

  1. The meter has its own impedance, which loads the circuit under test
  2. The meter's own tolerance alters the values of the circuit's components
  3. Analogue meters load the circuit, but digital meters never do
  4. The meter draws no current, so only the range chosen alters the reading

Answer: A, The meter has its own impedance, which loads the circuit under test. A test meter has its own input impedance, so connecting it loads the circuit under test and can change the reading. You must allow for this effect when choosing and using a meter (syllabus 9A1).

A power of 100 Watts is equivalent to:

  1. 40 dBW
  2. 10 dBW
  3. 30 dBW
  4. 20 dBW

Answer: D, 20 dBW. dBW = 10 × log₁₀(P/1 W). For 100 W: 10 × log₁₀(100) = 10 × 2 = 20 dBW. EX320 defines dBW as power in dB relative to 1 W; every 10 dBW step multiplies power by 10.

A resistor is marked as 470 ohms with a tolerance of 5%. What is the range of actual values you might expect?

  1. 460.0 to 480.0 ohms
  2. 423.0 to 517.0 ohms
  3. 446.5 to 493.5 ohms
  4. 465.0 to 475.0 ohms

Answer: C, 446.5 to 493.5 ohms. 5% of 470 ohms is 23.5 ohms, so the actual value should lie between 446.5 and 493.5 ohms (syllabus 9C1, component tolerances).