Questions on Basic Radio Theory
Decimetric waves correspond to the frequency range:
3000 to 30000 MHz
3000 to 30000 KHz
300
to 3000 MHz
300 to 3000 KHz
Decimeter: A metric unit of length, equal to one tenth of a
meter.
Decameter: A metric unit of length, equal to 10 meters.
According to ICAO Annex 10, in which frequency
band(s) does a locator normally transmit?
HF/VHF
MF/HF
HF
LF/MF
A locator outer marker, or LOM, is a navigation aid used as
part of an instrument landing system (ILS) instrument approach. LOM is a
non-directional beacon (NDB) co-located with the outer marker beacon in the ILS
approach, so that aircraft can navigate directly to the location using the NDB
as well as be alerted when they fly over it by the beacon.
A low or medium frequency radio beacon transmits
nondirectional signals whereby the pilot of an aircraft properly equipped can
determine bearings and "home" on the station. These facilities
normally operate in a frequency band of 190 to 535 kilohertz (kHz), according
to ICAO Annex 10 the frequency range for NDBs is between 190 and 1750 kHz, and
transmit a continuous carrier with either 400 or 1020 hertz (Hz) modulation.
All radio beacons except the compass locators transmit a continuous
three-letter identification in code except during voice transmissions.
A marker beacon is a particular type of VHF radio beacon used
in aviation, usually in conjunction with an instrument landing system (ILS), to
give pilots a means to determine position along an established route to a
destination such as a runway. Markers operate at 75 MHz (VHF).
An aircraft travelling at 330 metres a second
transmits a signal at 10 GHz to a stationary receiver. If the aircraft is flying
directly towards the receiver and they are approximately at the same height the
received frequency will be:
11 MHz
10,000011
GHz
9,999989 GHz
11 GHz
Doppler Shift (Hz) = Relative velocity (metres per second) /
Transmitted wavelength (metres)
Wavelength = Speed of Sound / Frequency
= 3 x 10^8 / 10 GHz or
= 300,000,000 / 10,000,000,000
= 0.03 meters
Dopler Shift = 330/0.03 = 11,000 Hz
Received frequecy = 10 GHz + 11,000 HZ = 10,000011000 Hz or
10.000011 GHz
Which of the following statements is correct in respect
of a RF signal:
the plane of polarisation is dictated by the oscillator unit
in the transmitter
the
electrical component of the signal is parallel to the aerial
the magnetic component of the signal is parallel to the aerial
both the electrical and magnetic components are parallel to
the aerial
A half wave dipole aerial suitable for transmitting
an RF signal at 18 MHz should have an effective length of:
16,67 metres
166,67 metres
83,33 metres
8,33
metres
RF signal at 18 MHz has a wavelength = Speed of sound /
frequency
= 3 x 10^8 / 18 MHz
= 16.66 meters
Half wave dipole = 16.66 / 2 = 8.33 meters
Which of the following statements is true?
A broad bandwidth gives a narrow beam width
A narrow bandwidth improves beam width
A transmission's bandwidth is affected by the design of the
aerial
Bandwidth
must be reduced in order to reduce noise
Diffraction of a RF signal is a displacement of its
propagation path due to:
reflection from the surface
passing over or through mediums of different conductivity
passing
over obstacles with dimensions close to the wavelength
passing through ionised regions of the upper atmosphere
One major consideration that determines if radio waves are
blocked involves the concept of diffraction. This depends on the wavelength of
the radiation and the size of the obstacle in which it is attempting to
penetrate. Low frequencies have a easier time passing over large objects such
as hills, while higher frequencies work better with small obstacles such as
rooftops. This can be very useful in making a radio wave blocked using the
knife-edge diffraction method. If a wave does not have a line of sight over an
object, a sharp edge can be created, which causes the radio wave to be blocked
and redirected to where the broadcast should go.
Refraction of an electro-magnetic radiation is:
The
bending of its propagation path as it passes through or over areas of different
electrical conductivity
The loss of power as it passes through or over areas of
different electrical conductivity
Is bending resultant from reflection from objects
Is loss of power through reflection from objects
With regards to radio waves, which statement is true?
They
are reflected by metallic objects with a size compatible to the wavelength
The longer the wavelength the greater the surface attenuation
They travel at 186,000 nm a second in a vacuum
High frequencies need large aerials
You are at an altitude of 9,000 feet. At a range of
200 nm from a VHF communications transmitter, and you are receiving a good
signal.
You should expect this since the transmitter is at a height
of 2.000 feet
You should have been receiving the signal from a range of 240
nm
You are probably receiving a sky wave signal
You
are probably receiving a duct propagation signal
To establish and maintain effective HF communications
the frequency used at a given range: "should be decreased at night"
sun down frequency down
Refer to figure.In the following diagram, the phase
difference between the two signals is: 180°
To determine phase difference find where your wave rises
through the neutral position and read the phase of the reference wave (solid
one in this fig) at that point.
Ground direction finding at aerodromes utilises what
frequencies?
"VHF at civil aerodromes and UHF at military aerodromes"
The emission characteristics A3E describe:
ILS
VHF
communications
HF communications
VOR
Type of Radio Emissions: http://en.wikipedia.org/wiki/Types_of_radio_emissions
Q codes
QDR - Magnetic Bearing From
QDM - Magnetic Bearing To
QTE - True Bearing From
QUJ - True Bearing To
The gain of an aerial is a measure of its ability to:
Focus power
Transmit intelligence
Overcome transmitter line resistance
Compensate
for attenuation
What is the lowest frequency where freedom from
static interference can be guaranteed?
3 MHz
30
MHz
300 MHz
3 GHz
Start of VHF
The VHF frequency band has a wavelength limit of:
100 m to 10 m
1 m to 100 cms
10
m to 1 m
100 cms to 10 cms
The wavelength of a radio signal is 200 metres. What
is the frequency?
1.5 KHz
1.5 GHz
1.5
MHz
15 MHz
Frequency = Speed of Sound / Wavelength
= 3 x 10^8 / 200
= 1,500,000 or 1.5 MHz
The approximate ground waves of LF and MF are (by
day) ___ and ___ respectively, with ___ suffering more from atmospheric
attenuation.
1500nm; 1000nm; MF
1000nm; 500nm; LF
1000nm;
300nm; MF
500nm; 100nm; LF
The distance a surface wave can travel is limited by surface
attenuation (which decreases at lower frequencies and over the sea). Surface
waves are around:
100 NM long in the HF band.
500 NM long in the MF band.
1000 NM long in the LF band.
4000+ NM long in the VLF bands.
Surface waves and space waves occur together and the
combination is called a ground wave
As the frequency of a transmitter is increased, the
range of the ground wave will:
Decrease
Decrease only at night
Increase only over the sea
A radio beacon has an operational range of 10 NM. By
what factor should the transmitter power be increased in order to achieve an
operational range of 20 NM?
Eight
Six
Four
Two
If an NDB with a transmitter power of 25 KW which has
a range of 50 nm is adjusted to give a power output of 100 KW the new range of
the NDB will be approximately: "100 nm"
An RMI indicates aircraft heading. To convert the RMI
bearings of NDBs and VORs to true bearings the correct combination for the
application of magnetic variation is: NDB: aircraft
position, VOR: beacon position
The speed of a radio wave in nm/sec is:
300,000
161,842
163,842
186,000
A radio wave with a horizontal magnetic component
would be best received by a ___ aerial.
magnetic
parabolic
horizontal
vertical
The question is about a horizontal magnetic component which
means the electrical component is vertical.
Post by OEH » Tue Jan 08,
2013 6:05 pm
An aircraft is HOMING
to a radio beacon whilst maintaining a relative bearing of zero. If the magnetic
heading decreases, the aircraft
is experiencing:
left drift
right drift
a wind from the west
zero drift
Magnetic heading decreases
= Wind from left = Drift towards right
An
NDB transmits a signal pattern in the horizontal plane which is:
a beam rotating at 20 Hz
bi-local circular
a cardioid balanced at 30
Hz
omnidirectional
Using
an NDB it is possible to experience
which of the following errors or limitations?
Coastal refraction, timing error and night effect
Night effect, station interference and latitude error
Night effect, station interference and lack of a failure warning system
Coastal refraction, timing error and lack of a failure warning system
Timing error and latitude error are the odd ones.
Factors liable to affect most NDB/ADF system performance and reliability include:
height error - station interference - mountain effect
static interference - station interference - latitude error
static interference
- night effect - absence of failure warning system
coastal refraction - lane slip - mountain effect
Which of the following are all errors
associated with ADF:
selective availability, coastal refraction, night effect
night effect, quadrantal error, lane slip
mountain effect, station interference, static interference
selective availability, coastal refraction, quadrantal error
Which of the following factors could cause
an error of an ADF bearing
of an NDB?
Scalloping
Atmospheric scatter
Phase interference
Night effect
Which of the following is likely to have the greatest effect on ADF accuracy?
Interference from other NDBs, particularly
during the day
Frequency drift at the ground
station
Interference from
other NDBs, particularly at night
Mutual interference between aircraft aerials
What gives the greatest error in ADF:
Coastal effect
Night effect
Static interference
from thunderstorms
Quadrantal error
Which combination gives the greatest reduction in reliability of ADF:
Station interference; Static
interference; Lane slip
Mountain effect; Station interference; site error
Night effect; Static interference; Quadrantal error
Nigh effect; Quadrantal error; Station interference
Which of the following may cause inaccuracies
in ADF bearings?
static interference, height effect, lack of failure warning
station interference, mountain effect, selective availability
coastal refraction, slant range, night
effect
lack of failure warning, station interference, static interference
With a transmission from an NDB aerial, the ___ component travels in the ___ plane and the signal is ___ polarised.
magnetic; horizontal;
vertically
electrical; horizontal; vertically
electrical; vertical; horizontally
magnetic; vertical; horizontally
The
BFO selector on an ADF receiver
is used to:
find the loop NULL position
stop loop rotation
hear the IDENT and must always be switched ON
hear the IDENT of some
NDB stations radiating a continuous wave signal
The purpose of the BFO switch
on the ADF receiver is to make
the signal audible
When receiving an NDB signal on an ADF receiver the BFO
can be selected OFF for the:
tone signal on N0NA1A
ident signal on N0NA1A
tone signal on N0NA2A
ident signal on
N0NA2A
The first symbol describes the transmitted
waveform, the second describes the modulation and
the third describes
the type of information carried. Emission Classification: http://www.theairlinepilots.com/forum/v ... .php?t=809
When considering the propagation of ADF transmissions night effect is most pronounced:
at dusk and dawn
during the long winter nights
at or near the coast
when flying at low altitude
In order to obtain
an ADF bearing the: "signal
must be received by both the sense and loop aerials"
An
ADF uses a sense aerial to:
"resolve ambiguous bearings".
An
NDB signal crossing from land to sea
will ___ speed and bend ___
the normal.
decrease, towards
increase, towards
decrease, away from
increase, away from
Errors caused by
the effect of coastal refraction on bearings at lower altitudes are maximum when
the NDB is:
inland and the bearing
crosses the coast at an acute angle
inland and the bearing crosses the coast at right angles
When ADF equipment which incorporates a sense aerial and a loop aerial is tuned to a NDB and the loop aerial is rotated so that
a sharp null is found the aerial is:
either at right angles or in line
with the incoming signals
in line with the incoming signals
at right angles to the incoming signals
aligned with the aircraft nose
Quadrantal errors associated with aircraft Automatic Direction Finding (ADF) equipment are caused
by:
misalignment of the loop aerial
signal bending caused by electrical
interference from aircraft wing
signal bending by the aircraft metallic surfaces
sky wave/ground wave contamination
Which one of the following disturbances is most likely to cause the greatest inaccuracy in ADF bearings?
Coastal effect
Local thunderstorm
activity
Quadrantal error
Precipitation interference
ICAO
allocated frequency band for ADF receivers is 190
- 1750 kHz
The
D layer of the ionosphere affects the accuracy of NDB bearings:
by day and night
by day only
by night only
never
Sky waves are refracted from the ionosphere. Where the ion density changes
radio waves will be refracted. Most of the refraction occurs about 125km up, at what is known as the E layer.
Night Effect which
causes loss of signal and fading, resulting in bearing errors from NDB transmissions, is due to:
sky wave distortion of the null position and is maximum at dawn and dusk
static activity increasing at night particularly in the lower frequency band
The accuracy of ADF within
the DOC (designated OP coverage)
by day is: +/-5 deg
The
95% accuracy for ADF bearings of an NDB by
day is:
±2°
±7°
±10°
±3°
The
signal to noise ratio for an NDB is ___ allowing
a maximum error of ___ on
95% of occasions during ___
3/1, ±5 degrees, daylight
hours only
3/1, ±5 degrees, 24 hours
In accordance with
Doc 8168, a pilot flying an
NDB approach must achieve a tracking accuracy within ___ of the published approach track.
+/-10°
+/- 5°
+/- 2.5°
+/- 2°
An
aircraft heading 315oM shows
an NDB bearing 180o on the RMI. Any
quadrantal error affecting the accuracy of this bearing is likely to be:
zero, as quadrantal errors are not found on the RMI
at a maximum
at a minimum
zero, as quadrantal errors affect only
the VOR
There are two
NDBs, one 20 NM inland, and the other 50 NM inland from the coast. Assuming that the error caused by coastal
refraction is the same for both propagations, the extend of the error is a position line plotted
by an aircraft that is over water
will be:
the same from both beacons when
the aircraft is on a relative
bearing of 180o and 360o
greater from the beacon that is 20 NM inland
the same from both beacons when
the aircraft is on a relative
bearing of 090o and 270o
greater from the beacon that is 50 NM inland
An
aircraft 10 nm from a north-south coastline takes two three-position line fixes from
an inland NDB. The aircraft's
indicated position is:
Nearer to the coastline
than its actual position
Further from the coastline than its actual position
Correct because the coastal refraction errors will cancel out because two
fixes have been taken
Correct in azimuth but false in range
Speed of a surface
wave is affected by the surface over which it travels
(faster over water than land).
This change of speed means the wave is refracted at low altitude as it passes over a coastline.
Refraction is always towards the coast. An aircraft receiving a refracted wave would give a false
indication of the beacon's position. It will place the aircraft nearer to the coast than it actually
is. This effect is worse the further back from the coast
the beacon is sited. It can
be avoided by:
1) Taking bearings at right angles to the coast.
2) Flying at a higher altitudes.
3) Choosing beacons closer to the coast
A
long range NDB is likely to transmit on ___ and be classified as ___ Select the answer
to complete this statement.
200 Khz; A2A
800 Khz; A2A
200 Khz; A1A
800 Khz; A1A
The first symbol describes the transmitted waveform. N = Unmodulated carrier, A = Double sideband.
The second describes the modulation. 0 = No modulation, 1
= Unmodulated digtal information, 2 = Modulated digital informaton.
The third describes the type of information carried. N = No information, A = Morse.
The common designators used in aviation are:
N0N = NDB carrier wave
A1A = NDB ident
A2A = Alternative NDB ident
For details on Emission Classification click here: http://www.theairlinepilots.com/forum/v ... .php?t=809
Older systems interrupt the carrier wave (Keying) to send an unmodulated (but also inaudible) Morse code ident
(classified as A1A). User will have also erratic indications
while the signal is interrupted in this type of beacon. To make the unmodulated parts of the signal audible, ADF equipment incorporates a Beat Frequency Oscillator (BF0). More modern systems imprint the ident onto the carrier wave by 'keying'
an audible, AM signal, in
time with the Morse Code ident (classified
A2A).
The
BFO:
creates the audio ident for an NDB
is used to make the ident from an A2A NDB audible
is used to make the ident from an A1A NDB audible
is used to determine the signal strength of an NDB
According to another question the answer is: "hear the IDENT of NDBs using N0N A1A transmissions"
What actually happens
in the ADF receiver when
the BFO position is selected?
The BFO circuit is activated,
and the receiver accepts only A1A modulated signals
The BFO circuit oscillates
at an increased frequency
in order to allow identification of A2A NDBs
The BFO circuit is de-activated
The BFO circuit imposes
a tone onto the carrier wave to make the NDB's ident audible
The
nominal maximum range of an
NDB with a transmitter power is 200 watts is:
50 to 60 nm
100 to 120 nm
150 to 170 nm
200 to 200 nm
In another question it is mentioned 40 to 45
NM. Anyway among the options its always
less than 100.
With regard to the following types of NDB which statements is correct?
Locators have 200 W power,
50 nm range and are N0N A2A
Locators have 15 W power,
10-25 nm range and are N0N A2A
Locators have 5000 W power,
50 nm range and are N0N A2A
Locators have 5000 W power,
50 nm range and are N0N A1A
An
NDB aerial is (i) so as to ensure the range is (ii) by minimising (iii) due to (iv):
(i) horizontal; (ii) maximum;
(iii) diffraction; (iv) the ground
wave
(i) vertical; (ii) maximum;
(iii) attenuation; (iv) energy
losses to the surface
(i) horizontal; (ii) maximum;
(iii) refraction; (iv) the D layer
(i) vertical; (ii) maximum;
(iii) attenuation; (iv) atmospheric
refraction
What according to ICAO Annex 10 is the range of a locator?
50 - 100 NM
25 - 50 NM
10 - 25 NM
100 - 300 NM
Which of the following is correct regarding the range of an
NDB?
The range is limited to the
line of sight
Aircraft height is not limiting
for the reception of signals
from the NDB
The range of an NDB will most
likely increase at day time
compared to night time
The transmitter power of
the NDB station has no affect on the range
When using ADF (i)BY
NIGHT, the accuracy is (ii)LESS than (iii)BY DAY, because
the surface wave is (iv)
CONTAMINATED BY SKY WAVES
Snow will NOT affect ADF.
An
NDB has a range of 50 nm with a power output
of 80 watts: The power required to increase the range to 75 nm is:
120 watts
150 watts
180 watts
320 watts
If an NDB signal is received at a range of 1000 nm:
The signal is a surface wave and is quite usable
It will be a ground wave
and will be inaccurate
It is a space wave and will
be inaccurate
It is a sky wave and is inaccurate
For
long range NDBs the most common
type is:
LF N0N A1A
LF N0N A2A
MF N0N A1A
MF N0N A2A
If an NDB has a published range of 30 nm, its accuracy is: "only guaranteed by day to that range"