Copyright: Copyright ® 1997 Tony Lacy.


ADVANCED Class.

FCC License Preparation

Element 4A

To go to an explanation press the Search button and select the required question designator.


A1A01
A1A01:

Advanced class privileges cover almost the entire 80-meter band except for two 25 kHz sections at the bottom of each sub band.  [97.301]


A1A02
A1A02:

Advanced class privileges cover almost the entire 40-meter band except for a 25 kHz section at the bottom.  [97.301]


A1A03
A1A03:

Advanced class operators have privileges over the entire 20-meter band except for two 25 kHz sections at the bottom of each sub band.  [97.301]


A1A04
A1A04:

Advanced class operators have privileges over the entire 15-meter band except for two 25 kHz sections at the bottom of each sub band.  [97.301]


A1A05
A1A05:

When you have your CSCE for Advanced privileges you can go home and use these privileges on the air if you add the /AA to your existing call sign.  [97.119]


A1A06
A1A06:

In this case the Advanced operators' privileges are being exceeded.  Therefore, the control operators call sign must be used.  [97.119]


A1A07
A1A07:

The Technician Plus privileges are being exceeded and so the Advanced class operator's call sign must be used.  [97.119]


A1A08
A1A08:

It must also be at least 40 dB below the main signal if the mean output power is greater than 5 watts.  [97.307]


A1A09
A1A09:

It must also be less than 50 mW.  [97.307]


A1A10
A1A10:

The spurious emission requirements for VHF and UHF transmitters are more stringent than for HF equipment.  [97.307]


A1A11
A1A11:

The spurious emission requirements for VHF and UHF transmitters are more stringent than for HF equipment.  [97.307]


A1B01
A1B01:

The control link is often a radio link and the control operator may operate the link from a hand held transceiver.  [97.3]


A1B02
A1B02:

An example of automatic control is a repeater.  [97.3]


A1B03
A1B03:

Obviously a model plane needs to be manually supervised! [97.201]


A1B04
A1B04:

Repeaters are the most common form of automatically controlled amateur stations.  [97.205]


A1B05
A1B05:

The correct option is almost a quotation from the FCC rules.  The point is that the control operator is not doing the actual operating of the station.  [97.3]


A1B06
A1B06:

Although a control operator is responsible for an automatic station, he/she is not actually controlling it.  [97.3]


A1B07
A1B07:

This is the only HF band that has repeaters.  They are permitted only between 29.5 - 29.7 MHz.  The input/output difference is 100 kHz with the output on the higher frequency.  [97.205]


A1B08
A1B08:

Most of the 6-meter band is available for repeater operation.  [97.205]


A1B09
A1B09:

There are two repeater sub-bands on the 2-meter band.  [97.205]


A1B10
A1B10:

All but the bottom 15 kHz is available for repeater use.  [97.205]


A1B11
A1B11:

Three sub-bands are available for repeater use on the 70 cm band..  [97.205]


A1B12
A1B12:

Repeater operation is permitted in any part of the 23-cm band.  The US national repeater plan for this band has the inputs from 1,270 - 1,276 and the corresponding outputs from 1,282 - 1,288 [97.301]


A1B13
A1B13:

The purpose of this rule is to avoid a malfunctioning station permanently transmitting, possibly for days at a time.  [97.213]


A1B14
A1B14:

The control point could be a hand held transceiver used by the local repeater control operator.  [97.3]


A1B15
A1B15:

In the case of radio control the control link consists of a receiver at the remotely controlled station plus a transmitter at the control point.  [97.3]


A1C01
A1C01:

A licensed amateur may build or modify one of each type of power amplifier per calendar year for operation below 144 MHz without having to obtain FCC type acceptance.  [97.315]


A1C02
A1C02:

Type acceptance has to be granted for each model.  [97.315]


A1C03
A1C03:

Selling on to amateurs is permitted.  [97.315]


A1C04
A1C04:

In other words, if the amplifier cannot satisfy amateur radio emission standards then it will not gain type approval.  [97.317]


A1C05
A1C05:

The amplifier must not generate spurious emissions when in standby mode.  [97.317]


A1C06
A1C06:

The FCC is trying to make it difficult for illicit high power transmissions to be made on these frequencies.  [97.317]


A1C07
A1C07:

At this drive level the amplifier will be working at close to its maximum output power.  [97.317]


A1C08
A1C08:

The aim is to prevent easy modification for illicit use.  [97.317]


A1C09
A1C09:

The aim is to prevent easy modification for illicit use.  [97.317]


A1C10
A1C10:

The aim is to prevent easy modification for illicit use.  [97.317]


A1C11
A1C11:

There should be no gain between these frequencies.  [97.317]


A1D01
A1D01:

Spread spectrum modulation involves "scattering" a signal over a wide bandwidth during transmitting and "de-scattering" the signal during receiving.  [97.3]


A1D02
A1D02:

In direct sequence transmission the carrier is modulated by a high speed pseudo-random code.  The same code sequence is used by the receiver to recover the signal.  In frequency hopping a fixed number of predetermined frequencies are used.  The carrier jumps from one to another in a predetermined sequence.  This occurs at a relatively low speed.  [97.311]


A1D03
A1D03:

This is because the FCC has chosen to take a cautious approach to this relatively new type of amateur transmission.  [97.311]


A1D04
A1D04:

The important term here is "point to point."  [97.3]


A1D05
A1D05:

The point to point communication is between the control operator and the repeater station.  [97.3]


A1D06
A1D06:

For a station to be an auxiliary station it must be a member of a group.  [97.3]


A1D07
A1D07:

The 70-cm sub-bands are satellite sub-bands.  [97.201]


A1D08
A1D08:

Any class except Novice can be in control of an auxiliary station.  [97.201]


A1D09
A1D09:

This limit is generally the maximum permitted for radio amateur identification purposes.  [97.1191]


A1D10
A1D10:

This is the same requirement as for normal amateur radio operation.  [97.1191]


A1D11
A1D11:

If it is a data transmission then the station identification can also be sent in digital form.  [97.119]


A1E01
A1E01:

An exact description of Line A can be found in the FCC Part 97 definitions [97.3] while the frequencies permitted North of Line A are stated in [97.303].


A1E02
A1E02:

An exact description of Line A can be found in the FCC Part 97 definitions [97.3] while the frequencies permitted North of Line A are stated in [97.303].


A1E03
A1E03:

The precise location is in FCC Part 97 definitions:  [97.3]


A1E04
A1E04:

The precise location is in FCC Part 97 definitions:  [97.203]


A1E05
A1E05:

This type of transmission involves safety of life or property and so it is permitted.  [97.113]


A1E06
A1E06:

The FCC rules do not permit you to run a business over the air.  However, you are allowed to mention that you have radio equipment for sale if this is not done regularly. [97.113]


A1E07
A1E07:

You may not accept payment for passing third-party messages over the air.  [97.113]


A1E08
A1E08:

A control operator may not accept payment for passing on third-party messages.  [97.113]


A1E09
A1E09:

You can communicate with an employee so long as it is not for business purposes.  [97.113]


A1E10
A1E10:

Special rules apply near airports.  [97.15]


A1E11
A1E11:

Approval may be required for lower structures near airports.  [97.15]


A1F01
A1F01:

A holder of an Amateur License is assumed to have passed the examinations required for that license class.  [97.505]


A1F02
A1F02:

The code test is for a minimum of five (5) minutes.  [97.503]


A1F03
A1F03:

There are dozens of prosigns but these are the only ones required for examination purposes.  [97.503]


A1F04
A1F04:

This is the average size of a word in normal text.  [97.507]


A1F05
A1F05:

There is no lower age limit for obtaining an Amateur radio license, but there is a lower limit on VE age.  [97.509]


A1F06
A1F06:

There are no exceptions to this rule.  [97.509]


A1F07
A1F07:

VEs must have no conflict of interest and they must be impartial.  [97.509]


A1F08
A1F08:

The principle of this rule is that VEs must have attained a higher level of amateur radio knowledge than the applicants.  [97.509]


A1F09
A1F09:

Volunteer examiners are volunteers! [97.509]


A1F10
A1F10:

If your license is revoked you will never be allowed to be a VE again.  [97.509]


A1F11
A1F11:

If your license is revoked you will never be allowed to be a VE again.  [97.509]


A1F12
A1F12:

It is possible to go to an examination center, take and pass the examination and get on the air with your new privileges all in a few hours.  [97.509]


A1F13
A1F13:

This is a maximum and it would normally be much less than this.  [97.509]


A1F14
A1F14:

The FCC used to receive applications but it is now done by the coordinating VEC.  [97.509]


A2A01
A2A01:

The important word here is "printed."  Facsimiles (Faxes) may be transmitted by telephone or by radio.


A2A02
A2A02:

At this rate it takes about 3.3 minutes to receive a FAX.


A2A03
A2A03:

One frame is 3.3 minutes at 240 lines per minute.


A2A04
A2A04:

The important word here is "printed."  Facsimiles (Faxes) may be transmitted by telephone or by radio.


A2A05
A2A05:

As a dot of light scans across the document the scattered light varies.  These light variations are converted to voltage variations and sent to the FAX receiver.


A2A06
A2A06:

In slow scan television the picture is received on a screen instead of being printed.  It is common to send photographs by slow scan television (SSTV).  Because the scan rate is relatively slow SSTV has a narrow bandwidth and can be used on the HF bands.


A2A07
A2A07:

A slow scan television black and white picture can take tens of seconds for a single frame.  Color pictures take longer.


A2A08
A2A08:

The two frequencies used in SSTV transmissions are well within the 3 kHz passband of an SSB receiver.  Shades of gray will be at intermediate frequencies.


A2A09
A2A09:

The two frequencies used in SSTV transmissions are well within the 3 kHz passband of an SSB receiver.  Shades of gray will be at intermediate frequencies.


A2A10
A2A10:

The receiver is averaging over a number of frequencies.  Interference will tend to cancel out and become background noise while the required signal will be additive.


A2A11
A2A11:

The receive frequency will have to hop using the same pseudo random sequence.


A2A12
A2A12:

This is lower than the rates used on telephone lines, but radio air time is free!


A3A01
A3A01:

Sporadic-E propagation occurs mainly in summer at times of highest solar activity.  Long distance skywave propagation becomes possible on the 6-meter band and at even higher frequencies.


A3A02
A3A02:

Sporadic-E propagation occurs mainly in summer at times of highest solar activity.  Long distance skywave propagation becomes possible on the 6-meter band and at even higher frequencies.


A3A03
A3A03:

The equatorial regions are subject to higher solar flux and so more ionization will be present in the E-layer above these regions.


A3A04
A3A04:

Sporadic-E propagation occurs mainly in summer at times of highest solar activity.  Long distance skywave propagation becomes possible on the 6-meter band and at even higher frequencies.


A3A05
A3A05:

The auroral curtains are constantly moving and of irregular shape so there will be multi-path reflections.  The fluttery sound is due to these paths causing cancellations and additions of the signal.


A3A06
A3A06:

The particles are emitted from disturbances on the surface of the sun.  They take several days to reach earth and are then trapped by the earth's magnetic field.  They are funneled down into the atmosphere near the North and South poles.  When the solar particles strike the atmosphere it glows just like the gas in fluorescent tubes.  This is what produces auroral displays.


A3A07
A3A07:

This is where the auroral displays can be seen.


A3A08
A3A08:

This is the height at which the atmosphere becomes dense enough to significantly interact with the solar particles to produce visible light.


A3A09
A3A09:

Narrow band signals are the most tolerant to the rapid fading effects of auroral propagation


A3A10
A3A10:

Propagation by tropospheric ducting is not a ground wave phenomenon.


A3A11
A3A11:

The ground wave vertical polarization component is propagated more efficiently than the horizontal component.


A3B01
A3B01:

The phase differences are caused because the signal is arriving at the receiver by several different paths.  It can be highly frequency dependent.


A3B02
A3B02:

The phase differences are caused because the signal is arriving at the receiver by several different paths.  It can be highly frequency dependent.


A3B03
A3B03:

The phase differences are caused because the signal is arriving at the receiver by several different paths.  It can be highly frequency dependent.  Therefore, wide band signals such as FM and double sideband are more affected.


A3B04
A3B04:

Selective fading can be highly frequency dependent.  Therefore, wide band signals such as FM and double sideband are more affected.


A3B05
A3B05:

The bending is in a downward direction due to the change in air density with height.  Because of this the radio path extends beyond the line of sight horizon.  The extra path length is about 15%


A3B06
A3B06:

The bending is in a downward direction due to the change in air density with height.  Because of this the radio path extends beyond the line of sight horizon.  The extra path length is about 15%


A3B07
A3B07:

Low angle radiation gives a greater communication range.  This is another reason to mount an antenna as high as possible.


A3B08
A3B08:

So the best site for a Yagi is as high as possible on the side of a hill sloping in the direction most often used!


A3B09
A3B09:

This is a sort of ducting effect that can happen at HF.


A3B10
A3B10:

This assumes high powers and high gain antennas properly aligned.


A3B11
A3B11:

Under normal conditions up to 500 miles is the best that can be expected for normal tropospheric propagation at VHF.  When the range is greater than this extra effects are likely to be responsible, such as the tropospheric ducting effect.


A3B12
A3B12:

There is a very slight heating effect on the air or particles.  The loss of energy is not to be confused with losses in field strength due to the wave spreading out as it propagates.


A4A01
A4A01:

A frequency standard is checked against more accurate frequency standards.  The ultimate frequency standards are called "atomic clocks."


A4A02
A4A02:

A frequency counter counts how many cycles occur in a given time.  The frequency is calculated from this and then displayed.


A4A03
A4A03:

A Lissajous pattern is always displayed when the two signals are roughly similar in amplitude and are simple multiples of each other.  If there is a small discrepancy in one of the frequencies then the whole pattern will appear to slowly rotate.

Unknown frequency = (n2/n1) x known frequency

Where:

n1 is the loops along a vertical edge

n2 is the loops along a horizontal edge


A4A04
A4A04:

A dip meter is useful for measuring resonant frequencies of tuned circuits.  At resonance, power is "sucked out" of the dip meter oscillator by the circuit being tested.  This can be seen on a meter.


A4A05
A4A05:

A dip meter is useful for measuring resonant frequencies of tuned circuits.  At resonance, power is "sucked out" of the dip meter oscillator by the circuit being tested.  This can be seen on a meter.


A4A06
A4A06:

A dip meter is useful for measuring resonant frequencies of tuned circuits.  At resonance, power is "sucked out" of the dip meter oscillator by the circuit being tested.  This can be seen on a meter.


A4A07
A4A07:

Antenna traps are tuned circuits placed in a multi-band antenna system to isolate parts of the elements.


A4A08
A4A08:

The coupling should be the smallest possible that gives a visible reading on the meter.  It should be loosely coupled.  If you over-couple the dip meter, you may affect the tuned circuit under test.


A4A09
A4A09:

The coupling should be the smallest possible that gives a visible reading on the meter.  It should be loosely coupled.  If you over-couple the dip meter, you may affect the tuned circuit under test.


A4A10
A4A10:

The coupling should be the smallest possible that gives a visible reading on the meter.  It should be loosely coupled.  If you over-couple the dip meter, you may affect the tuned circuit under test.


A4A11
A4A11:

Apart from the cost of manufacturing it, the circuit will have less stray capacitance and inductance.  It will also be smaller and more reliable.  On the other hand, surface mounting of components does not lend itself to easy repair and modification.


A4B01
A4B01:

This is the older style of mechanical meter with jeweled movement and "hair spring."


A4B02
A4B02:

Accuracy and linearity of the time base are relevant to frequency measurements.  Linearity and bandwidth of the deflection amplifiers are relevant to amplitude measurements and maximum operating frequency.


A4B03
A4B03:

As these parameters improve, the cost increases


A4B04
A4B04:

Frequency counter time bases are usually derived from crystals in temperature controlled ovens.  Sometimes there is a facility for feeding an external frequency standard into the frequency counter.


A4B05
A4B05:

This is the one important limit on accuracy.  Frequency counter time bases are usually derived from crystals in temperature controlled ovens.  Sometimes there is a facility for feeding an external frequency standard into the frequency counter.


A4B06
A4B06:

One (1) p.p.m. means one (1) part per million.  This is simply the frequency divided by 1,000,000.

In the question this comes to 146,520,000 / 1,000,000 = 146.52 Hz


A4B07
A4B07:

0.1 p.p.m. means 0.1 parts per million.  This is simply the frequency divided by 10,000,000.

In the question this comes to 146,520,000 / 10,000,000 = 14.652 Hz


A4B08
A4B08:

Ten (10) p.p.m. means ten (10) parts per million.  This is simply the frequency divided by 100,000.

In the question this comes to 146,520,000 / 100,000 = 1,465.20 Hz


A4B09
A4B09:

One (1) p.p.m. means one (1) part per million.  This is simply the frequency divided by 1,000,000.

In the question this comes to 432,100,000 / 1,000,000 = 432.1 Hz


A4B10
A4B10:

0.1 p.p.m. means 0.1 parts per million.  This is simply the frequency divided by 10,000,000.

In the question this comes to 432,100,000 / 10,00,000 = 43.21 Hz


A4B11
A4B11:

Ten (10) p.p.m. means ten (10) parts per million.  This is simply the frequency divided by 100,000.

In the question this comes to 432,100,000 / 100,000 = 4,321 Hz


A4C01
A4C01:

The local oscillator is modulated by the strong signals and this modulation is then superimposed on the required signal.


A4C02
A4C02:

Receiver desensitization is a potential problem with repeaters.  High quality filters on the repeater input are used to reduce it.  Careful shielding of your receiver and transmitter also help.


A4C03
A4C03:

Receiver desensitization is a potential problem with repeaters.  High quality filters on the repeater input are used to reduce it.  Careful shielding of your receiver and transmitter also help.


A4C04
A4C04:

Receiver desensitization is a potential problem with repeaters.  High quality filters on the receiver input are used to reduce it.  Careful shielding of your receiver and transmitter shielding also help.


A4C05
A4C05:

This is a consequence of the way that FM discriminators work.


A4C06
A4C06:

This is a consequence of the way that FM discriminators work.


A4C07
A4C07:

This is a consequence of the way that FM discriminators work.


A4C08
A4C08:

Most of the internal noise comes from the first active device at the receiver front end.


A4C09
A4C09:

The noise due to thermal movement at the receiver input is p = KTB watts.

Where:

K=1.37 E - 23 (Boltzmanns constant)

T = absolute temperature in degrees Kelvin.

B = bandwidth in Hz

Giving p = 1.9923 E-18 watts = -177 dB0 = -147 dBm

At a bandwidth of 500 Hz a receiver has a minimum theoretical noise floor of -147 dBm.  The active device at the front end will contribute some noise, in this question this noise is 8 dBm.  So the noise power is -147 + 8 = -139 dBm.

A signal of -139 dB level could just be detected.  This is the bottom end of the dynamic range.  The top end (for 1 dB blocking) is given as -20 dBm.  Therefore, the blocking dynamic range is 139 - 20 = 119 dB.


A4C10
A4C10:

One way to obtain a good image rejection ratio is to have good RF selectivity before the mixer so those image frequencies at (IF frequency x 2) away from the required signal are suppressed.  If the chosen IF frequency is high then it will be easier for the RF filters to reject image frequencies since they will be farther away from the required signal.  So correct answers would be RF filter selectivity and IF filter frequency.  However, RF filter selectivity is not presented as an option and so this leaves option D, IF filter.


A4C11
A4C11:

The noise due to thermal movement on a receiver input is p = KTB watts.

Where:

K=1.37 E - 23 (Boltzmanns constant).

T = absolute temperature in degrees Kelvin.

B = bandwidth in Hz

Let's assume a bandwidth of 500 Hz and a temperature of 20 degrees Centigrade that is 290 degrees Kelvin..

This gives p = 1.37 E -23 x 290 x 500 = 1.9923 E-18 watts = -177 dB0

dB0 is a power level relative to 1 watt.  We usually use dBm that is a power level relative to 1 mw.

Power level in dBm = Power level in dB0 + 30

-177 dB0 = -147 +30 = -147 dBm

At a bandwidth of 500 Hz a receiver has a minimum theoretical noise floor of -147 dBm.

The minimum discernible signal will be above the theoretical noise floor of 147 dBm (at 500 Hz bandwidth).  Typical figures quoted by manufacturers are around 145 dBm for VHF receivers.


A4D01
A4D01:

"Inter" means "between".  The products due to mixing between two frequencies are intermodulation products.


A4D02
A4D02:

"Inter" means "between".  The products due to mixing between two frequencies are intermodulation products.  This question is checking for knowledge that transmitter power amplifiers are capable of acting as mixers if strong external signals are applied to their outputs.  Solid state power amplifiers are particularly prone to this effect.


A4D03
A4D03:

A power amplifier can produce cross modulation products if several strong signals are present at the output.  This can be a problem at radio sites with several transmitters sharing a common tower with antennas close together.  Isolators are directional devices that prevent unwanted signals going down the feedline to the power amplifier.  They also attenuate the forward signal to some extent.


A4D04
A4D04:

Immunity to cross modulation is an important factor in receiver performance.


A4D05
A4D05:

Immunity to cross modulation is an important factor in receiver performance.


A4D06
A4D06:

The best place for filtering to reduce cross modulation is before the first active device in the receiver.


A4D07
A4D07:

Intermodulation and cross modulation occur due to non linearity in the early stages of a receiver.  The strong signals modulate the weaker ones.  To correct this condition, either eliminate the strong unwanted signals with filters or improve the linearity of the RF stages of the receiver.


A4D08
A4D08:

Intermodulation and cross modulation occur due to non linearity in the early stages of a receiver.  The strong signals modulate the weaker ones.  To correct this condition, either eliminate the strong unwanted signals with filters or improve the linearity of the RF stages of the receiver.


A4D09
A4D09:

The difference between the two frequencies given is 0.18 MHz and so the other interfering signal could be at 146.52 - 0.18 = 146.34 MHz.

If the interfering signal were at 146.61 MHz then there would be a difference of:

146.61 - 146.52 = 0.09 MHz.

This could produce an intermodulation product at 146.61+0.09 = 146.70 MHz


A4D10
A4D10:

The high pass filter would allow the VHF television transmissions but would block the lower frequency HF amateur transmissions.


A4D11
A4D11:

Intermodulation requires a none linear device.  In this case the none linear device is an over modulated SSB transmitter and the intermodulation products are derived from the frequencies present in the speech waveform.


A5A01
A5A01:

At resonance the circulating currents and resulting voltages can be much larger than the applied voltage of current.  Think of a child's swing that has built up much energy from a series of small pushes.


A5A02
A5A02:

At resonance XL = XC and the circuit appears resistive to the AC current.


A5A03
A5A03:

At resonance XL = XC and the circuit appears resistive to the AC current.


A5A04
A5A04:

At resonance XL = XC and the circuit appears resistive to the AC current.


A5A05
A5A05:

The impedance of a series L-C circuit is at a minimum at resonance and the circuit appears resistive because no reactance is present.


A5A06
A5A06:

The inductor capacitor combination will present a very high impedance at resonance.  Therefore, the parallel resistor will determine the total circuit impedance.


A5A07
A5A07:

The impedance of a series L-C circuit is at a minimum at resonance and the circuit appears resistive.


A5A08
A5A08:

Be careful here.  The impedance of the circuit is at a maximum and little current will flow through it.  However, the circulating current between L and C will be at a maximum.


A5A09
A5A09:

The circulating current will be at a maximum, but the current flowing across the circuit will be at a minimum.


A5A10
A5A10:

The current will be in phase because a resonant circuit is resistive.


A5A11
A5A11:

The current will be in phase because a resonant circuit is resistive.


A5B01
A5B01:

Use an approximate value for pi of 3.14

Use the formula Fr = 1 / (2 x pi x square root ( L x C))

We get 1 / (2 x 3.14 x square root (50E-6 x 40E-12))

= 3,560,617 Hz = 3.56 MHz


A5B02
A5B02:

Use an approximate value for pi of 3.14

Use the formula Fr = 1 / [2 x pi x square root ( L x C)]

We get 1 / [2 x 3.14 x square root (40E-6 x 200E-12)]

= 1,780,308 Hz = 1.78 MHz


A5B03
A5B03:

Use an approximate value for pi of 3.14

Use the formula Fr = 1 / [2 x pi x square root ( L x C)]

We get 1 / [2 x 3.14 x square root (50E-6 x 10E-12)]

= 7,121,235 Hz = 7.21 MHz


A5B04
A5B04:

Use an approximate value for pi of 3.14

Use the formula Fr = 1 / [2 x pi x square root ( L x C)]

We get 1 / [2 x 3.14 x square root (25E-6 x 10E-12)]

= 1,007,094 Hz = 10.1 MHz


A5B05
A5B05:

Use an approximate value for pi of 3.14

Use the formula Fr = 1 / [2 x pi x square root ( L x C)]

We get 1 / [2 x 3.14 x square root (3E-6 x 40E-12)]

= 14,536,161 Hz = 14.5 MHz


A5B06
A5B06:

Use an approximate value for pi of 3.14

Use the formula Fr = 1 / [2 x pi x square root ( L x C)]

We get 1 / [2 x 3.14 x square root (4E-6 x 20E-12)]

= 17,803,088 Hz = 17.8 MHz


A5B07
A5B07:

Use an approximate value for pi of 3.14

Use the formula Fr = 1 / [2 x pi x square root ( L x C)]

We get 1 / [2 x 3.14 x square root (8E-6 x 7E-12)]

= 21,278,761 Hz = 27.3 MHz


A5B08
A5B08:

Use an approximate value for pi of 3.14

Use the formula Fr = 1 / [2 x pi x square root ( L x C)]

We get 1 / [2 x 3.14 x square root (3E-6 x 15E-12)]

= 23,737,451 Hz = 23.7 MHz


A5B09
A5B09:

Use an approximate value for pi of 3.14

Use the formula Fr = 1 / [2 x pi x square root ( L x C)]

We get 1 / [2 x 3.14 x square root (4E-6 x 8E-12)]

= 28,149,155 Hz = 28.1 MHz


A5B10
A5B10:

Use an approximate value for pi of 3.14

Use the formula Fr = 1 / [2 x pi x square root ( L x C)]

We get 1 / (2 x 3.14 x square root (1E-6 x 9E-12))

= 53,078,556 Hz = 53.1 MHz


A5B11
A5B11:

We have to re arrange the formula:

Fr = 1 / [2 x pi x square root ( L x C)] to obtain C

C = [(1/(2 x pi x Fr))squared] / L

Giving:

C = [(1/(2 x 3.14 x 14.25E6))squared] / 44E-6

We get C = 4.39E-11 Farads = 43 picofarads.


A5C01
A5C01:

Use an approximate value for pi of 3.14

Use the formula Fr = 1 / [2 x pi x square root ( L x C)]

We get 1 / [2 x 3.14 x square root (1E-6 x 10E-12)]

= 50,354,739 Hz = 50.3 MHz


A5C02
A5C02:

Use an approximate value for pi of 3.14.

Use the formula Fr = 1 / [2 x pi x square root ( L x C)].

We get 1 / [2 x 3.14 x square root (2E-6 x 15E-12)]

= 29,072,322 Hz = 29.1 MHz.


A5C03
A5C03:

Use an approximate value for pi of 3.14.

Use the formula Fr = 1 / [2 x pi x square root ( L x C)].

We get 1 / [2 x 3.14 x square root (5E-6 x 9E-12)]

= 23,737,745 Hz = 23.7 MHz.


A5C04
A5C04:

Use an approximate value for pi of 3.14.

Use the formula Fr = 1 / [2 x pi x square root ( L x C)].

We get 1 / [2 x 3.14 x square root (2E-6 x 30E-12)]

= 20,557,723 Hz = 20.5 MHz.


A5C05
A5C05:

Use an approximate value for pi of 3.14.

Use the formula Fr = 1 / [2 x pi x square root ( L x C)].

We get 1 / [2 x 3.14 x square root (15E-6 x 5E-12)]

= 18,386,951 Hz = 23.7 MHz.


A5C06
A5C06:

Use an approximate value for pi of 3.14.

Use the formula Fr = 1 / [2 x pi x square root ( L x C)].

We get 1 / [2 x 3.14 x square root (3E-6 x 40E-12)]

= 14,536,161 Hz = 14.5 MHz.


A5C07
A5C07:

Use an approximate value for pi of 3.14.

Use the formula Fr = 1 / [2 x pi x square root ( L x C)].

We get 1 / [2 x 3.14 x square root (40E-6 x 6E-12)]

= 10,278,618 Hz = 10.3 MHz.


A5C08
A5C08:

Use an approximate value for pi of 3.14.

Use the formula Fr = 1 / [2 x pi x square root ( L x C)].

We get 1 / [2 x 3.14 x square root (10E-6 x 50E-12)]

= 7,121,235 Hz = 7.12 MHz.


A5C09
A5C09:

Use an approximate value for pi of 3.14.

Use the formula Fr = 1 / [2 x pi x square root ( L x C)].

We get 1 / [2 x 3.14 x square root (200E-6 x 10E-12)]

= 3,560,617 Hz = 3.56 MHz.


A5C10
A5C10:

Use an approximate value for pi of 3.14.

Use the formula Fr = 1 / [2 x pi x square root ( L x C)].

We get 1 / [2 x 3.14 x square root (90E-6 x 100E-12)]

= 1,678,491 Hz = 1.68 MHz.


A5C11
A5C11:

We have to re arrange the formula:

Fr = 1 / [2 x pi x square root ( L x C)] to obtain L.

L = [(1/(2 x pi x Fr)) squared] / C.

Giving:

L = [(1/(2 x 3.14 x 14.25E6) squared] / 44E-12.

We get L = 2.8E-6 Henrys = 2.8 microhenrys.


A5D01
A5D01:

The rapidly changing fields in a conductor carrying RF cause the current to flow mostly in the outer parts.  Metal tubing would be just as good a conductor as solid wire at RF frequencies.  If low RF resistance is important then wire is sometimes silver plated because silver is a better electrical conductor than copper.


A5D02
A5D02:

The rapidly changing fields in a conductor carrying RF cause the current to flow mostly in the outer parts.  Metal tubing would be just as good a conductor as solid wire at RF frequencies.  If low resistance is important then wire is sometimes silver plated because silver is a better electrical conductor than copper.


A5D03
A5D03:

This is called the skin effect.  The rapidly changing fields in a conductor carrying RF cause the current to flow mostly in the outer parts.  Metal tubing would be just as good a conductor as solid wire at RF frequencies.  If low resistance is important then wire is sometimes silver plated because silver is a better electrical conductor than copper.


A5D04
A5D04:

The rapidly changing fields in a conductor carrying RF cause the current to flow mostly in the outer parts.  Metal tubing would be just as good a conductor as solid wire at RF frequencies.  If low resistance is important then wire is sometimes silver plated because silver is a better electrical conductor than copper.


A5D05
A5D05:

The skin effect results in an increase in resistance because the effective conductor area is less.  The rapidly changing fields in a conductor carrying RF cause the current to flow mostly in the outer parts.  Metal tubing would be just as good a conductor as solid wire at RF frequencies.  If low resistance is important then wire is sometimes silver plated because silver is a better electrical conductor than copper.


A5D06
A5D06:

The electrostatic field is in the insulator between the two electrodes.  This insulator may be air, plastic film or even a vacuum in certain specialized capacitors.


A5D07
A5D07:

The Joule is a measure of total energy.  It is equivalent to a watt of power flowing for a second.


A5D08
A5D08:

A magnetic field can be used to store energy.


A5D09
A5D09:

The left hand rule:

Grip a conductor with your left hand.  Straighten out your thumb.  If your thumb points in the direction of the electron flow, then your fingers will point in the direction of the resulting magnetic field.


A5D10
A5D10:

Double the current and the magnetic field becomes twice as strong.


A5D11
A5D11:

Potential energy is "frozen" energy ready to be released.  It has the potential to do work.


A5E01
A5E01:

For half power bandwidth use the formula:

B = Fr/Q Hz.

Where "Fr" is the resonant frequency and "Q" is the "quality" (or goodness) of the circuit.

We have:

B = 1.8E6/95 = 18947 Hz = 18.9 kHz.


A5E02
A5E02:

For half power bandwidth use the formula:

B = Fr/Q Hz.

Where "Fr" is the resonant frequency and "Q" is the "quality" (or goodness) of the circuit.

We have:

B = 3.6E6/218 = 16,513 Hz = 16.5 kHz.


A5E03
A5E03:

For half power bandwidth use the formula:

B = Fr/Q Hz.

Where "Fr" is the resonant frequency and "Q" is the "quality" (or goodness) of the circuit.

We have:

B = 7.1E6/150 = 47,333 Hz = 47.3 kHz.


A5E04
A5E04:

For half power bandwidth use the formula:

B = Fr/Q Hz.

Where "Fr" is the resonant frequency and "Q" is the "quality" (or goodness) of the circuit.

We have:

B = 12.8E6/218 = 58,715 Hz = 58.7 kHz.


A5E05
A5E05:

For half power bandwidth use the formula:

B = Fr/Q Hz.

Where "Fr" is the resonant frequency and "Q" is the "quality" (or goodness) of the circuit.

We have:

B = 14.25E6 / 150 = 95,000 Hz = 95 kHz.


A5E06
A5E06:

For half power bandwidth use the formula:

B = Fr/Q Hz.

Where "Fr" is the resonant frequency and "Q" is the "quality" (or goodness) of the circuit.

We have:

B = 21.15E6 / 95 = 222,631 Hz = 222.6 kHz.


A5E07
A5E07:

For half power bandwidth use the formula:

B = Fr/Q Hz.

Where "Fr" is the resonant frequency and "Q" is the "quality" (or goodness) of the circuit.

We have:

B = 10.1E6 / 225 = 44,888 Hz = 44.9 kHz.


A5E08
A5E08:

For half power bandwidth use the formula:

B = Fr/Q Hz.

Where "Fr" is the resonant frequency and "Q" is the "quality" (or goodness) of the circuit.

We have:

B = 18.1E6 / 195 = 92,8205 Hz = 92.8 kHz.


A5E09
A5E09:

For half power bandwidth use the formula:

B = Fr/Q Hz.

Where "Fr" is the resonant frequency and "Q" is the "quality" (or goodness) of the circuit.

We have:

B = 3.7E6 / 118 = 31,355 Hz = 31.4 kHz.


A5E10
A5E10:

For half power bandwidth use the formula:

B = Fr/Q Hz.

Where "Fr" is the resonant frequency and "Q" is the "quality" (or goodness) of the circuit.

We have:

B = 14.25E6 / 187 = 76,203 Hz = 76.2 kHz.


A5E11
A5E11:

For half power bandwidth use the formula:

B = Fr/Q Hz.

Where "Fr" is the resonant frequency and "Q" is the "quality" (or goodness) of the circuit.


A5F01
A5F01:

For the Q of a parallel R-L-C circuit we use the formula:

Q = Rp / X.

Where:

X is the reactance of the inductor or capacitor at resonance.

R is the resistance in parallel with the tuned circuit.

In this circuit the inductive reactance is obtained from:

Xl = 2 x pi x Fr x L = 2 x 3.14 x 14.128E6 x 2.7E-6 = 239.55 ohms.

So Q = 18E3 / 239.55 = 75.1


A5F02
A5F02:

For the Q of a parallel R-L-C circuit we use the formula:

Q = Rp / X.

Where:

X is the reactance of the inductor or capacitor at resonance.

R is the resistance in parallel with the tuned circuit.

In this circuit the inductive reactance is obtained from:

Xl = 2 x pi x Fr x L = 2 x 3.14 x 14.128E6 x 4.7E-6 = 417 ohms.

So Q = 18E3 / 417 = 43.1


A5F03
A5F03:

For the Q of a parallel R-L-C circuit we use the formula:

Q = Rp / X.

Where:

X is the reactance of the inductor or capacitor at resonance.

R is the resistance in parallel with the tuned circuit.

In this circuit the inductive reactance is obtained from:

Xl = 2 x pi x Fr x L = 2 x 3.14 x 4.468E6 x 47E-6 = 1316 ohms.

So Q = 180 / 1316 = 0.136


A5F04
A5F04:

For the Q of a parallel R-L-C circuit we use the formula:

Q = Rp / X

Where:

X is the reactance of the inductor or capacitor at resonance.

R is the resistance in parallel with the tuned circuit.

In this circuit the inductive reactance is obtained from:

Xl = 2 x pi x Fr x L = 2 x 3.14 x 14.225E6 x 3.5E-6 = 312.6 ohms.

So Q = 10E3 / 312.6 = 31.9


A5F05
A5F05:

For the Q of a parallel R-L-C circuit we use the formula:

Q = Rp / X

Where:

X is the reactance of the inductor or capacitor at resonance.

R is the resistance in parallel with the tuned circuit.

In this circuit the inductive reactance is obtained from:

Xl = 2 x pi x Fr x L = 2 x 3.14 x 7.125E6 x 8.2E-6 = 366.9 ohms.

So Q = 1000 / 366.9 = 2.73


A5F06
A5F06:

For the Q of a parallel R-L-C circuit we use the formula:

Q = Rp / X

Where:

X is the reactance of the inductor or capacitor at resonance.

R is the resistance in parallel with the tuned circuit.

In this circuit the inductive reactance is obtained from:

Xl = 2 x pi x Fr x L = 2 x 3.14 x 7.125E6 x 10.1E-6 = 451.9 ohms.

So Q = 100 / 451.9 = 0.221


A5F07
A5F07:

For the Q of a parallel R-L-C circuit we use the formula:

Q = Rp / X

Where:

X is the reactance of the inductor or capacitor at resonance.

R is the resistance in parallel with the tuned circuit.

In this circuit the inductive reactance is obtained from:

Xl = 2 x pi x Fr x L = 2 x 3.14 x 7.125E6 x 12.6E-6 = 563 ohms.

So Q = 22E3 / 563 = 39


A5F08
A5F08:

For the Q of a parallel R-L-C circuit we use the formula:

Q = Rp / X

Where:

X is the reactance of the inductor or capacitor at resonance.

R is the resistance in parallel with the tuned circuit.

In this circuit the inductive reactance is obtained from:

Xl = 2 x pi x Fr x L = 2 x 3.14 x 3.625E6 x 3E-6 = 68.295 ohms.

So Q = 2.2E3 / 68.295 = 32.2


A5F09
A5F09:

For the Q of a parallel R-L-C circuit we use the formula:

Q = Rp / X

Where:

X is the reactance of the inductor or capacitor at resonance.

R is the resistance in parallel with the tuned circuit.

In this circuit the inductive reactance is obtained from:

Xl = 2 x pi x Fr x L = 2 x 3.14 x 3.625E6 x 42E-6 = 956.1 ohms.

So Q = 220 / 956.1 = 0.23


A5F10
A5F10:

For the Q of a parallel R-L-C circuit we use the formula:

Q = Rp / X

Where:

X is the reactance of the inductor or capacitor at resonance.

R is the resistance in parallel with the tuned circuit.

In this circuit the inductive reactance is obtained from:

Xl = 2 x pi x Fr x L = 2 x 3.14 x 3.625E6 x 43E-6 = 978.895 ohms.

So Q = 1.8E3 / 978.895 = 1.84


A5F11
A5F11:

Since Q = Rp / X then reducing "Rp" will reduce "Q".


A5G01
A5G01:

To calculate phase angle we use:

Tan (Phase angle) = X / R.

Where X = total reactance = Xl - Xc.

R = total series resistance.

If X is negative then the circuit is capacitive and the voltage lags the current.

If X is positive then the circuit is inductive and the voltage leads the current.

To obtain Phase angle from Tan (Phase angle) you need a set of trigonometry tables or a calculator that supports inverse trig functions.  The Tan (Phase angle) function can usually be accessed using the sequence INV TAN after entering a number.

In the question we have:

Tan (Phase angle) = (Xl - Xc) / R = (100-25) / 100 = 0.75

So Phase angle = 36.9 degrees

Xl - Xc is positive so the voltage leads the current.


A5G02
A5G02:

To calculate phase angle we use:

Tan (Phase angle) = X / R

Where X = total reactance = Xl - Xc

R = total series resistance.

If X is negative then the circuit is capacitive and the voltage lags the current.

If X is positive then the circuit is inductive and the voltage leads the current.

To obtain Phase angle from Tan (Phase angle) you need a set of trigonometry tables or a calculator that supports inverse trig functions.  The Tan (Phase angle) function can usually be accessed using the sequence INV TAN after entering a number.

In the question we have:

Tan (Phase angle) = (Xl - Xc) / R = (50-25) / 100 = 0.25

So Phase angle = 14 degrees

Xl - Xc is positive so the voltage leads the current.


A5G03
A5G03:

To calculate phase angle we use:

Tan (Phase angle) = X / R

Where:

X = total reactance = Xl - Xc

R = total series resistance.

If X is negative then the circuit is capacitive and the voltage lags the current.

If X is positive then the circuit is inductive and the voltage leads the current.

To obtain Phase angle from Tan (Phase angle) you need a set of trigonometry tables or a calculator that supports inverse trig functions.  The Tan (Phase angle) function can usually be accessed using the sequence INV TAN after entering a number.

In the question we have:

Tan (Phase angle) = (Xl - Xc) / R = (250-500) / 1000 = -0.25

So Phase angle = 14 degrees

Xl - Xc is negative so the voltage lags the current.


A5G04
A5G04:

To calculate phase angle we use:

Tan (Phase angle) = X / R

Where:

X = total reactance = Xl - Xc

R = total series resistance.

If X is negative then the circuit is capacitive and the voltage lags the current.

If X is positive then the circuit is inductive and the voltage leads the current.

To obtain Phase angle from Tan (Phase angle) you need a set of trigonometry tables or a calculator that supports inverse trig functions.  The Tan (Phase angle) function can usually be accessed using the sequence INV TAN after entering a number.

In the question we have:

Tan (Phase angle) = (Xl - Xc) / R = (100-75) / 100 = 0.25

So Phase angle = 14 degrees

Xl - Xc is positive so the voltage leads the current.


A5G05
A5G05:

To calculate phase angle we use:

Tan (Phase angle) = X / R

Where:

X = total reactance = Xl - Xc

R = total series resistance.

If X is negative then the circuit is capacitive and the voltage lags the current.

If X is positive then the circuit is inductive and the voltage leads the current.

To obtain Phase angle from Tan (Phase angle) you need a set of trigonometry tables or a calculator that supports inverse trig functions.  The Tan (Phase angle) function can usually be accessed using the sequence INV TAN after entering a number.

In the question we have:

Tan (Phase angle) = (Xl - Xc) / R = (25-50) / 100 = -.25

So Phase angle = 14 degrees

Xl - Xc is negative so the voltage lags the current.


A5G06
A5G06:

To calculate phase angle we use:

Tan (Phase angle) = X / R

Where:

X = total reactance = Xl - Xc

R = total series resistance.

If X is negative then the circuit is capacitive and the voltage lags the current.

If X is positive then the circuit is inductive and the voltage leads the current.

To obtain Phase angle from Tan (Phase angle) you need a set of trigonometry tables or a calculator that supports inverse trig functions.  The Tan (Phase angle) function can usually be accessed using the sequence INV TAN after entering a number.

In the question we have:

Tan (Phase angle) = (Xl - Xc) / R = (50-75) / 100 = -.25

So Phase angle = 14 degrees

Xl - Xc is negative so the voltage lags the current.


A5G07
A5G07:

To calculate phase angle we use:

Tan (Phase angle) = X / R

Where:

X = total reactance = Xl - Xc

R = total series resistance.

If X is negative then the circuit is capacitive and the voltage lags the current.

If X is positive then the circuit is inductive and the voltage leads the current.

To obtain Phase angle from Tan (Phase angle) you need a set of trigonometry tables or a calculator that supports inverse trig functions.  The Tan (Phase angle) function can usually be accessed using the sequence INV TAN after entering a number.

In the question we have:

Tan (Phase angle) = (Xl - Xc) / R = (75-75) / 100 = -.25

So Phase angle = 14 degrees

Xl - Xc is negative so the voltage lags the current.


A5G08
A5G08:

To calculate phase angle we use:

Tan (Phase angle) = X / R

Where:

X = total reactance = Xl - Xc

R = total series resistance.

If X is negative then the circuit is capacitive and the voltage lags the current.

If X is positive then the circuit is inductive and the voltage leads the current.

To obtain Phase angle from Tan (Phase angle) you need a set of trigonometry tables or a calculator that supports inverse trig functions.  The Tan (Phase angle) function can usually be accessed using the sequence INV TAN after entering a number.

In the question we have:

Tan (Phase angle) = (Xl - Xc) / R = (500-250) / 1000 = 0.25

So Phase angle = 14.04 degrees

Xl - Xc is negative so the voltage leads the current.


A5G09
A5G09:

To calculate phase angle we use:

Tan (Phase angle) = X / R

Where:

X = total reactance = Xl - Xc

R = total series resistance.

If X is negative then the circuit is capacitive and the voltage lags the current.

If X is positive then the circuit is inductive and the voltage leads the current.

To obtain Phase angle from Tan (Phase angle) you need a set of trigonometry tables or a calculator that supports inverse trig functions.  The Tan (Phase angle) function can usually be accessed using the sequence INV TAN after entering a number.

In the question we have:

Tan (Phase angle) = (Xl - Xc) / R = (75-50) / 100 = 0.25

So Phase angle = 14 degrees

Xl - Xc is negative so the voltage leads the current.


A5G10
A5G10:

In a pure capacitor, the voltage lags the current by 90 degrees.


A5G11
A5G11:

In a pure inductor, the voltage leads the current by 90 degrees.


A5H01
A5H01:

This is the power that is being transferred to and from a reactive component, but is not dissipated.


A5H02
A5H02:

This is the power that is being transferred from and to a reactive component, but is not dissipated.


A5H03
A5H03:

It is returned to the rest of the circuit and can not be dissipated in the reactive components.


A5H04
A5H04:

The apparent power is the product of the current times the voltage as measured with a meter.  The true power will be less than this.


A5H05
A5H05:

Power factor can be obtained from cos (phase angle)

We have power factor = cos (60) = 0.5


A5H06
A5H06:

Power factor can be obtained from cos (phase angle)

We have power factor = cos (45) = 0.707


A5H07
A5H07:

Power factor can be obtained from cos (phase angle)

We have power factor = cos (30) = 0.866


A5H08
A5H08:

True power = apparent power x power factor

And so true power = current x voltage x power factor

We get 4 x 100 x 0.2 = 80 watts.


A5H09
A5H09:

The apparent power is given by voltage x current.

True power = apparent power x power factor

This gives 5 x 200 x 0.6 = 600 watts.


A5H10
A5H10:

In this question apparent power means current x voltage.

So, true power = apparent power x power factor.

We get 500 x 0.71 = 355 watts.


A5H11
A5H11:

Power factor = cos (phase angle) and this will always be equal to or less than one (1).  Therefore, in a reactive circuit the true power will be less than (voltage times current).


A5I01
A5I01:

First we need to calculate the total loss between the transmitter and the antenna.  Remember that antenna gain is given in dBd

- 4 -2 -1 + 6 = -1 dB.

To convert to a power ratio we divide decibels by 10 and take the antilog of the result.  For this question, we have antilog (-1 /10) = 0.794

So effective radiated power = 0.794 x 50 = 39.7 watts.


A5I02
A5I02:

First we need to calculate the total loss between the transmitter and the antenna.  Remember that antenna gain is given in dBd

- 5 -3 -1 + 7 = -2 dB.

To convert to a power ratio we divide decibels by 10 and take the antilog of the result.  For this question we have antilog(-2 /10) = 0.63

So effective radiated power = 0.63 x 50 = 31.5 watts.


A5I03
A5I03:

First we need to calculate the total loss between the transmitter and the antenna.  Remember that antenna gain is given in dBd

- 4 + 10 = 6 dB.

To convert to a power ratio we divide decibels by 10 and take the antilog of the result.  For this question we have antilog(6 /10) = 3.98

So effective radiated power = 3.98 x 75 = 299 watts.


A5I04
A5I04:

First we need to calculate the total loss between the transmitter and the antenna.  Remember that antenna gain is given in dBd

- 5 -3 -1 + 6 = -3 dB.

To convert to a power ratio we divide decibels by 10 and take the antilog of the result.  For this question we have antilog(-3 /10) = 0.501

So effective radiated power = 0.501 x 75 = 37.6 watts.


A5I05
A5I05:

First we need to calculate the total loss between the transmitter and the antenna.  Remember that antenna gain is given in dBd

- 1 + 6 = 5 dB.

To convert to a power ratio we divide decibels by 10 and take the antilog of the result.  For this question we have antilog( 5 /10) = 3.16

So effective radiated power = 3.16 x 100 = 316 watts.


A5I06
A5I06:

First we need to calculate the total loss between the transmitter and the antenna.  Remember that antenna gain is given in dBd

- 5 -3 -1 + 10 = 1 dB.

To convert to a power ratio we divide decibels by 10 and take the antilog of the result.  For this question we have antilog(1 /10) = 1.26

So effective radiated power = 1.26 x 100 = 126 watts.


A5I07
A5I07:

First we need to calculate the total loss between the transmitter and the antenna.  Remember that antenna gain is given in dBd

- 5 -3 -1 + 6 = -3 dB.

To convert to a power ratio we divide decibels by 10 and take the antilog of the result.  For this question we have antilog(-3 /10) = 0.501

So effective radiated power = 0.501 x 120 = 60 watts.


A5I08
A5I08:

First we need to calculate the total loss between the transmitter and the antenna.  Remember that antenna gain is given in dBd

- 2 - 2.2 + 7 = 2.8 dB.

To convert to a power ratio we divide decibels by 10 and take the antilog of the result.  For this question we have antilog(2.8 /10) = 1.905

So effective radiated power = 1.905 x 150 = 286 watts.


A5I09
A5I09:

First we need to calculate the total loss between the transmitter and the antenna.  Remember that antenna gain is given in dBd

- 4 -3.2 - 0.8+ 6 = 2 dB.

To convert to a power ratio we divide decibels by 10 and take the antilog of the result.  For this question we have antilog(2 /10) = 1.585

So effective radiated power = 1.585 x 200 = 317 watts.


A5I10
A5I10:

First we need to calculate the total loss between the transmitter and the antenna.  Remember that antenna gain is given in dBd

- 2 - 2.8 -1.2 + 7 = 1 dB.

To convert to a power ratio we divide decibels by 10 and take the antilog of the result.  For this question we have antilog(1 /10) = 1.259

So effective radiated power = 1.259 x 200 = 252 watts.


A5I11
A5I11:

Effective radiated power can be greater than the actual output power of the transmitter.


A5J01
A5J01:

This problem requires the use of Thevenin's Theorem.  The equivalent series resistance is (8,000 x 8,000) / (8,000 + 8,000) = 4,000 ohms = 4 kilohms.  The equivalent series voltage is 8 x 8,000 / (8,000 + 8,000) = 4 volts.


A5J02
A5J02:

This problem requires the use of Thevenin's Theorem.  The equivalent series resistance is (8,000 x 16,000) / (8,000 + 16,000) = 5,300 ohms = 5.3 kilohms.  The equivalent series voltage is 8 x 8,000 / (8,000 + 16,000) = 2.6 volts.


A5J03
A5J03:

This problem requires the use of Thevenin's Theorem.  The equivalent series resistance is (16,000 x 8,000) / (16,000 + 8,000) = 5,300 ohms = 5.3 kilohms.  The equivalent series voltage is 8 x 16,000 / (16,000 + 8,000) = 5.3 volts.


A5J04
A5J04:

This problem requires the use of Thevenin's Theorem.  The equivalent series resistance is (10,000 x 10,000) / (10,000 + 10,000) = 5,000 ohms = 5 kilohms.  The equivalent series voltage is 10 x 10,000 / (10,000 + 10,000) = 5 volts.


A5J05
A5J05:

This problem requires the use of Thevenin's Theorem.  The equivalent series resistance is (10,000 x 20,000) / (10,000 + 20,000) = 6,666 ohms = 6.67 kilohms.  The equivalent series voltage is 10 x 10,000 / (10,000 + 20,000) = 3.33 volts.


A5J06
A5J06:

This problem requires the use of Thevenin's Theorem.  The equivalent series resistance is (10,000 x 20,000) / (10,000 + 20,000) = 6,666 ohms = 6.67 kilohms.  The equivalent series voltage is 10 x 20,000 / (10,000 + 20,000) = 6.66 volts.


A5J07
A5J07:

This problem requires the use of Thevenin's Theorem.  The equivalent series resistance is (10,000 x 10,000) / (10,000 + 10,000) = 5,000 ohms = 5 kilohms.  The equivalent series voltage is 12 x 10,000 / (10,000 + 10,000) = 6 volts.


A5J08
A5J08:

This problem requires the use of Thevenin's Theorem.  The equivalent series resistance is (10,000 x 20,000) / (10,000 + 20,000) = 6,666 ohms = 6.7 kilohms.  The equivalent series voltage is 12 x 10,000 / (10,000 + 20,000) = 4 volts.


A5J09
A5J09:

This problem requires the use of Thevenin's Theorem.  The equivalent series resistance is (20,000 x 10,000) / (20,000 + 10,000) = 6,666 ohms = 6.7 kilohms.  The equivalent series voltage is 12 x 20,000 / (20,000 + 10,000) = 8 volts.


A5J10
A5J10:

This problem requires the use of Thevenin's Theorem.  The equivalent series resistance is (20,000 x 20,000) / (20,000 + 20,000) = 10,000 ohms = 10 kilohms.  The equivalent series voltage is 12 x 20,000 / (20,000 + 20,000) = 6 volts.


A5J11
A5J11:

Thevenin's Theorem states:

A two terminal complex circuit containing any number of resistors and voltage sources can be replaced by a single series resistor and voltage source.


A6A01
A6A01:

Germanium is a gray, brittle, metallic element.  It is purified to 1 part in 10E10.  Then, special impurities are added (doping) to make it a useful semiconductor.

Silicon makes up 25% of the earth's crust.  It has a gray metallic sheen and is generally prepared from sand.


A6A02
A6A02:

Semiconductor devices depend on the movement of charge carriers (holes and electrons) one factor limiting the maximum frequency that can be used is the mobility of the charge carriers.  Gallium arsenide is a semiconductor compound that has highly mobile charge carriers.


A6A03
A6A03:

The Antimony atoms contribute electrons.


A6A04
A6A04:

Gallium is an acceptor of electrons.  The resulting holes are positive charge carriers.


A6A05
A6A05:

The free electrons are contributed by a donor impurity.  Antimony is a donor element.


A6A06
A6A06:

The Arsenic atoms contribute electrons.


A6A07
A6A07:

Indium is an acceptor of electrons.  The resulting holes are positive charge carriers.


A6A08
A6A08:

The charge carriers in P-type semiconductors are holes.  Holes are regions of positive charge.


A6A09
A6A09:

The charge carriers in P-type semiconductors are holes.  Holes are regions of positive charge.


A6A10
A6A10:

The charge carriers in N-type semiconductors are electrons.  Electrons are carriers of negative charge.


A6A11
A6A11:

The free electrons are contributed by a donor impurity.  Antimony is a donor element.


A6A12
A6A12:

For example indium is an acceptor of electrons.  The resulting holes are positive charge carriers.


A6B01
A6B01:

The constant voltage characteristic of Zener diodes is exploited in voltage regulation.  The Zener effect occurs during reverse biasing of the diode junction.


A6B02
A6B02:

The elongated "Z" in the symbol is a clue to the correct option.


A6B03
A6B03:

Over part of the voltage/current characteristic, the voltage decreases as the current increases.  This is a negative resistance region.


A6B04
A6B04:

Over part of the voltage/current characteristic, the voltage decreases as the current increases.  This is a negative resistance region.  A circuit containing negative resistance is capable of exhibiting a power gain.


A6B05
A6B05:

A tunnel diode cannot rectify.  It has bi-directional properties.  The symbol shows two diodes pointing in opposite directions.


A6B06
A6B06:

Varactors are often employed to vary the resonant frequency of tuned circuits.  The capacitance across the junction varies as the reverse bias voltage changes.


A6B07
A6B07:

Can you see a capacitor symbol and a diode symbol combined in the varactor symbol?  A varactor diode is used as a voltage controlled capacitor.


A6B08
A6B08:

Hot carrier diodes have low noise figures, good conversion efficiency, good dynamic range and excellent high frequency characteristics.


A6B09
A6B09:

If the maximum junction temperature is exceeded then the junction will be destroyed.  To limit the maximum temperature many high power semiconductor devices have heat sinks.


A6B10
A6B10:

If these ratings are exceeded the junction will heat up beyond the maximum temperature.


A6B11
A6B11:

In junction diodes the junction is fabricated using the standard semiconductor techniques of doping and etching.  In point contact diodes a small conductor is placed in contact with the surface of a semiconductor and the junction is a small area around the contact.


A6B12
A6B12:

Point contact diodes have low capacitance and are therefore suitable for high frequency applications.


A6B13
A6B13:

A diode is a directional component.  Its symbol looks like an arrow pointing at a line.  The arrow points in the direction of conduction.


A6B14
A6B14:

A pin diode is used as an RF switch by appearing as a capacitor or a high impedance depending on the bias conditions.


A6B15
A6B15:

The symbol looks like an ordinary diode except that arrows representing rays of light are added.


A6B16
A6B16:

This is because LED junctions glow when they are forward biased.  Appreciable current flow occurs through the junction and it lights up.


A6C01
A6C01:

Inductor cores increase the inductance by decreasing the "resistance" of the magnetic circuit.


A6C02
A6C02:

Some core materials are designed for high permeability.  Low frequency inductors use laminated iron cores.  Mid range and RF inductors use ferrite materials.  High frequency inductors are air cored.


A6C03
A6C03:

Ferrite is a ceramic material containing iron compounds.  It is mechanically superior to powdered iron toroids but powdered iron toroids are more temperature stable.


A6C04
A6C04:

Ferrite is a ceramic material containing iron compounds.  It is mechanically superior to powdered iron toroids but powdered iron toroids are more temperature stable.


A6C05
A6C05:

The permeability of ferrite materials is generally higher.


A6C06
A6C06:

For this application we need the highest possible inductance so that the choke will present a high impedance to the RFI.  A high permeability will help achieve this.


A6C07
A6C07:

Ferrite has a high permeability.  Therefore, ferrite beads increase the inductance of any wire passing through them.  The extra reactance due to this inductance helps to prevent parasitic oscillations.


A6C08
A6C08:

This useful property is exploited in power supplies in audio equipment.  The external magnetic field of a toroidal power transformer is very small and unlikely to affect the rest of the circuit.


A6C09
A6C09:

Bifilar wound transformers are used when the maximum possible coupling is desired.


A6C10
A6C10:

The formula to obtain the number of turns required on a toroidal core is:

N= 1000 x square root (L/A)

A is the inductance index

N is the number of turns

L is inductance in mH

For this question:

N = 1000 x square root (1/523) = 43


A6C11
A6C11:

The formula to obtain the number of turns required on a toroidal core is:

N= 1000 x square root (L/A)

A is the inductance index

N is the number of turns

L is inductance in mH

For this question:

N = 1000 x square root (5/40) = 35


A6D01
A6D01:

Memorize these three names.


A6D02
A6D02:

The difference between collector and emitter current is the base current.


A6D03
A6D03:

This parameter, together with other circuit parameters determines the gain of a transistor amplifier stage,


A6D04
A6D04:

The alpha cut off frequency is defined as the frequency at which the gain is 0.707 times the gain at 1000 Hz in grounded base mode.


A6D05
A6D05:

The direction of the arrow is what determines the type of transistor.  Memorize this symbol.


A6D06
A6D06:

The direction of the arrow is what determines the type of transistor.  Memorize this symbol.


A6D07
A6D07:

The alpha cut off frequency is defined as the frequency at which the gain is 0.707 times the gain at 1000 Hz in grounded base mode.


A6D08
A6D08:

The beta cut off frequency is defined as the frequency at which the gain is 0.707 times the gain at 1000 Hz in grounded emitter mode.


A6D09
A6D09:

The transition region is electrically equivalent to the dielectric of a capacitor.  The thickness (and therefore the junction capacitance) will vary with junction voltage.


A6D10
A6D10:

A transistor being used as a switch and in the conducting state is saturated.


A6D11
A6D11:

For a transistor that is being used as a switch and in the non-conducting state, it is said to be "cut off."


A6D12
A6D12:

A unijunction transistor has two connected bases and an emitter.  If you compare the symbol with the normal transistor symbol you will see how this symbol is derived.


A6D13
A6D13:

A common application for unijunction transistors is in relaxation oscillator circuits.


A6E01
A6E01:

An SCR is a semiconductor device that is non conducting until a trigger pulse arrives at the gate.  The device then switches on and remains on until the supply is interrupted.  SCRs are used in over voltage crowbar circuits.


A6E02
A6E02:

An SCR is a semiconductor device that is non conducting until a trigger pulse arrives at the gate.  The device then switches on and remains on until the supply is interrupted.  SCRs are used in over voltage crowbar circuits.


A6E03
A6E03:

It has a maximum forward current and a peak inverse voltage.


A6E04
A6E04:

An SCR is a semiconductor device that is non conducting until a trigger pulse arrives at the gate.  The device then switches on and remains on until the supply is interrupted.  SCRs are used in over voltage crowbar circuits.


A6E05
A6E05:

The symbol for a silicon controlled rectifier is like a normal rectifier with its anode and cathode connections.  The extra controlling connection is called the gate.


A6E06
A6E06:

A TRIAC can be used to control an AC circuit.  The use of TRIACs has revolutionized the control of AC machinery.  A typical domestic application is in lighting dimmer switches.


A6E07
A6E07:

A TRIAC can be used to control an AC circuit.  The use of TRIACs has revolutionized the control of AC machinery.  A typical domestic application is in lighting dimmer switches.


A6E08
A6E08:

A TRIAC is equivalent to two SCRs connected in parallel.  One is forward biased and the other one is reverse biased.  You can see the two SCRs in the schematic symbol.


A6E09
A6E09:

An unconnected neon lamp can be used to detect strong RF fields.


A6E10
A6E10:

Once a neon lamp is "struck" its resistance drops to a low level and the current through it must be limited.


A6E11
A6E11:

The large circle represents the glass envelope and the two small circles are the electrodes.  The dot indicates that the envelope is gas filled.


A6F01
A6F01:

This is about the minimum bandwidth for reasonable quality speech signals.


A6F02
A6F02:

Each sideband is about 2 kHz wide, but the sidebands lie on each side of the carrier position and a little distance away from it.  So the whole signal occupies about 6 kHz.


A6F03
A6F03:

Most receivers for radio amateur use have this type of filter for receiving SSB transmissions.


A6F04
A6F04:

It is a minority of radio amateurs who make their own crystal filters.


A6F05
A6F05:

The crystal frequency spacing affects the bandwidth as well as the frequency response in the passband.


A6F06
A6F06:

This deformation can be used to generate sound as in piezoelectric alarm transducers.


A6F07
A6F07:

A discrete device, such as the MFR901 bipolar transistor, gives worthwhile noise figures.  They are about 2.5 dB at a power gain of about 10 dB.  However, the MSA-0135 MMIC has a higher gain/bandwidth product.  It also has good stability at 1,296 MHz.


A6F08
A6F08:

The use of MMICs has made the construction of low noise microwave receivers much easier.


A6F09
A6F09:

The device sits in a small hole separating the input and output enclosures.  The ground leads are soldered to the sides of the hole.  Meanwhile, the input and output leads are soldered to copper or silver strips.  The strips function as input and output tuned circuits.


A6F10
A6F10:

The device sits in a small hole separating the input and output enclosures.  The ground leads are soldered to the sides of the hole.  Meanwhile, the input and output leads are soldered to copper or silver strips.  The strips function as input and output tuned circuits.


A6F11
A6F11:

This is the power supply to the device.  The RF choke will typically be 8 turns of thin wire close wound with an air core of about 0.1 inches.


A7A01
A7A01:

Class A and Class B amplifiers are linear amplifiers.  They are the only types suitable for amplifying SSB signals.  Class A amplifiers have the best linearity, but are the least efficient.  They have only about 20% efficiency.


A7A02
A7A02:

Class A and Class B amplifiers are linear amplifiers.  They are the only types suitable for amplifying SSB signals.  Class A amplifiers have the best linearity, but are the least efficient.  The average DC input current is constant regardless of the input drive.


A7A03
A7A03:

Class AB amplifiers are biased so that they are a compromise between class A and Class B amplifiers.


A7A04
A7A04:

In a class B amplifier only half of the input signal cycle appears at the output.  Under zero drive conditions little or no current is drawn from the power supply.  Class B amplifiers are often run in pairs.  Each one will be one working for half of the signal cycle.


A7A05
A7A05:

A Class C amplifier is non-linear.  It is also more efficient than the other amplifier classes.  It has a typical efficiency of 80%.  The amplifier device is biased in the zero current condition.  In a no drive condition, there is no current drawn from the power supply.  When driven, the output of a Class C amplifier is a series of short pulses.  The pulses are less than half of the signal cycle.


A7A06
A7A06:

A Class C amplifier is non-linear.  It is also more efficient than the other amplifier classes.  It has a typical efficiency of 80%.  The amplifier device is biased in the zero current condition.  In a no drive condition, there is no current drawn from the power supply.  When driven, the output of a Class C amplifier is a series of short pulses.  The pulses are less than half of the signal cycle.


A7A07
A7A07:

The saturation point is where no additional output occurs when the drive is increased.


A7A08
A7A08:

This is fairly straightforward.  The RF power out could be measured with an oscilloscope and a dummy load.  The DC power in is calculated as (amplifier current times amplifier voltage).


A7A09
A7A09:

Neutralization is the application of a small amount of negative feedback to cancel out the positive feedback that can occur within a power amplifier active device.  The trick is to make sure that the cancellation is maintained over a range of signal frequencies and amplifier operating conditions.


A7A10
A7A10:

As the procedure is carried out, the RF output power increases.


A7A11
A7A11:

The even-order harmonics cancel out if the two halves of the push-pull amplifier are evenly matched.


A7A12
A7A12:

Intermodulation products will be generated by a none linear amplifier resulting in distortion and broadening of the signal.


A7B01
A7B01:

Neutralization is the application of a small amount of negative feedback.  This cancels out the positive feedback that can occur.  Positive feedback is usually caused by the interelectrode capacitance of the power amplifier tube.  The trick is to make sure that the cancellation is maintained over a range of signal frequencies and amplifier operating conditions.


A7B02
A7B02:

Energy is stored in an L-C circuit.  An analogy is a flywheel that stores energy.


A7B03
A7B03:

A higher tank Q will result in high current losses within the tank coil.  A lower tank Q will result in greater harmonic output.


A7B04
A7B04:

The emitter is at ground potential for ac due to C3.  The input signal is applied across base and emitter.  The output signal appears across collector and emitter.  The emitter is common to both input and output.


A7B05
A7B05:

The resistors form a "potential divider".  This biases the transistor into the required operating region.


A7B06
A7B06:

C1 allows the AC signal to get to the transistor.  However, it prevents DC from the previous stage from affecting the bias.


A7B07
A7B07:

C3 keeps the emitter at ground potential as far as AC signals are concerned.


A7B08
A7B08:

A DC voltage exists across R3.  It raises the emitter voltage relative to the base voltage.  R3 contributes toward the transistor bias.


A7B09
A7B09:

Because of C1, the collector is at ground potential for AC.  The input signal is applied across base and collector (by way of ground and C1).  The output signal appears across collector and emitter.  The collector is common to both input and output.


A7B10
A7B10:

As the emitter current varies the voltage across R will vary and this voltage is available as an output.


A7B11
A7B11:

C1 connects the AC potential of the power rail to ground.  This allows the collector to function as the common electrode..


A7B12
A7B12:

The AC signal across R is passed to the next stage by C2.  This prevents DC from disturbing the bias conditions of the next stage.


A7B13
A7B13:

A circuit like this is used in most electronics equipment.  Sometimes the components are fabricated as a single device (except for the capacitors).  An alternative type of regulator, called a switching regulator, is more efficient.  However, it is much more complex to design and build.


A7B14
A7B14:

D1 is a Zener diode.  The voltage drop across it is constant over a range of current.


A7B15
A7B15:

A Zener diode on its own can function as a shunt regulator but the current range of such a regulator is limited to the constant voltage portion of the Zener diode current versus voltage characteristic.


A7B16
A7B16:

Direct current is obtained from alternating current using rectifiers.  Rectifiers give a pulsing DC output.  C1 filters these pulses, leaving a steady DC voltage.  In a practical circuit, C1 will be an electrolytic capacitor.


A7B17
A7B17:

We need the current through D1 to be as steady as possible since the voltage across it is used as a reference by Q1.  C1 reduces any hum that may be present on the input to the regulator.


A7B18
A7B18:

Along with C3, there will also be a number of decoupling capacitors.  They will be distributed around the load circuit.


A7B19
A7B19:

The value should be such as to supply current to D1 that is on the constant voltage part of its current versus voltage characteristic.


A7B20
A7B20:

Having a constant current portion in the regulator load helps the voltage regulation.


A7C01
A7C01:

A pi network is often used for coupling a transmitter RF amplifier to the antenna feedline.


A7C02
A7C02:

The plate impedance of a tube amplifier could be as high as 8,000 ohms.  The pi network couples this to the 50 ohm impedance feedline.


A7C03
A7C03:

In a schematic diagram, the components of a pi-network form the Greek letter "pi".

A pi network:


A7C04
A7C04:

In a schematic diagram of an L-network, the components form an inverted letter "L".  The capacitor forms the vertical leg and the inductor forms the horizontal leg.

An L network:


A7C05
A7C05:

A pi network gives a greater impedance transformation range than an L-network.


A7C06
A7C06:

This is a pi-network with an L-network on the output (antenna) side.


A7C07
A7C07:

The capacitors are in series with the signal and present a smaller reactance at higher frequencies.  The inductor is a low reactance at low frequencies and is in parallel with the signal.  Both of these effects combine to make the T-filter a high pass filter.


A7C08
A7C08:

The extra section aids in harmonic suppression.  If a 14 MHz signal has a significant harmonic on 28 MHz then this will be efficiently radiated by a 14/21/28 MHz multiband antenna.


A7C09
A7C09:

The extra section aids in harmonic suppression.  If a 14 MHz signal has a significant harmonic on 28 MHz then this will be efficiently radiated by a 14/21/28 MHz multiband antenna.


A7C10
A7C10:

They are generally called tank circuits.  Another important function is harmonic suppression.


A7C11
A7C11:

Ideally a load should be resistive and of the correct resistance.  A matching network can achieve both of these requirements.


A7D01
A7D01:

All three filters are widely used in amateur radio.


A7D02
A7D02:

We have to re arrange the formula:

Fr = 1 / [2 x pi x square root ( L x C)] to obtain C

Where Fr is the resonant frequency

L is inductance in Henrys

C is capacitance in Farads.

C = [(1/(2 x pi x Fr)) squared] / L

If we take "Fr" as 3.5 MHz for the 80-meter band then:

C = [(1/(2 x 3.14 x 3.5E6)) squared] / 20E-6

We get C = 1.03E-10 Farads = 100 picofarads.


A7D03
A7D03:

We have to re arrange the following formula to obtain L:

Fr = 1 / [2 x pi x square root ( L x C)]

Where Fr is the resonant frequency

L is inductance in Henrys

C is capacitance in Farads

L = [(1 / (2 x pi x Fr))squared] / C

If we take Fr as 7.0 MHz for the 40-meter band then:

L = [(1 / (2 x 3.14 x 7.0E6))squared] / 100E-12

We get L = 5.1E-6 henrys = 5 microhenrys.


A7D04
A7D04:

We have to re arrange the formula:

Fr = 1 / [2 x pi x square root ( L x C)] to obtain C

Where Fr is the resonant frequency

L is inductance in Henrys

C is capacitance in Farads

C = [(1/(2 x pi x Fr)) squared] / L

If we take Fr as 14 MHz for the 20-meter band then:

C = [(1/(2 x 3.14 x 14E6)) squared] / 2E-6

We get C = 6.4-11 Farads = 64 picofarads.


A7D05
A7D05:

We have to re arrange the formula:

Fr = 1 / [2 x pi x square root ( L x C)] to obtain L

Where Fr is the resonant frequency

L is inductance in Henrys

C is capacitance in Farads

L = [(1/(2 x pi x Fr)) squared] / C

If we take Fr as 21 MHz for the 15-meter band then:

L = [(1/(2 x 3.14 x 21E6))squared] / 15E-12

We get L = 3.8E-6 henrys = 4 microhenrys.


A7D06
A7D06:

We have to re arrange the formula:

Fr = 1 / [2 x pi x square root ( L x C)] to obtain C

Where Fr is the resonant frequency

L is inductance in Henrys

C is capacitance in Farads

C = [(1/(2 x pi x Fr))squared] / L

If we take Fr as 1.8 MHz for the 160-meter band then:

C = [(1/(2 x 3.14 x 1.8E6))squared] / 100E-6

We get C = 7.8E-11 Farads = 78 picofarads.


A7D07
A7D07:

The pass band of a Butterworth filter is free of ripple.


A7D08
A7D08:

A Chebyshev filter will typically have steeper skirts than a Butterworth filter with a similar number of components.


A7D09
A7D09:

The sharp cutoff is achieved at a cost of some ripple in the passband.


A7D10
A7D10:

An elliptical filter design can be regarded as an extreme version of the Chebyshev filter.  This is where the passband ripples are infinitely deep and the cut-off is even sharper.


A7D11
A7D11:

An elliptical filter design can be regarded as an extreme version of the Chebyshev filter where the passband ripples are infinitely deep and the cut-off is even sharper.


A7E01
A7E01:

The control element is often a high power transistor.  The line voltage is compared to a reference voltage (often this is a Zener diode) to derive a difference signal that is fed to the control element.


A7E02
A7E02:

The switching is done at a fairly high frequency and the resultant pulses can be smoothed to DC fairly easily without the need for large filter components.  Since the control device is being rapidly switched a relatively small amount of heat is generated.


A7E03
A7E03:

A Zener diode has a junction that, when reverse biased across a certain current range, has a fairly constant voltage across it.  Zener diodes come in a variety of voltage and current ratings.


A7E04
A7E04:

The control element is in series with the load.


A7E05
A7E05:

The voltage across the load is varied by varying the amount of current bypassed into the control element.  The element is in parallel with the load.


A7E06
A7E06:

We need an operating voltage that is high enough to provide a good reference voltage.  However, it must not be too high for reasons of temperature stability.


A7E07
A7E07:

Remote sensing allows voltage drops in the wiring to a high current load to be sensed.  It can then be compensated for by adjusting the voltage of the power supply.


A7E08
A7E08:

This device contains all the components needed for a complete regulator circuit.  One terminal is usually (but not always) grounded.  One is the unregulated input and the other is the regulated output.  Three terminal regulators come in many voltage and current ratings.


A7E09
A7E09:

If you were choosing a three terminal voltage regulator, you would have to consider the amount of regulation given by the device and whether any overload protection was present.


A7E10
A7E10:

A Zener diode can be used as a two terminal, shunt type voltage regulator.


A7E11
A7E11:

This device contains all the components needed for a complete regulator circuit.  One terminal is usually (but not always) grounded.  One is the unregulated input and the other is the regulated output.  Three terminal regulators come in many voltage and current ratings.


A7F01
A7F01:

Oscillators all rely on positive feedback.  The three oscillators in this question use different methods to apply this feedback.


A7F02
A7F02:

For maximum purity of output the positive feedback should be just sufficient to overcome losses, but no more.


A7F03
A7F03:

The tap is at the lowest position to maintain oscillation.  Less than 25% is usually sufficient, depending on the gain of the active device.


A7F04
A7F04:

The capacitors in the divider must be selected to give the least possible amount of feedback consistent with reliable oscillation.  The extra capacitance in the divider appears across the oscillator tuned circuit, which means that a smaller inductance is used in the parallel tuned version of the Colpitts oscillator.


A7F05
A7F05:

The feedback capacitor is actually the capacitance of a series connected crystal


A7F06
A7F06:

In the Pierce oscillator the feedback path is by way of a crystal that has capacitance.  The reactance of this capacitance is low enough for feedback to occur.


A7F07
A7F07:

Using a quartz crystal to determine the frequency results in a more frequency stable oscillator.  For a given crystal the frequency of the oscillator can only be varied slightly.  For wider frequency coverage there must be several switched crystals.


A7F08
A7F08:

Colpitts and Hartley oscillators are both controlled by L-C tuned circuits.  The frequency can be varied over a range by varying either the L or the C.  It is usually the C that is varied.  Since L and C are determined by mechanical dimensions the L-C part of a VFO must be constructed for maximum mechanical stability.


A7F09
A7F09:

The construction of a stable VFO oscillator is something of an art.  In the case of a Colpitts oscillator it is important to carefully select the capacitors for temperature stability.  Polystyrene capacitors are good in this respect.


A7F10
A7F10:

Frequency control by use of a varactor diode is a quite coarse method when used by itself.  It is often used as part of a phase locked loop circuit.


A7F11
A7F11:

This is because the output signal is phase locked to the reference oscillator signal.


A7G01
A7G01:

In amateur radio communication the information is speech or data.


A7G02
A7G02:

FM is generated by varying a reactance in the oscillator tuned circuit.  The reactance is always capacitive for practical reasons.  If the final amplifier is reactance modulated then phase modulation will result.


A7G03
A7G03:

FM is generated by varying a reactance in the oscillator tuned circuit.  The reactance could be inductive, but is always capacitive for practical reasons.  If the final amplifier is reactance modulated then phase modulation will result.


A7G04
A7G04:

FM is generated by varying a reactance in the oscillator tuned circuit.  The reactance could be inductive, but is always capacitive for practical reasons.  If the final amplifier is reactance modulated then phase modulation will result.


A7G05
A7G05:

Phase modulation sounds and behaves much like frequency modulation.


A7G06
A7G06:

If the final amplifier is reactance modulated then phase modulation will result.  The reactance could be inductive, but is always capacitive for practical reasons.


A7G07
A7G07:

The balancing refers to cancellation of the carrier.  Balanced modulators can provide 30-50 dB of carrier suppression.


A7G08
A7G08:

One of the sidebands is filtered out.  The suppressed carrier is further reduced by the filter skirt attenuation that may be 20 - 30 dB.  Total carrier suppression of 80 dB is not uncommon.


A7G09
A7G09:

This is the classical method of producing a double sideband with carrier AM signal.


A7G10
A7G10:

A pre-emphasis network in the transmitter attenuates lower frequencies and emphasizes the higher ones. The resulting audio band has an even spread of energy. A de-emphasis network in the receiver restores the original signal. A result of this processing is a better overall signal to noise ratio.


A7G11
A7G11:

A pre-emphasis network in the transmitter attenuates lower frequencies and emphasizes the higher ones. The resulting audio band has an even spread of energy. A de-emphasis network in the receiver restores the original signal. A result of this processing is a better overall signal to noise ratio.


A7H01
A7H01:

This is the opposite of modulation.  An alternate term for detection is de-modulation.


A7H02
A7H02:

A diode is the simplest type of detector.  A tuned circuit diode is the simplest type of radio receiver.  It is called a "crystal set".


A7H03
A7H03:

The locally generated carrier replaces the carrier that was removed in the transmitter.  It must be very close to the original carrier position for proper reproduction of the original modulating signal.


A7H04
A7H04:

A discriminator is another name for an FM detector.


A7H05
A7H05:

A discriminator is another name for an FM demodulator.


A7H06
A7H06:

Signal to noise ratio is not relevant since an active filter is used at a point in the receiver where the signal level is high.


A7H07
A7H07:

If the interfering signal is a plain carrier it will have a very narrow bandwidth and can be "notched out" with little effect on the SSB transmission.


A7H08
A7H08:

Typical SSB filters are built from a number of elements to produce the required passband and skirt shape.  The resulting phase response in the filter passband is not important for speech work.


A7H09
A7H09:

The adaptive filter will analyze the signal and remove parts of it that contain signals that do not change with time.


A7H10
A7H10:

Notch and elliptical filters are not necessarily digital signal processing filters.  An adaptive filter is used to reduce extraneous signals.


A7H11
A7H11:

A cavity filter will have a very high Q value.  This will give the filter a narrow bandwidth.  It will have little loss at the center of the passband.


A7I01
A7I01:

For new frequencies to be produced there must be a non linear element in the mixer circuit.


A7I02
A7I02:

For new frequencies to be produced there must be a non linear element in the mixer circuit.


A7I03
A7I03:

A wide range of input frequencies is converted to a standard frequency.  Tuned circuits and filters can be more easily optimized when designed for a fixed frequency.


A7I04
A7I04:

The spurious products will be heard as extra signals.


A7I05
A7I05:

A frequency synthesizer has the best of both worlds.  It has the high stability of a fixed reference source and the ability to output a wide range of frequencies.


A7I06
A7I06:

A direct digital synthesizer uses digital computer techniques to synthesize a sine-wave.  The amplitude values at regular intervals of a sine wave are stored in a lookup table.  These amplitude values are read out in succession to produce a signal.  The anti-alias filter removes the "steps" produced between each amplitude value.


A7I07
A7I07:

A frequency synthesizer has the best of both worlds.  It has the high stability of a fixed reference source and the ability to output a wide range of frequencies.


A7I08
A7I08:

A direct digital synthesizer uses digital computer techniques to synthesize a sine-wave.  The amplitude values at regular intervals of a sine wave are stored in a lookup table.  These amplitude values are read out in succession to produce a signal.  The anti-alias filter removes the "steps" produced between each amplitude value.


A7I09
A7I09:

A direct digital synthesizer uses digital computer techniques to synthesize a sine-wave.  The amplitude values at regular intervals of a sine wave are stored in a lookup table.  These amplitude values are read out in succession to produce a signal.  The anti-alias filter removes the "steps" produced between each amplitude value.


A7I10
A7I10:

The spurs correspond to the steps that are produced during digital synthesis of the signal.


A7I11
A7I11:

Some of this noise comes from random phase variations in the reference source.  Another source is the junction in the varactor diode that controls the output signal as well as noise on the varactor control signal.


A7J01
A7J01:

This upper limit is what determines the bandwidth of the transmitted signal.


A7J02
A7J02:

This answer is fairly self-explanatory.  RMS values for each harmonic are obtained and then the values are summed.


A7J03
A7J03:

Gains of more than 100 dB are common with this configuration.


A7J04
A7J04:

The output from a microphone will be a few millivolts, much too low for a modulator.


A7J05
A7J05:

Because it is fixed-tuned the frequency response can be tailored exactly to communications requirements.


A7J06
A7J06:

Good selectivity is easier to achieve at low IF frequencies but better RF image rejection can be obtained at high IF frequencies.


A7J07
A7J07:

This stage may include crystal or other types of filters.


A7J08
A7J08:

This concept provides the ultimate in image rejection performance.


A7J09
A7J09:

Mixers are inherently quite noisy devices and so the input signal must be raised to a higher level if weak signals are not to be swamped by mixer noise.  If too much RF signal is present then the mixer will generate spurious products.


A7J10
A7J10:

Mixers are inherently quite noisy devices and so the input signal must be raised to a higher level if weak signals are not to be swamped by mixer noise.  If too much RF signal is present then the mixer will generate spurious products.


A7J11
A7J11:

Mixers are inherently quite noisy devices and so the input signal must be raised to a higher level by an RF amplifier if weak signals are not to be swamped by mixer noise.  The receiver noise figure will be mainly determined by the noise performance of the RF amplifier.


A8A01
A8A01:

Facsimile (FAX) is the transmission of printed pictures.  As a dot of light scans across the document the scattered light varies.  These light variations are converted to a voltage variation and sent to the FAX receiver.  AM facsimile is designated by A3C while FM FAX is designated by F3C (see ITU Emission Designators).


A8A02
A8A02:

Facsimile (FAX) is the transmission of printed pictures.  As a dot of light scans across the document the scattered light varies.  These light variations are converted to a voltage variation and sent to the FAX receiver.  AM facsimile is designated by A3C while FM FAX is designated by F3C (see ITU Emission Designators).


A8A03
A8A03:

Facsimile (FAX) is the transmission of printed pictures.  As a dot of light scans across the document the scattered light varies.  These light variations are converted to a voltage variation and sent to the FAX receiver.  AM facsimile is designated by A3C while FM FAX is designated by F3C (see ITU Emission Designators).


A8A04
A8A04:

Facsimile (FAX) is the transmission of printed pictures.  As a dot of light scans across the document the scattered light varies.  These light variations are converted to a voltage variation and sent to the FAX receiver.  AM facsimile is designated by A3C while FM FAX is designated by F3C (see ITU Emission Designators).


A8A05
A8A05:

Facsimile (FAX) is the transmission of printed pictures.  As a dot of light scans across the document the scattered light varies.  These light variations are converted to a voltage variation and sent to the FAX receiver.  AM facsimile is designated by A3C while FM FAX is designated by F3C (see ITU Emission Designators).


A8A06
A8A06:

Television is designated by A3F in the case of amplitude modulated television signals and F3F for FM modulated television (see ITU Emission Designators).


A8A07
A8A07:

Television is designated by A3F in the case of amplitude modulated television signals and F3F for FM modulated television (see ITU Emission Designators).


A8A08
A8A08:

Television is designated by A3F in the case of amplitude modulated television signals and F3F for FM modulated television (see ITU Emission Designators).


A8A09
A8A09:

Television is designated by A3F in the case of amplitude modulated television signals and F3F for FM modulated television (see ITU Emission Designators).


A8A10
A8A10:

J3F is the ITU emission designator for slow scan television signals (see ITU Emission Designators).


A8A11
A8A11:

The key word here is "telegraphy."


A8B01
A8B01:

(See ITU Emission Designators.)


A8B02
A8B02:

(See ITU Emission Designators.)


A8B03
A8B03:

(See ITU Emission Designators.)


A8B04
A8B04:

(See ITU Emission Designators.)


A8B05
A8B05:

(See ITU Emission Designators.)


A8B06
A8B06:

(See ITU Emission Designators.)


A8B07
A8B07:

(See ITU Emission Designators.)


A8B08
A8B08:

(See ITU Emission Designators.)


A8B09
A8B09:

(See ITU Emission Designators.)


A8B10
A8B10:

(See ITU Emission Designators.)


A8B11
A8B11:

(See ITU Emission Designators.)


A8C01
A8C01:

FM is generated by varying a reactance in the oscillator tuned circuit.  The reactance could be inductive but is always capacitive for practical reasons.  If the final amplifier is reactance modulated then phase modulation will result.


A8C02
A8C02:

One of the sidebands is filtered out.  The suppressed carrier is further reduced by the filter skirt attenuation that may be 20 - 30 dB.  Total carrier suppression of 80 dB is not uncommon.


A8C03
A8C03:

The formula for modulation index is:

(Maximum carrier frequency deviation Hz) / (Modulation frequency Hz).


A8C04
A8C04:

The formula for modulation index is:

(Maximum carrier frequency deviation Hz) / (Modulation frequency Hz).


A8C05
A8C05:

The formula for modulation index is:

(Maximum carrier frequency deviation Hz) / (Modulation frequency Hz).

There is no term for carrier frequency in this formula.


A8C06
A8C06:

The formula for modulation index is:

(Maximum carrier frequency deviation Hz) / (Modulation frequency Hz).

We have:

Modulation index = 3000 / 1000 = 3


A8C07
A8C07:

The formula for modulation index is:

(Maximum carrier frequency deviation Hz) / (Modulation frequency Hz).

We have:

Modulation index =  63000 / 2000 = 3


A8C08
A8C08:

Deviation ratio is given by the formula:

Deviation ratio = (Maximum carrier deviation Hz) / (Highest modulation frequency Hz).


A8C09
A8C09:

Deviation ratio is given by the formula:

Deviation ratio = (Maximum carrier deviation Hz) / (Highest modulation frequency Hz).


A8C10
A8C10:

Deviation ratio is given by the formula:

Deviation ratio = (Maximum carrier deviation Hz) / (Highest modulation frequency Hz).

In the question we have:

Deviation ratio = 5000 / 3000 = 1.66


A8C11
A8C11:

Deviation ratio is given by the formula:

Deviation ratio = (Maximum carrier deviation Hz) / (Highest modulation frequency Hz).

In the question we have:

Deviation ratio = 7,500 / 3,500 = 2.14


A8D01
A8D01:

When we refer to polarization, it is always the electric field that is referenced.


A8D02
A8D02:

At this velocity it takes about 8 minutes for light from the sun to reach us.  It can take millions of years for light from some stars to reach us! If you have full break in keying and are using a long distance band under good conditions you may be able to hear your own "dits" after they have traveled around the earth.


A8D03
A8D03:

This principle is the basis of RF shielding.


A8D04
A8D04:

Electromagnetic waves do not require a transmission medium.  They can propagate across a perfect vacuum.


A8D05
A8D05:

When we describe polarization it is always the electric field that is referenced.


A8D06
A8D06:

Circularly polarized radio waves can be generated by helical antennas.  The waves can have a left hand or right hand twist…honestly!


A8D07
A8D07:

The electromagnetic field is always at right angles to the electric field.  Normally, it is the electric field that we refer to when describing polarization.


A8D08
A8D08:

The electromagnetic field is always at right angles to the electric field.  Normally, it is the electric field that we refer to when describing polarization.


A8D09
A8D09:

The electromagnetic field is always at right angles to the electric field.  Normally, it is the electric field that we refer to when describing polarization.


A8D10
A8D10:

The electromagnetic field is always at right angles to the electric field.  Normally, it is the electric field that we refer to when describing polarization.


A8D11
A8D11:

At HF atmospheric noise is always greater than receiver generated front-end noise.


A8D12
A8D12:

At VHF atmospheric noise is at low levels and receiver front-end noise places a lower limit on weak signal reception.


A8E01
A8E01:

A clue to the correct answer is the reference to trigonometry, which relates to circles or circular movement.  If you feed two sine waves to the horizontal and vertical plates of an oscilloscope and they are of equal frequency (and 90 degrees out of phase) then a circle will be seen.


A8E02
A8E02:

Once at the 180 degree halfway point then as it finishes at the 360 degree point.


A8E03
A8E03:

Just like on a protractor.


A8E04
A8E04:

A sine wave cycle has a positive half and a negative half.


A8E05
A8E05:

The wave gets its name from the shape when seen on an oscilloscope.


A8E06
A8E06:

The wave gets its name from the square shape when seen on an oscilloscope.


A8E07
A8E07:

All wave shapes can be analyzed in terms of combinations of sine waves.  This technique is called Fourier analysis.


A8E08
A8E08:

All wave shapes can be analyzed in terms of combinations of sine waves.  This technique is called Fourier analysis.


A8E09
A8E09:

The wave gets its name from the shape when seen on an oscilloscope.  It looks like a close-up of the cutting edge of a saw blade.


A8E10
A8E10:

The wave gets its name from the shape when seen on an oscilloscope.  It looks like a close-up of the cutting edge of a saw blade.


A8E11
A8E11:

All wave shapes can be analyzed in terms of combinations of sine waves.  This technique is called Fourier analysis.


A8F01
A8F01:

Peak voltage = RMS voltage x 1.414

For this question:

Peak voltage = 117 x 1.414 = 170 volts (approximately).


A8F02
A8F02:

Peak-to-peak voltage is how you would measure an ac waveform on an oscilloscope.  It is measured from the bottom to the top of the displayed waveform.  It is twice the peak voltage.  In this question peak-to-peak voltage from a household electrical outlet is:

RMS voltage x 1.414 x 2 = 117 x 1.414 = 340 volts (approximately).


A8F03
A8F03:

The RMS voltage is the equivalent DC voltage that would produce the same heating effect if connected to a resistor.  RMS voltage = peak voltage / 1.414


A8F04
A8F04:

In this question we halve the peak-to-peak voltage to get the peak voltage.  We must then divide by 1.414 to get the RMS voltage.

RMS voltage = (Peak-to-peak voltage / 2) / 1.414 = 120 volts


A8F05
A8F05:

The RMS voltage is the equivalent DC voltage that would produce the same heating effect as the AC voltage if it were connected to a resistor.  RMS voltage = peak voltage / 1.414


A8F06
A8F06:

The RMS value would be the voltage of a DC waveform that produced the same heating effect.


A8F07
A8F07:

The 1.414 value used in RMS to peak voltage calculations only applies to sine waves.  A speech waveform is nothing like a sine wave and often results in a ratio of about 2.5 for many voices.


A8F08
A8F08:

The 1.414 value used in RMS to peak voltage calculations only applies to sine waves.  A speech waveform is nothing like a sine wave and often results in a ratio of about 2.5 for many voices.


A8F09
A8F09:

In a Class B amplifier efficiency is about 60%.

RF output = DC input x efficiency

So DC input = RF output / efficiency = 1500 / 0.6 = 2,500 watts.


A8F10
A8F10:

In a Class C amplifier efficiency is about 80%.

RF output = DC input x efficiency

So DC input = RF output / efficiency = 1000 / 0.8 = 1,250 watts.


A8F11
A8F11:

In a Class AB amplifier efficiency is about 50%.

RF output = DC input x efficiency

So DC input = RF output / efficiency = 500 / 0.5 = 1000 watts.


A9A01
A9A01:

An antenna with a 50 ohm radiation resistance would be a good match to a 50 ohm feedline.


A9A03
A9A03:

An antenna with a 50 ohm radiation resistance would be a good match to a 50 ohm feedline.


A9A04
A9A04:

An antenna with a 50 ohm radiation resistance would be a good match to a 50 ohm feedline.  Factors such as the presence of nearby objects causes the radiation resistance to vary and result in reflected power due to the mismatch.


A9A05
A9A05:

Antenna efficiency = Radiation resistance / (radiation resistance + ohmic resistance) x 100


A9A06
A9A06:

Antenna efficiency = Radiation resistance / (radiation resistance + ohmic resistance) x 100


A9A07
A9A07:

On VHF a folded dipole is made of parallel rigid conductors, and may be fabricated from a single, folded rod.  On HF a folded dipole could be made from a length of twin lead transmission line.  The conductors would have to be joined at the ends.


A9A08
A9A08:

A folded dipole can be fed with 300 ohm twin feeder.  A matching network will be needed at the transmitter end.  Folded dipoles have greater bandwidth than simple dipoles, since the effective element thickness is greater.


A9A09
A9A09:

The "other antenna" is sometimes a dipole or it may be a theoretical isotropic (point source) antenna.


A9A10
A9A10:

Higher gain antennas tend to have smaller bandwidths.  Also, lower frequency antennas tend to have narrower bandwidths.


A9A11
A9A11:

This can be done using your receiver S-meter, a signal of constant amplitude and your antenna rotator.


A9A12
A9A12:

Antenna efficiency = Radiation resistance / (radiation resistance + ohmic resistance) x 100


A9A13
A9A13:

The radials can be on, or just below, ground level.  Good connections at the central point are important.


A9B01
A9B01:

The electromagnetic field is always at right angles to the electric field.  Normally, it is the electric field that we refer to when describing polarization.


A9B02
A9B02:

Azimuth is an angle measured clockwise from the North.  The angles around an azimuth chart are directions.


A9B03
A9B03:

The main lobe is the one that extends to the edge of the chart.  It is in the upward position on this chart.  At approximately 25 degrees each side of the upward position the lobe is about 3 dB from the edge, giving a total beamwidth of 50 degrees.


A9B04
A9B04:

The lobe at 180 degrees from the main lobe is 18 dB away from the edge of the chart.


A9B05
A9B05:

The radiation resistance determines how much power is transferred from the feedline.  The field patterns indicate directions of maximum/minimum radiation.  The loss resistance affects how much power is absorbed by the antenna structure.


A9B06
A9B06:

Dielectric constant is a term used to describe insulators.  It does not vary with frequency and is not relevant to antenna design.  The other options are all antenna characteristics that will vary with frequency.


A9B07
A9B07:

Antenna design is a compromise between many different requirements.


A9B08
A9B08:

As well as a gain increase, there is an increase in antenna bandwidth as the element spacing increases.


A9B09
A9B09:

A "moment" is a single value that is derived from a set of values.  In antenna modeling programs the antenna elements are modeled as a set of segments, each having a specific current.  The total effect of an element at any given point in space is a "moment" of the effects of all the segments in that element.


A9B10
A9B10:

A "moment" is a single value that is derived from a set of values.  In antenna modeling programs the elements are modeled as a set of segments, each having a specific current.


A9B11
A9B11:

The side lobes, at 90 degrees from the main lobe in this example, are 14 dB away from the edge of the chart.


A9C01
A9C01:

The angles around the edge of an elevation pattern represent upward tilt.  An elevation of 90 degrees is vertical while 0 degrees elevation is horizontal.


A9C02
A9C02:

The electric field lines are in a direction parallel to the elements and so it is horizontally polarized.


A9C03
A9C03:

Perfectly conducting soil would result in a pattern like a vertical dipole in free space.  The elevation pattern would show a maximum in the horizontal direction.


A9C04
A9C04:

The conductivity of the soil is not relevant to the RF characteristics of an antenna that uses an adequate system of ground radials.  The optimum length of the radials is the same length as the height of the antenna.  In the antenna example here 60 ground radials is already ample!


A9C05
A9C05:

A perfectly conducting ground would result in a pattern like a vertical dipole in free space.  The elevation pattern would show a maximum in the horizontal direction.  Sea water is a better conductor than rocky soil and so the antenna would behave more like a vertical dipole in free space.


A9C06
A9C06:

At antenna heights of one wavelength and above effects due to ground proximity become insignificant.


A9C07
A9C07:

In many elevated radial designs, the radials slope downwards.  They double as guy wires for the supporting mast.  A further advantage of this design is that the downward slope of the radials results in a feed impedance of about 50 ohms.


A9C08
A9C08:

The size of the mesh should be approximately equal to the height of the antenna.


A9C09
A9C09:

The main (largest) lobe touches the edge of the chart at 7.5 degrees.


A9C10
A9C10:

The small lobe pointing away from the main lobe is 28 dB away from the edge of the chart.


A9C11
A9C11:

There are four lobes in the same direction as the main lobe.


A9D01
A9D01:

A folded dipole has a feed impedance four times greater than a simple dipole.


A9D02
A9D02:

The position of the loading coil is a compromise.  If the coil is placed near the top then the high current portion of the antenna will be longer, resulting in greater radiation efficiency.  However the coil will require more inductance for resonance.  More inductance requires more turns.  This adds to the total antenna resistance and reduces the efficiency.  A base-loaded vertical requires the least coil inductance.  However, it is the least efficient due to the current carrying portion of the antenna being limited to the coil.


A9D03
A9D03:

The loading coil should be air cored and be wound with thick wire to reduce ohmic losses.  You might be able to fabricate a self supporting coil from copper microbore heating tube!


A9D04
A9D04:

An antenna that is shorter than a quarter wavelength will be capacitive.


A9D05
A9D05:

For example, a dual band antenna for 14 and 28 MHz will efficiently radiate second harmonic emissions from a transmitter working on 14 MHz.


A9D06
A9D06:

The traps act as isolators that shorten the elements when the antenna is used on the higher frequency bands.  The traps are parallel resonant tuned circuits.


A9D07
A9D07:

So the coil inductance has to be increased and the matching coil of L-network has to be adjusted.  In practice, a mobile antenna will be preset for a preferred working frequency.


A9D08
A9D08:

Both the transmission-line impedance and the antenna feedpoint impedance are needed to determine what impedance transformation is needed.


A9D09
A9D09:

The impedance matching section is not a quarter wave in length and this will introduce reactance at the feedline connection.  Shortening the driven element slightly will introduce enough reactance to cancel the reactance of the matching section.


A9D10
A9D10:

It is equivalent to a series inductor, which is why the driven element needs to be slightly capacitive.


A9D11
A9D11:

The bandwidth tends to become dominated by the Q of the loading coil.


A9D12
A9D12:

It is the current section of the antenna that does most of the radiating.  A top loaded antenna will have a longer current section.


A9E01
A9E01:

The velocity factor is influenced mainly by the dielectric of a transmission line.


A9E02
A9E02:

The velocity factor is influenced mainly by the dielectric of a transmission line.


A9E03
A9E03:

So the length of a quarter wave section using this type of feeder is 0.66 x (quarter wave length in free space)


A9E04
A9E04:

Less lossy dielectrics tend to have velocity factors closer to one (1).


A9E05
A9E05:

The velocity of electromagnetic waves is given as 299,792,458 meters per second in free space.  When the electromagnetic energy encounters matter the velocity is always reduced.  The velocity is usually approximated to 300,000,000 meters per second in radio calculations.


A9E06
A9E06:

At 14.1 MHz a quarter wavelength in free space is:

(3e8 / 14.1E6) / 4 = 5.3 meters

So the physical length of a quarter wave transmission line at this frequency (velocity factor = 0.66) is:

5.3 x 0.66 = 3.51 meters.


A9E07
A9E07:

At 14.1 MHz a quarter wavelength in free space is:

(3e8 / 7.2E6) / 4 = 10.4 meters

So the physical length of a quarter wave transmission line at this frequency (velocity factor = 0.66) is:

10.4 x 0.66 = 6.88 meters.


A9E08
A9E08:

At 14.1 MHz a quarter wavelength in free space is:

(3e8 / 14.1E6) / 2 = 10.6 meters

So the physical length of a quarter wave transmission line at this frequency (velocity factor = 0.66) is:

10.6 x 0.95 = 10.1 meters.


A9E09
A9E09:

No calculation is required for this question.  The frequency is not relevant.  A velocity factor of 0.8 means that the physical length is (electrical length) x 0.8


A9E10
A9E10:

The reflection coefficient is defined as:

(voltage in the reflected wave) / (Voltage in forward wave).


A9E11
A9E11:

In practice SWRs of up to 2:1 may be considered tolerable.


A9E12
A9E12:

At 50 MHz RG-58 cable has a loss of 3.5 dB per 100 feet while 450-ohm ladder line has a loss of less than 0.3 dB per 100 feet.


ITU emission designators

First Symbol	Type of main modulation.
N A J F G P	Unmodulated Amplitude modulation SSB suppressed carrier Frequency modulation Phase modulation Unmodulated pulses
Second Symbol	Modulating signal
0 1 2 3 7 8 9	No modulation Unmodulated digital channel Modulated digital channel Single analog channel Two or more channels, digital Two or more channels, analog Combined digital and analog
Third Symbol
N A B C D E F W	No information present Telegraphy, human readable Telegraphy, machine readable Facsimile Telemetry and telecommand data Voice telephony Video A combination of above


Thevenin
Thevenin's Theorem

Any network of voltage sources and resistors can be replaced by an equivalent circuit containing a single voltage source and a single series resistor.  For example

 can be replaced by

You are not required to work with complex circuits like the one above.  All the questions are based on converting circuit A to the equivalent circuit B:

The questions ask you to find V2 and R3.  To determine the equivalent single voltage source V2 in circuit B we calculate the open circuit voltage of circuit A using Ohms Law:

In Circuit B we have V2 = V1 x R2 / (R1 + R2).

To determine the equivalent single series resistance R3 in circuit B we replace the voltage source in circuit A with a short circuit and calculate the resistance at the output as shown below:

In Circuit B R3 = (R1 x R2) / (R1 + R2).

The formulae for V2 and R3 allow you to answer all the Thevenin theory questions in the Advanced Class question pool.