CW may be used from 3525 to 3750 kHz and phone may be usedÿfromÿ3775ÿtoÿ4000ÿkHz.
The band is from 7000 to 7300 kHz.  The first 25 kHz is reservedÿfor Amateur Extra only.
14025 - 14150 kHz and 14175 - 14350 kHz
21025 - 21200 kHz and 21225 - 21450 kHz
Give your call sign followed by the slant mark "/" followed by the identifier "AA".
The control operator's call sign must be used.
Withÿeither his or her own call sign only, or the Technician Plus call sign followed by his or her own call sign
50 mW
At least 40 dB
At least 60 dB
25 microwatts
A station controlled indirectly through a control link
Automatic control
Control of a model craft
Stations in repeater operation
The use of devices and procedures for control so that a control operator does not have to be present at a control point
Under automatic control a control operator is not required to be present at a control point
29.5 - 29.7 MHz
51.00 - 54.00 MHz
144.5 - 145.5 and 146 - 148 MHz
222.15 - 225.00 MHz
420 - 431, 433 - 435 and 438 - 450 MHz
1240 - 1300 MHz
The limit is three minutes.
The means of control between a control point and a remotely controlled station
A control link
None
Only the one type accepted may be marketed
Sold to an amateur operator
Spurious emission standards must be met
Spurious emission standards must be met in the "off" position
Must not work in the CB band
It must satisfy the spurious emission standards when driven withÿat least 50 W mean RF power.
Too easy modified
Too easy to operate on the CB band
The inclusion of instructions for operation or modification of the amplifier in a manner contrary to the FCC Rules
The amplifier produces 3 dB of gain for input signals between 26 MHz and 28 MHz (CB Band)
Spread spectrum
Frequency hopping and direct sequence
One hundred watts
Communications sent point-to-point within a system of cooperating amateur stations
Remote control of a station in repeater operation
Those within a system of cooperating amateur stations
222.00 - 222.15 MHz, 431 - 433 MHz and 435 - 438 MHz
Tech & above
20 words per minute maximum
At least once every ten minutes during and at the end ofÿtheÿQSOÿ
When the code is used for the communication
A line roughly parallel to, and south of, the US-Canadian borderÿ
420-430 MHz
An Area surrounding the National Radio Astronomy Observatory
The National Radio Astronomy Observatory
Calling a commercial tow truck service
You may sell amateur equipment over the airwaves if it is notÿyour business(or your employer's business) and you do notÿsell equipment over the airwaves on a regular basis.
Never
Never
When it does not involve your business or your employer's business
Prior FCC approval
FCC & FAA
Credit for the elements required for the license
To send and receive text at 5 WPM
The slant mark and prosigns AR, BT and SK
Five
18 years old and General class or above
Never
When the employee does not normally communicate with the manufacturing or distribution part of the company
Three accredited Volunteer Examiners at least 18 years old       and holding at least a General class license
Never
Revocation of amateur station license and suspension of operator's license
Revocation of amateur station license and suspension of operator's license
Immediately
Ten days
To the VEC
FAX is the transmission of printed pictures for permanent display on paper
240 lines per minute
3.3 minutes
Facsimile
Photodetector
Still pictures
128 or 256 lines
1500 Hz
2300 Hz
Signals not using the spectrum-spreading algorithm are suppressed in the receiver
The frequency of an RF carrier is changed very rapidly according to a particular pseudo-random sequence
300 bauds
Patches of dense ionization at E-region height
Sporadic E
Down Mexico way
Six meters
CW signals have a fluttery tone
The emission of charged particles from the sun
Toward the north
At E-region height
CW and SSB
Decreases
Vertical
Phase differences between radio-wave components of the same transmission, as experienced at the receiving station
Phase differences
Wider modes
It is more pronounced at wide bandwidths
Radio waves bend
Fifteen percent
Vertical angle decreases. This is a reason why Yagi antennas should be « wavelength above ground.
It decreases as the slope gets steeper
Pedersen ray
Five hundred miles
Ducting
Do microwaves heat food and other particles?
A device used to produce a highly accurate reference frequency
It makes frequency measurements
A looping pattern with 3 loops horizontally and 2 loops vertically
A variable LC oscillator with metered feedback current
When used properly, it gives the resonant frequency of a circuit.
Power coupled from an oscillator causes a decrease in meter current
To measure resonant frequency
Inductive and capacitive
As loosely as possible
A less accurate reading results
Surface mounting
Calibration, mechanical tolerance and coil impedance
The frequency response is limited by the bandwidth of the deflection amplifiers.
The frequency response can be improved by increasing the vertical amplifier bandwidth (frequency response).
Time base accuracy, speed of the logic and time base stability
By increasing the accuracy of the time base
One part-per-million (ppm) is one millionth (1E-6). 1E-6*146.52E+6 = 146.52 Hz.
One part-per-million (ppm) is one millionth (1E-6). .1ÿppmÿ=ÿ1/10 of 1E-6 = 1E-7.  1.4652E8*1E-7ÿ=ÿ1.4652E1ÿorÿ14.652 Hz
One part-per-million (ppm) is one millionth (1E-6). 10ÿppmÿ=ÿ10*1E-6 = 1E-5. 1E-5*146.52E6 = 1465.2 Hz
One part-per-million (ppm) is one millionth (1E-6). 1E-6*432.1E6ÿ=ÿ432.1E0 or 432.1 Hz
One part-per-million (ppm) is one millionth (1E-6). .1ÿppmÿ=ÿ.1*1E-6 = 1E-7.  1E-7*432.1E6 = 432.1E-1 = 43.21 Hz
One part-per-million (ppm) is one millionth (1E-6). 10ÿppmÿ=ÿ10*1E-6 = 1E-5.  1E-5*432.1E6 = 432.1E1 = 4321 Hz
It allows strong signals on nearby frequencies to interfere with reception of weak signals
Desensitization
Strong adjacent-channel signals
Shield the receiver from the transmitter causing the problem
The strongest signal received is the only demodulated signal
Capture effect
FM
The weakest signal that can be detected
First find the noise floor due to the bandwidth and the Noise Figure.  NFlr = -174 + 10*Log(BW) + NF = -174 + 10*Log(500) + 8  NFlr = -174 + 27 +8 = -139 dBm.   The dynamic range is the 1 dB compression point less the noise floor:  -20 dBm - (-139 dBm) = 119 dB.  (dB is used because a range is given - not an absolute level)  The bandwidth (BW) is in Hz.
The selectivity of the RF amplifier determines how well theÿimage response is rejected.
The minimum discernible signal (MDS) is the weakest signal that the receiver can detect.
Intermodulation distortion
When they are in close proximity and the signals mix in one orÿboth of their final amplifiers     
By installing isolators in the feed line
Modulation from an unwanted signal is heard in addition to theÿdesired signal
Cross-modulation interference
By installing a filter at the receiver
An unwanted signal is hear in addition to the desired signalÿÿÿÿÿ
Nonlinear devices
146.34 MHz and 146.61 MHz
A high-pass filter attached to the input of the television receiver                                    
Splatter from an SSB transmitter
Resonance
The frequency at which capacitive reactance equals inductive reactance
Inductive and capacitive reactances are equal
Resonance
Approximately equal to circuit resistance
Approximately equal to circuit resistance
Maximum
Maximum
Minimum
At resonance, the inductive and capacitive reactances are equalÿmaking the net reactance zero.  The current is in phaseÿwith the voltage.
The voltage and current are in phase
F = 1/(6.2832*û(50E-6*40E-12)) = 1/(6.2832*û(2000E-18)) Fÿ=ÿ1/(6.2832*4.472E-8) = 1/2.810E-7 = 3.56E6 or 3.56 MHz
F = 1/(6.2832*û(40E-6*200E-12)) = 1/(6.2832*û(8000E-18)) Fÿ=ÿ1/(6.2832*8.944E-8) = 1/5.620E-7 = 1.78E6 or 1.78 MHz
F = 1/(6.2832*û(50E-6*10E-12)) = 1/(6.2832*û(500E-18)) Fÿ=ÿ1/(6.2832*2.236E-8) = 1/1.505E-7 = 7.12E6 or 7.12 MHz
F = 1/(6.2832*û(25E-6*10E-12)) = 1/(6.2832*û(250E-18)) Fÿ=ÿ1/(6.2832*1.581E-8) = 1/9.935E-7 = 1.01E7 or 10.1 MHz
F = 1/(6.2832*û(3E-6*40E-12)) = 1/(6.2832*û(120E-18)) Fÿ=ÿ1/(6.2832*1.095E-8) = 1/6.883E-8 = 1.45E7 or 14.5 MHz
F = 1/(6.2832*û(4E-6*20E-12)) = 1/(6.2832*û(80E-18)) Fÿ=ÿ1/(6.2832*8.944E-9) = 1/5.620E-8 = 1.78E7 or 17.8 MHz
F = 1/(6.2832*û(8E-6*7E-12)) = 1/(6.2832*û(56E-18)) Fÿ=ÿ1/(6.2832*7.48E-9) = 1/4.70E-8 = 2.13E7 or 21.3 MHz
F = 1/(6.2832*û(3E-6*15E-12)) = 1/(6.2832*û(45E-18)) Fÿ=ÿ1/(6.2832*6.71E-9) = 1/4.210E-8 = 2.37E7 or 23.7 MHz
F = 1/(6.2832*û(4E-6*8E-12)) = 1/(6.2832*û(32E-18)) Fÿ=ÿ1/(6.2832*5.66E-9) = 1/3.55E-8 = 2.81E7 or 28.1 MHz
F = 1/(6.2832*û(1E-6*9E-12)) = 1/(6.2832*û(9E-18)) Fÿ=ÿ1/(6.2832*3E-9) = 1/1.88E-8 = 5.31E7 or 53.1 MHz
2ãF = 1/û(LC),  4ãýFý = 1/LC,  4ãýFýL = 1/C  or  C = 1/(4ãýFýL) Cÿ=ÿ1/(4*9.87*2.03E14*2.84E-6) = 1/(2.28E10) = 4.4E-11 or 44 pF
F = 1/(6.2832*û(1E-6*10E-12)) = 1/(6.2832*û(10E-18)) Fÿ=ÿ1/(6.2832*3.16E-9) = 1/1.99E-8 = 5.03E7 or 50.3 MHz
F = 1/(6.2832*û(2E-6*15E-12)) = 1/(6.2832*û(30E-18)) Fÿ=ÿ1/(6.2832*5.48E-9) = 1/3.44E-8 = 2.91E7 or 29.1 MHz
F = 1/(6.2832*û(5E-6*9E-12)) = 1/(6.2832*û(45E-18)) Fÿ=ÿ1/(6.2832*6.71E-9) = 1/4.21E-8 = 2.37E7 or 23.7 MHz
F = 1/(6.2832*û(2E-6*30E-12)) = 1/(6.2832*û(60E-18)) Fÿ=ÿ1/(6.2832*7.75E-9) = 1/4.87E-8 = 2.05E7 or 20.5 MHz
F = 1/(6.2832*û(15E-6*5E-12)) = 1/(6.2832*û(75E-18)) Fÿ=ÿ1/(6.2832*8.66E-9) = 1/5.44E-8 = 1.84E7 or 18.4 MHz
F = 1/(6.2832*û(3E-6*40E-12)) = 1/(6.2832*û(120E-18)) Fÿ=ÿ1/(6.2832*1.10E-8) = 1/6.88E-8 = 1.45E7 or 14.5 MHz
F = 1/(6.2832*û(40E-6*6E-12)) = 1/(6.2832*û(240E-18)) Fÿ=ÿ1/(6.2832*1.55E-8) = 1/9.73E-8 = 1.03E7 or 10.3 MHz
F = 1/(6.2832*û(10E-6*50E-12)) = 1/(6.2832*û(500E-18)) Fÿ=ÿ1/(6.2832*2.24E-8) = 1/1.40E-7 = 7.12E6 or 7.12 MHz
F = 1/(6.2832*û(200E-6*10E-12)) = 1/(6.2832*û(2000E-18)) Fÿ=ÿ1/(6.2832*4.472E-8) = 1/2.810E-7 = 3.56E6 or 3.56 MHz
F = 1/(6.2832*û(90E-6*100E-12)) = 1/(6.2832*û(9000E-18)) Fÿ=ÿ1/(6.2832*9.49E-8) = 1/5.97E-7 = 1.68E6 or 1.68 MHz
2ãF = 1/û(LC),  4ãýFý = 1/LC,  4ãýFýC = 1/L or L = 1/(4ãýFýC) Lÿ=ÿ1/(4*9.87*2.03E14*44E-12) = 1/(3.53E5) = 2.8E-6 or 2.8 æH
As frequency increases, RF current flows in a thinner layer of the conductor, closer to the surface  
Skin effect
Along the surface of the conductor
Because of skin effect
Because of skin effect
Capacitor
Joule
The space around a conductor, through which a magnetic force acts
In a direction determined by the left-hand rule
The amount of current
Potential energy
BW = F/Q = 1.8E6/95 = 1.89E4 or 18.9 kHz
BW = F/Q = 3.6E6/218 = 1.65E4 or 16.5 kHz
BW = F/Q = 7.1E6/150 = 4.73E4 or 47.3 kHz
BW = F/Q = 12.8E6/218 = 5.87E4 or 58.7 kHz
BW = F/Q = 14.25E6/150 = 9.5E4 or 95 kHz
BW = F/Q = 21.15E6/95 = 2.226E5 or 222.6 kHz
BW = F/Q = 10.1E6/225 = 4.49E4 or 44.9 kHz
BW = F/Q = 18.1E6/195 = 9.28E4 or 92.8 kHz
BW = F/Q = 3.7E6/118 = 3.14E4 or 31.4 kHz
BW = F/Q = 14.25E6/187 = 7.62E4 or 76.2 kHz
Half-power bandwidth
Q = 18E3/(6.2832*14.128E6*2.7E-6) = 75.1
Q = 18E3/6.2832*14.128E6*4.7E-6) = 43.1
Q = 180/(6.2832*4.468E6*47E-6) = .136
Q = 10E3/(6.2832*14.225E6*3.5E-6) = 32
Q = 1000/(6.2832*7.125E6*8.2E-6) = 2.7
Q = 100/(6.2832*7.125E6*10.1E-6) = .221
Q = 22E3/(6.2832*7.125E6*12.6E-6) = 39
Q = 2200/(6.2832*3.625E6*3E-6) = 32.2
Q = 220/(6.2832*3.625E6*42E-6) = .23
Q = 1800/(6.2832*3.625E6*43E-6) = 1.84
As can be seen in the equation on the above right, the lower the resistance, the lower the circuit Q (and the greater bandwidth).
Z = 100 +j100 -j25 = 100 +j75.  The +j indicates the voltage is leading.  é = ATN(X/R) = ATN(75/100) = ATN(.75) = 36.9ø
Z = 100 +j50 -j25 = 100 +j25.  The +j indicates the voltage is leading.  é = ATN(X/R) = ATN(25/100) = ATN(.25) = 14ø
Z = 1000 +j250 -j500 = 1000 -j250.  The -j indicates the voltage is lagging.  é = ATN(X/R) = ATN(-250/1000) = ATN(-.25) = -14ø
Z = 100 +j100 -j75 = 100 +j25.  é = ATN(X/R) = ATN(25/100) éÿ=ÿATN(.25) = 14ø.  Positive angle indicates leading.
Z = 100 +j25 -j50 = 100 -j25.  é = ATN(X/R) = ATN(-25/100) éÿ=ÿATN(-.25) = -14ø.  Negative angle indicates lagging.
Z = 100 +j50 -j75 = 100 -j25.  The -j indicates the voltage is lagging.  é = ATN(X/R) = ATN(-25/100) = ATN(-.25) = -14ø
Z = 100 +j75 -j100 = 100 -j25.  é = ATN(X/R) = ATN(-25/100) éÿ=ÿATN(-.25) = -14ø.  Negative angle indicates lagging.
Z = 1000 +j500 -j250 = 1000 +j250.  The +j indicates the voltage is leading.  é = ATN(X/R) = ATN(250/1000) = ATN(.250) = 14.0ø
Z = 100 +j75 -j50 = 100 +j25.  é = ATN(X/R) = ATN(25/100) éÿ=ÿATN(.25) = 14ø.  Positive angle indicates leading.
Current leads voltage by 90 degrees
Voltage leads current by 90 degrees
Wattless, nonproductive power
Reactive power
It goes back and forth between magnetic and electric fields, butÿis not dissipated  
By multiplying the apparent power times the power factor
The power factor is the cosine of the phase angle. PFÿ=ÿCOS(60)ÿ=ÿ0.5.
PF = COS(45) = .707
PF = COS(30) = .866
P(real) = Volts*Amperes*PF = 100*4*.2 = 80 watts
P(real) = Volts*Amperes*PF = 200*5*.6 = 600 watts
P(real) = P(apparent)*PF = 500*.71 = 355
The voltage and current are out of phase
First find the total gain (or loss) in dB. Gain(dB)ÿ=ÿ-4ÿ-2ÿ-1ÿ+6ÿ=ÿ-1 dB. NextÿconvertÿtheÿgainÿinÿdBÿtoÿnumericÿratio. P2/P1ÿ=ÿ10(dB/10)ÿ=ÿ10(-1/10)ÿ=ÿ10(-.1) = .794 P2ÿ=ÿ.794*P1ÿ=ÿ.794*50 = 39.7
First find the total gain (or loss) in dB. Gain(dB)ÿ=ÿ-5ÿ-3ÿ-1ÿ+7ÿ=ÿ-2 dB. NextÿconvertÿtheÿgainÿinÿdBÿtoÿnumericÿratio. P2/P1ÿ=ÿ10(dB/10)ÿ=ÿ10(-2/10)ÿ=ÿ10(-.2) = .631 P2ÿ=ÿ.631*P1ÿ=ÿ.631*50 = 31.5 watts
Gain(dB)ÿ=ÿ-4ÿ+10ÿ=ÿ6 dB. P2/P1ÿ=ÿ10(dB/10)ÿ=ÿ10(6/10)ÿ=ÿ10(.6) = 3.98 P2ÿ=ÿ3.98*P1ÿ=ÿ3.98*75 = 299 watts
Gain(dB)ÿ=ÿ-5ÿ-3ÿ-1ÿ+6ÿ=ÿ-3 dB. P2/P1ÿ=ÿ10(dB/10)ÿ=ÿ10(-3/10)ÿ=ÿ10(-.3) = .501 P2ÿ=ÿ.501*P1ÿ=ÿ.501*75 = 37.6 watts
Gain(dB)ÿ=ÿ-1ÿ+6ÿ=ÿ5 dB. P2/P1ÿ=ÿ10(dB/10)ÿ=ÿ10(5/10)ÿ=ÿ10(.5)ÿ= 3.16 P2ÿ=ÿ3.16*P1ÿ=ÿ3.16*100 = 316 watts
Gain(dB)ÿ=ÿ-5ÿ-3ÿ-1ÿ+10ÿ=ÿ1 dB. P2/P1ÿ=ÿ10(dB/10)ÿ=ÿ10(1/10)ÿ=ÿ10(.1)ÿ= 1.26 P2ÿ=ÿ1.26*P1ÿ=ÿ1.26*100 = 126
Gain(dB)ÿ=ÿ-5ÿ-3ÿ-1ÿ+6ÿ=ÿ-3 dB. P2/P1ÿ=ÿ10(dB/10)ÿ=ÿ10(-3/10)ÿ=ÿ10(-.3) = .50 P2ÿ=ÿ.50*P1ÿ=ÿ.50*120 = 60 watts
Gain(dB)ÿ=ÿ-2ÿ-2.2ÿ+7ÿ=ÿ2.8 dB. P2/P1ÿ=ÿ10(dB/10)ÿ=ÿ10(2.8/10)ÿ=ÿ10(.28) = 1.90 P2ÿ=ÿ1.90*P1ÿ=ÿ1.90*150 = 286 watts
Gain(dB)ÿ=ÿ-4ÿ-3.2ÿ-0.8ÿ+10ÿ=ÿ2 dB. P2/P1ÿ=ÿ10(dB/10)ÿ=ÿ10(2/10)ÿ=ÿ10(.2) = 1.585 P2ÿ=ÿ1.585*P1ÿ=ÿ1.585*200 = 317 watts
Gain(dB)ÿ=ÿ-2ÿ-2.8ÿ-1.2ÿ+7ÿ=ÿ1 dB. P2/P1ÿ=ÿ10(dB/10)ÿ=ÿ10(1/10)ÿ=ÿ10(.1) = 1.26 P2ÿ=ÿ1.26*P1ÿ=ÿ1.26*200 = 252 watts
Effective radiated power (ERP)
Let Rload = 100 Mê.   I(R2) ÷ V1/(R1+R2) = 8/(8000+8000) = .5 mA V2 ÷ V(R2) = I(R2)*R2 = .0005*8000 = 4 volts.  Let Rload = .01 ê I(load) ÷ V1/R1 = 8/8000 = 1 mA, R3 = V2/I(load) = 4/.001 = 4 kê
Let Rload = 100 Mê. I(R2) ÷ V1/(R1+R2) and V2 ÷ V(R2) = I(R2)*R2 Therefore V2 = V1*R2/(R1+R2) = 8*8000/(16000+8000) = 2.67 volts.
V2 = V1*R2/(R1+R2) = 8*16000/(8000+16000) = 5.33 volts.
The impedance of a voltage source (V1) is considered to be zero ohms.   Therefore, R1 and R2 are in parallel and R3 is equal to R1/2.  R3 = R1*R2/(R1+R2) = 1E4*1E4/2E4 = 1E8/2E4 = 5000 ê
V2 = V1*R2/(R1+R2) = 10*10000/(20000+10000) = 3.33 volts R3ÿ=ÿR1*R2/(R1+R2) = 20000*10000/(20000+10000) = 6.67 Kê
V2 = V1*R2/(R1+R2) = 10*20000/(10000+20000) = 6.67 volts R3ÿ=ÿR1*R2/(R1+R2) = 10000*20000/(10000+20000) = 6.67 Kê
V2 = V1*R2/(R1+R2) = 12*10000/(10000+10000) = 6 volts R3ÿ=ÿR1*R2/(R1+R2) = 10000*10000/(10000+10000) = 5 Kê
V2 = V1*R2/(R1+R2) = 12*10000/(20000+10000) = 4 volts R3ÿ=ÿR1*R2/(R1+R2) = 20000*10000/(20000+10000) = 6.67 Kê
V2 = V1*R2/(R1+R2) = 12*20000/(10000+20000) = 8 volts R3ÿ=ÿR1*R2/(R1+R2) = 10000*20000/(10000+20000) = 6.67 Kê
V2 = V1*R2/(R1+R2) = 12*20000/(20000+20000) = 6 volts R3ÿ=ÿR1*R2/(R1+R2) = 20000*20000/(20000+20000) = 10 Kê
Thevenin's Theorem (using open circuit voltage and short circuit current) is used in the solution of question 152.
Silicon and germanium
At microwave frequencies
N-type
P-type
N-type
N-type
P-type
P-type
Holes
Electrons
Donor
Acceptor
A zener diode has a constant voltage drop across it under conditions of varying current
Symbol 3 is given as the correct answer. You will also see symbol 6 used for a zener diode.
A negative resistance region exists in the forward conduction curve.  The forward current decreases with an increase in the applied voltage in the negative resistance region.
Tunnel diode
Look for a "tunnel"
VARiable-capACiTOR
Look for the two lines of a capacitor with an added arrow.
VHF and UHF mixers and detectors
If it gets too hot, it will melt, fuse, open etc.
Maximum forward current and Peak-Inverse-Voltage
Junction and point contact
As an RF detector

As an RF switch
Look for the light "rays"
Forward
Permeability
To a GHz
Ferrite and powdered-iron toroids
Better temperature stability
Fewer turns to produce a given inductance value
Type 43 mix ferrite
Ferrite beads
Most of the magnetic field is within the core material
A pair of wires is used to place two windings on the core. Thisÿmakes a type of transmission line transformer.
43 turns
35 turns
Base, emitter and collector
Alpha is the common base current gain
Beta is the common emitter current gain
Common base upper frequency limit for amplifiers.
Think of NPN as the arrow "Not Pointing iN".  Base (single lead on left side of drawing
"Pointing iN Part" with base (single lead) on left
Alpha cutoff frequency
Common emitter
Area near the junction
Maximum collector current
No collector current
Only the Emitter (always drawn at an angle) is on the right.
Two bases on the left side of the drawing and a single emitterÿon the right side of the drawing
Diodes have a cathode and an anode.
On and off
Junction diode
On
Diode drawing with a gate on the top right
A TRIAC will pass AC
Gate, anode 1 and anode 2
Back to back diodes to pass AC
It'll glow
Resistor in series
Two identical electrodes and a dot to indicate gas filled
Audio passband for SSB is about .3 to 3 kHz
DSB is twice as wide as SSB
A filter with narrow bandwidth and steep skirts made using quartz crystals
Measure crystal frequencies and carefully select the crystals for the proper frequencies
Frequencies of the individual crystals
Crystals (like quartz) change shape when a voltage is applied.
MRF901's have been used for this application. Better performanceÿand easier matching can be obtained with aÿMSA-0135.ÿ
MSA-0735
If MMIC is an answer, choose MMIC
Microstrip
Bias is to the output lead
The entire cycle (360ø)
Class "A" act
Between 180ø and 360ø
180ø
Less than 180ø
Effi"C"iency
Just below the saturation point
(RF power out / DC power in) x 100%
Neutralization
Set the plate current with the loading capacitor while keepingÿthe plate current dipped (at resonance) with theÿtuningÿcapacitor.
By using a push-pull amplifier
Distortion
Out of phase feed back
Back and forth oscillations in a tank circuit
12
Emitter is at signal ground. Therefore it is a common emitterÿcircuit.
Fixed base bias
Coupling cap
Emitter bypass
Self bias
Common collector (collector at signal ground) amplifier
Actually it is the emitter DC return (and self bias)
Grounds (for the signal) the collector
Coupling
Voltage regulator
Constant voltage reference
It provides the current handling capacity
It filters the supply voltage
It provides additional filtering for the reference zener diodeÿbias.
Prevents oscillation
It supplies current to D1
It provides a minimum load for Q1.
A three element network in the shape of the symbol ã.
ã
ÿÿÿÿÿÿoÄÄÄÂÄÄ/\/\/\/\ÄÄÂÄÄÄoÿÿÿÿÿÿÿThis is the ÿÿÿÿÿÿÿÿÿÄÁÄÿÿÿÿÿÿÿÿÿÿÄÁÄÿÿÿÿÿÿÿÿÿÿshape of a ÿÿÿÿÿC1ÿÿÄÂÄÿÿÿÿÿÿÿÿÿÿÄÂÄÿÿÿC2ÿÿÿÿÿã network. ÿÿÿÿÿÿÿÿÿÿ³ÿÿÿÿÿÿÿÿÿÿÿÿ³
A two element network
Limited impedance matching range
A four element network
A high pass network
Greater harmonic suppression
ã-L
L, ã and ã-L
It cancels the reactive part of an impedance and changes theÿresistive part
High-pass, low-pass and band-pass
F(MHz) = 300/WL(meters) = 300/80 = 3.75 MHz,   F = 1/2ãûLC) 2ãFÿ=ÿ1/û(LC),ÿ4ãýFýÿ= 1/LC, 4ãýFýL = 1/C  or  C = 1/(4ãýFýL) Cÿ=ÿ1/(4*9.87*1.41E13*20E-6) = 1/(1.11E10) = 9.0E-11 or 90 pF Pick the closest answer.
F(MHz) =  300/WL(meters) = 300/40 = 7.5 MHz,   F = 1/2ãûLC) 2ãFÿ=ÿ1/û(LC),  4ãýFý = 1/LC,  4ãýFýC = 1/L or L = 1/(4ãýFýC) Lÿ=ÿ1/(4*9.87*5.6E13*100E-12) = 1/(2.2E5) = 4.5E-6 or 4.5 æH Pick the closest answer.
Pick 14.1 MHz,  F = 1/2ãûLC),  2ãFÿ=ÿ1/û(LC),  4ãýFý = 1/LC 4ãýFýL = 1/C  or  C = 1/(4ãýFýL) =ÿ1/(4*9.87*2.0E14*2E-6) Cÿ=ÿ1/(1.6E10) = 6.3.E-11 or 63 pF. Pick the closest answer
Pick 21.1 MHz.  F = 1/2ãûLC),  2ãF = 1/û(LC),  4ãýFý = 1/LC 4ãýFýC = 1/L  or  L = 1/(4ãýFýC) =ÿ1/(4*9.87*4.5E14*15E-12) Lÿ=ÿ1/(2.6E5) = 3.8E-6 or 3.8 æH.  Pick the closest answer.
F(MHz) = 300/WL(meters) = 300/160 = 1.88 MHz,   F = 1/2ãûLC) 2ãFÿ=ÿ1/û(LC),ÿ4ãýFýÿ= 1/LC, 4ãýFýL = 1/C  or  C = 1/(4ãýFýL) Cÿ=ÿ1/(4*9.87*3.5E12*100E-6) = 1/(1.38E10) = 7.2E-11 or 72 pF Pick the closest answer.
It has a smooth "buttered" passband
Russians are always making ripples
Chebyshev filter
Sharp cutoff
Elliptical
The pass (control) element is always turned on
The control device is switched off or on.
Zener
Series
Shunt
Six volts
Feedback comes directly from the load - not from an internalÿpoint in the power supply
A three terminal regulator has an input (supply) terminal, a ground connection and an output terminal.
Supply voltage range, output voltage and current
Zener
Three terminal
Colpitts, Hartley and Pierce
Positive feedback
Through a tapped coil
Through a capacitive divider
Through capacitive coupling
The Hartley and Colpitts oscillators can be crystal controlled, but the Pierce oscillator must have a crystal.
The crystal replaces the LC tank circuit.
Colpitts and Hartley
It can be stable
A VARiable-capACiTOR
The output will only be as "clean" and "stable" and the reference.
Modulation is the process of adding information to a carrier.
With a reactance modulator on the oscillator
It acts as a variable inductance or capacitance to produce FM signals (frequency variations) in an oscillator.
A reactance modulator acts as a variable inductance or capacitance
It varies the tuning of an amplifier tank circuit to produce phase variations - which produces a signal like FM.
A phase modulator
Double sideband, suppressed carrier
The filter removes the undesired sideband
By modulating the plate voltage of a Class C amplifier
A pre-emphasis network is used in FM
A de-emphasis network is used in an FM receiver
Recovery of the "intelligence" from a modulated RF signal
Rectification and filtering of RF
It mixes an incoming signal with a locally generated carrierÿtoÿreplace the carrier in a suppressd carrier signalÿÿÿÿÿ
FM signals can be detected (demodulated) with a frequency discriminator circuit
FM signals can be detected (demodulated) with a frequency discriminator circuit
Active filters induce noise and require power
Notch filter
Linear phase response
An adaptive filter
A Hilbert-transform filter
A cavity filter has high Q and can handle high power
The combination of two signals to produce sum and difference frequencies  
The original frequencies and the sum and difference frequencies
Easier tuned circuit design at IF frequencies
Spurious mixer products can be generated
PLL
Direct
A VCO, a divider, phase detector, loop filter and a reference
A phase accumulator (counter), lookup table, D/A and a low-pass  filter                                   
Sine wave table
Spurs at discrete frequencies
Noise
Audio frequencies of 300 to 3000 Hz are generally used
Total harmonic distortion
A Darlington pair is a DC coupled pair of (bipolar) transistors with a high gain, high input impedance and low output impedance
Audio gain
Fixed-tuned pass-band amplifier at an Intermediate Frequency
In the distant past, the IF frequency was selected to provide aÿcompromise between image rejection and selectivity - about 10%ÿof the RF frequency was generally used.
To provide selectivity
To provide a greater tuning range
Enough gain to allow weak signals to overcome mixer noise
To prevent the generation of spurious mixer products
Improve the receiver noise figure
First symbol   A - DSB Amplitude Mod. Secondÿsymbolÿÿ3ÿ-ÿSingleÿChannel Analog ThirdÿsymbolÿÿÿCÿ-ÿFacsimile (FAX)
First symbol   A - DSB Amplitude Mod. Secondÿsymbolÿÿ3ÿ-ÿSingleÿChannel Analog ThirdÿsymbolÿÿÿCÿ-ÿFacsimile (FAX)
Printed pictures by electrical means
First symbol   F - Frequency Modulation SecondÿSymbolÿÿ3ÿ-ÿSingleÿChannel Analog ThirdÿSymbolÿÿÿCÿ-ÿFacsimile (FAX)
First symbol   F - Frequency Modulation SecondÿSymbolÿÿ3ÿ-ÿSingle Channel Analog ThirdÿSymbolÿÿÿCÿ-ÿFacsimile (FAX)
First symbol   A - DSB Amplitude Mod. SecondÿSymbolÿÿ3ÿ-ÿSingleÿChannel Analog ThirdÿSymbolÿÿÿFÿ-ÿTelevision
First symbol   A - DSB Amplitude Mod. SecondÿSymbolÿÿ3ÿ-ÿSingleÿChannel Analog ThirdÿSymbolÿÿÿFÿ-ÿTelevision
First symbol   F - Frequency Modulation SecondÿSymbolÿÿ3ÿ-ÿSingleÿChannel Analog ThirdÿSymbolÿÿÿFÿ-ÿTelevision
First symbol   F - Frequency Modulation SecondÿSymbolÿÿ3ÿ-ÿSingleÿChannel Analog ThirdÿSymbolÿÿÿFÿ-ÿTelevision
First symbol   J - Single Sideband SecondÿSymbolÿÿ3ÿ-ÿSingleÿChannel Analog ThirdÿSymbolÿÿÿFÿ-ÿTelevision
CW
Emission Designators
Type of modulation, nature of the modulating signal and type of  information to be transmitted
The type of modulation of the main carrier
The type of modulation of the main carrier
The type of modulation of the main carrier
The nature of signals modulating the main carrier
The nature of signals modulating the main carrier
The type of information to be transmitted
The type of information to be transmitted
The type of information to be transmitted
The type of information to be transmitted
By using a reactance modulator on an oscillator
By filtering
The ratio between the deviation and the modulating frequencyÿÿÿÿÿ
Modulation index
It does not depend on the RF carrier frequency
MI = D/Fm = 3000/1000 = 3
MI = D/Fm = 6000/2000 = 3
The ratio of the maximum carrier frequency deviation to the highest audio modulating frequency
Deviation ratio
DR = D(max)/Fm(max) = 5000/3000 = 1.67
DR = D(max)/Fm(max) = 7500/3500 = 2.14
A wave consisting of an electric field and a magnetic field atÿright angles to each other
300 million meters per second (as used in the wavelength toÿfrequency conversions)
The wave is short-circuited
Changing electric and magnetic fields propagate the energy
Waves with an electric field parallel to the Earth
Waves with a rotating electric field
Vertical
Polarization refers to the electric field, which is 90ø to the magnetic field.
Polarization refers to the electric field, which is 90ø to the magnetic field.
Horizontal
Atmospheric noise
Receiver noise
A wave whose amplitude follows the trigonometric sine function
sin(0ø) = 0 and sin(180ø) = 0
360ø
The time required to complete one cycle
A wave that changes back and forth between two voltage levels and remains an equal time at each level
Square wave
All odd harmonics
Square wave
A wave with a straight line rise and fall, one faster than the other
Sawtooth
Sawtooth
Household electrical voltage is 117 VAC RMS. Multiply by û2ÿ(1.414) to get peak voltage.  Vpeak = 1.414*117 = 165
Household electrical voltage is 117 VAC RMS.  Multiply by û2ÿ(1.414) to get peak voltage.  Vpeak = 1.414*117 = 165 Peakÿtoÿpeak is twice peak voltage.  Vpp = 2*165 =  330
117 VAC
Divide by 2 to get peak, then divide by û2 to get RMS. Vpÿ=ÿ340/2ÿ=ÿ170,  Vrms 170/1.414 = 120
Same heating as a DC voltage
Measure heating effect
2.5 to 1
Speech characteristics
Class B efficiency is about 65%.  Pin = Pout/Ef = 1500/.65 Pinÿ=ÿ2308 watts. Pick the closest.
Class C efficiency is about 80%.  Pin = Pout/Ef = 1000/.8 Pinÿ=ÿ1250 watts.
Class AB efficiency is about 50%.  Pin = Pout/Ef = 500/.5 Pinÿ=ÿ1000 watts
Equivalent resistance to the antenna's radiated power "resistance"
To match impedances for maximum power transfer
Antenna conductors' length/diameter ratio
Antenna efficiency
Radiation resistance plus ohmic resistance
A dipole made up of two close spaced parallel elements
Wider 
Gain is the increased signal strength of an antenna compared toÿanother (usually a dipole) antenna.
The frequency range over which an antenna can be expected toÿperform well
As you rotate the antenna, note the 3 dB points in degrees. Theÿbeamwidth is the difference between the two readings.
Radiation resistance / total resistance
By installing a good ground radial system
Orientation of its electric field
Azimuth pattern
Pattern crosses -3 dB line at -25ø and +25ø for a total of 50ø
Forward part of pattern is 0 dB and reverse is at -18 db point
All of the above is needed to evaluate the overall performance. The easiest way to determine antenna gain is to compare it to another (known) antenna.
Very few antennas depend on a constant dielectric constant forÿtheir performance.
The feedpoint impedance becomes very low - this can be overcomeÿby a matching network
Overall performance improves - including gain
Method of Moments
A wire is modeled as a series of segments, each having a distinct value of current
Front of pattern is 0 dB and side is near, but not at -12 dB.
Elevation pattern
The electric field plane is the same as the element plane inÿaÿYagi antenna
How good the ground is
If the "ground" was a ground radial system, watering the baseÿofÿthe antenna would be useless.  Impedance inverts atÿaÿquarter wavelength, making it the best length.
The saltwater improves the ground condition and increases theÿdesired low-angle radiation
The effect on the radiation pattern is minor at one wavelengthÿabove ground(pattern is distorted below one-halfÿwavelength)
The artificial ground works quite well and it reduces the near-field ground losses
Build a big capacitor with a wire-mesh screen
Between 0 and 15 degrees
Front of pattern is 0 dB and rear is close to -30 dB
Between 0 and 90 degrees there are four lobes
300 ê
Near the center - but you will find many near the bottom
To minimize losses
It's hard to drive under an underpass with a 40 foot high whip
The ham bands are harmonically related (to keep the trashÿ"inÿband")
Multiband
The capacitive reactance increases and a loading coil (an inductor) is needed to "tune" it out.
All of the components of antenna feed-point impedance and theÿfeed-line impedance are needed
The driven element reactance is capacitive
L network
Bandwidth decreases
Improved radiation efficiency
It is a number of 1 or less given by: Vf = V(line)/V(space)
Velocity factor
Two-thirds
Line dielectrics
RF energy moves slower along the coaxial cable
WL = 300E6/F(MHz) = 300E6/14.1E6 = 21.28 meters ¬ÿWLÿ=ÿ21.28/4ÿ=ÿ5.32.  ¬ WL(line) = .66*5.32 = 3.52 meters
WL = 300E6/F(MHz) = 300E6/7.2E6 = 41.67 meters ¬ÿWLÿ=ÿ41.67/4ÿ=ÿ10.42.  ¬ WL(line) = .66*10.42 = 6.88 meters
WL = 300E6/F(MHz) = 300E6/14.1E6 = 42.56 meters ¬ÿWLÿ=ÿ42.56/4ÿ=ÿ10.64.  ¬ WL(line) = .95*10.64 = 10.1 meters
Electrical length times 0.8
SWR can be calculated from the reflection coefficient
SWR greater than 1:1
In general, open wire lines have lower loss than coax
