ExtraClassSylalbus2009jan-AD7FO

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Rev 2.02 Admittance = 1/100 +(-j/100) or 0.01 +j 0.01 Angle =arc tan .01/.01 or 45° (-45° in polar coordinates) Impedance = 1/ (√ ( (.01)² x (.01)² )) or 1/(.0141) or 70.71 Ώ E5C08 In polar coordinates, the impedance of a network comprised of a 300-ohm-reactance inductor in series with a 400ohm resistor is 500 ohms at an angle of 37 degrees. Z= √(X² + (XL –Xc)²) or Z= √( 400² + (0-300)²) or Z= √(250,000) or Z= 500 Ω Angle is arc tan (reactance/resistance) or arc tan (300/400) or arc tan (.75) or 36.86° - j300 400Ω E5C09 When using rectangular coordinates to graph the impedance of a circuit, the horizontal axis represents the voltage or current associated with the resistive component. E5C10 When using rectangular coordinates to graph the impedance of a circuit, the vertical axis represents the voltage or current associated with the reactive component. E5C11 The two numbers used to define a point on a graph using rectangular coordinates represent the coordinate values along the horizontal and vertical axes. E5C12 If you plot the impedance of a circuit using the rectangular coordinate system and find the impedance point falls on the right side of the graph on the horizontal line, you know the circuit is equivalent to a pure resistance. E5C13 The Rectangular coordinate system is often used to display the resistive, inductive, and/or capacitive reactance components of impedance. E5C14 The Polar coordinate system is often used to display the phase angle of a circuit containing resistance, inductive and/or capacitive reactance. E5C15 (A) In polar coordinates, the impedance of a circuit of 100 -j100 ohms impedance is 141 ohms at an angle of -45 degrees. Z= √(X² + (XL –Xc)²) or Z= √( 100² + ( -100)²) or Z= √(20,000) or Z= 141.42 Ω Angle is arc tan (reactance/resistance) or arc tan (-100/100) or arc tan (1) or -45° Jack Tiley <strong>AD7FO</strong> Page 46 3/15/2009
Rev 2.02 Parallel circuit solutions Solving for parallel circuits for ac circuits is similar to the way we solved resistance parallel circuits. Remember the Equation: The solution involved finding the conductance (G) of each leg by dividing the resistances into 1 and summing them. This gave the total circuit conductance in Siemens. The Mho was the term previously used for Seimen. To find the resistance we divided the conductance into 1 and ended up with the parallel circuit resistance. We do the same thing to find the impedance of parallel ac circuits. The names of the circuit references change- Impedance becomes admittance, “Y”, (1/impedance). Resistance becomes conductance, “G”, (1/resistance) and susceptance , “B” (1/reactance) We start by finding the “conductance of the resistive and reactive components and just add them as we did in the resistance solution (remember we will be summing resistive (real) and reactive (imaginary) conductance. The rectangular coordinates for parallel circuit solutions are shown below. Note that the reactive axis direction is opposite that of the series circuit solutions in that that reactive conductance is + for capacity and – for inductance. E5C16 In polar coordinates, the impedance of a circuit that has an admittance of 7.09 milli-siemens at 45 degrees is 141 ohms at an angle of -45 degrees. Polar Impedance (Z) =1/admittance or Z= 1/.00709 or Z= 141.04Ω Polar angle = 1 / j (admittance angle) = 1/j(45°) or -j45° E5C17 (C) In rectangular coordinates, the impedance of a circuit that has an admittance of 5 millisiemens at -30 degrees is C. 173 + j100 ohms. Polar Impedance (Z) =1/admittance or Z= 1/.005 or Z= 200Ω Polar angle = 1/amdittance angle = 1/- 30° or +30° Cos θ= resistance(R) / Impedance(Z) or R = 200Ω x Cosine 30° or R = 200Ω x .866 or 173.2 Ω Sin θ= reactance(j) / Impedance(Z) or j = 200 x Sine 30° or j = 200Ω x .50 or j100Ω Jack Tiley <strong>AD7FO</strong> Page 47 3/15/2009 Susceptance
Page 1 and 2: Rev 2.02 Preparing for the Extra cl
Page 3 and 4: Rev 2.02 Guide to using this syllab
Page 5 and 6: Rev 2.02 SUBELEMENT E4 -- AMATEUR R
Page 7 and 8: Rev 2.02 ELEMENT 4 - EXTRA CLASS QU
Page 9 and 10: Rev 2.02 E1A09 (Band Plan Chart) Th
Page 11 and 12: Rev 2.02 E1B13 (FCC Rules) Communic
Page 13 and 14: Rev 2.02 E1D08 (FCC Rules) The 2 me
Page 15 and 16: Rev 2.02 E1E20 (FCC Rules) You must
Page 17 and 18: Rev 2.02 SUBELEMENT E2 - OPERATING
Page 19 and 20: Rev 2.02 E2A13 Geosynchronous satel
Page 21 and 22: Rev 2.02 E2B17 Immunity from fading
Page 23 and 24: Rev 2.02 E2D02 The definition of
Page 25 and 26: Rev 2.02 SUBELEMENT E3 -- RADIO WAV
Page 27 and 28: Rev 2.02 E3C01 Auroral activity cau
Page 29 and 30: Rev 2.02 E4A06 A spectrum analyzer
Page 31 and 32: Rev 2.02 E4B09 High impedance input
Page 33 and 34: Rev 2.02 E4C06 If t he thermal nois
Page 35 and 36: Rev 2.02 E4D11 Third-order intermod
Page 37 and 38: Rev 2.02 SUBELEMENT E5 -- ELECTRICA
Page 39 and 40: Rev 2.02 E5B Time constants and pha
Page 41 and 42: Rev 2.02 Example 2: To find the ang
Page 43 and 44: Rev 2.02 j 50 -j 25 E5B12 The phase
Page 45: Rev 2.02 j 600 -j300 E5C04 In polar
Page 49 and 50: Rev 2.02 R=300 Ω XL= (2 π FL) or
Page 51 and 52: Rev 2.02 Power Factor represents th
Page 53 and 54: Rev 2.02 SUBELEMENT E6 -- CIRCUIT C
Page 55 and 56: Rev 2.02 Field-effect transistors e
Page 57 and 58: Rev 2.02 E6B12 Junction diodes are
Page 59 and 60: Rev 2.02 E6D02 Cathode ray tube (CR
Page 61 and 62: Rev 2.02 E6E04 The technique used t
Page 63 and 64: Rev 2.02 E6F07 Cadmium sulfide will
Page 65 and 66: Rev 2.02 E7A07 An AND gate produces
Page 67 and 68: Rev 2.02 When a Class C rather than
Page 69 and 70: Rev 2.02 E7C01 In a low-pass filter
Page 71 and 72: Rev 2.02 E7D02 One characteristic o
Page 73 and 74: Rev 2.02 E7E05 A pre-emphasis netwo
Page 75 and 76: Rev 2.02 E7F08 One purpose of a mar
Page 77 and 78: Rev 2.02 E7G17 The typical output i
Page 79 and 80: Rev 2.02 E7H19 A phase-locked loop
Page 81 and 82: Rev 2.02 E8A06 The approximate rati
Page 83 and 84: Rev 2.02 E8B13 In time division mul
Page 85 and 86: Rev 2.02 E8D07 An electromagnetic w
Page 87 and 88: Rev 2.02 E9A10 Antenna bandwidth is
Page 89 and 90: Rev 2.02 E9B10 The “Method of Mom
Page 91 and 92: Rev 2.02 E9C09 The elevation angle
Page 93 and 94: Rev 2.02 E9D09 An advantage of usin
Page 95 and 96: Rev 2.02 E9E08 An SWR greater than
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Rev 2.02 E9F11 A 1/8-wavelength tra
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Rev 2.02 E9H05 The main drawback of
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Rev 2.02 SUBELEMENT E0 -- SAFETY [1
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Rev 2.02 Jack Tiley AD7FO Page 103
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Rev 2.02 Capacitors in series C= __
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Rev 2.02 Impedance of complex serie
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Rev 2.02 Calculating Component valu
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Rev 2.02 AC Voltage Calculations Th
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