An electromagnetic wave is a wave consisting of an electric field and a magnetic field oscillating at right angles to each other. (E8D07) Changing electric and magnetic fields propagate the energy is a phrase that best describes electromagnetic waves traveling in free space. (E8D08)

The polarization of an electromagnetic wave is related to the orientation of the wave’s electric field. If, for example, the electric field is oriented vertically, we say that the electromagnetic wave is vertically polarized. Waves with a rotating electric field are called circularly polarized electromagnetic waves.(E8D09)

Peak-to-peak voltage is the easiest voltage amplitude parameter to measure when viewing a pure sine wave signal on an analog oscilloscope. (E8D01) The relationship between the peak-to-peak voltage and the peak voltage amplitude of a symmetrical waveform is 2:1. (E8D02) Peak voltage is a valuable input-amplitude parameter for evaluating the signal-handling capability of a Class A amplifier.(E8D03)

For sinusoidal voltages, the peak voltage is 1.414 times the RMS voltage, and the peak-to-peak voltage is 2.828 times the RMS voltage. The peak voltage of a sinusoidal waveform would be 48 volts if an RMS-reading voltmeter reads 34 volts. (E8D12) If an RMS-reading AC voltmeter reads 65 volts on a sinusoidal waveform, the peak-to-peak voltage is 184 volts. (E8D05) 

120V AC is a typical value for the RMS voltage at a standard U.S. household electrical power outlet. (E8D15) 170 volts is a typical value for the peak voltage at a standard U.S. household electrical outlet. (E8D13) 340 volts is a typical value for the peak-to-peak voltage at a standard U.S. household electrical outlet. (E8D14) 120V AC is the RMS value of a 340-volt peak-to-peak pure sine wave. (E8D16)

The peak envelope power of a radio signal is equal to V²_peak/2 x 1/R. Consequently, the PEP output of a transmitter that develops a peak voltage of 30 volts into a 50-ohm load is 9 watts. (E8D04)

V_peak = 30 V, V²_peak = 900 V²

PEP = 900 V² / 2 x 50 = 9 W.

The average power of a radio signal is equal to V²_RMS/R. The average power dissipated by a 50-ohm resistive load during one complete RF cycle having a peak voltage of 35 volts is 12.2 watts. (E8D11)

V²_RMS = 35 V / 1.414 = 24.75V

V²_RMS = 612 V²

P_avg = 612 V² / 50 = 12.2 W.

Radio amateurs most often specify the output power of a single-sideband transmitter as peak envelope power and use a peak-reading wattmeter.  The advantage of using a peak-reading wattmeter to monitor the output of a SSB phone transmitter is that it gives a more accurate display of the PEP output when modulation is present. (E8D06) A peak-reading wattmeter should be used to monitor the output signal of a voice-modulated single-sideband transmitter to ensure you do not exceed the maximum allowable power. (E8D10)

Admittance is the inverse of impedance. So, in polar coordinates, the impedance of a circuit that has an admittance of 7.09 millisiemens at 45 degrees is 141 ohms at an angle of -45 degrees. (E5C16) You calculate it this way:

|Z| = 1/7.09×10^-3 = 141 ohms

The angle is the mirror image about the x axis:

θ = 0 – -45 degrees = 45 degrees

Let’s look at another example. In rectangular coordinates, the impedance of a circuit that has an admittance of 5 millisiemens at -30 degrees is 173 +j100 ohms. (E5C17)

|Z| = 1/5×10^-3 = 200 ohms

θ = 0 – -30 degrees = 30 degrees

R = |Z| × cos 30 degrees = 200 × .866 = 173 ohms

X (the reactance part of the impedance) = |Z| × sin 30 degrees = 200 × .5 = +j100

Now, let’s take a look at some actual circuits.

On Figure E5-2, the point that best represents the impedance of a series circuit consisting of a 400 ohm resistor and a 38 picofarad capacitor at 14 MHz is Point 4. (E5C19) Right off the bat, we know that the only choices are really Points 2, 4, and 6 because the resistance is 400 ohms. Next, we calculate the capacitive reactance:

X_C = 1/2πfC = 1/(2 × 3.14 × 14×10⁶ × 38×10^-12) ≈ 300 ohms

Because the reactance is capacitive, it’s plotted as a negative value.

On Figure E5-2, the point that best represents the impedance of a series circuit consisting of a 300 ohm resistor and an 18 microhenry inductor at 3.505 MHz is Point 3. (E5C20) The resistance is 300 ohms and the reactance is:

X_L = 2πfL = 2 × 3.14 × 3.505×10⁶ × 18×10^-6) ≈ 400 ohms

And, since the reactance is inductive, it’s plotted as a postive value.

On Figure E5-2, the point that best represents the impedance of a series circuit consisting of a 300 ohm resistor and a 19 picofarad capacitor at 21.200 MHz is Point 1. (E5C21) The resistance is 300 ohms, and the reactance is:

X_C = 1/2πfC = 1/(2 × 3.14 × 21.2×10⁶ × 19×10^-12) ≈ 400 ohms

Because the reactance is capacitive, it’s plotted as a negative value.

On Figure E5-2, the point that best represents the impedance of a series circuit consisting of a 300-ohm resistor, a 0.64-microhenry inductor and an 85-picofarad capacitor at 24.900 MHz is Point 8. (E5C23) This problem is a little tougher because it has both capacitive and inductive reactance.

X_C = 1/2πfC = 1/(2 × 3.14 × 29.4×10⁶ × 85×10^-12) ≈ 63.7 ohms

X_L = 2πfL = 2 × 3.14 × 29.4×10⁶ × 0.64×10^-6) ≈ 118.2 ohms

X = X_L – X_C = 118.2 – 63.7 = 55.5 ohms

Because the net reactance is inductive, it is plotted as a positive value, and because the resistance is 300 ohms, the answer is Point 8.

Most often when we plot values on a graph, we use the rectangular, or Cartesian, coordinate system. The two numbers that are used to define a point on a graph using rectangular coordinates are the coordinate values along the horizontal and vertical axes. (E5C11) In the graph above, point P is at x,y. Rectangular coordinates are often used to display the resistive, inductive, and/or capacitive reactance components of an impedance. (E5C13)

When thinking about how capacitive reactances, inductive reactances, and resistance combine, it’s useful to think in terms of polar coordinates. Polar coordinates are often used to display the phase angle of a circuit containing resistance, inductive and/or capacitive reactance. (E5C14) In a polar-coordinate system, each point on the graph has two values, a magnitude (shown by r in the figure above) and an angle (shown by θ in the figure above).

When using rectangular coordinates to graph the impedance of a circuit, the vertical axis represents the reactive component. (E5C10) To figure out the impedance of a circuit, you first plot the inductive reactance on the positive y-axis and the capacitive reactance on the negative y-axis. The net reactance, X, will be the sum of the two reactances.

When using rectangular coordinates to graph the impedance of a circuit, the horizontal axis represents the resistive component. (E5C09) After you’ve computed the net reactance, you plot the resistance on the x-axis and compute the magnitude of the impedance, shown by r in the graph above. If you consider that r is the third side of a right triangle made up of the sides r, x, and y, r is equal to the square root of x² and y².

Let’s take a look at an example. In polar coordinates, is the impedance of a network consisting of a 100-ohm-reactance inductor in series with a 100-ohm resistor is 141 ohms at an angle of 45 degrees. (E5C01) In this example, x=100 and y=100, so

r = sqrt (X² + R²) = sqrt (100² + 100²) = sqrt (20000) = 141 ohms.

The cosine of the phase angle θ is equal to x/r, or 100/141, or .707.  If you look up this value in a table of cosines, you’ll find that the angle is 45 degrees.

Here’s another thing to notice. When the value of the reactance is equal to the value of the resistance, the angle will be either 45 degrees or -45 degrees, depending on whether the net reactance is inductive or capacitive.

Now, let’s look at an example with both inductive and capacitive reactance. In polar coordinates, the impedance of a network consisting of a 100-ohm-reactance inductor, a 100-ohm-reactance capacitor, and a 100-ohm resistor, all connected in series is 100 ohms at an angle of 0 degrees. (E5C02) In this case, the inductive reactance and the capacitive reactance are the same, meaning that there is no net reactance. 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 axis, you know that the circuit impedance is equivalent to a pure resistance. (E5C12)

Here’s an example with unequal inductive and capacitive reactances. In polar coordinates, the impedance of a network consisting of a 300-ohm-reactance capacitor, a 600-ohm-reactance inductor, and a 400-ohm resistor, all connected in series is 500 ohms at an angle of 37 degrees. (E5C03) Here’s how we got that result:

X = 600 – 300 = 300 ohms

r = sqrt (X² + R²) = sqrt (300² + 400²) = sqrt (250000) = 500 ohms

θ = cos^-1(x/r) = cos^-1(400/500) = 37 degrees

Here are some more examples. I’ll leave the solutions up to you:

• In polar coordinates, the impedance of a network consisting of a 400-ohm-reactance capacitor in series with a 300-ohm resistor is 500 ohms at an angle of -53.1 degrees. (E5C04)
• In polar coordinates, the impedance of a network consisting of a 400-ohm-reactance inductor in parallel with a 300-ohm resistor is 240 ohms at an angle of 36.9 degrees. (E5C05)
• In polar coordinates, the impedance of a network consisting of a 100-ohm-reactance capacitor in series with a 100-ohm resistor is 141 ohms at an angle of -45 degrees. (E5C06)
• In polar coordinates, the impedance of a network comprised of a 100-ohm-reactance capacitor in parallel with a 100-ohm resistor is 71 ohms at an angle of -45 degrees. (E5C07)
• In polar coordinates, what is the impedance of a network comprised of a 300-ohm-reactance inductor in series with a 400-ohm resistor is 500 ohms at an angle of 53 degrees. (E5C08)
• In polar coordinates, the impedance of a series circuit consisting of a resistance of 4 ohms, an inductive reactance of 4 ohms, and a capacitive reactance of 1 ohm is 5 ohms at an angle of 37 degrees. (E5C18)