Extra Class question of the day: Waveforms and measurements

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 V2peak/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)

Vpeak = 30 V, V2peak = 900 V2

PEP = 900 V2 / 2 x 50 = 9 W.

The average power of a radio signal is equal to V2RMS/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)

V2RMS = 35 V / 1.414 = 24.75V

V2RMS = 612 V2

Pavg = 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)

Comments

  1. Charles AK4?? says:

    For what it’s worth, I passed my Extra exam this morning in no small part with the help of these Question of the Day articles. The math reviews helped during my study and yesterday I read all the articles one last time. Practice exams were a breeze and the real thing was pretty painless. Thanks Dan.

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