Ham Radio General Class Practice Test

What is the output PEP from a transmitter if an oscilloscope measures 200 volts peak-to-peak across a 50-ohm dummy load connected to the transmitter output?

1. 1.4 watts

2. 100 watts

3. 353.5 watts

4. 400 watts

The correct answer is: 100 watts

To determine the output PEP (Peak Envelope Power) of the transmitter, one must analyze the measured voltage across the 50-ohm dummy load. The voltage measured is peak-to-peak, so to calculate the PEP, the peak voltage is needed. First, the peak voltage (Vpk) can be derived from the peak-to-peak voltage. The relationship is that the peak voltage is half of the peak-to-peak voltage. By taking the 200 volts peak-to-peak and dividing it by 2, the peak voltage is found to be 100 volts. Next, to calculate the PEP, the formula used is: \[ PEP = \frac{(V_{pk})^2}{R} \] where: - Vpk is the peak voltage (100 volts) - R is the resistance (50 ohms) Inserting the calculated peak voltage into this formula: \[ PEP = \frac{(100)^2}{50} = \frac{10000}{50} = 200 \text{ watts} \] However, when considering the PEP in linear metrics, it is important to note that the value since power is given in terms of RMS voltage and peak power can be discussed