If a transmission line has a power loss of 6 dB per 100 feet, how much power is available at the feed point of a 200 foot line fed by a 100 watt transmitter?

To determine the available power at the feed point of a 200-foot transmission line with a loss of 6 dB per 100 feet, we must first understand how dB represents power loss. A loss of 6 dB corresponds to a reduction in power by a factor of 4 (because every 3 dB represents a power change by a factor of approximately 2, and thus, 6 dB represents 2 x 2 = 4).

Starting with the transmitter power of 100 watts, we need to calculate the power lost over 200 feet. Since the transmission line loss is given as 6 dB for every 100 feet, the total power loss for 200 feet is:

6 dB + 6 dB = 12 dB.

Now, we need to find out how much power remains after this loss. A 12 dB loss reduces the power by a factor of:

10^(12/10) = 10^(1.2) ≈ 15.85.

To find the available power at the feed point, we take the initial power output from the transmitter and divide it by the loss factor:

Available power = 100 watts / 15.85 ≈