How much signal is needed from the tap port @ 750 MHz to provide a minimum of +3.0 dBmV to each TV when there is 132 feet of drop and a splitter?

To determine how much signal is needed from the tap port to ensure that each TV receives a minimum of +3.0 dBmV, you must account for the losses introduced by the splitter and the coaxial cable drop over a distance of 132 feet.

In coaxial cable, the signal loses strength over distance due to attenuation, which is typically measured in dB per foot. At a frequency of 750 MHz, the attenuation might be around 1 dB for approximately every 100 feet of RG-6 coaxial cable. Therefore, over 132 feet, the expected attenuation would likely be in the range of about 1.3 dB.

Additionally, if there is a splitter involved in the setup, it usually introduces an additional loss. A typical 2-way splitter introduces a loss of about 3.5 dB.

So, to calculate the total signal level required from the tap port, you would take the desired signal level at each TV (+3.0 dBmV), add the splitter loss (approximately 3.5 dB), and add the cable loss (approximately 1.3 dB). This gives you:

• Desired signal at each TV: +3.0 dBmV