For a two-way splitter with 132 feet of Series 6 cable, how much signal is needed from the subscriber's tap port @ 750 MHz to ensure each TV set receives a minimum of +3.0 dBmV?

To determine how much signal is required from the subscriber's tap port at 750 MHz for each TV set to receive a minimum of +3.0 dBmV, we need to consider both the signal loss through the cable and the splitter.

Using Series 6 coaxial cable, which typically has a loss of approximately 2.5 dB per 100 feet at 750 MHz, we can calculate the total cable loss for 132 feet. The loss for this length of cable can be calculated as follows:

• Cable loss = (132 feet / 100 feet) * 2.5 dB = 3.3 dB.

Next, we must consider the additional loss from the two-way splitter, which is usually around 3.5 dB as it divides the signal between two outputs.

Therefore, the combined loss from the cable and splitter will be:

• Total loss = Cable loss (3.3 dB) + Splitter loss (3.5 dB) = 6.8 dB.

To ensure that each TV receives a minimum of +3.0 dBmV, we need to account for this total loss. We can calculate the required signal level at the tap port:

• Minimum signal