Final answer:
The frame size for pulse-stuffing TDM of two channels with bitrates of 190 kbps and 180 kbps is 34,200 bits.
Step-by-step explanation:
The size of a frame in pulse-stuffing time division multiplexing (TDM) can be determined by finding the greatest common divisor of the bitrates of the channels being multiplexed. If we have two channels, one with a bitrate of 190 kbps and another with a bitrate of 180 kbps, we must find a frame size that could accommodate both bitrates efficiently for synchronous transmission.Both 190 and 180 are divisible by 10, so we can express the speeds as 19 kbps and 18 kbps.
The lowest common multiple (LCM) of 19 and 18 totality is used to find a frame size that aligns with both bitrates. The LCM of 19 and 18 is 342. Since the rates are actually in hundreds (as per the original bitrates), we scale up the LCM by a hundred, which gives us a frame size of 34,200 bits.The size of a frame in bits can be calculated by adding the bitrates of the two channels together. In this case, the bitrate of the first channel is 190 kbps and the bitrate of the second channel is 180 kbps. Adding them together, we get a total bitrate of 370 kbps (190 + 180). To convert this bitrate to the size of a frame, we need to divide it by the frame rate. However, the frame rate is not provided in the question, so it is not possible to determine the exact size of a frame without that information.