Final answer:
To send data 3 bits at a time at 6 Mbps using FSK, 8 different levels or frequencies are required, with a baud rate of 2 Mbps. The bandwidth needed is greater than 2 times the baud rate, although the exact bandwidth depends on other specifics like frequency deviation, which is not provided.
Step-by-step explanation:
The question requires the calculation of the number of levels, baud rate, and bandwidth required to send data using frequency-shift keying (FSK). FSK is a modulation scheme where different frequencies represent different data values.
To send data 3 bits at a time at a bit rate of 6 Mbps, the number of different frequencies, also known as levels, can be calculated using the formula 2^n, where n is the number of bits sent at a time. Therefore, 2^3 equals 8 different levels or frequencies. The baud rate, which is the number of signal changes or symbols per second, needs to be derived from the bit rate. Because each symbol represents 3 bits, the baud rate is 2 Mbps (6 Mbps divided by 3 bits per symbol).
The bandwidth required for FSK can be estimated with Carson's rule: Bandwidth (BW) = 2 x (Delta f + baud rate). However, since the exact frequency deviation (Delta f) isn't provided, we cannot calculate the exact bandwidth. We can assume it needs to be greater than 2 x baud rate to prevent interference. Figures typically cannot be shown without use of graphics, so a drawing is not possible in this response.