It is possible.

Because C=BWlog2M

BW is the transmission bandwidth, M is the modulation, and C is the baud rate

It can be achieved when log2M>=3, which means that the minimum use of eight-phase modulation.

For example:

1. According to Nyquist Sampling Theorem:

For an ideal low-communication channel with a bandwidth of W, the maximum codeword transmission rate = 2×W = 2×3000 Hz = 6K Baud.

2. Under this channel, the maximum data rate = 2×W×log28 = 18K bps

3. According to Shannon's theorem: the maximum data rate of a noisy channel with a bandwidth of W and a signal-to-noise ratio of S/N is:

Maximum data rate (bps) = Wlog2(1+S/N)

So the channel's signal-to-noise ratio should be 1023=30db for a data rate of 30Kbps.

Extended information

Extended information:

In the information transmission channel, the signal unit carrying data information is called a code element, the number of code elements transmitted through the channel per second is called the code element transmission rate, referred to as the baud rate, whose unit is the baud (Baud,symbol/s), the baud rate is an indicator of the frequency width of the transmission channel.

The baud rate is an indicator of the frequency width of the transmission channel.

"Baud" (Baud) itself is the rate, so there is no need to write Baud Rate (Rate is superfluous). The unit "Baud" itself already represents the number of modulations per second, with "Baud per second" as the unit is a common error, but in the general Chinese colloquial communication or often to "Baud Rate

Baidu Encyclopedia - Baud Rate