A digital data signaling system is required to operate at 9600bps If a signal element encodes a 3-bit word what is the minimum required bandwidth of the channel?

To determine the minimum required bandwidth for a digital signaling system operating at 9600 bps with each signal element encoding a 3-bit Word, we can use the Nyquist formula for maximum data rate: ( R = 2B \log_2(M) ), where ( R ) is the data rate, ( B ) is the bandwidth, and ( M ) is the number of discrete signal levels. Given that each signal element encodes 3 bits, ( M = 2^3 = 8 ). Rearranging the formula gives ( B = R / (2 \log_2(M)) = 9600 / (2 \cdot 3) = 1600 ) Hz. Therefore, the minimum required bandwidth of the channel is 1600 Hz.