Final answer:
The required bandwidth for 64-QAM to send 2000 bps, assuming d equals 1, is approximately 333.33 Hz. This calculation is based on the formula bandwidth = bit rate / (log2M * d), where in this case M equals 64 for 64-QAM, resulting in a division by 6.
Step-by-step explanation:
To calculate the required bandwidth for 64-QAM (Quadrature Amplitude Modulation), you can use the formula bandwidth = (bit rate) / (log2M * d), where M is the modulation order and d is the minimum Euclidean distance between constellation points. In 64-QAM, M=64 which is log264 = 6 (since 26 = 64). Assuming d=1 (typically a normalized value), the formula simplifies to bandwidth = bit rate / 6. Thus, for a bit rate of 2000 bps (bits per second), the required bandwidth would be approximately 333.33 Hz. However, this theoretical bandwidth does not take into account any real-world inefficiencies or the need for guard bands.