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Wireless Communication Basics

With 16QAM at a symbol rate of 5 M symbols per second (5 Mbaud), which is the bit rate?

  1. A1.25 Mbit/s
  2. B5 Mbit/s
  3. C20 Mbit/s
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Answer: C. 20 Mbit/s

16QAM sends log₂ 16 = 4 bits per symbol. The bit rate is "symbol rate × bits per symbol", so it is 5 M × 4 = 20 Mbit/s. The symbol rate (baud rate) is the number of symbols sent per second, and with multilevel modulation it does not equal the bit rate. Because the occupied bandwidth is set mainly by the symbol rate, using more levels raises the bit rate without widening the bandwidth.

  • A. This divides the symbol rate by 4. Each symbol carries 4 bits, so the bit rate is four times the symbol rate.
  • B. This treats the symbol rate and bit rate as the same, which holds only for BPSK, which sends 1 bit per symbol.
  • C. Correct. 5 M × 4 = 20 Mbit/s.
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More Wireless Communication Basics questions

  1. Which is a general property when the modulation is changed from 16QAM to 256QAM at the same symbol rate?
  2. For a channel with a bandwidth of 20 MHz and an S/N (signal-to-noise ratio) of 30 dB, which is closest to the channel capacity (the theoretical upper limit on the error-free transmission rate) given by Shannon's theorem?
  3. Which describes OFDM (orthogonal frequency-division multiplexing)?
  4. In OFDM, which is the main purpose of the cyclic prefix (CP), formed by copying part of the end of each symbol to its beginning?
  5. In OFDM, the symbol length excluding the CP is the reciprocal of the subcarrier spacing. When the subcarrier spacing is widened from 15 kHz to 60 kHz, which is closest to the symbol length excluding the CP?