The current Wi-Fi 6 protocol modulation method already supports 1024 QAM. Compared with 256- QAM, how much is the physical layer link establishment rate increased by 1024-QAM?

Question

Options

• A1
• B0.5
• C0.25
• D0.2

• A
  7% (4)
• B
  3% (2)
• C
  76% (45)
• D
  14% (8)

Explanation

Option C (0.25) is correct because 256-QAM encodes 8 bits per symbol (2⁸ = 256) while 1024-QAM encodes 10 bits per symbol (2¹⁰ = 1024). The increase is (10 − 8) / 8 = 2/8 = 0.25, meaning a 25% throughput gain at the physical layer.

Why the distractors are wrong:

• A (1.0) implies a 100% increase (doubling), which would require moving from 8 to 16 bits/symbol - that's 65536-QAM, not 1024-QAM.
• B (0.5) implies a 50% increase (8 → 12 bits/symbol), which corresponds to no standard QAM level.
• D (0.2) implies 20% (8 → 9.6 bits/symbol), also not achievable with integer-power-of-2 QAM levels.

Memory tip: Express both QAM orders as powers of 2, take the exponents (the bit depths), and apply simple percentage-change math: (new bits − old bits) / old bits. For any QAM comparison, this gives you the exact physical-layer rate increase.

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