GATE 2022 EC Q.35: solution and answer

GATE 2022 · ECQ.35 · 1 mark · NAT

A symbol stream contains alternate QPSK and 16-QAM symbols. If symbols from this stream are transmitted at the rate of 1 mega-symbols per second, the raw (uncoded) data rate is _______ mega-bits per second (rounded off to one decimal place).

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Answer and solution

ANSWER

The raw data rate is 3.0 mega-bits per second.

Official answer: 2.99 to 3.01Source: Official answer key, GATE 2022 EC (IIT Kharagpur)

Bits per symbol: QPSK has $M = 4$ points, so it carries $lo g_{2} 4 = 2$ bits per symbol. 16-QAM has $M = 16$ points, so it carries $lo g_{2} 16 = 4$ bits per symbol.
The stream alternates QPSK, 16-QAM, QPSK, 16-QAM, and so on. Exactly half of the symbols are QPSK and half are 16-QAM.
Average bits per symbol: $\overset{ˉ}{b} = \frac{1}{2} (2) + \frac{1}{2} (4) = 3$ bits/symbol.
Raw (uncoded) data rate: $R_{b} = R_{s} \overset{ˉ}{b} = 1 \times 1 0^{6} symbols/s \times 3 bits/symbol = 3 \times 1 0^{6} bits/s$ .
So the raw rate is $3.0$ mega-bits per second. Check over one QPSK and 16-QAM pair: $2 + 4 = 6$ bits in $2 μ s$ , which is $3$ Mbit/s.

KEY IDEA

Bits per symbol of an $M$ -ary constellation is $lo g_{2} M$ . With two alternating constellations, average the bits per symbol over the stream, then multiply by the symbol rate.

WATCH OUT FOR

Using only one constellation, for example 4 bits per symbol for 16-QAM (giving 4 Mbit/s) or 2 bits per symbol for QPSK (giving 2 Mbit/s). Another slip is adding the two rates (2 + 4 = 6) instead of averaging them, because only one symbol is sent in each $1 μ s$ slot.

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GATE 2022 EC Q.35: solution and answer

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