GATE 2026 EC Q.32: solution and answer

GATE 2026 · ECQ.32 · 1 mark · NAT

A wireless digital transmission scheme is using 16-QAM over an additive white Gaussian noise channel and a maximum-likelihood receiver. Consider the information bit rate from source to be $4 \times 1 0^{6}$ bits per second. The minimum transmission bandwidth (in MHz) of the modulated signal necessary for optimum recovery of information at the receiver is ____. (rounded off to two decimal places)

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Answer and solution

ANSWER

The minimum transmission bandwidth is $1.00 MHz$ .

Official answer: 0.95 to 1.05Source: Official answer key, GATE 2026 EC (IIT Guwahati)

16-QAM has $M = 16$ signal points, so each symbol carries $k = lo g_{2} 16 = 4$ bits.
Symbol rate: $R_{s} = \frac{R _{b}}{k} = \frac{4 \times 1 0 ^{6}}{4} = 1 \times 1 0^{6} symbols/s$ .
The smallest bandwidth that allows distortion-free (ISI-free) recovery is the Nyquist limit, reached with zero roll-off. Each of the in-phase and quadrature baseband pulse streams then needs $R_{s} /2$ (one-sided, in Hz).
The modulated signal is a band-pass signal. The two quadrature carriers share the same band, and the band-pass spectrum extends $R_{s} /2$ on each side of the carrier, so $B_{m i n} = 2 \cdot \frac{R _{s}}{2} = R_{s}$ .
$B_{m i n} = 1 \times 1 0^{6} Hz = 1.00 MHz$ .

KEY IDEA

For M-ary QAM the symbol rate is $R_{s} = R_{b} / lo g_{2} M$ . The minimum (Nyquist, zero roll-off) band-pass bandwidth equals the symbol rate, $B_{m i n} = R_{s}$ .

WATCH OUT FOR

Using $B = R_{b}$ (ignoring that each 16-QAM symbol carries 4 bits), or using the baseband figure $R_{s} /2$ , which gives $0.5$ MHz. $R_{s} /2$ is the baseband bandwidth of each of the I and Q streams, not the band-pass bandwidth of the modulated signal.

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GATE 2026 EC Q.32: solution and answer

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