A QPSK modulator has an input data rate of 80 Mbps and a carrier frequency of 60 MHz. The minimum double sided Nyquist bandwidth is ________ Mbps.
80

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Solution

The correct option is A 80
$B = \frac{2 R_{b}}{{log}_{2} M} = \frac{2 \times 80}{{log}_{2} 4} = 80 M b p s$

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A QPSK modulator has an input data rate of 80 Mbps and a carrier frequency of 60 MHz. The minimum double sided Nyquist bandwidth is ________ Mbps.80

The correct option is A 80B=2Rblog2M=2×80log24=80 Mbps

A QPSK modulator has an input data rate of 80 Mbps and a carrier frequency of 60 MHz. The minimum double sided Nyquist bandwidth is ________ Mbps.
80

The correct option is A 80
$B = \frac{2 R_{b}}{{log}_{2} M} = \frac{2 \times 80}{{log}_{2} 4} = 80 M b p s$