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Question 2.6: For a QPSK modulator with an input data rate (fb) equal to 1......

For a QPSK modulator with an input data rate (f_{b}) equal to 10 Mbps and a carrier frequency of 70 MHz, determine the minimum double-sided Nyquist bandwidth (f_{N}) and the baud. Also, compare the results with those achieved with the BPSK modulator in Example 4. Use the QPSK block diagram shown in Figure 17 as the modulator model.

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The bit rate in both the I and Q channels is equal to one-half of the transmission bit rate, or


The highest fundamental frequency presented to either balanced modulator is

f_{a}=\frac{f_{bQ}}{2} or \frac{f_{bI}}{2}=\frac{5\,Mbps}{2}=2.5\,MHz

The output wave from each balanced modulator is

(sin 2πf_{a}t)(sin 2πf_{c}t)

\frac{1}{2}cos 2π(f_{c}\, – \,f_{a})t – \frac{1}{2}cos 2π(f_{c}\, +\, f_{a})t

\frac{1}{2}cos 2π[(70 – 2.5) MHz]t – \frac{1}{2}cos 2π[(70 + 2.5) MHz]t

\frac{1}{2}cos 2π(67.5 MHz)t – \frac{1}{2}cos 2π(72.5 MHz)t

The minimum Nyquist bandwidth is
B = (72.5 – 67.5) MHz = 5 MHz
The symbol rate equals the bandwidth; thus,
symbol rate = 5 megabaud
The output spectrum is as follows:

B = 5 MHz

It can be seen that for the same input bit rate the minimum bandwidth required to pass the output of the QPSK modulator is equal to one-half of that required for the BPSK modulator in Example 4. Also, the baud rate for the QPSK modulator is one-half that of the BPSK modulator.


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