For an 8-PSK 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 and QPSK modulators in Examples 4 and 6. Use the 8-PSK block diagram shown in Figure 23 as the modulator model.

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The bit rate in the I, Q, and C channels is equal to one-third of the input bit rate, or

f_{bC}=f_{bQ}=f_{bI}=\frac{10\,Mbps}{3}=3.33\,MbpsTherefore, the fastest rate of change and highest fundamental frequency presented to either balanced modulator is

f_{a}=\frac{f_{bC}}{2} or \frac{f_{bQ}}{2} or \frac{f_{bI}}{2}=\frac{3.33\,Mbps}{2}=1.667\,Mbps

The output wave from the balance modulators 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 – 1.667) MHz]t – \frac{1}{2}cos 2π[(70 + 1.667) MHz]t

\frac{1}{2}cos 2π(68.333 MHz)t – \frac{1}{2}cos 2π(71.667 MHz)t

The minimum Nyquist bandwidth is

B = (71.667 – 68.333) MHz = 3.333 MHz

The minimum bandwidth for the 8-PSK can also be determined by simply substituting into Equation 10:

B=\frac{f_{b}}{N} (10)

B=\frac{10\,Mbps}{3}

=3.33 MHz

Again, the baud equals the bandwidth; thus,

baud = 3.333 megabaud

The output spectrum is as follows:

B = 3.333 MHz

It can be seen that for the same input bit rate the minimum bandwidth required to pass the output of an 8-PSK modulator is equal to one-third that of the BPSK modulator in Example 4 and 50% less than that required for the QPSK modulator in Example 6. Also, in each case the baud has been reduced by the same proportions.

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