# Question 2.8: For an 8-PSK modulator with an input data rate (fb) equal to......

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

$f_{bC}=f_{bQ}=f_{bI}=\frac{10\,Mbps}{3}=3.33\,Mbps$

Therefore, 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 2π$f_{a}$t)(sin 2π$f_{c}$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.

Question: 2.15

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Question: 2.13

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Question: 2.11

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Question: 2.10

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Question: 2.9

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Question: 2.6

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Question: 2.17

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