## Q. 2.11

For a 16-QAM 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 frequency ($f_{N}$) and the baud. Also, compare the results with those achieved with the BPSK, QPSK, and 8-PSK modulators in Examples 4, 6, and 8. Use the 16-QAM block diagram shown in Figure 33 as the modulator model. ## Verified Solution

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

$f_{bI}=f_{bI^′}=f_{bQ}=f_{bQ^′}=\frac{f_{b}}{4}=\frac{10\,Mbps}{4}=2.5\,Mbps$

Therefore, the fastest rate of change and highest fundamental frequency presented to either balanced modulator is

$f_{a}=\frac{f_{bI}}{2}$ or $\frac{f_{bI^′}}{2}$ or  $\frac{f_{bQ}}{2}$ or $\frac{f_{bQ′}}{2}=\frac{2.5\,Mbps}{2}=1.25\,Mbps$

The output wave from the balanced modulator 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 – 1.25) MHz]t – $\frac{1}{2}$cos 2π[(70 + 1.25) MHz]t

$\frac{1}{2}$cos 2π(68.75 MHz)t – $\frac{1}{2}$cos 2π(71.25 MHz)t

The minimum Nyquist bandwidth is
B = (71.25 – 68.75) MHz = 2.5 MHz

The minimum bandwidth for the 16-QAM can also be determined by simply substituting into Equation 10:

$B=\frac{f_{b}}{N}$          (10)
$B=\frac{10\,Mbps}{4}$
=2.5 MHz

The symbol rate equals the bandwidth; thus,

symbol rate = 2.5 megabaud

The output spectrum is as follows:

B = 2.5 MHz
For the same input bit rate, the minimum bandwidth required to pass the output of a 16-QAM modulator is equal to one-fourth that of the BPSK modulator, one-half that of QPSK, and 25% less than with 8-PSK. For each modulation technique, the baud is also reduced by the same proportions. 