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.
The bit rate in both the I and Q channels is equal to one-half of the transmission bit rate, or
f_{bQ}=f_{bI}=\frac{f_{b}}{2}=\frac{10\,Mbps}{2}=5\,MbpsThe 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.
