# 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

$f_{bQ}=f_{bI}=\frac{f_{b}}{2}=\frac{10\,Mbps}{2}=5\,Mbps$

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 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 – 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.

Question: 2.15

## For a QPSK system and the given parameters, determine a. Carrier power in dBm. b. Noise power in dBm. c. Noise power density in dBm. d. Energy per bit in dBJ. e. Carrier-to-noise power ratio in dB. f. Eb/N0 ratio. C = 10^-12 W fb = 60 kbps N = 1.2 × 10^-14 W B = 120 kHz ...

a.  The carrier power in dBm is determined by subs...
Question: 2.13

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Question: 2.11

## For a 16-QAM modulator with an input data rate (fb) equal to 10 Mbps and a carrier frequency of 70 MHz, determine the minimum double-sided Nyquist frequency (fN) 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 ...

The bit rate in the I, I′ , Q, and Q′ channels is ...
Question: 2.10

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Question: 2.9

## For a tribit input of Q = 0, I = 0, and C = 0 (000), determine the output amplitude and phase for the 8-QAM transmitter shown in Figure 30a. ...

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Question: 2.8

## For an 8-PSK modulator with an input data rate (fb) equal to 10 Mbps and a carrier frequency of 70 MHz, determine the minimum double-sided Nyquist bandwidth (fN) 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 ...

The bit rate in the I, Q, and C channels is equal ...
Question: 2.5

## For the QPSK modulator shown in Figure 17, construct the truth table, phasor diagram, and constellation diagram. ...

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Question: 2.7

## For a tribit input of Q = 0, I = 0, and C = 0 (000), determine the output phase for the 8-PSK modulator shown in Figure 23. ...

The inputs to the I channel 2-to-4-level converter...
Question: 2.17

## Which system requires the highest Eb/N0 ratio for a probability of error of 10^-6, a four-level QAM system or an 8-PSK system? ...

From Figure 49, the minimum E_{b}/N_{0}[/la...
From Figure 48, the minimum E_{b}/N_{0}[/la...