9hw

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9-2. Determine the maximum bit rate for an FSK signal with a mark frequency of 48 kHz, a space frequency of 52 kHz, and an available bandwidth of 10 kHz. 3 kbps 9-4. Determine The maximum bit rate for an FSK signal with a mark frequency of 102 kHz, a space frequency of 104 kHz, and an available bandwidth of 8 kHz. 3 kbps 9-6. For the QPSK modulator shown in Figure 9-17, change the + 90 ° phase-shift network to - 90 ° and sketch the new constellation diagram. Q I Phase 0 0 +135 0 1 +45 1 0 -135 1 1 -45 9-8. For an 8-PSK modulator with an input data rate (fi) equel to 20 Mbps and a carrier frequency of 100 Mhz, determine the minimum double-sided Nyquist bandwidth (fN) and the baud. Sketch the output spectrum. 6.667 MHz, 6.667 Mbaud 9-10. For a 16-QAM modulator with an input bit rate ( f b ) equal to 20 Mbps and a carrier frequency of 100 MHz, determine the minimum double-sided Nyquist bandwidth (f N ) and the baud. 5 MHz, 5 Mbaud Sketch the output spectrum. 9-12. Determine the bandwidth efficiency for the following modulators, a. QPSK, fb = 10 Mbps 2 bps/Hz b. 8-PSK, fb = 21 Mbps 3 bps/Hz c. 16-PSK, fb = 20 Mbps 4 bps/Hz 9-14. 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. f. E b /N O ratio. C = 10-13 W f b = 30 kbps N = 0.06 × I0 -15 W B = 60kHz a. C = - 100 dBm b. N = -132.2 dBm c. N o = -180 dBm d. E b = -174.77 dBJ e. C/N = 32.23 dB f. E b /N o = 35.23 dB

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9hw

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Page 1: 9hw

9-2. Determine the maximum bit rate for an FSK signal with a mark frequency of 48 kHz, aspace frequency of 52 kHz, and an available bandwidth of 10 kHz.

3 kbps

9-4. Determine The maximum bit rate for an FSK signal with a mark frequency of 102 kHz, aspace frequency of 104 kHz, and an available bandwidth of 8 kHz.

3 kbps 9-6. For the QPSK modulator shown in Figure 9-17, change the + 90° phase-shift network to- 90° and sketch the new constellation diagram.

Q I Phase 0 0 +1350 1 +451 0 -1351 1 -45

9-8. For an 8-PSK modulator with an input data rate (fi) equel to 20 Mbps and a carrierfrequency of 100 Mhz, determine the minimum double-sided Nyquist bandwidth (fN) and thebaud. Sketch the output spectrum.

6.667 MHz, 6.667 Mbaud

9-10. For a 16-QAM modulator with an input bit rate (fb) equal to 20 Mbps and a carrierfrequency of 100 MHz, determine the minimum double-sided Nyquist bandwidth (fN) and thebaud.

5 MHz, 5 Mbaud

Sketch the output spectrum.

9-12. Determine the bandwidth efficiency for the following modulators, a. QPSK, fb = 10 Mbps 2 bps/Hzb. 8-PSK, fb = 21 Mbps 3 bps/Hzc. 16-PSK, fb = 20 Mbps 4 bps/Hz

9-14. For a QPSK system and the given parameters, determinea. 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. f. Eb/NO ratio.

C = 10-13 W fb = 30 kbpsN = 0.06 × I0-15 W B = 60kHz

a. C = - 100 dBmb. N = -132.2 dBmc. No = -180 dBmd. Eb = -174.77 dBJe. C/N = 32.23 dBf. Eb/No = 35.23 dB

Page 2: 9hw

9-16. Determine the minimum bandwidth and baud for a BPSK modulator with a carrierfreqency of 80 Mhz and an input bit rate Fb = 1Mbps. Sketch the output spectrum. B = 1 MHz, 1 Mbaud

9-18. For the QPSK demodulator shown in Figure 9-21, determine the I and Q bits for an inputsignal -sin ct + cos ct

I = 0, Q = 1

9-20. For an 8-PSK modulator shown in Figure 9-23, change the +90 phase-shift network to a-90 phase shifter and sketch the new constellation diagram.

Q I C Phase 0 0 0 +112.5 0 0 1 +157.5 0 1 0 +67.5 0 1 1 +22.5 1 0 0 -112.5 1 0 1 -157.5 1 1 0 -67.5

1 1 1 -22.5

9-22. For the 16-QAM modulator shown in Figure 9-33, change the 90° phase shift network to a- 90° phase shifter and determine the output expressions for the following I, I', Q, and Q' inputconditions: 0000, 1111, 1010, and 0101.

Q Q’ I I’ Phase 0 0 0 0 +1351 1 1 1 -451 0 1 0 -450 1 0 1 +135

9-24. For the DPSK modulator shown in Figure 9-40a, determine the output phase sequence forthe following input bit sequwnxw: 11001100101010 (assume the reference bit is a logic 1).