Digital Modulation.pdf
- 1. 1 Digital modulation: ASK, PSK, BPSK, QPSK, FSK, MSK
- 2. 2 Modulation of Digital Data ▪ Carrier Signal - High-frequency signal used for digital-to-analog or analog-to-digital modulation ▪ May change any aspect of an analog signal to represent digital data: amplitude, frequency, and phase ▪ Modulation - process of changing one of the characteristics of an analog signals based on information in digital signal
- 4. Types of methods ▪ Amplitude shift keying (ASK): some audio modem, optical fiber, printer ▪ Frequency shift keying (FSK): Computer Modem, HF and VHF radio ▪ Amplitude shift keying (PSK): Mobile, Satellite (microwave) etc Digital to Analog Modulation FSK QAM PSK ASK Bit rate is the number of bits transmitted per second. Baud rate is the number of signal elements transmitted per second. Baud rate is less than or equal to the bit rate
- 5. 5 Amplitude Shift Keying (ASK) ▪ Strength of carrier signal is varied to represent a 1 or 0 ▪ Frequency and phase remain constant while amplitude is changed ▪ Amplitude remains constant during bit duration ▪ Highly susceptible to noise interference since noise usually affects amplitude ▪ Minimum bandwidth required is equal to baud rate
- 6. 6 Amplitude Shift Keying (ASK)
- 7. 7 Relationship between baud rate and bandwidth in ASK BW = (1+d)×Nbaud
- 8. 8 Example 3 Find the minimum bandwidth for an ASK signal transmitting at 2000 bps. The transmission mode is half-duplex. Solution In ASK the baud rate and bit rate are the same. The baud rate is therefore 2000. An ASK signal requires a minimum bandwidth equal to its baud rate. Therefore, the minimum bandwidth is 2000 Hz.
- 9. 9 Frequency Shift Keying (FSK) ▪ Frequency is varied to represent a 1 or 0 ▪ Frequency during bit duration is constant ▪ Amplitude and phase remain constant ▪ Avoids most of noise problems of ASK; can ignore voltage spikes ▪ Limited by physical capabilities of medium ▪ Bandwidth required is equal to baud rate plus frequency shift
- 10. 10 Frequency Shift Keying (FSK)
- 11. 11 Relationship between baud rate and bandwidth in FSK BW = fc1 -fc0 +Nbaud
- 12. 12 Example 6 Find the minimum bandwidth for an FSK signal transmitting at 2000 bps. Transmission is in half-duplex mode, and the carriers are separated by 3000 Hz. Solution For FSK BW = baud rate + fc1 − fc0 BW = bit rate + fc1 − fc0 = 2000 + 3000 = 5000 Hz
- 13. 13 Phase Shift Keying (PSK) ▪ Phase of carrier is varied to represent 1 or 0 ▪ Peak amplitude and frequency remain constant ▪ Phase remains constant during each bit duration ▪ Not as susceptible to noise degradation as ASK or to bandwidth limitations of FSK
- 14. 14 Phase Shift Keying (PSK) 2-PSK or Binary PSK
- 15. 15 PSK constellation (Phase state diagram) Constellation is a graphical representation of the phase and amplitude of different bit combinations in digital-to-analog modulation.
- 16. 16 4-PSK; 8-PSK ▪ 4-PSK May utilize four variations of phase shift by 90 degrees Each phase shift represents 2 bits (dibit); technique is referred to as 4-PSK Allows data transmission two times as fast as 2-PSK ▪ 8-PSK Vary signal by shifts of 45 degrees; each shift may then represent three bits (tribit) and send data three times as fast
- 17. 17 The 4-PSK method (Q-PSK)
- 20. 20 Relationship between baud rate and bandwidth in PSK BW = (1+d)×Nbaud
- 21. 21 Example 8 Find the bandwidth for a 4-PSK signal transmitting at 2000 bps. Transmission is in half-duplex mode. Solution For 4-PSK the baud rate is one –half of the bit rate (1 baud : 2 bit). The baud rate is therefore 1000. A PSK signal requires a bandwidth equal to its baud rate . Therefore, the bandwidth is 1000 Hz
- 22. 22 Example 9 Given a bandwidth of 5000 Hz for an 8-PSK signal, what are the baud rate and bit rate? Solution For PSK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps.
- 23. 23 Quadrature Amplitude Modulation (QAM) ▪ Combination of ASK and PSK so that a maximum contrast between each signal unit (bit, dibit, tribit, and so on) is achieved ▪ x number of variations in phase and y variations in amplitude ▪ Number of phase shifts is always larger than number of amplitude shifts due to amplitude susceptibility to noise ▪ QAM is therefore less susceptible to noise than ASK ▪ Same bandwidth is required for ASK and PSK
- 24. 24 Credits ▪ All figures obtained from Data Communications and Networking, 3rd edition by Behrouz A. Forouzan. McGraw Hill Publishing, 2004