1. 1587: COMMUNICATION SYSTEMS 1 Digital Signals, modulation and noise Dr. George Loukas University of Greenwich, 2011-2012

2. Signals quick revision Music, physics, mechanics and communications depend on signals defined by: amplitude frequency phase V t

3. Signals quick revision Music, physics, mechanics and communications depend on signals defined by: amplitude frequency phase t A -A

4. Signals quick revision Music, physics, mechanics and communications depend on signals defined by: amplitude frequency phase V t

5. Signals quick revision Music, physics, mechanics and communications depend on signals defined by: amplitude frequency phase V t T

6. Signals quick revision Music, physics, mechanics and communications depend on signals defined by: amplitude frequency phase 20 kHz – 20 Hz = ~ 20 kHz ~ 4 kHz ~ 500 MHz

7. Signals quick revision Music, physics, mechanics and communications depend on signals defined by: amplitude frequency phase V t

8. Signals quick revision Music, physics, mechanics and communications depend on signals defined by: amplitude frequency phase V t

9. Converting Analogue to Digital: Sampling Quantisation 00 01 10 11 10 4 quantisation levels (2 bits) 0 1 2 3 01110011100010...

10. Converting Analogue to Digital: Sampling Quantisation 000 001 011010101000110011000100... 010 011 100 101 110 111 8 quantisation levels (3 bits)

11. Converting Analogue to Digital: Sampling Quantisation 000 001 010 011 100 101 110 111 010011011010100101001001100110110100010000010100101... Greater sampling rate

12. Sampling Rate: Nyquist’s theorem To accurately represent the original analogue signal the sampling rate has to be double the maximum frequency in the original signal. example: If is 1 Hz, you must sample at 2 Hz: one sample every

13. Transmission Impairment Attenuation Attenuation distortion Cross-talk noise Delay distortion Impulse noise Inter modulation noise Thermal noise

14. 5 minutes

15. Analogue transmission over a long distance 50 km100 km

16. Digital transmission over a long distance 50 km100 km

18. Multiplexing Multiple users sharing a communications medium FDM: Frequency division multiplexing TDM: Time division multiplexing

19. Frequency Division Multiplexing FDM is achieved by modulation. There are three techniques, all encoding the baseband information onto a high frequency sinusoidal carrier, thus moving the baseband up the frequency spectrum. There are three techniques. AM FM PM

20. Frequency Division Multiplexing: AM - analogue

21. Frequency Division Multiplexing: AM - digital

22. Frequency Division Multiplexing: FM - analogue

23. Frequency Division Multiplexing: FM - digital

24. Frequency Division Multiplexing: PM - analogue

25. Frequency Division Multiplexing: PM - digital

26. Time Division Multiplexing 1874

27. Time Division Multiplexing - Statistical

28. Synchronisation

29. Digital Encoding NRZ Coding NRZ- L :- Voltage never returns to zero, negative voltage used to represent 1. Short distances only, i.e. PC modem, PC printer NRZI :- NRZ (Invert on ones), more reliable than NRZI, easier to detect in noise With NRZ-L and NRZI it difficult to detect correctly after a long string of 1's or O's Biphase Coding Manchester Differential Manchester Modulation rate is twice that of NRZ schemes and hence the bandwidth is twice as large. But! The clock is embedded in the signal The absence of an expected transition can be used to detect errors

30. Decibels As a signal propagates along a transmission medium there will be a loss, or attenuation of signal strength. This loss is expressed in dB where P o and P i are the input and output power levels. Example A signal with a power of 10 mW is transmitted. At the receiver end, the signal is measured to be only 5 mW.

31. Shannon’s Theory In 1948, Claude Shannon developed a mathematical expression for digital data which relates the capacity of a channel to bandwidth, and signal to noise ratio. C = B log 2 (1+S/N) where S/N is the signal power to noise power ratio, and B is the bandwidth of the channel. Example A channel has a bandwidth of 3 kHz, and a S/N of 30dB. What is the capacity of the channel? First Find the S/N ratio NdB = 10log 10 (P o /P i ) 30 = 10 log 10 (S/N) 30/10 = log 10 (S/N) 3 = log 10 (S/N) 10 3 = S/N S/N = 1000 Now you can find the capacity C = B log 2 (1+S/N) C = 3000 log 2 (1+1000) Using the relationship log b x = log y x / log y b C = 3000 log 10 (1001) / log 10 2 C = 29.91 kbps