1. 1/15 KLKSK Pertemuan III Analog & Digital Data Shannon Theorem xDSL

2. 2/15 Analog and Digital Signaling of Analog and Digital Data Analog signals: represents data with continuously varying electromagnetic wave Digital signals: represents data with sequence of voltage pulses

3. 3/15 Data and Signals Usually use digital signals for digital data and analog signals for analog data Can use analog signal to carry digital data –Modem Can use digital signal to carry analog data –Compact Disc audio

4. 4/15 Analog & Digital Transmission Analog SignalDigital Signal Analog Data Two alternatives: (1) signal occupies the same spectrum as the analog data; (2) analog data are encoded to occupy a different portion of spectrum Analog data are encoded using a codec to produce a digital bit stream Digital Data Digital data are encoded using a modem to produce analog signal Two alternatives: (1) signal consists of two voltage levels to represent the two binary values; (2) digital data are encoded to produce a digital signal with desired properties Data and Signal

5. 5/15 Analog Signals Carrying Analog and Digital Data

6. 6/15 Digital Signals Carrying Analog and Digital Data

7. 7/15 Analog & Digital Transmission Analog Transmission Digital Transmission Analog Signal Is propagated through amplifiers; same treatment whether signal is used to represent analog data or digital data Assumes that the analog signal represents digital data. Signal is propagated through repeaters; at each repeater, digital data are recovered from inbound signal and used to generate a new analog outbound signal Digital Signal Not used Digital signal represents a stream of 1s and 0s, which may represent digital data or may be an encoding of analog data. Signal is propagated through repeaters; at each repeater, stream of 1s and 0s is recovered from inbound signal and used to generate a new digital outbound signal Treatment of Signals

8. 8/15 Channel Capacity Data rate –In bits per second –Rate at which data can be communicated Bandwidth –In cycles per second of Hertz –Constrained by transmitter and medium

9. 9/15 Nyquist Bandwidth If rate of signal transmission is 2B then signal with frequencies no greater than B is sufficient to carry signal rate Given bandwidth B, highest signal rate is 2B Given binary signal, data rate supported by B Hz is 2B bps Can be increased by using M signal levels C= 2B log 2 M M is the of discrete signal or voltage levels

10. 10/15 Shannon Capacity Formula Consider data rate, noise and error rate Faster data rate shortens each bit so burst of noise affects more bits –At given noise level, high data rate means higher error rate Signal to noise ratio (in decibels) SNR db = 10 log 10 (signal/noise) Capacity C=B log 2 (1+SNR)

11. 11/15 Consider a voice channel being used, via modem, to transmit digital data. Assume a bandwidth of 3100 Hz. A typical value of S/N for a voice-grade line is 30 dB; or a ratio of 1000:1 Thus C = 3100 log 2 (1 + 1000) = 30,894 bps

12. 12/15 Shannon proved that if the actual information rate on a channel is less than the error-free capacity, then it is theoretically possible to use a suitable signal code to achieve error – free transmission through the channel Unfortunately Shannon formula does not suggest a means for finding such codes

13. 13/15 E b /N o : ratio of signal energy per bit to noise power density per hertz Consider a signal, digital or analog, that contains binary digital data transmitted at a certain bit rate R 1 W = 1 J/s The energy per bit in a signal E b = ST b S : the signal power T b : the time required to send one bit The data rate R = 1/T b (E b /N o ) = (S/R)/N o = S/(kTR) E b /N o = S – 10log R + 228.6 dBW – 10 logT

14. 14/15 Ex. For binary phase-shift keying, E b /N o = 8.4 dB is required for a bit error rate of 10 -4 (probability of error = 10 -4 ). If the effective noise temperature is 290 o K (room temperature) and the data rate is 2400 bps, what received signal level is required?

15. 15/15 Ans. 8.4 = S (dBW) – 10 log 2400 + 228.6 dBW – 10 log 290 = S (dBW) – (10) (3.38) + 228.6 – (10)(2.46) S = -161.8 dBW