Stream Information. A real-time voice signal must be digitized & transmitted as it is produced Analog signal level varies continuously in time

Transcription

1 , German University in Cairo Stream Information A real-time voice signal must be digitized & transmitted as it is produced Analog signal level varies continuously in time Th e s p ee ch s i g n al l e v el v a r ie s w i th t i m(e)

2 , German University in Cairo 3 bits / sample Digitization of Analog Signal Sample analog signal in time and amplitude Find closest approximation 7D/2 5D/2 3D/2 D/2 -D/2-3D/2-5D/2-7D/2 Original signal Sample value Approximat R s = Bit rate = # bits/sample x # samples/second

3 , German University in Cairo Bit Rate of Digitized Signal Bandwidth W s Hertz: how fast the signal changes Higher bandwidth more frequent samples Minimum sampling rate = 2 x W s Representation accuracy: range of approximation error Higher accuracy smaller spacing between approximation values more bits per sample

4 , German University in Cairo Example: Voice & Audio Telephone voice W s = 4 khz 8000 samples/sec 8 bits/sample R s =8 x 8000 = 64 kbps Cellular phones use more powerful compression algorithms: 8-12 kbps CD Audio W s = 22 khertz samples/sec 16 bits/sample R s =16 x 44000= 704 kbps per audio channel MP3 uses more powerful compression algorithms: 50 kbps per audio channel

5 , German University in Cairo Video Signal Sequence of picture frames Each picture digitized & compressed Frame repetition rate frames/second depending on quality Frame resolution Small frames for videoconferencing Standard frames for conventional broadcast TV HDTV frames 30 fps Rate = M bits/pixel x (WxH) pixels/frame x F frames/second

6 , German University in Cairo Video Frames 176 QCIF videoconferencing at 30 frames/sec = 760,000 pixels/sec Broadcast TV at 30 frames/sec = x 10 6 pixels/sec HDTV 1080 at 30 frames/sec = 67 x 10 6 pixels/sec

7 , German University in Cairo Digital Video Signals Type Method Format Original Compressed Video Conference H x144 or 352x288 fr/sec 2-36 Mbps kbps Full Motion MPEG 2 720x480 fr/sec 249 Mbps 2-6 Mbps HDTV MPEG 2 fr/sec 1.6 Gbps Mbps

8 , German University in Cairo Transmission of Stream Information Constant bit-rate Signals such as digitized telephone voice produce a steady stream: e.g. 64 kbps Network must support steady transfer of signal, e.g. 64 kbps circuit Variable bit-rate Signals such as digitized video produce a stream that varies in bit rate, e.g. according to motion and detail in a scene Network must support variable transfer rate of signal, e.g. packet switching or rate-smoothing with constant bit-rate circuit

9 , German University in Cairo Stream Service Quality Issues Network Transmission Impairments Delay: Is information delivered in timely fashion? Jitter: Is information delivered in sufficiently smooth fashion? Loss: Is information delivered without loss? If loss occurs, is delivered signal quality acceptable? Applications & application layer protocols developed to deal with these impairments

10 , German University in Cairo Bandwidth of General Signals speech s (noisy ) p (air stopped) ee (periodic) t (stopped) sh (noisy) Not all signals are periodic E.g. voice signals varies according to sound Vowels are periodic, s is noiselike Spectrum of long-term signal Averages over many sounds, many speakers Involves Fourier transform Telephone speech: 4 khz CD Audio: 22 khz X(f) 0 W s f 1 Slides are modified version of Digital Transmission Fundamentals Slides by Leon-Garcia/Widjaja L 21 - Page 10

11 , German University in Cairo Bit Rates of Digital Transmission Systems System Bit Rate Observations Telephone twisted pair Ethernet twisted pair kbps 4 khz telephone channel 10 Mbps, 100 Mbps 100 meters of unshielded twisted copper wire pair Cable modem 500 kbps-4 Mbps Shared CATV return channel ADSL twisted pair kbps in, Mbps out Coexists with analog telephone signal 2.4 GHz radio 2-11 Mbps IEEE wireless LAN 28 GHz radio Mbps 5 km multipoint radio Optical fiber Gbps 1 wavelength Optical fiber >1600 Gbps Many wavelengths

12 , German University in Cairo Examples of Channels Channel Bandwidth Bit Rates Telephone voice channel 3 khz 33 kbps Copper pair 1 MHz 1-6 Mbps Coaxial cable 500 MHz (6 MHz channels) 5 GHz radio (IEEE ) 300 MHz (11 channels) 30 Mbps/ channel 54 Mbps / channel Optical fiber Many TeraHertz 40 Gbps / wavelength

13 , German University in Cairo Shannon Limit If transmitted power is limited, then as M increases spacing between levels decreases Presence of noise at receiver causes more frequent errors to occur as M is increased Shannon Channel Capacity: The maximum reliable transmission rate over an ideal channel with bandwidth W Hz, with Gaussian distributed noise, and with SNR S/N is C = W log 2 ( 1 + S/N ) bits per second Reliable means error rate can be made arbitrarily small by proper coding 1 Slides are modified version of Digital Transmission Fundamentals Slides by Leon-Garcia/Widjaja L 21 - Page 13

14 , German University in Cairo Fourier Series Periodic Signals Representation A periodic signal with period T can be represented as sum of sinusoids using Fourier Series: x(t) = a 0 + a 1 cos(2pf 0 t + f 1 ) + a 2 cos(2p2f 0 t + f 2 ) + + a k cos(2pkf 0 t + f k ) + DC longterm average fundamental frequency f 0 =1/T first harmonic kth harmonic a k determines amount of power in kth harmonic Amplitude specturm a 0, a 1, a 2, 1 Slides are modified version of Digital Transmission Fundamentals Slides by Leon-Garcia/Widjaja L 21 - Page 14

15 , German University in Cairo Example Fourier Series x 1 (t) x 2 (t) t t T 2 =0.25 ms T 1 = 1 ms 4 4 x 1 (t) = 0 + cos(2p4000t) x p 2 (t) = 0 + cos(2p1000t) p cos(2p3(4000)t) + cos(2p3(1000)t) 3p 3p cos(2p5(4000)t) + + cos(2p5(1000)t) + 5p 5p Only odd harmonics have power 1 Slides are modified version of Digital Transmission Fundamentals Slides by Leon-Garcia/Widjaja L 21 - Page 15

16 , German University in Cairo Spectra & Bandwidth Spectrum of a signal: magnitude of amplitudes as a function of frequency x 1 (t) varies faster in time & has more high frequency content than x 2 (t) Spectrum of x 1 (t) Spectrum of x 2 (t) frequency (khz) frequency (khz) Bandwidth W s : is defined as range of frequencies where a signal has non-negligible power, e.g. range of band that contains 99% of total signal power 1 Slides are modified version of Digital Transmission Fundamentals Slides by Leon-Garcia/Widjaja L 21 - Page 16

17 , German University in Cairo Multilevel Signaling Nyquist pulses achieve the maximum signalling rate with zero ISI, 2W c pulses per second or 2W c pulses / W c Hz = 2 pulses / Hz With two signal levels, each pulse carries one bit of information Bit rate = 2W c bits/second With M = 2 m signal levels, each pulse carries m bits Bit rate = 2W c pulses/sec. * m bits/pulse = 2W c m bps Bit rate can be increased by increasing number of levels r(t) includes additive noise, that limits number of levels that can be used reliably.

18 , German University in Cairo Example of Multilevel Signaling Four levels {-1, -1/3, 1/3, +1} for {00,01,10,11} Waveform for 11,10,01 sends +1, +1/3, -1/3 Zero ISI at sampling instants Composite waveform

19 , German University in Cairo Sampling Theorem Nyquist: Perfect reconstruction (a) x(t) if sampling rate 1/T > 2W s x(nt) t Sampler t (b) x(nt) x(t) t Interpolation filter t 1 Slides are modified version of Digital Transmission Fundamentals Slides by Leon-Garcia/Widjaja L 21 - Page 19

20 , German University in Cairo Digital Transmission of Analog Information 2W samples / sec m bits / sample Analog source Sampling (A/D) Quantization Original x(t) Bandwidth W 2W m bits/sec Approximation y(t) Transmission or storage Display or playout Interpolation filter Pulse generator 2W samples / sec 1 Slides are modified version of Digital Transmission Fundamentals Slides by Leon-Garcia/Widjaja L 21 - Page 20

21 3 bits / sample, German University in Cairo Quantization of Analog Samples output y(nt) - D -3D -2D -3.5D -2.5D -1.5D -D -0.5D 1.5D 0.5D 2.5D 3.5D D 2D 3D D input x(nt) Quantizer maps input into closest of 2 m representation values Quantization error: noise = x(nt) y(nt) 7D/2 5D/2 3D/2 D/2 -D/2-3D/2-5D/2-7D/2 Original signal Sample value Approximation 1 Slides are modified version of Digital Transmission Fundamentals Slides by Leon-Garcia/Widjaja L 21 - Page 21

22 , German University in Cairo 3 bits / sample Quantization Error Quantization error is can be though of as an error that was added to the original signal 7D/2 5D/2 3D/2 D/2 -D/2-3D/2-5D/2-7D/2 Original signal Sample value Approximation Error 1 Slides are modified version of Digital Transmission Fundamentals Slides by Leon-Garcia/Widjaja L 21 - Page 22

23 , German University in Cairo Quantizer Performance M = 2 m levels, Dynamic range( -V, V) Δ = 2V/M... D 2 error = y(nt)-x(nt)=e(nt) -2D D D 2D 3D... input -V - D 2 V x(nt) If the number of levels M is large, then the error is approximately uniformly distributed between (-Δ/2, Δ2) Average Noise Power = Mean Square Error: σ 2 e = x 2 1 dx = Δ2 Δ 12 Δ 2 Δ 2 1 Slides are modified version of Digital Transmission Fundamentals Slides by Leon-Garcia/Widjaja L 21 - Page 23

24 , German University in Cairo Quantizer Performance Figure of Merit: Signal-to-Noise Ratio = Avg signal power / Avg noise power Let x 2 be the signal power, then SNR = x 2 D 2 /12 = 12 x = 4V 2 /M 2 The ratio V/ x 4 2 x 3 ( ) 2 M 2 = x 3 ( V ) 2 2 2m V The SNR is usually stated in decibels: SNR db = 10 log 10 x2 / e 2 = log 10 3 x2 /V 2 SNR db = 6m db for V/ x = 4. 1 Slides are modified version of Digital Transmission Fundamentals Slides by Leon-Garcia/Widjaja L 21 - Page 24

25 , German University in Cairo Example: Telephone Speech W = 4KHz, so Nyquist sampling theorem 2W = 8000 samples/second Suppose error requirement 1% error SNR = 10 log(1/.01) 2 = 40 db Assume V/ x then 40 db = 6m 7 m = 8 bits/sample PCM ( Pulse Code Modulation ) Telephone Speech: Bit rate= 8000 x 8 bits/sec= 64 kbps 1 Slides are modified version of Digital Transmission Fundamentals Slides by Leon-Garcia/Widjaja L 21 - Page 25

26 , German University in Cairo Example Consider a 3 khz channel with 8-level signaling.compare bit rate to channel capacity at 20 db SNR 3KHz telephone channel with 8 level signaling Bit rate = 2*3000 pulses/sec * 3 bits/pulse = 18 kbps 20 db SNR means 10 log10 S/N = 20 Implies S/N = 100 Shannon Channel Capacity is then C = 3000 log ( ) = 19, 963 bits/second 1 Slides are modified version of Digital Transmission Fundamentals Slides by Leon-Garcia/Widjaja L 21 - Page 26

27 2015 Line Coding

28 , German University in Cairo What is Line Coding? Mapping of binary information sequence into the digital signal that enters the channel Ex. 1 maps to +A square pulse; 0 to A pulse Line code selected to meet system requirements: Transmitted power: Power consumption = $ Bit timing: Transitions in signal help timing recovery Bandwidth efficiency: Excessive transitions wastes bw Low frequency content: Some channels block low frequencies long periods of +A or of A causes signal to droop Waveform should not have low-frequency content Error detection: Ability to detect errors helps Complexity/cost: Is code implementable in chip at high speed?

29 , German University in Cairo Line coding examples Unipolar NRZ Polar NRZ NRZ-inverted (differential encoding) Bipolar encoding Manchester encoding Differential Manchester encoding

30 , German University in Cairo pow er density Spectrum of Line codes Assume 1s & 0s independent & equiprobable NRZ Bipolar NRZ has high content at low frequencies Bipolar tightly packed around T/2 Manchester wasteful of bandwidth Manchester ft

31 , German University in Cairo Unipolar & Polar Non-Return-to-Zero (NRZ) Unipolar NRZ Polar NRZ Unipolar NRZ 1 maps to +A pulse 0 maps to no pulse High Average Power 0.5*A *0 2 =A 2 /2 Long strings of A or 0 Poor timing Low-frequency content Simple Polar NRZ 1 maps to +A/2 pulse 0 maps to A/2 pulse Better Average Power 0.5*(A/2) *(-A/2) 2 =A 2 /4 Long strings of +A/2 or A/2 Poor timing Low-frequency content Simple

32 , German University in Cairo Bipolar Code Bipolar Encoding Three signal levels: {-A, 0, +A} 1 maps to +A or A in alternation 0 maps to no pulse Every +pulse matched by pulse so little content at low frequencies String of 1s produces a square wave Spectrum centered at T/2 Long string of 0s causes receiver to lose synch Zero-substitution codes

33 , German University in Cairo Manchester code & mbnb codes Manchester Encoding maps into A/2 first T/2, -A/2 last T/2 0 maps into -A/2 first T/2, A/2 last T/2 Every interval has transition in middle Timing recovery easy Uses double the minimum bandwidth Simple to implement Used in 10-Mbps Ethernet & other LAN standards mbnb line code Maps block of m bits into n bits Manchester code is 1B2B code 4B5B code used in FDDI LAN 8B10b code used in Gigabit Ethernet 64B66B code used in 10G Ethernet

34 , German University in Cairo Differential Coding NRZ-inverted (differential encoding) Differential Manchester encoding Errors in some systems cause transposition in polarity, +A become A and vice versa All subsequent bits in Polar NRZ coding would be in error Differential line coding provides robustness to this type of error 1 mapped into transition in signal level 0 mapped into no transition in signal level Same spectrum as NRZ Errors occur in pairs Also used with Manchester coding

35 , German University in Cairo Amplitude Modulation and Frequency Modulation Information Amplitude Shift Keying T 2T 3T 4T 5T 6T t Map bits into amplitude of sinusoid: 1 send sinusoid; 0 no sinusoid Demodulator looks for signal vs. no signal +1 Frequency Shift Keying -1 0 T 2T 3T 4T 5T 6T t Map bits into frequency: 1 send frequency f c + d ; 0 send frequency f c - d Demodulator looks for power around f c + d or f c - d 1 L 1 - Page 35

36 , German University in Cairo Phase Modulation Information Phase Shift Keying 0 T 2T 3T 4T 5T 6T t -1 Map bits into phase of sinusoid: 1 send A cos(2pft), i.e. phase is 0 0 send A cos(2pft+p), i.e. phase is p Equivalent to multiplying cos(2pft) by +A or -A 1 send A cos(2pft), i.e. multiply by 1 0 send A cos(2pft+p) = - A cos(2pft), i.e. multiply by -1 We will focus on phase modulation 1 L 1 - Page 36

37 , German University in Cairo Modulator & Demodulator Modulate cos(2pf c t) by multiplying by A k for T seconds: A k x Y i (t) = A k cos(2pf c t) cos(2pf c t) Transmitted signal during kth interval Demodulate (recover A k ) by multiplying by 2cos(2pf c t) for T seconds and lowpass filtering (smoothing): Y i (t) = A k cos(2pf c t) Received signal during kth interval x 2cos(2pf c t) Lowpass Filter (Smoother) X i (t) 2A k cos 2 (2pf c t) = A k {1 + cos(2p2f c t)} 1 L 1 - Page 37

38 , German University in Cairo Example of Modulation Information Baseband Signal +A -A 0 T 2T 3T 4T 5T 6T Modulated Signal x(t) +A -A 0 T 2T 3T 4T 5T 6T A cos(2pft) -A cos(2pft) 1 L 1 - Page 38

39 , German University in Cairo Example of Demodulation A {1 + cos(4pft)} -A {1 + cos(4pft)} After multiplication at receiver x(t) cos(2pf c t) +A -A 0 T 2T 3T 4T 5T 6T +A Baseband signal discernable after smoothing -A 0 T 2T 3T 4T 5T 6T Recovered Information L 1 - Page 39

40 , German University in Cairo Amplitude Modulation and Frequency Modulation Information Amplitude Shift Keying T 2T 3T 4T 5T 6T t Map bits into amplitude of sinusoid: 1 send sinusoid; 0 no sinusoid Demodulator looks for signal vs. no signal +1 Frequency Shift Keying 0 T 2T 3T 4T 5T 6T -1 t Map bits into frequency: 1 send frequency f c + d ; 0 send frequency f c - d Demodulator looks for power around f c + d or f c - d 1 L 1 - Page 40

41 , German University in Cairo Phase Modulation Information Phase Shift Keying 0 T 2T 3T 4T 5T 6T t -1 Map bits into phase of sinusoid: 1 send A cos(2pft), i.e. phase is 0 0 send A cos(2pft+p), i.e. phase is p Equivalent to multiplying cos(2pft) by +A or -A 1 send A cos(2pft), i.e. multiply by 1 0 send A cos(2pft+p) = - A cos(2pft), i.e. multiply by -1 We will focus on phase modulation 1 L 1 - Page 41

42 , German University in Cairo Modulator Modulate cos(2pf c t) by multiplying by A k for T seconds: A k x Y i (t) = A k cos(2pf c t) +A cos(2pf c t) Transmitted signal during kth interval X -A +A Carrier Information -A 0 T 2T 3T 4T 5T 6T Baseband Signal 1 L 1 - Page 42

43 , German University in Cairo Example Modulated Signal Modulated Signal x(t) +A -A 0 T 2T 3T 4T 5T 6T 1 L 1 - Page 43

44 , German University in Cairo DeModulator Demodulate (recover A k ) by multiplying by 2cos(2pf c t) for T seconds and lowpass filtering (smoothing): Y i (t) = A k cos(2pf c t) Received signal during kth interval x 2cos(2pf c t) Lowpass Filter (Smoother) X i (t) 2A k cos 2 (2pf c t) = A k {1 + cos(2p2f c t)} Cos(x+y)=Cos(x)Cos(y)-sin(x)sin(y) Cos(x+x)=Cos 2 (x)-sin 2 (x) Cos(2x)=2Cos 2 (x)-1 2Cos2(x)=1+ Cos(2x) 1 L 1 - Page 44

45 , German University in Cairo Example of Modulation Modulated Signal x(t) +A -A +A 0 T 2T 3T 4T 5T 6T A {1 + cos(4pft)} -A After multiplication at receiver x(t) cos(2pf c t) +A -A 0 T 2T 3T 4T 5T 6T -A {1 + cos(4pft)} 1 L 1 - Page 45

46 , German University in Cairo A {1 + cos(4pft)} Example of Demodulation +A -A 0 T 2T 3T 4T 5T 6T -A {1 + cos(4pft)} Input Signal to low pass filer Baseband signal discernable after smoothing Recovered Information +A -A 0 T 2T 3T 4T 5T 6T 1 L 1 - Page 46

47 , German University in Cairo Signaling rate and Transmission Bandwidth Fact from modulation theory: If then Baseband signal x(t) with bandwidth B Hz Modulated signal x(t)cos(2pf c t) has bandwidth 2B Hz B f c -B If bandpass channel has bandwidth W c Hz, Then baseband channel has W c /2 Hz available, so f c f f c +B modulation system supports W c /2 x 2 = W c pulses/second That is, W c pulses/second per W c Hz = 1 pulse/hz Recall baseband transmission system supports 2 pulses/hz 1 L 1 - Page 47 f

48 , German University in Cairo Quadrature Amplitude Modulation Why not use Amplitude modulation and phase modulation Y i (t) = A k cos(2pf c t+f k ) Amplitude modulation of symbol K Phase modulation of symbol K 1 L 1 - Page 48

49 , German University in Cairo Quadrature Amplitude Modulation (QAM) QAM uses two-dimensional signaling A k modulates in-phase cos(2pf c t) B k modulates quadrature phase cos(2pf c t + p/4) = sin(2pf c t) Transmit sum of inphase & quadrature phase components A k x Y i (t) = A k cos(2pf c t) cos(2pf c t) + Y(t) B k x Y q (t) = B k sin(2pf c t) Transmitted Signal sin(2pf c t) Y i (t) and Y q (t) both occupy the bandpass channel QAM sends 2 pulses/hz 1 L 1 - Page 49

50 , German University in Cairo QAM Demodulation Y(t) x Lowpass filter (smoother) A k 2cos(2pf c t) 2cos 2 (2pf c t)+2b k cos(2pf c t)sin(2pf c t) = A k {1 + cos(4pf c t)}+b k {0 + sin(4pf c t)} x 2sin(2pf c t) Lowpass filter (smoother) B k smoothed to zero 2B k sin 2 (2pf c t)+2a k cos(2pf c t)sin(2pf c t) = B k {1 - cos(4pf c t)}+a k {0 + sin(4pf c t)} smoothed to zero 1 L 1 - Page 50

51 , German University in Cairo Signal Constellations Each pair (A k, B k ) defines a point in the plane Signal constellation set of signaling points (-A,A) B k (A, A) B k A k A k (-A,-A) (A,-A) 4 possible points per T sec. 2 bits / pulse 16 possible points per T sec. 4 bits / pulse 1 L 1 - Page 51

52 , German University in Cairo Other Signal Constellations Point selected by amplitude & phase A k cos(2pf c t) + B k sin(2pf c t) = A k 2 + B k 2 cos(2pf c t + tan -1 (B k /A k )) B k B k A k A k 4 possible points per T sec. 16 possible points per T sec. 1 L 1 - Page 52

53 , German University in Cairo How Many Constellation Points More Constellation Points = more bits per Symbol More Constellation Points = more errors Why 1 L 1 - Page 53

54 , German University in Cairo Multilevel Signaling Nyquist pulses achieve the maximum signalling rate with zero ISI, 2W c pulses per second or 2W c pulses / W c Hz = 2 pulses / Hz With two signal levels, each pulse carries one bit of information Bit rate = 2W c bits/second With M = 2 m signal levels, each pulse carries m bits Bit rate = 2W c pulses/sec. * m bits/pulse = 2W c m bps Bit rate can be increased by increasing number of levels r(t) includes additive noise, that limits number of levels that can be used reliably.

55 , German University in Cairo Example of Multilevel Signaling Four levels {-1, -1/3, 1/3, +1} for {00,01,10,11} Waveform for 11,10,01 sends +1, +1/3, -1/3 Zero ISI at sampling instants Composite waveform

56 , German University in Cairo Noise Limits Accuracy Receiver makes decision based on transmitted pulse level + noise Error rate depends on relative value of noise amplitude and spacing between signal levels Large (positive or negative) noise values can cause wrong decision Noise level below impacts 8-level signaling more than 4-level signaling +A +A/3 -A/3 -A Typical noise +A +5A/7 +3A/7 +A/7 -A/7-3A/7-5A/7 -A 1 Four signal levels Eight signal L levels 1 - Page 56

57 , German University in Cairo Signal Constellations (errors) B k (-A,A) B k (A, A) A k A k (-A,-A) (A,-A) 1 L 1 - Page 57