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Chapter 3: Data Transmission

COE 341: Data & Computer Communications Dr. Radwan E. Abdel-Aal. Chapter 3: Data Transmission. Remaining Six Chapters:. Chapter 7: Data Link: Flow and Error control, Link management. Data Link. Chapter 8: Improved utilization: Multiplexing.

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Chapter 3: Data Transmission

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  1. COE 341: Data & Computer Communications Dr. Radwan E. Abdel-Aal Chapter 3: Data Transmission

  2. Remaining Six Chapters: Chapter 7: Data Link: Flow and Error control, Link management Data Link Chapter 8: Improved utilization: Multiplexing Chapter 6: Data Communication: Synchronization, Error detection and correction Physical Layer Chapter 4: Transmission Media Transmission Medium Chapter 5: Encoding: From data to signals Chapter 3: Signals, their representations, their transmission over media, Resulting impairments

  3. Agenda • Concepts & Terminology • Signal representation: Time and Frequency domains • Bandwidth and data rate • The decibels notation for signal strength (Appendix 3A) • Fourier Analysis (Appendix B) • Analog & Digital Data Transmission • Transmission Impairments • Channel Capacity

  4. Terminology (1) Transmission system: Components • Transmitter • Receiver • Medium • Guided media • e.g. twisted pair, coaxial cable, optical fiber • Unguided media • e.g. air, water, vacuum

  5. Terminology (2) Link Configurations: • Direct link • Nointermediate ‘communication’ devices (these exclude repeaters/amplifiers) Two types: • Point-to-point (A-B) • Only 2 devices share link • Multi-point (C-B) • More than two devices share the same link, e.g. Ethernet bus segment C Amplifier A B

  6. Terminology (3) Transmission Types (ANSI Definitions) • Simplex • Information flows only in one directionall the time e.g. Television, Radio broadcasting • Duplex • Information flows in both directions • Two types: • Half duplex • Only one direction at a time e.g. Walki-Talki • Full duplex • In both directions at the same time e.g. telephone

  7. Frequency, Spectrum and Bandwidth • Time domain concepts • Analog signal • Varies in a smooth, continuous way in both time and amplitude • Digital signal • Maintains a constant level for sometime and then changes to another constant level (i.e. amplitude takes only a finite number of discrete levels) • Periodic signal • Same pattern repeated over time, e.g. sine wave or a square wave • Aperiodic signal • Pattern not repeated over time

  8. Analogue & Digital Signals All values on the time and amplitude axes are allowed Analogue Only a few amplitude levels allowed - Binary signal: 2 levels Digital

  9. T PeriodicSignals Temporal Period … t t+1T t+2T For any periodic wave: S (t+nT) = S (t); 0  t T Where: t is time over first period T is the waveform period n is an integer Signal behavior over one period describes behavior at all times

  10. 1 0 t + X/2 - X/2 Aperiodic (non periodic) Signals in time s(t)

  11. Continuous Versus Discrete Availability of the signal over the horizontal axis (Time or Frequency) Continuous: Signal is defined at all points on the horizontal axis Sampling with a train of very narrow pulses (delta function) Discrete: Signal is defined Only at certain points on the horizontal axis

  12. T (Period) Sine Wave s(t) = A sin(2f t +) = A sin (F) A (Amplitude) w • Peak Amplitude (A) • Peak strength of signal, volts • Repetition Frequency (f), Cycles/s = Hz • Measures how fast the signal varies with time • Number of cycles per second (Hz) • f = 1/ T(xx sec/cycle) = yy cycles/sec = yy Hz • Angular Frequency (w), Radians/s w = radians per second = 2 f = 2 /T • Temporal (time) Period, T = 1/f • Phase Angle (), Radians • Determines relative position in time, radians (how to calculate?)

  13. Varying one of the three parameters of a sine wave carriers(t) = A sin(2ft +) = A sin(wt+F) Can be used to convey information…! M o d u l a t I o n AM Varying A Varying  FM PM Varying f

  14. Sine Wave Traveling in the +ive x directions(t) = A sin (k x -  t)  = Angular Frequency = 2 f = 2 / T k = Wave Number = 2 /  • = Spatial Period = Wavelength  • For point p on the wave: • Total phase at t = 0: kx -  (0) = kx • Total phase at t = t: k(x+ x) -  (t) • Same total phase, • kx = k(x+ x) -  (t) • k x =  t Wave propagation velocity v = x / t v = /k = /T = f x p x Distance, x t = 0 t = t + ive x direction Direction of wave travel, at velocity v  Show that the wave s(t) = A sin (k x +  t] travels in the negative x direction V is constant for a given wave type (e.g. sound) and medium (e.g. air) v = f

  15. Wave Propagation Velocity, v m/s • Constant for: • A given wave type (e.g. electromagnetic, seismic, ultrasound, ..) • and a given propagation medium (air, water, optical fiber) • For all types of waves: • v = l f • For a given wave type and medium (given v): higher frequencies correspond to shorter wavelengths and vise versa: Electromagnetic waves: long wave radio (km), short wave radio (m), microwave (cm)… light (nm) • For electromagnetic waves: • In free space, v  speed of light in vacuum v  c = 3x108 m/sec • Over other guided media (coaxial cable, optical fiber, twisted pairs): v is always lower than c Shorter wavelength  Higher frequency

  16. Wavelength, l (meters) • Is the Spatial period of the wave: i.e. distance between two points in space on the wave propagation path where the wave has the same total phase • Also: = Distance traveled by the wave during one temporal (time) cycle: dT = v T = (l f) T = l

  17. Frequency Domain Concepts • Response of systems to a sine waves is easy to analyze • But signals we deal with in practice are not all sine waves, e.g. Square waves • Can we relate waves we deal with in practice to sine waves? YES! • Fourier analysis shows that any signal can be treated as the sum of many sine wave components having different frequencies, amplitudes, and phases(Fourier Analysis: Appendix B) • This forms the basis for frequency domainanalysis • For a linear system, its response to a complex signal will be the sum of its response to the individual sine wave components representing that signal. • Dealing with functions in the frequency domain is simpler than in the time domain

  18. Fundamental Addition of Twofrequency Components A = 1*(4/) frequency = f + 1/3 rd the Amplitude 3 times the frequency 3rd harmonic A = (1/3)*(4/) frequency = 3f Frequency Spectrum = Approaching a square wave Fourier Series 3 t f Frequency Domain: S(f) vs f Time Domain: s(t) vs t Fourier Series Discrete Function in f Periodic function in t

  19. Asymptotically approaching a square wave by combining the fundamental + an infinite number of odd harmonics at prescribed amplitudes Topic for a programming assignment Adding more of the higher harmonics What is the highest Harmonic added?

  20. s(t) 1 0 t + X/2 - X/2 time More Frequency Domain Representations: A single square pulse (Aperiodic signal) Sinc(f) = sin(f)/f Fourier Transform X To  To  To  frequency 1/X Fourier Transform Frequency Domain: S(f) vs f Time Domain: s(t) vs t Continuous Function in f Aperiodic function in t • What happens to the spectrum as the pulse gets broader … DC ? • What happens to the spectrum as the pulse gets narrower … spike ?

  21. Spectrum & Bandwidth of a signal • Spectrum of a signal • Range of frequencies contained in a signal • Absolute (theoretical) Bandwidth (BW): • Is the full width of spectrum = fmax- fmin • But in many situations, fmax = ! (e.g. a square wave), so: • Effective Bandwidth • Often called just bandwidth • Narrow band of frequencies containing most of the signal energy • Somewhat arbitrary: what is “most”?  e.g. that contains say 95% of the energy of the signal S(f) …. 7f 5f f 3f f

  22. Signals with a DC Component NO DC Component, Signal average over a period = 0 + _ t + 1V DC Level + t 1V DC Component • DC Component: •  is the component at zero frequency  Determines if fmin = 0 or not

  23. = (fmax- fmin) Bandwidth for these signals:

  24. S(f) …. 7f 5f f 3f f Bandwidth of a transmission system • Is the Range of signal frequencies that are adequately passed by the system • Effectively, the transmission system (TX, medium, RX) acts as a filter • Poor transmission media, e.g. twisted pairs, have a narrow bandwidth • This effectively cuts off higher frequency signal components  poor signal quality at receive  limit the signal frequencies (Hz) that can be used for transmission  this limits the data rates that can be used (bps), examples: • Twisted pair: 4 KHz BW  100 Kbps • Optical fiber: 4 THz BW  10 Gbps

  25. Received Waveform Limiting Effect of System Bandwidth 1,3 Better reception requires larger BW BW = 2f More difficult reception with smaller BW f 3f 1 1,3,5 BW = 4f 5f f 3f 2 Varying System BW 1,3,5,7 BW = 6f 7f 5f f 3f 3 … BW =  1,3,5,7 ,9,… ……  7f 5f f 3f 4 Fourier Series for a Square Wave

  26. System Bandwidth and Achievable Data Rates • Any transmission system supports only a limited range of frequencies(bandwidth) for satisfactory transmission • For example, this bandwidth is largest for expensive optical fibers and smallest for cheap twisted pair wires • So, bandwidth is money  Economize in its use • Limited system bandwidth degrades higher frequency components of the signal transmitted  poorer received waveforms  more difficult to interpret the signal at the receiver (especially with noise)  Data Errors • More degradation occurs when higher data ratesare used(signal will have higher frequency components)

  27. Maximum Data Rate (= link channel capacity)Considerations: • Bandwidth of transmission system • Signal to noise ratio (SNR) • Receiver type • Specified acceptable error performance

  28. 5f f 3f Data Element = Signal Element Bandwidth and Data Rates Period T = 1/f T/2 Data rate = 1/(T/2) = (2/T) bits per sec = 2f bps B 0 0 1 1 Data B = 4f Given a bandwidth B, Data rate = 2f = B/2 To double the data rate you need to double f: Two ways to do this… 1. Double the bandwidth, same received waveform quality (same RX conditions & error rate) 2B = 4f’ 2B 1 0 0 1 1 1 0 0 1 0 0 1 1 1 0 0 New bandwidth: 2B, Data rate = 2f’ = 2(2f)= 4f = B X 2 f’ 3f’ 5f’ 2. Same bandwidth, B, but tolerate poorer received waveform (needs better receiver, higher S/N ratio, or tolerating more errors in data) 1 B = 2f’ 0 0 0 1 1 1 0 B X 2 Bandwidth: B, Data rate = 2f’ = 2(2f) = 4f = B 5f’ 3f’ f’

  29. Bandwidth & Data Rates: Tradeoffs… Compromises • Increasing the data rate (bps) while keeping BWthe same(to economize) means working with inferior (poorer) waveforms at the receiver, which may require: • Ensuring higher signal to noise ratio (SNR) at RX • Larger transmitted power (may cause interference to others!) • Limited (shorter) link distances • Use of more en-route repeaters/amplifiers • Better shielding of cables to reduce noise, etc. • More sensitive (& costly!) receiver • Suffering from higher bit error rates • Tolerate them? • Add more efficient means for error detection and correction- this also increases overhead!.

  30. Appendix 3A: Decibels and Signal Strength • The decibel notation (dB) is a logarithmic measure of the ratio between two signal power levels • NdB= number of decibels • P1 = input power level (Watts) • P2 = output power level (Watts) • e.g.  Amplifier gain  Signal loss (attenuation) over a link • Example: • A signal with power level of 10mW is inserted into a transmission line • Measured power some distance away is 5 mW • Power loss in dBs is expressed as NdB =10 log (5/10)=10(-0.3)= -3 dB • - ive dBs: P2 < P1 (Loss), • +ive dBs: P2 > P1 (Gain) P2 P3 P1 Lossy Link Amplifier

  31. Relationship Between dB Values and Power ratio (P2/P1)

  32. Decibels and Signal Strength • Decibel notation is a relative, not absolute, measure: • A loss of 3 dB halves the power (could be 100 to 50, 16 to 8, …) • A gain of 3 dB doubles the power (could be 5 to 10, 7.5 to 15, …) • Will see shortly how we can handle absolute levels • Advantages of using dBs: • The “log” allows replacing: • Multiplication with Addition C = A * B Log C = Log A + Log B • and Division with Subtraction A = C / B Log A = Log C - Log B

  33. Amplifier ? 4 mW Gain: 35 dB Transmitted Signal Received Signal Loss: 10 dB Loss: 12 dB Decibels and Signal Strength • Example: Transmission line with an intermediate amplifier (/10) (/15.6) (*3162) • Net power gain over transmission path: + 35 –12 – 10 =+13 dB(+ ive means there is net gain)  • Received signal power = (4 mW) log10-1(13/10) = 4 x 101.3 = 4 x 101.3 mW = 79.8 mW Still we use some multiplication!

  34. WK 4 How to represent absolute power levels?Decibel-Watt (dBW) and Decibel-mW (dBm) • As a ratio relative to a fixed reference power level • With 1 W used as a reference  dBW • With 1 mW used as a reference  dBm • Examples: • Power of 1000 W is 30 dBW, 1 W = ? dBW • –10 dBm represents a power of 0.1 mW, • 1 mW = ? dBm X dBW = (X + ?) dBm Caution!: Must be same units at top and bottom Caution!: Must be same units at top and bottom

  35. dBs & dBms are added algebraically P2 P1 G G is  Positive for gain  Negative for loss (attenuation) G = power ratio = G dBs = 10 log10 G = Similarly for dBs & dBWs

  36. Decibels and Signal Strength • If all ratios are in dBs and all levels are in dBm solve by algebraic addition Same for {dBs and dBWs} (No need for any multiplication/division) • Example: Transmission line with an intermediate amplifier ? 4 mW Gain: 35 dB Transmitted Signal Received Signal Loss: 10 dB Loss: 12 dB Amplifier • Net power gain over transmission path: + 35 –12 – 10 =+13 dB (+ ive means actual net gain) TX Signal Power in dBm = 4 mW = 10 log (4/1) = 6.02 dBm • RX signal power (dBm) = 6.02 + 13 =19.02 dBm • Check: 19.02 dBm = 10 log (RX signal in mW/1 mW)  RX signal = log-1 (19.02/10) = 79.8 mW As on Slide 33

  37. Note that this is still a power ratio… But expressed in terms of voltages Caution!: Must be same units at top and bottom Decibels & Voltage ratios • Power decibels can also be expressed in terms of voltage ratios • Power P = V2/R, assuming same R  Relative:  Absolute: dBV and dBmV • Decibel-millivolt (dBmV) is an absolute unit, with 0 dBmV being equivalent to 1mV. dBV is similarly defined

  38. Pitfalls with the Decibel Notation • Wrong to multiply dBs dBs x dBs ! 35 dBs x 5 dBs (what would be the units of the result!) • Wrong to divide dBs dBs / dBs ! (caution: dBs / (dBs/km) is OK) • Wrong to mix numerical levels with dBs 3.5 W + 24 dBs ….. • OK to mix dBs with dBms -35 dBm + 12 dBs – 18 dBs …. • But wrong to use dBm and dBW in the same expression -35 dBm + 12 dBs + X dBs = 3 dbW

  39. Appendix B: Fourier Analysis Signals in Time Aperiodic Periodic … Discrete Continuous Discrete Continuous DFS FS FT DFT Use Fourier Series Use Fourier Transform FS : Fourier Series DFS : Discrete Fourier Series FT : Fourier Transform DFT : Discrete Fourier Transform

  40. Fourier Series for periodic continuous signals • Any periodic signal x(t) of period T and repetition frequency f0 (f0 = 1/T) can be represented as an infinite sum of sinusoids of different frequencies and amplitudes – its Fourier Series. Expressed in Two forms: • 1. The sine/cosine form: Frequencies are multiples of the fundamental frequency f0 f0 = fundamental frequency = 1/T Where: DC Component = f(n) Two components at each frequency All integrals over one period only If A0 is not 0, x(t) has a DC component = f’(n)

  41. Fourier Series: 2. The Amplitude-Phase form: • Previous form had two components at each frequency (sine, cosine i.e. in quadrature) : An, Bn coefficients • The equivalent Amplitude-Phase representation has only one component at each frequency: Cn, qn • Derived from the previous form using trigonometry: cos (a) cos (b) - sin (a) sin (b) = cos [a +b] Now we have Only one component at each frequency nf0 Now components have different amplitudes, frequencies, and phases The C’s and ’s are obtained from the previous A’s and B’s using the equations:

  42. Fourier Series: General Observations Fourier Series Expansion Function Odd Function Even Function DC

  43. Correction

  44. 1 -3/2 -1 -1/2 1/2 1 3/2 2 -1 T Fourier Series Example x(t) Note: (1) x(– t)=x(t)  x(t) is an even function (2) f0 = 1 / T = ½ Hz Note: A0 by definition is 2 x the DC content

  45. 1 -3/2 -1 -1/2 1/2 1 3/2 2 -1 T Contd… = 0 for n even = (4/n) sin (n/2) for n odd f0 =1/2 a function of n only  Replace t by –t  Swap limits in the first integral - sin(2pnf t) - dt Then Bn = 0 for all n x(t), since x(t) is an even function

  46. Contd… f0 = ½, so 2 f0 =  A0 = 0, Bn = 0 for all n, An = 0 for n even: 2, 4, … = (4/n) sin (n/2) for n odd: 1, 3, … Original x(t) is an even function! Amplitudes, n odd Cosine is an even function 2 p (1/2) t 2 p 3 (1/2) t f0 = ½  Fundamental 3rd Harmonic

  47. Another Example Previous Example x1(t) 1 -2 -1 1 2 -1 T Note that x1(-t)= -x1(t)  so, x(t) is an odd function Also, x1(t)=x(t-1/2) This waveform is the previous waveform shifted right by 1/2

  48. Another Example, Contd… Sine is an odd function As given before for the square wave on slide 25. Because:

  49. Fourier Transform • For aperiodic (non-periodic) signals in time, the spectrum consists of a continuum of frequencies (not discrete components) • This spectrum is defined by the Fourier Transform • For a signal x(t) and a corresponding spectrum X(f), the following relations hold Imaginary 1 nf0 f T/2   Inverse FT (from frequency to time ) Forward FT (from time to frequency) Real  Express sin and cos • X(f) is always complex (Has both real & Imaginary parts), even for x(t) real.

  50. (Continuous in Frequency) (non-periodic in time) Sinc function Sinc2 function

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