Data Transmission: Data Rate Limits and Transmission Impairments

1. Lecture 3:Data Transmission 1st semester 1439 - 2017 By: Elham Sunbu

2. DataTransmission • DATA RATE LIMITS • Transmission Impairments • Examples Outline

3. The successful transmission of data depends on two factors:: DataTransmission characteristics of the transmission medium quality of the signal beingtransmitted

4. Data transmission occurs between transmitter and receiver over some transmissionmedium. Communication is in the formofelectromagnetic waves. Unguided media (wireless) Guided media Transmission Terminology twisted pair, coaxialcable, opticalfiber air,vacuum, seawater

5. No intermediate devices • Direct link. • Only 2 devices share link. TransmissionTerminology • More than 2 devices share link.

6. Spectrum & Bandwidth

7. Signal with dc Component

8. Data Rate and Bandwidth There is a direct relationship between data rate and bandwidth.

9. data • entities that conveyinformation • signals • electric or electromagnetic representations of data • signaling • physically propagates along a medium • transmission • communication of data by propagation and processing of signals Analog and Digital Data Transmission

10. Acoustic Spectrum (Analog)

11. Advantage and disadvantage of digital signals Advantages &Disadvantages of DigitalSignals

12. frequency range of typical speech is100Hz-7kHz • easily converted into electromagneticsignals • varying volume converted to varyingvoltage • can limit frequency range for voice channel to 300-3400Hz Audio Signals

13. Video Signals • to produce a video signal a TV camera is used • USA standard is 483 lines per frame, at a rate of 30 complete frames per second • actual standard is 525 lines but 42 lost during vertical retrace • horizontal scanning frequency is 525 lines x 30 scans = 15750 linesper second • max frequency if line alternates black and white • max frequency of 4.2MHz

14. Analog Signals

15. Digital Signals

16. DATA RATE LIMITS

17. A very important consideration in data communications is how fast we can send data, in bits per second, over a channel. Data rate depends on threefactors: • The bandwidthavailable • The level of the signals weuse • The quality of the channel (the level ofnoise) • Topics discussed in this section: • Channel Capacity. • Nyquist Bit Rate, Noisy Channel. • Shannon Capacity Using BothLimits DATA RATE LIMITS

18. Channel Capacity

19. ChannelcapacityisconcernedwiththeinformationhandlingcapacityofaChannelcapacityisconcernedwiththeinformationhandlingcapacityofa • given channel. It is affectedby: • The attenuation of a channel which varies with frequency as well as channellength. • The noise induced into the channel which increases withdistance. • Non-linear effects such as clipping on thesignal. • Some of the effects may change with time e.g. the frequency response of a copper cable changes with temperature andage. Channel Capacity

20. Obviouslyweneedawaytomodelachannelinordertoestimatehowmuchinformation can be passed through it. Although we can compensate for non linear effects and attenuation it is extremely difficult to removenoise. • Thehighestrateofinformationthat can be transmitted throughachannel is called the channel capacity,C. Channel Capacity

21. Assume a channel is noise free. • Nyquist formulation: • if the rate of signal transmission is 2B, then a signal with frequencies no greater than B is sufficient to carry the signalrate. • Given bandwidth B, highest signal rate is2B. • Why is there such alimitation? • due to intersymbol interference, such as is produced by delay distortion. • Given binary signal (two voltage levels), the maximum data rate supported by B Hz is 2B bps. • One signal represents onebit • • Nyquist Formula

22. • Signals with more than two levels can be used, i.e., each signal element can represent more than onebit. – E.g., if a signal has 4 different levels, then a signal can be used to represents two bits: 00, 01, 10,11 • With multilevel signalling, the Nyquist formulabecomes: • C = 2Blog2M • Mis the number of discrete signal levels, B is the given bandwidth, C is the channel capacity inbps. • How large can Mbe? • The receiver must distinguish one of M possible signalelements. • Noise and other impairments on the transmission line will limit the practical value ofM. – – – Nyquist Formula • Nyquist’s formula indicates that, if all other things are equal, doubling the bandwidth doubles the datarate.

23. Shannon’s Channel Coding Theorem states that if theinformation • rate, R (bits/s) is equal to or less than the channel capacity, C, (i.e. R < C) then there is, in principle, a coding technique which enables transmission over the noisy channel with noerrors. • The inverse of this is that if R > C, then the probability of error is close to 1 for everysymbol. • The channel capacity is defined as: the maximum rate of reliable (error- free) information transmission through thechannel. Shannon’s Channel Capacity Theorem

24. Shannon’s Channel Capacity Theorem (or theShannon-HartleyTheorem) statesthat: where Cis the channel capacity, Bis the channel bandwidth in hertz,Sis thesignalpowerandNisthenoisepower with being the two sided noisePSD). Note: S/N is the ratio watt/watt not dB. Shannon’s Channel Capacity Theorem (

25. Transmission Impairments

26. Signals travel through transmission media, which are not perfect. The imperfection causes signal impairment. This means that the signal at the beginning of the medium is not the same as the signal at the end of the medium. What is sent is not what is received. Three causesof impairment are attenuation, distortion, and noise. Topics discussed in this section: 3 1 2 Transmission Impairments ATTENUATION Distortion Noise

27. ATTENUATION • Attenuation is a general term that refers to any reduction in the strength of a signal. Attenuation occurs with any type of signal, whether digital or analog. • Sometimes called loss, attenuation is a natural consequence of signal transmission over long distances. • Receivedsignalstrength mustbe: • strong enough tobe detected • sufficiently higherthan noise to be received withouterror Equalize attenuationacross the band of frequencies used by using loading coils oramplifiers. Strength can be increased using amplifiers or repeaters.

28. (Following the previousslide) • Attenuation is often an increasing function of frequency. This leads to attenuationdistortion: • some frequency components are attenuated more than other frequencycomponents. • Attenuation distortion is the distortion of an analog signal that occurs during transmission when the transmission medium does not have a flat frequency response across the bandwidth of the medium or the frequency spectrum of the signal. Attenuation Distortion

29. Q1) Suppose a signal travels through a transmission medium anditspowerisreducedtoone-half.ThismeansthatP2is(1/2)P1. Solution: In this case, the attenuation (loss of power) can be calculated as Alossof3dB(–3dB)isequivalenttolosingone-half thepower. Example

30. Q2) Asignaltravelsthroughanamplifier, anditspoweris increased 10 times. This means that P2= 10P1 . Solution: In this case, the amplification (gain of power) can be calculated as Example

31. Q3) One reason that engineers use the decibel to measure the changes in the strength of a signal is that decibel numbers can be added (or subtracted) when we are measuring several points (cascading) instead of just two. In Figure a signal travels from point 1 to point 4. Solution: In this case, the decibel value can be calculated as Example Figure: Decibels for Example3

32. occurs because propagation velocity of a signal through a guided medium varies with frequency • various frequency components arrive at different times resulting in phase shifts between the frequencies • particularly critical for digital data since parts of one bit spill over into others causing intersymbol interference Delay Distortion

33. Noise

34. 1 2 Categories of Noise

35. Categories of Noise 3 4

36. Q1) Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. The maximum bit rate can be calculated as Q2) Consider the same noiseless channel transmitting a signal with four signal levels (for each level, we send 2 bits). The maximum bit rate can be calculated as Example

37. Q3) We need to send 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How many signal levels do we need? • Solution • We can use the Nyquist formula as shown: • Since this result is not a power of 2, we need to either increase the number of levels or reduce the bit rate. If we have 128 levels, the bit rate is 280 kbps. If we have 64 levels, the bit rate is 240 kbps. Example

38. Q1) Consider an extremely noisy channel in which the value of the signal-to-noise ratio is almost zero. In other words, the noise is so strong that the signal is faint. For this channel the capacity C is calculated as This means that the capacity of this channel is zero regardless of the bandwidth. In other words, we cannot receive any data through this channel. Example

39. Q1) We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000. The signal-to-noise ratio is usually 3162. This means that the highest bit rate for a telephone line is 34.860 kbps. If we want to send data faster than this, we can either increase the bandwidth of the line or improve the signal-to-noise ratio. Example

40. Q1) Thesignal-to-noiseratioisoftengivenindecibels. Assume that SNRdB = 36 and the channel bandwidth is 2MHz • Solution • The theoretical channel capacity can be calculated as Example

41. For practical purposes, when the SNR is very high, we can assume that SNR + 1 is almost the same as SNR. In these cases, the theoretical channel capacity can be simplified to Q1) For example, we can calculate the theoretical capacity of the previous example as Example Solution

42. Q1) We have a channel with a 1-MHz bandwidth. The SNR for this channel is 63. What are the appropriate bit rate and signal level? • Solution • First, we use the Shannon formula to find the upper limit. Example

43. Q1) The Shannon formula gives us 6 Mbps, the upper limit. For better performance we choose something lower, 4 Mbps, for example. Then we use the Nyquist formula to find the number of signal levels. Example (continued)

44. The Shannon capacity gives us the upper limit; the Nyquist formula tells us how many signal levels we need. Note:

45. transmission concepts andterminology • guided/unguidedmedia • frequency, spectrum andbandwidth • analog vs. digitalsignals • data rate and bandwidthrelationship • transmissionimpairments • attenuation/delaydistortion/noise • channelcapacity • Nyquist/Shannon Summary

46. Thank You