Lecture 3: Data Transmission

Transcription

1 Lecture 3: Data Transmission 1 st semester By: Elham Sunbu

2 OUTLINE Data Transmission DATA RATE LIMITS Transmission Impairments Examples

3 DATA TRANSMISSION The successful transmission of data depends on two factors:: 1. characteristics of the transmission medium 2. quality of the signal being transmitted

4 TRANSMISSION TERMINOLOGY Data transmission occurs between transmitter and receiver over some transmission medium. Communication is in the form of electromagnetic waves. Guided media twisted pair, coaxial cable, optical fiber Unguided media (wireless) air, vacuum, seawater

5 TRANSMISSIONTERMINOLOGY No intermediate devices Direct Link Direct link. Only 2 devices share link. Point to point More than 2 devices share link. Multi point

6 SPECTRUM & BANDWIDTH 1. Spectrum (Range of frequencies contained in signal) 2. Absolute bandwidth (Width of spectrum) 3. Effective bandwidth (*Often just bandwidth. * Narrow band of frequencies containing most energy.) 4. Dc component ( component of zero frequency)

7 SIGNAL WITH DC COMPONENT

8 DATA RATE AND BANDWIDTH any transmission system has a limited band of frequencies limiting bandwidth creates distortions this limits the data rate that can be carried on the transmission medium most energy in first few components square waves have infinite components and hence an infinite bandwidth There is a direct relationship between data rate and bandwidth.

9 ANALOG AND DIGITAL DATA TRANSMISSION data entities that convey information signals electric or electromagnetic representations of data signaling physically propagates along a medium transmission communication of data by propagation and processing of signals

10 ACOUSTIC SPECTRUM (ANALOG)

11 ADVANTAGE AND DISADVANTAGE OF DIGITAL SIGNALS * Cheaper. * Less susceptible to noise interference. * Suffer more from attenuation digital now preferred choice. ADVANTAGES & DISADVANTAGES OF DIGITAL SIGNALS

12 AUDIO SIGNALS frequency range of typical speech is 100Hz-7kHz easily converted into electromagnetic signals varying volume converted to varying voltage can limit frequency range for voice channel to Hz

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 = lines per 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 DATA RATE LIMITS 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 three factors: 1. The bandwidth available 2. The level of the signals we use 3. The quality of the channel (the level of noise) Topics discussed in this section: 1. Channel Capacity. 2. Nyquist Bit Rate, Noisy Channel. 3. Shannon Capacity Using Both Limits

18 CHANNEL CAPACITY - Maximum rate at which data can be transmitted over a given communications channel under given conditions data rate in bits per second Bandwidth in cycles per second or Hertz Noise average noise level over path Error rate rate of corrupted bits limitations due to physical properties main constraint on achieving efficiency is noise

19 CHANNEL CAPACITY Channel capacity is concerned with the information handling capacity of a given channel. It is affected by: The attenuation of a channel which varies with frequency as well as channel length. The noise induced into the channel which increases with distance. Non-linear effects such as clipping on the signal. Some of the effects may change with time e.g. the frequency response of a copper cable changes with temperature and age.

20 CHANNEL CAPACITY Obviously we need a way to model a channel in order to estimate how much information can be passed through it. Although we can compensate for non linear effects and attenuation it is extremely difficult to remove noise. The highest rate of information that can be transmitted through a channel is called the channel capacity, C.

21 Assume a channel is noise free. Nyquist formulation: NYQUIST FORMULA if the rate of signal transmission is 2B, then a signal with frequencies no greater than B is sufficient to carry the signal rate. Given bandwidth B, highest signal rate is 2B. Why is there such a limitation? 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 one bit

22 NYQUIST FORMULA Signals with more than two levels can be used, i.e., each signal element can represent more than one bit. 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 formula becomes: C = 2B log 2 M M is the number of discrete signal levels, B is the given bandwidth, C is the channel capacity in bps. How large can M be? The receiver must distinguish one of M possible signal elements. Noise and other impairments on the transmission line will limit the practical value of M. Nyquist s formula indicates that, if all other things are equal, doubling the bandwidth doubles the data rate.

23 SHANNON S CHANNEL CAPACITY THEOREM Shannon s Channel Coding Theorem states that if the information 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 no errors. The inverse of this is that if R > C, then the probability of error is close to 1 for every symbol. The channel capacity is defined as: the maximum rate of reliable (error- free) information transmission through the channel.

24 SHANNON S CHANNEL CAPACITY THEOREM Shannon s Channel Capacity Theorem (or the Shannon-Hartley Theorem) states that: where C is the channel capacity, B is the channel bandwidth in hertz, S is the signal power and N is the noise power with ( being the two sided noise PSD). Note: S/N is the ratio watt/watt not db.

25 TRANSMISSION IMPAIRMENTS

26 TRANSMISSION IMPAIRMENTS 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 causes of impairment are attenuation, distortion, and noise. Topics discussed in this section: ATTENUATION Distortion Noise

27 ATTENUATION It is a signal strength falls off with distance over any transmission medium. varies with frequency Equalize attenuation across the band of frequencies used by using loading coils or amplifiers. Received signal strength must be: strong enough to be detected sufficiently higher than noise to be received without error Strength can be increased using amplifiers or repeaters.

28 ATTENUATION DISTORTION (Following the previous slide) Attenuation is often an increasing function of frequency. This leads to attenuation distortion: some frequency components are attenuated more than other frequency components. Attenuation distortion is particularly noticeable for analog signals: the attenuation varies as a function of frequency, therefore the received signal is distorted, reducing intelligibility.

29 EXAMPLE Q1) Suppose a signal travels through a transmission medium and its power is reduced to one-half. This means that P 2 is (1/2)P 1. Solution: In this case, the attenuation (loss of power) can be calculated as A loss of 3 db ( 3 db) is equivalent to losing one-half the power.

30 EXAMPLE Q2) A signal travels through an amplifier, and its power is increased 10 times. This means that P2 = 10 P1. Solution: In this case, the amplification (gain of power) can be calculated as

31 EXAMPLE 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 Figure : Decibels for Example 3

32 DELAY DISTORTION 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

33 NOISE unwanted signals inserted between transmitter and receiver is the major limiting factor in communications system performance

34 CATEGORIES OF NOISE 1 Thermal noise Due to thermal agitation of electrons. Uniformly distributed across bandwidth. Referred to as white noise 2 Intermodulation noise produced by nonlinearities in the transmitter, receiver, and/or intervening transmission medium effect is to produce signals at a frequency that is the sum or difference of the two original frequencies

35 CATEGORIES OF NOISE 3 Crosstalk a signal from one line is picked up by another can occur by electrical coupling between nearby twisted pairs or when microwave antennas pick up unwanted signals 4 Impulse Noise caused by external electromagnetic interferences noncontiguous, consisting of irregular pulses or spikes short duration and high amplitude minor annoyance for analog signals but a major source of error in digital data

36 EXAMPLE Q1) Sometimes the decibel is used to measure signal power in milliwatts. In this case, it is referred to as db m and is calculated as db m = 10 log10 P m, where P m is the power in milliwatts. Calculate the power of a signal with db m = 30. Solution We can calculate the power in the signal as

37 EXAMPLE Q1) The power of a signal is 10 mw and the power of the noise is 1 μw; what are the values of SNR and SNRdB? Solution The values of SNR and SNR db can be calculated as follows:

38 EXAMPLE 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

39 EXAMPLE 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.

40 EXAMPLE 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.

41 EXAMPLE Q1) We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of The signal-to-noise ratio is usually This means that the highest bit rate for a telephone line is 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.

42 EXAMPLE Q1) The signal-to-noise ratio is often given in decibels. Assume that SNRdB = 36 and the channel bandwidth is 2 MHz Solution The theoretical channel capacity can be calculated as

43 EXAMPLE 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 Solution

44 EXAMPLE 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.

45 EXAMPLE (CONTINUED) 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.

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

47 SUMMARY transmission concepts and terminology guided/unguided media frequency, spectrum and bandwidth analog vs. digital signals data rate and bandwidth relationship transmission impairments attenuation/delay distortion/noise channel capacity Nyquist/Shannon

48 THANK YOU