data communication networks

Data Communication NetworksLec 8 and 9

Physical Layer and Media

• Bottom-most layer.• Interacts with transmission media.• Physical part of the network.• Provides services to data link layer.• 0s and 1s ready to be send on transmission

medium.• Must be converted into another entity-

SIGNALS

Data and Signals

• Move data in the form of electromagnetic signals across transmission medium.

Analog and Digital

• Analog is the data that is continuous.• Digital is the data that is discrete.• Sounds by human voice • Digital in the form of 1s and 0s

Analog and Digital Signals

Analog and Digital Signals

• Signals can be analog or digital. • Analog signals can have an infinite number of

values in a range.• digital signals can have only a limited number

of values.

Periodic signal

• That completes a pattern within a measurable time frame

• Repeats the pattern over subsequent identical periods.

• The completion of one full pattern is called a cycle.

Periodic Analog Signal

• Simple or composite.• Sine wave, a simple oscillating curve , changes

over course of time is smooth and consistent , a continuous , rolling flow.

A Sine Wave

Peak Amplitude

• Absolute value of its highest intensity. • Energy it carries.• Electrical signals , measured in volts.

Peak Amplitude

Period and frequency

• Period is the amount of time , in seconds , a signal need to complete one cycle.

• Frequency refer to number of periods in 1s • F = 1/T, T=1/F

Period and Frequency

3.14

To be transmitted, data must be transformed to electromagnetic signals.

Note

3.15

3-4 TRANSMISSION IMPAIRMENT

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.

AttenuationDistortionNoise

Topics discussed in this section:

3.16

Figure 3.25 Causes of impairment

3.17

Figure 3.26 Attenuation

3.18

Suppose a signal travels through a transmission medium and its power is reduced to one-half. This means that P2 is (1/2)P1. In this case, the attenuation (loss of power) can be calculated as

Example 3.26

A loss of 3 dB (–3 dB) is equivalent to losing one-half the power.

3.19

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

Example 3.27

3.20

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 3.27 a signal travels from point 1 to point 4. In this case, the decibel value can be calculated as

Example 3.28

3.21

Figure 3.27 Decibels for Example 3.28

3.22

Sometimes the decibel is used to measure signal power in milliwatts. In this case, it is referred to as dBm and is calculated as dBm = 10 log10 Pm , where Pm is the power in milliwatts. Calculate the power of a signal with dBm = −30.

SolutionWe can calculate the power in the signal as

Example 3.29

3.23

The loss in a cable is usually defined in decibels per kilometer (dB/km). If the signal at the beginning of a cable with −0.3 dB/km has a power of 2 mW, what is the power of the signal at 5 km?SolutionThe loss in the cable in decibels is 5 × (−0.3) = −1.5 dB. We can calculate the power as

Example 3.30

3.24

Figure 3.28 Distortion

3.25

Figure 3.29 Noise

3.26

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 ?

SolutionThe values of SNR and SNRdB can be calculated as follows:

Example 3.31

3.27

The values of SNR and SNRdB for a noiseless channel are

Example 3.32

We can never achieve this ratio in real life; it is an ideal.

3.28

Figure 3.30 Two cases of SNR: a high SNR and a low SNR

Digital Transmission

• To represent digital data by using digital signals.

• Involve three techniques– Line coding– Block coding– Scrambling

Line coding

• Process of converting digital data to digital signals.• Data in the form of text, numbers, audio, images are

stored in computer memory as sequence of bits.• Line coding convert sequence of bits to a digital

signal.• At the sender, digital data are encoded into digital

signal;• At the receiver, the digital data are recreated by

decoding the digital signal.

4.31

Line coding and decoding

• Difference between Data Element and Signal Element??????

4.32

4.33

Signal element versus data element

Data Rate Vs Signal Rate

• The data rate defines the number of data elements (bits) sent in 1s.

• Unit is bits per second.• The signal rate is the number of signal elements

sent in 1s.• Unit is baud.• Data rate is sometimes called bit rate, signal

rate is sometimes called pulse rate, modulation rate and baud rate.

Data Rate Vs Signal Rate

• Goal in DC is to increase data rate , while decreasing signal rate.

• Increase will increase speed of transmission.• Decrease will decrease bandwidth

requirements.

Relationship B/w Bit rate and baud rate

• Depend on the value of r.• Also depend on data pattern.• Three cases ; worst, best, and average.• Worst case is the maximum signal rate ; best

case is the minimum.• S=cxNx1/r baud

4.37

A signal is carrying data in which one data element is encoded as one signal element ( r = 1). If the bit rate is 100 kbps, what is the average value of the baud rate if c is between 0 and 1?

SolutionWe assume that the average value of c is 1/2 . The baud rate is then

Example

4.38

Although the actual bandwidth of a digital signal is infinite, the effective

bandwidth is finite.

Note

Baseline Wandering

• In decoding a digital signal, the receiver calculates a running average of the received signal power; called baseline.

• A long string of 0s and 1s can cause a drift in the baseline(baseline wandering)

• Difficult for receiver to decode correctly.• Good line coding scheme needs to prevent

baseline wandering.

DC Components

• Constant voltage level , creates frequencies around zero.

• Called DC components.• Telephone lines cannot pass frequency below

200 hz• Need scheme with no DC component.

4.41

Figure 4.3 Effect of lack of synchronization

4.42

In a digital transmission, the receiver clock is 0.1 percent faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is 1 kbps? How many if the data rate is 1 Mbps?

SolutionAt 1 kbps, the receiver receives 1001 bps instead of 1000 bps.

Example 4.3

At 1 Mbps, the receiver receives 1,001,000 bps instead of 1,000,000 bps.

Built-in Error Detection

• Capability to detect some or all errors that occurred in transmission.

Immunity to Noise and interference

• Should have capability of immunity to noise and interference.

Complexity

• Four level more costly than two levels.

4.46

Figure 4.4 Line coding schemes

Unipolar Scheme

• All signal levels are on the one side of the time axis, either above or below.

• NRZ(non-Return-to-zero): in which the positive voltage defines bit 1 and the zero voltage defines bit 0.

• Costly, power required is double than polar NRZ.

• Not used in data communication.

4.48

Figure 4.5 Unipolar NRZ scheme

Polar Scheme

• The voltage is at both side of the time axis.• Non-Return-to-Zero(NRZ):uses two levels of the

voltage amplitude.• NRZ-L:the level of the voltage determine the

value of the bit.• NRZ-I: the change or lack of change in the level of

the voltage determine the value of the bit.– If no change the bit is 0 ; If there is change the bit is 1.

4.50

Figure 4.6 Polar NRZ-L and NRZ-I schemes

4.51

In NRZ-L the level of the voltage determines the value of the bit.

In NRZ-I the inversion or the lack of inversion

determines the value of the bit.

Note

4.52

NRZ-L and NRZ-I both have a DC component problem.

Note

4.53

A system is using NRZ-I to transfer 10-Mbps data. What are the average signal rate and minimum bandwidth?

SolutionThe average signal rate is S = N/2 = 500 kbaud. The minimum bandwidth for this average baud rate is Bmin = S = 500 kHz.

Example 4.4

Return to Zero(RZ)

• Main problem is that sender and receiver are not synchronized.

• The receiver does not know when one bit has ended and next bit is starting.

• Solution id RZ, three values positive, negative and zero.

• The signal goes to zero in the middle of each bit.• Remains there until the beginning of the next bit.

Return to Zero(RZ)

• Main disadvantage , it require two signal changes to encode a signal bit , require greater bandwidth.

• No DC component.• More complex, three level of voltage , more

complex to create.

4.56

Figure 4.7 Polar RZ scheme

Biphase: Manchester

• The idea of RZ(transition at the middle of the bit). And idea at NRZ-L are combined into the Manchester scheme.

• Duration is divided into two halves.• The voltages remain at one level during the first

half and moves to the other level in the second half.

• Transition at the middle of the bit provides syncronization.

Differential Manchester

• Combines the ideas of RZ and NRZ-I.• Transition at the middle of the bit.• But the bit values are determined at the

beginning of the bit.• If next bit is 0, there is transition, if next bit is

1 , there is none.

4.59

Figure 4.8 Polar biphase: Manchester and differential Manchester schemes

4.60

In Manchester and differential Manchester encoding, the transition

at the middle of the bit is used for synchronization.

Note

Manchester and Differential Manchester

• No baseline wandering.• No DC component because each bit has a

positive and negative voltage contribution.• One , drawback is signal rate is double of NRZ.

– There is always one transition at the middle of the bit , and may be one transition at the end of the bit.

Bi-polar scheme

• There voltages levels, positive, negative and zero.

• The voltage level for one data element is at zero, while the voltage level for the other elements alternates between positive and negative.

4.63

In bipolar encoding, we use three levels: positive, zero, and negative.

Note

AMI and Pseudoternary

• AMI(alternate mark inversion).• The word mark comes from telegraphy and

means 1.• So AMI means alternate 1 inversion.• A neural zero voltage represents binary 0.• Binary 1 is represented by alternating positive

and negative voltage.• Variation of AMI is Pseudoternary.

AMI and Pseudoternary

• Developed alternate to NRZ.• Same signal rate as NRZ but no DC

component.• NRZ has most of the energy concentrated near

zero frequency, which make it unsuitable for transmission over channel with poor performance around this frequency.

4.66

Figure 4.9 Bipolar schemes: AMI and pseudoternary

No DC Component

• Long sequence of 1s, the voltage alternate between positive and negative , no constant .

• For long sequence of 0s, the voltage remain constant , but voltage is zero, which is same as no DC component.

• Used for long-distance communication, but has synchronization problem, when long sequence of 0s is present in the data.