DATA COMMUNICATION Lecture-28. Recap of Lecture 27  Frequency Ranges  Terrestrial Microwave Communication  Satellite Communication  Cellular Telephony.

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Presentation transcript:

DATA COMMUNICATION Lecture-28

Recap of Lecture 27  Frequency Ranges  Terrestrial Microwave Communication  Satellite Communication  Cellular Telephony  Transmission Impairments

Overview of Lecture 26  Transmission Impairments  Performance of Transmission Medium  Wavelength  Shannon Capacity  Media Comparison

Transmission Impairments

Attenuation  Attenuation means loss of energy  Some of electrical energy is converted to heat

Attenuation

Decibel (dB)  Relative strengths of two signals or a signal at two points dB = 10 log 10 (P 2 /P 1 ) P 2 and P 1 are signal powers  Negative dB means attenuation  Positive dB means amplification

Example 7.1  Imagine a signal travels through a transmission medium and its power is reduced to half. This means P 2 =(1/2)P 1. Calculate Attenuation?  Solution: – 10log 10 (P 2 / P 1 )= 10log 10 (0.5 P 1 / P 1 ) =10(-0.3)= -3 dB

Distortion  Distortion means that the signal changes its form or shape  Distortion occurs in a composite signal

Distortion

Noise  Thermal noise  Induced noise  Cross talk  Impulse noise

Noise

Performance  Performance of media can be measured in: – Throughput – Propagation speed – Propagation time

Throughput  How fast data can pass through a point in the medium?

Propagation speed  The distance a signal (or a bit) can travel through a medium in one second  Depends on – Medium – Frequency of Signal

Propagation Time  The time a signal (or a bit) to travel from one point of a medium to another  Propagation time =

Propagation Time  3.33  s/km for twisted pair  5  s/km for coaxial or fiber optic cable

Wavelength  Distance a simple signal can travel in one period  The wavelength depends on both the frequency and the medium – Wavelength = Propagation speed * period – Wavelength = Propagation speed / frequency

Wavelength

Shannon Capacity  Shannon introduces a formula to determine the theoretical highest data rate for a channel  C = B log 2 (1 + S/N) in bps – B: bandwidth of the channel – S/N: signal to noise ratio

Shannon Capacity (Example)  Extremely noisy channel – S/N  0 – C = B log 2 (1+0) = 0

Shannon Capacity (Example)  Telephone line – Bandwidth is 3000 Hz, S/N ratio is 3162 (35 dB) – C = 3000 log 2 (1+3162) = 34,860 bps

Media Comparison  Five factors should be kept in mind when evaluating a medium: – Cost – Speed – Attenuation – EM Interference – Security

Summary  Transmission Impairments  Performance of Transmission Medium  Wavelength  Shannon Capacity  Media Comparison

Suggested Reading  Section 7.4, 7.5, , “Data Communications and Networking” 2 nd Edition by Behrouz A. Forouzan