Introduction to Computers CS1029 Lecture 7

Introduction to ComputersCS1029

Lecture 7

Instructor: Ibrahim Tariq

Data Communication & Networks, Summer 2009

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Physical Layer Topics to Cover

Signals

Digital TransmissionAnalog Transmission

Multiplexing

Transmission Media

4.3

4-1 DIGITAL-TO-DIGITAL CONVERSION

In this section, we see how we can represent digital data by using digital signals. The conversion involves three techniques: line coding, block coding, and scrambling. Line coding is always needed; block coding and scrambling may or may not be needed.

Line CodingLine Coding SchemesBlock CodingScrambling

Topics discussed in this section:

Data Communication & Networks, Summer 2009

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Digital to Digital Conversion

• The conversion involves three techniques: line coding, block coding, and scrambling. Line coding is always needed; block coding and scrambling may or may not be needed.

• Line Coding• Line Coding Schemes• Block Coding• Scrambling

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Line Coding & Decoding

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Signal Levels (Elements) Vs Data Levels (Elements)

4.7

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 4.1

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Bit rate is the number of bits per second. Baud rate is the number of signal

elements per second.

In the analog transmission of digital data, the baud rate is less than

or equal to the bit rate.

Note

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An analog signal carries 4 bits per signal element. If 1000 signal elements are sent per second, find the bit rate.

SolutionIn this case, r = 4, S = 1000, and N is unknown. We can find the value of N from

Example

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Example

An analog signal has a bit rate of 8000 bps and a baud rate of 1000 baud. How many data elements are carried by each signal element? How many signal elements do we need?

SolutionIn this example, S = 1000, N = 8000, and r and L are unknown. We find first the value of r and then the value of L.

4.11

Figure 4.3 Effect of lack of synchronization

4.12

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.

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DC Component

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Line Coding Schemes

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In unipolar encoding, we use only one voltage level.

Note

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Unipolar Encoding

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In polar encoding, we use two voltage levels: positive & negative

Note

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Polar: NRZ-L and NRZ-I Encoding

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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

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Polar: RZ Encoding

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Polar: Manchester Encoding

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Polar: Differential Manchester Encoding

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In Manchester and differential Manchester encoding, the transition

at the middle of the bit is used for synchronization.

Note

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

Note