Introduction to Computers CS1029 Lecture 7

1. Introduction to ComputersCS1029Lecture 7 Instructor: Ibrahim Tariq

2. Physical Layer Topics to Cover Signals Digital Transmission Analog Transmission Multiplexing Transmission Media Data Communication & Networks, Summer 2009

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. Topics discussed in this section: Line Coding Line Coding SchemesBlock Coding Scrambling

4. Digital to Digital Conversion • The conversion involves three techniques:line coding,block coding, andscrambling. Line coding is always needed; block coding and scrambling may or may not be needed. • Line Coding • Line Coding Schemes • Block Coding • Scrambling Data Communication & Networks, Summer 2009

5. Line Coding & Decoding Data Communication & Networks, Summer 2009

6. Signal Levels (Elements) Vs Data Levels (Elements) Data Communication & Networks, Summer 2009

7. Example 4.1 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? Solution We assume that the average value of c is 1/2 . The baud rate is then

8. Note 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.

9. Example An analog signal carries 4 bits per signal element. If 1000 signal elements are sent per second, find the bit rate. Solution In this case, r = 4, S = 1000, and N is unknown. We can find the value of N from

10. 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? Solution In 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.

11. Figure 4.3 Effect of lack of synchronization

12. Example 4.3 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? Solution At 1 kbps, the receiver receives 1001 bps instead of 1000 bps. At 1 Mbps, the receiver receives 1,001,000 bps instead of 1,000,000 bps.

13. DC Component Data Communication & Networks, Summer 2009

14. Line Coding Schemes Data Communication & Networks, Summer 2009

15. Note In unipolar encoding, we use only one voltage level. Data Communication & Networks, Summer 2009

16. Unipolar Encoding Data Communication & Networks, Summer 2009

17. Note In polar encoding, we use two voltage levels: positive & negative Data Communication & Networks, Summer 2009

18. Polar: NRZ-L and NRZ-I Encoding Data Communication & Networks, Summer 2009

19. Note 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. Data Communication & Networks, Summer 2009

20. Polar: RZ Encoding Data Communication & Networks, Summer 2009

21. Polar: Manchester Encoding Data Communication & Networks, Summer 2009

22. Polar: Differential Manchester Encoding Data Communication & Networks, Summer 2009

23. Note In Manchester and differential Manchester encoding, the transition at the middle of the bit is used for synchronization. Data Communication & Networks, Summer 2009

24. Note In bipolar encoding, we use three levels: positive, zero, and negative. Data Communication & Networks, Summer 2009