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Data Communication

Data Communication. Lecture # 05 Course Instructor: Engr. Sana Ziafat. Digital Transmission. Digital to Digital Conversion.

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Data Communication

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  1. Data Communication Lecture # 05 Course Instructor: Engr. Sana Ziafat

  2. Digital Transmission

  3. 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.

  4. Line Coding & Decoding

  5. Signal Levels (Elements) Vs Data Levels (Elements) • Data element is defined as smallest entity to represent piece of information • Where as signal element is the shortest unit of digital signal. • Data element is actually what we need to end and signal element is what we can send

  6. Signal Levels (Elements) Vs Data Levels (Elements)

  7. “r” is the ratio of data element to signal element

  8. Data rate versus signal rate • Data rate: -Number of data elements sent in one second -The unit is bits per second(bps) • Signal Rate: -Number of signals elements sent in one second. -The unit is baud • Increase the data rate while decreasing the signal rate. • Increasing the data rate increase the speed of data transmission • Decreasing the signal rate decrease the bandwidth requirement.

  9. Data Rate vs. Signal Rate • Increasing the data rate increases the speed of transmission where as increasing the signal rate increases the bandwidth requirement • HOW???

  10. Relationship of data rate and signal rate • S= c* N * 1/r

  11. Pulse Rate Vs Bit Rate Example A signal has two data levels with a pulse duration of 1 ms. We calculate the pulse rate and bit rate as follows: Pulse Rate = 1/ 10-3= 1000 pulses/s Bit Rate = Pulse Rate x log2 L = 1000 x log2 2 = 1000 bps

  12. Example 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

  13. Important Characteristics of line coding • No of signal levels • Bit rate verses baud rate • DC components • Noise immunity • Error detection • Synchronization • Cost of implementation

  14. DC Component

  15. Lack of Synchronization

  16. Example 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: 1000 bits sent 1001 bits received1 extra bps At 1 Mbps: 1,000,000 bits sent 1,001,000 bits received1000 extra bps

  17. Line Coding Schemes

  18. Note In unipolar encoding, we use only one voltage level.

  19. UnipolarNRZ Encoding

  20. Characteristics of Unipolar Signal • It uses only one polarity of voltage level • Bit rate same as data rate • Dc component present • Lack of synchronization for long sequence of 0’s & 1’s. • Simple but obsolete

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

  22. Polar: NRZ-L and NRZ-I Encoding

  23. 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. Both have average signal rate of N/2 baud.

  24. Problems Associated with NRZ-L & NRZ-I • Synchronization problem • Dc component problem • Sudden change of polarity in system (for case of NRZ-L)

  25. Polar: RZ Encoding

  26. Polar: Manchester Encoding

  27. Polar: Differential Manchester Encoding

  28. Note In Manchester and differential Manchester encoding, the transition at the middle of the bit is used for synchronization.

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

  30. Bipolar schemes: AMI and pseudo ternary • AMI: Alternate Mark Inversion • It uses three voltage levels • Unlike RZ 0 level is used to represent a 0 • Binary 1’s are represented by alternating positive and negative. • Pseudoternary • variation of AMI encoding in which 1 bit is encoded as zero voltage level and 0 bit is encoded as alternating positive and negative voltages

  31. Summary

  32. Readings Chapter 4 (B. A Forouzan) Section 4.1

  33. Q & A

  34. Assignment#1 • Draw the graphs for all encoding techniques learnt in this lecture, using each of following data stream. a.11111111 b.00000000 c.00110011 d.01010101 • What is the Nyquist sampling rate for each of the following signals? a. A low pass signal with bandwidth of 300 KHz? b. A band pass signal with bandwidth of 300KHz if the lowest frequency is 100 KHz?

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