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# Lecture 09

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### Lecture 09

1. 1. DATA COMMUNICATIONS & NETWORKING LECTURE-09 Course Instructor : Sehrish Rafiq Department Of Computer Science University Of Peshawar
2. 2. LECTURE OVERVIEW RZ Manchester and Differential Manchester Bipolar Block coding Steps in Block coding 4B/5B substitution Analog to digital conversion Sampling PAM PCM Nyquist Theorem
3. 3. RZ OR RETURN TO ZERO ENCODING Any time the original data contain strings of consecutive 1s or 0s, the receiver can lose its place. A solution is to some how include synchronization in the encoded signal, something like the solution provided by NRZ-I but one capable of handling strings of 0’s as well as 1s. Solution: To ensure synchronization , there must be a signal change for each bit. The receiver can use these changes to synchronize its clock. NRZ-I accomplishes this for sequence of 1’s.
4. 4. RZ CONTINUED… But to change with every bit we need more than just two values. RZ encoding uses three values: positive, negative and zero. In RZ the signal changes not between bits but during each bit. A 1 bit is actually represented by positive-to-zero and a 0 bit by negative-to-zero rather than by positive and negative alone. The main disadvantage of RZ encoding is that it requires two signal changes to encode one bit and therefore occupies more bandwidth. But it is more effective than NRZ-L and NRZ-I.
5. 5. RZ ENCODING
6. 6. MANCHESTER ENCODING Manchester encoding uses an inversion at the middle of each bit interval for both synchronization and bit representation. A negative to positive transition sent binary 1 and a positive to negative transition represents binary 0. By using a single transition for a dual purpose, Manchester encoding achieves the same level of synchronization as RZ but with only two levels of amplitude.
7. 7. MANCHESTER ENCODING
8. 8. DIFFERENTIAL MANCHESTER ENCODING In differential Manchester encoding, the inversion at the middle of the bit interval is used for synchronization but the presence or absence of an additional transition at the beginning of the interval is used to identify the bit. A transition means binary 0 and no transition means binary 1. Differential Manchester encoding requires two signal changes to represent binary 0 but only one to represent binary 1.
9. 9. DIFFERENTIAL MANCHESTER ENCODING
10. 10. BIPOLAR ENCODING Bipolar like RZ uses three voltage levels: positive, negative and zero . However the zero level in bipolar encoding is used to represent binary 0. The 1’s are represented by alternating positive and negative voltages. This alternation occurs even when the one bits are not consecutive.
11. 11. AMI ENCODING A common bipolar encoding scheme is called bipolar alternate mark inversion (AMI). AMI means alternate 1 inversion. A neutral zero voltage represents binary zero. Binary 1s are represented by alternating positive and negative voltages.
12. 12. BIPOLAR AMI ENCODING
13. 13. 2B1Q(TWO BINARY 1 QUATERNARY) The 2B1Q uses four voltage levels. Each pulse can represent 2 bits, making each pulse more efficient.
14. 14. MLT-3 Multiline transmission, three level (MLT-3) is very similar to NRZ-I. But it uses three levels of signals (+1,0,-1). The signal transitions from one level to the next at the beginning of a 1 bit, there is no transition at the beginning of a zero bit.
15. 15. BLOCK CODING To improve the performance of line coding,block coding was introduced. We need some kind of redundancy to ensure synchronization. We need to include other redundandant bits to detect errors. Block coding can achieve to some extent these two goals.
16. 16. BLOCK CODING
17. 17. STEPS IN TRANSFORMATION Division Substitution Line coding
18. 18. DIVISION In this step the sequence of bits is divided in to groups of m bits. E.g. in 4B/5B encoding, the orignal bit sequence is divided in to 4-bit groups.
19. 19. SUBSTITUTION In this step we substitute m-bit code for an n-bit group. To achieve synchronization we can use the m-bit codes in such a way that for example we don’t have more than 3 consecutive 0’s and 1’s, Block coding can definitely help in error detection. Because only a subset of the 5-bit codes is used, If one or more of the bits in the block is changed in such a way that one of the unused codes is received, the receiver can easily detect the error.
20. 20. 4B/5B
21. 21. 4B/5BData Code Data Code0000 11110 1000 100100001 01001 1001 100110010 10100 1010 101100011 10101 1011 101110100 01010 1100 110100101 01011 1101 110110110 01110 1110 111000111 01111 1111 11101
22. 22. 4B/5B Data CodeQ (Quiet) 00000I (Idle) 11111H (Halt) 00100J (start delimiter) 11000K (start delimiter) 10001T (end delimiter) 01101S (Set) 11001R (Reset) 00111
23. 23. LINE CODING In Line coding the block codes are transformed in to the digital signals using one of the Line coding techniques.
24. 24. SAMPLING The process through which an Analog signal is changed to digital signal is called sampling. The idea of digitizing the analog signals started with telephone companies. Digital signals are less prone to noise and distortion. A small change in an analog signal can change the received voice substantially but it takes a considerable change to convert a 0 to 1 or a 1 to 0. PAM PCM
25. 25. PULSE AMPLITUDE MODULATION(PAM) The analog–to-digital conversion method is called pulse amplitude modulation. This technique takes an analog signal , samples it and generates a series of pulses based on the results of the sampling. Sampling: The term sampling means measuring the amplitude of the signal at equal intervals. PAM uses a technique called sample and hold. At a given moment the signal level is read and then held briefly.
26. 26. PAM
27. 27. PAM Pulse amplitude modulation has some applications, but it is not used by itself in data communication. However, it is the first step in another very popular conversion method called pulse code modulation. The sampled value occurs only instantaneously in the actual waveform but it is generalized over a still but measurable period in the PAM result. PAM is not useful to data communications because the pulses measured by PAM are still of any amplitude(still an analog signal not digital). To make them digital, we must modify them by
28. 28. PCM PCM modifies the pulses created by PAM to create a completely digital signal. PCM first quantizes the PAM pulses. Quantization is a method of assigning integral values in a specific range to sampled instances.
29. 29. QUANTIZED PAM SIGNAL
30. 30. PCM Then a sign and magnitude is assigned to quantized samples. Each value is translated in to its 7-bit binary equivalent. The eighth bit creates the sign.
31. 31. PCM CONTINUED… The binary digits are then transformed to a digital signal by using one of the line coding techniques.
32. 32. FROM ANALOG SIGNAL TO PCM DIGITAL CODE
33. 33. WHAT SHOULD BE THE SAMPLING RATE??? The accuracy of any digital reproduction of an analog signal depends on the number of samples taken. Using PAM and PCM we can reproduce the waveform exactly by taking infinite samples or we can reproduce the barest generalization of its direction change by taking three samples. How many samples are sufficient? Answer : Nyquist theorem
34. 34. NYQUIST THEOREM According to Nyquist theorem the sampling rate must be at least twice the highest frequency of the signal.
35. 35. Example 1What sampling rate is needed for a signal with abandwidth of 10,000 Hz (1000 to 11,000 Hz)?SolutionThe sampling rate must be twice the highestfrequency in the signal: Sampling rate = 2 x (11,000) = 22,000 samples/s
36. 36. NYQUIST THEOREM
37. 37. HOW MANY BITS PER SAMPLE??? This depends on the level of precision needed. The number of bits is chosen such that the original signal can be reproduced with the desired precision in amplitude.
38. 38. Example 2A signal is sampled. Each sample requires at least 12levels of precision (+0 to +5 and -0 to -5). How many bitsshould be sent for each sample?Solution We need 4 bits; 1 bit for the sign and 3 bits for the value. A 3-bit value can represent 23 = 8 levels (000 to 111), which is more than what we need. A 2-bit value is not enough since 22 = 4. A 4-bit value is too much because 24 = 16.
39. 39. Have a nice day!!!