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DATA COMMUNICATIONS &
          NETWORKING
           LECTURE-09

         Course Instructor : Sehrish Rafiq
         Department Of Computer Science
                  University Of Peshawar
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
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.
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.
RZ ENCODING
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.
MANCHESTER ENCODING
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.
DIFFERENTIAL MANCHESTER
       ENCODING
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.
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.
BIPOLAR AMI ENCODING
2B1Q(TWO BINARY 1
            QUATERNARY)
 The 2B1Q uses four voltage levels.
 Each pulse can represent 2 bits, making each
  pulse more efficient.
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.
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.
BLOCK CODING
STEPS IN TRANSFORMATION
 Division
 Substitution

 Line coding
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.
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.
4B/5B
4B/5B

Data   Code    Data   Code
0000   11110   1000   10010
0001   01001   1001   10011
0010   10100   1010   10110
0011   10101   1011   10111
0100   01010   1100   11010
0101   01011   1101   11011
0110   01110   1110   11100
0111   01111   1111   11101
4B/5B

        Data           Code
Q (Quiet)              00000
I (Idle)               11111
H (Halt)               00100
J (start delimiter)    11000
K (start delimiter)    10001
T (end delimiter)      01101
S (Set)                11001
R (Reset)              00111
LINE CODING
   In Line coding the block codes are transformed in
     to the digital signals using one of the Line coding
    techniques.
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
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.
PAM
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
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.
QUANTIZED PAM SIGNAL
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.
PCM CONTINUED…
   The binary digits are then transformed to a
    digital signal by using one of the line coding
    techniques.
FROM ANALOG SIGNAL TO PCM
      DIGITAL CODE
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
NYQUIST THEOREM
 According to Nyquist theorem the
 sampling rate must be at least twice
 the highest frequency of the signal.
Example 1
What sampling rate is needed for a signal with a
bandwidth of 10,000 Hz (1000 to 11,000 Hz)?
Solution
The sampling rate must be twice the highest
frequency in the signal:


   Sampling rate = 2 x (11,000) = 22,000
 samples/s
NYQUIST THEOREM
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.
Example 2
A signal is sampled. Each sample requires at least 12
levels of precision (+0 to +5 and -0 to -5). How many bits
should 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.
Have a nice day!!!

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Lecture 09

  • 1. DATA COMMUNICATIONS & NETWORKING LECTURE-09 Course Instructor : Sehrish Rafiq Department Of Computer Science University Of Peshawar
  • 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. 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. 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.
  • 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.
  • 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.
  • 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. 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.
  • 13. 2B1Q(TWO BINARY 1 QUATERNARY)  The 2B1Q uses four voltage levels.  Each pulse can represent 2 bits, making each pulse more efficient.
  • 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. 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.
  • 17. STEPS IN TRANSFORMATION  Division  Substitution  Line coding
  • 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. 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. 4B/5B
  • 21. 4B/5B Data Code Data Code 0000 11110 1000 10010 0001 01001 1001 10011 0010 10100 1010 10110 0011 10101 1011 10111 0100 01010 1100 11010 0101 01011 1101 11011 0110 01110 1110 11100 0111 01111 1111 11101
  • 22. 4B/5B Data Code Q (Quiet) 00000 I (Idle) 11111 H (Halt) 00100 J (start delimiter) 11000 K (start delimiter) 10001 T (end delimiter) 01101 S (Set) 11001 R (Reset) 00111
  • 23. LINE CODING  In Line coding the block codes are transformed in to the digital signals using one of the Line coding techniques.
  • 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. 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. PAM
  • 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. 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.
  • 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. PCM CONTINUED…  The binary digits are then transformed to a digital signal by using one of the line coding techniques.
  • 32. FROM ANALOG SIGNAL TO PCM DIGITAL CODE
  • 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. NYQUIST THEOREM  According to Nyquist theorem the sampling rate must be at least twice the highest frequency of the signal.
  • 35. Example 1 What sampling rate is needed for a signal with a bandwidth of 10,000 Hz (1000 to 11,000 Hz)? Solution The sampling rate must be twice the highest frequency in the signal: Sampling rate = 2 x (11,000) = 22,000 samples/s
  • 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. Example 2 A signal is sampled. Each sample requires at least 12 levels of precision (+0 to +5 and -0 to -5). How many bits should 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. Have a nice day!!!