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.

5. RZ ENCODING

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. MANCHESTER ENCODING

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. DIFFERENTIAL MANCHESTER
ENCODING

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.

12. BIPOLAR AMI ENCODING

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.

16. BLOCK CODING

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.

29. QUANTIZED PAM SIGNAL

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

36. NYQUIST THEOREM

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!!!