Lecture # 05
Digital Transmission
Course Instructor:
Engr. Sana Ziafat
Digital Transmission
Digital to Digital Conversion
 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.
Line Coding & Decoding
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
Signal Levels (Elements) Vs Data
Levels (Elements)
“r” is the ratio of data element to signal
element
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.
Data Rate vs. Signal Rate
 Increasing the data rate increases the speed of
transmission where as increasing the signal rate
increases the bandwidth requirement
Relationship of data rate and signal
rate
 S= c* N * 1/r
Pulse
Example
Rate Vs Bit Rate
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
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
Important Characteristics of line
coding
 No of signal levels
 Bit rate verses baud rate
 DC components
 Noise immunity
 Error detection
 Synchronization
 Cost of implementation
DC Component
Lack of Synchronization
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
Line Coding Schemes
Note
In unipolar encoding, we use only one
voltage level.
Unipolar NRZ Encoding
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
Note
In polar encoding, we use two voltage
levels: positive & negative
Polar: NRZ-L and NRZ-I Encoding
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.
Problems Associated with NRZ-L &
NRZ-I
 Synchronization problem
 Dc component problem
 Sudden change of polarity in system (for case of NRZ-
L)
Polar: RZ Encoding
Polar: Manchester Encoding
Polar: Differential Manchester
Encoding
Note
In Manchester and differential Manchester
encoding, the transition
at the middle of the bit is used for
synchronization.
Note
In bipolar encoding, we use three levels:
positive, zero, and negative.
Bipolar schemes: AMI and pseudo
ternary
AMI: Alternate Mark Inversion
o It uses three voltage levels
oUnlike RZ 0 level is used to represent a 0
oBinary 1’s are represented by alternating positive and negative.
Pseudoternary
o 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
Summary
Scrambling
 It is solution for the synchronization problem
associated with Bipolar Schemes.
 Two common scrambling techniques are:
1.
B8ZS (Eight consecutive zero levels are replaced by sequence
000VB0VB)
2. HDB3 (Four consecutive zero levels are replaced by 000V 0r B00V)
Figure Two cases of B8ZS scrambling technique
Note
B8ZS substitutes eight consecutive
zeros with 000VB0VB.
Figure Different situations in HDB3 scrambling technique
Note
HDB3 substitutes four consecutive
zeros with 000V or B00V depending
on the number of nonzero pulses after
the last substitution.
Analog To Digital Conversion
Sampling
Pulse Code Modulation
Sampling Rate: Nyquist Theorem
Figure Components of PCM encoder
Note
According to the Nyquist theorem, the
sampling rate must be
at least 2 times the highest frequency
contained in the signal.
Figure Nyquist sampling rate for low-pass and bandpass signals
Figure Quantization and encoding of a sampled signal
Figure Components of a PCM decoder
Figure The process of delta modulation
Figure Delta modulation components
Figure Delta demodulation components
Readings
 Chapter 4 (B. A Forouzan)
 Section 4.1, 4.2, 4.3
Q&A
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
o 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?