Transcript Document

UNIT - 2
McGraw-Hill
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©The McGraw-Hill Companies, Inc., 2004
Unit 2
Title: Data, Signals & Digital Transmission
Syllabus:
Analog & Digital Signals, Transmission impairments,
Data rate limits, Performance, Digital to analog
conversion, Analog to digital conversion, Transmission
modes.
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Analog & Digital data
Data can be in analog or digital.
i) Analog Data – refers to information that is continuous and
takes continuous values.
Example: Human voice.
ii) Digital data – refers to information that has discrete states
and take discrete values.
Example: Data stored in a computer memory.
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Analog & Digital signals
Signals can be Analog or Digital.
Before transmitting the data over a medium, the data
must be converted in to electromagnetic signals.
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Analog signal
An analog signal is a continuous signal. It has an infinite number
of values in a range.
Example - Human Voice.
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Digital signal
A digital signal is a discrete signal. It has a limited number of
values.
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A sine wave
Mathematically, a Sine wave can be represented by
S(t) = A sin(2Π f t + Φ)
Where, S – instantaneous amplitude, A – peak amplitude,
f – frequency
and
Φ - phase
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Amplitude
Highest intensity of a signal represent the peak
amplitude of the signal. The intensity is proportional to
the energy it carries.
Amplitude is measured in volts.
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Period & Frequency
Period refers to the time taken by a signal to
complete one cycle & expressed in seconds. It is
denoted by ‘ T ‘ .
Frequency refers to numbers of signals produced in
one second & and expressed in hertz ( Hz ). It is
denoted by ‘ f ‘.
The relation between period and frequency is given by
T x f = 1.
Frequency and period are inverses of each other.
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Period and frequency
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Units of periods and frequencies
Unit
Seconds (s)
Equivalent
1s
Unit
hertz (Hz)
Equivalent
1 Hz
Milliseconds (ms)
10–3 s
kilohertz (KHz)
103 Hz
Microseconds (ms)
10–6 s
megahertz (MHz)
106 Hz
Nanoseconds (ns)
10–9 s
gigahertz (GHz)
109 Hz
Picoseconds (ps)
10–12 s
terahertz (THz)
1012 Hz
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Example 1
Express a period of 100 ms in microseconds, and express the
corresponding frequency in kilohertz.
Solution
We Know that, 1 ms = 10 3 µs.
100 ms = 100  103 µs = 105 ms
Now we use the inverse relationship to find the frequency,
changing hertz to kilohertz
100 ms = 100  10-3 s = 10-1 s = T
f = 1/10-1 Hz = 10  10-3 KHz = 10-2 KHz
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Note:
• Frequency is the rate of change of the
signal with respect to time.
• Change in a short span of time means
high frequency.
• Change over a long span of time means
low frequency.
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Note:
• If a signal does not change at all, its
frequency is zero.
• If a signal changes instantaneously (it
jumps from one level to another in no
time), its frequency is infinite, because
its period is zero.
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Phase
Phase describes the position
of the waveform relative to
time zero. It is measured in
degrees or radians.
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Signals with different phases
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Example
A sine wave is offset one-sixth of a cycle with
respect to time zero. What is its phase in degrees
and radians?
Solution
We know that one complete cycle is 360 degrees.
Therefore, 1/6 cycle is
(1/6) 360 = 60 degrees = 60 x 2p /360 rad = 1.046 rad
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Sine wave examples
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Sine wave examples (continued)
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Sine wave examples (continued)
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Time and frequency domains
Time-domain plot shows changes in signal amplitude with
respect to time.
Frequency-domain plot show a relations between amplitude and
frequency .
A signal with peak amplitude= 5 volts, and frequency =0
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Time and frequency domains (continued)
A signal with peak amplitude= 5 volts, and frequency = 8
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Time and frequency domains (continued)
A signal with peak amplitude= 5 volts,
and frequency = 16
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Note:
A single-frequency sine wave is
not useful in data
communications; we need to
change one or more of its
characteristics to make it useful.
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Note:
When we change one or more
characteristics of a singlefrequency signal, it becomes a
composite signal made of many
frequencies.
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Note:
According to Fourier analysis, any
composite signal can be represented as a
combination of simple sine waves with
different frequencies, phases, and
amplitudes.
Any composite signal is a sum of set sine waves of different frequencies,
phases and amplitudes. Mathematically it is represented by
S(t) = A 1 sin(2Π f1 t + Φ1) + A2 sin(2Π f2 t + Φ2) + A3 sin(2Π f 3 t + Φ3 ) +…..
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Three harmonics
A graph with three harmonic waves
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Adding first three harmonics
Fig. Adding first three harmonics
To create a complete square wave sum up all the odd
harmonics up to infinity.
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Signal corruption & Bandwidth
A signal has to pass through a medium. One of the
characteristics of the medium is frequency. The medium needs to
pass every frequency and also preserve the amplitude and phase.
No medium is perfect. A medium passes some frequencies and
blocks some others.
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Bandwidth
Bandwidth of a composite signal is the difference
between the highest and the lowest frequencies
contained in that signal.
For example, Voice normally has a spectrum of 300 –
3300 Hz. Thus, requires a bandwidth of 3000 Hz.
The bandwidth is a property of a medium.
Bandwidth is the difference between the
highest and the lowest frequencies that the
medium can satisfactorily pass.
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Bandwidth (continued)
The medium can pass some frequencies above
5000 and below 1000, but the amplitude of
those frequencies are less than those in the
middle.
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Example: If a periodic signal is decomposed into five sine waves
with frequencies of 100, 300, 500, 700, and 900 Hz, what is the
bandwidth? Draw the spectrum, assuming all components have a
maximum amplitude of 10 V.
Solution
B = fh - fl = 900 - 100 = 800 Hz
The spectrum has only five spikes, at 100, 300, 500, 700, and 900
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Example
A signal has a bandwidth of 20 Hz. The highest frequency is 60
Hz. What is the lowest frequency? Draw the spectrum if the
signal contains all integral frequencies of the same amplitude.
Solution
B = fh - f l
→
20 = 60 - fl
→
fl = 60 - 20 = 40 Hz
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Example
A signal has a spectrum with frequencies between 1000
and 2000 Hz (bandwidth of 1000 Hz). A medium can pass
frequencies from 3000 to 4000 Hz (a bandwidth of 1000
Hz). Can this signal faithfully pass through this medium?
Solution
The answer is definitely no. Although the signal can
have the same bandwidth (1000 Hz), the range does
not overlap. The medium can only pass the
frequencies between 3000 and 4000 Hz. The signal
with frequency 1000
& 2000
Hzfor students
is totally
lost.
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Digital Signals
Data can be represented by a digital signal.
Bit 1 can be encoded by positive voltage and
bit 0 can be encoded by zero voltage.
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Bit rate & Bit interval
Bit interval - is the time required to send one bit.
Bit rate - is the number of bits sent in 1 second. It is
expressed in bps.
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Example
A digital signal has a bit rate of 2000 bps.
What is the duration of each bit (bit interval)
Solution
The bit interval is the inverse of the bit rate.
Bit interval = 1/ 2000 s = 0.000500 s
= 0.000500 x 106 ms = 500 ms
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Example
Assume we need to download text documents at the
rate of 100 pages per minute. What is the required
bit rate of the channel?.
Answer
A page is an average of 24 lines with 80 characters in
each line. If we assume that one character require 8
bits.
The bit rate of the channel =
100 x 24 x 80 x 8 = 1,636,000 bps = 1.636 Mbps
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Bit Length
It is the distance one bit occupies on the transmission medium.
Bit Length = Propagation speed x Bit interval.
Transmission of Digital Signals
A digital signal can transmit by using either baseband
transmission or Broadband transmission.
Baseband transmission -means sending a digital signal
with out changing it to an analog signal.
Baseband transmission requires a low-pass channel.
Digital transmission use a low-pass channel.
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Broadband transmission (modulation)- means
sending a digital signal after changing it to an
analog signal.
Broadband transmission requires a
bandpass channel.
Analog transmission use a band-pass
channel.
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Low-pass & Band-pass
A channel is either low pass or band pass.
A low – pass channel has a B/W With frequencies
between 0 & f.
A band pass has B/W with frequencies between f1 & f2.
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Data Rate Limit
The data rate over a channel depends on 3 factors.
i) The Band width available.
ii) The levels of signals that can use for data transmission.
iii) The quality of the channel.
Two theoretical formulas were developed to calculate the data
rate.
1) Nyquist Bit Rate - Noiseless Channel
2) Shannon Capacity - Noisy Channel
Nyquist Bit Rate
Maximum Bit rate = 2 x Bandwidth x log 2L where L is number of
levels used to represent data.
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Example
Consider a noiseless channel with a
bandwidth of 3000 Hz transmitting a signal
with two signal levels. Calculate the bit
rate.
Answer:
Bit Rate = 2  3000  log2 2 = 6000 bps
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Example
Consider the same noiseless channel,
transmitting a signal with four signal levels
(for each level, we send two bits). Calculate
the bit rate.
Answer:
Bit Rate = 2 x 3000 x log2 4 = 12,000 bps
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Shannon capacity of Bit rate
Maximum data rate of a noisy
channel,
C=Band width x log 2 (1+SNR)
Where SNR is the Signal to Noise
Ratio.
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Signal – to – Noise Ratio (SNR)
SNR is defined as
SNR = Average signal power / Average noise power.
It is described in decibel units, SNR dB.
SNR dB = 10 log 10 SNR
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Example:
Consider an extremely noisy channel in which the
value of the signal-to-noise ratio is almost zero. In
other words, the noise is so strong that the signal is
faint. Calculate the channel capacity.
Answer:
C = B log2 (1 + SNR) = B log2 (1 + 0)
= B log2 (1) = B  0 = 0
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Example:
A telephone line has a bandwidth of 3000 Hz (300
Hz to 3300 Hz). The signal-to-noise ratio is 3162.
Find the channel capacity.
Answer:
C = B log2 (1 + SNR) = 3000 log2 (1 + 3162)
= 3000 log2 (3163)
= 3000  11.62 = 34,860 bps
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Example
We have a channel with a 1 MHz bandwidth. The
SNR for this channel is 63. what is the
appropriate bit rate and signal level?
Solution
First, we use the Shannon formula to find our upper
limit.
C = B log2 (1 + SNR) = 106 log2 (1 + 63) = 106 log2 (64) = 6 Mbps
Then we use the Nyquist formula to find the
number of signal levels.
6 Mbps = 2  1 MHz  log2 L  L = 8
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Transmission Impairments
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Attenuation
Attenuation means loss of energy. When a signal
(simple /composite) travels through a medium, it loses
some of its energy.
To compensate the loss, amplifiers are used to amplify
the signal. It is measured in decibels (dB)
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Decibel (dB)
To show that a signal has lost or gained strength, we
use the unit called decibel. The decibel measures the
relative strengths of two signals or one signal at two
different points. The decibel is negative if a signal is
attenuated and positive if a signal is amplified.
db = 10 log 10 p2 / p1 where p1 & p2 are the powers of a
signal at point 1 & 2 respectively.
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Example
Imagine a signal travels through a transmission
medium and its power is reduced to half. This
means that P2 = 1/2 P1. Find the attenuation
(loss of power).
Solution
Attenuation Loss =10x log10 (P2/P1)
= 10x log10 (0.5P1/P1)
= 10x log10 (0.5) = 10x(–0.3) = –3 dB
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Example
Imagine a signal travels through an amplifier and
its power is increased 10 times. This means that P2
= 10 x P1. find the amplification (gain of power).
Answer:
Amplification= 10 log10 (P2/P1)
= 10 log10 (10P1/P1)
= 10 log10 (10) = 10 (1) = 10 dB
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Example
The loss in a cable is measured in decibels per kilometer
( dB/km). If the signal at the beginning of a cable with -0.3
db/km has a power of 2 mW. What is the power of the signal at
5 km?.
Solution
The loss of energy in the cable in decibels is = 5 x (-0.3) = -1.5 dB.
Power of the signal at 5 kms= dB = 10 x log p2/p1 = - 1.5
or
p2/p1 = 10 -0.15 = 0.71
P2 = 0.71 P1 = 0.7 x 2 = 1.4 mW
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Distortion
Distortion means the signal changes its shape.
Distortion occurs in a composite signal because a it is
made up of signals of different frequencies. Each
signal has its own propagation speed. So, its own
delay in arriving at the destination.
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Noise
Several types of noise such as thermal noise, induced
noise, crass talk and impales noise may corrupt the
signals.
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Example
The power of a signal is 10 mW and the power of the noise is 1 µW.
What are the values of SNR & SNR dB?.
Solution
SNR = Average signal power / Average noise power.
= 10 x 10 -6 W / 10 -9 W = 10,000
SNR dB = 10 x log 10 SNR = 10 x log 10 10000 = 40
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Example
What is the channel capacity for a teleprinter channel
with a 300 Hz bandwidth and a signal to noise ratio of 3
dB.
Solution
Using Shannon's equation: C = B log 2 (1 + SNR)
We have B = 300 Hz,
(SNR) dB = 3
C = 300 log 2 (1 +3)
= 300 log 2 (4)
= 600 bps
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Example
A digital signaling system is required to operate at 9600 bps.
a) If a signal element encodes a 4 bit word, what is the minimum
required bandwidth of the channel?
b) Repeat part (a) for the case of 8 – bit words.
Solution
By Nyquist's Theorem: C = 2 x B x log2M
We have C = 9600 bps
a) log 2M = 4, because a signal element encodes a 4-bit word
Therefore, C = 2 x B x 4
9600 = 2xBx4
Hence, B = 1200 Hz
b)
9600 = 2 x B x 8 and
B = 600 Hz
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Example
Given a channel with an intended capacity of 20 Mbps, the
bandwidth of the channel is 3 MHz. Assuming thermal noise, what
signal to noise ratio is required to achieve this capacity?
Solution
C = B log2 (1 + SNR)
20 x 106 = 3 x 106 x log 2 (1 + SNR)
log 2 (1 + SNR) = 6.67
1 + SNR = 2 6.67
SNR = 2 6.67 - 1
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Performance
Bandwidth is a potential measure of a link.
Bandwidth in Hz – The range of frequencies that a
channel can pass.
Bandwidth in bps – Number of bits per second that a
channel can transmit.
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Throughput
Throughput is an actual measurement of
data received at the receiver.
Example
A network with bandwidth 10 Mbps can pass only an
average of 12,000 frames per minute with each frame
carrying an average of 10,000 bits. What is the
throughput of this network?
Solution
Throughput = ( 12000 x 10000) /60 = 2 Mbps
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Latency ( Delay )
The latency the time taken for an entire message to
arrive completely at the destination taken from the
time the first bit is sent out from the source.
Latency = propagation time + transmission time +
queuing time + Processing delay.
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Propagation time
Propagation time is the time required for a bit to
travel from source to the destination.
Propagation time = Distance / Propagation speed.
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Wavelength
Wave length is the distance a simple
signal can travel in one period.
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Examples
Question paper: January / February 2005
Using Shannon’s theorem, compute the maximum bit
rate for a channel having a bandwidth of 3100 Hz and
signal to noise ratio is 20 dB.
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Digital
Transmission
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Line coding
Line Coding is the process of converting binary data into digital
data.
Characteristics of line coding techniques are
1) Signal level Vs Data level
2) DC ( Direct Current ) – component & self synchronization
3)
Pulse rate Vs Bit rate
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Signal level versus data level
The number of values allowed in a signal are called Signal
Levels.
The number of values used to represent data are called Data
Levels.
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DC component
Some line coding schemes leave have residual dc component. It is undesirable because,
i) it causes distortion and may create errors in the output. ii) This component is the
extra energy residing on the line and useless. a) + voltage is not cancelled by – voltage
b) + voltage are cancelled by - voltage.
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Pulse Rate & Bit Rate
A pulse is the minimum amount of time required to
transmit a symbol.
The number of pulses per second is called pulse rate.
Bit Rate- The number of bits transmitted per second is
called bit rate.
If a pulse carries a bit, then the pulse rate = bit rate. If
pulse carries more than one bit, then bit rate > pulse
rate.
Bit Rate = Pulse Rate x log 2 L
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Example
A signal has two data levels with a pulse duration of 1
ms. Calculate the pulse rate and bit rate.
Answer:
Pulse Rate = 1/ 10-3= 1000 pulses/s
Bit Rate = Pulse Rate x log2 L = 1000 x log2 2 = 1000 bps
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Example
A signal has four data levels with a pulse duration of 1
ms. Calculate the pulse rate and bit rate.
Answer:
Pulse Rate = = 1000 pulses/s
Bit Rate = PulseRate x log2 L = 1000 x log2 4 = 2000 bps
Example
Can the bit rate be less than pulse rate.
Solution
The bit rate is always greater than or equal to the pulse rate
because a pulse contains one or more bits.
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Synchronization
At the receiver, to interpret the signal correctly , the receiver bit interval
must match exactly to the senders bit interval. If the receiver clock is faster or
slower, the bit intervals are not matched. So the output will be wrong.
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Line coding schemes
There are three classifications of line coding
1) Unipolar 2) Polar & 3) Bipolar
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Unipolar encoding
The signal levels are on one side of the time axis either above or below.
The 1’s are encoded as positive voltage value and 0’s are encoded as zero voltage value.
Disadvantages
The average amplitude of a unipolar encoded signal is nonzero. This method
lacks synchronization.
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Polar Encoding
Note:
Polar encoding uses two voltage levels
(positive and negative).
The average voltage level on the line is reduced.
The D.C component problem is alleviated.
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polar encoding
There are 4 types of polar encoding schemes
i) NonReturn to Zero ( NRZ )
ii) Returned to Zero ( RZ )
iii) Manchester
iv) Differential Manchester
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NRZ – L ( NonReturn Zero – Level ) Encoding
A positive voltage represent the bit 0,
a negative voltage represent the bit 1.
Disadvantage:
When the data contains a long stream of 0’s or 1’s cases the
problem of synchronization with the sender clock.
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NRZ – I ( NonReturn Zero – Invert ) Encoding
The inversion is the transition between a positive and a negative
voltage.
A 1 bit is represented when there is a transition.
A 0 bit is represented when no change.
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NRZ-L and NRZ-I encoding
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Example
How does NRZ – L differs from NRZ – I ?
Solution
In NRZ-L the signal depends on the state of the bit: a positive
voltage is a 0, and the negative a 1.
In NRZ-I the signal is inverted when a 1 is encountered.
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RZ ( Return to Zero ) encoding
RZ encoding uses three values- positive, negative & zero. Signal
changes for each bit. At the mid of each bit interval, the signal
returns to zero. A transition from positive to zero represents bit 1.
A transition from negative to zero represents bit 0.
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Manchester encoding
Manchester encoding uses two level of amplitude. Signal
changes for each bit. Inversion takes place at the middle of each
bit interval. A transition from negative to positive represents bit
1. A transition from positive to negative represents bit 0.
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Differential Manchester encoding
The differential Manchester encoding uses two level of amplitude.
Signal changes for each bit. Inversion takes place at the middle of each bit
interval is used for synchronization. The additional transition at the beginning of
the interval represents bit 0. The absence of additional transition at the
beginning of the interval represents bit 1.
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Bipolar Encoding
Bipolar Encoding
Note:
In bipolar encoding, we use three
levels: positive, zero,
and negative.
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Bipolar AMI (Alternate Mark Inversion) encoding
The AMI encoding uses three voltage levels, positive, negative &
zero. The zero voltage level represents bit 0. The alternate
positive and negative voltage levels represent bit 1.
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2B1Q ( two binary, one quaternary)
2BIQ encoding uses four voltage levels. Each pulse represent 2 bits.
-
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MLT-3(Multi Line Transmission three level) signal
MLT-3 encoding uses three level signals, +1,0 & -1.
The signal transitions from one level to the next at the beginning
of a bit interval represent bit 1 , there is no transition at the
beginning of a bit 0
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Example
Assume a data stream is made of ten 0’s. Encode this stream, using the following encoding
schemes. How many changes (vertical line) can you find for each scheme?
a) Unipolar b) NRZ- L c) NRZ-I d) RZ e) Manchester f) Differential Manchester g) AMI
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Example
Assume a data stream is made of ten alternate 0’s & 1’s. Encode this stream, using the
following encoding schemes. How many changes (vertical line) can you find for each
scheme?
a) Unipolar b) NRZ- L c) NRZ-I d) RZ e) Manchester f) Differential Manchester g) AMI
Solution
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EXAMPLE Jan/Feb 2005
Sketch the signal waveforms when 00110101 is transmitted in the
following signal codes.
i) NRZ – L ii) Manchester Code
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Block Coding
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Block coding
Block coding is used to improve the performance of line coding. The method has
three steps- Division, Substitution & Line coding.
Division – The sequence of bits is divided into group of m bits.
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Substitution in block coding
Substitute an m-bit code for an n-bit group. For example in the
figure, 4B/5B ,a 4-bit group is substituted to a 5-bit code. While
substituting choose a policy such that it helps in synchronization
and error detection.
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Table : 4B/5B encoding
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
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Analog to Digital conversion
SAMPLING
•Pulse Amplitude Modulation
•Pulse Code Modulation
•Sampling Rate: Nyquist Theorem
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PCM Technique
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PAM ( Pulse Amplitude Modulation)
PAM is a method used to convert analog signal level in to
discrete digital signal value at constant interval of time called
sample. PAM uses a method called sample and hold. At a given
interval, the signal level is read, then held briefly.
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Quantized PAM signal
Quantization is a method of assigning integral values in a
specific range to the sampled instances.
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Binary Encoding
The quantized samples are assigned by sign and
magnitude. Each value is translated into 8 bit
representation. In eight bits first bit is used to indicate
the sign and the other seven bits to represent the
quantized value.
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PCM
The binary digits are transformed to a digital signal by
using a line coding method called Pulse Coding Method.
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Sampling Rate : Nyquist Theorem
Note:
According to the Nyquist theorem, the
sampling rate must be at least 2 times
the highest frequency.
Example:We want to sample telephone voice with a
maximum frequency of 3000 Hz, we need a sampling
rate of 6000 samples.
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Nyquist theorem
For voice-over –phone –lines, take a sample for every 1/6000 s.
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Example
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
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Bit Rate –
Bit Rate = Sampling rate x Number of bits per sample
Example
We want to digitize the human voice. What is the bit rate,
assuming 8 bits per sample?
Solution
The human voice normally contains frequencies from 0 to
4000 Hz.
Sampling rate = 4000 x 2 = 8000 samples/s
Bit rate = sampling rate x number of bits per sample
= 8000 x 8 = 64,000 bps = 64 Kbps
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Transmission Mode
Parallel Transmission
Serial Transmission
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Data transmission
The transmission of binary data across a medium can be
made in either parallel or serial mode.
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Parallel transmission
In parallel mode, a group of n bits are sent with each clock tick. It needs n
lines
Advantage - Speed.
Disadvantage – Cost.
So, parallel transmission restricted to short distance.
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Serial transmission
In serial mode, 1 bit is sent with each clock tick. There are two classes of serial
transmission – i) Synchronous & ii) Asynchronous
Advantage – Less cost.
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Asynchronous transmission
The start bits are 0’s and the stop bits are 1’s.The gap is represented by an
idle line.
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Synchronous transmission
In Synchronous transmission, the bit stream is a combination of many bytes.
Each byte is introduced with out a gap. The receiver reconstruct the
information.
The advantage of synchronous transmission is speed.
Synchronous transmission is more faster than
Asynchronous transmission. So, Synchronous
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for high-speed
applications.
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