LECTURE 3.pptx

Concepts and Terminologies
Transmission Impairment
Signal-to-Noise Ratio (SNR)
Bandwidth
Channel Capacity
TRANSMISSION IMPAIRMENTS AND CHANNEL CAPACITY

TRANSMISSION TERMINOLOGY
 Two devices communicate with each other by sending and receiving data. Data transmission
occurs between transmitter and receiver over some transmission medium.
Communication between two devices can be classified as:
Simplex
Half-duplex
Full-duplex
 Simplex communication is one-way (unidirectional). Communication is possible in only one
direction. Example: Broadcasting, paging system, etc.
Figure 1: Simplex mode of Communication

TRANSMISSION TERMINOLOGY
Half Duplex is two-way communication but not simultaneous. It uses the same radio channel
for both transmission and reception. This means that at any given time, a user can only transmit or
receive information. An example is a walkie-talkie.
In full Duplex, both stations can transmit and receive simultaneously. One common example of
full-duplex communication is the telephone network. When two people are communicating by a
telephone line, both can talk and listen at the same time.
Figure 2: Half Duplex mode of Communication
Figure 3: Full Duplex mode of Communication

TRANSMISSION IMPAIRMENT
 With any communications system, the signal received may differ from the signal transmitted
due to various transmission impairments.
 Signals travel through transmission media, which are not perfect. The imperfection causes
signal impairment. This means that the signal at the beginning of the medium is not the same as
the signal at the end of the medium. What is sent is not what is received. Three causes of
impairment are attenuation, distortion, and noise.
 For analog signals, these impairments can degrade the signal quality. For digital signals, bit
errors may be introduced, such that a binary 1 is transformed into a binary 0 or vice versa.
Figure 4: Transmission Impairment

 The strength of the signal falls with distance over a transmission medium. The extent of
attenuation is a function of distance, transmission medium, as well as the frequency of the
underlying transmission.
 Even in free space, with no other impairment, the transmitted signal attenuates over distance
simply because the signal is being spread over a larger area. The received signal must have
sufficient strength so that the circuitry in the receiver can interpret the signal. The signal must
maintain a level sufficiently higher than noise to be received without error. Attenuation is
greater at higher frequencies, causing distortion.
 When a signal travels through a medium it loses energy overcoming the resistance of the
medium. Attenuation is the loss of energy as it travels through the medium. Amplifiers are
used to compensate for this loss of energy by amplifying the signal.
TRANSMISSION IMPAIRMENT (ATTENUATION)
Figure 5: Signal Attenuation

To show the loss or gain of energy, the unit “decibel” is used. It measures the relative strength
of two signals.
P2: Power at the receiver(output signal)
P1: Power at the sender(input signal)
dB
P
P
 10
2
1
10
log
Figure 5: Signal Attenuation

Attenuation introduces three considerations for the transmission engineer.
 First, a received signal must have sufficient strength so that the electronic circuitry in the
receiver can detect the signal.
 Second, the signal must maintain a level sufficiently higher than noise to be received without
error.
 Third, attenuation varies with frequency. (equalize attenuation across a band of frequencies
used)
Note:
Some books define dB in terms of the voltage of the signal. This is because power is proportional
to the square of the voltage.

1. Suppose a signal travels through a transmission medium and its power is reduced to one-half.
This means that P2 is (1/2) P1. In this case, the attenuation (loss of power) can be calculated as:
A loss of 3 dB (–3 dB) is equivalent to losing one-half the power. (what does a gain of 3dB
means?)
2. A signal travels through an amplifier, and its power is increased 10 times. This means that P2 =
10P1. In this case, the amplification (gain of power) can be calculated as:
TRANSMISSION IMPAIRMENT (ATTENUATION)-EXAMPLE

 One reason that engineers use the decibel to measure the changes in the strength of a signal is
that decibel numbers can be added (or subtracted) when we are measuring several points
(cascading) instead of just two. In the figure below, a signal travels from point 1 to point 4. In
this case, the decibel value can be calculated as:
 Sometimes the decibel is used to measure signal power in milliwatts. In this case, it is referred
to as dBm and is calculated as dBm = 10 log10 Pm, where Pm is the power in milliwatts.
Calculate the power of a signal with dBm = −30.
 Solution:
We can calculate the power in the signal as

The loss in a cable is usually defined 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 in the cable in decibels is 5 × (−0.3) = −1.5 dB. We can calculate the power as:

TRANSMISSION IMPAIRMENT (NOISE)
Noise is a major limiting factor in communications system performance. Noise may be divided
into four categories namely: Thermal Noise, Intermodulation noise, Crosstalk, Impulse Noise
Thermal noise
Thermal noise due to agitation of electrons and is uniformly distributed across the
frequency spectrum.
It is present in all electronic devices and transmission media. It is a function of temperature.
It is uniformly distributed across the bandwidths typically used in communications systems
and hence is often referred to as white noise. It cannot be eliminated
The amount of thermal noise found in a bandwidth of 1Hz in any device or conductor is:
FORMULA IS KBT.

Example 1: Given a receiver with an effective noise temperature of 294 K and a 10-MHz
bandwidth, calculate the thermal noise level at the receiver’s output.
Example 2: Calculate the amount of thermal noise to be found in a bandwidth of 1 Hz in a device
at room temperature, T = 17.
Noise is assumed to be independent of frequency. Thermal noise present in a bandwidth of B
Hertz (in watts) is shown below.

Intermodulation noise
Produced by nonlinearities in the transmitter, receiver, and/or intervening transmission
medium.
caused by signals produced at frequencies that are sums or differences of carrier frequencies.
Occurs if signals with different frequencies share the same medium. XPM, SPM, FWM, etc.
Crosstalk
 Crosstalk is an unwanted coupling between signal paths (interference between two or more
signals).
 A signal from one line is picked up by another
 It can also occur when microwave antennas pick up unwanted signals
Impulse noise
 It is non-continuous, consisting of irregular pulses or noise spikes of short duration and of
relatively high amplitude. It is generated from a variety of causes, including external
electromagnetic disturbances, such as lightning, and faults and flaws in the communications
system.
 While an impulse noise may not have a significant impact on analog data, it has a noticeable
effect on digital data, causing burst errors.

SIGNAL-TO-NOISE RATIO (SNR)
 To measure the quality of a system, the SNR is often used.
 It indicates the strength of the signal with respect to the noise power in the system. It is
the ratio between two powers. It is usually given in dB and referred to as SNRdB.
 Engineers like to express signal-to-noise ratio in decibels (dB) using the following quantity:
SNR SNR
dB  10 10
log
10log10(S/N) or
Example: a signal-to-noise ratio of 100 is expressed as 20dB
Example: a signal-to-noise ratio of 30dB is the same as 10(30/10) or 1000
avsp = average signal power
avnp = average noise power
SNR
avsp
avnp

10
In terms of voltage, SNR 20log s
dB
n
v
v
 
  
 
 A ratio higher than 1 and greater zero indicates more signal than noise.

 The noisier the signal, the lower the SNR and the higher the SNR the better the system.
SIGNAL-TO-NOISE RATIO (SNR)
Figure 6: Two Cases of SNR: A High and a Low SNR Figure 7: Eye Diagram for Low and High SNR

 In communications and electronics, distortion is the alteration of the waveform of an
information-bearing signal, such as an audio signal representing sound or a video signal
representing images, in an electronic device or communication channel.
 Distortion occurs in composite signals (collection of other component signals). Each frequency
component has its own propagation speed traveling through a medium.
 Delay distortion occurs because the velocity of propagation of a signal through a guided
medium varies with frequency. Thus various frequency components of a signal will arrive at
the receiver at different times, resulting in phase shift (a change in the phase of a waveform)
between different frequencies.
 It only occurs in guided media. Some of the signal components of a one-bit position will spill
over into other bit positions, causing intersymbol interference.
DISTORTION

DISTORTION
This is because each frequency signal has its own propagation speed through a medium
Figure 8: Diagram of Distortion

1. What are the propagation time and the transmission time for a 5 Mbyte message (an image) if
the bandwidth of the network is 1 Mbps? Assume that the distance between the sender and the
receiver is 12,000 km and that light travels at 2.4 × 108 m/s.
Solution:
DISTORTION-EXAMPLE
 Note that in this case, because the message is very long and the bandwidth is not very high, the
dominant factor is the transmission time, not the propagation time. The propagation time can be
ignored in most cases.
tan
Pr
dis ce
opagationTime
propagation speed

TransmisisonTime
MessageSize
Bandwidth


OTHER IMPAIRMENTS
Atmospheric absorption: water vapor and oxygen contribute to attenuation
Multipath: obstacles reflect signals so that multiple copies with varying delays are received
Refraction: bending of radio waves as they propagate through the atmosphere
Fading: Variation over time or distance of received signal power caused by changes in the
transmission medium or path(s)

CHANNEL CAPACITY
A very important consideration in data communications is how fast we can send data, in bits per
second, over a channel. Data rate depends on three factors: the bandwidth available, the level of
the signals we use, and the quality of the channel (the level of noise)
 The maximum rate at which data can be transmitted over a given communication channel under
given conditions is referred to as the channel capacity. There are four concepts here that relate
to one another; Data rate, Bandwidth, Noise, and Error rate. Data rate (bps) is the rate at which
data can be communicated. Impairments such as noise limit the data rate that can be achieved.
Bandwidth (B) is defined as the difference between the highest (f2) and the lowest frequency
(f1) that a channel can transmit. Noise is the impairment on the communication path. Error rate
is the rate at which errors occur.
Bit rate is defined as the speed information is transferred between the transmitter and receiver.
It is measured in bit/sec (bps). Example 1: Calculate the bit rate if a pattern 1010010101 is
transmitted in 10 ms.
3
Number of bits transmitted 10
Bit rate = 1000
time taken 10*10
bps

 

BANDWIDTH
 Bandwidth is defined as the amount of information that can flow through a network connection
in a given period. It is important to understand the concept of bandwidth for the following
reasons.
 Bandwidth is finite. Regardless of the media used to build a network, there are limits on the
network’s capacity to carry information. Bandwidth is limited by the laws of physics and by the
technologies used to place information on the media.
 For example, the bandwidth of a conventional modem is limited to about 56 kbps by both the
physical properties of twisted-pair phone wires and by modem technology. DSL uses the same
twisted-pair phone wires.
 However, DSL provides much more bandwidth than conventional modems. So, even the limits
imposed by the laws of physics are sometimes difficult to define.

BANDWIDTH
Optical fiber has the physical potential to provide virtually limitless bandwidth. Even so, the
bandwidth of optical fiber cannot be fully realized until technologies are developed to take full
advantage of its potential.
Bandwidth is not free. It is possible to buy equipment for a LAN that will provide nearly
unlimited bandwidth over a long period of time. For WAN connections, it is usually necessary
to buy bandwidth from a service provider.
In either case, individual users and businesses can save a lot of money if they understand
bandwidth and how the demand will change over time.
A network manager needs to make the right decisions about the kinds of equipment and
services to buy.
Bandwidth is an important factor that is used to analyze network performance, design new
networks, and understand the Internet.
A networking professional must understand the tremendous impact of bandwidth and
throughput on network performance and design. Information flows as a string of bits from
computer to computer throughout the world. These bits represent massive amounts of
information flowing back and forth across the globe in seconds or less.

BANDWIDTH
 The demand for bandwidth continues to grow. As soon as new network technologies and
infrastructures are built to provide greater bandwidth, new applications are created to take
advantage of the greater capacity. The delivery of rich media content such as streaming video
and audio over a network requires tremendous amounts of bandwidth.
 IP telephony systems are now commonly installed in place of traditional voice systems, which
further adds to the need for bandwidth.
 The successful networking professional must anticipate the need for increased bandwidth and
act accordingly. Bandwidth is not free, Bandwidth requirements are growing at a rapid rate,
and Bandwidth is critical to network performance.

BANDWIDTH OF A SIGNAL
Bandwidth can be defined as the portion of the electromagnetic spectrum occupied by the signal
OR
Bandwidth (transmission capacity) is defined as the difference between the highest (f2) and the
lowest frequency (f1) that a channel can transmit (It is the frequency range over which a signal
is transmitted). BW= f2-f1
Bandwidth of an Analog Signal:
 Bandwidth of an analog signal is expressed in terms of its frequencies.
 It is defined as the range of frequencies that the composite analog signal carries.
 It is calculated by the difference between the maximum frequency and the minimum
frequency.
 Bandwidth of a Digital Signal:
 It is defined as the maximum bit rate of the signal to be transmitted.
 It is measured in bits per second.

Figure 9: Bandwidth of a Signal in Time Domain and Frequency Domain
BANDWIDTH OF A SIGNAL
The signal shown in the diagram is a composite
analog signal with many component signals
It has a minimum frequency of F1=30 Hz and a
maximum frequency of F2=90Hz.
The bandwidth is given by F2-F1=90-30=60Hz
Example 2: What is the lowest frequency if a signal
has a BW of 200 Hz and its highest frequency is
600Hz?
Solution: f1 = f2 -BW=600-200=400 Hz

PROPAGATION AND TRANSMISSION DELAY
Propagation speed - speed at which a bit travels though the medium from source to destination.
Transmission speed - the speed at which all the bits in a message arrive at the destination.
(difference in arrival time of first and last bit)
Propagation Delay = Distance/Propagation speed
Transmission Delay = Message size/bandwidth bps
Latency = Propagation delay + Transmission delay + Queueing time + Processing time
Example: What is the propagation time if the distance between the two points is 12,000 km?
Assume the propagation speed to be 2.4 × 108 m/s in cable.
Solution
We can calculate the propagation time as
 The example shows that a bit can go over the Atlantic Ocean in only 50 ms if there is a direct
cable between the source and the destination.

EXAMPLE
What are the propagation time and the transmission time for a 2.5-kbyte message (an e-mail) if
the bandwidth of the network is 1 Gbps? Assume that the distance between the sender and the
Solution
We can calculate the propagation and transmission time as shown
Note that in this case, because the message is short and the bandwidth is high, the dominant factor
is the propagation time, not the transmission time. The transmission time can be ignored.

What are the propagation time and the transmission time for a 5-Mbyte message (an image) if
the bandwidth of the network is 1 Mbps? Assume that the distance between the sender and the
Solution
We can calculate the propagation and transmission times as shown
EXAMPLE
Note that in this case, because the message is very long and the bandwidth is not very high, the
dominant factor is the transmission time, not the propagation time. The propagation time can be
ignored.

BANDWIDTH OF A CHANNEL
A channel is the medium through which the signal carrying information will pass.
In terms of an analog signal, the bandwidth of the channel is the range of frequencies that the
channel can carry.
In terms of digital signal, the bandwidth of the channel is the maximum bit rate supported by
the channel. i.e. the maximum amount of data that the channel can carry per second.
The bandwidth of the medium should always be greater than the bandwidth of the signal to be
transmitted else the transmitted signal will be either attenuated or distorted or both leading to
loss of information.
The channel bandwidth determines the type of signal to be transmitted i.e. analog or digital.

 The Shannon formula represents the theoretical maximum that can be achieved. In practice,
however, only much lower rates are achieved, in part because the formula only assumes white
noise (thermal noise).
 Capacity C=B log2(1+SNR)
C is the capacity of the channel in bits per second and B is the bandwidth of the channel in
Hertz.
 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. For this channel the capacity
C is calculated as:
 This means that the capacity of this channel is zero regardless of the bandwidth. In other
words, we cannot receive any data through this channel.
 THE MAXIMUM AMOUNT OF INFORMATION WHICH CAN BE SENT OVER A
MEDIUM.
SHANNON CAPACITY

Example 1: A telephone line has a bandwidth of 3000Hz. The signal-to-noise ratio is 3162.
Calculated the channel capacity.
This means that the highest bit rate for a telephone line is 34.860 kbps. If we want to send data
faster than this, we can either increase the bandwidth of the line or improve the signal-to-noise
ratio.
SHANNON CAPACITY

Example 2: The signal-to-noise ratio is often given in decibels. Assume that SNRdB = 36 dB and
the channel bandwidth is 2 MHz. The theoretical channel capacity can be calculated as:
SHANNON CAPACITY
Example 3: If the SNR of a wireless communication link is 20 dB and the RF bandwidth is 30
kHz. Determine the maximum theoretical data rate that can be transmitted.
Solution:
Given: S/N=20 dB = 100
RF= Bandwidth (B) = 30000 Hz
Using Shannon’s channel capacity formula;
C= Blog2(1+S/N) = 30000log2(1+100) = 199.75 kbps

SHANNON CAPACITY
Example 4: Assume that the TV picture is to be transmitted over a channel with 4.5 MHz
bandwidth and a 35 dB signal-to-noise ratio. Find the capacity of the channel (bps).

SHANNON CAPACITY
Example 5: If the SNR of a wireless communication link is 20 dB and the RF bandwidth is 30
kHz, determine the maximum theoretical data that can be transmitted.
Example 6: The SNR for a wireless channel is 30 dB and the RF bandwidth is 200 kHz. Compute
Shannon’s maximum data rate that can be transmitted over this channel.

NYQUIST BANDWIDTH
 Consider a noise-free channel where the limitation on data rate is simply the bandwidth of the
signal. Nyquist states that if the rate of signal transmission is 2B, then a signal with frequencies
no greater than B is sufficient to carry the signal rate. Conversely, given the bandwidth of B, the
highest signal rate that can be carried is 2B.
 This limitation is due to the effect of intersymbol interference, such as is produced by delay
distortion.
 If the signals to be transmitted are binary (two voltage levels), then the data rate that can be
supported by B Hz is 2B bps.
 We can increase the rate by using M or L signal levels
 Nyquist Formula is: C = 2B log2M or L
 C is the capacity of the channel in bits per second and B is the bandwidth of the channel in
Hertz.
 Increasing the levels of a signal may reduce the reliability of the system. This is because the
receiver needs better hardware and a reception signal to recover the data.

TWO DIGITAL SIGNALS: ONE WITH TWO SIGNAL LEVELS AND THE OTHER
WITH FOUR SIGNAL LEVELS
Figure 10: Digital Signals with Different Levels

1. Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two
signal levels. The maximum bit rate can be calculated as:
2. Consider the same noiseless channel transmitting a signal with four signal levels (for each
level, we send 2 bits). The maximum bit rate can be calculated as:
3. Find the channel capacity of a noiseless 3-kHz channel. Assume binary signals are used.
Solution: C=2Blog2L =2*3000log22 =6000 bps
4. Consider a voice channel being used via a modem of 8 discrete signal levels to transmit digital
data. Assume a bandwidth of 3100 Hz. Find the Nyquist capacity of the channel.
NYQUIST BANDWIDTH-EXAMPLE

NYQUIST BANDWIDTH-EXAMPLE
 Suppose that the spectrum of a channel is between 3 MHz and 4 MHz and the SNRdB =24 dB,
find the maximum channel based on Shannon’s formula. How many signal levels are required
based on Nyquist’s formula?
What is the maximum bit rate of a noiseless channel with a bandwidth of 5000 Hz transmitting
a signal with two signal levels.

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