Signal Propagation

This is the movement of radio waves from a transmitter to a receiver. When the waves travel
(propagate) from one point to another, they are, like light waves, affected by different phenomena
such as light reflection, absorption, or scattering.
Signal
 A ‘signal’ is an electrical or electromagnetic current that is used for carrying data from one
device or network to another.
 For example, you cannot roll up a photograph, insert it into a wire, and transmit it across
town.
 You can transmit, however, an encoded description of the photograph. Instead of sending
the actual photograph, you can use an encoder to create a stream of 1’s and 0’s that tells a
receiving device how to reconstruct the image of the photograph.
 But even 1’s and 0’s cannot be sent as such across network links. They must be further
converted into a form that transmission media can accept.
 Transmission media work by conducting energy along a physical path. So a data stream of
1’s and 0’s must be turned into energy in the form of electromagnetic signals.
Types of Signals :
 The classification of Signals:
 Analog and Digital Signals
 Periodic and Aperiodic Signals
 Continuous and Discrete Signals
 Deterministic and Random Signals
 Energy and Power Signals
 Even and Odd Signals.
Analog and Digital Signals :
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
An ANALOG signal is a continuous wave form that changes smoothly over time. As the wave
moves from value A to value B, it passes through and includes an infinite number of values
along its path.
 An example of Analog data is the Human voice.
 A DIGITAL signal, on the other hand,is discrete. It can have only a limited number of defined
values, often as simple as ‘1’ and ‘0’.
 An example of Digital data is data stored in the memory of a computer in the form of 0’s
and 1’s.
Periodic and Aperiodic Signals :
Fig.2 (a)




Both Analog and Digital Signals can be of two forms: Periodic and Aperiodic (Nonperiodic).
A PERIODIC signal consists of a continuously repeated pattern.
The period of a signal (T) is expressed in seconds.
Whereas, an APERIODIC, or NONPERIODIC, signal changes constantly without exhibiting a
pattern or cycle that repeats over time.
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Fig.2 (b)
Continuous and Discrete Signals :
Fig.3

A signal of continuous amplitude and time is known as a CONTINUOUS signal or an ANALOG
signal.
 This signal will have some “value” at every instant of time.
 The electrical signals derived in proportion with the physical quantities such as
temperature, pressure, sound, sine wave, etc. Are generally continuous signals.
 Whereas, DISCRETE signal, if the signal is represented only at “Discrete instants of time”,
then it is known as a DISCRETE time signal.
 The discrete time signals have “value” only at certain instant of time.
Deterministic and Random Signals :
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Fig.4

A signal which can be described by s mathamaical expression, look up table or some well
defined rule is known as the DETERMINISTIC signal. Examples- sine wave, cosine wave,
square wave etc.
 Mathamatically, X(t) = A sin (2πft)
 Whereas, a signal which cannot be described by any mathamaical expression is called as a
RANDOM signal. Example- ‘Noise’.
Energy and Power Signals :
Fig.5



A signal is called as a POWER signal if it’s “average normalized power” is non-zero and
finite.
It has been observed that almost all the PERIODIC signals are POWER signals.
Whereas, a signal having a finite non-zero total normalized energy is called as an ENERGY
signal.
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
It is observed that almost all the NON-PERIODIC signals defined over a finite period, are
energy signals.
Even and Odd Signals :



EVEN signals are symmetric around vertical axis, and ODD signals are symmetric about
origin.
EVEN signal, a signal is referred to as an EVEN if it is identical to its time-reversed
counterparts; g(x) = g(-x).
ODD signal, a signal is ODD if h(-x) = -h(x).
FREQUENCY
 Frequency is a electromagnetic wave using AC (Alternating Current).
Just as the name implies, "frequency", its something that happens over and over and over
again. It is very frequent, consistent, and repetitive.
 There are different types of frequency; light, sound and in our case radio frequency
"Frequency is the number of times a specified event occurs within a specified time interval. A
standard measure of frequency is hertz (Hz)"
This specified event mentioned in the CWNA Study Guide is the cycle.
CYCLE
"An oscillation, or cycle, of this alternating current is defined as a single change from up to down
to up, or as a change from positive, to negative to positive.”
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Lets look at a few examples of a Cycle.
Example 1 - (1) Cycle
 One cycle, specified event, is measured 1 second in time which equals 1 Hz. "alternating
current is defined as a single change from up to down to up, or as a change from
positive, to negative to positive"
Example 2 - (5) Cycles
Five cycles, specified events, measured 1 second in time which equals 5 Hz.
High frequency simple means there are more cycles per second.
Example 3 - Low and High frequency example
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So — Remember —- Frequency is simply something that repeats itself over and over again. It is
measured in cycles per seconds. The more cycles per second, the more frequency or referenced
as higher frequency.
WAVELENGTH
"Wavelength is the distance between similar points on two back-to-back waves."
Waves can be measured at different points. In the below example, reference #1 is the most
often way wavelength is measured.
Example 4
AMPLITUDE
 "Amplitude is the height, force or power of the wave"
What is important to remember — frequency, cycle and wavelength remain constant, however,
the hight of the wave form is dynamic based on the power of the wave. The higher power, or
amplitude, the higher the wave form peeks. The lower the power, or amplitude, the lower the
wave form peeks all while frequency, cycle and wavelength remain the same.
Example 5 - Amplitude shown by the hight or peeks of the wave form.
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PHASE
 Phase is the same frequency, same cycle, same wavelength, but are 2 or more wave
forms not exactly aligned together.
"Phase is not a property of just one signal but instead involves the relationship between two or
more signals that share the same frequency. The phase involves the relationship between the
position of the amplitude crests and troughs of two waveforms.
Phase can be measured in distance, time, or degrees. If the peaks of two signals with the same
frequency are in exact alignment at the same time, they are said to be in phase. Conversely, if
the peaks of two signals with the same frequency are not in exact alignment at the same time,
they are said to be out of phase."
Example 6
Below is an example of 2 wave forms 90 degree out of phase.
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Attenuation
 Attenuation means a loss of energy when a signal, simple or composite, travels through
a medium, it loses some of its energy in overcoming the resistance of the medium.
That is why a wire carrying electric signals gets warm, if not hot, after a while. Some of the
electrical energy in the signal is converted to heat. To compensate for this loss, amplifiers are
used to amplify the signal.
Transmission impairment
 Transmission impairment occurs when the received signal is different from the
transmitted signal.
 As we know, a signal can be transmitted as Analog signal or it can be transmitted as a
digital signal.
In Analog signals due to transmission impairment the resulting received signal gets different
amplitude or the shape. In the case of digitally transmitted signals at the receiver side we get
changes in bits (0's or 1's).
Causes
There are various causes of transmission impairments, which are as follows −

Noise
 Delay Distortion
 Attenuation Distortion
Let us discuss Delay Distortion.
Delay Distortion
This kind of distortion is mainly appearing in case of composite signals in which a composite
signal has various frequency components in it and each frequency component has some time
constraint which makes a complete signal.
But while transmitting this composite signal, if a certain delay happens between the
frequencies components, then there may be the chance that the frequency component will
reach the receiver end with a different delay constraint from its original which leads to the
change in shape of the signal. The delay happens due to environmental parameters or from the
distance between transmitter and receiver etc.
The delay distortion is diagrammatically represented as follows −
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Refer the diagram given below −
Step 1 − In the above diagram, the delay distortion is due to velocity of propagation the
frequency varies. Thus, various frequency components of a signal arrive at the receiver at
different times.
Step 2 −: For digital data, fast components from one bit may catch up and overtake slow
components from bit ahead, mixing the two bits and increasing the probability of incorrect
reception.
Step 3 − Thus, it is difficult for digital data because signal components of bit positions spill into
other bit positions, and so limit the allowed rate of transmission.
Data Rate Limit:
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Data rate governs the speed of data transmission. A very important consideration in data
communication is how fast we can send data, in bits per second, over a channel. Data rate
depends upon 3 factors:



The bandwidth available
Number of levels in digital signal
The quality of the channel – level of noise
Two theoretical formulas were developed to calculate the data rate: one by Nyquist for a
noiseless channel, another by Shannon for a noisy channel.
1. Noiseless Channel : Nyquist Bit Rate –
For a noiseless channel, the Nyquist bit rate formula defines the theoretical maximum bit rate
BitRate = 2 * Bandwidth * log2(L)
In the above equation, bandwidth is the bandwidth of the channel, L is the number of signal
levels used to represent data, and BitRate is the bit rate in bits per second.
Bandwidth is a fixed quantity, so it cannot be changed. Hence, the data rate is directly
proportional
to
the
number
of
signal
levels.
Note –Increasing the levels of a signal may reduce the reliability of the system.
Examples:
Input1 : Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with
two
signal
levels.
What
can
be
the
maximum
bit
rate?
Output1 : BitRate = 2 * 3000 * log2(2) = 6000bps
Input2 : We need to send 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How
many signal levels do we need?
Output2 : 265000 = 2 * 20000 * log2(L)
log2(L) = 6.625
L = 26.625 = 98.7 levels
2. Noisy Channel : Shannon Capacity –
In reality, we cannot have a noiseless channel; the channel is always noisy. Shannon capacity is
used, to determine the theoretical highest data rate for a noisy channel:
Capacity = bandwidth * log2(1 + SNR)
In the above equation, bandwidth is the bandwidth of the channel, SNR is the signal-to-noise
ratio, and capacity is the capacity of the channel in bits per second.
Bandwidth is a fixed quantity, so it cannot be changed. Hence, the channel capacity is directly
proportional to the power of the signal, as SNR = (Power of signal) / (power of noise).
The signal-to-noise ratio (S/N) is usually expressed in decibels (dB) given by the formula:
10 * log10(S/N)
so for example a signal-to-noise ratio of 1000 is commonly expressed as:
10 * log10(1000) = 30 dB.
Examples:
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Input1 : A telephone line normally has a bandwidth of 3000 Hz (300 to 3300 Hz) assigned for
data communication. The SNR is usually 3162. What will be the capacity for this channel?
Output1 : C = 3000 * log2(1 + SNR) = 3000 * 11.62 = 34860 bps
Input2 : The SNR is often given in decibels. Assume that SNR(dB) is 36 and the channel
bandwidth is 2 MHz. Calculate the theoretical channel capacity.
Output2 : SNR(dB) = 10 * log10(SNR)
SNR = 10(SNR(dB)/10)
SNR = 103.6 = 3981
Hence, C = 2 * 106 * log2(3982) = 24 MHz
Performance of Channel(Network)




Bandwidth
Throughput
Latency (Delay)
Jitter
BANDWIDTH
 Simple Bandwidth means how much data transmit.
 Bandwidth determines how rapidly the webserver is able to upload the requested
information.
 Bandwidth is characterized as the measure of data or information that can be
transmitted in a fixed measure of time.
 The term can be used in two different contexts with two distinctive estimating values.
In the case of digital devices, the bandwidth is measured in bits per second(bps) or
bytes per second. In the case of analogue devices, the bandwidth is measured in
cycles per second, or (when we measure the frequency i.e wireless) Hertz (Hz).
“Bandwidth” means “Capacity” and “Speed” means “Transfer rate”.
 Bandwidth in Hertz: It is the range of frequencies contained in a composite signal or
the range of frequencies a channel can pass. For example, let us consider the
bandwidth of a subscriber telephone line as 4 kHz.
 Bandwidth in Bits per Seconds: It refers to the number of bits per second that a
channel, a link, or rather a network can transmit. For example, we can say the
bandwidth of a Fast Ethernet network is a maximum of 100 Mbps, which means that
the network can send 100 Mbps of data.
THROUGHPUT
 Simple, How fast data can be transmitted.
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 Throughput is the number of messages successfully transmitted per unit time.
 It is controlled by available bandwidth, the available signal-to-noise ratio and
hardware limitations. The maximum throughput of a network may be consequently
higher than the actual throughput achieved in everyday consumption.
 The terms ‘throughput’ and ‘bandwidth’ are often thought of as the same, yet they are
different. Bandwidth is the potential measurement of a link, whereas throughput is an
actual measurement of how fast we can send data.
Throughput is measured by tabulating the amount of data transferred between multiple
locations during a specific period of time, usually resulting in the unit of bits per second(bps),
which has evolved to bytes per second(Bps), kilobytes per second(KBps), megabytes per
second(MBps) and gigabytes per second(GBps). Throughput may be affected by numerous
factors, such as the hindrance of the underlying analogue physical medium, the available
processing power of the system components, and end-user behaviour. When numerous
protocol expenses are taken into account, the use rate of the transferred data can be
significantly lower than the maximum achievable throughput.
Let us consider: A highway which has a capacity of moving, say, 200 vehicles at a time. But at
a random time, someone notices only, say, 150 vehicles moving through it due to some
congestion on the road. As a result, the capacity is likely to be 200 vehicles per unit time and
the throughput is 150 vehicles at a time.
Example:
Input:A network with bandwidth of 10 Mbps can pass only an average of 12, 000 frames
per minute where each frame carries an average of 10, 000 bits. What will be the
throughput for this network?
Output: We can calculate the throughput asThroughput = (12, 000 x 10, 000) / 60 = 2 Mbps
The throughput is nearly equal to one-fifth of the bandwidth in this case.
For the difference between Bandwidth and Throughput, refer.
LATENCY
 Simple, How much time to take reach source to destination.
 latency(also known as delay) is defined as the total time taken for a complete
message to arrive at the destination, starting with the time when the first bit of the
message is sent out from the source and ending with the time when the last bit of the
message is delivered at the destination.
 The network connections where small delays occur are called “Low-LatencyNetworks” and the network connections which suffer from long delays are known as
“High-Latency-Networks”.
 Latency is also known as a ping rate and is measured in milliseconds(ms).
In simpler terms: latency may be defined as the time required to successfully send a packet
across a network.
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

It is measured in many ways like round trip, one way, etc.
It might be affected by any component in the chain which is utilized to vehiculate data,
like workstations, WAN links, routers, LAN, servers and eventually may be limited for large
networks, by the speed of light.
Latency = Propagation Time + Transmission Time + Queuing Time + Processing Delay
Propagation Time: It is the time required for a bit (first bit) to travel from the source to the
destination. Propagation time can be calculated as the ratio between the link length
(distance) and the propagation speed over the communicating medium. For example, for an
electric signal, propagation time is the time taken for the signal to travel through a wire.
Propagation time = Distance / Propagation speed
Example:
Input: What will be the propagation time when the distance between two points is
12, 000 km? Assuming the propagation speed to be 2.4 * 10^8 m/s in cable.
Output: We can calculate the propagation time asPropagation time = (12000 * 10000) / (2.4 * 10^8) = 50 ms
Transmission Time: Transmission time is a time based on how long it takes to send the signal
down the transmission line. It consists of time costs for an EM signal to propagate from one
side to the other, or costs like the training signals that are usually put on the front of a packet
by the sender, which helps the receiver synchronize clocks. The transmission time of a
message relies upon the size of the message and the bandwidth of the channel.
Transmission time = Message size / Bandwidth
Example:
Input:What will be the propagation time and the transmission time for a 2.5-kbyte
message when the bandwidth of the network is 1 Gbps? Assuming the distance between
sender and receiver is 12, 000 km and speed of light is 2.4 * 10^8 m/s.
Output: We can calculate the propagation and transmission time asPropagation time = (12000 * 10000) / (2.4 * 10^8) = 50 ms
Transmission time = (2560 * 8) / 10^9 = 0.020 ms
Note: Since the message is short and the bandwidth is high, the dominant factor is the
propagation time and not the transmission time(which can be ignored).
Queuing Time: Queuing time is a time based on how long the packet has to sit around in the
router. Quite frequently the wire is busy, so we are not able to transmit a packet
immediately.
Processing delay:
Now the packet will be taken for the processing which is called processing delay.
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Time is taken to process the data packet by the processor that is the time required by
intermediate routers to decide where to forward the packet, update TTL, perform header
checksum calculations.
JITTER
 Jitter is another performance issue related to delay. In technical terms, jitter is a
“packet delay variance”.
 It can simply mean that jitter is considered as a problem when different packets of
data face different delays in a network and the data at the receiver application is timesensitive, i.e. audio or video data. Jitter is measured in milliseconds(ms).
 It is defined as an interference in the normal order of sending data packets. For
example: if the delay for the first packet is 10 ms, for the second is 35 ms, and for the
third is 50 ms, then the real-time destination application that uses the packets
experiences jitter.
In the above image, it can be noticed that the time it takes for packets to be sent is not the
same as the time in which he will arrive at the receiver side. One of the packets faces an
unexpected delay on its way and is received after the expected time. This is jitter.
Bit Error Rate(BER)
 Bit Error Rate, BER is used as an important parameter in characterizing the performance
of data channels.
 When transmitting data from one point to another, either over a radio/ wireless link or
a wired telecommunications link, the key parameter is how many errors will appear in
the data that appears at the remote end.
As such Bit Error Rate, BER is applicable to everything from fiber optic links, to ADSL, Wi-Fi,
cellular communications, IoT links and many more.
Even though the data links may utilize very different types of technology, the basics of the
assessment of the bit error rate are exactly the same.
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Numerical Assisgnment of Nyquist Theorem and Shanon Theorem
For Nyquist Theorem
Formula: Bit rate= 2 X bandwidth X log2 L
Q1. A noiseless channel with a bandwidth of 3000Hz transmitting a signal with two signal
levels. Calculate the maximum bit rate.
Bitrate=2*3000* log2 2=6000bps
Q2. The noiseless channel with a bandwidth of 3000Hz transmitting a signal with four signal
levels.Calculate maximum bit rate.
Bitrate=2*3000* log2 4=12000bps
Q3. Suppose we need to send 265 kbps over a noiseless channel with a bandwidth of 20KHz.
How many signal levels do we need ?
265000=2*20000* log2 L
log2 L=6.625
or, L=26.625 =98.7 levels
Shannon Capacity:
Formulla: Capacity=bandwidth X log2(1+SNR) where SNR is signal to noise ratio.
Q1. Consider an extreamly noisy channel in which the value of the signal to noise ratio is almost
0 .In other words, the noise is no strong that the signal is faint. Calculate the capacity.
C=Blog2(1+SNR)= Blog2(1+0)=B*0=0
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Q2. We can calculate the theoretical highest bitrate of a regular telephone line. A telephone line
normally has a bandwidth of 3000Hz(300-3300Hz) assigned for data communication. The signal
to noise ratio is usually 3162. Calculate capacity for this channel.
C=Blog2(1+SNR)=3000* log2(1+3162)=3000 log23163=3000*11.62=34860bps
Q3. The signal-to-noise ratio is often given in decibels. Assume that SNRDb = 36 and the channel
bandwidth is 2 MHz. Calculate The theoretical channel capacity
SNRDb = 10log10 SNR => SNR= lOSNRdb/10 =>SNR=10 3.6 =3981
C =Blog2 (1+ SNR)= 2 X 106 X log2 3982 =24 Mbps
Q4. For practical purposes, when the SNR is very high, we can assume that SNR + I is almost
the same as SNR. In these cases, simplified the theoretical channel capacity.
C= B X
𝑆𝑁𝑅 𝐷𝑏
3
C=2MHz X
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36
3
For example, we can calculate the theoretical capacity of the previous
=24Mbps