Unit-1-03-signal-in data-communication.pdf

Signal Transmission and
Impairments
CS44 Data Communications
Dr. Shilpa chaudhari
Department of Computer Science and Engineering
RIT, Bangalore

2
Outline Session- 3
• TRANSMISSION IMPAIRMENT
◦ Attenuation
◦ Distortion
◦ Noise
• DATA RATE LIMITS
◦ Noiseless Channel: Nyquist Bit Rate
◦ Noisy Channel: Shannon Capacity
◦ Using Both Limits
• PERFORMANCE
◦ Bandwidth
◦ Throughput
◦ Latency (Delay)
◦ Bandwidth-Delay Product
◦ Jitter

3
Transmission Impairments
•Signals travel through transmission media not perfect
•Imperfection  causes signal impairment
◦ 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
◦ Attenuation
◦ Distortion
◦ Noise

4
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
◦ Why wire carrying electric signals gets warm 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
• The higher the frequency, the higher the attenuation

5
Attenuation…
•How to show that a signal has lost or gained strength?
◦ engineers use the unit of the decibel
•The decibel (dB) measures the relative strengths of two
signals or one signal at two different points
◦ Positive dB  signal is amplified (gains strength)
◦ Negative dB  signal is attenuated (loses strength)
•Decibel  defined in terms of voltage or power
◦ because power is proportional to the square of the voltage

6
Attenuation - Relative Signal Strength
• Measured in Decibel (dB)
Point 1 Point 2
dB in terms
of
Formula variables
Power dB = 10 log10 (P2/P1) P1 and P2  powers of a signal at points 1 and
2, respectively
voltage dB = 20 log10 (V2/V1) V1 and V2  voltage of a signal at points 1 and
2, respectively

7
Attenuation Example 1
•Suppose a signal travels through a transmission medium
and its power is reduced to one-half
◦ 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

8
•A signal travels through an amplifier, and its power is
increased 10 times
◦ Means that P2 = 10P1
•In this case, the amplification (gain of power) can be
calculated as

9
• One reason that engineers use the decibel to measure the
changes in the strength of a signal  decibel numbers can be
added (or subtracted) when we are measuring several points
(cascading) instead of just two
• A signal travels from point 1 to point 4
◦ The signal is attenuated by the time it reaches point 2
◦ Between points 2 and 3, the signal is amplified
◦ Again, between points 3 and 4, the signal is attenuated
• In this case, the decibel value can be calculated as
The signal has gained in power.

10
Attenuation - Link Budget
Accounting of all gains and losses of signal power throughout
the signal's path
𝑅𝑥 𝑃𝑜𝑤𝑒𝑟 𝑑𝐵 = 𝑇𝑥 𝑃𝑜𝑤𝑒𝑟 𝑑𝐵 + 𝐺𝑎𝑖𝑛𝑠 𝑑𝐵 − 𝐿𝑜𝑠𝑠𝑒𝑠 (𝑑𝐵)
Cable
Sender Receiver
Tx Power Cable loss Rx Power
TX
Amplifier
RX
Amplifier
Sender Receiver
Tx Power Rx Power
Tx amp
gain
Rx amp
gain
Tx antenna gain Rx antenna gain
path loss

11
• Sometimes the decibel is used to measure signal power in
milliwatts
◦ referred as dBm
◦ calculated as dBm = 10 log10 Pm
◦ where Pm is the power in milliwatts
• Calculate the power of a signal with dBm = −30
• We can calculate the power in the signal as

12
•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

13
Distortion
•Distortion means that the signal changes its form or shape
◦ occur in a composite signal made of different frequencies.
•Each signal component has its own propagation speed
through a medium
◦ its own delay in arriving at the final destination
•Propagation velocity varies with frequency

14
Distortion …
• If the delay is not exactly the same as the period duration, it may create a
difference in phase
◦ Signal components at the receiver have phases different from what they had at the sender
◦ The shape of the composite signal is therefore not the same

15
Noise
• Noise  Undesirable signals added between the transmitter and the receiver
◦ may corrupt the signal
Types of noise Description
thermal noise • random motion of electrons in a wire
◦ creates an extra signal not originally sent by the transmitter
induced noise • comes from sources such as motors and appliances
◦ Devices act as a sending antenna
◦ Transmission medium acts as the receiving antenna
crosstalk • effect of one wire on the other
◦ One wire acts as a sending antenna
◦ Other wire as the receiving antenna
impulse noise a spike (a signal with high energy in a very short time) that comes from power lines,
lightning, and so on
Wire 1
Wire 2

16
Noise : Signal-to-Noise Ratio
•SNR is the ratio of what is wanted (signal) to what is not
wanted (noise).
◦ A measurement of signal reception's quality
◦ often described in decibel units, SNRdB
•high SNR means the signal is less corrupted by noise.
•Low SNR means the signal is more corrupted by noise
SNRdB = 10 log10 SNR

17
•The power of a signal is 10 mW and the power of the noise
is 1 μW; what are the values of SNR and SNRdB ?
w
w
w
SNR 


10000
1
10000

 40
10
10000
10 10
log
log 4
10
10
dB



SNR

18
•The values of SNR and SNRdB for a noiseless channel are
•We can never achieve this ratio in real life; it is an ideal

19
Data Rate Limits
•Data rate: How fast we can send data in bits per second,
over a channel?
•Data rate depends on three factors:
1. The bandwidth available
2. The level of the signals we use
3. The quality of the channel (the level of noise)
•Two theoretical formulas were developed to calculate the
data rate:
◦ Nyquist for a noiseless channel
◦ Shannon for a noisy channel.

20
Noiseless Channel: Nyquist Bit Rate
• Nyquist bit rate formula defines the theoretical
maximum bit rate
◦ Bandwidth is the bandwidth of the channel
◦ L is the number of signal levels used to represent data
◦ BitRate is the bit rate in bits per second.
• theoretically , given a specific bandwidth, any bit rate can be
calculated by increasing the number of signal levels
• Practically there is a limit
◦ Increase in the number of signal levels, a burden on the receiver is imposed
◦ If the number of levels in a signal is just 2, the receiver can easily distinguish between a 0 and a 1
◦ If the level of a signal is 64, the receiver must be very sophisticated to distinguish between 64 different
levels.
◦ increasing the levels of a signal reduces the reliability of the system.
Harry Nyquist
(1889-1976)
Bit Rate = 2 × Bandwidth × log2L

21
•Does the Nyquist theorem bit rate agree with the intuitive
bit rate described in baseband transmission?
◦ They match when we have only two levels.
◦ In baseband transmission, the bit rate is 2 times the bandwidth if
we use only the first harmonic in the worst case
◦ However, the Nyquist formula is more general and can be applied
to baseband transmission and modulation
◦ It can be applied when we have two or more levels of signals.

22
•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
•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

23
•We need to send 265 kbps over a noiseless channel with a
bandwidth of 20 kHz. How many signal levels do we need?
◦ use the Nyquist formula as shown:
◦ Since this result (L) is not a power of 2, we need to either increase
the number of levels or reduce the bit rate
◦ If we have 128 levels, the bit rate is 280 kbps
◦ If we have 64 levels, the bit rate is 240 kbps

24
Noisy Channel: Shannon's Capacity
• In reality, the channel is always noisy.
• Claude Shannon introduced a formula
◦ determine the theoretical highest data rate for a
noisy channel: Shannon's Capacity
◦ bandwidth is the bandwidth of the channel (Hz)
◦ SNR is the signal-to-noise ratio
◦ capacity is the capacity of the channel in bits per second
• No indication of the signal level - means that no matter how many
levels we have, we cannot achieve a data rate higher than the capacity
of the channel
• It is a characteristic of the channel, not the method of transmission
Claude Elwood Shannon
(1916-2001)
Capacity = Bandwidth × log2(1+SNR)

25
Example 1 – Shannon's Capacity
•Consider an extremely noisy channel in which the value of
the signal-to-noise ratio is almost zero.
◦ 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
◦ We cannot receive any data through this channel

26
• We can calculate the theoretical highest bit rate of a regular
telephone line. A telephone line normally has a bandwidth of
3000. The signal-to-noise ratio is usually 3162. For this channel
the capacity is calculated as
• 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.

27
•The signal-to-noise ratio is often given in decibels. Assume
that SNRdB = 36 and the channel bandwidth is 2 MHz. The
theoretical channel capacity can be calculated as

28
•For practical purposes, when the SNR is very high, we can
assume that SNR + 1 is almost the same as SNR. In these
cases, the theoretical channel capacity can be simplified to
•For example, we can calculate the theoretical capacity of
the previous example as

29
Using Both Limits
• In practice, both methods used to find the limits and signal levels
• Example 1: We have a channel with a 1-MHz bandwidth. The SNR
for this channel is 63. What are the appropriate bit rate and
signal level?
◦ Use the Shannon formula to find the upper limit.
◦ The Shannon formula gives us 6 Mbps, the upper limit
◦ For better performance we choose something lower, 4 Mbps, for example.
◦ Use the Nyquist formula to find the number of signal levels
The Shannon capacity gives us the upper limit; the Nyquist formula tells us
how many signal levels we need.

30
Network Performance
•important issue in networking  performance of the
network
•Performance parameters:
◦ Bandwidth
◦ Analog – Hertz
◦ Digital – Bits per second (bps)
◦ Throughput
◦ Actual data rate
◦ Latency (delay)
◦ Time it takes for an entire message to completely arrive at the destination

31
Bandwidth
• used in two different contexts with two different measuring values:
• bandwidth in hertz
◦ range of frequencies contained in a composite signal or the range of frequencies
a channel can pass
◦ Ex. bandwidth of a subscriber telephone line is 4 kHz
• bandwidth in bits per second
◦ number of bits per second that a channel, a link, or even a network can transmit
◦ Ex. bandwidth of a Fast Ethernet network (or the links in this network) is a maximum of 100 Mbps -
means that this network can send 100 Mbps.
• relationship between the bandwidth in hertz and bandwidth in bits
per second
◦ increase in bandwidth in hertz means an increase in bandwidth in bits per
second
◦ depends on whether we have baseband transmission or transmission with modulation

32
Bandwidth – examples
•The bandwidth of a subscriber line is 4 kHz for voice or
data.
•The bandwidth of this line for data transmission can be up
to 56,000 bps using a sophisticated modem to change the
digital signal to analog.
•If the telephone company improves the quality of the line
and increases the bandwidth to 8 kHz, we can send
112,000 bps.

33
Throughput
• Measure of how fast we can actually send data through a
network
• Bandwidth in bits per second and throughput are different
◦ A link may have a bandwidth of B bps, but we can only send T bps
through this link with T always less than B.
◦ Bandwidth is a potential measurement of a link; the throughput is an
actual measurement of how fast we can send data.
◦ For example, we may have a link with a bandwidth of 1 Mbps, but the
devices connected to the end of the link may handle only 200 kbps
◦ cannot send more than 200 kbps through this link.
◦ Imagine a highway designed to transmit 1000 cars per minute from one
point to another.
◦ Due to congestion on the road, 100 cars per minute are transmitted
◦ The bandwidth is 1000 cars per minute; the throughput is 100 cars per minute.

34
Throughput - Example
•A network with bandwidth of 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?
•The throughput is almost one-fifth of the bandwidth in this
case.

35
Latency
• Defines how long it takes for an entire message to completely
arrive at the destination from the time the first bit is sent out
from the source
• Composed of
◦ Propagation time
◦ Transmission time
◦ Queuing time
◦ Processing time
Entire message
Transmission time
Propagation time
Latency = propagation time
+ transmission time
+ queuing time
+ processing delay

36
Latency - Propagation time
•time required for a bit to travel from the source to the
destination.
•Propagation time = Distance / (Propagation Speed)
•The propagation speed of electromagnetic signals depends
on the medium and on the frequency of the signal
◦ For example, in a vacuum, light is propagated with a speed of 3 ×
108 m/s. It is lower in air; it is much lower in cable.

37
Example Latency - Propagation time
•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.
•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.

38
Latency - Transmission time
•In data communications , message is sent not just 1 bit
◦ The first and last bit may take a time equal to the propagation time
to reach its destination
◦ The first bit leaves earlier and arrives earlier; the last bit leaves later and arrives later.
•Transmission time  time between the first bit leaving the
sender and the last bit arriving at the receiver
•The transmission time of a message depends on the size
of the message and the bandwidth of the channel
Transmission time = (Message size) / Bandwidth

39
Data bits
Latency - Transmission time
Time Time
First bit leaves
Last bit leaves
First bit arrives
Last bit arrives
Sender Receiver
Propagation time
Transmission time

40
Example 1 - Transmission time
•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 receiver is 12,000 km and that light travels
at 2.4 × 108 m/s.
Note : Message is short and the bandwidth is high so the dominant factor is the propagation
time, not the transmission time. The transmission time can be ignored.

41
Example 2 - Transmission time
•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.
Note: Message is very long and the bandwidth is not very high so the dominant factor is the
transmission time, not the propagation time. The propagation time can be ignored.

42
Latency – Queuing time
•time needed for each intermediate or end device to hold
the message before it can be processed.
◦ not a fixed factor - changes with the load imposed on the network.
◦ When there is heavy traffic on the network, the queuing time increases
◦ An intermediate device, such as a router, queues the arrived
messages and processes them one by one
◦ If there are many messages, each message will have to wait

43
Bandwidth-Delay Product
•The product of bandwidth and delay is the number of bits
that can fill the link
•two hypothetical cases as examples
◦ Filling the links with bits
◦ Filling the pipe with bits
•Two cases show that product of bandwidth and
delay is the number of bits that can fill the link

44
• Filling the links with bits
◦ assuming that we have a link with a bandwidth of 1 bps (unrealistic).
◦ We also assume that the delay of the link is 5 s (also unrealistic).
◦ We want to see what the bandwidth-delay product means in this case?
◦ This product 1 × 5 is the maximum number of bits that can fill the link.
◦ There can be no more than 5 bits at any time on the link.

45
•Filling the pipe with bits
◦ Assuming we have a bandwidth = 5 bps, and delay of the link = 5 s
◦ There can be maximum 5 × 5 = 25 bits on the line.
◦ The reason is that, at each second, there are 5 bits on the line
◦ The duration of each bit is: 1/5 = 0.20 s.

46
•The number of bits that can fill the link is important if we
need to send data in bursts and wait for the
acknowledgment of each burst before sending the next
one
•To use the maximum capability of the link, we need to
make the size of our burst 2 times the product of
bandwidth and delay
◦ sender should send a burst of data of (2 × bandwidth × delay) bits
◦ sender then waits for receiver acknowledgment for part of the
burst before sending another burst
◦ amount 2 × bandwidth × delay is the number of bits that can be in
transition at any time

47
Jitter
•Related to delay is jitter
•Jitter is a problem if different packets of data encounter
different delays and the application using the data at the
receiver site is time-sensitive (audio and video data, for
example)
•If the delay for the first packet is 20 ms, for the second is
45 ms, and for the third is 40 ms, then the real-time
application that uses the packets endures jitter.

48
Summary
• Signals get impaired by attenuation, distortion, and noise
• For a noiseless channel, the Nyquist bit rate formula defines the theoretical
maximum bit rate.
• For a noisy channel, we need to use the Shannon capacity to find the
maximum bit rate.
• Attenuation, distortion, and noise can impair a signal.
• Attenuation is the loss of a signal’s energy due to the resistance of the
medium.
• Distortion is the alteration of a signal due to the differing propagation
speeds of each of the frequencies that make up a signal.
• Noise is the external energy that corrupts a signal.
• The bandwidth-delay product defines the number of bits that can fill the
link.

49
Homeworks
•Solve the chapter 3 problems: P3-15 to P3-33

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