Chp3_Physical Layer computer networking and communication

Chapter 3
Introduction to Physical Layer

3.2
Analog and Digital Data
Data
- Analog: refers to information that is continuous
e.g., analog clock
- Digital: refers to information that has discrete states
e.g., digital clock

3.3
Analog and Digital Signals
Like the data they represent, signals can be either
analog or digital.
- An analog signal has infinitely many levels of
intensity over a period of time
- A digital signal can have only a limited number of
defined values.

3.4
Figure 3.2: Comparison of analog and digital signals

3.5
Periodic and Nonperiodic
- A periodic signal completes a pattern within a
measurable time frame, called a period, and repeats
that pattern over subsequent identical periods.
The completion of one full pattern is called a cycle.
- A nonperiodic signal changes without exhibiting a
pattern or cycle that repeats over time.

3.6
PERIODIC ANALOG SIGNALS
Periodic analog signals can be classified as simple
or composite.
- A simple periodic analog signal, a sine wave,
cannot be decomposed into simpler signals.
- A composite periodic analog signal is composed
of multiple sine waves.

3.7
Figure 3.3: A sine wave
Time
Value
• • •

3.8
Figure 3.4: Two signals with two different amplitudes
Peak
amplitude
Peak
amplitude
Peak
amplitude
Peak
amplitude

3.9
Figure 3.5: Two signals with the same phase and frequency, but
different amplitudes

3.10
Table 3.1: Units of period and frequency

3.11
Phase
The term phase, or phase shift, describes the position of the
waveform relative to time 0.

3.12
Figure 3.6: Three sine waves with different phases

3.13
Wavelength
Wavelength binds the period or the frequency of a
simple sine wave to the propagation speed of the
medium.

3.14
Figure 3.7: Wavelength and period
Direction of
propagation

3.15
Time and Frequency Domains
A sine wave is comprehensively defined by its
amplitude, frequency, and phase.
We have been showing a sine wave by using what is
called a time domain plot. The time-domain plot
shows changes in signal amplitude with respect to
time (it is an amplitude-versus-time plot).
Phase is not explicitly shown on a time-domain plot.

3.16
Figure 3.8: The time-domain and frequency-domain plots of a sine wave

3.17
The frequency domain is more compact and useful when we
are dealing with more than one sine wave. For example,
Figure 3.9 shows three sine waves, each with different
amplitude and frequency. All can be represented by three
spikes in the frequency domain.
Example

3.18
Figure 3.9: The time domain and frequency domain of three sine waves

3.19
Bandwidth
The bandwidth is normally a difference between two
numbers. For example, if a composite signal contains
frequencies between 1000 and 5000, its bandwidth is
5000 − 1000, or 4000.

3.20
If a periodic signal is decomposed into five sine waves with
frequencies of 100, 300, 500, 700, and 900 Hz, what is its
bandwidth?
Example
Solution
Let fh be the highest frequency, fl the lowest frequency, and
B the bandwidth. Then

3.21
A periodic 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 frequencies of the same
amplitude.
Example
Solution
Let fh be the highest frequency, fl the lowest frequency, and
B the bandwidth. Then
The spectrum contains all integer frequencies. We show this
by a series of spikes (see Figure 3.15).

3.22
Figure 3.15: The bandwidth for example 3.11

3.23
DIGITAL SIGNALS
- In addition to being represented by an analog
signal, information can also be represented by a
digital signal.
For example, a 1 can be encoded as a positive
voltage and a 0 as zero voltage.
- A digital signal can have more than two levels. In
this case, we can send more than 1 bit for each
level.

3.24
Figure 3.17: Two digital signals: one with two signal levels and the
other with four signal levels

3.25
A digital signal has eight levels. How many bits are needed
per level? We calculate the number of bits from the
following formula. Each signal level is represented by 3 bits.
Example

3.26
A digital signal has nine levels. How many bits are needed
per level?
The number of bits sent per level needs to be an integer as
well as a power of 2. For this example, 4 bits can represent
one level.
Example

3.27
Bit Rate
Most digital signals are nonperiodic, and thus period
and frequency are not appropriate characteristics.
Another term— bit rate (instead of frequency)—is used
to describe digital signals.
The bit rate is the number of bits sent in 1s, expressed in
bits per second (bps).

3.28
TRANSMISSION IMPAIRMENT
- What is sent is not what is received
- Three causes of impairment are
attenuation, distortion, and noise

3.29
3.4.1 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.
- To compensate for this loss, amplifiers are used to
amplify the signal.

3.30
Figure 3.27: Attenuation and amplification

3.31
Suppose a signal travels through a transmission medium and
its power is reduced to one half. This means that P2 = 0.5
P1. In this case, the attenuation (loss of power) can be
calculated as
Example
A loss of 3 dB (−3 dB) is equivalent to losing one-half the
power.

3.32
Distortion
- Signal changes its form or shape.
- Distortion can occur in a composite signal made
of different frequencies.
- Each signal component has its own propagation
speed through a medium and, therefore, its own
delay in arriving at the final destination.
- Differences in delay may create a difference in
phase if the delay is not exactly the same as the
period duration.

3.34
Noise
- Noise is another cause of impairment.
- Several types of noise, such as thermal noise,
induced noise, crosstalk, and impulse noise, may
corrupt the signal.
- Thermal noise is the random motion of electrons in
a wire, which creates an extra signal not originally
sent by the transmitter.
- Induced noise comes from sources such as motors.
- Crosstalk is the effect of one wire on the other.

3.36
DATA RATE LIMITS
- How fast we can send data, in bits per second,
over a channel.
- Two theoretical formulas were developed to
calculate the data rate:
Nyquist for a noiseless channel
Shannon for a noisy channel.

3.37
Noiseless Channel: Nyquist Rate
For a noiseless channel, the Nyquist bit rate formula
defines the theoretical maximum bit rate.

3.38
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
Example

3.39
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
Example

3.40
We need to send 265 kbps over a noiseless channel with a
bandwidth of 20 kHz. How many signal levels do we need?
Example
Solution
We can use the Nyquist formula as shown:
Since this result 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.

3.41
Noisy Channel: Shannon Capacity
In reality, we cannot have a noiseless channel; the
channel is always noisy. In 1944, Claude Shannon
introduced a formula, called the Shannon capacity, to
determine the theoretical highest data rate for a noisy
channel:

3.42
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
Example
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.

3.43
We can calculate the theoretical highest bit rate of a regular
telephone line. A telephone line normally has a bandwidth of
3000 Hz (300 to 3300 Hz) assigned for data
communications. The signal-to-noise ratio is usually 3162.
For this channel the capacity is calculated as
Example
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.

3.44
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
Example

3.45
Using Both Limits
In practice, we need to use both methods to find the
limits and signal levels.

3.46
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?
Example
Solution
First, we 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.
Then we use the Nyquist formula to find the number of
signal levels.

3.47
PERFORMANCE
One important issue in networking is the
performance of the network— how good is it?

3.48
Bandwidth
- One characteristic that measures network performance
- can be used in two different contexts with two different
measuring values:
bandwidth in hertz
bandwidth in bits per second

3.49
Throughput
a measure of how fast we can actually send data
through a network.

3.50
Throughput
- The latency or delay 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.
- We can say that latency is made of four
components: propagation time, transmission time,
queuing time and processing delay.

3.51
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?
Example
Solution
We can calculate the throughput as
The throughput is almost one-fifth of the bandwidth in this
case.

3.52
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.
Example
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.

3.53
What are the propagation time and the transmission time for
a 2.5-KB (kilobyte) message 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.
Example
Solution
We can calculate the propagation and transmission time as
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.

3.54
What are the propagation time and the transmission time for
a 5-MB (megabyte) 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.
Example
Solution
We can calculate the propagation and transmission times as
We can calculate the propagation and transmission times as

3.55
Bandwidth-Delay Product
- Bandwidth and delay are two performance metrics
of a link.
- What is very important in data communications is
the product of the two, the bandwidth-delay product.

3.56
Figure 3.32: Filling the links with bits for Case 1

3.57
Figure 3.33: Filling the pipe with bits for Case 2

3.58
- We can think about the link between two points as a pipe.
- The cross section of the pipe represents the bandwidth,
and the length of the pipe represents the delay.
- We can say the volume of the pipe defines the
bandwidth-delay product
Example

3.59
Figure 3.34: Concept of bandwidth-delay product

3.60
Jitter
- Another performance issue that is related to delay
is jitter.
- We can roughly say that 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).

Chp3_Physical Layer computer networking and communication

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