Chapter 3: Introduction to Physical

Layer
Revision
Question 1

 Given the frequency 140 KHz, calculate the

corresponding period.
Solution
 To calculate the period from the frequency, we use the formula:

 T= 1/f, ​where:

• T is the period in seconds,

• f is the frequency in Hz.

 Given f=140 KHz =140,000 Hz

 T=1/140,000 = 7.14×10−6 seconds = 7.14μs

 So, the period is 7.14 microseconds.

Question 2

 The period is equal 220 ns, calculate the

corresponding frequency.
Solution
 To calculate the frequency from the period, we use the formula:

 f=1/T , ​where:

• f is the frequency in Hz,

• T is the period in seconds.

 Given T=220 ns=220×10−9 seconds, the frequency f is:

 f=1/220×10−9 = 4.545 MHz

 So, the frequency is approximately 4.545 MHz.

Question 3
 What is the bandwidth of a signal that can be
decomposed into five sine waves with
frequencies at 0, 20, 50, 100, and 200 Hz? All
peak amplitudes are the
Solution
 The bandwidth of a signal is the difference between the
highest and lowest frequencies.

 Given the frequencies 0 Hz, 20 Hz, 50 Hz, 100 Hz, and

200 Hz, the bandwidth is:

 Bandwidth=200 − 0= 200 Hz
Question 4
 A periodic composite signal with a bandwidth
of 2000 Hz is composed of two sine waves.
 The first one has a frequency of 100 Hz with a
maximum amplitude of 20 V; the second one
has a maximum amplitude of 5 V. Draw the
bandwidth.
Solution
 A periodic composite signal composed of two sine waves
with the following information:
• First sine wave: frequency 100 Hz, amplitude 20 V
• Second sine wave: unknown frequency, amplitude 5 V
 Since the bandwidth is 2000 Hz, and the first sine wave is
at 100 Hz, the second sine wave must be at:
 Second frequency=100 +2000 =2100 Hz
Question 5

 Which signal has a wider bandwidth, a sine wave with a

frequency of 100 Hz or a sine wave with a frequency of

200 Hz?
Solution

 The bandwidth of a sine wave is typically zero because

a pure sine wave contains only a single frequency.

Therefore, neither signal has a "wider" bandwidth. Both

have the same bandwidth (zero).

Question 6

 What is the bit rate for each of the following signals?

 a. A signal in which 1 bit lasts 0.001 s

Solution

 The bit rate R :

 R=1/ (bit duration)

 1 bit lasts 0.001 s

 R=1(1 / 0.001 s )=1000 bps = 1 kbps

Question 7

 A device is sending out data at the rate of 1000 bps.

 How long does it take to send out 10 bits?

Solution
Question 8
 A signal travels from point A to point B.

 At point A, the signal power is 100 W.

 At point B, the power is 90 W.

 What is the attenuation in decibels?

Solution
Question 9

 The attenuation of a signal is −10 dB.

 What is the final signal power if it was

originally 5 W?
Solution
 Attenuation=10log10(p2/5) = −10 dB.
 Agenda
1. DATA AND SIGNALS

2. PERIODIC ANALOG SIGNALS

3. DIGITAL SIGNALS

4. TRANSMISSION IMPAIRMENT

5. DATA RATE LIMITS

6. PERFORMANCE
DATA RATE LIMITS
 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:

1. The bandwidth available

2. The level of the signals we use

3. The quality of the channel (the level of noise)

Increase the number of signal levels
 When we increase the number of signal levels, we impose
a burden on the receiver.
 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.
 In other words, increasing the levels of a signal reduces
the reliability of the system
DATA RATE LIMITS

 Two theoretical formulas were developed to calculate the

data rate:

 Nyquist for a noiseless channel,

 Shannon for a noisy channel.

Noiseless Channel: Nyquist Bit Rate
 In this formula,
 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.

Example
 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
 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
 We need to send 265 kbps over a noiseless channel with a
bandwidth of 20 kHz. How many signal levels do we need?
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
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:
Noisy Channel: Shannon Capacity
 In this formula,
 bandwidth is the bandwidth of the channel
 SNR is the signal-to noise ratio,
 capacity is the capacity of the channel in bits per second.

 Note that in the Shannon formula there is no indication of the

signal level, which means that no matter how many levels we
have, we cannot achieve a data rate higher than the capacity of
the channel.
 In other words, the formula defines a characteristic of the
channel, not the method of transmission.
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. 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.
Example
 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

 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.
Example
 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
 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?

Solution First, we use the Shannon formula to find the
upper limit
Example
 The Shannon formula gives us 6 Mbps, the upper limit.

 For better performance we choose something lower, 4

Mbps, for example.

 Then we use the Nyquist formula to find the number of

signal levels.
PERFORMANCE
 One important issue in networking is the performance of
the network: how good is it?

 We discuss quality of service (QoS) an overall

measurement of network performance.

 QoS measured through:

1. Bandwidth

2. Throughput

3. Delay times
Bandwidth
 In networking, we use the term bandwidth in two
contexts.
 The first, bandwidth in hertz: when refers to the range of
frequencies in analogue signals of the range of frequencies
that a channel can pass.

 The second, bandwidth in bits/second: when refers to the

speed of bit transmission in a channel or link. Often referred
to as Capacity.
Bandwidth
 Relationship - Bandwidth in hertz and bandwidth in
bits per second are related.

 Increasing bandwidth in hertz leads to an increase in

bandwidth in bits per second.

 The specific relationship between them depends on the

type of transmission, whether it's baseband or with
modulation.
Throughput
 The throughput is a measure of how fast we can actually send
data through a network.
 Although, at first glance, bandwidth in bits per second and
throughput seem the same, they 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.
 In other words, the bandwidth is a potential measurement of
a link; the throughput is an actual measurement of how fast
we can send data
Throughput
 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.
 This means that we cannot send more than 200 kbps through this
link.
 Imagine a highway designed to transmit 1000 cars per minute from
one point to another.
 However, if there is congestion on the road, this figure may be
reduced to 100 cars per minute.
 The bandwidth is 1000 cars per minute; the throughput is 100
cars per minute.
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?
 Solution We can calculate the throughput as

 The throughput is almost one-fifth of the bandwidth in this

case.
Latency (Delay)
 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.
 Or Latency (or delay) is the time it takes for a message to
reach its destination from the moment it starts being sent.
 We can say that latency is made of four components:
propagation time, transmission time, queuing time and
processing delay
Propagation Time
 Propagation time measures the time required for a bit to
travel from the source to the destination.

 The propagation time is calculated by dividing the

distance by the propagation speed.

 Propagation time = Distance / (Propagation Speed)

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

Transmission Time
 In data communications we don’t send just 1 bit, we send a message.
 The first bit may take a time equal to the propagation time to reach
its destination; the last bit also may take the same amount of time.
 However, there is a time between the first bit leaving the sender and
the last bit arriving at the receiver.
 The first bit leaves earlier and arrives earlier; the last bit leaves later
and arrives later.
 The transmission time of a message depends on the size of the
message and the bandwidth of the channel.
 Transmission time = (Message size) / Bandwidth
Example
 What are the propagation time and the transmission
time for a 2.5-KB (kilobyte) 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.
 Solution: We can calculate the propagation and
transmission time as
Example

 Note that in this case:

 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.

Example
 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. Solution We can calculate the propagation and
transmission times as
Example
 Note that in this case,

 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.

Queuing Time
 Is the time needed for each intermediate or end device to
hold the message before it can be processed.
 The queuing time is not a fixed factor; it 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.
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).

 If the delay for the first packet is 20 ms, for the second
is 45 ms, and for the third is 40ms, then the real-time
application that uses the packets endures jitter.
Exercises
True or False
1. The Nyquist bit rate formula applies to both noisy and
noiseless channels.

2. In a noiseless channel, increasing the signal levels increases

the bit rate.

3. In Shannon’s formula, the number of signal levels directly

affects the calculated channel capacity.

4. Bandwidth refers to the actual amount of data transmitted

over a network in a given time period.

5. Throughput can never exceed bandwidth.

True or False
1. The Nyquist bit rate formula applies to both noisy and
noiseless channels.

2. In a noiseless channel, increasing the signal levels increases

the bit rate.

3. In Shannon’s formula, the number of signal levels

directly affects the calculated channel capacity.

4. Bandwidth refers to the actual amount of data

transmitted over a network in a given time period.

5. Throughput can never exceed bandwidth.

 In Shannon’s capacity formula, what does
the term SNR stand for?

A) Signal Noise Resistance

B) Signal Non-Response
C) Signal-to-Noise Ratio
D) Signal Numeric Ratio
 Which formula is used to calculate the data
rate in a noisy channel?

A) Shannon capacity formula

B) Nyquist bit rate formula
C) Fourier transform formula
D) Hartley’s law
Which component of latency measures the time required
for a bit to travel from source to destination?

o A) Transmission time

o B) Queuing time

o C) Propagation time

o D) Processing delay
The transmission time of a message depends on:
o A) The distance between sender and receiver
o B) The size of the message and the bandwidth of
the channel
o C) The type of network connection
o D) The number of packets sent
Task
Task :
 5G Network Performance Metrics:
• Explore the performance characteristics of 5G networks,
including latency, throughput, and reliability, and compare them
with previous generations (4G, 3G).

 Quality of Service (QoS) in IoT Networks:

• Study the challenges of ensuring QoS in IoT environments,
focusing on latency, bandwidth, and energy efficiency.
Practical
Project Assignment: Network Performance
Measurement Tool
 Objective:
 Your goal is to develop a Python program that accurately
measures and reports key metrics related to network
performance.
 These metrics will include latency, bandwidth, throughput,
and jitter.
 The final product should be user-friendly and present
information in a clear and understandable format.