Principles of Modern Communications
Digital Communications II
based on 2011 lecture series by Dr. S. Waharte.
Department of Computer Science and Technology,
University of Bedfordshire.
15th January 2013
Outline
Modern
Communications
David Goodwin
University of
Bedfordshire
1 Frequency Modulation
Frequency Modulation
Data Communications
Baseband
transmission
2 Data Communications
Broadband
transmission
Baseband transmission
Broadband transmission
Modulation
Transmission impairments
Nyquist Theorem
Shannon’s Law
Performance
Modulation
Transmission
impairments
Nyquist Theorem
Shannon’s Law
Performance
57
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation 3
Data Communications
Frequency Modulation
Baseband
transmission
Broadband
transmission
Modulation
Transmission
impairments
Nyquist Theorem
Shannon’s Law
Performance
57
Carrier wave equation
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation 4
Data Communications
Baseband
transmission
ˆ Given a carrier wave whose equation is given by:
v(t) = Vc cos (2πfc t + φc )
Broadband
transmission
ˆ where
ˆ Vc = Amplitude
ˆ fc = carrier frequency and
ˆ φc = phase angle of the wave
Modulation
Transmission
impairments
Nyquist Theorem
Shannon’s Law
Performance
ˆ The modulating wave can be defined as above.
57
Modulating wave equation
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation 5
ˆ A modulating wave equation is given by:
Data Communications
e(t) = Vm cos (2πfm t + φm )
Baseband
transmission
Broadband
transmission
ˆ where
ˆ Vm = Amplitude
ˆ fm = modulating frequency and
ˆ φm = phase angle of the modulating wave
Modulation
Transmission
impairments
Nyquist Theorem
Shannon’s Law
ˆ The above equations can be used to define Amplitude
Performance
Modulation (AM), Frequency Modulation (FM) and Phase
Modulation (PM).
57
Defining AM, FM and PM
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation 6
Data Communications
ˆ In Amplitude Modulation, the modulating signal is used to
Baseband
transmission
modify the amplitude Vc of the carrier wave.
Broadband
transmission
ˆ In Frequency Modulation the modulating signal is used to
Modulation
modify the carrier frequency fc .
Transmission
impairments
ˆ In Phase Modulation, the modulating signal is used to modify
Nyquist Theorem
Shannon’s Law
the phase φc of the carrier signal.
Performance
57
FM and PM analysis
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation 7
Data Communications
Baseband
transmission
Broadband
transmission
ˆ FM and PM signals
Modulation
(a)
(b)
(c)
(d)
Transmission
impairments
Nyquist Theorem
Shannon’s Law
Performance
57
Carrier signal
Modulating signal
FM signal
PM signal
FM and PM analysis
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation 8
Data Communications
Baseband
transmission
ˆ The complete mathematical analysis of FM and PM is
Broadband
transmission
beyond the scope this class.
Modulation
ˆ For our purposes, it will suffice to simply give solutions and
Transmission
impairments
discuss them.
Nyquist Theorem
Shannon’s Law
Performance
57
FM signal
Modern
Communications
David Goodwin
University of
Bedfordshire
ˆ An FM signal is given by the equation:
Frequency Modulation 9
Data Communications
v(t) = A cos (ωc t + mf sin ωm t)
Baseband
transmission
ˆ Where:
Broadband
transmission
ˆ A= amplitude of the carrier wave
ˆ ωc , ωm = angular frequencies for the carrier and modulating
Modulation
Transmission
impairments
signals respectively.
Nyquist Theorem
ˆ mf = modulation index for FM
ˆ mf = fδ where δ = max frequency shift caused by the
m
Shannon’s Law
Performance
modulating signal
ˆ fm : highest frequency component in the modulating signal
57
FM mathematical solution
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation 10
ˆ The equations for FM and PM are more complex than they
look as they contain cosine of a sine.
Data Communications
ˆ Therefore, to solve the frequency components of an FM wave
Baseband
transmission
we need use a mathematical tool known as Bessel functions.
Broadband
transmission
Modulation
Transmission
impairments
vs (t) = Vc
Nyquist Theorem
∞
X
Jn (β) cos (ωc + nωm )t
n=−∞
Shannon’s Law
ˆ The Bessel functions gives us a way of calculating the
Performance
sidebands of an FM signal when the modulating index is
known.
57
Bessel functions
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation 11
Data Communications
Baseband
transmission
Broadband
transmission
Modulation
Transmission
impairments
Nyquist Theorem
Shannon’s Law
Performance
57
Sidebands
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation 12
ˆ In FM and PM as in AM sum and difference sidebands are
Data Communications
Baseband
transmission
produced.
Broadband
transmission
ˆ Sidebands are determined in FM and PM in accordance with
Bessel functions.
Modulation
Transmission
impairments
ˆ FM and PM use a wider spectrum than in AM.
Nyquist Theorem
ˆ All the modulations systems produce infinite number of
Shannon’s Law
Performance
sidebands but only the significant ones are relevant.
57
Example to illustrate use of Bessel chart
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation 13
Data Communications
Baseband
transmission
Broadband
transmission
Modulation
Transmission
impairments
Nyquist Theorem
Shannon’s Law
Performance
57
Phase modulation
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation 14
Data Communications
ˆ Phase modulation is achieved when the amount of phase shift
Baseband
transmission
of a constant-frequency carrier is varied in accordance with a
modulating signal.
ˆ This can be achieved with a phase shifter that changes the
phase of a carrier signal as the amplitude of the modulating
signal changes.
Broadband
transmission
Modulation
Transmission
impairments
Nyquist Theorem
Shannon’s Law
Performance
57
Frequency-Shift Keying (FSK)
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation 15
Data Communications
Baseband
transmission
Broadband
transmission
Modulation
Transmission
impairments
Nyquist Theorem
Shannon’s Law
Performance
ˆ Frequency modulating a carrier with binary data produces
FSK
57
Phase-shift keying (PSK)
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation 16
Data Communications
Baseband
transmission
Broadband
transmission
Modulation
Transmission
impairments
Nyquist Theorem
Shannon’s Law
Performance
ˆ Phase modulating a carrier with a binary data produces PSK
57
Directed Learning and Self Study
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation 17
Data Communications
Baseband
transmission
Example
Broadband
transmission
1
Use the internet to find what Carson’s rule is about and
explain the formula that defines the rule.
2
What is the meaning of ASK? What is its relationship with
FSK and PSK?
Modulation
Transmission
impairments
Nyquist Theorem
Shannon’s Law
Performance
57
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications 18
Data Communications
Baseband
transmission
Broadband
transmission
Modulation
Transmission
impairments
Nyquist Theorem
Shannon’s Law
Performance
57
Overview
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications 19
ˆ Baseband transmission
Baseband
transmission
ˆ Broadband transmission
Broadband
transmission
ˆ Modulation
Modulation
ˆ Transmission impairment (attenuation, distortion, and noise)
Transmission
impairments
ˆ Nyquist Theorem
Nyquist Theorem
ˆ Shannon’s law
Shannon’s Law
Performance
ˆ Performance (Bandwidth, Throughput, Latency)
57
Baseband Transmission
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Baseband
transmission
20
Broadband
transmission
ˆ Sends digital signal over a channel.
ˆ The signal is not changed to analogue signal.
Modulation
ˆ Require a channel with a low pass bandwidth. ( i.e. 0 –
Transmission
impairments
infinity) Hz
Nyquist Theorem
Shannon’s Law
Performance
57
Practical Baseband Bandwidth
Modern
Communications
David Goodwin
University of
Bedfordshire
ˆ A low pass channel with a narrow bandwidth.
Frequency Modulation
ˆ A bandwidth from 0 - h Hz where h lies between 0 and
Data Communications
Baseband
transmission
21
Broadband
transmission
Modulation
Transmission
impairments
Nyquist Theorem
Shannon’s Law
Performance
57
infinity.
Broadband Transmission
Modern
Communications
David Goodwin
University of
Bedfordshire
ˆ We change the digital signal to analogue signal by
modulating the digital signal with a carrier frequency.
Frequency Modulation
ˆ This allows us to use a band pass channel.
Data Communications
Baseband
transmission
Broadband
transmission
22
Modulation
Transmission
impairments
Nyquist Theorem
Shannon’s Law
Performance
57
ˆ A band pass channel does not start from 0 Hz. E.g. F1
Modulation technique
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Baseband
transmission
Broadband
transmission
Modulation
23
Transmission
impairments
Nyquist Theorem
Shannon’s Law
Performance
57
Signal in transmission medium
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
ˆ They suffer transmission impairments.
Data Communications
Baseband
transmission
ˆ For example:
Broadband
transmission
ˆ Attenuation (i.e. loss of energy).
Modulation
Transmission
impairments
24
ˆ Distortion (due to transmission delays in the components of
the composite signal and
Nyquist Theorem
Shannon’s Law
ˆ Noise (due to external and internal influences).
Performance
57
Attenuation
Modern
Communications
David Goodwin
University of
Bedfordshire
ˆ Reduction in power when signals pass through a medium.
Frequency Modulation
Data Communications
ˆ If a signal with power P1 passes through a medium and the
Baseband
transmission
Broadband
transmission
Modulation
Transmission
impairments
25
Nyquist Theorem
Shannon’s Law
Performance
signal is measured as power P2 at the end of the medium
then the gain of the signal is:
P2
Gain = 10log10
P1
ˆ A reduction in the gain may require an amplifier to boost the
signal up
57
Distortion
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
ˆ Different components in a composite signal reacts differently
Baseband
transmission
to the medium.
Broadband
transmission
ˆ Thus, signals arriving at a point may be out of phase and
Modulation
Transmission
impairments
26
with different amplitudes.
ˆ The cumulative effect is a distorted signal of the composite
Nyquist Theorem
Shannon’s Law
signal.
Performance
57
Noise (causes)
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Baseband
transmission
ˆ Thermal noise: Random noise generated in the medium.
Broadband
transmission
ˆ Induced noise: Generated from motors and electrical
Modulation
Transmission
impairments
27
Nyquist Theorem
appliances. These serve as transmitters and the medium
serves as receivers.
ˆ Cross-talk: This is the effect one wire on another.
Shannon’s Law
Performance
57
Signal-to-noise Ratio
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
ˆ This ratio helps us to estimate the effect noise on a signal.
Data Communications
Baseband
transmission
ˆ It is measured as:
Broadband
transmission
Modulation
Transmission
impairments
28
Nyquist Theorem
SN R = 10log10
Signal Power
Noise Power
ˆ This ratio is important as it is one of the determining factors
Shannon’s Law
Performance
of the channel capacity.
57
Exercise
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Baseband
transmission
Broadband
transmission
Example
Modulation
Transmission
impairments
29
Nyquist Theorem
Shannon’s Law
Performance
57
ˆ Calculate the signal to noise ratio of a lossless channel
Solution
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Example
Baseband
transmission
ˆ Solution:
Broadband
transmission
ˆ SNR=Signal Power/Noise Power
Modulation
Transmission
impairments
30
ˆ Noise Power = 0 (for a lossless channel)
ˆ Thus, SNR=Signal Power/0 = ∞
Nyquist Theorem
Shannon’s Law
ˆ Therefore, SNR=1og10 (∞) = ∞
Performance
57
Data rate limits
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
ˆ A very important consideration in data communications is
Data Communications
how fast we can send data, in bits per second, over a channel.
Data rate depends on three factors:
Baseband
transmission
Broadband
transmission
1
Modulation
Transmission
impairments
31
2
3
Nyquist Theorem
The bandwidth available
The level of the signals we use
The quality of the channel (the level of noise)
ˆ Noiseless Channel: Nyquist Bit Rate
Shannon’s Law
Performance
ˆ Noisy Channel: Shannon Capacity
57
Question
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Baseband
transmission
Example
Broadband
transmission
ˆ Increasing the levels of a signal increases the probability of
Modulation
Transmission
impairments
32
Nyquist Theorem
Shannon’s Law
Performance
57
an error occurring, in other words it reduces the reliability of
the system. Why??
Capacity of a system
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
ˆ The bit rate of a system increases with an increase in the
Data Communications
Baseband
transmission
number of signal levels we use to denote a symbol.
Broadband
transmission
ˆ A symbol can consist of a single bit or “n” bits.
Modulation
Transmission
impairments
33
ˆ The number of signal levels = 2n.
ˆ As the number of levels goes up, the spacing between level
Nyquist Theorem
decreases → increasing the probability of an error occurring in
the presence of transmission impairments.
Shannon’s Law
Performance
57
Nyquist Lossless Channel
Modern
Communications
David Goodwin
University of
Bedfordshire
ˆ Nyquist gives the upper bound for the bit rate of a
Frequency Modulation
transmission system by calculating the bit rate directly from
the number of bits in a symbol (or signal levels) and the
bandwidth of the system (assuming 2 symbols/per cycle and
first harmonic).
Data Communications
Baseband
transmission
Broadband
transmission
Modulation
ˆ Nyquist theorem states that for a noiseless channel:
Transmission
impairments
Nyquist Theorem
34
C = 2Blog2 (2n)
Shannon’s Law
Performance
ˆ C= capacity in bps
ˆ B = bandwidth in Hz
57
Transmitting over a lossless channel
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Baseband
transmission
ˆ Let us suppose we want to transmit 500kbps over a noiseless
Broadband
transmission
channel with a 30Khz bandwidth. How many signal levels do
we require?
Modulation
Transmission
impairments
Nyquist Theorem
35
Shannon’s Law
Performance
57
Solution
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Example
Baseband
transmission
ˆ Using Nyquist Theorem:
Broadband
transmission
Modulation
ˆ Bit Rate = 2 × Bandwidth × log2 (L)
Transmission
impairments
ˆ 500 × 1000 = 2 × (30 × 1000) × log2 (L)
Nyquist Theorem
36
Shannon’s Law
ˆ Log2 (L) = 50/6 = 8.333
ˆ L = Log10 (8.333)/Log10 (2)
Performance
57
Shannon’s Law
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Baseband
transmission
ˆ We can never have a lossless channel.
Broadband
transmission
ˆ Shannon(1944) proposed that the maximum bit rate (i.e.
channel capacity) depends on the signal-to-noise ratio. i.e.
Modulation
Transmission
impairments
Bit Rate = Bandwidth × log2 (1 + S/N )
Nyquist Theorem
Shannon’s Law
37
Performance
57
Observations of Shannon’s Law
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Baseband
transmission
ˆ Shannon’s Law does not mention signal levels.
Broadband
transmission
ˆ This means that no matter what signal levels you have, you
Modulation
Transmission
impairments
cannot achieve a data rate higher than the channel capacity.
Nyquist Theorem
Shannon’s Law
38
Performance
57
Problem
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Baseband
transmission
Example
Broadband
transmission
ˆ What is the channel capacity of a faint signal that is
Modulation
Transmission
impairments
completely swamped by noise?
Nyquist Theorem
Shannon’s Law
39
Performance
57
Answer
Modern
Communications
David Goodwin
University of
Bedfordshire
Example
Frequency Modulation
Data Communications
ˆ Using Shannon’s Law:
Baseband
transmission
ˆ C = BLog2 (1 + S/N )
Broadband
transmission
ˆ Noise >> S
Modulation
Transmission
impairments
ˆ Therefore, S/N=0
Nyquist Theorem
Shannon’s Law
40
ˆ Log2 (1 + S/N ) = Log2 (1 + 0) = 0
ˆ Hence, C = 0
Performance
ˆ This mean that the channel cannot transmit any signal.
57
Performance metrics of a communication
system
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
ˆ Bandwidth: Defines the channel or the bit rate
Baseband
transmission
ˆ Throughput: Refers to the actual bit rate measured at the
Broadband
transmission
end of a transmission
Modulation
ˆ Latency (Delay): Refers to delays caused during the
Transmission
impairments
transmission of a signal.
Nyquist Theorem
Shannon’s Law
Performance
41
57
ˆ These metrics are important in communication systems.
Bandwidth (Hz)
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Baseband
transmission
ˆ This means a range of channel frequencies that allows a
Broadband
transmission
signal to pass.
Modulation
ˆ E.g. The bandwidth of a telephone line is 4KHz.
Transmission
impairments
Nyquist Theorem
Shannon’s Law
Performance
42
57
Bandwidth (Bits Per Second)
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Baseband
transmission
ˆ This is the number of bits per second that a link, channel or a
Broadband
transmission
network can transmit.
Modulation
ˆ E.g. A fast Ethernet network can transmit at 100 Mbps (i.e.
Transmission
impairments
Mega bits per second)
Nyquist Theorem
Shannon’s Law
Performance
43
57
Bandwidth in Hz Versus Bandwidth in Bps
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Baseband
transmission
ˆ Bandwidth in Hz refer to a range of frequencies.
Broadband
transmission
ˆ Bandwidth in Bps refers to the speed of transmission in a
channel or a link.
Modulation
Transmission
impairments
ˆ The relationship between them is that, the bandwidth in Bps
Nyquist Theorem
is proportional to bandwidth in Hz
Shannon’s Law
Performance
44
57
Throughput
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
ˆ A measure of how fast we actually send data through a
Data Communications
Baseband
transmission
network.
Broadband
transmission
ˆ Suppose you have a link designated as C bps but you may be
Modulation
able to measure the signal through the link as T bps where
T<<C.
Transmission
impairments
Nyquist Theorem
ˆ This is because intermediary devices connected to the channel
Shannon’s Law
Performance
45
57
have slower bit rates affecting the transmission.
Example
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Baseband
transmission
Example
Broadband
transmission
ˆ A network with a bandwidth of 12 Mbps can only pass an
Modulation
average 15,000 frames per minute. If each frame carries an
average of 10,000 bits, calculate the throughput of the
network
Transmission
impairments
Nyquist Theorem
Shannon’s Law
Performance
46
57
Solution
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Example
Data Communications
Baseband
transmission
ˆ Throughput is that actual bit rate measured.
Broadband
transmission
ˆ It is independent of the channel speed but depends on
Modulation
devices located on the channel.
Transmission
impairments
ˆ Therefore,
Nyquist Theorem
Shannon’s Law
Performance
47
ˆ Throughput= 15, 000 × 10, 000/60=???
ˆ Answer is less than 12 Mbps channel speed
57
Latency (Delays)
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Baseband
transmission
ˆ This defined as the time time it takes for a message to arrive
Broadband
transmission
at the destination i.e. from the time it was first sent.
Modulation
ˆ Latency = Propagation Time + Transmission Time +
Transmission
impairments
Queuing Time + Processing Delay.
Nyquist Theorem
Shannon’s Law
Performance
48
57
Propagation Time
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
ˆ This is a measure of the time it takes a bit to travel from the
Baseband
transmission
source to the destination.
Broadband
transmission
Modulation
ˆ Propagation Time = Distance/Propagation Speed
Transmission
impairments
ˆ This depends on the transmission medium (in a vacuum
Nyquist Theorem
Shannon’s Law
Performance
49
57
electro magnetic waves travels at 3 × 108 m/s but in a
coaxial cable it is far less.
Transmission Time
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Baseband
transmission
ˆ The Transmission Time is defined as: Transmission Time =
Broadband
transmission
Message Size/Bandwidth
Modulation
ˆ Note that the Bandwidth is in Bps.
Transmission
impairments
Nyquist Theorem
Shannon’s Law
Performance
50
57
Exercise
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Example
Baseband
transmission
ˆ Calculate:
Broadband
transmission
ˆ The propagation time and Transmission time, for a 2.5
Modulation
Transmission
impairments
Nyquist Theorem
Shannon’s Law
Performance
51
57
kbyte email message if the bandwidth is 1.0 Gbps. Assume
the distance between sender and receiver is 12,000km and
that light travels at 2.4 × 108 m/s.
Solution
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Example
Baseband
transmission
Broadband
transmission
ˆ Propagation Time = 12, 000 × 1000/2.4 × 108 =50 ms
Modulation
ˆ Transmission Time =2500 × 8/109 =0.020 ms
Transmission
impairments
ˆ Note that the dominant factor is the propagation time and
Nyquist Theorem
not transmission time.
Shannon’s Law
Performance
52
57
Queuing Time
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Baseband
transmission
ˆ This is the time required for devices (i.e. intermediate or end
Broadband
transmission
devices) to hold data before it can be processed.
Modulation
ˆ QT changes with load on the network.
Transmission
impairments
Nyquist Theorem
Shannon’s Law
Performance
53
57
Bandwidth-Delay Product
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Baseband
transmission
ˆ An important metric in data transmission.
Broadband
transmission
ˆ It is used by network engineers to define the number of bits
Modulation
Transmission
impairments
that can fill a link.
Nyquist Theorem
Shannon’s Law
Performance
54
57
Jitter
Modern
Communications
David Goodwin
University of
Bedfordshire
Frequency Modulation
Data Communications
Baseband
transmission
ˆ This is the delay caused by different packets arriving at a
Broadband
transmission
destination at different times e.g. time sensitive data such as
audio and video.
Modulation
Transmission
impairments
ˆ This can cause serious effects on the received signal.
Nyquist Theorem
Shannon’s Law
Performance
55
57
Summary (1)
Modern
Communications
David Goodwin
University of
Bedfordshire
ˆ Data communication uses periodic analogue and non-periodic
digital signals.
Frequency Modulation
ˆ Fourier analysis shows that a composite signal is a
Data Communications
Baseband
transmission
combination of simple sine waves with different frequencies,
amplitudes an phases.
Broadband
transmission
ˆ A digital signal is a composite analogue signal with infinite
Modulation
Transmission
impairments
bandwidth.
Nyquist Theorem
ˆ Baseband transmission transmits digital signals through a low
Shannon’s Law
Performance
56
pass channel with an infinite bandwidth.
ˆ Nyquist Theorem defines the theoretical maximum bit rate a
digital signal.
ˆ Signal impairment are: attenuation, distortion and noise.
57
Directed Reading and Self Study
Modern
Communications
David Goodwin
University of
Bedfordshire
Example
Frequency Modulation
ˆ Given a list of frequencies shown below calculate their
Data Communications
corresponding periods.
Baseband
transmission
Broadband
transmission
ˆ a) 24 Hz, b) 8 MHz, c) 140 GHz, d) 100 GHz
Modulation
ˆ What are the corresponding frequencies of following
Transmission
impairments
periods:
Nyquist Theorem
Shannon’s Law
Performance
57
ˆ a) 5s, b) 12 µs, c) 200 ms.
ˆ A device is sending out data at the rate of 1000 bps. How
long does it take to send:
ˆ a) 10 bits, b) 1 single character, c) a thesis containing
20,000 words? The average word length is 6 characters.
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