Transcript CMPT 371: Chapter 1 - Simon Fraser University

Basics of Data Transmission
Our Objective is to understand …
 Signals, bandwidth, data rate concepts
 Transmission impairments
 Channel capacity
 Data Transmission
1-1
Signals
 A signal is



generated by a transmitter and
transmitted over a medium
function of time
function of frequency, i.e.,
composed of components of
different frequencies
 Analog signal


varies smoothly with time
E.g., speech
 Digital signal


maintains a constant level for
some period of time, then
changes to another level
E.g., binary 1s and 0s
1-2
Periodic vs. Aperiodic Signals
 Periodic signal
 Pattern repeated over
time
 s(t+T) = s(t)
 Aperiodic signal
 Pattern not repeated
over time
1-3
Sine Wave
 The fundamental periodic
signal
 Peak Amplitude (A)


maximum strength of signal
volts
 Frequency (f)




Rate of change of signal
Hertz (Hz) or cycles per
second
Period = time for one
repetition (T)
T = 1/f
 Phase ()

Relative position in time
1-4
Signals in Frequency Domain
 Signal is made up of many components
 Components are sine waves with different frequencies
 In early 19th century, Fourier proved that
 Any periodic function can be constructed as the sum of
a (possibly infinite) number of sines and cosines
1
s(t )  c 
2
T

a
n 1
n
sin 2nft 

b
n 1
n
cos2nft
T
T
2
2
2
an   s(t ) sin(2 nft)dt, bn   s(t ) cos(2 nft)dt, c   s(t )dt
T0
T0
T0
 This decomposition is called Fourier series
 f is called the fundamental frequency
 an, bn are amplitude of nth harmonic
 c is a constant
1-5
Frequency Domain (cont’d)
 Fourier Theorem
enables us to
represent signal in
Frequency Domain

i.e., to show
constituent
frequencies and
amplitude of signal at
these frequencies
 Example 1: sine wave:
S(f)
s(t) = sin(2πft)
1f
Frequency, f
1-6
Time and Frequency Domains: Example 2
Time
domain s(t)
Frequency
domain S(f)
1-7
Frequency Domain (cont’d)
 So, we can use Fourier theorem to represent a
signal as function of its constituent frequencies,
 and we know the amplitude of each constituent
frequency. So what?
 We know the spectrum of a signal, which is the
range of frequencies it contains, and
 Absolute bandwidth = width of the spectrum
 Q: What is the bandwidth of the signal in the
previous example? [sin(2πft) + sin(2π3ft)]
 A: 2f Hz
1-8
Frequency Domain (cont’d)
 Q. What is the absolute bandwidth of square wave?


1
Hint: Fourier tells you that s(t )   
sin 2kft 
 k odd , k 1 k
4
 Absolute BW
= ∞
(ooops!!)
 But, most of the energy is contained within a
narrow
band (why?) we refer to this band as effective
bandwidth, or just bandwidth
1-9
Approximation of Square Wave
Using the first 3
harmonics, k=1, 3, 5
A. BW = 4*f Hz
Using the first 4
harmonics, k=1, 3, 5, 7
A. BW = 6*f Hz
Q. What is BW in each
case?
Cool applet on Fourier Series
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Signals and Channels
 Signal
 can be decomposed to components (frequencies)
 spectrum: range of frequencies contained in signal
 (effective) bandwidth: band of frequencies containing
most of the energy
 Communications channel (link)
 has finite bandwidth determined by the physical
properties (e.g., thickness of the wire)
 truncates (or filters out) frequencies higher than its BW
• i.e., it may distort signals

can carry signals with bandwidth ≤ channel bandwidth
1-11
Bandwidth and Data Rate
 Data rate: number of bits per second (bps)
 Bandwidth: signal rate of change, cycles per sec (Hz)
 Well, are they related?
 Ex.: Consider square wave with high = 1 and low = 0 
 We can send two bits every cycle (i.e., during T = 1/f sec)
 Assume f =1 MHz (fundamental frequency)  T = 1 usec
 Now, if we use the first approximation (3 harmonics)
 BW of signal = (5 f – 1 f) = 4 f = 4 MHz
 Data rate = 2 / T = 2 Mbps

So we need a channel with bandwidth 4 MHz to send at date
rate 2 Mbps
1-12
Bandwidth and Data Rate (cont’d)
 But, if we use the second approx. (4 harmonics)
 BW of signal = (7 f – 1 f) = 6 f = 6 MHz
 Data rate = 2 / T = 2 Mbps
 Which one to choose? Can we use only 2 harmonics
(BW = 2 MHz)?
 It depends on the ability of the receiver to discern
the difference between 0 and 1
 Tradeoff: cost of medium vs. distortion of signal
and complexity of receiver
1-13
Bandwidth and Data Rate (cont’d)
 Now, let us agree that the first appox. (3 harmonics)
is good enough

Data rate of 2 Mbps requires BW of 4 MHz
 To achieve 4 Mbps, what is the required BW?
 data rate = 2 (bits) / T (period) = 4 Mbps  T = 1 /2 usec
  f (fundamental freq) = 1 /T = 2 MHz 
 BW = 4 f = 8 MHz
 Bottom line: there is a direct relationship between
data rate and bandwidth


Higher data rates require more bandwidth
More bandwidth allows higher data rates to be sent
1-14
Bandwidth and Data Rate (cont’d)
 Nyquist Theorem: (Assume noise-free channel)
 If rate of signal transmission is 2B then signal with
frequencies no greater than B is sufficient to carry signal
rate, OR alternatively
 Given bandwidth B, highest signal rate is 2B
 For binary signals
 Two levels  we can send one bit (0 or 1) during each period
 data rate (C) = 1 x signal rate = 2 B
 That is, data rate supported by B Hz is 2B bps
 For M-level signals
 M levels  we can send log2M bits during each period 
 C= 2B log2M
1-15
Bandwidth and Data Rate (cont’d)
 Shannon Capacity:
 Considers data rate, (thermal) noise and error rate
 Faster data rate shortens each bit so burst of noise affects
more bits
 At given noise level, high data rate means higher error rate
 SNR ≡ Signal to noise ration
 SNR = signal power / noise power
 Usually given in decibels (dB): SNRdB= 10 log10 (SNR)
 Shannon proved that: C = B log2(1 + SNR)
 This is theoretical capacity, in practice capacity is much
lower (due to other types of noise)
1-16
Bandwidth and Data Rate (cont’d)
 Ex.: A channel has B = 1 MHz and SNRdB =
24 dB, what is the channel capacity limit?
SNRdB = 10 log10 (SNR)  SNR = 251
 C = B log2(1 + SNR) = 8 Mbps

 Assume we can achieve the theatrical C,
how many signal levels are required?

C = 2 B log2M  M = 16 levels
1-17
Transmission Impairments
 Signal received may differ from signal
transmitted
 Analog - degradation of signal quality
 Digital - bit errors
 Caused by
Attenuation and attenuation distortion
 Delay distortion
 Noise

1-18
Attenuation
 Signal strength falls off with distance
 Depends on medium
 Received signal strength:
 must be enough to be detected
 must be sufficiently higher than noise to be
received without error
 Attenuation is an increasing function of
frequency  attenuation distortion
1-19
Delay Distortion
 Only in guided media
 Propagation velocity varies with frequency
 Critical for digital data
 A sequence of bits is being transmitted
 Delay distortion can cause some of signal
components of one bit to spill over into other
bit positions 
 intersymbol interference, which is the major
limitation to max bit rate
1-20
Noise (1)
 Additional signals inserted between
transmitter and receiver
 Thermal
Due to thermal agitation of electrons
 Uniformly distributed across frequencies 
 White noise

 Intermodulation
 Signals that are the sum and difference of
original frequencies sharing a medium
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Noise (2)
 Crosstalk

A signal from one line is picked up by another
 Impulse
 Irregular pulses or spikes, e.g. external
electromagnetic interference
 Short duration
 High amplitude
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Data and Signals
 Data
 Entities that convey meaning
 Analog: speech
 Digital: text (character strings)
 Signals
 electromagnetic representations of data
 Analog: continuous
 Digital: discrete (pulses)
 Transmission
 Communication of data by propagation and
processing of signals
1-23
Analog Signals Carrying Analog
and Digital Data
1-24
Digital Signals Carrying Analog
and Digital Data
1-25
Analog Transmission
 Analog signal transmitted without regard
to content
 May be analog or digital data
 Attenuated over distance
 Use amplifiers to boost signal
 But, it also amplifies noise!
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Digital Transmission
 Concerned with content
 Integrity endangered by noise, attenuation
 Repeaters used
 Repeater receives signal
 Extracts bit pattern
 Retransmits
 Attenuation is overcome
 Noise is not amplified
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Advantages of Digital
Transmission
 Digital technology
 Low cost LSI/VLSI technology
 Data integrity
 Longer distances over lower quality lines
 Capacity utilization
 High bandwidth links economical
 High degree of multiplexing easier with digital techniques
 Security & Privacy
 Encryption
 Integration
 Can treat analog and digital data similarly
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Summary
 Signal: composed of components (Fourier
Series)

Spectrum, bandwidth, data rate
 Shannon channel capacity
 Transmission impairments
 Attenuation, delay distortion, noise
 Data vs. signals
 Digital vs. Analog Transmission
1-29