Transcript Comp 445 - Aiman Hanna

Analog & Digital Signals
© Prof. Aiman Hanna
Department of Computer Science
Concordia University
Montreal, Canada
A nalog & Digital Signals
Figure 3.1 – Digital & Analog Signals
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D igital Encoding Schemes
There are 10 types of people; those who know binary and those who do not.

Digital data are represented by a sequence of 1s & 0s

1 refer to a high electrical voltage, and 0 refers to a low electrical voltage

Two major digital encoding schemes exist:
• NonReturn to Zero (NRZ) Encoding
• Manchester Encoding
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D igital Encoding Schemes
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NRZ Encoding
 A 0 voltage is transmitted by raising the voltage level high, while 1 is
transmitted by using a low voltage
Figure 3.2(a) – NRZ Encoding
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D igital Encoding Schemes
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NRZI Encoding
 An alternative to NRZ is NRZI (Inverted)

The voltage changes only when a 1 is to be sent
Figure 3.2(b) – NRZI Encoding
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D igital Encoding Schemes
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
Both NRZ and NRZI have problems; for example what is the
exact sequence being transmitted in the sequence below?

Is time synchronization possible?
Figure 3.3 – NRZ Encoding of a Sequence of 0s
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D igital Encoding Schemes
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Manchester Encoding
 Also called Self-Synchronizing Code

Uses signal changes to keep the sending and receiving devices synchronized
 0 is represented by a change from high to low in the middle of transmission
and 1 is represented by a low to high change in the middle of transmission

Are there any disadvantages?
Figure 3.4 – Manchester Encoding
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D igital Encoding Schemes
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Differential Manchester Encoding

Similar to Manchester encoding, the signal will change in the
middle, however

1 causes the signal to remain the same, while 0 causes the signal
to change
Figure 3.5 – Differential Manchester Encoding
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A nalog Signals

Adds complexity to data communication

Phone lines carry analog signals

Digital computers need to convert their signal to analog before placing it on the wire; this is called
Modulation

Similarly, all received signals must be converted to digital; this is called Demodulation

A modem is hence needed
An analog signal is not that simple; in general, an analog signal is characterized by its frequency, amplitude
and phase shift

Different Amplitude & phase shift
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B it Rate

Describes a medium capacity; that is how many bits can be transferred per unit of time

Measured as Bits Per Second (bps)

A higher bandwidth medium is capable of a higher bit rate

Transmitter sends a signal representing a bit sting (a component) , while the receiver listens to
the medium and creates a bit string based on what it receives

Once the component is sent, the transmitter gets another bit string (another component) and the
process repeats
Figure 3.9 – Sending Data via Signals
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B it Rate
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The Nyquist Theorem & Noiseless Channels

Baud Rate: the frequency with which components change

Each bit string
is composed of n bits, and hence the signal component may
n
have up to 2 different amplitudes (one for each unique combination for b1,
b2, …bn)

Are bit rate and baud rate the Same?

No, bit rate depends on the number of bits (n) as well as the baud rate; more
precisely:
Bit Rate = n * Baud Rate

Bit rate can then be increased by either increasing the baud rate or n; however
only up to a point
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B it Rate
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Bit Rate = n * Baud Rate

This result is surprisingly old, back to 1920s, when Harry Nyquist developed his classic theory

Nyquist theory showed that if f is the maximum frequency a medium can transmit, then the receiver can
reconstruct the signal by sampling it 2f times per second

In other words, the receiver can construct the signal by sampling it at intervals of 1 / 2f seconds, or twice each
period (one period is 1 / f ).

For example, if the maximum frequency is 4000 Hz, then the receiver can completely construct it by sampling
it every 1/8000th of a second

Assuming that the transmitter baud rate is 2 f , in other words changes signal each 1 / 2f intervals, we can state
Bit Rate = n * Baud Rate = n * 2 * f

This can also be stated based on component; if B is the number of different components, then
n
B=2
Or
n = log2(B)

Hence,
Bit Rate = 2 * f * log2(B)
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B it Rate
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Noisy Channels
We interject that Nyquist theory assumed that there is absolutely no noise in the
channel to alter the signal
The result of Nyquist Theorem for a maximum frequency of 3300Hz, the
approximate upper limit of the telephone system, can then be summarized as
follows:


Result of Nyquist Theorem for a maximum of 3300 Hz

N, number of bits
B, number of signal components
Maximum bit rate (bps)
1
2
6,600
2
4
13,200
3
8
19,800
4
16
26,400
This result seem to imply that there is no upper bound for a bit rate given the
maximum frequency; unfortunately this is not the case due to two main reasons:
1.
2.
More components mean subtler change among them
Channels are subject to noise
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B it Rate
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Noisy Channels
1.
More components mean subtler change among them
2.
Channels are subject to noise

The transmitted signal can be distorted due to the channel noise

If distortion is too large, the receiver may not be able to reconstruct
the signal at all
Figure 3.10 – Effect on Noise on Digital Signals
(The same applies to analog signals)
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B it Rate
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Shannon’s Result

How much noise is bad? This depends on its ratio to the signal

We define S/N (Signal-to-Noise-Ratio)
A higher S/N (less significant noise) indicates higher quality
Because S >> N, the ratio is often scaled down as


R = log10(S/N) bels
// bels is the measurement unit
For example,
If S is 10 times larger than N, then
R = log10(10N/N) = 1 bel
If S is 100 times larger than N, then
R = log10(100N/N) = 2 bels
Perhaps, a more familiar measurement is the decibel (dB)
1 dB = 0.1 bel
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B it Rate
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Shannon’s Result


In 1940, Claude Shannon went beyond Nyquist’s results and considered noisy channels
Shannon related the maximum bit rate not only to the frequency but also to the S/N ratio;
specifically he showed that:
Bit Rate = Bandwidth * log2(1 + S/N) bps


The formula states that a higher BW and S/N ratio allow higher bit rate
Hence, for the telephone system, which has a frequency of about 3000 Hz and S/N ≈ 35 dB, or
3.5 bels, Shannon’s result yields the following
3.5 = log10(S/N)  S = 103.5N  S ≈ 3162 N  S/N ≈ 3162
Bit Rate = Bandwidth * log2(1 + S/N)
= 3000 * log2(1 + 3162)
≈ 3000 * 11.63 bps
≈ 34,880 bps ≈ 35 kbps
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B it Rate
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Shannon’s Result




Shannon’s result is not just a theoretical result; rather it has its
very real implication
During the 1980s, 2400 & 9600 bps modems became common
Early 1990s, modems with a rate of 28.8 and 33.6 kbps became
common (this matched the maximum bit rate that Shannon’s
result indicated)
Not much longer, modems that supported 56.6 kbps rates were a
reality. Did this violate Shannon’s results?
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D igital-to-Analog-Conversion

Computers are digital, however phone lines connecting
them to remote machines/servers are analog
Figure 3.11 – Computer Data Transmitted Over Telephone Lines
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D igital-to-Analog-Conversion


New phone lines use optical fiber, which carry digital signal
Codec (Coder/Decoder) is needed in such cases
Figure 3.12 – Voice Information Transmitted Digitally
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D igital-to-Analog-Conversion
Frequency Modulation (FM)





Simple method of digital to analog conversion,
Also known as Frequency Shift Keying (FSK)
Assigns digital 0 to one signal and 1 to another based on some rule
In general, if n bits are sent per baud, then we can have 2n signal combination
The signal is transmitted for a fixed period of time
Figure 3.13 – FSK (Two Frequencies), One Bit per Baud – Analog signal for 01001
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D igital-to-Analog-Conversion
Amplitude Modulation (AM)



Also known as Amplitude Shift Keying (ASK)
Each bit group is assigned to an analog signal of given
magnitude
The signal is transmitted for a fixed period of time as with FM
Figure 3.14 – ASK (Four Amplitudes), Two Bits per Baud – Analog signal for 00110110
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D igital-to-Analog-Conversion
Phase Modulation (PM)



Also known as Phase Shift Keying (PSK)
1 results in a phase change of 180°, while 0 results in 0° (no
change)
In such case, one bit of information can be transmitted per baud
PSK (One Shift), One Bit per Baud – Analog signal for 0010110
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D igital-to-Analog-Conversion
Differential Phase Modulation (DPM)



Also known as Differential Phase Shift Keying (DPSK)
Allows transmission of more than 1 bit per baud
For example, if the phase is shifted by multiples of 90°, two bits
at a time can be transmitted
DPSK, Two Bits per Baud – Analog signal for 10001110
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D igital-to-Analog-Conversion
Quadrature Amplitude Modulation (QAM)




A greater signal variety means a greater bit rate per
baud rate
The problem is that the higher the number of signal,
the smaller the difference among them
One common approach to avoid that is to use a
combination of frequencies, amplitudes or phase shifts
to allow a large group of bits while marinating a larger
differences among them
QAM is a common technique in which a group of bits
is assigned a signal defined by its amplitude and phase
shift
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D igital-to-Analog-Conversion
Quadrature Amplitude Modulation (QAM)
Bit Value: Amplitude, Phase Shift
Bit Value: Amplitude, Phase Shift
000: A1, 00 phase shift
001: A2, 00 phase shift
010: A1, 900 shift
011: A2, 900 phase shift
100: A1, 1800 phase shift
101: A2, 1800 phase shift
110: A1, 2700 phase shift
111: A2, 2700 phase shift
Figure 3.15 – QAM (Two Amplitudes & Four Phases), Three Bits per Baud
Analog Signal for 001 010 100 011 101 000 011 110
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A nalog-to-Digital-Conversion

Some analog-to-digital conversion is nothing but the
opposite to what has been discussed, however,

Other conversions are more complex; for example
those that are required to convert analog data for voice
or music
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A nalog-to-Digital-Conversion
Pulse Amplitude Modulation (PAM)
 An analog signal is sampled at a regular interval, then a
pulse with amplitude equal to the sampled signal is
generated
Figure 3.17 – Pulse Amplitude Modulation
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A nalog-to-Digital-Conversion
Pulse Code Modulation (PCM)
 PAM-generated signals look digital, but the signal has analog
characteristics
 PCM allows the signal to truly be digital by assigning amplitude
from a predefined set of digital codes
 By sampling s times per second, we obtain a bit rate of n * s bps
Figure 3.18 – Pulse Code Modulation, n=3, 2n codes
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A nalog-to-Digital-Conversion
Pulse Code Modulation (PCM)
 At the receiving end, the accuracy of the construction depends on:
• Sampling frequency: a slower sampling frequency may result in some oscillations
to be missed entirely



higher sampling rate will produce a higher accuracy; but up to a point
Based on Nyquist theorem, if the original signal has f maximum frequency, then
sampling at 2f is enough
That is, sampling at a higher rate than 2f will not produce any better accuracy than 2f
• Number of amplitude (binary codes) from which to choose


The higher the difference between the sampled signal and the pulse/code, the higher
the chances that the reconstructed signal become distorted
This is called, quantization noise
It should be noted however that higher sampling and higher number of pulse amplitude comes with a price
Figure 3.19 –
Sampling at Too Low Frequency
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A nalog-to-Digital-Conversion
Pulse Code Modulation (PCM)
 PAM has several common applications; which include
• Digitizing of voice signal over long-distance telephone lines

8000 samples, with 8-bit sample  bit rate of 64 kbps
• Compact discs (CDs)



Varies from one device to another
As an example: 44.1 kHz sampling frequency, with a D-A conversion
of 16-bit linear
16-bit would allow 64,000 amplitude (linear indicates that the
differences between these amplitudes are the same)
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M odems
Two important issues are there with modems





• Software (deliver signals to and from modem)
• Compatibility (demodulate what other modems modulated)
Several standards were defined for modems
CCITT (ITU) defined V.xx series of modem standards
V.21 uses FSK with 1 frequency for 1 bit, and the resulting bit
rate is equal to the baud rate. (supports up to 300 bps.)
V.22 uses PSK with 2 bits for each phase shift; resulting bit rate
is twice the baud rate. 600 bauds/sec = 1200bps
Sending and receiving modems use different frequency; that
enable full-duplex communication
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M odems
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Signal Constellation



Many modems work by changing more than one analog signal’s components,
typically phase shift & amplitude (QAM)
QAM can visually be described a signal constellation graph
Distance represents amplitude, while angel represents phase shift
Figure 3.20 – Quantifying a Point on a signal Constellation
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M odems
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Signal Constellation




A shift with x° corresponds to x/360° of a period
2
V.22 uses fixed amplitude with 4 (2 ) points and 600 bauds
 600 * 2 = 1200 bps
4
V.22 bis uses 16 (2 ) points and 600 bauds
 600 * 4 = 2400 bps
V.32 uses 5 bits per baud but one of them is for Trellis Coding error detection
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 16 (2 ) points are counting and 2400 bauds
 2400 * 4 = 9600 bps
Figure 3.21 – Quantifying a Point on a signal Constellation
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M odems
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Signal Constellation

Distortion is possible, which may change the amplitude or the
phase shift
Figure 3.22 – Distortion of Signal Constellation Points
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M odems
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Signal Constellation

Depending on the distortion level, the modem may or may not
be able to interpret the distorted signal correctly
Figure 3.23 – Interpreting Constellation Points for a distorted signal
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M odems
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Modem Standards




Few modem standards exist!
They differ in baud rate, bit per baud, modulation techniques, error detection,
compressions, ..etc.
Autobaud modems exist and convenient since they are capable of
automatically choose the appropriate standard; they also allow users to
communicate using any of several standards
56 kbps modems were achievable when connecting directly with ISP
Figure 3.24 – Connections using a Modem
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M odems
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Cable Modems

Driven by many facts
including the inability
of modem over phone
lines to go beyond
56kbps

They connect with the
analog component of a
cable TV (CATV)
instead of the analog
component of the
telephone system
Figure 3.25 – Cable Modem Placement
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M odems
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Cable Modems

Utilizes the much higher frequency of CATV, so few Mbps
speed is achieved

However bit rate may differ depending on the number of users
(neighbors usually!) that are using the cable line
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M odems
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Cable Modems


Usually has a frequency of about 750 MHz, divided into many
6-MHz for different channels
Information from the Internet can be downloaded onto a 6MHz band somewhere between 42 and 750 MHz
Figure 3.26 –
Cable Modem & Carrier Signals
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M odems
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Cable Modems

A number of techniques can be used for modulating &
demodulating, but two are more popular:
• Quaternary Phase Shift Keying (QPSK)
• QAM64, a variation of QAM


QAM64 is typical for high bandwidth requirement – data
download rate can be 36 Mbps, which many PCs today are
not capable of handling
A more realistic figure is between 1 and 11 Mbps
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M odems
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Cable Modems

Can also be used to transmit information on the
opposite direction (Uploading)

Frequency used for uploading is usually between 5 &
40 MHz
• Uploading has lower bit rate, which might be okay.
Why?
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D SL

DSL - Digital Subscriber Line

Fast and does not require cable wiring or dialing up to ISP
Uses telephone lines!!!!
Local loop (last mile) is capable of transmitting signals of up
to 1 MHz frequency, that Local exchange can receive


Figure 3.27 – Local Loop - POTS (Plain Old Telephone Service) Configuration
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D SL
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Various forms of DSL exist
Asymmetric DSL
Figure 3.28 – ADSL Connection
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ADSL Lite

D SL
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Designed for residential customers
Figure 3.30 – ADSL Lite
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D SL
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VDSL – Very High Data Rate DSL



The local loop causes many DSL problems
Signal degrades over long distance, do DSL is not available if the house is
more than 3.5 miles from the local office
New improved local loop using fiber/copper hybrid is used, which
improves quality, reduces the copper length and makes VDSL possible
Figure 3.31 – Local Loop: Fiber/Copper Hybrid
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