chapter 3: data transmission coe 341: data & computer communications (t061) dr. radwan e....
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Chapter 3: Data Transmission
COE 341: Data & Computer Communications (T061)Dr. Radwan E. Abdel-Aal
2
Remaining Six Chapters:
Physical Layer
Transmission Medium
Data Link
Chapter 4: Transmission Media
Chapter 3: Signals, their representations, their
transmission over media, Resulting impairments
Chapter 5: Encoding: From data to signals
Chapter 7: Data Link: Flow and Error control,
Link management
Chapter 6: Data Communication: Synchronization,
Error detection and correction
Chapter 8: Improved utilization: Multiplexing
3
Agenda Concepts & Terminology Signal representation:
Time and Frequency domains Bandwidth and data rate Decibels and Signal Strength (Appendix 3A ) Fourier Analysis (Appendix B ) Analog & Digital Data Transmission Transmission Impairments Channel Capacity
4
Terminology (1)
Transmission system: Transmitter Receiver Transmission Media
Guided media e.g. twisted pair, coaxial cable, optical fiber
Unguided media e.g. air, water, vacuum
5
Terminology (2)
Link Configurations: Direct link
No intermediate ‘communication’ devices (these exclude repeaters/amplifiers)
Two types: Point-to-point
Only 2 devices share link Multi-point
More than two devices share the same link, e.g. Ethernet bus segment
Amplifier
A
B
C
6
Terminology (3)Transmission Types (ANSI Definitions) Simplex
In one direction only all the timee.g. Television, Radio broadcasting
Duplex In both directions Two types:
Half duplex Only one direction at a time e.g. Walki-Talki
Full duplex In both directions at the same time e.g. telephone
7
Frequency, Spectrum and Bandwidth Time domain concepts Analog signal
Varies in a smooth, continuous way in both time and amplitude
Digital signal Maintains a constant level for sometime and then changes to
another constant level (i.e. amplitude takes only a finite number of discrete levels)
Periodic signal Same pattern repeated over time
Aperiodic signal Pattern not repeated over time
8
Analogue & Digital Signals
9
PeriodicSignals
S (t+nT) = S (t); 0 t TWhere:t is time over first periodT is the waveform periodn is an integer
T
Temporal Period
t t+2Tt+1T
Signal behavior over one perioddescribes behavior at all times
…
10
Aperiodic (non periodic) Signals in time
0
1
+ X/2- X/2 t
s(t)
11
Sine Wave s(t) = A sin(2ft +)
Peak Amplitude (A) Peak strength of signal, volts
T = Period = time for one repetition (cycle) Repetition Frequency (f)
Measure of signal variations with time In Hertz (Hz) (cycles per second) f = 1/T(sec) = Hz
Angular Frequency () = radians per second = 2 f = 2 /T
Phase () Relative position in time, radians (how to determine?)
T
A= A sin ()
12
Varying one of the three parameters of a sine wave carriers(t) = A sin(2ft +) = A sin(t+)
Varying A
Varying
Varying f
Can be used to convey information…!
M o d u l a t I o n
13
Traveling Sine Waves(t) = A sin (k x - t]
Direction of Wave Travel, Velocity v
k = Wave Number= 2 /
= Angular Frequency= 2 f = 2 / T
x
Distance, x
t = 0
t = t
Spatial Period = Wavelength
xFor point p on the wave:
Total phase at t = 0: kx - (0) = kx
Total phase at t = t: k(x+ x) - (t)
Same total phase, kx = k(x+ x) - (t)k x = t
Wave propagation velocity v = x / t v = /k = /T = f
v = f
p
What is the expression for a wave traveling in the negative x direction?
V is constant for a given wave type and medium
14
Wave Propagation Velocity, v m/s Constant for a given wave type (e.g. electromagnetic, seismic, sound, ...) and propagation medium (air, water)
For all types of waves: v = f
For a given wave type (given v), higher frequencies correspond to shorter wavelengths and vise versa:
Radio: long wave (km), short waves, … light (nm), etc.. For electromagnetic waves:
In free space, v speed of light in vacuum
= c = 3x108 m/sec Over other guided media v is lower than c
15
Wavelength, meters) Is the Spatial period of the wave:
i.e. distance between two points in space where the wave has the same total phase
Also: Distance traveled by the wave during one temporal (time) cycle:
dT = v T = (f) T =
16
Continuous and Discrete RepresentationsAvailability of the signal over the horizontal axis
Continuous:
Signal is defined at all points on the horizontal axis
Discrete:
Signal is defined Only at certain points on the horizontal axis
Sampling with a train of delta function
(Time or Frequency)
17
Frequency Domain Concepts
System response to sine waves is easy to analyze Signals we deal with in practice are not all sine waves,
e.g. Square waves Can we relate practical waves to sine waves? YES Fourier analysis shows that any signal can be treated
as the sum of many sine wave components having different frequencies, amplitudes, and phases (Fourier Analysis: Appendix B)
This forms the basis for frequency domain analysis For a linear system, its response to a complex signal
will be the sum of its response to the individual sine wave components of the signal.
Dealing with functions in the frequency domain is simpler than in the time domain
18
Addition of Twofrequency Components
A = 1*(4/)frequency = f
A = (1/3)*(4/)frequency = 3f
+
=
Fundamental
3rd harmonic
Approaching a square wave
Frequency Domain: S(f) vs f Time Domain: s(t) vs t
Discrete Function in f
t3
Periodic function in t
Fourier Series
f
Frequency Spectrum
19
Asymptotically approaching a square wave by combining the fundamental + an infinite number of odd harmonics at prescribed amplitudes
-1.5
-1
-0.5
0
0.5
1
1.5
0 2 4 6 8 10t
sq
ua
re s
ign
al,
sw
(t)
What is the highestHarmonic added?
Topic of a programming assignment
20
More Frequency Domain Representations: A single square pulse (Aperiodic signal)
Time Domain: s(t) vs tFrequency Domain: S(f) vs f
• What happens to the spectrum as the pulse gets broader …-> DC ?
• What happens to the spectrum as the pulse gets narrower …-> spike ?
Continuous Function in f Aperiodic function in t
Fourier Transform
0
1
+ X/2- X/2t
s(t)
timefrequency
1/X
21
Spectrum & Bandwidth Spectrum
range of frequencies contained in a signal Absolute (theoretical) Bandwidth (BW)
Is the width of spectrum = fmax- fmin In many situations fmax =
Effective Bandwidth Often just bandwidth Narrow band of frequencies containing most of
the signal energy Somewhat arbitrary: what is “most”?
DC Component Component at zero frequency
f 3f 5f 7f f
S(f)
22
Signal with a DC Component
t
1V DC Level
1V DC Component
t
+
+
+
_
23
Bandwidth for these signals:fmin fmax Absolute
BWEffective
BW
1f 3f 2f 2f
0 3f 3f 3f
0 1/X ?
= (fmax- fmin)
24
Limiting Effect of System Bandwidth
)2sin(k
14)(
1 k odd,
kfttsk
f 3f
f 3f 5f
f 3f 5f 7f
f 3f 5f 7f ……
BW = 2f
BW = 4f
BW = 6f
BW =
…
Better w
aveforms require larger B
W
Mo
re d
iffic
ult
rece
ptio
n w
ith s
ma
ller
BW
1,3
1,3,5
1,3,5,7
1,3,5,7 ,9,…
1
2
3
4Fourier Series for a Square Wave
Var
ying
Sys
tem
BW
Received Waveform
25
System Bandwidth and Achievable Data Rates Any transmission system supports only a limited band of frequencies (bandwidth) for satisfactory transmission
“system” includes: TX, RX, and Transmission medium For example, this bandwidth is largest for optical fibers
and smallest for twisted pair wires. This limited system bandwidth degrades higher frequency
components of the signal transmitted poorer received waveforms more difficult to interpret the signal at the receiver (especially with noise) Data Errors
More degradation occurs when higher data rates are used (signal will have more components at higher frequency )
This puts a limit on the data rate that can be used with a given signal to noise requirement, receiver type, and a specified error performance Channel capacity issues
26
Bandwidth and Data Rates
2B = 4f’
B = 4f
f’ 3f’
f 3f 5f
Period T = 1/f
1 0 1 0
Data Element =Signal Element
T/2Data rate = 1/(T/2) = 2/T bits per sec = 2f
Given a bandwidth B,Data rate = 2f = B/2
To double the data rate you need to double f: Two ways to do this…
1. Double the bandwidth with same received waveform (same RX conditions & error rate)
f’ 3f’ 5f’
New bandwidth: 2B,Data rate = 2f’ = 2(2f)= 4f = B
2. Same bandwidth, B, but tolerate poorer received waveform (needs better receiver, higher S/N ratio, or tolerating more errors in data)
B = 2f’
Bandwidth: B,Data rate = 2f’ = 2(2f) = 4f = B
1 1 1 10 0 0 0
1 1 1 10 0 0 0
B
2B
B
Data
5f’
1 1 1 10 0 0 0
27
Bandwidth & Data Rates: Tradeoffs… Compromises Increasing the data rate (bps) while keeping BW the same means working with inferior (poorer) waveforms at the receiver, which may require: Ensuring higher signal to noise ratio at RX (larger
signal relative to noise): Shorter link distances Use of more en-route repeaters/amplifiers Better shielding of cables to reduce noise, etc.
More sensitive (& costly!) receiver Getting higher bit error rates
Tolerate them? Add more efficient means for error detection and correction-
this also increases overhead!.
28
Appendix 3A: Decibels and Signal Strength The decibel notation (dB) is a logarithmic measure of
the ratio between two signal power levels NdB = number of decibels P1 = input power level P2 = output power level
Example: A signal with power level of 10mW is inserted into a
transmission line Measured power some distance away is 5mW Power loss in dBs is expressed as
NdB =10 log (5/10)=10(-0.3)= -3 dB - ive dBs: P2 < P1 (Loss), +ive dBs: P2 > P1 (Gain)
1
210log10
P
PNdB
29
Decibels and Signal Strength
Decibel notation is a relative, not absolute, measure: A loss of 3 dB halves the power (could be 100 to 50, 16 to 8, …) A gain of 3 dB doubles the power (could be 5 to 10, 7.5 to 15, …)
Will see shortly how we can handle absolute levels Advantage:
The “log” allows replacing: Multiplication with addition
C = A * B
Log C = Log A + Log B and division with subtraction
30
Decibels and Signal Strength
Gain: 35 dB4 mW
Loss: 12 dB Loss: 10 dBTransmitted Signal
Amplifier
ReceivedSignal
?
Net power gain over transmission path:+ 35 – 12 – 10 = + 13 dB (+ ive means there is actual net gain)
mW4
Power Signal Receivedlog1013 10
Received signal power = (4 mW) log10-1(13/10)
= 4 x 101.3 mW = 79.8 mW
Example: Transmission line with an intermediate amplifier
31
Relationship Between dB Values and Power ratio (P2/P1)Power RatioPower Ratio dBdB Power RatioPower Ratio dBdB
1 0
101 10 10-1 -10
102 20 10-2 -20
103 30 10-3 -30
104 40 10-4 -40
105 50 10-5 -50
106 60 10-6 -60
2 3 1/2 -3
32
How to represent absolute power levels?Decibel-Watt (dBW) and Decibel-mW (dBm)
As a ratio relative to a fixed reference power level Value of 1 W is a reference defined as 0 dBW
Value of 1 mW is a reference defined as 0 dBm
Examples: Power of 1000 W is 30 dBW –10 dBm represents a power of 0.1 mW
W
PowerPower W
dBW 1log10 10
mW
PowerPower mW
dBm 1log10 10
Caution!: Must be same units attop and bottom
Caution!: Must be same units attop and bottom
WK 4
X dBW = (X + ?) dBm
33
Decibels and Signal Strength Example: Transmission line with an intermediate amplifier
Gain: 35 dB4 mW
Loss: 12 dB Loss: 10 dBTransmitted Signal
Amplifier
ReceivedSignal
?
Net power gain over transmission path:+ 35 – 12 – 10 = + 13 dB (+ ive means actual net gain)
RX signal power (dBm) = 6.02 + 13 =19.02 dBm Check: 19.02 dBm = 10 log (RX signal in mW/1 mW)
RX signal = log-1 (19.02/10) = 79.8 mW
TX Signal Power = 4 mW = 10 log (4/1) = 6.02 dBm
Everything in terms of dBs and dBm (or dBW) Levels {dBs and dBms} can be added and subtracted. Same for {dBs and dBWs}
34
Decibels & Voltage ratios Power decibels can also be expressed in terms
of voltage ratios Power P = V2/R, assuming same R
Decibel-millivolt (dBmV) is an absolute unit, with 0 dBmV being equivalent to 1mV.
1
22
1
22
1
2 log20/
/log10log10
V
V
RV
RV
P
PNdB
mV
VoltageN mV
dBmV 1log20
Note that this is still a power ratio… But expressed in voltages
Caution!: Must be same units attop and bottom
35
Appendix B: Fourier AnalysisSignals in Time
Periodic Aperiodic
Discrete Continuous Discrete Continuous
DFS FS FTDFT
FS : Fourier SeriesDFS : Discrete Fourier SeriesFT : Fourier TransformDFT : Discrete Fourier Transform
Use Fourier TransformUse Fourier Series
…
36
Fourier Series for periodic continuous signals Any periodic signal x(t) of period T and repetition
frequency f0 (f0 = 1/T) can be represented as an infinite sum of sinusoids of different frequencies and amplitudes – its Fourier Series. Expressed in Two forms:
1. The sine/cosine form:
1
000 )2sin()2cos(
2)(
nnn tnfBtnfA
Atx
T
dttxT
A0
0 )(2
T
n dttnftxT
A0
0 )2cos()(2
T
n dttnftxT
B0
0 )2sin()(2
If A0 is not 0,x(t) has a DC component
DC Component
f0 = fundamental frequency = 1/T Where:
Two components at each frequency
Frequencies are multiples
of the fundamentalfrequency f0
= f(n)
= f’(n)
All integrals overone period only
37
Fourier Series: 2. The Amplitude-Phase form: Previous form had two components at each frequency (sine, cosine i.e. in quadrature) : An, Bn coefficients
The equivalent Amplitude-Phase representation has only one component at each frequency: Cn, n
Derived from the previous form using trigonometry:
cos (a) cos (b) - sin (a) sin (b) = cos [a +b]
1
00 )2cos(
2)(
nnn tnfC
Ctx
00 AC 22nnn BAC
n
nn A
B1tan
Now components have different amplitudes, frequencies, and phases
Now one component at each frequency nf0
The C’s and ’s are obtained from the previous A’s and B’s using these equations.They are functionsof n
38
Fourier Series: General Observations
FunctionFunction SeriesSeries
No DC A0 = 0
Even Function x(t) = x(-t)
Symmetric about Y axis
Bn = 0;
for all n
Odd Function x(t) = - x(-t)
Symmetric about the origin
An = 0;
for all n
1
000 )2sin()2cos(
2)(
nnn tnfBtnfA
Atx
0)(1
0
T
dttxT
39
Correction
40
Fourier Series Example
1
-1
1/2-1/2 1 3/2-3/2 -1 2
T
0}][]{[2
1212)(2)(2
2)(
2
12/1
2/10
1
2/1
2/1
0
1
0
2
00
0
tt
dtdtdttxdttxdttxT
AT
x(t)
Note: (1) x(– t)=x(t) x(t) is an even function(2) f0 = 1 / T = ½ Hz
Note: A0 by definition is 2 x the DC content
41
1
0
0
2/
0
0
0
0 )2cos()(2)2cos()(4
)2cos()(2
dttnftxdttnftxT
dttnftxT
ATT
n
2/
2/
0
0
0 )2sin()(2
)2sin()(2 T
T
T
n dttnftxT
dttnftxT
B
2/
0
0
0
2/
0 )2sin()(2
)2sin()(2 T
T
dttnftxT
dttnftxT
0)2sin()(2
)()2sin()(2 2/
0
0
2/
0
0 TT
dttnftxT
tdtnftxT
Replacing t by –tin the first integralsin(-2nf t)=- sin(2nf t)
Contd…
Since x(– t)=x(t) as x(t) is an even function, then Bn = 0 for all n
= 0 for n even
= (4/n) sin (n/2) for n odd
1
-1
1/2-1/2 1 3/2-3/2 -1 2
T
a function of n only
f0 =1/2
42
1
000 )2sin()2cos(
2)(
nnn tnfBtnfA
Atx
tnn
ntx
oddn
cos2
sin4
)(,1
4 4 4 4( ) cos cos3 cos5 cos 7 ...
3 5 7x t t t t t
4 1 1 1( ) cos cos3 cos5 cos 7 ...
3 5 7x t t t t t
Contd…
A0 = 0, Bn = 0 for all n,
An = 0 for n even: 2, 4, … = (4/n) sin (n/2) for n odd: 1, 3, …
Cosine is an even function
Amplitudes, n odd
Original x(t) is an even function!
f0 = ½, so 2 f0 =
2 3 (1/2) t3rd Harmonic
43
Another Example
1
-1
1-1 2
T
-2
x1(t)
Note that x1(-t)= -x1(t) x(t) is an odd function
Also, x1(t)=x(t-1/2)
2
17 cos
7
1
2
15 cos
5
1
2
13 cos
3
1
2
1 cos
4)(1 tttttx
This waveform is the previous waveform shifted right by 1/2
Previous Example
44
Another Example, Contd…
2
77 cos
7
1
2
55 cos
5
1
2
33 cos
3
1
2 cos
4)(1
tttttx
...7in
7
1 5sin
5
1 3in
3
1 in
4)(1 tsttststx
tt sin2
cos
tt 3sin
2
33 cos
tt 5sin2
55 cos
tt 7sin
2
77 cos
Because:
Sine is an odd function
)2sin(k
14)(1 0
1 k odd,
tkftxk
As given before on slide 23.
45
Fourier Transform For aperiodic (non-periodic) signal in time, the spectrum
consists of a continuum of frequencies (not discrete components) This spectrum is defined by the Fourier Transform For a signal x(t) and a corresponding spectrum X(f), the
following relations hold
sincos je j
dfefXtx ftj 2 )()(
dtetxfX ftj 2 )()(
Forward FT (from time to frequency) Inverse FT (from frequency to time )
X(f) is always complex (Has both real & Imaginary parts), even for x(t) real.
sincos je j
je
Real
Imaginary
nf0 f
46
Sinc function
Sinc2 function
(non-periodic in time)
(Continuous in Frequency)
47
Fourier Transform Example
x(t)A
22
dtetxfX ftj 2 )()(
2/
2/
22/
2/
2
2 )(
ftjftj efj
AdteAfX
f
fA
f
ff
f
A
j
ee
f
A fjfj )sin()sin(
12
2/22/2
j
ee jj
2sin
2cos
jj ee
Sin (x) / xi.e. “sinc” function
48
Fourier Transform Example, contd.
f
fAfX
)sin()(
A
1/
= A
f
Study the effect of the pulse width
Sin (x) / x“sinc” functionLim x0 (sin x)/x = (cos x)x=0/1 =1
49
The narrower a function is in one domain, the wider its transform is in the other domain
The Extreme Cases
50
Power Spectral Density (PSD) & Bandwidth
Absolute bandwidth of any time-limited signal is infinite But luckily, most of the signal power will be
concentrated in a finite band of lower frequencies Power spectral density (PSD) describes the
distribution of the power content of a signal as a function of frequency
Effective bandwidth is the width of the spectrum portion containing most of the total signal power
We estimate the total signal power in the time domain
51
Signal Power in the time domain Signal is specified as a function x(t) representing signal voltage or
current Assuming R = 1
Instantaneous signal power (t) = v(t)2/1= i(t)2*1 = |x(t)|2 = f(t) Signal power can be obtained as the average of the instantaneous
signal power over a given interval of time = constant
For periodic signals, this averaging is taken over one period, i.e.
This measure in the time domain gives the total signal power Effective BW is then determined such that it contains an adequate
portion (percentage) of this total signal power
22
1
1( )
2 1
t
t
x t dtt t
T
Total dttxT
P0
2)(
1
TotalP (1)
52
Signal Power in the Frequency Domain: Periodic signals For periodic signal we have a discrete spectrum (the F Series):
For a DC component, Power = Vdc2
For AC components Power = Vrms2= Vpeak
2 (use eqn. 1 on prev. slide)
Power spectral density (PSD) is a discrete function of frequency:
Where (f) is the Dirac delta function: and total signal power (watts) in the first j harmonics is:
2
1
j
nnjUpto CCP
1
220Component th the 2
1
4
1
1
00 )2cos(
2)(
nnn tnfC
Ctx
220
1
1( ) ( )
4 2 n on
CPSD f C f nf
1 =00 0( ) f
ff
1
0 f0 2f0 3f0
…
(f-nf0)
f
(A function of frequency)
(A quantity, summation of PSD components- not a function of a frequency)
53
Example
Consider the following signal
The PSD is: (A function of Frequency)
The signal power is: (A quantity)
7in
7
1 5sin
5
1 3in
3
1 in1)( tsttststx
watt586.0 49
1
25
1
9
11
2
1] 7,5,3,1
harmonicsthPower
)]5.3(]7
1[)5.2(]
5
1[)5.1(]
3
1[)5.0(1[
2
1)( 2222 fffffPSD
(No DC)
54
Continuous (not discrete) frequency spectrum PSD (Power spectrum density) function is a
continuous function of frequency S(f) Power contained in the frequency band
f1< f < f2 is given by:
(Integration, instead of summation, over frequency)
2
1
)(2f
f
dffSP
Signal Power in the Frequency Domain: Aperiodic signals
55
Complete Fourier Analysis Example Consider the half-wave rectified cosine signal, Figure B.1 on page 793:
1. Write a mathematical expression for s(t) over its period T
2. Compute the Fourier series for s(t) (Amplitude & Phase form)
3. Get an expression for the power spectral density function for s(t)
4. Find the total power of s(t) from the time domain
5. Find the order of the highest harmonic n such that the Fourier series for s(t) contains at least 95% of the total signal power
6. Determine the corresponding effective bandwidth for the signal
56
Example (Cont.)
1. Mathematical expression for s(t):
cos(2 ) , -T/4 T/40 , T/4 3T/4( ) oA f t t
ts t
-T/4-3T/4 +3T/4+T/4
T/2
Where f0 is the fundamental frequency, f0 = (1/T)
57
Example (Cont.)
Before we start… what to expect?
-T/4-3T/4 +3T/4+T/4
• DC Component?• Even or odd function?• A0 ?• An ?• Bn ?
1
000 )2sin()2cos(
2)(
nnn tnfBtnfA
Ats
Sine/cosine form of the Fourier Series
58
Example (Cont.)Fourier Analysis:
1 )2/sin( as , 2
)2/sin(2)2/sin()2/sin(
)2/sin()2/sin(/2
)/2sin(2
)2cos(2
)(2
4/
4/
4/
4/
4/
4/
0
A
AA
A
T
Tt
T
A
dttfT
Adtts
TA
Tt
Tt
T
T
o
T
T
T
dttxT
A0
0 )(2
-T/4-3T/4 +3T/4+T/4
f0 = (1/T)
DC = ?
59
Example (Cont.)
2. Fourier Analysis (cont.):
/ 4 / 4
/ 4 / 4
/ 4
/ 4
2 2( )cos(2 ) cos(2 )cos(2 )
sin(2 ( 1) ) sin(2 ( 1) )2 , for 1
4 ( 1) 4 ( 1)
cos( / 2) cos( / 2) , for
( 1) ( 1)
T T
n o o o
T T
T
o o
o o T
AA s t nf t dt f t nf t dt
T T
n f t n f tAn
T n f n f
A n nn
n n
1
2
sin( ) sin( ) cos( )cos( ) , and
2( ) 2( )
sin( ) cos(
Note:
)
ax bx ax bxax bx dx
a b a b
x x
T
n dttnftxT
A0
0 )2cos()(2
f0 = (1/T)
n = 1 will be treatedSeparately later
From integral tables
60
Example (Cont.)Fourier Analysis (cont.):
n 1
2 2
2 2
2
( ) ( )
( ) ( )
( )
2
0 , for and 1
( 1) ( 1)
( 1) ( 1)
( 1) ( 1) ( 1)( 1) ( 1)
( 1)( 1)
( 1) ( 1) ( 1)( 1)
( 1)
oddn n
n n
n
n
n
A n n
AA
n n
A n n
n n
An n
n
2(1 )
2
2 ( 1) , for
( 1)even
n
An
n
, for n even
61
Example (Cont.)Fourier Analysis (cont.):For n = 1, A1 is obtained separately
Note: cos2 = ½(1 + cos 2)
62
Example (Cont.)
Fourier Analysis (cont.):
/ 4 / 4
/ 4 / 4
/ 4
/ 4
2 2( )sin(2 ) cos(2 )sin(2 )
cos(2 ( 1) ) cos(2 ( 1) )2 , for 1
4 ( 1) 4 ( 1)
0
T T
n o o o
T T
T
o o
o o T
AB s t nf t dt f t nf t dt
T T
n f t n f tAn
T n f n f
, for 1n
cos( ) cos( ) sin( )cos(Note
): )
2( 2( )
ax bx ax bxax bx dx
a b a b
T
n dttnftxT
B0
0 )2sin()(2
-
63
Example (Cont.)Fourier Analysis (cont.):For n = 1, B1 is obtained separately
/ 4 / 4
1
/ 4 / 4
/ 4
/ 4
/ 4
/ 4
2 2( )sin(2 1 ) cos(2 )sin(2 )
sin(4 )
cos(4 ) cos( ) cos( )4 4
0
T T
n o o o
T T
T
o
T
T
o T
AB s t f t dt f t f t dt
T T
Af t dt
T
A Af t
i.e. Bn = 0 for all n (our function is even!)
64
Example (Cont.)Fourier Analysis (cont.):
00 AC 0 ince ,22 nnnnn BsABAC
Note: n are not required for PSD and power calculations
65
Example (Cont.)
Power Spectral Density function (PSD):
220
1
2 2 2
2 2 2 22,4,6,...
1( ) ( )
4 2
( )2 ( ) ( )
8 ( 1)
n on
oo
n
CPSD f C f nf
f nfA A Af f f
n
n = 1n = 0 (DC) n = Even
1
00 )2cos(
2)(
nnn tnfC
Ctx
For large n, power decays (1/n4)… Good or bad?
66
Example (Cont.)Total Power:
(From the time domain)3 / 4 / 42
2 2
/ 4 / 4
/ 42
/ 4
2
1( ) cos (2 )
sin(4 )
2 8
4
T T
s o
T T
T
o
o T
AP s t dt f t dt
T T
f tA t
T f t
A
Note: cos2 = ½(1 + cos 2)
= Half the power of a full
sine wave
= 0.25 A2
67
Example (Cont.)Finding n such that we get at least 95% of the
total power:
2 2 220
0 2 2
2
2
For
40.1014
4 4
0.1014% 40.5%
0.25
0
n
C A APSD A
APower
A
n
(Only the DC component)
Power
% of total power in this component
68
Example (Cont.)Finding n such that we get at least 95% of the
total power:
2 2 2 220 1
1 2
2
2
For
0.2264 2 8
0.226% 90.5%
0.
1
25
n
C C A APSD A
APower
n
A
(DC + first harmonic)
Power
% of total power in these two components
69
Example (Cont.)Finding n such that we get at least 95% of the
total power:
2 2 2 2 2 220 1 2
2 2 2
2
2
For
20.2485
4 2 2 8 9
0.2485% 99.41
2
2
0. 5%
n
C C C A A APSD A
AP wer
A
n
o
(DC + first harmonic + second harmonic)
n = 2, and the effective bandwidth is: Beff = fmax – fmin
Beff = 2f0 – 0 = 2f0
0 f0 2f0 3f0
…
f
Beff
DC
Power
OK! 95%
70
Bandwidth and the Center Frequency Data are often sent as variations in a high frequency carrier signal having a frequency fc
So, bandwidth (BW) provided by this carrier signal occupies a range of frequencies centered about fc
The larger fc, the larger the BW obtainable Maximum BW obtainable for a given center
frequency fc is 2 fc
0 fc f
BW
])(2cos[])(2cos[)2cos()2cos(2 tfftfftftf mcmcmc
Carrier
CarrierModulating
Signal
71
Analog and Digital Data Transmission Data
Entities that convey meaning Signals
Electric or electromagnetic representations of data
Data Transmission Communication of data
through propagation and processing of signals
WK5
72
Data types in nature: Analog and Digital Data Analog Data Continuous values within some interval Examples: audio, video Typical bandwidths:
Speech: 100Hz to 7kHz Telephone: 300Hz to 3400Hz Video: 4MHz
Digital Data Discrete values (not necessarily binary) Examples: integers, text characters, mixture:
2347, “text”, SDR054
73
Analog and Digital Signals
Means by which data get transmitted over various media, e.g. wire, fiber optic, space
Analog signal: Continuously variable in time and amplitude
Digital signal: Uses a few (two or more) DC levels
74
Analog Signal Example 1: Speech
Frequency range for human hearing: 20Hz-20kHz Almost fully utilized by music Human speech: 100Hz-7kHz Telephone voice channel: Spectrum is further limited to
300-3400Hz (why?) Mechanical sound waves are easily converted into
electromagnetic signal for transmission: Mechanical waves (Sound) of varying pitch and
loudness (Data) is represented as:Electromagnetic signals of different frequencies and voltages (Signal)
75
Analog Example: 1. The Acoustic Spectrum
Frequency Range Dyn
amic
Ran
ge o
f S
igna
l Pow
er
Source Data
Hearing Spectrum
Log Scale
SPEECH
Dynamic range of the human ear can be as high as 120 dBs!
-70
76
Conventional Telephony: Analog data – Analog Signal
Telephone mouthpiece converts mechanical voice analog data into electromagnetic analog electrical signal
Signal travels on telephone lines At receiver, speaker re-converts received electrical
signal to voice
77
Analog Signal Example 2. Video Electrical signal proportional to the brightness of
image spot on a raster scanned phosphor screen
Interlaced Scan
Line Scan
Frame Scan
52.5 s (Active)11 s
78
Bandwidth of a Video Signal USA Specification: 525 lines per frame scanned at the rate
of 30 frames per second 525 lines = 483 active scan lines + 42 lost during vertical retrace
So 525 lines x 30 scans/second = 15750 lines per second Line scan interval = 1/15750 = 63.5s 11s go for horizontal retrace, so 52.5 s for active video per line
Effective vertical resolution = 0.7 x 483 = 338 lines Horizontal resolution = 338 x aspect ratio
= 338 x (4/3) = 450 dots Max frequency is when black and white dots alternate 450 picture dots correspond to 225 cycles in 52.5 s
Time period = 52.5/225 s fmax = 1/Period = 4.2 MHz
fmin (DC) = 0 Bandwidth = fmax - fmin = 4.2 MHz
79
Digital Signals
Advantages: Cheaper and easier to generate: No extra processing
needed Less susceptible to noise
(The threshold effect)
Disadvantages: When noise is above threshold Total data reversal
(1 0, 0 1) Greater attenuation
Line capacitances make pulses rounded and smaller in amplitude, leading to loss of information
More so at higher data rates and longer distances So, use at low data rates over short distances
80
Attenuation of Digital Signals
Effect of line capacitances
Worse at higher data rates (narrower pulses)
Pulse shaping
81
Digital Binary Signal Example: Between keyboard and computer Two bipolar dc levels (+ and – : Why?) Bandwidth required depends on the signal
frequency, which depends on: The data rate (bps) and The actual data sequence transmitted
_
-Data rate = ?
- Maximum f = ?
- Minimum f = ? Signal
Data
1
-
Data element
82
Data and Signal combinations We have seen above: (data and signal of same type)
Analog signals carrying analog data: Telephony, Video Digital signals carrying digital data: Keyboard to PC
Simple- only needs a transducer/transceiver But we may also have: (data and signal of different type)
Analog signal representing digital data: Data over telephone wires (using a modem)
Digital signal representing analog data: CD Audio, PCM (pulse code modulation) (using a codec) More complex- Needs a converter
So, all the four data-signal combinations are possible!
83
Analog Signals can carry Analog Data or Digital Data
(Base band)
(Converter)
We need a converter when the signal type is different from the data type
(Transducer)i.e. in its original form
84
Digital Signals can carry Analog Data or Digital Data
Coder-Decoder
Transmitter-Receiver
(Converter)
We need a converter when the signal type is different from the data type
85
Four Data/Signal Combinations Signal
Analog Digital
Data
Analog
Two ways: Signal has:- Same spectrum as data (base band): e.g. Conventional Telephony
- Different spectrum (modulation): e.g. AM, FM Radio
Use a (converter): codec, e.g. for PCM
(pulse code modulation)
Digital Use a (converter): modem e.g. the V.90
standard
- Simple two signal levels: e.g. NRZ code- Special Encoding: e.g. Manchester code (Chapter 5)
86
Two Modes of Transmitting Signals: 1. The Analog Mode (associated with FDM)
Treats the signal as “analog” regardless of what it represents (No interest in the data content of signal)
Following attenuation over distance, signal level is boosted using “amplifiers”
Unfortunately, this also amplifies in-band noise With cascaded amplifiers (i.e. one after the other at locations
along the link), effect on noise and distortion is cumulative, i.e. they get amplified again and again
Effect of noise and distortion on analog systems may be tolerated, e.g. with telephony you can still manage to get it! (Humans are good at filling-in the spaces!)
But digital systems are more sensitive to the effects of excessive noise and distortion unacceptable errors
So… Do not transmit digital signals the analog way!
87
Two Modes of Transmitting Signals: 2. The Digital Mode (Associated with TDM)
Concerned with the data content of the signal It assumes that the signal carries digital data Uses “repeaters” (not amplifiers), which:
Receive the signal Extract the data bit stream from it Retransmit a fresh, strong signal representing the
extracted data This way:
Effect of attenuation is overcome Noise and distortion are not cumulative
88
Four Signal/Transmission Mode Combinations Transmission mode
Analog- Uses amplifiers- Not concerned with what data the signal represents- Noise and distortion are cumulative-- Associated with FDM
Digital- Uses repeaters- Assumes signal represents digital data, recovers this data and represents it as a new outbound signal- This way, noise and distortion are not cumulative- Associated with TDM
Signal
Analog
OK
Makes sense only if the analog signal represents digital data! (Ask yourself: What digital is the repeater going to extract?!)
Digital Avoid OK
FDM: Frequency Division MultiplexingTDM: Time Division Multiplexing
Which transmission mode is more versatile
and useful for integrating different signal types?
89
Advantages of Digital Mode of Transmission Use of digital technology
Lower cost, smaller size, and high speed VLSI technology Higher data integrity (transmission quality) as noise effects are not
cumulative (fresh signal restoration en-route) Cover longer distances, at higher data rate, at low error rates, over lower
quality lines: Easier to implement multiplexing for improved utilization of link capacity
High bandwidth links are now economical (Fiber, Satellite…) To utilize them efficiently we need to do a lot of multiplexing This is done more efficiently using digital (TDM) rather than analog (FDM)
(Chapter 8) Easier encryption (digitally) for data
security and privacy Easier to integrate different data types
Convert analog data to digital signals…and use one same system to handle all voice, video, and data, e.g. one network for all types of traffic
Time Division Multiplexing
Frequency Division
Multiplexing
90
Transmission Impairments Signal received is often a degraded form of the signal
transmitted Why? What happens en-route?... Impairments:
Attenuation: Limits bandwidth In-band signals arrive weaker Attenuation distortion (Attenuation is not uniform over bandwidth)
Delay Delay distortion Noise (including crosstalk)
Effect: On analog data - Some degradation in signal quality On digital data – Fatal bit errors (total bit reversals)
91
Attenuation Signal strength falls off with distance traveled Nature of loss in signal power depends on medium:
Guided (Wires, etc.): Exponential drop is signal power with distance: Pd = P0 e-d
10 ln (Pd/P0) = -d
10 log (Pd/P0) = -’d
Loss: ’ dBs per km (’ depends on medium type e.g. fiber) Unguided (Open space):
Inverse square law spread with distance: P = P0 /d2
Loss: 6 dBs for each distance doubling Absorption, scattering May also depend on weather, e.g. rain, sunspots,
Signal power after traveling distance d
92
Effects of Attenuation Received signal strength must be:
Large enough to be detected Sufficiently higher than noise to be interpreted correctly
(without error) To overcome these problems:
Use amplifiers (analog transmission mode) or repeaters (digital transmission mode) en-route
Amplifier gains should not be too large as this may cause signal distortion due to saturation (nonlinearities)
Problem with networks: distance actually traveled (hence attenuation) will depend on actual route taken through the network!
93
Attenuation Distortion Attenuation usually increases with frequency This causes bandwidth limitation (understood) Moreover, over the transmitted bandwidth itself:
Different frequency components of the signal get attenuated differently Signal distortion
Affects analog signals more To overcome this problem:
Use Equalizers that reverse the effect of frequency-dependent attenuation distortion: Passive: e.g. loading coils in telephone circuits Active: Amplifier gain designed specifically for this purpose
94
Attenuation DistortionEqualizationTo Reduce Attenuation Distortion
Q. What is my signal ?
95
Delay Distortion
Happens only on guided media Wave propagation velocity varies with frequency:
Highest at the center frequency (minimum delay) Lower at both ends of the bandwidth (larger delay)
Effect: Different frequency components of the signal arrive at slightly different times! (Dispersion in time)
Affects digital data more: due to bit spill-over (timing is more critical here than for analog data)
Again, equalization can help overcome the problem
96
Delay DistortionEqualizationTo Reduce Delay Distortion
Without Equalizer
With Equalizer
97
Noise (1) Definition: Any additional unwanted signal inserted
between transmitter and receiver The most limiting factor in communication systems Noise Types:
Thermal Noise Inter-modulation Noise Crosstalk Noise Impulse Noise
98
Noise (2) Thermal Noise
Due to thermal agitation of electrons
(Increases with temperature) Uniformly distributed over frequency (White noise)
Difficult to eliminate
(exists even in the same bandwidth as your signal!) Effect becomes more significant when received signal is weak,
e.g. from satellites
PSD
f
99
Thermal Noise, Contd. Thermal noise power density in 1 Hz of bandwidth, N0
(Constant, Independent of frequency):
k Boltzmann’s constant = 1.3810-23 J/K T temperature in degrees Kelvin (= 273 + t C)
Thermal noise power in a bandwidth of B Hz:
)Hz/W(kTN 0
10 log k
Example: at t = 21 C (T = 294 K) and for a bandwidth of 10 MHz:
N = -228.6 + 10 log 294 + 10 log 107
= - 133.9 dBW
f
PSD
N0
1 Hz
B
Can you see some disadvantage now in having a larger
BW?
100
Noise (3) Inter-modulation Noise
Signals having the sum and difference (frequency mixing) of original frequencies sharing a transmission system
f1, f2 (f1+f2) and (f1-f2) Caused by nonlinearities in the medium and equipment,
e.g. due to overdrive and saturation of amplifiers Danger: Resulting new frequency components may fall
within valid signal bands, thus causing interference
A cos 1 + B cos 2 Linear System
K(A cos 1 + B cos 2)
A’ cos 1 + B’ cos 2
A cos 1 + B cos 2 Non-Linear System
K(A cos 1 + B cos 2) + K(A cos 1 + B cos 2)2
A’ cos 1 + B’ cos 2+ f(21)+f(22)+f(1-2)+f(1+2)
Inter-modulation componentsInput
Ou
tpu
t
New spurious components can fall within genuine signal bands causing interference
WK 6
101
Noise (4)
Crosstalk Noise A signal from one channel picked up by another channel in
close proximity Examples:
Physical proximity: coupling between adjacent twisted pair channels
Shield properly Directional proximity: antenna pick up from other directions
Use directional antennas Spectral proximity: leakage between adjacent channels in
frequency division multiplexing (FDM) systems
Use guard bands between adjacent channels
102
Noise (5)
Impulse Noise Pulses (spikes) of irregular shape and high amplitude
lasting short durations Causes: External electromagnetic interference due to
switching large currents, car ignition, lightning, … Minor effect on analog signals (e.g. crackling noise in
voice channels) Major effect on digital signals- Bit reversal error! More damage at higher data rates
(a given noise pulse can destroy a larger block of bits)
103
Effect of Impulse Noise on a Digital Signal
Q: What is the effect of the same noise at 10 times the data rate?
Impulse
+
=
RX
104
Channel Capacity Channel capacity: Maximum data rate usable under given
communication conditions How channel BW (B), signal level, noise and impairments,
and the amount of data error that can be tolerated limit the channel capacity?
In general, Max possible data rate, C, on a given channel = Function (B, Signal wrt noise, Bit error rate allowed) Max data rate: Max rate at which data can be communicated on
the channel, bits per second (bps) Bandwidth: BW of the transmitted signal as constrained by the
transmission system, cycles per second (Hz) Signal relative to Noise, SNR = signal power/noise power ratio
(Higher SNR better communication conditions higher C) Bit error rate (BER) allowed: in (bits received in error)/(total bits
transmitted). Equal to the bit error probability. e.g. Higher allowed higher usable data rates higher C
105
Channel Capacity, C: So, in general: C = F(B, SNR, BER) bits/s Three Formulations under different assumptions:
Assumptions Formulation
Ideal: Noise-free, Error-free: C = F(B) Nyquist
Noisy, Error-free: C = F(B, SNR) Shannon
Practical: Noisy, Error: C = F(B, SNR, BER) Eb/N0 Vs Error Rate
Realistic
Idealistic
106
1. Nyquist Channel capacity: (Noise-free, Error-free) Idealized, theoretical Assumes a noise-free error-free channel Nyquist showed that (without noise, without errors): If rate of
signal transmission is 2B then a signal with frequencies no greater than B is sufficient to carry that signalling rate
In other words: Given bandwidth B, highest signalling rate possible is 2B signals/s
How much data rate does this represent? Given a binary signal (1,0), data rate is same as signal rate
Data rate supported by a BW of B Hz is 2B bps C = 2B For the same B, data rate can be increased by sending one
of M different signals (symbols): as each signal level now represents log2M bits
Generalized Nyquist Channel Capacity, C = 2B log2M bits/s (bps)
Bandwidth efficiency = C/B = 2 log2M (bits/s)/Hz : Dimensionless quantity
Signals/s bits/signal
107
Nyquist Bandwidth: Example C = 2B log2M bits/s
C = Nyquist Channel Capacity B = Bandwidth M = Number of discrete signal levels (symbols) used
Data on telephone Channel: B = 3400-300 = 3100 Hz
With a binary signal (M = 2 symbols, e.g. 2 amplitudes):
C = 2B log2 2 = 2B = 6200 bps With a quadnary signal (M = 4 symbols):
C = 2B log2 4 = 2B x 2 = 4B = 12,400 bps
Channel capacity increased, but disadvantage: Larger number of signal levels (M) makes it more difficult for the receiver to determine data correctly in the presence of noise
0
1
00
01
10
11
108
2. Shannon Capacity Formula: (Noisy, Error-Free) Highest error-free data rate in the presence of noise
Signal to noise ratio SNR = signal / noise levelsSNRdB
= 10 log10 (SNR ratio) Errors are less likely with lower noise (larger SNR ratios).
This allows higher error-free data rates i.e. larger Shannon channel capacities
Shannon Capacity C = B log2(1+SNR):
Highest data rate transmitted error-free with a given noise level For a given BW, the larger the SNR the higher the data
rate I can use without introducing errors C/B: Spectral (bandwidth) efficiency, BE, (bps/Hz) (>1) Larger BEs mean better utilization of a given B for
transmitting data fast.
Caution! Log2 Not Log10
Caution! Ratio- Not dBs
109
Shannon Capacity Formula: Comments Formula says: for data rates calculated C, it is
theoretically possible to find an encoding scheme that achieves error-free transmission at the given SNR… But it does not say howAlso:
It is a theoretical approach based on thermal (white) noise only. But in practice, we also have impulse noise and attenuation and delay distortions, etc… So, maximum error-free data rates obtained in practice are
expected to be lower than the C predicted by the Shannon formula
However, maximum error-free data rates can be used to compare practical systems: The higher that rate the better the system…
110
Shannon Capacity Formula: Comments Contd. Formula suggests that changes in B and SNR can
be done arbitrarily and independently… but In practice, this may not be the case!
Higher SNR obtained through excessive amplification may introduce nonlinearities increased distortion and inter modulation noise … which reduces SNR!
High Bandwidth B opens the system for more thermal noise (kTB), and therefore reduces SNR!
111
Shannon Capacity Formula: Example
Spectrum of communication channel extends from 3 MHz to 4 MHz SNR = 24dB Then B = 4MHz – 3MHz = 1MHz
SNRdB = 24dB = 10 log10 (SNR)
SNR (ratio) = 251 Using Shannon’s formula: C = B log2 (1+ SNR)
C = 106 * log2(1+251) ~ 106 * 8 = 8 Mbps Based on Nyquist’s formula, determine M that gives the above
channel capacity:
C = 2B log2 M
8 * 106 = 2 * (106) * log2 M
4 = log2 M
M = 16
112
3. Eb/N0 Vs Error Rate Formulation (Noise and Error specified Together)
Handling both noise and a quantified error rate simultaneously
Introduce Eb/N0: A standard quality measure of channel parameters (B, SNR, R)
R is the data rate. Max value of R is the channel capacity Eb/N0 can also be independently related to the error rate It expresses SNR in a manner related to the data rate, R
Eb = Signal energy in one bit interval (Joules)
= Signal power (Watts) x bit interval Tb (second) = S x (1/R) = S/R
N0 = Noise power (watts) in 1 Hz = kT. Two formulations:
0 0
/b bE ST S R S
N N kT kTR
0 0
/b T TE B BS R SSNR
N N N R R
= SNR/BE
113
Eb/N0 Example 1:
If a minimum Eb/No of 8.4 dB is needed to achieve a bit error rate of 10-4
Given: The effective noise temperature, T, is 290oK The data rate, R, is 2400 bps
What is the minimum signal level required for the received signal?
8.4 = S(dBW) – 10 log 2400 + 228.6 dBW – 10 log290 = S(dBW) – (10)(3.38) + 228.6 – (10)(2.46)
S = -161.8 dBW
0
10log 10log 10log
10 log 228.6 10log
bdBW
dB
dBW
ES R k T
N
S R dBW T
114
Eb/N0 (Cont.) Bit error rate for digital data is a
decreasing function of Eb/N0 for a given signal encoding scheme
Which encoding scheme is better: A or B?
Analysis: For a given system (Eb/N0), determine error rate
Design: Given a desired error rate, get Eb/N0 to achieve it, then determine other parameters from formula, e.g. S, SNR, R, etc.
Effect of S, R, T on error performance
0
10log 10log 10log
10log 228.6 10log
bdBW
dB
dBW
ES R k T
N
S R dBW T
Lower E
rror Rate: larger E
b/N0
A B
0 0
/b T TE B BS R SSNR
N N N R R
BetterEncoding
= SNR
BE
Max R = C, BE = C/B
115
Shannon Capacity in terms of Eb/N0 From Shannon’s formula:
C = B log2(1+SNR)
We have:
From the Eb/N0 formula:
C/B (bps/Hz) is the spectral (bandwidth) efficiency BE
)12()12( / BEBCSNR
)12(1
0
BEb
BEBE
SNR
N
E
116
Eb/N0 (Cont.) Example 2
Find the minimum Eb/N0 required to achieve a bandwidth efficiency (BE=C/B) of 6 bps/Hz:
Substituting in the equation above:
Eb/N0 = (1/6) (26 - 1) = 10.5 = 10.21 dB
)12(1
0
BEb
BEN
E