Chapter 3: Data Transmission COE 341: Data & Computer Communications (T071) Dr. Radwan E. Abdel-Aal
Chapter 3: Data Transmission
COE 341: Data & Computer Communications (T071)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: Components Transmitter Receiver Medium
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
Information flows in one direction only all the timee.g. Television, Radio broadcasting
Duplex Information flows 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
Only a few amplitude levels allowed
- Binary signal: 2 levels
All values on the time and amplitude axes are allowed
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
s(t)
0
1
+ X/2- X/2 t
11
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)
12
Sine Wave s(t) = A sin(2f t +)
Peak Amplitude (A) Peak strength of signal, volts
Repetition Frequency (f) Measures how fast the signal varies with time Number of waveform cycles per second (Hz) f = 1/ T(xx sec/cycle) = yy cycles/sec = yy Hz
Angular Frequency () = radians per second = 2 f = 2 /T
Temporal (time) Period, T = 1/f Phase ()
Determines relative position in time, radians (how to calculate?)
T (Period)
A (Amplitude)= A sin ()
13
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
FMPM
AM
14
Sine Wave Traveling in the +ive x directions(t) = A sin (k x - t]
Direction of wave travel, at 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
Show that the wave s(t) = A sin (k x + t]travels in negative x direction
V is constant for a given wave type and medium
+ ive x direction
15
Wave Propagation Velocity, v m/s Constant for:
A given wave type (e.g. electromagnetic, seismic, ultrasound, ..) and a given propagation medium (air, water, optical fiber)
For all types of waves: v = f
For a given wave type and medium (given v): higher frequencies correspond to shorter wavelengths and vise versa:Electromagnetic waves:
long wave radio (km), short wave radio (m), microwave (cm)… light (nm) For electromagnetic waves:
In free space, v speed of light in vacuum v = c = 3x108 m/sec Over other guided media (coaxial cable, optical fiber, twisted
pairs): v is lower than c
Shorter wavelength .. Higher frequency
16
Wavelength, meters) Is the Spatial period of the wave:
i.e. distance between two points in space on the wave propagation path where the wave has the same total phase
Also: Distance traveled by the wave during one temporal (time) cycle:
dT = v T = (f) T =
17
Frequency Domain Concepts Response of systems to a sine waves is easy to analyze But signals we deal with in practice are not all sine
waves, e.g. Square waves Can we relate waves we deal with in practice 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
1/3 rd the Amplitude3 times the frequency
Fourier Series
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 for a programming assignmentAdding more
higherharmonics
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
Sinc(f) = sin(f)/f
To
To To
Fourier Transform
21
Spectrum & Bandwidth of a signal Spectrum of a signal
Range of frequencies contained in a signal Absolute (theoretical) Bandwidth (BW)
Is the width of spectrum = fmax- fmin But in many situations, fmax = !
(e.g. a square wave), so: Effective Bandwidth
Often called bandwidth Narrow band of frequencies containing most of
the signal energy Somewhat arbitrary: what is “most”?
f 3f 5f 7f f
S(f)
….
22
Signals with a DC Component
t
1V DC Level
1V DC Component
t
+
+
+
_
NO DC Component, Signal average over a period = 0
DC Component Component at zero frequency
Determines if fmin = 0 or not
23
Bandwidth for these signals:fmin fmax Absolute
BWEffective
BW
1f 3f 2f 2f
0 3f 3f 3f
0 1/X ?
= (fmax- fmin)
24
Bandwidth of a transmission system
Is the Range of signal frequencies that are adequately passed by the system
Effectively, the transmission system (TX, medium, RX) acts as a filter Poor transmission media, e.g. twisted pairs, have
a narrow filter bandwidth This cuts off higher frequency signal components
poor signal quality at receiver And limits the signal frequencies (Hz) that can be
used for transmission
limits the data rates used (bps)
f 3f 5f 7f f
S(f)
….
25
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 reception requires 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
26
System Bandwidth and Achievable Data Rates Any transmission system supports only a limited range of frequencies (bandwidth) for satisfactory transmission
For example, this bandwidth is largest for expensive optical fibers and smallest for cheap twisted pair wires
So, bandwidth is money Economize in its use 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
27
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 bps
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 0X 2
X 2
28
Bandwidth & Data Rates: Tradeoffs… Compromises Increasing the data rate (bps) while keeping BW the same (to economize) 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 Suffering from higher bit error rates
Tolerate them? Add more efficient means for error detection and correction- this
also increases overhead!.
29
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 (Watts) P2 = output power level (Watts)
e.g. Amplifier gain Signal loss over a link
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
Amplifier
P1 P2 P3
Lossy Link
30
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
31
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
A = C / B
Log A = Log C - Log B
32
Decibels and Signal Strength
]4
PowerSignal Received[log1013 10 mW
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 net gain)
Received signal power = (4 mW) log10-1(13/10) = 4 x 101.3
= 4 x 101.3 mW = 79.8 mW
Example: Transmission line with an intermediate amplifier
]10
13[log]
4
PowerSignal Received[ 10
1mW
Still we use some multiplication!
33
How to represent absolute power levels?Decibel-Watt (dBW) and Decibel-mW (dBm)
As a ratio relative to a fixed reference power level With 1 W used as a reference dBW
With 1 mW used as a reference dBm
Examples: Power of 1000 W is 30 dBW, 1 W = ? dBW –10 dBm represents a power of 0.1 mW, 1 mW = ? dBm X dBW = (X + ?) dBm
W
PowerLevelPower W
dBW 1log10_ 10
mW
PowerLevelPower mW
dBm 1log10_ 10
Caution!: Must be same units attop and bottom
Caution!: Must be same units attop and bottom
WK 4
34
dBs & dBms are added algebraically
mW
mW
P
P
1
2
GP1 P2
G = power ratio =
G dBs = 10 log10 G =
dBmdBm
mWmW
mWmW
mW
mW
mW
mW
PP
mWPmWP
mWPmWP
mWP
mWP
P
P
12
)]1/1([log10)]1/2(log[ 10
)]1/1(log)1/2(log[ 10
1/1
1/2log10
1
2log10
1010
1010
1010
Similarly for dBs & dBWs
G is Positive for gain Negative for attenuation
35
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 in dBm = 4 mW = 10 log (4/1) = 6.02 dBm
• If all ratios are in dBs and all levels are in dBm solve by algebraic addition Same for {dBs and dBWs} (No need for any multiplication/division)
As inSlide 31
36
Decibels & Voltage ratios Power decibels can also be expressed in terms
of voltage ratios Power P = V2/R, assuming same R
Relative:
Absolute: dBV and dBmV Decibel-millivolt (dBmV) is an absolute unit,
with 0 dBmV being equivalent to 1mV. Also dBV
1
22
1
22
1
2 log20/
/log10log10
V
V
RV
RV
P
PNdB Note that this is still a power ratio…
But expressed in terms of voltages
mV
VoltageN mV
dBmV 1log20 Caution!: Must be same units at
top and bottom
37
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
…
38
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
39
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 we have Only one component at each frequency nf0
The C’s and ’s are obtained from the previous A’s and B’s using the equations:
40
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
DC Even Function Odd FunctionFunction
Fourier Series Expansion
41
Correction
42
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
43
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
Replace t by –t Swap limitsin the first integral
Contd…
x(t), since x(t) is an even function
= 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
- sin(2nf t) dt Then Bn = 0 for all n
44
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) t
3rd Harmonic
2 (1/2) t
Fundamentalf0 = ½
45
Another Example
1
-1
1-1 2
T
-2
x1(t)
Note that x1(-t)= -x1(t) so, 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
46
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 for the square wave on slide 25.
47
Fourier Transform For aperiodic (non-periodic) signals 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 1T/2
Express sin and cos je
48
Sinc function
Sinc2 function
(non-periodic in time)
(Continuous in Frequency)
49
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
Area of pulseIn time domain
50
Fourier Transform Example, contd.
f
fAfX
)sin()(
A
2/
= A
f
Study the effect of the pulse width
Sin (x) / x“sinc” functionLim x0 (sin x)/x = (cos x)x=0/1 =1
First zero in the Frequency spectrum: sin f = 0ff =
51
The narrower a function is in one domain, the wider its transform is in the other domain
The Extreme Cases
0
52
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
53
Signal Power in the time domain Signal is specified as a function s(t) representing signal voltage or
current Assuming resistance R = 1
Instantaneous signal power (t) = v(t)2/1= i(t)2*1 = |s(t)|2
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 a specified
portion (percentage) of this total signal power
22
1
1( )
2 1
t
t
x t dtt t
T
Total dttsT
P0
2)(
1
TotalP (1)s
54
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: Total signal power (watts) up to the j th harmonic 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)
55
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)
56
Continuous (not discrete) frequency spectrum PSD (Power spectrum density) function, in Watts/Hz,
is a continuous function of frequency: S(f), Total signal power contained in the frequency band
f1< f < f2 (in Watts) is given by:
(Integration, instead of summation, over frequency)
2
1
)(2f
f
dffSP
Signal Power in the Frequency Domain: Aperiodic signals
Components exist in both negative and positive frequencies
Watts/Hz
57
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
58
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)
59
Example (Cont.)2. Fourier series
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
To get to the amplitude-phase form of the Fourier series, we must first obtain the sine-cosine form
60
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 = ?
61
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
62
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
63
Example (Cont.)Fourier Analysis (cont.):For n = 1, A1 is obtained separately
Note: cos2 = ½(1 + cos 2)
64
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
-
65
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!)
66
Example (Cont.)Fourier Analysis (cont.):
00 AC n 0 ince ,22 allforBsABAC nnnnn
Note: n are not required for PSD and power calculations
67
Example (Cont.)
3. 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?
68
Example (Cont.)4. 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
-T/4-3T/4 +3T/4+T/4
Zero
69
Example (Cont.)5. 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
70
Example (Cont.)Finding n such that we get at least 95% of the
total power, contd.:
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
71
Example (Cont.)Finding n such that we get at least 95% of the
total power, Contd.:
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
6. the effective bandwidth is: Beff = fmax – fmin
Beff = 2f0 – 0 = 2f00 f0 2f0 3f0
…
f
Beff
DC
Power
OK! 95%
72
Bandwidth about a Center Frequency So far we have considered signals in their base band form (without modulation)
Data is often sent as variations in a high frequency carrier signal having a frequency fc (modulation)
So, bandwidth (BW) of this signal occupies a range of frequencies centered about fc
The larger fc, the larger the BW obtainable
Largest BW obtainable for a given center frequency fc is 2
fc
With Amplitude Modulation, For each component of the modulating signal:
0 fc f
BW
])(2cos[])(2cos[)2cos()2cos(2 tfftfftftf mcmcmc
CarrierModulating
Signal
Carrier
73
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 that represent them
WK5
74
Data types in nature: Analog and Digital Data Analog Data Continuous values within some interval Examples: audio, video Typical bandwidths:
Speech: 100Hz to 7kHz Voice over telephone: 300Hz to 3400Hz Video: 4MHz
Digital Data Discrete values (not necessarily binary) Examples: integers, text characters, mixture:
2347, “text”, SDR054
75
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
76
Analog Signal Example 1: Speech Data
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 (data) are easily converted into
electromagnetic signal for processing and transmission: Mechanical waves (Sound) of varying pitch and loudness (Data)
is represented as:
Electromagnetic signals of different frequencies and amplitudes (Signal)
77
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
dBs
78
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
79
Analog Signal Example 2. Video Data 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
80
Bandwidth of a Black & White 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 frames/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
81
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
(Bit error) (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
82
Attenuation of Digital Signals
Effect of line capacitances
Worse at higher data rates (narrower pulses)
Pulse shapingDue to line capacitances:Worse over longer distances
1 1 1 1 . . .
0 0 0 0 . . .
83
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
84
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- one only needs a transducer/transceiver But we may also have: (data and signal of different types)
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- We Need a converter
So, all the four data-signal combinations are possible!
85
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
86
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
Digitized Analog Samples
e.g. using PCM(Pulse Code Modulation)
87
Four Data/Signal Combinations Signal
Analog Digital
Data
Analog
Two ways: Signal has:- Same spectrum as data (base band): e.g. Telephony to exchange
- Different spectrum (through modulation): e.g. AM Radio, FDM
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)
88
Two Modes of Transmitting Signals: 1. The Analog Mode (associated with FDM)
Treats the signal as “analog” regardless of what it represents (Not interested 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 gaps!)
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!
89
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 bit stream This way:
Effect of attenuation is overcome Noise and distortion are not cumulative
90
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 data 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?
91
Advantages of Digital Mode of Transmission Use of digital technology
Lower cost, smaller size, and high speed VLSI technology Higher data integrity (reliability) 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) Encryption for data security
and confidentiality is digital Easier to integrate different data types
Convert analog data to digital signals…and use one system to handle all voice, video, and data, e.g. one network for all types of traffic
Time Division Multiplexing
Frequency Division
Multiplexing
92
Transmission Impairments Signal received is often a degraded form of the signal
transmitted Why? What happens en-route?... Impairments:
Attenuation: Limits the bandwidth of the received signal In-band signals arrive weaker Attenuation distortion (Attenuation is not uniform over bandwidth)
Delay Delay distortion Noise and interference (including crosstalk)
Effect: On analog data - Some degradation in signal quality On digital data – Fatal bit errors (total bit reversals)
93
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, twisted pair, cable)
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
94
Effects of Attenuation Received signal strength must be:
Sufficiently 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!
95
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
96
Attenuation DistortionEqualizationTo Reduce Attenuation Distortion
Q. What is the signal ?
97
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
98
Delay DistortionEqualizationTo Reduce Delay Distortion
Without Equalizer
With Equalizer
99
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
100
Noise (2) Thermal (White) 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 is more significant on weak received signals,
e.g. from satellites
PSD
f
101
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?
102
Noise (3) Inter-modulation Noise
Signals having the sum and difference (frequency mixing) of original frequencies sharing a transmission system
(e.g. in FDM systems)
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 interferenceA 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
Input
103
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 cables 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
104
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 noise pulse of a given width can destroy a larger block of bits)
105
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
106
Channel Capacity Channel capacity: Maximum data rate usable under a
given set of 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
107
Channel Capacity, C: So, in general: C bps = F(B, SNR, BER) 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
108
Bandwidth (or Spectral) Efficiency (BE):
Measures how well we are utilizing a given bandwidth to send data at a high rate….
Can be greater than 1 (not like engineering efficiencies) The larger the better
HzbpsBBandwidth
CCapacityChannelBE / ,
109
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 frequency components up to B Hz is sufficient to carry that signalling rate
In other words: Given bandwidth B, highest signalling rate possible is 2B signal elements/s
How much data rate does this represent? (depends on how many bits are represented by each signal element!) 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
110
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 x 1 = 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
2 bits/Symboli.e. 2 bits /signal element
Signal Element
111
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 bandwidth
B for transmitting data fast.
Caution! Log2 Not Log10
Caution! Ratio- Not dBs
112
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 how!Also:
It is a theoretical approach based on thermal (white) noise only. But in practice, we also have impulse noise, attenuation and delay distortions, etc… So, maximum error-free data rates measured in practice
are expected to be lower than the C predicted by the Shannon formula due to the greater noise
However, maximum error-free data rates can be used to compare practical systems: The higher that rate the better the system…
113
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 also introduce nonlinearities increased distortion and inter-modulation noise … which reduces SNR!
High Bandwidth B opens the system up for more thermal noise (kTB), and therefore reduces SNR!
114
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) = log-110 (24/10) = 1024/10 = 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
115
3. Eb/N0 Vs Error Rate Formulation (Noise and Error are both specified Together) Handling both noise and a quantified error rate simultaneously
We introduce Eb/N0: A standard quality measure of three channel parameters (B, SNR, R) and can also be independently related to the error rateR is the data rate. Max value of R is the channel capacity C
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
Tb = 1/R
116
Eb/N0 (Cont.) Bit error rate for digital data is a
decreasing function of Eb/N0 for a given signal encoding scheme
Analysis: For a given system (SNR, B, R) (Eb/N0), determine error rate BER
Design: Given a desired error rate BER, 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
Which encoding scheme is better: A or B?
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
BER vs Eb/N0 curve for a given encoding scheme
117
Example:
Given: The effective noise temperature, T, is 290oK The data rate, R, is 2400 bps Would like to operate with a bit error rate of 10-4 (e.g. 1 error in 104 bits)
What is the minimum signal level required for the received signal?
From curve, a minimum Eb/No needed to achieve a bit error rate of 10-4 = 8.4 dB
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
Design or Analysis?
118
Eb/N0 in terms of BE, assuming Shannon channel capacity
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 based on Shannon channel capacity
)12()12( / BEBCSNR
)12(1
0
BEb
BEBE
SNR
N
E
119
Example
Find the minimum Eb/N0 required to achieve a Shannon bandwidth efficiency (BE=CShannon/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