Transcript COE 341: Data & Computer Communications (T062) Dr. Marwan Abu-Amara

COE 341: Data & Computer Communications (T062)
Dr. Marwan Abu-Amara
Chapter 3: Data Transmission
Agenda






Concepts & Terminology
Decibels and Signal Strength
Fourier Analysis
Analog & Digital Data Transmission
Transmission Impairments
Channel Capacity
COE 341 – Dr. Marwan Abu-Amara
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Terminology (1)



Transmitter
Receiver
Medium

Guided medium


e.g. twisted pair, optical fiber
Unguided medium

e.g. air, water, vacuum
COE 341 – Dr. Marwan Abu-Amara
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Terminology (2)

Direct link


Point-to-point



No intermediate devices
Direct link
Only 2 devices share link
Multi-point

More than two devices share the link
COE 341 – Dr. Marwan Abu-Amara
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Terminology (3)

Simplex

One direction


Half duplex

Either direction, but only one way at a time


e.g. Television
e.g. police radio
Full duplex

Both directions at the same time

e.g. telephone
COE 341 – Dr. Marwan Abu-Amara
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Frequency, Spectrum and Bandwidth

Time domain concepts

Analog signal


Digital signal


Maintains a constant level then changes to another
constant level
Periodic signal


Varies in a smooth way over time
Pattern repeated over time
Aperiodic signal

Pattern not repeated over time
COE 341 – Dr. Marwan Abu-Amara
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Analogue & Digital Signals
COE 341 – Dr. Marwan Abu-Amara
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T
Periodic
Signals
Temporal Period
S (t+nT) = S (t);
Where:
t is time
T is the waveform period
n is an integer
COE 341 – Dr. Marwan Abu-Amara
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Sine Wave – s(t) = A sin(2ft +)

Peak Amplitude (A)



maximum strength of signal
unit: volts
Frequency (f)



rate of change of signal
unit: Hertz (Hz) or cycles per second
Period = time for one repetition (T) = 1/f

Phase ()

relative position in time
 unit: radians
Angular Frequency (w)



w = 2 /T = 2 f
unit: radians per second
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Varying Sine Waves
s(t) = A sin(2ft +)
COE 341 – Dr. Marwan Abu-Amara
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Wavelength ()



Distance occupied by one cycle
Distance between two points of corresponding
phase in two consecutive cycles
Assuming signal velocity v



 = vT
f = v
For an electromagnetic wave,
v = speed of light in the medium

In free space, v = c = 3*108 m/sec
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Frequency Domain Concepts




Signal usually made up of many frequencies
Components are sine waves
Can be shown (Fourier analysis) that any
signal is made up of component sine waves
Can plot frequency domain functions
COE 341 – Dr. Marwan Abu-Amara
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Addition of
Frequency
Components
(T=1/f)
Fundamental Frequency
COE 341 – Dr. Marwan Abu-Amara
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Frequency
Domain
Representations
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Spectrum & Bandwidth

Spectrum


Absolute bandwidth


width of spectrum
Effective bandwidth
 Often just bandwidth


range of frequencies contained in signal
Narrow band of frequencies containing most of
the energy
DC Component

Component of zero frequency
COE 341 – Dr. Marwan Abu-Amara
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Signal with a DC Component
t
1V DC Level
t
1V DC
Component
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Bandwidth for these signals:
Absolute Effective
BW
BW
fmin
fmax
1f
3f
2f
2f
0
3f
3f
3f
0


1/x ?
COE 341 – Dr. Marwan Abu-Amara
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Bandwidth and Data Rate







Any transmission system supports only a limited band of
frequencies for satisfactory transmission
“system” includes: TX, RX, and Medium
Limitation is dictated by considerations of cost, number
of channels, etc.
This limited bandwidth degrades the transmitted signals,
making it difficult to interpret them at RX
For a given bandwidth: Higher data rates
More
degradation
This limits the data rate that can be used with given
signal and noise levels, receiver type, and error
performance
More about this later!!!
COE 341 – Dr. Marwan Abu-Amara
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Bandwidth
Requirements
BW = 2f
f
3f
1
1,3,5
BW = 4f
f
3f
5f
2
1,3,5,7
BW = 6f
f
3f
5f 7f
…
More difficult reception with more limited BW
Larger BW needed for better representation
4
1,3
3
BW = 
1,3,5,7
,9,…

1
s(t ) 
sin( 2kft)

 k odd, k  1 k
f
3f 5f 7f …… 
COE 341 – Dr. Marwan Abu-Amara
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Fourier Series for a Square Wave
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Decibels and Signal Strength

Decibel is a measure of ratio between two
signal levels




NdB = number of decibels
P1 = input power level
P2 = output power level
N dB  10 log 10
P2
P1
Example:



A signal with power level of 10mW inserted onto a
transmission line
Measured power some distance away is 5mW
Loss expressed as NdB =10log(5/10)=10(-0.3)=-3 dB
COE 341 – Dr. Marwan Abu-Amara
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Decibels and Signal Strength

Decibel is a measure of relative, not absolute, difference




A loss from 1000 mW to 500 mW is a loss of 3dB
A loss of 3 dB halves the power
A gain of 3 dB doubles the power
Example:





Input to transmission system at power level of 4 mW
First element is transmission line with a 12 dB loss
Second element is amplifier with 35 dB gain
Third element is transmission line with 10 dB loss
Output power P2


(-12+35-10)=13 dB = 10 log (P2 / 4mW)
P2 = 4 x 101.3 mW = 79.8 mW
COE 341 – Dr. Marwan Abu-Amara
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Relationship Between Decibel Values
and Powers of 10
Power
Ratio
101
dB
10
Power
Ratio
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
COE 341 – Dr. Marwan Abu-Amara
dB
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Decibel-Watt (dBW)


Absolute level of power in decibels
Value of 1 W is a reference defined to be 0 dBW
PowerdBW  10 log 10

PowerW
1W
Example:


Power of 1000 W is 30 dBW
Power of 1 mW is –30 dBW
COE 341 – Dr. Marwan Abu-Amara
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Decibel & Difference in Voltage


Decibel is used to measure difference in
voltage.
Power P=V2/R
2
N dB

P2
V2 / R
V2
 10 log
 10 log 2
 20 log
P1
V1
V1 / R
Decibel-millivolt (dBmV) is an absolute unit with
0 dBmV equivalent to 1mV.

Used in cable TV and broadband LAN
VoltagemV
VoltagedBmV  20 log
1mV
COE 341 – Dr. Marwan Abu-Amara
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Fourier Analysis
Signals
Aperiodic
Periodic (fo)
Discrete
DFS
FT
DFT
DTFT
FS
DFS
Continuous
Discrete
Continuous
FS
FT
: Fourier Transform
: Discrete Fourier Transform
: Discrete Time Fourier Transform
: Fourier Series
: Discrete Fourier Series
Infinite time
DTFT
COE 341 – Dr. Marwan Abu-Amara
Finite time
DFT
25
Fourier Series (Appendix B)

Any periodic signal of period T (f0 = 1/T) can be
represented as sum of sinusoids, known as
fundamental
Fourier Series
frequency

A0
x(t ) 
  An cos( 2nf 0t )  Bn sin( 2nf 0t )
2 n 1
DC
Component
T
2
A0   x(t )dt
T 0
If A0 is not 0,
x(t) has a DC
component
T
2
An   x(t ) cos( 2nf 0t )dt
T 0
T
2
Bn   x(t ) sin( 2nf 0t )dt
T 0
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Fourier Series

Amplitude-phase representation

C0
x(t ) 
  Cn cos( 2nf 0t   n )
2 n 1
C0  A0
Cn 
An2  Bn2
  Bn 

 n  tan 
 An 
1
COE 341 – Dr. Marwan Abu-Amara
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COE 341 – Dr. Marwan Abu-Amara
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Fourier Series Representation of
Periodic Signals - Example
x(t)
1
-3/2
-1
-1/2
1/2
1
3/2
2
-1
T
Note: (1) x(– t)=x(t)  x(t) is an even function
(2) f0 = 1 / T = ½
T
A0 
2
1
1/ 2
1
2
2
x
(
t
)
dt

x(t )dt  2 x(t )dt  2  1dt  2   1dt  1  1  0


T 0
20
0
0
1/ 2
COE 341 – Dr. Marwan Abu-Amara
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Fourier Series Representation of
Periodic Signals - Example
T
2
4
An   x(t ) cos( 2nf 0t )dt 
T0
T
T /2
1
 x(t ) cos(2nf t )dt  2 x(t ) cos(2nf t )dt
0
0
0
0
4
n
 2  cos( 2nf 0t )dt  2   cos( 2nf 0t )dt 
sin
n
2
0
1/ 2
1/ 2
1
T
T /2
2
2
Bn   x(t ) sin( 2nf 0t )dt 
x(t ) sin( 2nf 0t )dt

T0
T T / 2
0
2
2

x
(
t
)
sin(
2

nf
t
)
dt

0
T T/ 2
T
2

T
T /2
T /2
 x(t ) sin( 2nf t )dt
0
0
2
0 x(t ) sin( 2nf0t )dt  T
T /2
 x(t ) sin( 2nf t )dt
0
Replacing t by –t
in the first integral
sin(-2nf t)=
- sin(2nf t)
0
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Fourier Series Representation of
Periodic Signals - Example
Since x(– t)=x(t) as x(t) is an even function, then
Bn = 0 for n=1, 2, 3, …

A0
x(t ) 
  An cos( 2nf 0t )  Bn sin( 2nf 0t )
2 n 1

4
n
x(t )  
sin
cos nt
2
n 1 n
4
4
4
4
x (t )  cos  t 
cos 3 t 
cos 5 t 
cos 7 t  ...

3
5
7
4
1
1
1

x (t )  cos  t  cos 3 t  cos 5 t  cos 7 t  ...

3
5
7

COE 341 – Dr. Marwan Abu-Amara
Cosine is an
even function
31
Another Example
x(t)
x1(t)
1
-2
-1
1
2
-1
T
Note that x1(-t)= -x1(t)  x(t) is an odd function
Also, x1(t)=x(t-1/2)
4
x 1(t )  cos

 1 1
 1 1
 1 1
 1 
  t    cos 3  t    cos 5  t    cos 7  t    
 2 3
 2 5
 2 7
 2 
COE 341 – Dr. Marwan Abu-Amara
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Another Example
4
x 1(t )  cos

  1
  t    cos
2 3

3  1

 3 t    cos
2 5

5  1

 5 t    cos
2 7

7  

 7 t    
2 

4
1
1
1
x 1(t )  sin t  sin3 t  sin 5 t  sin7 t 

3
5
7


Where: cos  t    sin t
2

5 

cos  5t 
  sin 5t
2 

3 

cos  3t 
   sin 3t
2 



Sine is an
odd function
7 

cos  7t 
   sin 7t
2 

COE 341 – Dr. Marwan Abu-Amara
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Fourier Transform


For a periodic signal, spectrum consists of
discrete frequency components at fundamental
frequency & its harmonics.
For an aperiodic signal, spectrum consists of a
continuum of frequencies (non-discrete
components).


Spectrum can be defined by Fourier Transform
For a signal x(t) with spectrum X(f), the following
relations hold

x(t ) 
 X(f )e

j 2ft

df
X ( f )   x(t ) e
 j 2ft
dt

COE 341 – Dr. Marwan Abu-Amara
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COE 341 – Dr. Marwan Abu-Amara
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Fourier Transform Example
x(t)
A
 2
 2

X(f ) 
 j 2ft
x
(
t
)
e
dt


 /2
X( f ) 

 /2
Ae
 j 2ft
A  j 2ft  / 2
dt  
e
 / 2
j 2f
COE 341 – Dr. Marwan Abu-Amara
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Fourier Transform Example
2A

2f
 e j 2f / 2  e  j 2f / 2  2 A  2f  sin( 2f / 2) 





2j

 2f  2  2f / 2 
sin( 2f / 2)
sin( f )
X ( f )  A
 A
2f / 2
f
 e j  e  j 
sin   

 2j

 e j  e  j 
cos  

2


Sin (x) / x
“sinc” function
A


COE 341 – Dr. Marwan
Abu-Amara
Study
the effect
 1/
of the pulse width 
f
37
The narrower a function is
in one domain,
the wider its transform is
in the other domain
The Extreme Cases
COE 341 – Dr. Marwan Abu-Amara
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Power Spectral Density & Bandwidth




Absolute bandwidth of any time-limited signal is
infinite
However, most of the signal power will be
concentrated in a finite band of frequencies
Effective bandwidth is the width of the spectrum
portion containing most of the signal power.
Power spectral density (PSD) describes the
distribution of the power content of a signal as a
function of frequency
COE 341 – Dr. Marwan Abu-Amara
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Signal Power



A function x(t) specifies a signal in terms of
either voltage or current
Assuming R = 1 W,
Instantaneous signal power = V2 = i2 = |x(t)|2
Instantaneous power of a signal is related to
average power of a time-limited signal, and is
t2
defined as
1
2
t 2 t1

 x (t )
dt
t1
For a periodic signal, the averaging is taken
over one period to give the total signal power
T
1
2
PTotal 
x
(
t
)
dt

COE 341 –T
Dr. Marwan Abu-Amara
0
40
Power Spectral Density & Bandwidth

For a periodic signal, power spectral density
is
PSD (f ) 


n 
where (f) is
C n  (f  nf 0 )
2
 (f )   10 ff
=0
0
Cn is as defined before on slide 27, and f0
being the fundamental frequency
COE 341 – Dr. Marwan Abu-Amara
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Power Spectral Density & Bandwidth

For a continuous valued function S(f), power
contained in a band of frequencies f1 < f < f2
f2
P  2  S ( f )df
f1

For a periodic waveform, the power through
the first j harmonics is
j
1
P  0.25C02   Cn2
2 n 1
COE 341 – Dr. Marwan Abu-Amara
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Power Spectral Density & Bandwidth Example

Consider the following signal
1
1
1


x(t )   sin t  sin 3t  sin 5t  sin 7t 
3
5
7



The signal power is
1 1 1
1 
Power  1  

 0.586 watt

2  9 25 49 
COE 341 – Dr. Marwan Abu-Amara
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Fourier Analysis Example
Consider the half-wave rectified cosine signal from
Figure B.1 on page 793:

1.
2.
3.
4.
5.
6.
Write a mathematical expression for s(t)
Compute the Fourier series for s(t)
Write an expression for the power spectral density function
for s(t)
Find the total power of s(t) from the time domain
Find a value of n such that Fourier series for s(t) contains
95% of the total power in the original signal
Determine the corresponding effective bandwidth for the
signal
COE 341 – Dr. Marwan Abu-Amara
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Example (Cont.)
1.
Mathematical expression for s(t):
s (t ) 

A cos(2 f o t )
0
, -T/4  t  T/4
, T/4  t  3T/4
Where f0 is the fundamental frequency,
f0 = (1/T)
-3T/4
-T/4
+T/4
COE 341 – Dr. Marwan Abu-Amara
+3T/4
45
Example (Cont.)
2.
T
2
A0   x(t )dt
T 0
Fourier Analysis:
T /4
f0 = (1/T)
T /4
2
2A
A0 
s (t )dt 
cos( 2f ot )dt


T T / 4
T T / 4
2 A  sin( 2t / T ) 
A


  sin(  / 2)  sin(  / 2)

T  2 / T  T / 4 
T /4


A

 sin(  / 2)  sin(  / 2) 
2A

A

 2  sin(  / 2)
, where sin(  / 2)  1
COE 341 – Dr. Marwan Abu-Amara
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Example (Cont.)
2.
T
2
An   x(t ) cos( 2nf 0t )dt
T0
Fourier Analysis (cont.):
2
An 
T
T /4
2A
s (t ) cos(2 nf o t )dt 

T
T / 4
f0 = (1/T)
T /4

cos(2 f o t ) cos(2 nf o t )dt
T / 4
T /4
2A  sin(2 ( n  1)f o t ) sin(2 ( n  1)f o t ) 



, for n  1

T
4 (n  1)f o  T / 4
 4 (n  1)f o
 cos(n  / 2)  cos(n  / 2) 
 

, for n  1

  (n  1)
( n  1)

A
sin(ax  bx ) sin(ax  bx )
Note:  cos(ax ) cos(bx )dx 

, and
2(a  b )
2(a  b )
sin( x )  cos( x   2 )
COE 341 – Dr. Marwan Abu-Amara
47
Example (Cont.)
2.
Fourier Analysis (cont.):
An  0
, for n odd and n  1
 (1)
(1) 
An   


  (n  1)
(n  1) 
(n 2)
A
A  (1)(
 
 
n
2)
(n 2)
(n  1)  ( 1)( 1)
(n  1)(n  1)
(n 2)
( n  1) 


A (1)

  (n  1)  (1)( n  1) 
2
 (n  1)
(n 2)
2A (1)(1 2 )

 (n 2  1)
n
, for n even
COE 341 – Dr. Marwan Abu-Amara
48
Example (Cont.)
2.
An 1
Fourier Analysis (cont.):
2

T
T /4
T /4
2A
s (t ) cos(2 1 f ot )dt 
cos(2 f ot ) cos(2 f ot )dt


T T / 4
T / 4
T /4
2A
2

cos
(2 f ot ) dt

T T / 4
T /4
2 A  t sin(4 f ot ) 
2 A  T sin( )  sin(  ) 

 

 


T 2
4  2 f o  T / 4
T 4
8 f o

A

2
Note: cos2 = ½(1 + cos 2)
COE 341 – Dr. Marwan Abu-Amara
49
Example (Cont.)
T
2
Bn   x(t ) sin( 2nf 0t )dt
T 0
Fourier Analysis (cont.):
2.
Bn 
2
T
T /4

T / 4
s (t ) sin(2 nf o t )dt 
2A
T
T /4

cos(2 f o t ) sin(2 nf ot )dt
T / 4
T /4
2A  cos(2 ( n  1)f o t ) cos(2 ( n  1)f o t ) 



, for n  1

T
4 (n  1)f o
 4 (n  1)f o
 T / 4
0
, for n  1
 cos(ax  bx ) cos(ax  bx )
Note:  sin(ax ) cos(bx )dx 

2(a  b )
2(a  b )
COE 341 – Dr. Marwan Abu-Amara
50
Example (Cont.)
2.
B n 1
Fourier Analysis (cont.):
2

T
A

T
T /4
2A
s (t ) sin(2 1 f o t )dt 

T
T / 4
T /4

T /4

cos(2 f o t ) sin(2 f o t )dt
T / 4
sin(4 f o t )dt
T / 4
A
A
T /4

  cos(4 f o t ) T / 4 
  cos( )  cos(  ) 
4
4
0
COE 341 – Dr. Marwan Abu-Amara
51
Example (Cont.)
2.
Fourier Analysis (cont.):

Ao
s (t ) 
   An cos(2 nf o t )  Bn sin(2 nf ot ) 
2 n 1
A
2 A  ( 1)(1 2 )
  cos(2 f o t ) 
cos(2 nf ot )

2
 2
 n  2,4,6,... n  1
n
A
2A
A
Co 
, C1 

2
Cn  0
Cn 
(1 n 2 )
2 A(1)
 (n 2  1)
, n is odd and n  1
, n  2, 4, 6,...
C0  A0
Cn 
COE 341 – Dr. Marwan Abu-Amara
An2  Bn2
52
Example (Cont.)
3.
Power Spectral Density function (PSD):
C 02 1  2
PSD 
 C n
4
2 n 1
Or more accurately:
C 02
1  2
PSD 
  (f )  C n   (f  nf o )
4
2 n 1
COE 341 – Dr. Marwan Abu-Amara
53
Example (Cont.)
3.
Power Spectral Density function (PSD):
C 02
1  2
PSD 
  (f )  C n   (f  nf o )
4
2 n 1
A2
A2
2A 2
 2   (f ) 
  (f  f o )  2

8

COE 341 – Dr. Marwan Abu-Amara


n  2,4,6,...
 (f  nf o )
(n 2  1) 2
54
Example (Cont.)
4.
Total Power:
1
Ps 
T
3T / 4

T / 4
2
T /4
A
s (t ) dt 
  cos 2 (2 f ot )dt
T T / 4
2
T /4
A  t sin(4 f ot ) 

 

T 2
8 f ot  T / 4
2
A2

4
Note: cos2 = ½(1 + cos 2)
COE 341 – Dr. Marwan Abu-Amara
55
Example (Cont.)
5.
Finding n such that we get 95% of total power:
For n  0
2
0
2
2
C
4A
A
2
 PSD n 0 

 2  0.1014A
2
4
4

2
0.1014A
 Power % 
 40.5%
2
0.25A
COE 341 – Dr. Marwan Abu-Amara
56
Example (Cont.)
5.
Finding n such that we get 95% of total power:
For n  1
2
0
C
C 12 A 2 A 2
 PSD n 1 

 2 
 0.226A 2
4
2

8
2
0.226A
 Power % 
 90.5%
2
0.25A
COE 341 – Dr. Marwan Abu-Amara
57
Example (Cont.)
5.
Finding n such that we get 95% of total power:
For n  2
C 02 C 12 C 22 A 2 A 2 2A 2
2
 PSD n  2 


 2 


0.2485
A
4
2
2

8
9 2
0.2485A 2
 Power % 
 99.41%
2
0.25A
6.
Effective bandwidth with 95% of total power:
B
Beff = fmax – fmin
…
= 2f0 – 0 = 2f0
eff
0 f0
COE 341 – Dr. Marwan Abu-Amara
2f0 3f0
f
58
Data Rate and Bandwidth



Any transmission system has a limited band
of frequencies
This limits the data rate that can be carried
Example on pages 64 & 65
COE 341 – Dr. Marwan Abu-Amara
59
Data Element,
Signal Element
Bandwidth and Data Rates
T/2
B = 4f
Data rate = 1/(T/2)
= 2/T bits per sec
= 2f
Given a bandwidth B,
Data rate = 2f = 2(B/4) = B/2
Period T = 1/f
B
0
1
f
3f
0
1
5f
Two ways to double the data rate… To double the data rate you need to double f
1. Double the bandwidth, with the same receiver and error rate (same received waveform)
2B = 4f
New bandwidth: 2B,
Data rate = 2f = 2(2B/4) = B
2B
f
3f
1
0
1
0
1
0
1
0
5f
2. Same bandwidth, B, with a better receiver, higher S/N, or by tolerating more error
(poorer received waveform)
B = 2f
Bandwidth: B,
Data rate = 2f = 2(B/2) = B
B
1
fCOE 341 3f
– Dr. Marwan Abu-Amara
0
1
0
1
0
1
0
60
Bandwidth and Data Rates

Increasing the data rate (bps) with the same BW
means working with inferior waveforms at the receiver,
which may require:

Better signal to noise ratio at RX (larger signal relative to
noise):





Shorter link spans
Use of more repeaters/amplifiers
Better shielding of cables to reduce noise, etc.
More sensitive (& costly!) receiver
Dealing with higher error rates


Tolerating them
Adding more efficient means for error detection and correctionthis also increases overhead!.
COE 341 – Dr. Marwan Abu-Amara
61
Analog and Digital Data Transmission

Data


Signals


Entities that convey meaning
Electric or electromagnetic representations of data
Transmission

Communication of data by propagation and
processing of signals
COE 341 – Dr. Marwan Abu-Amara
62
Analog and Digital Data

Analog



Continuous values within some interval
e.g. sound, video
Digital


Discrete values
e.g. text, integers
COE 341 – Dr. Marwan Abu-Amara
63
Acoustic Spectrum (Analog)
COE 341 – Dr. Marwan Abu-Amara
64
Analog and Digital Signals


Means by which data are propagated
Analog


Continuously variable
Various media





wire, fiber optic, space
Speech bandwidth 100Hz to 7kHz
Telephone bandwidth 300Hz to 3400Hz
Video bandwidth 4MHz
Digital

Use two DC components
COE 341 – Dr. Marwan Abu-Amara
65
Advantages & Disadvantages of Digital



Cheaper
Less susceptible to noise
Greater attenuation


Pulses become rounded and smaller
Leads to loss of information
COE 341 – Dr. Marwan Abu-Amara
66
Attenuation of Digital Signals
COE 341 – Dr. Marwan Abu-Amara
67
Components of Speech

Frequency range (of hearing) 20Hz-20kHz




Speech 100Hz-7kHz
Easily converted into electromagnetic signal
for transmission
Sound frequencies with varying volume
converted into electromagnetic frequencies
with varying voltage
Limit frequency range for voice channel

300-3400Hz
COE 341 – Dr. Marwan Abu-Amara
68
Conversion of Voice Input into
Analog Signal
COE 341 – Dr. Marwan Abu-Amara
69
Video Components

USA - 483 lines scanned per frame at 30 frames per
second


So 525 lines x 30 scans = 15750 lines per second





525 lines but 42 lost during vertical retrace
63.5s per line
11s for retrace, so 52.5 s per video line
Max frequency if line alternates black and white
Horizontal resolution is about 450 lines giving 225
cycles of wave in 52.5 s
Max frequency of 4.2MHz
COE 341 – Dr. Marwan Abu-Amara
70
Binary Digital Data



From computer terminals etc.
Two dc components
Bandwidth depends on data rate
COE 341 – Dr. Marwan Abu-Amara
71
Conversion of PC Input to Digital
Signal
COE 341 – Dr. Marwan Abu-Amara
72
Data and Signals


Usually use digital signals for digital data and
analog signals for analog data
Can use analog signal to carry digital data


Modem
Can use digital signal to carry analog data

Compact Disc audio
COE 341 – Dr. Marwan Abu-Amara
73
Analog Signals Carrying Analog and
Digital Data
COE 341 – Dr. Marwan Abu-Amara
74
Digital Signals Carrying Analog and
Digital Data
COE 341 – Dr. Marwan Abu-Amara
75
Four Data/Signal Combinations
Signal
Analog
Same spectrum as
Analog data (base band): e.g.
Digital
-
Conventional Telephony
Different spectrum
(modulation): e.g.
AM, FM Radio
-
Data
Use a (converter):
codec, e.g. for PCM
(pulse code modulation)
-Simple
Digital
two signal
levels: e.g. NRZ code
Use a (converter):
modem e.g. with the -Special Encoding:
V.90 standard
e.g. Manchester code
76
COE 341 – Dr. Marwan Abu-Amara
(Chapter 5)
Analog Transmission





Analog signal transmitted without regard to
content
May be analog or digital data
Attenuated over distance
Use amplifiers to boost signal
Also amplifies noise
COE 341 – Dr. Marwan Abu-Amara
77
Digital Transmission



Concerned with content
Integrity endangered by noise, attenuation
etc.
Repeaters used





Repeater receives signal
Extracts bit pattern
Retransmits
Attenuation is overcome
Noise is not amplified
COE 341 – Dr. Marwan Abu-Amara
78
Four Signal/Transmission Mode Combinations
Transmission mode
Analog
Uses amplifiers
- Not concerned with
what data the signal
represents
- Noise is cumulative
-
Analog
OK
Signal
Digital
Avoid
Digital
Uses repeaters
- Assumes signal represents
digital data, recovers it and
represents it as a new
outbound signal
- This way, noise is not
cumulative
-
Makes sense only if the
analog signal represents
digital data
COE 341 – Dr. Marwan Abu-Amara
OK
79
Advantages of Digital Transmission

Digital technology


Data integrity



High bandwidth links economical
High degree of multiplexing easier with digital techniques
Security & Privacy


Longer distances over lower quality lines
Capacity utilization


Low cost LSI/VLSI technology
Encryption
Integration

Can treat analog and digital data similarly
COE 341 – Dr. Marwan Abu-Amara
80
Transmission Impairments




Signal received may differ from signal
transmitted
Analog - degradation of signal quality
Digital - bit errors
Caused by



Attenuation and attenuation distortion
Delay distortion
Noise
COE 341 – Dr. Marwan Abu-Amara
81
Attenuation



Signal strength falls off with distance
Depends on medium (guided vs. unguided)
Received signal strength:



must be enough to be detected
must be sufficiently higher than noise to be
received without error
Attenuation is an increasing function of
frequency


Different frequency components of a signal get
attenuated differently  Signal distortion
Particularly significant with analog signals
COE 341 – Dr. Marwan Abu-Amara
82
Delay Distortion


Only in guided media
Propagation velocity varies with frequency




Highest at 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)
Badly affects digital data due to bit spill-over
(timing is more important than for analog
data)
COE 341 – Dr. Marwan Abu-Amara
83
Noise (1)



Additional unwanted signals inserted between
transmitter and receiver
The most limiting factor in communication
systems
Thermal



Due to thermal agitation of electrons
Uniformly distributed
White noise
COE 341 – Dr. Marwan Abu-Amara
84
More on Thermal (White) Noise

Power of thermal noise present in a
bandwidth B (Hz) is given by
N = kT B = N 0B (wat t s)
= - 228.6 + 10 logT + 10 log B (dBw)

T is absolute temperature in kelvin and k is
Boltzmann’s constant (k = 1.3810-23 J/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
COE 341 – Dr. Marwan Abu-Amara
85
Noise (2)

Intermodulation




Signals that are the sum and difference of original
frequencies sharing a medium
f1, f2  (f1+f2) and (f1-f2)
Caused by nonlinearities in the medium and
equipment, e.g. due to overdrive and saturation
of amplifiers
Resulting frequency components may fall within
valid signal bands, thus causing interference
COE 341 – Dr. Marwan Abu-Amara
86
Noise (3)

Crosstalk



A signal from one channel picked up by another channel
e.g. Coupling between twisted pairs, antenna pick up,
leakage between adjacent channels in FDM, etc.
Impulse





Irregular pulses or spikes
Short duration
High amplitude
e.g. External electromagnetic interference
Minor effect on analog signals but major effect on digital
signals, particularly at high data rates
COE 341 – Dr. Marwan Abu-Amara
87
Channel Capacity



Channel capacity: Maximum data rate usable under given
communication conditions
How BW, signal level, noise and other impairments, and the amount
of error tolerated limit the channel capacity?
Max data rate
= Function (BW, Signal wrt noise, Error rate allowed)




Max data rate: Max rate at which data can be communicated, 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)
Error rate: bits received in error/total bits transmitted. Equal to the bit error
probability
COE 341 – Dr. Marwan Abu-Amara
88
1. Nyquist Bandwidth: (Noise-free, Error-free)







Idealized, theoretical
Assumes a noise-free, error-free channel
Nyquist: 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
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
For the same B, data rate can be increased by sending one of
M different signal levels (symbols): as a signal level now
represents log2M bits
Generalized Nyquist Channel Capacity, C = 2B log2M bits/s
(bps)
Signals/s
COE 341 – Dr. Marwan Abu-Amara
bits/signal
89
Nyquist Bandwidth: Examples

C = 2B log2M bits/s






C = Nyquist Channel Capacity
B = Bandwidth
M = Number of discrete signal levels (symbols) used
Telephone Channel: B = 3400-300 = 3100 Hz
1
With a binary signal (M = 2):
0
C = 2B log2 2 = 2B = 6200 bps
With a quandary signal (M = 4):
01
C = 2B log2 4 = 2B x 2 = 4B = 12,400 bps

11
10
00
Practical limit: larger M makes it difficult for the receiver
to operate, particular with noise
COE 341 – Dr. Marwan Abu-Amara
90
2. Shannon Capacity Formula: (Noisy, Error-Free)





Assumes error-free operation with noise
Data rate, noise, error: A given noise burst affects more bits at
higher data rates, which increases the error rate
So, maximum error-free data rate increases with reduced noise
Signal to noise ratio SNR = signal / noise levels
SNRdB= 10 log10 (SNR) dBs
Caution! Log2 Not Log10
Shannon Capacity C = B log2(1+SNR):
Caution! Ratio- Not log
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 errors
C/B: Spectral (bandwidth) efficiency, BE, (bps/Hz) (>1)
Larger BEs mean better utilizing of a given B for transmitting data
fast.
COE 341 – Dr. Marwan Abu-Amara
91
Shannon Capacity Formula: Comments





Formula says: for data rates  calculated C, it is theoretically
possible to find an encoding scheme that gives error-free
transmission.
But it does not say how…
It is a theoretical approach based on thermal (white) noise
only
However, in practice, we also have impulse noise and
attenuation and delay distortions
So, maximum error-free data rates obtained in practice are
lower than the C predicted by this theoretical formula
However, maximum error-free data rates can be used to
compare practical systems: The higher that rate the better
the system is
COE 341 – Dr. Marwan Abu-Amara
92
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!
 High SNR obtained through excessive amplification
may introduce nonlinearities: distortion and
intermediation noise!
 High Bandwidth B opens the system for more
thermal noise (kTB), and therefore reduces SNR!
COE 341 – Dr. Marwan Abu-Amara
93
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 = 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
COE 341 – Dr. Marwan Abu-Amara
94
3. Eb/N0 (Signal Energy per Bit/Noise Power
density per Hz) (Noise and Error Together)






Handling both noise and error together
Eb/N0: A standard quality measure for digital
communication system performance
Eb/N0 Can be independently related to the error rate
Expresses SNR in a manner related to the data rate, R
Eb = Signal energy per bit (Joules)
= Signal power (Watts) x bit interval Tb (second)
= S x (1/R) = S/R
N0 = Noise power (watts) in 1 Hz = kT
E b ST b S / R
S
E b S / R S BT
 BT 





 SNR 

95
COE
341
–
Dr.
Marwan
Abu-Amara
N0
N0
N R
N0 N0
kT
kTR
 R 
Eb/N0 Example 1:
 Eb 

  S dBW  10 log R  10 log k  10 logT
 N 0 dB
 S dBW  10 log R  228.6 dBW  10 logT




Given: Eb/No = 8.4 dB (minimum) is needed to achieve a bit error rate of 10-4
Given:

The effective noise temperature is 290oK (room temperature)

Data rate 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
COE 341 – Dr. Marwan Abu-Amara
96
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?
 Get Eb/N0 to achieve a desired
error rate, then determine other
parameters from formula, e.g. S,
SNR, R, etc. (Design)
Lower Error Rate: larger Eb/N0

A
B
Better
Encoding
Error performance of a given
 Eb 
system (Analysis)

  S dBW  10 log R  10 log k  10 logT
 N 0 dB
Effect of S, R, T on error
 S dBW  10 log R  228.6 dBW  10 logT
performance
Eb
S /R
S BT
B 


 SNR  T 
N0
N0
N R
 R 
COE 341 – Dr. Marwan Abu-Amara
97
Eb/N0 (Cont.)

From Shannon’s formula:
C = B log2(1+SNR)
We have:

From the Eb/N0 formula:
With R = C, substituting for SNR we get:

Relates achievable spectral efficiency C/B (bps/Hz) to Eb/N0
COE 341 – Dr. Marwan Abu-Amara
98
Eb/N0 (Cont.) Example 2

Find the minimum Eb/N0 required to achieve a
spectral efficiency (C/B) of 6 bps/Hz:

Substituting in the equation above:
Eb/N0 = (1/6) (26 - 1) = 10.5 = 10.21 dB
COE 341 – Dr. Marwan Abu-Amara
99