### Chapter3

```Datornätverk A – lektion 3
MKS B – lektion 3
Kapitel 3: Fysiska signaler.
Kapitel 4: Digital transmission.
Repetition: The TCP/IP model
H – header (pakethuvud): control data added at the front end of the data unit
T – trailer (svans): control data added at the back end of the data unit
Trailers are usually added only at layer 2.
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PART II
Physical Layer
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Chapter 3 – Time and Frequency Domain Concept,
Transmission Impairments
Figure 3.1
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Comparison of analog and digital signals
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Note:
Signals can be analog or digital.
Analog signals can have an infinite
number of values in a range; digital
signals can have only a limited
number of values.
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Periodic vs. Non Periodic Signals
 Periodic signal
 repeat over and over again,
once per period
 The period ( T ) is the time
it takes to make one
complete cycle
 Non periodic signal
 don’t repeat according to
any particular pattern
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Sinusvågor
Periodtid T = t2 - t1. Enhet: s.
Frekvens f = 1/T. Enhet: 1/s=Hz.
T=1/f.
Amplitud eller toppvärde Û. Enhet: Volt.
Fasläge: θ = 0 i ovanstående exempel. Enhet: Grader eller
radianer.
Momentan spänning: u(t)= Ûsin(2πft+θ)
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Figure 3.4
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Period and frequency
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Tabell 3.1 Enheter för periodtid och frekvens
Enhet
Sekunder (s)
Ekvivalent
1s
Enhet
Hertz (Hz)
Ekvivalent
1 Hz
Millisekunder (ms)
10–3 s
Kilohertz (kHz)
103 Hz
Mikrosekunder (μs)
10–6 s
Megahertz (MHz)
106 Hz
Nanosekunder (ns)
10–9 s
Gigahertz (GHz)
109 Hz
Pikosekunder (ps)
10–12 s
Terahertz (THz)
1012 Hz
Exempel: En sinusvåg med periodtid 1 ns har frekvens 1 GHz.
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Figure 3.6
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Sine wave examples
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Figure 3.6
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Sine wave examples (continued)
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Exempel
Vilken frekvens i kHz har en sinusvåg med periodtid 100
ms?
Lösning
Alternativ 1: Gör om till grundenheten. 100 ms = 0.1 s
f = 1/0.1 Hz = 10 Hz = 10/1000 kHz = 0.01 kHz
Alternativ 2: Utnyttja att 1 ms motsvarar 1 kHz.
f = 1/100ms = 0.01 kHz.
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Figure 3.5
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Relationships between different phases
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Measuring the Phase
 The phase is measured in degrees or in radians.
 One full cycle is 360o
360o (degrees) = 2p (radians)
p  3.14
Example: A sine wave is offset one-sixth of a cycle with
respect to time 0. What is the phase in radians?
Solution: (1/6) 360 = 60 degrees = 60 x 2p /360 rad =
1.046 rad
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Figure 3.6
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Sine wave examples (continued)
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Example 2
A sine wave is offset one-sixth of a cycle with respect
to time zero. What is its phase in degrees and radians?
Solution
We know that one complete cycle is 360 degrees.
Therefore, 1/6 cycle is
(1/6) 360 = 60 degrees = 60 x 2p /360 rad = 1.046 rad
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Figure 3.7
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Time and frequency domains (continued)
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Figure 3.7 A DC signal (likspänning), i.e. a
signal with frequency 0 Hz
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Figure 3.8
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Square wave
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Figure 3.9
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Three harmonics
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Figure 3.10
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Adding first three harmonics
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Figure 3.11 Frequency spectrum comparison
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Example: Square Wave
Square wave with frequency fo
4A
1
1
s(t ) 
{cos ot  cos 3ot  cos 5ot  ...}
p
3
5
Component 1:
Component 3:
.
.
.
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Component 5:
.
.
.
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s1 (t ) 
4A
cos o t
p
4A
s3 (t ) 
cos 3ot
3p
4A
s5 (t ) 
cos 5o t
5p
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Characteristic of the Component Signals in
the Square Wave
 Infinite number of components
 Only the odd harmonic components are present
 The amplitudes of the components diminish with
increasing frequency
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Figure 3.12
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Signal corruption
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Figure 3.13
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Bandwidth
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Example 3
If a periodic signal is decomposed into five sine waves
with frequencies of 100, 300, 500, 700, and 900 Hz,
what is the bandwidth? Draw the spectrum, assuming all
components have a maximum amplitude of 10 V.
Solution
B = fh  fl = 900  100 = 800 Hz
The spectrum has only five spikes, at 100, 300, 500, 700,
and 900 (see Figure 13.4 )
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Figure 3.14
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Example 3
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Example 5
A signal has a spectrum with frequencies between 1000
and 2000 Hz (bandwidth of 1000 Hz). A medium can pass
frequencies from 3000 to 4000 Hz (a bandwidth of 1000
Hz). Can this signal faithfully pass through this medium?
Solution
The answer is definitely no. Although the signal can have
the same bandwidth (1000 Hz), the range does not
overlap. The medium can only pass the frequencies
between 3000 and 4000 Hz; the signal is totally lost.
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Figure 3.16
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A digital signal
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Note:
A digital signal is a composite signal
with an infinite bandwidth.
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Figure 3.17
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Bit rate and bit interval
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Example 6
A digital signal has a bit rate of 2000 bps. What is the
duration of each bit (bit interval)
Solution
The bit interval is the inverse of the bit rate.
Bit interval = 1/ 2000 s = 0.000500 s
= 0.000500 x 106 ms = 500 ms
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Media Filters the Signal
Media
INPUT
OUTPUT
Certain frequencies
do not pass through
What happens when you limit frequencies?
Square waves (digital values) lose their edges -> Harder to read correctly.
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Table 3.12 Bandwidth Requirement
Bit
Rate
Harmonic
1
Harmonics
1, 3
Harmonics
1, 3, 5
Harmonics
1, 3, 5, 7
1 Kbps
500 Hz
1,5 KHz
3 KHz
4,5 KHz
10 Kbps
5 KHz
15 KHz
30 KHz
45 KHz
100 Kbps
50 KHz
150 KHz
300 KHz
450 KHz
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Note:
The bit rate and the bandwidth are
proportional to each other.
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Note:
The analog bandwidth of a medium is
expressed in hertz; the digital
bandwidth, in bits per second.
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Figure 3.19
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Low-pass and band-pass
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Filtering the Signal
 Filtering is equivalent to cutting all the frequiencies outside the
band of the filter
 Types of filters
Low pass
 Low pass
H(f)
INPUT
S1(f)
OUTPUT
S2(f)= H(f)*S1(f)
H(f)
f
Band pass
 Band pass
H(f)
INPUT
S1(f)
OUTPUT
S2(f)= H(f)*S1(f)
H(f)
f
High pass
 High pass
H(f)
INPUT
S1(f)
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H(f)
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f
OUTPUT
S2(f)= H(f)*S1(f)
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Note:
Digital transmission (without
modulation) needs a
low-pass channel.
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Note:
Analog transmission (by means of
modulation) can use a band-pass
channel.
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Figure 3.21
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Attenuation
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Förstärkning mätt i decibel (dB)
1 gång effektförstärkning = 0 dB.
2 ggr effektförstärkning = 3 dB.
10 ggr effektförstärkning = 10 dB.
100 ggr effektförstärkning = 20 dB.
1000 ggr effektförstärkning = 30 dB.
Osv.
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Dämpning mätt i decibel
 Dämpning 100 ggr = Dämpning 20 dB
= förstärkning 0.01 ggr = förstärkning med – 20
dB.
 Dämpning 1000 ggr = 30 dB dämpning = -30dB
förstärkning.
 En halvering av signalen = dämpning med 3dB =
förstärkning med -3dB.
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Measurement of Attenuation
 Signal attenuation is measured in units called
decibels (dB).
 If over a transmission link the ratio of output power
is Po/Pi, the attenuation is said to be –10log10(Po/Pi) =
10log10(Pi/Po) dB.
 In cascaded links the attenuation in dB is simply a
sum of the individual attenuations in dB.
 dB is negative when the signal is attenuated and
positive when the signal is amplified
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What is dB?
 A decibel is 1/10th of a Bel, abbreviated dB
 Suppose a signal has a power of P1 watts, and a
second signal has a power of P2 watts. Then the
power amplitude difference in decibels, is:
10 log10 (P2 / P1)
 As a rule of thumb:
10dB means power ratio 10/1
20dB means power ratio 100/1
30dB means power ratio 1000/1
40dB means power ratio 10000/1
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Example 12
Imagine a signal travels through a transmission medium
and its power is reduced to half. This means that P2 = 1/2
P1. In this case, the attenuation (loss of power) can be
calculated as
Solution
10 log10 (P2/P1) = 10 log10 (0.5P1/P1) = 10 log10 (0.5)
= 10(–0.3) = –3 dB
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Example 13
Imagine a signal travels through an amplifier and its
power is increased ten times. This means that
P2 = 10· P1.
In this case, the amplification (gain of power) can be
calculated as
10 log10 (P2/P1) = 10 log10 (10P1/P1)
= 10 log10 (10) = 10 (1) = 10 dB
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cascaded links the
amplification in dB is simply a sum of the
individual amplifications in dB.
Figure 3.22
Example 14: In
Total amplification:
–3dB + 7dB – 3dB = +1dB
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Figure 3.23
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Symbol distortion
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Figure 3.24
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Noise
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Noise and Interference
 Noise is present in the form of random motion of
electrons in conductors, devices and electronic
systems (due to thermal energy) and can be also
picked up from external sources (atmospheric
disturbances, ignition noise etc.)
 Interference (cross-talk) generally refers to the
unwanted signals, picked up by communication link
due to other transmissions taking place in adjacent
frequency bands or in physically adjacent
transmission lines
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Signal-brus-förhållande
 Ett signal-brus-förhållande på 100 dB innebär
att den starkaste signalen är 100 dB starkare än
bruset.
 Ljud som är svagare än bruset hörs inte utan
dränks i bruset.
 Ljudets dynamik skillnaden mellan den starkaste
ljudet och det svagaste ljudet som man kan höra,
och är vanligen ungefär detsamma som signalbrus-förhållandet.
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Throughput
(Genomströmningshastighet)
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Delay (Time, Latency)
 When data are sent from one node to next node (without
intermediate points), two types of delays are experienced:
 transmission time (Paketsändningstid)
 propagation delay (Utbredningsfördröjning)
 When data pass through intermediate nodes four types of
delay (latency) are experienced:




transmission time
propagation delay
queue time
processing time
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Figure 3.26
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Propagation time
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Transmission Time (Paketsändningstid)
 The transmission time is the time necessary to put
the complete message on the link (channel).
 The transmission time depends on the length of
the message and the bit rate of the link and is
expressed as:
length of packet (bits)
bit rate (bits/sec)
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Propagation Delay (Time)
 The propagation delay is the time needed for the signal to
propagate (travel) from one end of a channel to the other.
 The transmition time depends on the distance between the two
ends and the speed of the signal and is expressed as
distance (m) / speed of propagation (m/s)
 Through free space signals propagate at the speed of light
which is 3 * 108 m/s
 Through wires signals propagate at the speed of
2 * 108 m/s
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