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Datornätverk A – lektion 3MKS B – lektion 3
Kapitel 3: Fysiska signaler.Kapitel 4: Digital transmission.
Summer 2006 Computer Networks 2
Repetition: The TCP/IP model
H – header (pakethuvud): control data added at the front end of the data unitT – trailer (svans): control data added at the back end of the data unitTrailers are usually added only at layer 2.
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Physical LayerPhysical Layer
PART IIPART II
Chapter 3 – Time and Frequency Domain Concept, Transmission
Impairments
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Figure 3.1 Comparison of analog and digital signals
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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.
Note:Note:
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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
Periodic vs. Non Periodic Signals
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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 Period and frequency
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Tabell 3.1 Enheter för periodtid och frekvensTabell 3.1 Enheter för periodtid och frekvens
Enhet Ekvivalent Enhet Ekvivalent
Sekunder (s) 1 s Hertz (Hz) 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 Sine wave examples
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Figure 3.6 Sine wave examples (continued)
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ExempelExempel
Vilken frekvens i kHz har en sinusvåg med periodtid 100 ms?
LösningLösning
Alternativ 1: Gör om till grundenheten. 100 ms = 0.1 sf = 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 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) = 2 (radians)
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 Sine wave examples (continued)
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Example 2Example 2
A sine wave is offset one-sixth of a cycle with respect to time zero. What is its phase in degrees and radians?
SolutionSolution
We know that one complete cycle is 360 degrees.
Therefore, 1/6 cycle is
(1/6) 360 = 60 degrees = 60 x 2 /360 rad = 1.046 rad
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Figure 3.7 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 Square wave
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Figure 3.9 Three harmonics
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Figure 3.10 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
Component 1:
Component 5:
Component 3:
.
.
.
.
.
.
...}5cos5
13cos
3
1{cos
4)( ttt
Ats ooo
tA
ts o
cos4
)(1
tA
ts o
3cos3
4)(3
tA
ts o
5cos5
4)(5
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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 Signal corruption
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Figure 3.13 Bandwidth
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Example 3Example 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.
SolutionSolution
B = fh fl = 900 100 = 800 HzThe spectrum has only five spikes, at 100, 300, 500, 700, and 900 (see Figure 13.4 )
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Figure 3.14 Example 3
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Example 5Example 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?
SolutionSolution
The answer is definitely no. Although the signal can have The answer is definitely no. Although the signal can have the same bandwidth (1000 Hz), the range does not the same bandwidth (1000 Hz), the range does not overlap. The medium can only pass the frequencies overlap. The medium can only pass the frequencies between 3000 and 4000 Hz; the signal is totally lost.between 3000 and 4000 Hz; the signal is totally lost.
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Figure 3.16 A digital signal
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A digital signal is a composite signal A digital signal is a composite signal with an infinite bandwidth.with an infinite bandwidth.
Note:Note:
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Figure 3.17 Bit rate and bit interval
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Example 6Example 6
A digital signal has a bit rate of 2000 bps. What is the duration of each bit (bit interval)
SolutionSolution
The bit interval is the inverse of the bit rate.
Bit interval = 1/ 2000 s = 0.000500 s = 0.000500 x 106 s = 500 s
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Media Filters the Signal
Media
INPUT OUTPUT
Certain frequenciesdo 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 RequirementTable 3.12 Bandwidth Requirement
BitRate
Harmonic1
Harmonics1, 3
Harmonics1, 3, 5
Harmonics1, 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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The bit rate and the bandwidth are The bit rate and the bandwidth are proportional to each other.proportional to each other.
Note:Note:
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The analog bandwidth of a medium is The analog bandwidth of a medium is expressed in hertz; the digital expressed in hertz; the digital bandwidth, in bits per second.bandwidth, in bits per second.
Note:Note:
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Figure 3.19 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
High pass
INPUTS1(f)
H(f)
H(f)
OUTPUT S2(f)= H(f)*S1(f)
Low pass
INPUTS1(f)
H(f)
H(f)
f
OUTPUT S2(f)= H(f)*S1(f)
Band pass
INPUTS1(f)
H(f)
H(f)
OUTPUT S2(f)= H(f)*S1(f)
Types of filters Low pass
Band pass
High pass
f
f
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Digital transmission (without Digital transmission (without modulation) needs a modulation) needs a
low-pass channel.low-pass channel.
Note:Note:
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Analog transmission (by means of Analog transmission (by means of modulation) can use a band-pass modulation) can use a band-pass
channel.channel.
Note:Note:
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Figure 3.21 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 10/120dB means power ratio 100/130dB means power ratio 1000/140dB means power ratio 10000/1
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Example 12Example 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
SolutionSolution
10 log10 log1010 (P2/P1) = 10 log (P2/P1) = 10 log1010 (0.5P1/P1) = 10 log (0.5P1/P1) = 10 log1010 (0.5) (0.5)
= 10(–0.3) = –3 dB = 10(–0.3) = –3 dB
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Example 13Example 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 log1010 (P2/P1) = 10 log (P2/P1) = 10 log1010 (10P1/P1) (10P1/P1)
= 10 log= 10 log1010 (10) = 10 (1) = 10 dB (10) = 10 (1) = 10 dB
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Figure 3.22 Example 14: In cascaded links the amplification in dB is simply a sum of the individual amplifications in dB.
–3dB + 7dB – 3dB = +1dB
Total amplification:
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Figure 3.23 Symbol distortion
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Figure 3.24 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 signal-brus-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 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