datornätverk a – lektion 3 kapitel 3: fysiska signaler. kapitel 4: digital transmission

Datornätverk A – lektion 3

Kapitel 3: Fysiska signaler.Kapitel 4: Digital transmission.

Computer Networks

Chapter 3 – Time and Frequency Domain Concept, Transmission

Impairments

Summer 2006 Computer Networks 3

Figure 3.1 Comparison of analog and digital 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 signals don’t repeat

according to any particular pattern

Periodic vs. Non Periodic Signals


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)


Figure 3.6 Sine wave examples


Figure 3.6 Sine wave examples (continued)


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.


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.


Figure 3.5 Relationships between different phases


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


Figure 3.6 Sine wave examples (continued)


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


Figure 3.7 Time and frequency domains (continued)


Figure 3.7 Time and frequency domains


Example: Sine waves

Time domain

t

t

t

f

f

T1=1/f1

T5=1/f5

T2=1/3f1

3f1

5f1

Frequency domain

ff1


Example: A signal with frequency 0

Time domain

t

Frequency domain

f0

. . .


Figure 3.8 Square wave


Figure 3.9 Three harmonics


Figure 3.10 Adding first three harmonics


Figure 3.11 Frequency spectrum comparison


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


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


Examples

1. If digital signal has bit rate of 2000 bps, what is the duration of each bit?

bit interval = 1/2000 = 0.0005 = 500s

2. If a digital signal has a bit interval of 400 ns, what is the bit rate?

bit rate = 1/(400 ·10-9) = 25 ·106 = 25 Mbps


Bandwidth Requirements for a Digital Signal

BitRate

Harmonic1

Harmonics1, 3

Harmonics1, 3, 5

Harmonics1, 3, 5, 7

1 Kbps 500 Hz 1.5 KHz 2.5 KHz 3.5 KHz

10 Kbps 5 KHz 15 KHz 25 KHz 35 KHz



Figure 3.12 Signal corruption


Figure 3.13 Bandwidth


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 )


Figure 3.14 Example 3


Example 4Example 4

A signal has a bandwidth of 20 Hz. The highest frequency is 60 Hz. What is the lowest frequency? Draw the spectrum if the signal contains all integral frequencies of the same amplitude.

SolutionSolution

B = fB = fhh f fll

20 = 60 20 = 60 ffll

ffll = 60 = 60 20 = 40 Hz20 = 40 Hz


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.


Figure 3.16 A digital signal


A digital signal is a composite signal A digital signal is a composite signal with an infinite bandwidth.with an infinite bandwidth.

Note:Note:


Figure 3.17 Bit rate and bit interval


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


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


Band pass

INPUTS1(f)

H(f)

H(f)


Types of filters Low pas

Band pass

High pass

f

f


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.


Figure 3.18 Digital versus analog


Table 3.12 Bandwidth RequirementTable 3.12 Bandwidth Requirement

BitRate

Harmonic1

Harmonics1, 3

Harmonics1, 3, 5

Harmonics1, 3, 5, 7

1 Kbps 500 Hz 2 KHz 4.5 KHz 8 KHz




The bit rate and the bandwidth are The bit rate and the bandwidth are proportional to each other.proportional to each other.

Note:Note:


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:


Figure 3.19 Low-pass and band-pass


Digital transmission needs a Digital transmission needs a low-pass channel.low-pass channel.

Note:Note:


Analog transmission can use a band-Analog transmission can use a band-pass channel.pass channel.

Note:Note:


Figure 3.20 Impairment types


Figure 3.21 Attenuation


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.


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.


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


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, symbolized SdBP, is:

SdBP = 10 log10 (P2 / P1)

As a rule of thumb:

S/N ratio of 10dB means 10/1S/N ratio of 20dB means 100/1S/N ratio of 30dB means 1000/1S/N ratio of 40dB means 10000/1


Examples:

1. A signal that travels through a transmission medium is reduced to half. This means that P2 = (1/2)P1

The attenuation can be calculated as follows:10log10(P2/P1)=10 log10 (0.5 P1/P1)=10log10 (0.5)= 3 dB

2. Imagine a signal goes through an amplifier and its power is increased 10 times. This means that P2 = 10P1

The amplification is: 10 log10 (10 P1/P1)=

=10log1010=10dB


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



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



One reason that engineers use the decibel to measure the changes in the strength of a signal is that decibel numbers can be added (or subtracted) when we are talking about several points instead of just two (cascading). In Figure 3.22 a signal travels a long distance from point 1 to point 4. The signal is attenuated by the time it reaches point 2. Between points 2 and 3, the signal is amplified. Again, between points 3 and 4, the signal is attenuated. We can find the resultant decibel for the signal just by adding the decibel measurements between each set of points.



dB = –3 + 7 – 3 = +1


Figure 3.23 Distortion


Figure 3.24 Noise


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


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.


Genomströmningshastighet (throughput)


Figure 3.26 Propagation time


Delay (Time, Latency)

When data are sent from one point to the other point (without intermediate points), two types of delays are experienced: transmission delay (time) propagation delay (time)

When data pass through intermediate points four types of delay (latency) are experienced: transmission delay (time) propagation delay (time) queue time processing time


Transmission Delay (Time)

The transmission time is the time necessary to put the message on the link (chanel).

The transmission time depends on the length of the message and the throughput (bit rate) of the link and is expressed as:

length of message (bits)

bit rate (bits/sec)


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


Queue and Processing time

Queue time When the intermediate nodes are busy

processing other data, the data arrived at the node are queued. Queue time is the time spent waiting in the queue.

Processing time This is the time needed for the data to be

processed at the intermediate nodes.

datornätverk a – lektion 3 kapitel 3: fysiska signaler. kapitel 4: digital transmission

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