1

Data Communications

Physical Layer – Signals

By: Faiza Tariq

26th Sep,2012

Position of the physical layer

2

Services

3.1 Analog and DigitalTo be transmitted, data must be

transformed to electromagnetic signals.

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.

Analog and Digital Data

Analog and Digital Signals

Periodic and Aperiodic Signals

3

Analog/Digital Data and Signals

Data can be analog or digital. Analog data are continuous and take continuous values.

Digital data have discrete states and take discrete values.

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.

Analogue Signals

A continuous or sine wave

Simple or composite

A simple signal can not be decomposed into simple signal

Visualized as simple oscillating curve.

described by three characteristics

4

Periodic Signal Completes a pattern within a measurable time

frame called ‘Period’ and repeats pattern over identical subsequent periods

The completion of one full pattern is called a cycle.

Analog Signaling

represented by sine waves

time(sec)

amp

litu

de

(vo

lts)

1 cycle

frequency (hertz)= cycles per second

phase difference

5

Some Terms

PeriodAmount of time (in seconds) a signal takes to

complete one cycle

Period (T) = 1/f where f is frequncy of the signal

FrequencyNumber of cycles per second

Measured in Hertz (Hz)

Frequency (f) = 1/T where T is period

Frequency is the rate of change with respect to time.

Change in a short span of timemeans high frequency.

Change over a long span of time means low frequency.

If a signal does not change at all, its frequency is zero.

If a signal changes instantaneously, its frequency is infinite.

6

The power we use at home has a frequency of 60 Hz.The period of this sine wave can be determined asfollows:

Example 3.3

Figure 3.4 Two signals with the same amplitude and phase,but different frequencies

7

Frequency and periodFrequency and period are inverses of each

other.

If a signal does not change at all, its frequency is zero. If a signal changes

instantaneously, its frequency is infinite

Table 3.1 Units of periods and frequencies

Unit Equivalent Unit Equivalent

Seconds (s) 1 s hertz (Hz) 1 Hz

Milliseconds (ms) 10–3 s kilohertz (KHz) 103 Hz

Microseconds (ms) 10–6 s megahertz (MHz) 106 Hz

Nanoseconds (ns) 10–9 s gigahertz (GHz) 109 Hz

Picoseconds (ps) 10–12 s terahertz (THz) 1012 Hz

8

Questions

Q. 1. A sine wave has a frequency of 6 Hz. What is period?

Q. 2. A sine wave completes its one cycle in 4 seconds. What is its frequency

A. 1. Period (T) = 1/f

T = 1/6 = 0.17 seconds

A2 Frequency (f) = 1/T

f = ¼ = 0.25 Hz

Aperiodic Signal

Signal changes constantly without exhibiting a pattern or cycle that repeats over time.

Can be decomposed into periodic signals

9

Analogue Signal

In data communication, we commonly use periodic analog signals and aperiodic digital signals

What makes sound?

Vibration of air

Some Common Terminologies

Channel The physical link or line used to join two or more

communicating devices (like computers, network printer and network scanner) together e.g UTP Cable.

Spectrum: The range of frequencies that a signal spans from

minimum to maximum is called the spectrum.

10

Characteristics Of Waveforms

Analogue waveforms are transmitted over acarrier (the channel) in the form of a sine wave.

Time

+v

-v

Amplitude

Frequency

11

Waveform Characteristics Amplitude

The strength of the signal, which is normally measured in volts, also in amperes and watts.

Equal to the vertical distance from a given point on waveform to the horizontal axis

Two signals with the same phase and frequency, but different amplitudes

12

Waveform Characteristics

Frequency• This is the number of cycles which a signal goes

through in a given second.

• Amount of time, in seconds, a signal needs tocomplete one cycle. Measured in Hz.

Waveform CharacteristicsPhase

Phase is the relative starting point of the carrier signal.

Position of the waveform relative to time zero

Phase is measured in degrees or radians (3600

= 2л radian)

13

Figure 3.5 Three sine waves with the same amplitude and frequency,but different phases

A sine wave is offset 1/6 cycle with respect to time 0.What is its phase in degrees and radians?

Example 3.6

SolutionWe know that 1 complete cycle is 360°. Therefore, 1/6cycle is

14

15

0 degree

½ Cycle 180 degrees

¼ Cycle 90 degrees

¾ Cycle 270 degrees

Question

A sine wave is offset 1/3 of a cycle with respect to time zero. What is its phase?

One complete cycle = 360 degrees

1/3 Cycle is: 1/3 X 360 = 120 degrees

16

Frequency Conversion Unit of frequency

Unit of frequency Equivalent

Hertz (Hz) 1 Hz

Kilohertz (KHz) 103 Hz

Megahertz (MHz) 106 Hz

Gigahertz (GHz) 109 Hz

Terahertz (THz) 1012 Hz

High & Low Frequencies

17

Question

The telephone system uses analogue signal (human voices) for transmission, whereas data is digitally represented in the computer using binary digits.

How can we link two different systems together?

Simple Sine Wave

S(t) = A sin (2π ft + Ф)

Where A is amplitude, sin is trigonometric function, ft is frequency over a given period of time and Ф is phase

18

Sine Wave Parameters

s(t ) = A sin(2ft + )

Figure 3.6 Sine wave examples (continued)

19

Figure 3.6 Sine wave examples (continued)

Figure 3.7 Time and frequency domains

An analog signal is best represented in the frequency domain.

20

Figure 3.7 Time and frequency domains (continued)

Figure 3.7 Time and frequency domains (continued)

A complete sine wave in the time domain can be represented by one

single spike in the frequency domain.

21

Figure 3.7 The time-domain and frequency-domain plots of a sine wave

The frequency domain is more compact anduseful when we are dealing with more than onesine wave. For example, Figure 3.8 shows threesine waves, each with different amplitude andfrequency. All can be represented by threespikes in the frequency domain.

Example 3.7

22

Figure 3.8 The time domain and frequency domain of three sine waves

A single-frequency sine wave is not useful in data communications;

we need to send a composite signal, a signal made of many simple sine waves.

According to Fourier analysis, any composite signal is a combination of

simple sine waves with different frequencies, amplitudes, and phases.

23

Figure 3.9 shows a periodic composite signal withfrequency f. This type of signal is not typical of thosefound in data communications. We can consider it to bethree alarm systems, each with a different frequency.The analysis of this signal can give us a goodunderstanding of how to decompose signals.

Example 3.8

Figure 3.9 A composite periodic signal

24

Figure 3.10 Decomposition of a composite periodic signal in the time andfrequency domains

Figure 3.6 Wavelength and period

25

Figure 3.11 The time and frequency domains of a nonperiodic signal

Single vs. Multiple frequenciesA single-frequency sine wave may be helpful n carrying

electric energy from one place to another e.g. a power company may use 60 Hz sine wave to distribute electric energy, alarm security etc. ,

but it is not useful in voice/data communications; we need to change one or more of its characteristics to make it useful. E.g. if used single frequency sine wave to convey voice over phone, we would always hear a buzz

Through usage of multiple frequencies, we can distinguish one person’s voice from another or one music from another

26

Bandwidth The range of frequencies used to transmit the

data (measured in Hertz i.e. cycle per second).

Each transmission system has a limited bandwidth, which is maximum rate that the hardware can change a signal.

Bandwidth

The bandwidth is a property of a medium:

It is the difference between the highest and the lowest frequencies that the medium can satisfactorily pass.

27

Bandwidth If sender attempts to transmit changes faster than

bandwidth, the hardware will not keep up, because it will not have sufficient time to complete one change before the sender attempts to make another. Therefore some of the changes will be lost

Bandwidth = highest frequency of the range – lowest frequency

Biological system has bandwidth e.g. you can not hear sounds that are beyond human bandwidth limitations of ears.

Passband Range of frequencies passed through the

equipment. Normally specified as the lower limit and the upper limit frequency.

Question If a periodic signal is decomposed into five sine

waves with frequencies 100, 300, 500, 700 and 900 Hz. What is the band width? Draw the spectrum assuming all components have maximum amplitude of 10 volts:

Bw = Fh – Fl = 900 – 100 = 800 Hz

28

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 HzThe spectrum has only five spikes, at 100, 300, 500, 700, and 900 (see Figure 13.4 )

Figure 3.14 Example 3

29

Example 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.

Solution

B = fh fl

20 = 60 fl

fl = 60 20 = 40 Hz

Figure 3.15 Example 4

30

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.

A nonperiodic composite signal has a bandwidth of 200kHz, with a middle frequency of 140 kHz and peakamplitude of 20 V. The two extreme frequencies have anamplitude of 0. Draw the frequency domain of thesignal.

SolutionThe lowest frequency must be at 40 kHz and the highestat 240 kHz. Figure 3.15 shows the frequency domainand the bandwidth.

Example 3.12

31

Figure 3.15 The bandwidth for Example 3.12

Another example of a nonperiodic composite signal isthe signal received by an old-fashioned analog black-and-white TV. A TV screen is made up of pixels. If weassume a resolution of 525 × 700, we have 367,500pixels per screen. If we scan the screen 30 times persecond, this is 367,500 × 30 = 11,025,000 pixels persecond. The worst-case scenario is alternating black andwhite pixels. We can send 2 pixels per cycle. Therefore,we need 11,025,000 / 2 = 5,512,500 cycles per second, orHz. The bandwidth needed is 5.5125 MHz.

Example 3.15

32

The Telephone System

Was one of the earliest methods of transmitting data electronically and was invented to transmit human voice.

The signals transmitted over the telephone lines are analogue by nature.

The human ear is capable of capturing frequencies between the range of 20 Hz to 20,000 Hz.

The pass band of the telephone, on the other hand, is 300 Hz and 3300 Hz (which is an international standard).

Figure 3.8 Square wave

According to Fourier analysis, any composite signal can be represented as a combination of simple sine waves with different frequencies,

phases, and amplitudes.

S(t) = A1sin (2πf1t+Ф1)+ A2sin (2πf2t+Ф2)+ A3sin (2πf3t+Ф3)+…………

33

Addition of FrequencyComponents

When we change one or more characteristics of a single-frequency signal, it becomes a

composite signal made of many frequencies.

Three HarmonicsA digital signal can be decomposed into an

infinite number of simple sine waves, each with different amplitude, frequency and phase

34

Low – pass vs. Band – pass

Low pass channelA channel or link which has a bandwidth

with frequencies between 0 and f. the lower limit is zero and upper limit can be any frequency (including infinity)

Band pass channelA channel or link which has a bandwidth

with frequencies between f1 and f2

Figure 3.27 Wavelength

35

Digital SignalData can be represented by a digital signal

e.g. “1” can be encoded as positive voltage and “0” as zero voltage, as shown below in the diagram.

Figure 3.16 Two digital signals: one with two signal levels and the otherwith four signal levels

36

A digital signal has eight levels. How many bits areneeded per level? We calculate the number of bits fromthe formula

Example 3.16

Each signal level is represented by 3 bits.

A digital signal has nine levels. How many bits areneeded per level? We calculate the number of bits byusing the formula. Each signal level is represented by3.17 bits. However, this answer is not realistic. Thenumber of bits sent per level needs to be an integer aswell as a power of 2. For this example, 4 bits canrepresent one level.

Example 3.17

37

Assume we need to download text documents at the rateof 100 pages per minute. What is the required bit rate ofthe channel?SolutionA page is an average of 24 lines with 80 characters ineach line. If we assume that one character requires 8bits, the bit rate is

Example 3.18

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 s = 500 s

38

Figure 3.18 Digital versus analog

Bit rate and bit interval

39

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 s = 500 s

A digitized voice channel, as we will see in Chapter 4, ismade by digitizing a 4-kHz bandwidth analog voicesignal. We need to sample the signal at twice the highestfrequency (two samples per hertz). We assume that eachsample requires 8 bits. What is the required bit rate?

SolutionThe bit rate can be calculated as

Example 3.19

40

What is the bit rate for high-definition TV (HDTV)?

SolutionHDTV uses digital signals to broadcast high qualityvideo signals. The HDTV screen is normally a ratio of16 : 9. There are 1920 by 1080 pixels per screen, and thescreen is renewed 30 times per second. Twenty-four bitsrepresents one color pixel.

Example 3.20

The TV stations reduce this rate to 20 to 40 Mbpsthrough compression.

Figure 3.17 The time and frequency domains of periodic and nonperiodicdigital signals

41

Figure 3.18 Baseband transmission

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

Note

42

Figure 3.19 Bandwidths of two low-pass channels

Figure 3.20 Baseband transmission using a dedicated medium

43

Baseband transmission of a digital signal that preserves the shape of the

digital signal is possible only if we have a low-pass channel with an infinite or

very wide bandwidth.

Note

An example of a dedicated channel where the entirebandwidth of the medium is used as one single channelis a LAN. Almost every wired LAN today uses adedicated channel for two stations communicating witheach other. In a bus topology LAN with multipointconnections, only two stations can communicate witheach other at each moment in time (timesharing); theother stations need to refrain from sending data. In astar topology LAN, the entire channel between eachstation and the hub is used for communication betweenthese two entities. We study LANs in Chapter 14.

Example 3.21

44

Figure 3.23 Bandwidth of a bandpass channel

If the available channel is a bandpass channel, we cannot send the digital

signal directly to the channel; we need to convert the digital signal to an analog signal before transmission.

Note

45

Figure 3.24 Modulation of a digital signal for transmission on a bandpass channel

Noiseless Channel – Nyquist’s Algorithm For noiseless channel, the Nyquist bit rate formula

defines the theoretical maximum bit rate i.e.

Bit Rate = 2 X (Bandwidth X log2L)

L is number of signal levels used to represent data and bit rate unit is bits per second (bps)

For binary signals (two voltage levels) C = 2 X Bw

With multilevel signaling C = 2B log2 M

M = number of discrete signal or voltage levels

46

Examples

Example 1: Consider a noiseless Channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. What s maximum bit- rate Bit rate = 2 X 3000 X log22 = 6000 bps

Example 2: Consider same noiseless channel transmitting a signal with four signal levels (for each level we transmit two bits). Calculate maximum data rate Bit rate = 2 X 3000 log24 = 12000 bps

Signal-to-Noise Ratio

Ratio of the power in a signal to the power contained in the noise that’s present at a particular point in the transmission

Typically measured at a receiver Signal-to-noise ratio (SNR, or S/N)

A high SNR means a high-quality signal, low number of required intermediate repeaters

SNR sets upper bound on achievable data rate

power noise

power signallog10)( 10dB SNR

47

Noisy Channel – Shannon’s Algorithm In reality, we can not have a noiseless

channel. The channel is always noisy.

SNR is signal to noise ratio

Shannon capacity formula can be used to calculate the theoretical highest data rate for the noisy channel:

Bit rate = Bandwidth X log2(1+SNR)

Where SNR is statistical ratio of the power of the signal to the power of noise.

In Shannon’s formula we do not care about no. of levels we use

Shannon Capacity Formula

Equation:

Represents theoretical maximum that can be achieved

In practice, only much lower rates achieved Formula assumes white noise (thermal

noise)

Impulse noise is not accounted for

Attenuation distortion or delay distortion not accounted for

SNR1log2 BwC

48

Examples

Example 1: Consider extremely noisy channel in which the value of signal to noise ratio is almost zero. In other words noise is so strong that the signal is faint. Calculate the channel capacityC = Bw log2 (1+SNR) = Bw log2 (1+0) = Bw

log2 (1) = Bw X (0) = 0

Which means that the capacity of this channel is zero regardless of band width

Example 2:

Calculate the theoretical highest bit rate of a normal telephone line. A telephone line has a bandwidth of 3000 Hz (300 – 3300 Hz). The SNR is normally 3162.

Channel Capacity (C) = 3000 X log2 (1+3162) = 3000 X log23163 = 3000 X 11.62

=34860 bps

49

Example of Nyquist and Shannon Formulations Spectrum of a channel between 3 MHz and 4 MHz

; SNRdB = 24 dB

Using Shannon’s formula

251SNR

SNRlog10dB 24SNR

MHz 1MHz 3MHz 4

10dB

B

Mbps88102511log10 62

6 C

Example of Nyquist and Shannon Formulations How many signaling levels are required?

16

log4

log102108

log2

2

266

2

M

M

M

MBC

50

Example

We have a channel with a 1 MHz bandwidth. The SNR for this channel is 63; what is the appropriate bit rate and signal level?

Solution

C = B log2 (1 + SNR) = 106 log2 (1 + 63) = 106 log2 (64) = 6 Mbps

Then we use the Nyquist formula to find the number of signal levels.

4 Mbps = 2 1 MHz log2 L L = 4

First, we use the Shannon formula to find our upper limit.

Signal Strength

All signals experience loss (attenuation).

Attenuation is denoted as a decibel (dB) loss.

Decibel losses (and gains) are additive.

51

Signal StrengthSo if a signal loses 3 dB, is that a lot?

A 3 dB loss indicates the signal lost half of its power.

dB = 10 log10 (P2 / P1)

-3 dB = 10 log10 (X / 100)

-0.3 = log10 (X / 100)

10-0.3 = X / 100 (since if log10 X = Y, then X = 10Y)

0.50 = X / 100

X = 50

52

Attenuation When travelling through any type of medium, a

signal always looses some of its power due to friction/resistance. This loss of power or loss of signal strength is called attenuation – use amplifiers

Knowing the amount of attenuation in a signal (how much power signal lost) allows you to determine signal strength.

Decibel (dB) is relative measure of signal loss or gain & is used to measure the relative strengths of two signals or a signal at two different points (logarithmic change of a signal)

Attenuation

53

Signal Strength dB = 10 log10 (P1/P0) where P1 and P0 are ending and

beginning power levels of a signal expressed in watts.

Example1: If a signal starts at a transmitter with 10 watts of power and arrives at receiver with 5 watts of power, calculate the signal’s dB

dB = 10 log10 (P1/P0) = 10 log10 (5/10) = 10 log10 (0.5) = 10 X (- 0.3) = -3

There is a 3 dB loss between transmitter and receiver

Example2: If a signal starts at a transmitter with 5 watts of power and arrives at receiver with 10 watts of power, calculate the signal’s dB

There is a 3 dB gain between transmitter and receiver

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

54

Example 13

Imagine a signal travels through an amplifier and its power is increased ten times. This means that P2 = 10 X 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

Example 14

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.

55

dB = –3 + 7 – 3 = +1

Distortion

56

Distortion

Figure 3.25 Throughput

57

Figure 3.26 Propagation time

Figure 3.26 Attenuation

58

Suppose a signal travels through a transmission mediumand its power is reduced to one-half. This means that P2

is (1/2)P1. In this case, the attenuation (loss of power)can be calculated as

Example 3.26

A loss of 3 dB (–3 dB) is equivalent to losing one-halfthe power.

A signal travels through an amplifier, and its power isincreased 10 times. This means that P2 = 10P1 . In thiscase, the amplification (gain of power) can be calculatedas

Example 3.27

59

One reason that engineers use the decibel to measure thechanges in the strength of a signal is that decibelnumbers can be added (or subtracted) when we aremeasuring several points (cascading) instead of just two.In Figure 3.27 a signal travels from point 1 to point 4. Inthis case, the decibel value can be calculated as

Example 3.28

Figure 3.27 Decibels for Example 3.28

60

Sometimes the decibel is used to measure signal powerin milliwatts. In this case, it is referred to as dBm and iscalculated as dBm = 10 log10 Pm , where Pm is the powerin milliwatts. Calculate the power of a signal with dBm =−30.

SolutionWe can calculate the power in the signal as

Example 3.29

The loss in a cable is usually defined in decibels perkilometer (dB/km). If the signal at the beginning of acable with −0.3 dB/km has a power of 2 mW, what is thepower of the signal at 5 km?SolutionThe loss in the cable in decibels is 5 × (−0.3) = −1.5 dB.We can calculate the power as

Example 3.30

61

Figure 3.28 Distortion

Figure 3.29 Noise

62

The power of a signal is 10 mW and the power of thenoise is 1 μW; what are the values of SNR and SNRdB ?

SolutionThe values of SNR and SNRdB can be calculated asfollows:

Example 3.31

The values of SNR and SNRdB for a noiseless channelare

Example 3.32

We can never achieve this ratio in real life; it is an ideal.

63

Figure 3.30 Two cases of SNR: a high SNR and a low SNR

3-5 DATA RATE LIMITS

A very important consideration in data communicationsis how fast we can send data, in bits per second, over achannel. Data rate depends on three factors:

1. The bandwidth available2. The level of the signals we use3. The quality of the channel (the level of noise)

64

We need to send 265 kbps over a noiseless channel witha bandwidth of 20 kHz. How many signal levels do weneed?SolutionWe can use the Nyquist formula as shown:

Example 3.36

Since this result is not a power of 2, we need to eitherincrease the number of levels or reduce the bit rate. If wehave 128 levels, the bit rate is 280 kbps. If we have 64levels, the bit rate is 240 kbps.

Consider an extremely noisy channel in which the valueof the signal-to-noise ratio is almost zero. In otherwords, the noise is so strong that the signal is faint. Forthis channel the capacity C is calculated as

Example 3.37

This means that the capacity of this channel is zeroregardless of the bandwidth. In other words, we cannotreceive any data through this channel.

65

We can calculate the theoretical highest bit rate of aregular telephone line. A telephone line normally has abandwidth of 3000. The signal-to-noise ratio is usually3162. For this channel the capacity is calculated as

Example 3.38

This means that the highest bit rate for a telephone lineis 34.860 kbps. If we want to send data faster than this,we can either increase the bandwidth of the line orimprove the signal-to-noise ratio.

The signal-to-noise ratio is often given in decibels.Assume that SNRdB = 36 and the channel bandwidth is 2MHz. The theoretical channel capacity can be calculatedas

Example 3.39

66

For practical purposes, when the SNR is very high, wecan assume that SNR + 1 is almost the same as SNR. Inthese cases, the theoretical channel capacity can besimplified to

Example 3.40

For example, we can calculate the theoretical capacity ofthe previous example as

We have a channel with a 1-MHz bandwidth. The SNRfor this channel is 63. What are the appropriate bit rateand signal level?

SolutionFirst, we use the Shannon formula to find the upperlimit.

Example 3.41

67

The Shannon formula gives us 6 Mbps, the upper limit.For better performance we choose something lower, 4Mbps, for example. Then we use the Nyquist formula tofind the number of signal levels.

Example 3.41 (continued)

Data Communication

For how long should the sender hold a voltage on the wire for single bit?

How can a user know that transmitting hardware purchased from one vendor will function correctly with receiving hardware purchased from another vendor?

What is the maximum rate at which hardware can change signals?

68

Normal Ethernet 10/100 Base T Cable

NamePin No. Wire's Color Pin No. Name

TX+ 1White -Orange

1 TX+

TX- 2 Orange 2 TX-

RX+ 3White -Green

3 RX+

Optional 4 Blue 4 Optional

Optional 5 White - Blue 5 Optional

RX- 6 Green 6 RX-

Optional 7White -Brown

7 Optional

Optional 8 Brown 8 Optional