copyright 2001, s.d. personick. all rights reserved.1 telecommunications networking i topic 4...

Copyright 2001, S.D. Personick. All Rights Reserved.

1

Telecommunications Networking I

Topic 4Point-to-Point Communication over

Metallic Cables Dr. Stewart D. Personick

Drexel University


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Metallic Cables

Insulated Copper Wires

Coaxial Cable Center Conductor

Shield Insulator


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Metallic Cables• Copper wire (“wire pair”) cable is a very

important medium in the commercial sector, because-essentially all homes and small businesses in the United States, and in many other parts of the world, currently access the worldwide telecommunications infrastructure using a pair of wires-wire pairs are very commonly used for local area networks in offices and homes, including tactical operations vehicles/clusters of vehicles


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Metallic Cables (cont’d)• Coaxial cable is a very important medium

in the commercial sector, because-more than 80% of residences in the U.S., and a large fraction of residences in many other parts of the world can access the telecommunications infrastructure using coaxial cable-many local area networks utilize coaxial cable


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Metallic Cables (cont’d)• While optical fibers and wireless carry a

growing share of telecommunications traffic in the commercial sector, metallic cables will continue to carry the majority of local area network and access network traffic for many years to come

• The cost of replacing all of the metallic access cable in the U.S. with fiber would be around $1000/home x 100 million homes


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Metallic Cable System

Information Transmitter

Signal (e.g., 1 volt peak)

Cable

Cable (continued)

s(t)

r(t)Receiver

Information


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What can we say about r(t)?

• r(t) = s(t)*h(t) + n(t) + i(t), where:

s(t) is the signal that enters the cable,h(t) is the impulse response of the cable,n(t) is noise associated with the finite temperature of the cable,i(t) is interference from other signals, anda*b means: the convolution of a(t) and b(t)


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What can we say about h(t)?

• h(t) is the cable’s impulse response, and is equal to the Fourier transform of the cable’s frequency response: H(f)

• H(f), the cable’s frequency response, approximately takes the form:20 log H(f)= -aL[f**(1/2)], where L is the cable length (km) and a is a constant [dB/km-Hz**(1/2)]


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Typical Coaxial Cable Losses

0

2

4

6

8

10

12

dB/100 feet

1 MHz 10MHz

100MHz

1 GHz

RG-174RG-58RG-8


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What can we say about h(t)? (cont’d)

• The frequency response, H(f), “rolls off” as 10**[- square root of the frequency: f], due to the “skin effect” in a metallic cable (either wire pair cable or coaxial cable)

• This roll off, by definition, attenuates higher frequencies more than lower frequencies; and causes “dispersion” of the signal: s(t)


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“Dispersion”

Input pulse volts vs. time (microseconds)

Output pulse millivolts vs. time (microseconds)


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What can we say about n(t)?

• Noise, as we define it here, is an unwanted signal, or the sum of several unwanted signals….each of which is caused by a natural phenomenon

• Examples of sources of noise are:-thermally induced, random fluctuations of the properties of materials (thermal noise)-lightning


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What can we say about n(t)?

• In cables, n(t) is usually well modeled as white Gaussian noise

• This additive noise is called “thermal noise”, and results from the combination of the finite temperature of the cable (e.g., 293K) and the finite loss of the cable


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What can we say about i(t)?

• i(t) is a man made interference signal that adds to the desired signal at the receiver

• Interference can be caused by:-signals on other pairs of wire in the same wire pair cable (called “crosstalk”)-signals generated outside of the cable that “leak” into the cable (e.g., nearby, strong radio signals)


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Our challenge:

• Figure out how to do the best job we can of recovering the underlying information being communicated, given the received signal r(t)

• Understand what the basic limitations of cable systems are: e.g.; how far can we transmit and still recover the underlying information with adequate fidelity?


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Example: Dispersion-Limited Operation; 6dB maximum roll off

s(t) r(t)

cable

s(t) is a 1 volt pulse, 100 ns wide;

The cable is RG8X, with a loss of 1.8 dB/100ft @ 20MHz;

Suppose that the maximum allowable loss at 1/(100ns) = 10 MHz is 6dB

What is the maximum allowable cable length?


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Example: Dispersion Limited Operation (cont’d)

• If the cable loss at 20 MHz is 1.8 dB/100ft, then the loss at 10 MHz is 1.8 x [(10/20)**0.5)] =1.8 [.707] ~ 1.3dB/100ft

• If the maximum allowable roll off, at 10MHz, between the transmitter and the receiver is 6dB, then the maximum allowable cable length is (6/1.3) (100ft)~ 460 feet


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Example: Noise Limited Operation

s(t)

Cable: H(f) Noise=4kTB

Receiver, includesequalizer: G(f)

H(f):Equalizer: G(f)


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Noise Limited Operation (cont’d)

• The “equalizer” in the receiver amplifies higher frequencies more than lower frequencies, to compensate for the “roll off” in the cable

• The equalizer also amplifies the higher frequency noise at its input

• kT= Boltzman’s constant x temperature (K) = approximately 4 x 10**-21 (J) @ 293K


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Example: Noise Limited Operation

• Suppose the receiver has input noise, within band B, whose variance is 4kTB (watts), where B= 1/100ns

• Suppose the signal-to-noise ratio (SNR) at the receiver input must be > 100:1 = 20dB; where:

SNR=[{s(max)**2}/R] [10**-(aL/10)]/[4kTB]• Suppose: s(max) = 1 volt; a = 5dB/100ft; T=

293K, and R= 50 (ohms)• What is the maximum allowable cable length L?


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Example:Noise Limited Operation (cont’d)

• [s(max)**2]/R= [(1)**2]/50 = 0.02 watts

• 4kTB = 4 (4 x10**-21)(10**7)=1.6x10**-13 watts

• SNR= [.02/(1.6x10**-13)][10**-(aL/10)]= 1.25x10**11 [10**-(aL/10)]

• If the SNR must be greater than 100 (I.e., 20dB), then [10**(-aL/10)] must be greater than 1.25x10**-9; I.e., aL<89dB


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Example: Noise Limited Operation (cont’d)

• If aL must be less than 89 dB, and a equals 5 dB/100ft, then the maximum cable length is 89/5 (100 ft) ~ 1800 ft


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Line Amplifiers

• We can extend the distance of transmission in a noise-limited situation, by the use of line amplifiers

• The underlying concept is the equalize and amplify the signal before it has become attenuated by the cable to the point where the signal-to-noise ratio drops below an acceptable level:


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Line Amplifiers (cont’d)

Transmitter Cable section #1 Line Amplifier

#1

Cable section #2 Line Amplifier

#2

Cable section #3


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• Each line amplifier equalizes and amplifies its input signal to compensate the the roll off (frequency dependent loss) of the previous section of cable

• Each line amplifier also amplifies the noise arriving at its input, and adds some additional noise of its own


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frequencyGain= G(f)~ 1/[H(f)]

Cable section: H(f)

G(f

)


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• If the gain vs. frequency of the line amplifier approximately cancels the loss vs. frequency of the cable section preceding it…I.e., if G(f)~[1/H(f)] ….. Then the net gain through the combination of the cable section and the line amplifier is 1 (0 dB) over the range of frequencies where the above relationship holds


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• If the noise power per unit frequency (watts per Hz) arriving at the input of the first line amplifier is defined as Nin1(f), then the noise power per unit frequency at the output of that amplifier is: Nout1 (f) = {Nin1(f) [G(f)]**2} + Ninamp(f)[G(f)**2]}; whereNinamp(f) is noise added by the amplifier


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• The noise power at the output of the next cable section, which is also the noise arriving at the input to the following amplifier is (approximately):Nin2(f)=Nout1(f)[H(f)]**2 + Ncable(f)=Nin1(f) + Ninamp(f) + Ncable(f);where Ncable(f) is noise added by the cable section


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• Furthermore, if we extend this approach to a cascade of K amplifiers and K+1 sections of cable, then the noise arriving at the input to the Kth line amplifier is:NinK(f) = Nin1(f) + [K-1] [ ninamp(f) + ncable(f) ];

• Thus in a cascade of K amplifiers, the noise arriving at the input of the Kth amplifier is approximately proportional to K


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• If cable noise from the first section of cable dominates Nin1(f), and if the receiver adds input noise comparable to that of a line amplifier….then the total noise after K sections of cable, including the noise of the receiver is:NtotalK(f) = K [ Ncable(f) + Ninamp(f)]


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• The result of the above is:If the signal-to-noise ratio at the input of the final line amplifier (preceding the extraction of the arriving information from the arriving signal) must be greater than some threshold, SNRmin… then the allowable loss in each section of cable decreases as more sections are added


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• Quantitatively…If the allowable loss between the transmitter and the receiver in a noise-limited single segment transmission system is X dB, then the allowable loss in each segment of a transmission system with K sections, and K amplifiers/equalizers (including the receiver) is X-[10logK] dB


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Line Amplifiers (continued)

• In a previous example of a loss limited, single segment transmission system, the allowable cable loss (at 10 MHz) was 89dB If we try to extend the reach of the system by using 10 cable segments, then the allowable loss in each segment is 79dB; and the total loss of all the segments can be as much as 790dB (less than 10 x 89dB)


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Traditional Cable Television System

Head end Super trunk

Trunk

Line amplifier

Feeder

Drop

Splitter

Home


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Frequency Division Multiplexing (FDM):

Typical Cable System

f (MHz)500

Ch 2: 50 MHz +/- 3MHz

Ch 75: 525MHz

Video signals (each 6MHz wide) spaced ~every 6 MHz. 75 x 6 MHz = 450 MHz


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FDM: Typical Cable System

• If there are 75 television channels sharing the same composite signal (stacked up in frequency), and if distortion considerations limit the output of the transmitter or a line amplifier to 1 volt (for example)….then each of the component signals that represent a single channel must have a peak amplitude of about 1 volt/[75**(1/2)]~0.1V


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Cable TV FDM: Issues

• Distortion: If the transmitter and/or the line amplifiers are not linear, then we can end up with square law and higher order terms that result in interference among the FDM channels

• Noise; and interference from signals leaking into the cable system


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Impact of Distortion on FDM

• Distortion (simplified example):

s(t) is an FDM composite signal (e.g., 6 MHz TV channels stacked up in frequency)

r(t) = s(t) + b [s(t)]**2 + c[s(t)]**3


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Impact of Distortion (cont’d)

• If s(t) contains components centered at around 50MHz (for example) then b[s(t)**2] will contain frequencies at around 100MHz…interfering with the components of s(t) at 100MHz

• If s(t) contains components centered at 100MHz and at 150MHz, then b[s(t)**3] will contain frequencies at 50 MHz, 200 MHz, 300MHz, 350 MHz, 400 MHz, and 450 MHz…causing interference at all of those frequencies


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Impact of Noise

• Thermal noise associated with the finite temperatures of the cable sections, plus noise associated with the line amplifiers and the receivers connected to the system set a lower bound or baseline of system noise. This noise tends to appear as “snow” on a television receiver

• “Ingress noise” (interference) from radio signals, electric motors, etc, represent the more serious problem in most cable TV systems


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Radio Frequency Interference

• Another consideration, of importance in all metallic cable systems, is the possibility that the cable system will radiate signals that may interfere with other communications or information systems/appliances (e.g., medical devices in a hospital environment)

• Radiation can occur due to imperfect shielding on coaxial cables, bad connectors, unterminated splitter output ports, and imbalances in wire pair cable systems


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“Wiring” Concerns

Poorly designed splitter

Unused output port not terminated properly

Wrong kind of “wire”

Bad or loose connector

copyright 2001, s.d. personick. all rights reserved.1 telecommunications networking i topic 4...

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