1. Copyright 1998, S.D. Personick. All Rights Reserved. Telecommunications Networking I Lectures 6-8 Point-to-Point Communication over Metallic Cables

2. Copyright 1998, S.D. Personick. All Rights Reserved. Metallic Cables Insulated Copper Wires Coaxial Cable Center Conductor Shield Insulator

3. Copyright 1998, S.D. Personick. All Rights Reserved. Metallic Cables Copper wire (“wire pair”) cable is a very important medium, because -essentially all homes and small businesses in the United States, and in most 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

4. Copyright 1998, S.D. Personick. All Rights Reserved. Metallic Cables (cont’d) Coaxial cable is a very important medium 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

5. Copyright 1998, S.D. Personick. All Rights Reserved. Metallic Cables (cont’d) While optical fibers and wireless carry a growing share of telecommunications traffic, 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

6. Copyright 1998, S.D. Personick. All Rights Reserved. Metallic Cable System Information Transmitter Signal (e.g., 1 volt peak) Cable Cable (continued) s(t) r(t) Receiver Information

7. Copyright 1998, S.D. Personick. All Rights Reserved. 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, and a*b means: the convolution of a(t) and b(t)

8. Copyright 1998, S.D. Personick. All Rights Reserved. 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)]

9. Copyright 1998, S.D. Personick. All Rights Reserved. Typical Coaxial Cable Losses

10. Copyright 1998, S.D. Personick. All Rights Reserved. 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 rolloff, by definition, attenuates higher frequencies more than lower frequencies; and causes “dispersion” of the signal: s(t)

11. Copyright 1998, S.D. Personick. All Rights Reserved. “Dispersion” Input pulse volts vs time (microseconds) Output pulse millivolts vs time (microseconds)

12. Copyright 1998, S.D. Personick. All Rights Reserved. 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

13. Copyright 1998, S.D. Personick. All Rights Reserved. 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

14. Copyright 1998, S.D. Personick. All Rights Reserved. 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)

15. Copyright 1998, S.D. Personick. All Rights Reserved. Our challenge: Figure out how to do the best job we can of estimating 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?

16. Copyright 1998, S.D. Personick. All Rights Reserved. Example: Dispersion-Limited Operation; 6dB maximum rolloff 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?

17. Copyright 1998, S.D. Personick. All Rights Reserved. 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 rolloff, at 10MHz, between the transmitter and the receiver is 6dB, then the maximum allowable cable length is (6/1.3) (100ft)~ 460 feet

18. Copyright 1998, S.D. Personick. All Rights Reserved. Example: Noise Limited Operation s(t) cable: H(f) Noise=4kTB Receiver, includes equalizer: G(f) H(f): Equalizer: G(f)

19. Copyright 1998, S.D. Personick. All Rights Reserved. Noise Limited Operation (cont’d) The “equalizer” in the receiver amplifies higher frequencies more than lower frequencies, to compensate for the “rolloff” 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

20. Copyright 1998, S.D. Personick. All Rights Reserved. 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**-(a/10)]/[4kTB] Suppose: s(max) = 1 volt; a = 5dB/100ft; T= 293K, and R= 50 (ohms) What is the maximum allowable cable length?

21. Copyright 1998, S.D. Personick. All Rights Reserved. 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**-(a/10)] = 1.25x10**11 [10**-(a/10)] If the SNR must be greater than 100 (I.e., 20dB), then [10**(-a/10)] must be greater than 1.25x10**-9; I.e., a<89dB

22. Copyright 1998, S.D. Personick. All Rights Reserved. Example: Noise Limited Operation (cont’d) If a must be less than 89 dB, and a equals 5 dB/100ft, then the maximum cable length is 89/5 (100 ft) ~ 1800 ft

23. Copyright 1998, S.D. Personick. All Rights Reserved. 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:

24. Copyright 1998, S.D. Personick. All Rights Reserved. Line Amplifiers (cont’d) Transmitter Cable section #1 Line Amplifier #1 Cable section #2 Line Amplifier #2 Cable section #3

25. Copyright 1998, S.D. Personick. All Rights Reserved. Line Amplifiers (Con’t) Each line amplifier equalizes and amplifies its input signal to compensate the the rolloff (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

26. Copyright 1998, S.D. Personick. All Rights Reserved. Line Amplifiers (cont’d) frequency Gain= G(f)~ 1/[H(f)] Cable section: H(f) G(f)

27. Copyright 1998, S.D. Personick. All Rights Reserved. Line Amplifiers (cont’d) 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

28. Copyright 1998, S.D. Personick. All Rights Reserved. Line Amplifiers (cont’d) 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]}; where Ninamp(f) is noise added by the amplifier

29. Copyright 1998, S.D. Personick. All Rights Reserved. Line Amplifiers (cont’d) 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

30. Copyright 1998, S.D. Personick. All Rights Reserved. Line Amplifiers (cont’d) 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

31. Copyright 1998, S.D. Personick. All Rights Reserved. Line Amplifiers (cont’d) 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)]

32. Copyright 1998, S.D. Personick. All Rights Reserved. Line Amplifiers (cont’d) 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

33. Copyright 1998, S.D. Personick. All Rights Reserved. Line Amplifiers (cont’d) 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

34. Copyright 1998, S.D. Personick. All Rights Reserved. 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)

35. Copyright 1998, S.D. Personick. All Rights Reserved. Traditional Cable Television System Head end Super trunk Trunk Line amplifier Feeder Drop Splitter Home

36. Copyright 1998, S.D. Personick. All Rights Reserved. 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

37. Copyright 1998, S.D. Personick. All Rights Reserved. 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

38. Copyright 1998, S.D. Personick. All Rights Reserved. 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

39. Copyright 1998, S.D. Personick. All Rights Reserved. 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

40. Copyright 1998, S.D. Personick. All Rights Reserved. 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

41. Copyright 1998, S.D. Personick. All Rights Reserved. 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, electrical appliances, etc, represent the more serious problem in most cable TV systems

42. Copyright 1998, S.D. Personick. All Rights Reserved. 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

43. Copyright 1998, S.D. Personick. All Rights Reserved. Home “Wiring” Concerns Poorly designed splitter Unused output port not terminated properly Wrong kind of “wire” (zip cord) Bad or loose connector