PowerPoint PresentationSlide 2IntroductionSlide 4Wire as a Transmission MediumSlide 6Slide 7Cable characteristicsSlide 9Slide 10The DecibelExampleLogarithm (log)Slide 14Note..ExercisesFiber Optics as a Transmission MediumCross section of optical fiber cableSlide 19Slide 20Advantages of fiber opticsSIMS-201Wire and Fiber Transmission Systems2OverviewChapter 15Wire and Fiber Transmission SystemsWire as a transmission mediumFiber optics as a transmission medium3IntroductionThere are fundamentally two mediums for information transmission:Guided electromagnetic (EM) waves - wire, fiber optics, etc. Unguided EM waves - airThe past two lectures have concentrated on radio communications using air as the transmission mediumNext, we will learn about some important aspects of the forms of wire and fiber optics used for information transmission4AirCableFiber opticsGuidedUnguided5Wire as a Transmission MediumWire is currently the most common and versatile medium of transmissionAll wire-based transmission media are called cablesWire based transmission schemes guide electromagnetic waves either between a pair of separate wires or inside a coaxial (coax) arrangementA coax cable has both a center conductor and a second shield conductorThese conductors are separated by an insulating material, such that the shield conductor entirely surrounds the center conductor6In the case of non coaxial transmission, the pair of wires may be held either parallel to each other by a stiff insulating material, or individually insulated and twisted around each otherA surrounding shield may be placed around the resulting twisted pair to form a shielded twisted pair (STP)If a surrounding shield is not placed around the twisted pair, then this arrangement is called an unshielded twisted pair (UTP)7Parallel wiresUTPSTPCoax8Cable characteristicsA cable moves EM waves by providing a channel. The EM waves traverse the cable moving through the conductors. The EM waves are confined in this way, as they interact with the free electrons in the conductor, which are responsible for guiding the waves. While traversing through the cable however, due to physical effects, the wave loses energy and the intensity of the wave diminishes, the farther it goes.This results in a decrease of the signal amplitude at the receiving end – called attenuationIn other words, the magnitude of the signal diminishes as it reaches the end of the cable Original signal Attenuated signal9The longer the cable, the larger the attenuationThe larger the conductor in the cable (radius), the lower the attenuation (up to some extent)It is desirable to use larger, more expensive cables in situations that require high transmission quality over long distancesHigh transmission quality means that the receiver is able to detect correctly if a 1 or a 0 is transmittedIf a signal is highly attenuated at the receiving end, the receiver will not be able to distinguish between the levels of 1 and 0, and this will lead to erroneous transmission of information10Typical attenuation figures for various cables:Cable type Signal attenuation per 1000 ft @100 MHzUTP 56 dBSTP 37.5 dBCoax (thin ethernet)60 dBCoax (thick ethernet)20 dBCheapExpensive11What is a decibel?In electrical engineering, the decibel (abbreviated as dB) is a logarithmic unit used to describe the ratio between two power levels Power: unit of measurement is watts (W)dB = 10 log10 P1/P2 (power ratio)The Decibel12If the input signal power is 2 W and the output signal power is measured to be 2 milliWatts, calculate the power attenuation in dB of the cableOriginal signalAttenuated signal2W input power2mW output powerExampleThe input signal power is: P1=2 WThe output signal power is: P2=2x10-3 WThe Power attenuation is: dBP = 10 log10 P1/P2 =10 log10 2/(2x10-3)= 10 log10 1000=10 x 3 =30 dBThe signal power has attenuated by 30dB while passing through the cableNote: Since P1/P2 = 1000, we can say that the signal has suffered a power attenuation of 1000 times, or in other words, by 30 dBLength of cable13Logarithm (log)How do we know that 10 log10 1000 = 10 x 3 ?Logarithmic and Exponential Functions Logarithmic and exponential functions are inverses of each other: If y = logbx then x = byIn words, logb x is the exponent you put on base b to get x. lolSo, If x = log10 1000 then 1000 = 10x and how much is x here? 10 to the power of what is equal to 1000? It would be 10 to the power of 3 (10x10x10) = 1000.14If the power attenuation of a length of cable is given to be 15 dB, find the ratio of the input/output power 15 = 10log10 P1/P2 1.5 = log10 P1/P2 P1/P2 = 101.5 = 31.62P1/P2 = 31.62The calculation above illustrates that signals passing through this length of cable suffer a power attenuation by 31.62 times. Note that this is a ratio! There are no units. If, for example, the input power is 1W, the power at the output of the cable would be 1/31.62=0.0316 W Logarithm: if b = logax then x = abExample 1.5 = log10 P1/P2 then P1/P2 = 101.515Note that the dB scale is a logarithmic scale, and is a convenient method to express large ratiosIn the first example, the ratio 1000 was expressed as 30 dBFor example, if the ratio is: 800,000,000, then this expressed in dB is: 89 dB (a much smaller number)Ratio dB1 010 10100 201000 3010,000 40100,000 501,000,000 60Note..16ExercisesIf the input signal power is 10mW (milliWatts) and the output signal power is measured to be 5 μW (micro watts), calculate the power attenuation in dB of the cableIf the power attenuation of a length of cable is given to be 65 dB, find the ratio of the input/output power17Fiber Optics as a Transmission MediumInformation is carried through a fiber optic cable by transmitting pulses of light (which is also an EM wave)!A fiber optic cable is a coaxial arrangement of glass or plastic material of immense clarity (i.e., highly transparent)A clear cylinder of optical material called the core is surrounded by another clear wrapper of optical material called the claddingThese two materials are selected to have different indices of refractionThe fiber is surrounded by a plastic or teflon jacket to protect and stiffen the fiberLight is guided through the optical fiber by continual reflection from the core-cladding boundaryThis is made possible due to the different refractive indices of