Data Communications Networking Devices - 4th Ed.pdf

286 ________________________________________________________ LOCAL AREA NETWORKSIn examining the entries in Table 3.5 note that categories 3 through 5 supporttransmission with respect to indicated signaling rates. This means that the ability ofthose categories of UTP to support different types of LAN transmission willdepend upon thee signaling method used by different LANs. For example, considera LAN encoding technique which results in six bits encoded into four signalingelements that have a 100 MHz signaling rate. Through the use of category 5 cable adata transmission rate of 150 Mbps 6/4) 100) could be supported.Cateogory 3 cable is typically used for Ethernet and 4 Mbps Token-Ring LANs.Category 4 is normally used for 16 Mbps Token-Ring LANs, while category 5 cablesupports 100 Mbps Ethernet LANs, such as 100VGAny-LAN and 100BASE-T aswell as ATM to the desktop at both 25 Mbps and 155 Mbps operating rates.UTP speci®cations: generalThe requirement to qualify a segment of installed cable and attached connectorsresulted in the EIA/TIA-568 standard de®ning a series of link performanceparameters. These parameters cover attenuation and Near End CrossTalk NEXT)and are speci®ed for UTP cable categories in Annex E to the standard. Thus, let'sturn our attention to the manner by which attenuation and NEXT are measuredprior to examining the speci®cation limits for those two parameters on differenttypes of cable.UTP speci®cations: attenuationAttenuation represents the loss of signal power as a signal propagates from atransmitter at one end of the cable towards a receiving device located at the distantend of the cable. Attenuation is measured in decibels dB) as indicated below:Transmit voltageAttenuation ˆ 20 log 10Receive voltageFor those of us a little rusty with logarithms let's examine a few examples ofattenuation computations. First, let's assume the transmit voltage was 100, whilethe receive voltage was 1. Then, 100Attenuation ˆ 20 log 10 ˆ 20 log110 100The value of log 10 100 can be obtained by determining 10 to the appropriate powerto equal 100. Since the answer is 2 10 2 ˆ 100), log 10 100 has the value of 2 and 20log 10 100 then has a value of 40.Now assume the transmit voltage was 10 while the receive voltage was 1. Then, 10Attenuation ˆ 20 log 10 ˆ 20 log110 10