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Cross section of a cat 5e cable. The Category 5e specification improves upon the Category 5 specification by further mitigating crosstalk. [9] The bandwidth (100 MHz) and physical construction are the same between the two, [10] and most Cat 5 cables actually happen to meet Cat 5e specifications even though they are not certified as such. [11]
ANSI/TIA-568 is a technical standard for commercial building cabling for telecommunications products and services. The title of the standard is Commercial Building Telecommunications Cabling Standard and is published by the Telecommunications Industry Association (TIA), a body accredited by the American National Standards Institute (ANSI).
The Smith chart (sometimes also called Smith diagram, Mizuhashi chart (水橋チャート), Mizuhashi–Smith chart (水橋スミスチャート), [1] [2] [3] Volpert–Smith chart (Диаграмма Вольперта—Смита) [4] [5] or Mizuhashi–Volpert–Smith chart), is a graphical calculator or nomogram designed for electrical and electronics engineers specializing in radio ...
In a straight-through cable, pins on one end correspond exactly to the corresponding pins on the other end (pin 1 to pin 1, pin 2 to pin 2, etc.). Using the same wiring scheme at each end yields a straight-through cable (a given color wire connects to a given number pin, the same at both ends).
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If both devices being connected support 1000BASE-T according to the standards, they will connect regardless of whether a straight-through or crossover cable is used. [ 14 ] A 10BASE-T transmitter sends two differential voltages, +2.5 V or −2.5 V.
By reducing the original signal rate to 1 ⁄ 4 or 1 ⁄ 2, the link speed drops to 2.5 or 5 Gbit/s, respectively. [5] The spectral bandwidth of the signal is reduced accordingly, lowering the requirements on the cabling, so that 2.5GBASE-T and 5GBASE-T can be deployed at a cable length of up to 100 m on Cat 5e or better cables. [6] [7]
An observed value of 81 would use the tick mark above 81 in range E, and curved scale E would be used for the expected value. This allows five different nomograms to be incorporated into a single diagram. In this manner, the blue line demonstrates the computation of: (9 − 5) 2 / 5 = 3.2 and the red line demonstrates the computation of: