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The history of modems is the attempt at increasing the bit rate over a fixed bandwidth (and therefore a fixed maximum symbol rate), leading to increasing bits per symbol. For example, ITU-T V.29 specifies 4 bits per symbol, at a symbol rate of 2,400 baud, giving an effective bit rate of 9,600 bits per second.
Since most such codes correct only bit-flips, but not bit-insertions or bit-deletions, the Hamming distance metric is the appropriate way to measure the number of bit errors. Many FEC coders also continuously measure the current BER.
RIMM modules used by RDRAM are 16-bit- or 32-bit-wide. [49] DIMM modules connect to the computer via a 64-bit-wide interface. Some other computer architectures use different modules with a different bus width. In a single-channel configuration, only one module at a time can transfer information to the CPU.
As the description implies, is the signal energy associated with each user data bit; it is equal to the signal power divided by the user bit rate (not the channel symbol rate). If signal power is in watts and bit rate is in bits per second, E b {\displaystyle E_{b}} is in units of joules (watt-seconds).
The symbol rate is related to gross bit rate expressed in bit/s. The term baud has sometimes incorrectly been used to mean bit rate , [ 3 ] since these rates are the same in old modems as well as in the simplest digital communication links using only one bit per symbol, such that binary digit "0" is represented by one symbol, and binary digit ...
The 201A Data-Phone was a synchronous modem using two-bit-per-symbol phase-shift keying (PSK) encoding, achieving 2,000 bit/s half-duplex over normal phone lines. [10] In this system the two tones for any one side of the connection are sent at similar frequencies as in the 300 bit/s systems, but slightly out of phase.
The packet transmission time in seconds can be obtained from the packet size in bit and the bit rate in bit/s as: Packet transmission time = Packet size / Bit rate. Example: Assuming 100 Mbit/s Ethernet, and the maximum packet size of 1526 bytes, results in Maximum packet transmission time = 1526×8 bit / (100 × 10 6 bit/s) ≈ 122 μs
bit time = 1 / (10 * 10^6) = 10^-7 = 100 * 10^-9 = 100 nanoseconds The bit time for a 10 Mbit/s NIC is 100 nanoseconds. That is, a 10 Mbit/s NIC can eject 1 bit every 0.1 microsecond (100 nanoseconds = 0.1 microseconds). Bit time is distinctively different from slot time, which is the time taken for a pulse to travel through the longest ...