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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.
The table below shows values for PC memory module types. These modules usually combine multiple chips on one circuit board. SIMM modules connect to the computer via an 8-bit- or 32-bit-wide interface. 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.
The BER is the likelihood of a bit misinterpretation due to electrical noise ().Considering a bipolar NRZ transmission, we have = + for a "1" and () = + for a "0".Each of () and () has a period of .
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 theoretical maximum throughput for end user is clearly lower than the peak data rate due to higher layer overheads. Even this is never possible to achieve unless the test is done under perfect laboratory conditions. The typical throughput is what users have experienced most of the time when well within the usable range to the base station.
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 ...
In this case, the maximum throughput is often called net bit rate or useful bit rate. To determine the actual data rate of a network or connection, the "goodput" measurement definition may be used. For example, in file transmission, the "goodput" corresponds to the file size (in bits) divided by the file transmission time.
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