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Example of calculation of Average Per-Bit Delivery Cost for a small data-flow. Average Per-Bit Delivery Cost divides the cost of however many bits were actually modulated across a network or component of a network over a period of time by the total operational and amortized capital expenses associated with the network or component over the same time-period, to produce an average cost for the ...
The consumed bandwidth in bit/s, corresponds to achieved throughput or goodput, i.e., the average rate of successful data transfer through a communication path.The consumed bandwidth can be affected by technologies such as bandwidth shaping, bandwidth management, bandwidth throttling, bandwidth cap, bandwidth allocation (for example bandwidth allocation protocol and dynamic bandwidth ...
The Rayleigh bandwidth of a simple radar pulse is defined as the inverse of its duration. For example, a one-microsecond pulse has a Rayleigh bandwidth of one megahertz. [1] The essential bandwidth is defined as the portion of a signal spectrum in the frequency domain which contains most of the energy of the signal. [2]
z Uses 8b/10b encoding, meaning that 20% of each transfer is used by the interface instead of carrying data from between the hardware components at each end of the interface. For example, a single link PCIe 1.0 has a 2.5 Gbit/s transfer rate, yet its usable bandwidth is only 2 Gbit/s (250 MB/s).
For file sizes, it is usual for someone to say that they have a '64 k' file (meaning 64 kilobytes), or a '100 meg' file (meaning 100 megabytes). When talking about circuit bit rates , people will interchangeably use the terms throughput , bandwidth and speed, and refer to a circuit as being a '64 k' circuit, or a '2 meg' circuit — meaning 64 ...
The accounting equation plays a significant role as the foundation of the double-entry bookkeeping system. The primary aim of the double-entry system is to keep track of debits and credits and ensure that the sum of these always matches up to the company assets, a calculation carried out by the accounting equation.
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If the requirement is to transmit at 50 kbit/s, and a bandwidth of 10 kHz is used, then the minimum S/N required is given by 50000 = 10000 log 2 (1+S/N) so C/B = 5 then S/N = 2 5 − 1 = 31, corresponding to an SNR of 14.91 dB (10 x log 10 (31)). What is the channel capacity for a signal having a 1 MHz bandwidth, received with a SNR of −30 dB ?