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Bandwidth is the difference between the upper and lower frequencies in a continuous band of frequencies. It is typically measured in unit of hertz (symbol Hz). It may refer more specifically to two subcategories: Passband bandwidth is the difference between the upper and lower cutoff frequencies of, for example, a band-pass filter , a ...
Throughput is usually measured in bits per second (bit/s, sometimes abbreviated bps), and sometimes in packets per second (p/s or pps) or data packets per time slot. The system throughput or aggregate throughput is the sum of the data rates that are delivered over all channels in a network. [1] Throughput represents digital bandwidth consumption.
This is also sometimes described as the slew-rate-limited bandwidth. The full-power bandwidth B W {\displaystyle BW} is then related to the slew rate S R {\displaystyle SR} in volts per second and the peak-to-peak voltage swing V a m p {\displaystyle V_{amp}} by
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 ...
Throughput is controlled by available bandwidth, as well as the available signal-to-noise ratio and hardware limitations. Throughput for the purpose of this article will be understood to be measured from the arrival of the first bit of data at the receiver, to decouple the concept of throughput from the concept of latency.
Reasons for measuring throughput in networks. People are often concerned about measuring the maximum data throughput in bits per second of a communications link or network access. A typical method of performing a measurement is to transfer a 'large' file from one system to another system and measure the time required to complete the transfer or ...
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 ?
For transistors, the current-gain–bandwidth product is known as the f T or transition frequency. [4] [5] It is calculated from the low-frequency (a few kilohertz) current gain under specified test conditions, and the cutoff frequency at which the current gain drops by 3 decibels (70% amplitude); the product of these two values can be thought of as the frequency at which the current gain ...