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In audio, 0 dBm often corresponds to approximately 0.775 volts, since 0.775 V dissipates 1 mW in a 600 Ω load. [16] The corresponding voltage level is 0 dBu , without the 600 Ω restriction. Conversely, for RF situations with a 50 Ω load, 0 dBm corresponds to approximately 0.224 volts, since 0.224 V dissipates 1 mW in a 50 Ω load.
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As a reagent in organic chemistry, DBU is used as a ligand and base. As a base, protonation occurs at the imine nitrogen. [6] Lewis acids also attach to the same nitrogen. [7] These properties recommend DBU for use as a catalyst, for example as a curing agent for epoxy resins and polyurethane.
A power level of 0 dBm corresponds to one milliwatt, and 1 dBm is one decibel greater (about 1.259 mW). In professional audio specifications, a popular unit is the dBu. This is relative to the root mean square voltage which delivers 1 mW (0 dBm) into a 600-ohm resistor, or √ 1 mW × 600 Ω ≈ 0.775 V RMS.
The decibel unloaded reference voltage, 0 dBu, is the AC voltage required to produce 1 mW of power across a 600 Ω impedance (approximately 0.7746 V RMS). [2] This awkward unit is a holdover from the early telephone standards, which used 600 Ω sources and loads, and measured dissipated power in decibel-milliwatts ( dBm ).
When the test impedance is 600 Ω resistive, 0 dBm can be referred to a voltage of 775 mV, which results in a reference active power of 1 mW. Then 0 dBm0 corresponds to an overload level of approximately 3 dBm in the analog-to-digital conversion. Given a sinusoid signal of 0.775 V RMS, the power at a zero transmission level point is:
Some tools, such as Jmol and KiNG, [2] could read PDB files in gzipped format. The wwPDB maintained the specifications of the PDB file format and its XML alternative, PDBML. There was a fairly major change in PDB format specification (to version 3.0) in August 2007, and a remediation of many file problems in the existing database. [3]
A dBm is referenced to 600 ohms and is a power measurement. 0 dBm = 1mW across 600 ohms. (as defined by the I.E.E.E.) Combining Kirchoff's voltage law (V=IR)and the power formula (P=IV) you get: V = sqrt(PR); for a 600 ohm circuit, V = sqrt (0.001 * 600) = 0.7746 volts For a 50 ohm circuit, V = sqrt (0.001 * 50) = 0.2236 volts A dBu is not ...