What is the distance from an RF Transmitter that output power is determined?
142441
Output power is typically measured at the output of the final amplifier, before the feedline and the antenna.   Field strength is a measurement of radiated power at some point away from the antenna. There is no standard distance for every application when it comes to measuring field strength.\n\nRF power is typically expressed in terms of peak envelope power (PEP) at the output of the amplifier, but can also be expressed in terms of effective radiated power (ERP) when considering the overall effects in the transmitter's system, such as feedline loss and antenna gain.
765230
EER = Q in (BTU/hr) divided by W in (watts)\n\n600/500= 1.2
939675
the range fo the connection, the quality of it.
206010
you should use the same type of transmitter module as you are using for reciever
602903
You don't, any more than you convert between volts and watts.  You can read about dBm in the article below, but simply put, it is a measure of power.
623597
IEEE 802.11g dictates that a wireless g router oppertating at 2.4 ghz will trasmit to ranges of 300 meters (approxiamtely 100 feet). However, normal enviromental conditions can cause this distance to change. usually, for the worse. Besides before you do this, think about both the ethical, and legal consquences related to what you are thinkin bout doing, also, the person on the other side of that connection might have enough knowlegde to hack your computer.
409414
G will give you a little more distance, but not much.\n\nYour first fix would be to vary the channels.  Some 802.11 channels end up with reflective interference with cordless phones and other RF devices.  Play around with the channels to see if you can get the range you need.  if you can, then this was free!\n\n2nd measure is to improve your antenna (if that is an option).  If you are using a Linksys or D-Link router, odds are that there are replacement antennas for your unit that will boost your signal strength anywhere from 7 to 12dB.  This is less expensive than a new router.\n\nI wouldn't waste my time with a wireless extender unless the space you are trying to paint is HUGE (3000 sq ft or more), or if you have gobs of metal in your walls.  if you can vary the location of the router any, try not to put a wet wall between the router and where you want the signal.\n\nIf all else fails, there is always a spread-spectrum solution like Linksys' SRX routers and cards.  I've gotten pretty decent range with mine -- though for maximum effect, you have to change out both the router and the card.  I only had to change out 2 of my cards (both desktop units the farthest away from the router) with the "pickle fork" design external antenna.
561
8 amps 900(880) watts
401656
Damage to the human organism is dependent on frequency, distance to the source, and transmit power (in Watts)\n\nSee the following:
161029
A dual radio frequency interferometer aided by laser rangefinder and terrain database system utilizes two linear rays of RFI sensors to determine the position of a source of RF energy. The dual RFI system is very attractive in that it is a passive system and cannot be detected by other sensors or systems while it is going through its position determination operation. Furthermore, use of two linear arrays of sensors allows for the elimination of chronic effects inherent in radio frequency interferometers. The two linear arrays can be placed anywhere on a vehicle as long as they are in the same common geometric plane and in-flight mechanisms are provided to transform the coordinate systems from the in-flight platform coordinates to the local geodetic coordinates. This provides positioning and source identification which can be beneficial to not only the present system but other systems in the area. With the two linear arrays, a line-of-sight vector can be determined and other systems can be provided with this information in order to provide further details of the source position. Once a pointing vector has been determined, this information can be used with other active systems in order to provide a precise range and location of the source of the RF signals.
416586
DeciBels below 1 Milliwatt) A measurement of power loss in decibels using 1 milliwatt as the reference point. A signal received at 1 milliwatt yields 0 dBm. A signal at .1 milliwatt is a loss of 10 dBm.\n\nOR\n\nDataBase Manager
428054
To what distance? Either buy an extension cable (Radio Shack) or buy a longer cable (Radio Shack).
761790
Your transmitter will work, your receiver won't.\nbut you could bounce the signal onto a simple solar cell\nconnected to a set of headphones and you should be\nable to receive a weak signal...\nYour receiver won't work because light won't move the paper and the grains of sand.....
14994
Check out this site here It gives step by step directions. He says it will give you a 3-5km range. That should give you roughly 1.5 miles or so.\n\nhttp://www.usbwifi.orcon.net.nz/\n\nBy the way it isn't omnidirectional, so you have to have a pretty good line of sight between antenna 1 and antenna 2.
763237
Have you boosted the power all the way?  Generally by default it is set at the minimum for saving power.  You will need to go to the device manager to access this.  Find your wireless card in your Device Manager, then open it.  Click on the Advanced tab then click on power management.  If this is set to the lowest, you can raise it to help a little.  This will increase the power on the transmit end.
763289
You need to post some details - then I can post an answer...\n\nIn general, P = V*I = V^2/Z\nso calculate the complex impedance Z (= R+jX)\nthen take the real component of the complex power.
878170
Let us remember what is dB. It is the logarithm of relaton, at which 10dB = 1B = Lg(I/Io)=1 when I = 10 Io, That is when we enhance intensity 10 times, we get 10 dB more, when we enhance intensity 100 times, we get 20 dB more, and when we enhance intensity 10000 times, we get 40 dB more.\n\nIn acoustics, it is usually accepted to take the Io as the intensity at threshhold sound preddure of 20 microPascal. \n\nEarnestly speaking, I do not remember what is 120, though I belive this is the deciBell Intensity of 1Watt/sq.m. \n\nBut you do not need this figure at all to solve the problem. You only need two figures for sound meter figures and one for the first distance to find the second distance.\n\nAt the initial distance from the center of loudspeakers Ri of 2.8 meters you have intensity Ii of 130 dB, and the total sound power would be W = 4PiRi^2Ii , the same power will be at the second distance (we assume that the sound power is not dissipated in air significantly). So, at distance Rx (this is the distance which we shall have to determine), the power W = 4PiRx^2Ix = 4PiRi^2Ii.\nSo, the relation Ii/Ix = Rx^2/Ri^2.\n\nLet us now take into accont that\n 10 lg(Ii/Ix) = Bi - Bx = 130 - 90 = 40 dB\n from this      lg(Ii/Ix) = 4,\nand from this     Ii = 10000 Ix.\n\nThat is, the intensity at 90 dB is 10000 times less than at 130 dB.\nAnd from here, \nRx^2 = 10000Ri^2,\nhence  Rx = 100 Ri = 280m\nAnd this is the answer